pde1-1 Alber Green's Helmholtz

Lecture Notes on Linear Partial Dierential Equations (PDE 1) Hans-Dieter Alber Technische Universit at Darmstadt SS 2012 1 Contents 1 The wave equation as mathematical model for the vibrating string and the vibrating membrane. 4 1.1 Potential energy of the linear elastic string . . . . . . . . . . . . 4 1.2 The Hamiltonian principle . . . . . . . . . . . . . . . . . . . . . 7 1.3 Initial-boundary value problems for the one-dimensional wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.4 Initial-boundary value problems for the wave equation in higher space dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2 The Helmholtz equation obtained by reduction of the wave equation 13 2.1 Separation of variables and boundary value problems for the Helmholtz equation . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.2 Linear partial dierential equations of order m . . . . . . . . . . 15 3 Tools from functional analysis. Weak solutions of one dimen- sional boundary value problems. 18 3.1 The Hilbert space L2( ;C) . . . . . . . . . . . . . . . . . . . . . 18 3.2 The Riesz representation theorem and the projection theorem . 19 3.3 Complete orthonormal systems . . . . . . . . . . . . . . . . . . 23 3.4 Eigenfunctions of the Dirichlet boundary value problem in R1. . 25 3.5 Weak derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.6 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.7 Weak solution of the Dirichlet boundary value problem to the Helmholtz equation . . . . . . . . . . . . . . . . . . . . . . . . . 35 4 Boundary value problems in circular domains. Bessel functions 39 4.1 The Laplace operator in polar coordinates . . . . . . . . . . . . 39 4.2 Solution of the potential equation in circular domains. . . . . . . 41 4.3 Bessel's dierential equation. Solution of the Helmholtz equation in circular domains. . . . . . . . . . . . . . . . . . . . . . . . . . 44 5 Maximum principle, subsolutions, Perron's method 51 5.1 Maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . 51 5.2 Consequences of the maximum principle for the Helmholtz equa- tion in R2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 5.3 Subsolutions, supersolutions, comparison . . . . . . . . . . . . . 59 5.4 Perron's method . . . . . . . . . . . . . . . . . . . . . . . . . . 63 5.5 Boundary value problems, regular points . . . . . . . . . . . . . 65 2 6 Fundamental solution, Green's function 73 6.1 Convolution integrals . . . . . . . . . . . . . . . . . . . . . . . . 73 6.2 Fundamental solution . . . . . . . . . . . . . . . . . . . . . . . . 74 6.3 Green's function . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 6.4 The Green's function for the potential equation in a ball. Pois- son's representation formula. . . . . . . . . . . . . . . . . . . . . 80 6.5 Green's function for the half space . . . . . . . . . . . . . . . . . 84 7 Integral equation method 87 7.1 The boundary integral equations . . . . . . . . . . . . . . . . . 87 7.2 Properties of the double layer potential . . . . . . . . . . . . . . 89 7.3 Properties of the single layer potential . . . . . . . . . . . . . . 97 7.4 Compact operators on a Banach space . . . . . . . . . . . . . . 99 7.5 Solution of the Neumann problem . . . . . . . . . . . . . . . . . 102 7.6 Solution of the Dirichlet problem . . . . . . . . . . . . . . . . . 106 8 Hilbert space methods 108 8.1 Elliptic dierential operators, weak solutions . . . . . . . . . . . 108 8.2 Coercivity of sesquilinear forms to elliptic operators . . . . . . . 110 8.3 Existence of weak solutions to elliptic equations . . . . . . . . . 112 9 Eigenvalue problems, spectral theory 116 9.1 The Friedrichs' extension of the operator L. . . . . . . . . . . . 116 9.2 Existence of eigenvalues in bounded domains . . . . . . . . . . . 118 9.3 Spectral theorem and resolvent set . . . . . . . . . . . . . . . . 121 10 Linear hyperbolic equations of second order 127 10.1 Hyperbolic dierential operators . . . . . . . . . . . . . . . . . . 127 10.2 Energy estimate for the wave equation, uniqueness of solutions . 128 10.3 Existence of weak solutions of initial-boundary value problems to hyperbolic equations . . . . . . . . . . . . . . . . . . . . . . . 130 A Appendix: Bessel and Neumann functions 136 3 1 The wave equation as mathematical model for the vi- brating string and the vibrating membrane. 1.1 Potential energy of the linear elastic string We want to formulate mathematical equations, which allow to compute the vibrations in time of an elastic string, which is xed at both ends. To this we rst compute the potential energy stored in the string at time t. This requires to make an assumption for the elastic material properties of the string. Let b x au(x;t) Actual conguration of the string 1< a < b <1be given numbers. Imagine rst that the string is linearly stretched along the x-axis with ends xed at ( a;0) and (b;0) and is at rest. We call this the reference conguration of the string. Now consider an actual conguration, where the string is displaced from this reference conguration. Hypothesis: LetP(1); P(2)be material points of the string, whose positions are (x1;0), (x2;0) in the reference conguration and ( x(1);y(1)), (x(2);y(2)) in the actual conguration. Assume that in the actual conguration the string is linearly and uniformly stretched between P(1)andP(2). Then the force K acting onP(2)is K=jP(1)P(2)j jx1x2jP(1)P(2) jP(1)P(2)j=P(1)P(2) jx1x2j; with a material constant >0. Here we identied P(1); P(2)with the positions in the actual conguration. (Linearly elastic material behavior.) K P(1)P(2) Note that by this law of force a variation of the actual positions of the points P(1);P(2)in direction orthogonal to the x-axis does not alter the component of 4 the forceKparallel of the x-axis, and a variation of the positions parallel to this axis does not alter the force component orthogonal to this axis. Therefore the movements of the material points of the string in directions parallel and orthogonal to the x-axis are not coupled. We thus can and shall assume in the following that the material points of the string move only in the direction orthogonal to the x-axis. This implies that at time t0 the string can be represented by the graph of a function x7!u(x;t) : [a;b]!R: To compute the potential energy we approximate the graph of this function by a piecewise linear function: u(x;t)(x5;u(x5;t)) x1x2x3x4x5x6 a=x0 Withh=ba nlet xi=ih+a; i = 0;1;:::;n be thex-coordinates of the node points of the polygonal arc. We rst determine the potential energy stored in the polygonial arc. To this end we successively deform the string from the reference conguration to the polygon and compute the work done in every step. In the rst step we move all material points of the string vertically and in parallel from the x-axis to the horizontal line passing through the point ( x0;u(x0;t)). No work is done in this step. We then x the endpoint of the string at ( x0;u(x0;t)) and move all points on the line segmentf(x;u(x0;t))jx1xbgvertically and in parallel to the line segment f(x;u(x1;t))jx1xbg. During this movement an amount of energy V1is stored in the straight line segment above the interval [ x0;x1], which is equal to the work done in moving the material point at the position ( x1;u(x0;t)) along a vertical path to the position ( x1;u(x1;t)) against the vertical component K2 of the elastic force Kin this straight line segment. A parametrization of this path is s7!P(s) = (x1;s) : [u(x0;t); u(x1;t)]!R2: Since in the reference conguration the position of the point P(s) = (x1;s) is (x1;0) and the position of ( x0;u(x0;t)) is (x0;0), our hypothesis implies that the elastic force K(s) at the point P(s) is given by K(s) = (K1(s);K2(s)) =(x0;u(x0;t))(x1;s) x1x0; 5 whence K2(s) =su(x0;t) x1x0: We thus have V1=Zu(x1;t) u(x0;t)K2(s)ds= x1x0Zu(x1;t) u(x0;t)(su(x0;t))ds = x1x01 2(su(x0;t))2s=u(x1;t) s=u(x0;t)= 2(u(x1;t)u(x0;t))2 x1x0 = 2u(x1;t)u(x0;t) x1x02 (x1x0) = 2@ @xu(x 1;t)2 h; wherex 1is a point between x0andx1. Here we used the mean value theorem. We proceed in the same way and obtain for the elastic energy Vistored in the straight line segment above [ xi1;xi] that Vi= 2@ @xu(x i;t)2 h: For the total energy V(h)(t) of the polygonal arc we thus have V(h)(t) =nX i=1Vi(t) = 2nX i=1@ @xu(x i;t)
2h: Forh!0 the polygon converges to the string. Therefore one denes the potential energy V(t) of the string at time tby V(t) = lim h!0V(h)(t): On the other hand,Pn i=1(@ @xu(x i;t))2his a Riemann sum. If x7!@ @xu(x;t) is continuous we thus obtain by Riemann integration theory that lim n!0 2nX i=1@ @xu(x i;t)
2h= 2Zb a@ @xu(x;t)
2dx: Therefore we conclude that the stored energy of the string at time tis V(t) = 2Zb a ux(x;t)
2dx: 6 1.2 The Hamiltonian principle The velocity of the material point x;u(x;t)
of the string at time tin the direction orthogonal to the x-axis isd dtu(x;t). Therefore the kinetic energy E(t) of the string at time tis E(t) =Zb a1 2(x) ut(x;t)
2dx; where(x) is the mass of the string per unit length. To formulate Hamilton's principle I use the following notations: For a con- tinuously dierentiable function v: [a;b][0;T]!Rlet Vv(t) =Zb a 2 vx(x;t)
2dx Ev(t) =Zb a(x) 2 vt(x;t)
2dx: Hamilton's principle: LetT >0, let the movement of the string be given by the continuously dierentiable function u: [a;b][0;T]!R; and letw: [a;b][0;T]!Rbe a continuously dierentiable function satisfying w(x;0) =w(x;T) =w(a;t) =w(b;t) = 0 (1.1) for allaxband all 0tT. Letsdenote real numbers. Hamilton's principle states that the movement is such that d dsZT 0Eu+sw(t)Vu+sw(t)dtjs=0= 0: (1.2) Remark. Ifjsjis a small number, then v(x;t) =u(x;t) +sw(x;t) is a small perturbation of the movement of the string, which because of (1.1) does not change the boundary, initial and nal values. Therefore Hamilton's principle states that the material points of the string move such that the integral ZT 0E(t)V(t)dt is stationary when the movement of the string is perturbed such that the initial, nal and boundary values are not changed. 7 The equation (1.2) can be used to derive an equation for the movement of the string. For, (1.2) yields 0 =d dsZT 0Zb a(x) 2 ut(x;t) +swt(x;t)
2 2 ux(x;t) +swx(x;t)
2dxdtjs=0 =ZT 0Zb a(x) ut(x;t) +swt(x;t)
wt(x;t)  ux(x;t) +swx(x;t)
wx(x;t)dxdtjs=0 =ZT 0Zb a (x)ut(x;t)wt(x;t)ux(x;t)wx(x;t)
dxdt =:I: Ifuis two times continuously dierentiable, then the last integral can be trans- formed using partial integration. Since wvanishes at the boundary of the rectangle [a;b][0;T] we obtain 0 =I=ZT 0Zb a (x)utt(x;t)uxx(x;t)
w(x;t)dxdt: This must hold for all continuously dierentiable functions wvanishing at the boundary. If (x)utt(x;t)uxx(x;t) is continuous, this can only hold if (x)utt(x;t)uxx(x;t) = 0 (1.3) for all (x;t)2[a;b][0;T]. 1.3 Initial-boundary value problems for the one-dimensional wave equation SinceTis an arbitrary chosen positive number, we conclude that the vibrating string must satisfy the equation (1.3) in the whole domain [ a;b][0;1). We thus have (x)utt(x;t) =uxx(x;t);(x;t)2[a;b][0;1): This is a linear partial dierential equation of second order for u, the wave equation in one space dimension. Since the ends of the string at x=aorx=b can be xed or can be subjected to arbitrarily given motions, and since at time t= 0 the material points of the string can be displaced arbitrarily and can be submitted to arbitrarily given velocities, one wants to solve the following initial-boundary value problem to determine the motion of the string: (x)utt(x;t) =uxx(x;t); (x;t)2[a;b][0;1); (BD)u(a;t) =u(a)(t); u(b;t) =u(b)(t); t2[0;1); (IC)u(x;0) =u(0)(x); ut(x;0) =u(1)(x); x2[a;b]; 8 with given functions u(a);u(b): [0;1)!R,u(0);u(1): [a;b]!R:This is theDirichlet initial-boundary value problem for the wave equation. The Neumann initial-boundary value problem is obtained if instead of the valuesu(a;t) andu(b;t) the values ux(a;t) andux(b;t) for thexderivatives are prescribed: (x)utt(x;t) =uxx(x;t); (x;t)2[a;b][0;1); (BC)ux(a;t) =v(a)(t); ux(b;t) =v(b)(t); t2[0;1); (IC)u(x;0) =u(0)(x); ut(x;0) =u(1)(x); x2[a;b]: Ifa=1 andb=1and no boundary conditions are posed, then one speaks of the Cauchy problem : (x)utt(x;t) =uxx(x;t); (x;t)2(1;1)[0;1); (IC)u(x;0) =u(0)(x); ut(x;0) =u(1)(x); x2(1;1): 1.4 Initial-boundary value problems for the wave equation in higher space dimensions Consider an elastic membrane, which at the boundary is xed to a wire forming a closed loop. The projection of the membrane to the plane R2is a closed bounded set, the interior of which is . We assume that the boundary @ is continuously dierentiable, and that the wire is given by the graph of a continuously dierentiable function :@ !R. Letu(x;t)2Rbe the height of the membrane above the point x2 at timet0. Thus, at time tthe membrane is represented by the graph of the function x7!u(x;t) : !R: Since the membrane is attached at the boundary to the wire, we have the Dirichlet boundary condition u(x;t) =(x); x2@ ; t0: To determine a partial dierential equation for the function uwe again apply Hamilton's principle. We rst need to make assumptions for the elastic prop- erties of the membrane, or equivalently for the form of the potential energy stored in the membrane. Generalizing the one-dimensional potential energy we assume here that the potential energy Vu(t) of the membrane at time tis given by Vu(t) = 2Z jrxu(x;t)j2dx; with the gradient rxu(x;t) = @ @x1u(x;t) @ @x2u(x;t)! : 9 n(x) @ x The exterior unit normal vector n(x) The kinetic energy of the membrane is Eu(t) =Z (x) 2 ut(x;t)
2dx: IfT >0 and if w: [0;T]!R is continuously dierentiable with w(x;t) = 0;(x;t)2@ [0;T] w(x;0) =w(x;T) = 0; x2 ; then Hamilton's principle yields 0 =d dsZT 0Eu+sw(t)Vu+sw(t)dtjs=0 =d dsTZ 0Z (x) 2 ut(x;t) +swt(x;t)
2 2jrxu(x;t) +srxw(x;t)j2dxdtjs=0 =ZT 0Z (x)ut(x;t)wt(x;t)rxu(x;t)
rxw(x;t)
dxdt: Ifuis two times continuously dierentiable then the rst Green's formula yields 0 =ZT 0Z (x)utt(x;t)
xu(x;t)
w(x;t)dxdt +Z (x)ut(x;T)w(x;T)(x)ut(x;0)w(x;0)dx ZT 0Z @ @ @nxu(x;t)
w(x;t)dxdt =ZT 0Z (x)utt(x;t)
xu(x;t)
w(x;t)dxdt; (1.4) 10 with the Laplace operator
xu(x;t) =2X i=1@2 @x2 iu(x;t) and the normal derivative @ @nu(x;t) =n(x)
rxu(x;t); wheren(x)2R2is the unit normal vector to the boundary @ atx2@ pointing into the exterior R2n of . (1.4) must be satised for all wwith the stated properties. This is only possible if the bracketed expression in the integrand on the right hand side vanishes identically, whence umust satisfy (x)utt(x;t) =
xu(x;t);(x;t)2 [0;T]: This is the wave equation in two space dimensions. Since Tis arbitrary it follows that umust satisfy the wave equation for all ( x;t)2 [0;1). We already noted that umust satisfy the Dirichlet boundary condition. Therefore umust be a solution of the Dirichlet initial-boundary value problem, which we immediately formulate for the n-dimensional wave equation. Thus, for n2N let
xu(x;t) =nX i=1@2 @x2 iu(x;t); x = (x1;:::;xn)2Rn be then-dimensional Laplace operator. With this operator the inhomogeneous Dirichlet initial-boundary value problem in a domain Rnis (x)utt(x;t) =
xu(x;t) +f(x;t); (x;t)2 [0;1); (BC)u(x;t) =(x;t); (x;t)2@ [0;1); (IC)u(x;0) =u(0)(x); ut(x;0) =u(1)(x); x2 ; with given functions f: [0;1)!R;  :@ [0;1)!R; u(0);u(1): !R: The vibrations of the membrane can be determined by solving this problem forn= 2. Physically, fis a surface force acting on the membrane, for example the gravitational force. The Neumann initial-boundary value problem for the wave equation in  Rnis (x)utt(x;t) =
xu(x;t) +f(x;t); (x;t)2 [0;1); (BC)@ @nu(x;t) =(x;t); (x;t)2@ [0;1); (IC)u(x;0) =u(0)(x); ut(x;0) =u(1)(x); x2 ; 11 and the Cauchy problem is (x)utt(x;t) =
xu(x;t) +f(x;t); (x;t)2Rn[0;1); (IC)u(x;0) =u(0)(x); ut(x;0) =u(1)(x); x2Rn: 12 2 The Helmholtz equation obtained by reduction of the wave equation 2.1 Separation of variables and boundary value problems for the Helmholtz equation Letn1 be an integer, and let Rnbe an open set. Dene c(x) =r (x)>0: With this notation the homogeneous wave equation becomes utt(x;t) =c(x)2
xu(x;t);(x;t)2 [0;1): (2.1) Complex valued solution. Up to now we considered solutions of (2.1) with values in the real numbers. One denes complex valued solutions u:  [0;1)!Cwithu=u1+iu2,u1;u2: [0;1)!R, by setting
xu(x;t) =
xu1(x;t)+i
xu2(x;t) and@2 tu(x;t) =@2 tu1(x;t)+i@2 tu2(x;t). Of course, since c(x)2in (2.1) is real, in this special case a complex valued function is a solution, if both the real part u1and the imaginary part u2solve the wave equation. Though complex valued solutions seem to be more complicated than real valued solutions, it turns out that allowing complex valued solutions elucidates the situation considerably. Henceforth we consider complex valued solutions. Separation of variables. To nd a solution of the wave equation (2.1) it suggests itself to try the product ansatz u(x;t) =w(t)v(x): Insertion into the wave equation yields wtt(t)v(x) =c(x)2w(t)
v(x); hence wtt(t) w(t)=c(x)2
v(x) v(x): This equation must hold for all x2 and allt2[0;1). Since the left hand side only depends on tand the right hand side on x, this is only possible if the fractions on both sides have a constant value 2R. Thus, wtt(t) +w(t) = 0; t2[0;1) (2.2) c(x)2
v(x) +v(x) = 0; x2 : (2.3) The rst equation is a linear, homogeneous ordinary dierential equation of second order. The second equation is called Helmholtz equation or reduced wave 13 equation, a linear second order partial dierential equation. More precisely, these names are usually reserved to the equations obtained for c= 1, but we use them also in the case when the coecient c(x)2>0 is variable. With = 0 one obtains the potential equation
v(x) = 0; x2 : For6= 0 the general solution of (2.2) is w(t) =C1ep t+C2ep t =C1eRep t cos(Imp t) +isin(Imp t)
+C2eRep t cos(Imp t)isin(Imp t)
; whereas for = 0 the general solution is given by w(t) =C1t+C2: By choosing the constant suitably we can thus construct solutions of the wave equation with special behavior in time. For example, if >0 and ifvis a solution of the Helmholtz equation to this , then u(x;t) = C1cos(p t) +C2sin(p t)
v(x) is a solution representing an undamped oscillation with an amplitude jv(x)j depending on the position x. If<0 then u(x;t) =C1ep tv(x) is a solution, which increases exponentially in time, and u(x;t) =C2ep tv(x) is an exponentially decreasing solution. The method to solve the wave equation with the product ansatz u(x;t) = w(t)v(x) is called method of separation of variables. Of course, with this ansatz the Dirichlet boundary condition w(t)v(x) =u(x;t) =(x;t);(x;t)2@ [0;1) can only be solved if also the given boundary data are of the form (x;t) =w(t) (x) with a function :@ !R. In this case u(x;t) =w(t)v(x) solves the wave equation and the Dirichlet boundary condition if wsolves (2.2) and vsolves the Dirichlet boundary value problem c(x)2
v(x) +v(x) = 0; x2 ; v(x) = (x); x2@ 14 for the Helmholtz equation. Also, u(x;t) =w(t)v(x) solves the wave equation and the Neumann boundary condition @ @nu(x;t) =(x;t) =w(t) ^ (t);(x;t)2@ [0;1); ifvsatises the Neumann boundary value problem c(x)2
v(x) +v(x) = 0; x2 ; @ @nv(x) = ^ (x); x2@ for the Helmholtz equation. 2.2 Linear partial dierential equations of order m More general solutions of the wave equation can be obtained by adding two solutionsu1(x;t) =w1(t)v1(x) andu2(x;t) =w2(t)v2(x) of the wave equation constructed with the method of separation of variables, for example by choosing dierent constants 1and2. More precisely, any linear combination a1u1(x;t) +:::+amum(x:t); aj2C; of solutions ui(x;t) of the wave equation is itself a solution of the wave equation. Even innite series of solutions of the wave equation can yield new solutions. This is shown by the following Theorem 2.1 Letfumg1 m=1be a sequence of two times continuously dieren- tiable solutions of the wave equation (2.1) in the domain [0;1). If the function series 1X m=1@j1+:::+jn+k @xj1 1:::@xjnn@tkum(x1;:::;xn;t); k +j1+:::+jn2; converge uniformly in every compact subset of [0;1), then u(x;t) =1X m=1um(x;t) is a two times dierentiable solution of the wave equation. in [0;1). Theproof follows from the well known result of calculus, that under the as- sumptions of the theorem the function uis two-times continuously dierentiable. We leave the proof to the reader. Every linear combination of solutions of the wave equation is a solution since the unknown function uand its derivatives appear only linearly in the wave 15 equation. Partial dierential equations with this property are called linear. To dene precisely the notion of a linear partial dierential equation I introduce the following notations: For a multi-index = (1;:::;n)2Nn 0and forx= (x1;:::;xn)2Rnlet jj=1+:::+n(length of the multi-index) ; Dv(x) =@jj @1x1:::@nxnv(x1;:::;xn); ! =1!2!:::n!: Denition 2.2 Letm2Nbe a given number and let x= (x1;:::;xn) denote points in Rn. The expression X 2Nn 0 jjma(x)Dv(x) =f(x) is called linear partial dierential equation for the function vwith given (real or complex valued) coecient functions aand given right hand side f. This equation is called of order mat the point xif at least one of the coecient functionsawithjj=mdoes not vanish at x. The partial dierential equation is called homogeneous if f0. Partial dierential equations are grouped into various classes comprising equa- tions with similar properties. Most important are the classes of elliptic, parabolic and hyperbolic equations. The Helmholtz equation is the prototype of a linear elliptic equation, the wave equation is the prototype of a linear hyperbolic equation, the heat equation ut(x;t) =c(x)
xu(x;t);(x;t)2 [0;1); is the prototype of a linear parabolic equation. The precise denitions of elliptic and hyperbolic equations are given in Sections 8.1 and 10.1. Well posed problem. A boundary value problem, initial value problem or initial-boundary value problem is called well posed, if it has the following three properties: 1. a solution of the problem exists, 2. the solution is unique, 3. the solution depends continuously on the right hand side, on the boundary data and on the initial data. 16 The meaning of continuos dependence has to be made precise in the context of the particular problem studied. These lecture notes are mostly devoted to the study of questions of well posedness of linear elliptic partial dierential equations. Only in Section 10 we return to the wave equation, where we show how the solution theory for the Helmholtz equation, and more generally, for elliptic equations developed in Sections 8 and 9 can be used to solve initial- boundary value problems for the wave equation and for more general hyperbolic equations. 17 3 Tools from functional analysis. Weak solutions of one dimensional boundary value problems. 3.1 The Hilbert space L2( ;C) Let Rnbe a nonempty, open or closed set. L2( ) =L2( ;C) is the space of all quadratically integrable functions: L2( ) =ff: !CZ jf(x)j2dx<1g: We show that L2( ) is a vector space: Theorem 3.1 (Cauchy-Schwarz inequality) Letf;g2L2( ). Then the productf
gis integrable and satises jZ f(x)g(x)dxjZ jf(x)j2dx1=2Z jg(x)j2dx1=2 : Proof: Leta;b0. From 0(ab)2=a22ab+b2we infer that ab 1 2(a2+b2). Setting a=jf(x)j R jf(x)j2dx
1=2; b =jg(x)j R jg(x)j2dx
1=2; we conclude that jf(x)g(x)j R jf(x)j2dx
1=2R jg(x)j2dx
1=2jf(x)j2 2R jf(x)j2dx+jg(x)j2 2R jg(x)j2dx: Since the right hand side is integrable we see from this inequality that f
gis integrable and that R jf(x)g(x)jdx (R jf(x)j2dx)1=2(R jg(x)j2dx)1=2R jf(x)j2dx 2R jf(x)j2dx+R jg(x)j2dx 2R jg(x)j2dx= 1: This shows that the Cauchy-Schwarz inequality holds. Corollary 3.2 (Minkowski inequality) Letf;g2L2( ). Thenf+g2 L2( )and Z jf(x) +g(x)j2dx1=2 Z jf(x)j2dx1=2 +Z jg(x)j2dx1=2 : 18 Proof: The Cauchy-Schwarz inequality implies Z jf(x) +g(x)j2dx=Z (f(x) +g(x))(f(x) +g(x))dx =Z jf(x)j2+g(x)f(x) +f(x)g(x) +jg(x)j2dx =Z jf(x)j2dx+ 2 ReZ f(x)g(x)dx+Z jg(x)j2dx; Z jf(x)j2dx+ 2Z jf(x)j2dx1=2Z jg(x)j2dx1=2 +Z jg(x)j2dx =Z jf(x)j2dx1=2 +Z jg(x)j2dx1=22 : This implies Minkowski's inequality. Forf;g2L2( ) let kfk=kfk =Z jf(x)j2dx1=2 ; (f;g) = (f;g) =Z f(x)g(x)dx: Corollary 3.3 L2( ) is a vector space, kfkis a norm and (f;g)is a scalar product on this vector space with kfk= (f;f)1=2: ThereforeL2( )is a pre-Hilbert space. Theorem 3.4 (of Fischer-Riesz.) L2( ) is a Hilbert space, i.e. the pre- Hilbert space L2( )is complete with respect to the norm kfk. Theproof can be found in the book "Lineare Funktionalanalysis\ of H.W. Alt, Springer Verlag Berlin, 1999, pp. 49, 50. 3.2 The Riesz representation theorem and the projection theorem LetXbe an abstract Hilbert space over Cwith the scalar product ( u;v) and the normkuk= (u;u)1=2. LetF:X!Cbe a continuous linear functional (linear mapping). Fis continuous if and only if Fis bounded, i.e. if a constant Cexists such that jF(u)jCkuk for allu2X. Dene the mapping JF:X!Rby JF(u) =1 2kuk2ReF(u); for allu2X. 19 Theorem 3.5 LetYbe a closed subspace of X. Thenu2Ysatises JF(u) = min v2YJF(v) if and only if for all v2Y (v;u) =F(v): Proof: LetJF(u) = minv2YJF(v). Then for all v2Ythe function 7!JF(u+v) :R!R has the minimum at = 0, hence 0 =d dJF(u+v)j=0=d d1 2(u+v;u +v)ReF(u+v) =d d1 2(u;u) +Re (v;u) +21 2(v;v)ReF(u)ReF(v) = Re (v;u) +(v;v)ReF(v)
j=0 = Re (v;u)ReF(v): Therefore we have Re (v;u) = ReF(v) for allv2Y. Thus, we also have for v2Y Im (v;u) =Rei(v;u) =Re (iv;u) =ReF(iv) =ReiF(v) = ImF(v): Together it follows for all v2Y (v;u) = Re (v;u) +iIm (v;u) = ReF(v) +iImF(v) =F(v): Assume next that (v;u) =F(v) for allv2Y. We have for all v2Y JF(u+v) =1 2(u;u) + Re (v;u) +1 2(v;v)ReF(u)ReF(v) =1 2(u;u)ReF(u) +1 2(v;v)JF(u); whence JF(u) = min v2YJF(u+v) = min w2YJF(w): Theorem 3.6 The mapping JFassumes the minimum on the subspace Yat a uniqueu2Y. 20 Proof: We use the parallelogram equality ku+vk2+kuvk2= 2kuk2+ 2kvk2; which holds for all u;v2X. Note also that for all a;b0 and all">0 0p"a1p"b2 ="a22ab+1 "b2; whence ab" 2a2+1 2"b2: Therefore we have with "=1 2 JF(u) =1 2kuk2ReF(u)1 2kuk2jF(u)j 1 2kuk2Ckuk1 2kuk21 2"C2" 2kuk2 =1 4kuk2C2C2: Consequently the inmum of JFexists onYand satises d= inf v2YJF(v)C2: Choose a sequence fungnYsuch that lim n!1JF(un) =d: The parallelogram equality yields kumunk2= 2kumk2+ 2kunk24k1 2(um+un)k2 = 41 2kumk2ReF(um) +1 2kunk2ReF(un)
81 2k1 2(um+un)k2ReF1 2(um+un)
 = 4JF(um) + 4JF(un)8JF1 2(un+um)
4JF(um) + 4JF(un)8d!0; form;n!1 . Consequently, fungnis a Cauchy sequence and has a limit u. SinceYis closed,ubelongs to Y. From the Cauchy-Schwarz inequality j(v;w)jkvkkwkit follows that the mapping w7!kwk2:X!Ris continuous, henceJFis continuous. We thus obtain inf v2YJF(v) = lim n!1JF(un) =JF(u): 21 Thereforeuis the minimum of JFonY. To see that the minimum is unique, letuandvbe two minima on Y. The calculation above yields kuvk2= 4JF(u) + 4JF(v)8JF1 2(u+v)
4d+ 4d8d= 0; whenceu=v. This completes the proof. Corollary 3.7 (i)(Riesz representation theorem) To every bounded linear mappingF:X!Cthere is a unique u2Xsuch that (v;u) =F(v) for allv2X. (ii)Projection theorem) LetYbe a closed subspace of X. To every v2X there is a unique u2Ysuch that kvuk= min w2Ykvwk: uis the unique element in Ywhich satises (vu;w) = 0; (3.1) for allw2Y. Proof: (i) For the subspace in Theorems 3.5 and 3.6 choose Y=X, letube the minimum of JFand apply Theorem 3.5. (ii) Dene the bounded linear functional F:X!Cby F(w) = (w;v): By Theorem 3.6 the mapping JFhas a unique minimum uonY. Since JF(w) =1 2kwk2ReF(w) =1 2kwk2Re (w;v) +1 2kvk2 1 2kvk2=1 2kvwk21 2kvk2; uis also the unique minimum of w7!kvwkonY. By Theorem 3.5, u2Y is the unique element satisfying ( w;u) =F(w) = (w;v) for allw2Y. This implies (3.1). Remark 3.8 The spaceX0of bounded linear functionals on Xis called dual space ofX. The Riesz representation theorem shows that for the Hilbert space Xthere is a mapping T:X0!X, which assigns to every F2X0a unique elementTF2X, which allows to represent Fby the mapping (
;TF). We see immediately that Tis injective. It is also surjective: To see this, consider the linear mapping G:X!Cdened byG(v) = (v;u). The Cauchy-Schwarz inequality implies jG(v)jkukkvk. HenceGis a bounded linear functional withu=TG. This shows that X0is isomorphic to X. 22 3.3 Complete orthonormal systems Denition 3.9 Letfvmg1 m=1be a sequence in a Hilbert space X. (i) If (vm;v`) = 0 form6=`andkvmk= 1 for all m, thenfvmgmis called a (countable) orthonormal system in X. (ii) The orthonormal system fvmgmis called complete if the linear subspace spanfvmgm=nkX m=1amvmk2N; a1;:::ak2Co is dense in X. Theorem 3.10 Letfvmgmbe an orthonormal system. Equivalent are (i)fvmgmis complete. (ii) For all f2Xthe seriesP1 m=1(f;vm)vmconverges to finX: f=1X m=1(f;vm)vm; i.e. lim k!1kfkX m=1(f;vm)vmk= 0: (iii) (Parseval identity) For allf2Xwe have kfk2=1X m=1j(f;vm)j2: For a proof cf. pp. 274, 275 of the book of Alt.P1 m=1(f;vm)vmis called Fourier series of fand (f;vm) is the m-th Fourier coecient. Theorem 3.11 An orthonormal system fvmgmis complete if and only if for allf2X,f6= 0, there isvk2fvmgmsuch that (f;vk)6= 0: Proof. LetV= spanfvmgm. It is obvious that there is w2Vwith (f;w)6= 0 if and only if there is vk2fvmgmwith (f;vk)6= 0. Therefore it suces to show thatV=Xif and only if to all f2Xwithf6= 0 there is w2Vsuch that (f;w)6= 0. Now, ifV=Xthen for all f2X,f6= 0, choose w=f. This yields (f;w) = (f;f)>0. On the other hand, if V6=Xchooseg2XnV. SinceVis a closed subspace it follows by Corollary 3.7 (projection theorem) that there is g02Vsuch thatf=gg06= 0 satises ( f;w) = 0 for all w2V. Hence, the statement of the theorem follows. 23 Example 3.12 Form2Zletvm: (0;2)!Cbe dened by vm(x) =1p 2eimx: fvmg1 m=1is a complete orthonormal system in L2(0;2). Proof.fvmgmis an orthonormal system, since (v`;vm) =Z2 0v`(x)vm(x)dx=1 2Z2 0ei(`m)xdx=`m; with the Kronecker symbol `m=( 1; `=m; 0;otherwise: To show that the orthonormal system is complete, we need Theorem 3.13 (Fej er) Letg:R!Cbe continuous and 2-periodic. For k;m;n2Z,m0,n1dene ak=1 2Z2 0g(x)eikxdx=1p 2(g;vk); sm(x) =mX k=makeikx; n(x) =1 n s0(x) +:::+sn1(x)
: Then the sequence fng1 n=1converges to guniformly on [0;2]. With this theorem we can prove that fvmgmis complete: Let f2L2(0;2) and " >0 be given arbitrarily. By a well known result from Lebesgue integration theory, the set of continuous functions on [0 ;2] vanishing at x= 0 andx= 2 is dense in L2(0;2). We can therefore choose such a function gwith kfgk<": Sincegvanishes at the boundary points of the interval [0 ;2], it follows that the 2-periodic extension of gtoRis continuous. By the Theorem of Fej er it thus follows that there is n2Nwith sup 0x2jg(x)n(x)j<": Thus kfnk  kfgk+kgnk "+Z2 0jg(x)n(x)j2dx1=2 "(1 +p 2): 24 Sincenis a linear combination of functions from fvmg1 m=1, we conclude from this estimate that span fvmg1 m=1is dense inL2(0;2). Consequently the orthonormal system is complete. Remark 3.14 Sinceeimxis 2{periodic, the family f1p 2eimxg1 m=1is obvi- ously a complete orthonormal system on every interval ( a;2+a) obtained by translation of the interval (0 ;2) bya2R. This remark holds also for the or- thonormal system of the next example, which is often considered on the interval (;). Example 3.15 A complete orthonormal system in L2(0;2)of real functions is given by1pcos(mx);1psin(mx)jm= 0;1;2;::: : Proof: A well known computation shows that this system is orthonormal. To prove completeness it suces to remark that for the functions vmfrom Example 3.12 vm(x) =1p 2cos(mx) +i1p 2sin(mx): Hence, the linear span of this system is equal to the dense subspace span fvmgm. 3.4 Eigenfunctions of the Dirichlet boundary value problem in R1. The Helmholtz equation in R1is an ordinary dierential equation. Therefore the solution of the boundary value problems to the Helmholtz equation is consider- ably simpler in one space dimension than in higher dimensions. Nevertheless, the solution properties of the one dimensional and higher dimensional problems are similar. Since it is helpful to know these properties when studying higher dimensional problems, we investigate in this section the one-dimensional prob- lem. Thus, let = ( a;b), let a; b2Cand2C. We search a two times continuously dierentiable solution u: [a;b]!Cof u00(x) +u(x) = 0; x2[a;b]; u(a) = a; u(b) = b: For= 0 the general solution of the ordinary dierential equation is u(x) =C1x+C2; C 1;C22C: The boundary conditions yield the linear system C1a+C2= a; C1b+C2= b: 25 It follows that for = 0 the boundary value problem has a unique solution given by u(x) = a b abx+1 ab(a bb a): For6= 0 the general solution of the ordinary dierential equation is u(x) =C1ep x+C2ep x withC1;C22C. The boundary conditions imply C1ep a+C2ep a= a C1ep b+C2ep b= b: This is a linear system of equations for C1andC2with the coecient matrix A=ep aep a ep bep b : Therefore the boundary value problem is uniquely solvable for all a; bif and only if detA6= 0. Now detA=ep aep bep bep a=ep (ab)(1e2p (ba)): Thus, detA= 0 if and only if 2p (ba) = 2im; m2Z; which is equivalent to =m=m ba
2: Together we obtain Theorem 3.16 (i)The boundary value problem u00(x) +u(x) = 0; axb u(a) = a; u(b) = b is uniquely solvable for all a; b2Cif6=mfor allm2N, where m=m ba
2; m2N: In particular, u= 0 is the only solution to the homogeneous boundary value problem ( a= b= 0) . (ii)If there ism2Nsuch that=m, then the boundary value problem is not solvable for all a; b, and the solution is not unique. In particular, for every C6= 0 the function um(x) =Csinp m(xa)
=Csinm ba(xa)
is a nonzero solution of the homogeneous boundary value problem. 26 For the proof it only remains to show that umsolves the homogeneous boundary value problem. Yet, obviously um(a) = 0; um(b) =Csin(m) = 0: Denition 3.17 The numbers m=m ba
2,m2N, are called eigenvalues of the boundary value problem u00(x) +u(x) = 0; u(a) = a; u(b) = b: Every nonvanishing solution of this boundary value problem with =mand a= b= 0 is called eigenfunction to the eigenvalue m. Theorem 3.18 Let um(x) =r 2 basinm ba(xa) : fumg1 m=1is a complete orthonormal system in L2([a;b])of eigenfunctions to the Dirichlet boundary value problem. Proof. Above we showed that umis an eigenfunction for the Dirichlet boundary value problem, and a simple computation yields that fumgmis orthonormal. To prove completeness, we scale and translate umto dene the odd function wm: [;]!Cby wm(x) =8 >>< >>:umba x+a =r 2 basin(mx);0x; wm(x) =r 2 basin(mx);x0: By Remark 3.14 and Example 3.15, span fwmgmis dense in the space ff2L2(;)jf(x) =f(x)g; since for odd functions the Fourier coecients of the cosine functions vanish. From this we conclude immediately that span fumgmis dense in L2(a;b). This result suggests to construct solutions of the Dirichlet boundary value prob- lem u00(x) +u(x) =f(x); axb; u(a) =u(b) = 0 with a given function f2L2([a;b]) as follows: 27 Letfmgmbe the sequence of eigenvalues to the Dirichlet boundary value problem and assume that 6=mfor allm. With the complete orthonormal systemfumgmof eigenfunctions consider the series 1X m=11 m(f;um)um: This series converges in L2([a;b]). To see this, note that `X m=k1 m(f;um)um 2 =`X m;j=k(f;um) m(f;uj) j(um;uj) =`X m=k(f;um) m2 ; hence the series is a Cauchy sequence, and therefore converges, if and only ifP1 m=1(f;um) mj2<1. Now,m=m ba
2!1 form!1 implies that there is a constantC > 0 such that 1 mC m2 for allm2N. Thus, 1X m=1(f;um) m21X m=1C2 m4j(f;um)j2C21X m=1j(f;um)j2<1; since the Fourier seriesP1 m=1(f;um)umconverges to finL2([a;b]). Conse- quently, the seriesP1 m=11 m(f;um)umconverges. Denote the limit function byu: u=1X m=11 m(f;um)um: We want to show that uis a solution of the inhomogeneous boundary value problem. To this end note that if uis two-times dierentiable and if the deriva- tives can be interchanged with the summation sign it follows that u00+u=d2 dx21X m=1(f;um) mum+1X m=1(f;um) mum =1X m=1(f;um) m(u00 m+um) =1X m=1(f;um) m(m)um =1X m=1(f;um)um=f: Moreover, if in addition the seriesP1 m=1(f;um)um(x) converges for all x2[a;b] tou(x), then u(a) =1X m=1(f;um) mum(a) = 0; u(b) =1X m=1(f;um) mum(b) = 0; 28 because of um(a) =um(b) = 0. Thus, under the assumed properties of the seriesP1 m=1(f;um) mumthe limit function uis a solution of the inhomogeneous boundary value problem. However, in general these assumptions are not satised for f2L2([a;b]). Namely, a precise investigation shows that the boundary value problem is solv- able in the classical sense only if fsatises certain regularity properties, for example if fis continuous. Yet, if the boundary value problem has a classical solution, then it coincides with the function ugiven by the series. From there the idea originates to generalize the notion of a solution of the boundary value problem and to dene weak solutions. The weak solution has the property to coincide with the classical solution if it exists. I introduce weak solutions in the following. 3.5 Weak derivatives First I dene weak derivatives. I need the following standard notations: Denition 3.19 (i) Let Rnbe open. For m2N0[f1g let Cm( ) =Cm( ;C) =ff: !CjDfexists and is continuous for all2Nn 0such thatjjmg be the space of all m-times continuously dierentiable functions. One also writes C( ) =C0( ). (ii)C m( ) =ff2Cm( )jDf2L2( ) for alljjmg, (iii)Cm( ) =ff2Cm( )jDfcan be extended continuously up to the boundary g: (iv) Forf2C(Rn) let suppf=fx2Rnjf(x)6= 0gbe the support of f. (v) C1( ) =f'2C1(Rn)jsupp'is a compact subset of g. Of courseCm( );C m( );Cm( ) and C1( ) are vector spaces. Theorem 3.20 The space C1( ) is a dense subset of L2( ), i.e. C1( ) = L2( ). Aproof can be found in the book of H.W. Alt, pp. 74, 75. Denition 3.21 Letv2L2( ) and2Nn 0. If there is a function w2L2( ) such that (1)jj(v;D') = (w;') for all'2 C1( ), thenwis called the -th weak derivative of v. 29 Theorem 3.22 (i)The-th weak derivative is uniquely determined. (ii)Forv2C m( ) andjjmthe-th weak derivative coincides with the classical derivative Dv. Proof. (i) Letw1andw2be weak-th derivatives of v2L2( ). Then, for all '2 C1( ) (w1;') = (1)jj(v;D') = (w2;'); hence (w1w2;') = 0. Since C1( ) =L2( ), there is a sequence f'mgm  C1( ) such that lim m!1k(w1w2)'mk= 0. Thus (w1w2;w1w2) = lim m!1[(w1w2;(w1w2)'m) + (w1w2;'m)] lim m!1kw1w2kk(w1w2)'mk= 0; whencew1=w2. Here I used Cauchy-Schwarz' inequality. Therefore vhas at most one weak derivative. (ii) To'2 C1( ) there is a neighborhood of the boundary @ where'van- ishes. Thus, for v2C m( ) it follows by partial integration (1)jj(v;D') = (1)jjZ v(x)D'(x)dx =Z Dv(x)'(x)dx= (Dv;'): Consequently, Dv2L2( ) is the weak derivative of v. Because of this theorem one uses the notation Dvalso for weak derivatives of v. Confusion is not possible, since the weak derivative is equal to the classical derivative, if the latter exists. Examples (a) Let = (1;1) and letv2L2 (1;1)
be dened by v(x) =jxj. This function has the weak derivative v0(x) =( 1;1<x< 0 1;0x<1: For, if'2 C1 (1;1)
then (v;'0) =Z1 1v(x)'0(x)dx=Z0 1x'0(x)dxZ1 0x'0(x)dx =Z0 1'dx+Z1 0'(x)dx=Z1 1v0(x)'(x)dx= (v0;'): 30 (b)vdoes not have a second weak derivative. For, if v002L2( ) is the second weak derivative then for all '2 C1( ) (v00;') = (v;'00) =(v0;'0) =Z0 1'0(x)dxZ1 0'0(x)dx ='(0)'(1)'(1) +'(0) = 2'(0): Now choose '2 C1 (1;1)
with'(0)6= 0 and dene '`by'`(x) ='(`x), for`2N. Then'`2 C1 (1;1)
, and by the preceding equation 2j'(0)j= 2j'`(0)j= 2 lim `!1j'`(0)j= lim `!1j(v00;'`)j lim `!1kv00kk'`k=kv00klim `!1Z1 1j'(`x)j2dx1=2 =kv00klim `!1Z1 1j'(y)j21 `dy1=2 = 0: This contradicts '(0)6= 0, hence vcannot have the second weak derivative v00. 3.6 Sobolev spaces Denition 3.23 For an open set Rnandm2N0let Hm( ) =fv2L2( )jthe weak derivative Dvexists for alljjmg: Hm( ) is called Sobolev space. For u;v2Hm( ) we dene (u;v)m= (u;v)m; =X jjm(Du;Dv) ;kukm=kukm; = (u;u)1=2 m; : Hm( ) is a vector space. We even have: Theorem 3.24 Hm( ) is a Hilbert space with the scalar product (u;v)mand the normkukm. Proof. It is immediately seen that ( u;v)mhas the properties of a scalar product. Therefore it remains to show that Hm( ) is complete. Thus, let fu`g1 `=1be a Cauchy sequence in Hm( ). Since ku`ukk2 m= (u`uk; u`uk)m=X jjmkDu`Dukk2; it follows thatfDu`g`is a Cauchy sequence in L2( ) forjjm. Because L2( ) is complete,fDu`g`has a limit function u()2L2( ). I write u=u(0) 31 and show that u()=Dufor all 0<jjm. To this end let '2 C1( ). Then (1)jj(u;D') = lim `!1(1)jj(u`;D') = lim `!1(Du`;') = (u();'): This implies u()=Du. Consequently, u2Hm( ) andkuu`km!0 for `!1 , whenceHm( ) is complete. Theorem 3.25 (i)C m( )is dense in Hm( ): Hm( ) =C m( ): (ii) If has Lipschitz boundary, then Cm( )is dense in Hm( ): Hm( ) =Cm( ): Aproof of this theorem can be found for example in the book of Alt, pp. 108-109, and also in my lecture notes: H.-D. Alber, Variationsrechnung und Sobolevr aume, p. 33 and Chapter 31. Denition 3.26 Let Rnbe an open set. The closure of the linear subspace  C1( ) inHm( ) is denoted by Hm( ).  Hm( ) is a closed linear subspace of Hm( ), hence Hm( ) is complete as a closed subspace of the complete space Hm( ). Therefore Hm( ) is a Hilbert space with the scalar product ( u;v)mand the normkukm. In general Hm( ) is a proper subspace of Hm( ). This subspace consists of all functions of Hm( ), which in a generalized sense vanish on the boundary @ . Another important property of Sobolev functions is that if m >n 2, then u2Hm( ) is continuous and all weak derivatives Duwithjj< mn 2are classical, hence Hm( )C [mn 2]( ), where [ r] denotes the largest integer not greater than r. This property is called Sobolev imbedding theorem. The investigation of these properties of Sobolev functions is an extended topic. Fortunately, in this introductory course we almost exclusively need those properties of Sobolev functions which immediately follow from the denitions of the Sobolev spaces given above. Yet, to familiarize the reader with Sobolev spaces we prove now two of these properties in the case R1: Theorem 3.27 Let = (a;b)Rbe an open interval and u;v2H1 (a;b)
. Then ju(y)u(x)j  ku0k(a;b)jyxj1=2; (3.2) ju(x)j r1=2ku0k(a;b)+r1=2kuk(a;b); (3.3) (u0;v)(a;b)+ (u;v0)(a;b)=u(b)v(b)u(a)v(a); (3.4) for almost all x;y2(a;b)and for all 0<rba. 1www.mathematik.tu-darmstadt.de/ags/ag6/Skripten/Skripten Alber/Vorlesungen.html 32 Remark 3.28 This means that there is a set M (a;b) with meas (a;b)nM
= 0, which consequently is dense in [ a;b], such that (3.2) and (3.3) hold for all x;y2M. By (3.2), uis H older continuous on M with exponent1 2. Hence,uis uniformly continuous on Mand can be modi- ed on@M, such that the modied function ~ uis H older continuous on all of M=M[@M= [a;b]. There can be no other continuous function in the equiv- alence class of u. Therefore we can single out this continuous function and iden- tify the equivalence class with ~ u. With this identication every u2H1 (a;b)
belongs to the space C1=2([a;b]) of H older continuous functions with exponent 1 2, andH1 (a;b)
is embedded in this space. In (3.4) we use this identication, sou(a);u(b);v(a);v(b) are the values of the continuous representatives. (3.4) shows that partial integration is allowed for weak derivatives. Proof. Choose a sequence fu`g`C1 [a;b]
such thatkuu`k1;(a;b)!0 for`!1 . Thenfu`g`converges in L2 (a;b)
tou. Thus, by a well known theorem from Lebesgue integration theory we can select a subsequence fu`kgk such that lim k!1u`k(x) =u(x) for almost all x2[a;b]. Letx < y be two points with this property and let ">0. Then there is k0such that ju(x)u`k(x)j<";ju(y)u`k(y)j<" forkk0. The fundamental theorem of calculus yields for kk0 ju(y)u(x)j  ju(y)u`k(y)j+ju`k(y)u`k(x)j+ju(x)u`k(x)j 2"+Zy xu0 `k(z)dz2"+Zy xdz1=2Zy xju0 `k(z)j2dz1=2 2"+jyxj1=2ku0 `kk(a;b) 2"+jyxj1=2 ku0k(a;b)+ku0 `ku0k(a;b)
: Because ofku0 `ku0k(a;b)ku`kuk1;(a;b)< "forkk1withk1suciently large, we deduce from this inequality by choosing kmax(k0:k1) that ju(y)u(x)j" 2 + (ba)1=2
+ku0k(a;b)jyxj1=2: Since">0 was arbitrary, (3.2) follows. To prove (3.3), let ( c;d) withx2(c;d)(a;b) be an interval of nite length. We integrate (3.2) with respect to yfromctodand obtain ju(x)j(dc) ku0k(a;b)Zd cjxyj1=2dx+Zd cju(y)jdy  ku0k(a;b)(dc)3=2+ (dc)1=2Zd cju(y)j2dy : 33 Division by ( dc) yields ju(x)j(dc)1=2ku0k(a;b)+ (dc)1=2kuk(a;b): This implies (3.3) with r=dc. To prove (3.4) we can assume that uandvare continuous. Choose sequences fu`g`,fv`g`C1([a;b]) such thatkuu`k1;(a;b)!0,kvv`k1;(a;b)!0 for `!1 . From (3.3) we obtain lim `!1ju(x)u`(x)jlim `!1 r1=2ku0u0 `k+r1=2kuu`k
= 0 for almost all x2[a;b]. This relation shows that fu`g`andfv`g`converge uniformly on Mtouandv, respectively, with the set Mdened in Remark 3.28. Sinceu;v;u`;v`are continuous, we infer that the function sequences converge pointwise everywhere to uandv. In particular, we have lim `!1u`(a) =u(a), lim`!1u`(b) =u(b);and similarly lim `!1v`(a) =v(a), lim`!1v`(b) =vb. Using the continuity of the scalar product we obtain by partial integration (u0;v) + (u;v0) = lim `!1 (u0 `;v`) + (u`;v0 `)
= lim `!1 u`(b)v`(b)u`(a)v`(a)
=u(b)v(b)u(a)v(a): Lemma 3.29 The orthogonal space  H? 1 (a;b)
=fu2H1 (a;b)
j(u;v)1= 0 for allv2 H1 (a;b)
g is given by  H? 1 (a;b)
=fC1ex+C2exjC1;C22Cg: Hence, the orthogonal space is of dimension 2. Proof.ubelongs to H? 1 (a;b)
if and only if for all v2 H1 (a;b)
(u0;v0) =(u;v): Since C1 (a;b)
 H1 (a;b)
, this equation holds if and only if uhas a second weak derivative which satises u00=u: All solutions of this ordinary dierential equation are of the form u(x) =C1ex+ C2exwith arbitrary constants C1;C22C. Theorem 3.30 u2 H1 (a;b)
if and only if u2H1 (a;b)
andu(a) =u(b) = 0. 34 Proof. Letu2 H1 (a;b)
. By denition of H1 (a;b)
there is a sequence fu`g` C1 (a;b)
withkuu`k1!0 for`!1 . We apply (3.3) to the dierenceuu`and note that u`(a) = 0 to obtain ju(a)jr1=2ku0u0 `k(a;b)+r1=2kuu`k(a;b)!0; for`!1 , whenceu(a) = 0. In the same way we conclude that u(b) = 0. To prove the converse let u2H1 (a;b)
satisfyu(a) =u(b) = 0. By Lemma 3.29 there is a unique v2 H1 (a;b)
andC1;C22Csuch that u(x) =v(x) +C1ex+C2ex: Sincev(a) =v(b) = 0, we obtain from this equation by setting x=aandx=b that C1ea+C2ea= 0 C1eb+C2eb= 0: This is a system of two linear equations for C1andC2with determinant of the coecient matrix eaebebea=eab(1e2(ba))6= 0, since ba > 0. Consequently we have C1=C2= 0. Thus, u=v2 H1 (a;b)
. 3.7 Weak solution of the Dirichlet boundary value problem to the Helmholtz equation We begin with the denition of weak solutions of the Helmholtz equation and of weak solutions to the homogeneous Dirichlet boundary value problem to this equation in n{dimensional space: Denition 3.31 (i) Let Rnbe a nonempty open set, let 2Cand assume thatf2L2( ;C). (i) A function u2 H1( ;C) is called weak solution of the partial dierential equation
u(x) +u(x) =f(x) (3.5) in , if for all '2 C1( ;C) the equation (ru;r') +(u;') = (f;') (3.6) holds, where (ru;r') =Z ru(x)
r'(x)dx=3X i=1Z @ @xiu(x)@ @xi'(x)dx: 35 (ii) A weak solution of the homogeneous Dirichlet boundary value problem
u(x) +u(x) =f(x); x2 ; uj@ = 0; is by denition a weak solution uof the partial dierential equation (3.5) be- longing to H1( ;C). Formally the equation (3.6) is obtained by multiplication of both sides of the equation
u+u=fby', integration and application of the rst Green's formula. The advantage is that weak solutions need to have only rst derivatives and not second. Every classical solution is also a weak solution, but not vice versa. However, if a weak solution belongs to C 2( ), then it is also a classical solution. Again we restrict ourselves to n= 1 and assume that = ( a;b) is a bounded open interval. In this case u2 H1 (a;b)
is a weak solution if (u0;'0) +(u;') = (f;') for all'2 C1 (a;b)
. Theorem 3.32 Letfmgmbe the eigenvalues of the Dirichlet boundary value problem in the bounded interval (a;b)R, and letfumgmbe a complete or- thonormal system of eigenfunctions. Assume that 6=mfor allm. Then the Dirichlet boundary value problem u00(x) +u(x) =f(x); a<x<b; u(a) =u(b) = 0; has a unique weak solution to every f2L2 (a;b)
, which is given by u=1X m=11 m(f;um)um: Proof. At rst it must be shown that ubelongs to the space H1 (a;b)
. Since the eigenfunction um(x) =r 2 basinm ba(xa)
satisesum(a) =um(b) = 0, we infer from Theorem 3.30 that um2  H1 (a;b)
. To prove that u2 H (a;b)
it therefore suces to show thatP1 m=11 m(f;um)umconverges in the norm of H1 (a;b)
. Since we already proved in Section 3.4 thatP1 m=1(f;um) mumconverges in L2 (a;b)
, it suces to 36 verify that alsoP1 m=1(f;um) mu0 mconverges in L2 (a;b)
. Proceeding as in Section 3.4 we compute kX m=`(f;um) mu0 m 2 =kX m=`(f;um) mu0 m;kX j=`(f;uj) ju0 j =kX m=`kX j=`(f;um) m(f;uj) j(u0 m;u0 j): Since (u0 m;u0 j) =(u00 m;uj) = (mum;uj) =( m; m =j 0; m6=j; we conclude kX m=`(f;um) mu0 m 2 =kX m=`(f;um) m2 mCkX m=`j(f;um)j2; with the constant C= supm2Nm jmj2<1. This inequality and the equation P1 m=1j(f;um)j2=kfk2<1together imply that the seriesP1 m=1(f;um) mu0 m satises the Cauchy convergence criterion, hence it converges in the complete spaceL2 (a;b)
. This proves that u2 H1 (a;b)
and that u0=1X m=1(f;um) mu0 m: In the next step of the proof we use this equation. Namely, for '2 C1 (a;b)
we have (u0;'0) =1X m=1(f;um) m(u0 m;'0) =1X m=1(f;um) m(u00 m;') =1X m=1(f;um) m(mum;') =1X m=1(m) m (f;um)um;'
=(u;') +1X m=1(f;um)um;' =(u;') + (f;'): Consequently, uis a weak solution. It remains to show that uis the only weak solution. Assume that v2  H1 (a;b)
is a second weak solution. Then for every '2 C1 (a;b)
(u0v0;'0) +(uv;') = (ff;') = 0: 37 Since every eigenfunction umbelongs to H1 (a;b)
, we can choose a sequence f'kgk C1 (a;b)
such thatkum'kk1!0 fork!1 , by denition of  H1 (a;b)
, and obtain from this equation and from the continuity of the scalar product that (u0v0;u0 m) +(uv;um) = lim k!1[(u0v0;'0 k) +(uv;'k)] = 0: Sinceu(a) =v(a) =u(b) =v(b) = 0, we obtain from the partial integration formula (3.4) that (uv;um) = (u0v0;u0 m) =(uv;u00 m) = (uv;mum) =m(uv;um): Since by assumption 6=mit follows from this equation that ( uv;um) = 0 for allm. Because the orthonormal system fumgmis complete, we infer from Theorem 3.11 that uv= 0, whence u=v. For boundary value problems to the Helmholtz equation
u+u=fin higher dimensions a result holds, which is completely analogous to the result for the boundary value problem to the ordinary dierential equation u00+u=f discussed here. This will be shown in Sections 8 and 9. However, in the following investigations of higher dimensional problems we rst study classical solutions and return to weak solutions only later. 38 4 Boundary value problems in circular domains. Bessel functions 4.1 The Laplace operator in polar coordinates Let =fx2R2R1<jxj<R 2g with 0R1<R 21 , or let =fx2R2jxj<Rg: To nd solutions of
u(x) +u(x) = 0 in we want to use polar coordinates (r;') inR2and apply separation of variables. To this end we must determine the form of the Laplace operator in polar coordinates. Thus, let x= (x1;x2) and r=r(x) =q x2 1+x2 2=jxj '='(x) = arctanx2 x1: -6 .................................................................................................................................r ....................................... ' x1x2x r ....................................... Then @ @x1=@r @x1@ @r+@' @x1@ @'=x1 jxj@ @r1 1 + (x2 x1)2x2 x2 1@ @'=x1 jxj@ @rx2 jxj2@ @'; @ @x2=x2 jxj@ @r+x1 jxj2@ @': Also, @2 @x2 1=@ @x1x1 jxj@ @rx2 jxj2@ @' =1 jxjx2 1 jxj3@ @r+x1 jxjx1 jxj@2 @r2x2 jxj2@2 @r@' + 2x1x2 jxj4@ @'x2 jxj2x1 jxj@2 @'@rx2 jxj2@2 @'2 =x2 1 jxj2@2 @r2+1 jxjx2 1 jxj3@ @r+ 2x1x2 jxj4@ @' 2x1x2 jxj3@2 @'@r+x2 2 jxj4@2 @'2; 39 @2 @x2 2=x2 2 jxj2@2 @r2+1 jxjx2 2 jxj3@ @r2x1x2 jxj4@ @' + 2x1x2 jxj3@2 @'@r+x2 1 jxj4@2 @'2: Thus, ifu(x) = ~u(r(x);'(x)) then
xu(x) =2X i=1@2 @x2 i~u(r(x);'(x)) =x2 1+x2 2 jxj2@2 @r2~u(r(x);'(x)) +2 jxjx2 1+x2 2 jxj3@ @r~u(r(x);'(x)) +x2 2+x2 1 jxj4@2 @'2~u(r(x);'(x)) =@2 @r2~u(r(x);'(x)) +1 r(x)@ @r~u(r(x);'(x)) +1 r(x)2@2 @'2~u(r(x);'(x)): Consequently
(r;')=@2 @r2+1 r@ @r+1 r2@2 @'2: We next expand uin a Fourier series with respect to 'on every circlejxj=r withR1<r<R 2. Thus, assume that u2C2( ) is a solution of
u+u= 0. As usual, we drop the tilde and use the notation u(x) =u(r;'): Sincen 1p 2eim'o1 m=1is a complete orthonormal system in L2([0;2];C) we obtain u(r;') =1X m=1um(r)eim'; with um(r) =1p 2 u(r;
);1p 2eim' [0;2]=1 2Z2 0u(r;')eim'd': If we can interchange partial derivatives up to order 2 with the summation sign, we obtain 0 = (
+ )u(x) =@2 @r2+1 r@ @r+1 r2@2 @'2 u(r;') +u(r;') =1X m=1h@2 @r2+1 r@ @r um(r) + m2 r2 um(r)i eim': 40 Fixr. The Fourier series vanishes identically for all 0 ' < 2only if all coecients vanish. Thus d2 dr2um(r) +1 rd drum(r) + m2 r2 um(r) = 0; for allR1<r<R 2and allm2Z. This is a linear ordinary dierential equation forumof second order. 4.2 Solution of the potential equation in circular domains. We rst consider the case = 0. In this case the general solution of this dierential equation is u0(r) =C01+C02lnr um(r) =Cm1rm+Cm2rm; m6= 0: Thus, the general solution of the potential equation
u(x) = 0 in a circular domain = fx2R2jR1<x< R2gis u(x) =u(r;') =C01+C02lnr+1X m=1 m6=0(Cm1rm+Cm2rm)eim' with arbitrary constants Cm1;Cm22C. These coecients must be determined from boundary conditions and possibly from conditions at innity (radiation conditions). Example 4.1 Let =fx2R2jjxj<Rgbe a ball with center at 0. We want to solve the Dirichlet boundary value problem
u(x) = 0; in ; u(x) =u(b)(x); x2@ : We try to nd a classical solution u2C2( ). This requires that Cm1rm+Cm2rm=um(r) =1 2Z2 0u(r;')eim'd' and C01+C02lnr=u0(r) =1 2Z2 0u(r;')d' 41 must be bounded at x= 0, hence Cm2= 0 form0 andCm1= 0 form< 0. Thus u(r;') =C01+1X m=1rm(Cm1eim'+Cm2eim'): The Fourier series expansion of the boundary data is u(b)(') =1X m=1ameim': Fromu(R;') =u(b)(') and from the uniqueness of the Fourier expansion we therefore obtain am=RmCm1; m0; am=RjmjCm2; m< 0: Theorem 4.2 Let =BR(0)and letu(b)2L2(@ ;C). Then u(x) =u(r;') =1X m=1amr Rjmj eim'(4.1) is the unique solution u2C1( )of
u(x) = 0; x2 lim r!Ru(r;') =u(b)('); 0'<2; where am=1 2Z2 0u(b)eim'd' and where the limit is understood in the L2-sense: lim r!Rku(r;
)u(b)k[0;2]= 0: (4.2) Proof: Let 0<r 1<R. From 2P1 m=1jamj2=ku(b)k2 [0;2]we obtainjamj2 Cfor allm. Thus, for all = (1;2)2N2 0and all 0rr1, 0'<2we conclude 1X m=1@1 @r1@2 @'2amr Rjmj eim' X jmj1jamjR1jmj1r Rjmj1jmj22p CR11X k=1kjjr1 Rk1 = 2p CR11X k=0(k+1)jjr1 Rk <1; 42 which shows that the seriesP1 m=1D (r;')amr R
jmjeim'converges uniformly in every closed ball Br1(0) =frr1;0'<2g. From calculus we thus obtain that the classical derivative Duof the function udened in (4.1) exists in all of =BR(0) and can be computed under the summation sign. This yields u2C1( ). Moreover, it shows that our assumption made in the construction ofuis satised, whence
u= 0 in . To verify equation (4.2) let " > 0 and choose m0large enough such that 2P jmjm0jamj2<". Then lim r!Rku(r;
)u(b)k2 [0;2]= lim r!RZ2 01X m=1amr R
jmj1 eim'2 d' = lim r!R21X m=1amr R
jmj12 2lim r!RX jmj<m0jamj2 1r Rjmj2 + 2lim r!RX jmjm0jamj2<": This proves (4.2), since ">0 was chosen arbitrarily. Example 4.3 Let =fx2R2jjxj>Rgbe an exterior domain. We want to nd a solution of the Dirichlet boundary value problem
u(x) = 0;in ; u(x) =u(b)(x); x2@ : In this case we cannot conclude that half of the coecients in the expansion u(x) =C01+C02lnr+1X m=1 m6=0(Cm1rm+Cm2rm)eim' must vanish, and the boundary condition is not enough to determine all coe- cients uniquely. Therefore the solution of the problem is not unique. To get a unique solution one must pose suitable conditions for the asymptotic behavior ofuat innity. Normally one requires that the solution is bounded: ju(x)jC; x2 : As above we then obtain Cm2= 0 for allm0 andCm1= 0 form> 0. Thus u(r;') =C01+1X m=1rm(Cm2eim'+Cm1eim'): 43 From the Fourier expansion of the boundary data u(b)(') =1X m=1ameim' we then conclude u(x) =u(r;') =1X m=1amr Rjmj eim': 4.3 Bessel's dierential equation. Solution of the Helmholtz equa- tion in circular domains. To solve the Helmholtz equation
u+u= 0 in circular domains for 6= 0, rst consider Bessel's dierential equation d2 dx2w(x) +1 xd dxw(x) + (12 x2)w(x) = 0: Here2Cis a constant and x2C. This equation cannot be solved by elemen- tary functions. Instead, the solutions are the Bessel- and Neumann functions. These functions belong to a set of functions called special functions of math- ematical physics. The Bessel function or cylinder function of order 2C, 6=1;2;3;:::; is J(x) =x 21X k=0(1)k k!(+k+ 1)x 22k ; where is the Gamma function. The power series converges for all x2C. If is not a nonnegative integer, then the termx 2
and therefore also the function Jare only dened on the set Cn(1;0]. To be precise,x 2
andJare dened on a Riemannian manifold. If is equal to a nonnegative integer m, then the formula for the Bessel function becomes Jm(x) =1X k=0(1)k k!(m+k)!x 22k+m ; where we used the equation (`+ 1) =`!; which holds for integers `0. Therefore Jmis represented by a power series converging on all of C. Hence,Jmis an entire function. Since Bessel's equation is a linear dierential equation of second order there must exist other solutions of Bessel's equation which are linearly independent ofJ. In fact, if is not an integer, then one sees immediately that also Jis 44 a solution of Bessel's equation, which is linearly independent of J. Hence, also the Neumann function N(x) =J(x) cos()J(x) sin() is a solution of Bessel's equation linearly independent of J. If=mis an inte- ger this formula cannot be used to dene Nm, since the denominator vanishes. Instead, in this case the Neumann function is Nm(x) = lim !mN(x): A series expression for Nmis given in the appendix. The general solution of Bessel's dierential equation therefore is w(x) =C1J(x) +C2N(x) with arbitrary constants C1;C22C. If=mis a nonnegative integer the functionJ(x) is innitely dierentiable at x= 0, whereas in the appendix it is shown that lim x!0jNm(x)j=1: We can now solve the Helmholtz equation in circular domains. For, using Bessel's equation we immediately see that if 2Cwith6= 0 then um(r) =C1Jm(p r) +C2Nm(p r) satises d2 dr2um(r) +1 rd drum(r) + (m2 r2)um(r) = 0: Remembering the results of 4.1 we therefore see that a solution of the Helmholtz equation
u+u= 0 in circular domains must be of the form u(x) =u(r;') =1X m=1 Cm1Jjmj(p r) +Cm2Njmj(p r) eim': (4.3) The constants Cm1;Cm2must be determined from the boundary and radiation conditions. Example 4.4 Let =BR(0) be a ball, let 2C,6= 0 and assume that u(b)2L2(@ ). We want to solve
u(x) +u(x) = 0; x2 u(x) =u(b)(x); x2@ : 45 Sinceumust be two times continuously dierentiable at x= 0, it follows that in the expansion (4.3) of uwe must have Cm2= 0 for all m2Z, sinceJjmjis regular and Njmjis singular at r= 0. Thus, u(r;') =1X m=1Cm1Jjmj(p r)eim': Let u(b)(') =1X m=1ameim' be the Fourier series of u(b). Since u(R;') =1X m=1Cm1Jjmj(p R)eim'=u(b)(') =1X m=1ameim'; the uniqueness of the Fourier expansion implies am=Cm1Jjmj(p R): Theorem 4.5 (i)Let2C. If6= 0 assume that Jm(p R)6= 0 for all m2N0. Then the Dirichlet boundary value problem
u(x) +u(x) = 0; x2BR(0) lim r!Ru(r;') =u(b)(R)(in the sense of L2) has a unique solution u2C1(BR(0)) for allu(b)2L2(@BR(0)). This solution is given by u(x) =u(r;') =8 >>>>< >>>>:1X m=1am Jjmj(p R)Jjmj(p r)eim'; 6= 0 1X m=1am(r R)jmjeim';  = 0: In particular, the only solution to homogeneous boundary data u(b)= 0isu= 0. (ii)Assume that 6= 0 and that =fm2N0Jm(p R) = 0g is not empty. Then the Dirichlet boundary value problem is only solvable if in the Fourier expansion u(b)(') =1X m=1ameim' 46 of the boundary data we have am=am= 0 for allm2. On the other hand, ifuis a solution of the homogeneous Dirichlet boundary value problem (u(b)= 0) , then the Fourier expansion is of the form u(r;') =X m2Jm(p r)(Cmeim'+Cmeim'): Moreover, any function with such a Fourier expansion where only nitely many Cmdier from zero is a solution of the homogeneous boundary value problem. Henceis an eigenvalue. Statement (i) of this theorem is proved as in the case of = 0 using estimates for the Bessel functions Jm. We omit this proof. Statement (ii) is obvious from the Fourier expansion of the solution discussed above. Ifu1andu2are eigenfunctions to the eigenvalue of the Dirichlet problem then alsoC1u1+C2u2is an eigenfunction, if this function is not zero. Therefore the set of eigenfunctions together with the zero function forms a vector space V, the eigenspace of . The dimension of the eigenspace is called the geometric multiplicity of . In the next theorem we show that for every the set  is nite. The preceding theorem then implies that uis an eigenfunction to 6= 0 if and only if u(r;') =X m2Jm(p r)(Cmeim'+Cmeim'); hence dimV2jj; wherejjdenotes the number of elements of  . Since the functions eim'and eik'are linearly independent for m6=k, it follows that fJm(p r)eim'; Jm(p r)eim'gm2 is a linearly independent set of functions. Hence it is a basis of V. Therefore dimV= 2jj: Theorem 4.6 (i)Assume that m2N0and thaty2Cnf0gis a zero of Jm. Thenyis real and satises y2>m2: (ii)The zeros of Jmdo not have an accumulation point in C. Hence, the set of zeros is countable. 47 Proof: (ii) Form2N0the Bessel function Jmis entire. Therefore, if the zeros would accumulate in Cwe would have Jm0. Consequently the zeros do not have an accumulation point. (i) Let um(r) =Jm(yr): Thenumsatises um(1) = 0 and d2 dr2um(r) +1 rd drum(r) + (y2m2 r2)um(r) = 0: Multiply this equation by rand observe that rd2 dr2um(r) +d drum(r) =d dr rd drum(r) : Thus d dr rd drum(r)
+ y2m2 r2 rum(r) = 0: We multiply this equation by um(r) and integrate: Z1 0d dr rd drum(r) um(r) + y2m2 2 rjum(r)j2dr= 0: Partial integration yields Z1 0rd drum(r)d drum(r)dr+Z1 0 y2m2 r2 rjum(r)j2dr =rd drum(r)um(r)r=1 r=0= 0: Sinced drum(r)d drum(r) =d drum(r)d drum(r) =d drum(r)2, it follows Z1 0 d drum(r)2 + y2m2 r2 jum(r)j2 rdr= 0: (4.4) Since the imaginary part of this integral is (Imy2)Z1 0jum(r)j2rdr= 0; and sinceR1 0jum(r)j2rdr> 0, it follows that Im y2= 0, hencey22R. Moreover, we must have y2>m2: 48 For, otherwise Z1 0 d drum(r)2+ y2m2 r2
jum(r)j2 rdr Z1 0 d drum(r)2+ m2m2 r2
jum(r)j2 rdr< 0; which contradicts (4.4). The proof is complete. Corollary 4.7 The set of eigenvalues of the Dirichlet boundary value prob- lem inBR(0)is contained in the positive real axis. If 2then f0;1;:::; [Rp ]g; where [Rp ]denotes the largest integer not greater than Rp . Thus, the geo- metric multiplicity of is not greater than 2([Rp ] + 1) . The set does not have an accumulation point, hence it is a countable set. Proof: We already showed that u= 0 is the only solution of
u(x) = 0; x2BR(0) u(x) = 0;jxj=R; hence= 0 is not an eigenvalue. Also we showed that 2Cnf0gis an eigenvalue if and only if =fm2N0Jm(p R) = 0g6=;: Consequently, is an eigenvalue if and only if there is m2N0such thatp R is a zero of Jm. Therefore, iffy(m) ig1 i=1Rnf0gis the (countable) set of non- vanishing zeros of Jm, it follows  =ny(m) i R2m2N0;i2No (0;1): This is a countable set. All non-vanishing zeros of Jmsatisfy (y(m) i)2> m2, hence (y(m) i)22(m2;1); i= 1;2;:::; which implies that every interval (0 ;s] only contains zeros of those nitely many Bessel functions Jmwithm2< s. Since the zeros of a Bessel function do not accumulate, the set of zeros of Jmin (0;s] is nite, hence  \(0;s] is nite. Consequently,  does not have an accumulation point. Ifm2, theny=p Ris a zero of Jm, hencey2> m2impliesR2= y2>m2, and therefore f0;1;:::; [Rp ]g: It follows that the geometric multiplicity dim Vofsatises dimV= 2jj2([Rp ] + 1): The proof is complete. 49 Denition 4.8 The set  of eigenvalues is called the spectrum of the Dirichlet problem. We have not yet answered the question whether  6=;, i.e. whether eigenvalues exist. This problem will be investigated later in full generality. It will be shown that in fact there exist countably innitely many eigenvalues and that one can choose a complete orthonormal system in L2( ;C) consisting of eigenfunctions. An easy corollary of this result is that every Bessel function Jmwithm2N0has countably innitely many nonnegative zeros. Thus the situation is completely analogous to the situation in one space dimension. 50 5 Maximum principle, subsolutions, Perron's method In this section we only consider real valued solutions of the Helmholtz equa- tion. Of course, the results can also be applied to complex valued solutions by considering the real and imaginary parts separately. 5.1 Maximum principle Theorem 5.1 Let Rnbe a bounded open set, let g: !R;f: !R and assume that for the function u2C( ;R)the partial derivatives@u @xi,@2u @x2 iexist in fori= 1;:::n and thatusatises
u(x)g(x)u(x) =f(x); x2 : (i)Ifg0in , then for all x2 u(x)max(0;max y2@ u(y));iff0in ; u(x)min(0;min y2@ u(y));iff0in : (weak maximum principle) (ii)Ifg>0in , then for all x2 u(x)0oru(x)<max y2@ u(y);iff0in ; u(x)0oru(x)>min y2@ u(y);iff0in : (strong maximum principle) Proof: (i) We rst consider the case f0. Assume that the statement is false. Then there is x02 such that u(x0)>0 and max x2 u(x) =u(x0)>max y2@ u(y): Denev: !Rby v(x) =u(x) +"jxj2; x2 ; where">0 is chosen small enough such that v(x0) =u(x0) +"jx0j2>max y2@ (u(y) +"jyj2) = max y2@ v(y); max y2 "jyj2< u (x0): 51 vis continuous on the compact set and therefore assumes its maximum in a pointz2 . By the choice of "we havez62@ and u(z) =v(z)"jzj2>v(x0)u(x0) ="jx0j20: Thus, the maximum zbelongs to the open set , which implies that @v @xi(z) = 0;@2v @x2 i(z)0; i= 1;:::;n; whence
v(z) =nX i=1@2v @x2 i(z)0: On the other hand
v(z) =
u(z) +
("jxj2)jx=z=g(z)u(z) +"nX i=1@2 @x2 ix2 i+f(z) =g(z)u(z) + 2n"+f(z)>0; because of g(z)0;u(z)>0;f(z)0 and" > 0. This is a contradiction, hence max x2 u(x) =u(x0)max(0;max y2@ u(y)): Iff0 denew=u. Then
w(x)g(x)w(x) =f(x)0 for allx2 , hence min x2 u(x) = max x2 w(x)max(0;max y2@ w(y)) = max(0;min y2@ u(y)) =min(0;min y2@ u(y)); which implies the statement for the minimum. (ii) Letg >0 andf0, and assume that the statement for the maximum is false. Then there is x02 such that u(x0)>0 and u(x0) = max x2 u(x)max y2@ u(y): Consequently, x0is a local maximum of uin the open set , hence @u @xi(x0) = 0;@2u @x2 i(x0)0; i= 1;:::;n; whence 0
u(x0) =g(x0)u(x0) +f(x0)>0; which is a contradiction, and the statement for the maximum must be true. The statement for the minimum is proved by considering u. We note some consequences of the maximum principle: 52 Corollary 5.2 Let Rnbe a bounded open set, let g: !R+ 0;f: !R be given. Assume that u;v2C( ;R)and that for w=uandw=vthe partial derivatives@w @xi,@2w @x2 iexist in fori= 1;:::n and the equation
w(x)g(x)w(x) =f(x); x2 holds. Then the following statements are true: (i) Ifu(y)v(y)for ally2@ thenu(x)v(x)for allx2 : (ii) For allx2 , ju(x)v(x)jmax y2@ ju(y)v(y)j: (iii) Letg >0andf= 0 in . Ifuassumes the maximum in , thenu0. Ifuassumes the minimum in , thenu0. Proof: (i)w=uvsatises
wgw= 0 in . Hence, by the weak maximum principle 0 = min(0;min y2@ w(y))w(x) =u(x)v(x); x2 : (ii) Again we apply the weak maximum principle to w=uvand obtain min y2@ (jw(y)j)min(0;min y2@ w(y))w(x) max(0;max y2@ w(y))max y2@ jw(y)j: Thus w(x)max y2@ jw(y)j;w(x) min y2@ (jw(y)j) = max y2@ jw(y)j; whencejw(x)jmaxy2@ jw(y)j. (iii) Letx02 and assume that u(x0) = maxx2 u(x). By the strong maximum principle this can only be if u(x0)0. The statement for the minimum is proved in the same way. Corollary 5.3 (Uniqueness) Let Rnbe a bounded open set, let g: ! R+ 0,f: !Rand let the functions u;v2C( ;R)have the dierentiability properties stated in Corollary 5.2. Assume that uj@ =vj@ and thatuandv both satisfy the dierential equation
w(x)g(x)w(x) =f(x); x2 : Thenu=v. Proof: The preceding corollary yields ju(x)v(x)jmax y2@ ju(y)v(y)j= 0; x2 : 53 5.2 Consequences of the maximum principle for the Helmholtz equa- tion in R2 We showed that the Dirichlet problem for the Helmholtz equation in a ball B inR2has a solution, which is innitely dierentiable in the interior of Band satises the boundary condition in the L2-sense. However, up to now we do not know whether the solution is continuous on Bif the boundary data u(b)are continuous. The maximum principle can be used to show that this is in fact true. As preparation we need the following result: Theorem 5.4 LetBR(0)R2and let0. Letu(b) m,u(b)2L2(@B)and let um;ube the solutions of
v(x) +v(x) = 0; x2BR(0) lim r%Rkv(r;
)v(b)k[0;2]= 0 to the data v(b)=u(b) m,v(b)=u(b). If lim m!1ku(b) mu(b)k[0;2]= 0; then in every ball B^r(0)with ^r<R the sequencefumg1 m=1converges uniformly tou. Proof: Sincefein'p 2gn2Nis a complete orthonormal system in L2((0;2)), we have for the Fourier expansions u(b) m(') =1X n=1a(m) nein'; u(b)(') =1X n=1anein' by Theorem 3.10 that 21X n=1ja(m) nanj2=1X n=1jp 2(a(m) nan)j2=ku(b) mu(b)k2 [0;2]!0; form!1 . We rst consider the case = 0. Then um(r;') =1X n=1a(m) nr Rjnj ein' u(r;') =1X n=1anr Rjnj ein': 54 Let" >0 and choose m0such thatku(b) mu(b)k< "for allmm0. Cauchy- Schwarz inequality yields for r^r, 0'<2andmm0that jum(r;')u(r;')j1X n=1(a(m) nan)r Rjnj 1X n=1ja(m) nanj21=21X n=1r R2jnj1=2 1p 2ku(b) mu(b)k2 1(r R)21=2 1p 1^r R21=2 ":(5.1) Since">0 was arbitrary, it follows that umconverges uniformly to uinB^r(0). To prove the statement for <0 we use that by Theorem 4.5 the represen- tations um(r;') =1X n=1a(m) n Jjnj(p R)Jjnj(p r)ein'; u(r;') =1X n=1an Jjnj(p R)Jjnj(p r)ein': hold forumandu. Withx=p randy=p Rwe thus obtain jum(r;')u(r;')j1X n=1ja(m) nanjjJjnj(ix)j jJjnj(iy)j: (5.2) The fractionjJjnj(ix)j jJjnj(iy)jcan be estimated using Lemma A.1 in the appendix. For, 0rRand > 0 implyx;y2Rand 0xy. Therefore the assumptions of the lemma are satised. The estimate from that lemma and (5.2) together yield jum(r;')u(r;')jja(m) 0a0jjJ0(ix)j jJ0(iy)j+1X n=1 n6=0ja(m) nanjr Rjnj : With this estimate we can proceed as above and obtain that (5.1) also holds for <0 with the right hand side multiplied by C= max 1;max 0xyfjJ0(ix)j jJ0(iy)jg , which shows that umconverges uniformly to uinB^r(0) also in this case. Theorem 5.5 LetBR2be a bounded open ball and let 0. Then for everyu(b)2C(@B;R)there is a unique solution u2C(B;R)\C1(B;R)of the Dirichlet problem
u(x) +u(x) = 0; x2B uj@B=u(b): 55 Proof: Without restriction of generality we can assume that B=BR(0) with R> 0. The uniqueness follows from Corollary 5.3. To prove that a continuous solution exists, let u(b)=1X m=1ameim' be the Fourier expansion of u(b). Form0 andn1 let sm(') =mX k=makeik'; n(') =1 n s0(') +:::+sn1(')
: Since'7!u(b)(') : [0;2)!Ris continuous and can be extended to a contin- uous, periodic function on R, it follows from Theorem 3.13, that the sequence fng1 n=1converges uniformly on [0 ;2) tou(b). Let vm(r;') =8 >>>>< >>>>:mX k=mak Jjkj(p R)Jjkj(p r)eik'; < 0 mX k=makr Rjkj eik';  = 0:(5.3) By the results of Section 4, every term in these nite sums is an innitely dierentiable solution of
v(x) +v(x) = 0 in all of R2, hencevmis an innitely dierentiable solution of this equation in all ofR2. We remark that if u(b)is real then ak=1p 2Z2 0u(b)(')eik'd'=1p 2Z2 0u(b)(')eik'd'=ak; whenceakeik'+akeik'= 2Re(akeik') is real. Moreover, it is seen from the power series expansion of Jjkjthat Jjkj(p r) Jjkj(p R)2R; since < 0, hencep is imaginary. From (5.3) we consequently see that vm has real values. For n1 we set n(r;') =1 n v0(r;') +:::+vn1(r;')
: 56 The function n2C1(R2;R) is a solution of the Dirichlet boundary value problem
n(x) +n(x) = 0; x2BR(0) n(x) =n(x);jxj=R: Therefore the maximum principle yields jn(x)`(x)jmax jyj=Rjn(y)`(y)j; for allx2BR(0). Sincefngnconverges uniformly to u(b), it follows from this estimate thatfng1 n=1converges in BR(0) to a limit function u2C BR(0)
: The limit function satises u(R;') =u(b);0'<2: Moreover, since fng1 n=1converges uniformly to u(b), it also converges in L2 @BR(0)
tou(b). From the preceding theorem we thus conclude that the sequencefng1 n=1converges pointwise to the solution ^ uof
^u(x) +^u(x) = 0; x2BR(0) lim r!Rk^u(r;
)u(b)k[0;2]= 0; which satises ^ u2C1 BR(0)
. Since the pointwise limit coincides with the uniform limit, we obtain that u= ^u, henceubelongs toC BR(0)
\C1 BR(0)
and is a solution of the Helmholtz equation in BR(0). The proof is complete. We next show that solutions of the Helmholtz equation have the mean value property: Theorem 5.6 LetBR(0)R2, let0and letu2C(BR(0);R)\ C1(BR(0);R)solve
u(x) +u(x) = 0; x2BR(0): Then u(0) =1 2RJ 0(p R)Z jxj=Ru(x)ds: SinceJ0(0) = 1 , this formula becomes for = 0 u(0) =1 2RZ jxj=Ru(x)ds: Proof: Let<0. From the preceding investigations we know that if u(R;') =1X m=1ameim'; 57 then u(r;') =1X m=1am Jjmj(p R)Jjmj(p r)eim';0<rR: The power series expansion of Jmshows that Jm(0) =(1; m = 0 0; m2N; hence u(0) =a0 J0(p R)=1 2J0(p R)Z2 0u(R;')d'=1 2RJ 0(p R)Z jxj=Ru(x)ds: To prove the statement for = 0 we proceed in the same way, using the Fourier expansion of u. Corollary 5.7 Let R2be a open, connected set, let 0and letu2 C1( ;R)be a solution of
u(x) +u(x) = 0; x2 : Assume that x02 exists such that u(x0) = 0 . Then, ifu0oru0in it follows that u= 0 in . Proof: Let M=fx2 ju(x) = 0g: By assumption, Mis not empty since x02 . We prove that Mis closed and open in , which implies M= , since is connected. Sinceuis continuous, Mis obviously closed. To verify that Mis open, let y2M. Since is open there exists a ball BR(y) with center ycontained in . By the mean value property we have for all 0 <r<R 1 2rZ jxyj=ru(x)dsx=J0(p r)u(y) = 0: Ifu0 oru0 in this can only be if u(x) = 0 for all xwithjxyj=r. This holds for all 0 <r <R , henceu(x) = 0 for all x2BR(y). Thus,BR(y)M, henceMis open. The proof is complete. Corollary 5.8 Let be a bounded, open, connected set and let u2C( ;R)be a solution of
u(x) +u(x) = 0; x2 : (i) If<0, thenu= 0 or, for allx2 , min(0;min y2@ u(y))<u(x)<max(0;max y2@ u(y)): 58 (ii) If= 0, thenu= const or, for allx2 , min y2@ u(y)<u(x)<max y2@ u(y): Proof: (i) Combine the strong maximum principle with the preceding result. (ii) Assume that there is x02 such that u(x0) = maxy2 u(y). Then the functionv2C( )\C1( ) dened by v(x) =u(x)u(x0) satises
v= 0, v0 andv(x0) = 0. Since x0is an interior point of , Corollary 5.7 implies thatv= 0, hence u= const =u(x0). In the same way it follows that if there is x12 such that u(x1) = miny2 u(y), thenu= const =u(x1). Statement (ii) follows from these results. 5.3 Subsolutions, supersolutions, comparison Up to now we only know how to solve boundary value problems in circular domains. Our goal is to develop a method to solve boundary value problems in very general domains R2. To this end we need subsolutions and superso- lutions, which we dene and discuss in this section. Thus, let R2be a bounded open domain and let 0. Denition 5.9 A function v2C( ;R) is called subsolution (supersolution) of the equation
u+u= 0 in , if to every open ball BwithB the uniquely determined solution u2C(B) of
u(x) +u(x) = 0; x2B uj@B=vj@B; satises v(x)u(x);(v(x)u(x)); for allx2B. Remark 5.10 In the case = 0 a subsolution is also called a subharmonic function, since a solution uof
u(x) = 0 is called a harmonic function. Theorem 5.11 Letv1;:::;vmbe subsolutions of
u+u= 0. Then v= max(v1;:::;vm) is a subsolution. Proof: Letv(2)= max(v1;v2), letB be a closed ball and let u1;u2;u(2)2 C(B) be solutions of
w+w= 0 59 inBsatisfying u1j@B=v1j@B; u 2j@B=v2j@B; u(2) j@B=v(2) j@B: Fory2@Bwe have u(2)(y) =v(2)(y)vi(y) =ui(y); i= 1;2: Using the maximum principle and noting that v1;v2are subsolutions we thus obtain forx2B v1(x)u1(x)u(2)(x) v2(x)u2(x)u(2)(x); which yields v(2)(x) = max(v1(x);v2(x))u(2)(x): This shows that v(2)is a subsolution. Since v(k)= max(v1;:::;vk) = max(vk;max(v1;:::;vk1)) = max(vk;v(k1)); the statement follows by induction. Theorem 5.12 Letvbe a subsolution of
u+u= 0 in , letBbe an open ball withB and letu2C(B)\C1(B)be the solution of
u(x) +u(x) = 0; x2B uj@B=vj@B: Then the function ~v: !R; ~v(x) =(v(x); x62B u(x); x2B is a subsolution. Proof: We have ~vvsincevis a subsolution. Now let B1be an open ball withB1 and let ~u2C(B1) be the solution of
~u(x) +~u(x) = 0; x2B1 ~uj@B1= ~vj@B1: We are nished if we can show that ~ u~vinB1. By denition of ~ vthis holds if ~u(x)v(x); x2B1nB (5.4) ~u(x)u(x); x2B1\B: (5.5) 60 B B1 B1\B BallB1in arbitrary position To prove these relations, let ^ u2C(B1) be the solution of
^u(x) +^u(x) = 0; x2B1 ^uj@B1=vj@B1: Sincevis a subsolution, it follows that ^ uvinB1. Also, ~uand ^usolve the Helmholtz equation in B1and satisfy ~uj@B1= ~vj@B1vj@B1= ^uj@B1; hence the maximum principle yields ~ u^uvinB1, which proves (5.4). To verify (5.5), note that @(B1\B) = (B1\@B)[(B\@B1) and that ~ujB\@B1= ~vjB\@B1=ujB\@B1 ~ujB1\@BvjB1\@B=ujB1\@B; where we used (5.4) to get the last relation. Since both ~ uandusatisfy the Helmholtz equation in B1\B, it follows from these relations and from the maximum principle that ~ uuinB1\B. This is (5.5). The proof is complete. Theorem 5.13 (Comparison) Let0and letv2C( ;R)be a subso- lution,w2C( ;R)be a supersolution with vj@ wj@ . Thenvwin . Proof: Assume that there is x02 such that v(x0)> w(x0). Then since h=vwis less or equal to zero on @ , it follows that hassumes the positive maximum at a point z2 . We can assume that zis a boundary point of the closed set M=fx2 jh(x) = maxy2 h(y)g . It follows that every neighborhood of zcontain points, where hassumes values smaller than h(z). Therefore we can choose a ball Bwith center zand withB such that @B contains such a point. We thus have h(z)>0; h(z)max x2@Bh(x); h(z)>min x2@Bh(x): (5.6) 61 Now let ^v;^wbe the solutions of
^v(x) +^v(x) = 0 in B; ^vj@B=vj@B;
^w(x) +^w(x) = 0 in B; ^wj@B=wj@B: Then ^vvandw^winB. Therefore the function u= ^v^w satisesu= ^v^wvw=hinB, hence u(z)h(z)>0; (5.7) and uj@B= ^vj@B^wj@B=vj@Bwj@B=hj@B: This equation and (5.6), (5.7) imply u(z)max(0;max y2@Bu(y)); u(z)>min y2@Bu(y): (5.8) Since
u(x) +u(x) = 0 inB, it follows in the case  < 0 from the rst of the inequalities (5.8) and from Corollary 5.8(i) that u= 0, which contradicts (5.7). In the case = 0 it follows from the rst of the inequalities (5.8) and from Corollary 5.8(ii) that u= const, which contradicts the second of the inequalities (5.8). Consequently, in both cases we must have vwin . Theorem 5.14 Let0and letw2C( )\C1( ) be a solution of the potential equation
w= 0. Ifwis non-negative, then wis a supersolution, if wis non-positive, then wis a subsolution of the equation
u+u= 0. Proof: For a ballBwithB letube the solution of
u(x) +u(x) = 0; x2B uj@B=wj@B: Since
w= 0, the function h=wusatises
h(x) +h(x) =w(x)( 0;ifw0; 0;ifw0: Since alsohj@B= 0, we conclude from the maximum principle in the rst case thath0, henceuw, which shows that wis a supersolution. In the second case the maximum principle yields h0, henceuw, which implies that w is a subsolution. 62 Corollary 5.15 (Maximum principle for sub- and supersolutions) Let 0. (i) Any non-negative constant function is a supersolution and any non-positive constant function is a subsolution of
u+u= 0. (ii) Ifvis a subsolution and wis a supersolution, then v(x)max 0;max y2@ v(y)
; w(x)min 0;min y2@ v(y)
: Proof: (i) A non-negative constant function is a supersolution and a non- positive constant function is a subsolution, since they satisfy the potential equa- tion. (ii) For a subsolution vdene ^v: !Rby ^v(x) = const = max 0;max y2@ v(y)
; x2 : Then ^v0 is a supersolution satisfying vj@ ^vj@ , whencev^v, by Theo- rem 5.13. Similarly, for a supersolution wdene ^w: !Rby ^w(x) = const = min 0;min y2@ w(y)
; x2 : Then ^w0 is a subsolution satisfying ^ wj@ wj@ , whence ^ww. 5.4 Perron's method For a bounded open set R2, for0 and for a function f2C(@ ;R) dene Sf=fv2C( )jvis a subsolution of
u+u= 0 withvj@ fg: Note that by the preceding corollary every v2Sfsatises v(x)max 0;max y2@ f(y)
: Theorem 5.16 (Oskar Perron (1880 { 1975)) IfSf6=;then uf(x) = sup v2Sfv(x); x2 ; satisesuf2C1( )and
uf(x) +uf(x) = 0; x2 : 63 Proof: The proof is in two steps. In the rst step we construct in a neigh- borhood of an arbitrary y2 a solution uof the Helmholtz equation with u(y) =uf(y). In the second step we show that u=ufin this neighborhood, henceufis a solution of the Helmholtz equation in this neighborhood. This proves the theorem, since ywas arbitrary. I.) Lety2 and choose a sequence fvmg1 m=1Sfwith lim m!1vm(y) =uf(y). We can assume that v1v2v3::: : (5.9) Otherwise we consider the sequence fvmg1 m=1dened by vm= max(v1;:::;vm): This is a monotonically increasing sequence of subsolutions with vm2Sfand vmvm, hence lim m!1vm(y) =uf(y). Thus, let (5.9) be satised. We choose an open bounded ball Bwithy2BandB . Consider the subsolution wm(x) =( vm(x); x2 nB; um(x); x2B; whereum2C(B)\C1(B) is the solution of
um(x) +um(x) = 0; x2B umj@B=vmj@B: Thenwm2Sfsatisesvmwm. Moreover, since umj@B=vmj@Bit follows that fumj@Bgmis monotonically increasing, hence the maximum principle implies thatfumgmand therefore also fwmgmis monotonically increasing. Because the last sequence is bounded above by the function uf, it follows thatfwmgm converges pointwise everywhere on to a limit function. Let ube the restriction of this limit function to B. The function uis the pointwise limit of fumgmand satises uuf; u(y) =uf(y) (5.10)
u(x) +u(x) = 0; x2B: (5.11) (5.10) follows from uf(y) = lim m!1vm(y)lim m!1wm(y) = lim m!1um(y)uf(y): To see (5.11) note that u(x)um(x)
2decreases pointwise monotonically to zero. The monotone convergence theorem of Beppo Levi therefore yields lim m!1kuumk2 @B= lim m!1Z @B u(x)um(x)
2dx= 0: 64 Sinceumsatises the Helmholtz equation in B, it thus follows from Theorem 5.4 thatfumgmconverges uniformly in every compact subset of Bto the solution of the Helmholtz equation with Dirichlet boundary data given by uj@B. This solution equals u, since the pointwise and uniform limit functions coincide, whence (5.11) holds. II.) To verify that ujB=ufjBlety12Band choose as above a monotonically increasing sequence fv0 mgmSfsatisfying lim m!1v0 m(y1) =uf(y1). We can assume that v0 mwm; (5.12) since we otherwise replace v0 mby max(wm;v0 m). With the sequence fv0 mgmwe construct in the same way as above a solution u02C(B) of the Helmholtz equation in the Ball Bsatisfyingu0(y1) =uf(y1). Because uis the pointwise limit offwmgand because u0v0 mfor everym, it follows from (5.12) that u0u; which implies that uf(y) =u(y)u0(y)uf(y), whenceu0(y) =u(y). There- foreu0uis a solution of the Helmholtz equation in Bsatisfying (u0u)0 and (u0u)(y) = 0. Since yis an interior point of B, we conclude from Corollary 5.7 thatu0u= 0 inB, thence uf(y1) =u0(y1) =u(y1): Sincey12Bwas arbitrary, we obtain ufjB=ujB, henceufis a solution of the Helmholtz equation in the neighborhood Bofy. Sincey2 was arbitrary, we conclude that ufis a solution of the Helmholtz equation in , as asserted by the theorem. 5.5 Boundary value problems, regular points Corollary 5.17 Let the assumptions of Theorem 5.16 be satised. If w2 C( ;R)is a supersolution with fwj@ , then the solution uconstructed in this theorem satises u(x)w(x); x2 : Proof: Any subsolution v2Sfsatisesvj@ =fwj@ , hencevwon , by comparison. Consequently u(x) = sup v2Sfv(x)w(x); x2 : Corollary 5.18 Let R2be a bounded open set, let 0and letf2 C(@ ;R). If there is a subsolution v2C( )and a supersolution w2C( ) satisfying vj@ =f=wj@ ; 65 then there is a unique solution u2C( )\C1( )of
u(x) +u(x) = 0; x2 uj@ =f: The solution satises vuwon . Proof: By assumption the set Sfof all subsolutions ^ vsatisfying ^vj@ f contains the function v, hence is nonempty. Consequently, by Theorem 5.16 there is a solution u2C1( ) of
u(x) +u(x) = 0; x2 ; which by Corollary 5.17 satises vuwon . Extend ufrom to a function on by dening u(x) =f(x); x2@ : To see that the extended function satises u2C( ), letx2@ . Sincev;w2 C( ) satisfyvj@ =wj@ =f, we obtain f(x) = limy!x y2 v(y)limy!x y2 u(y)limy!x y2 w(y) =f(x); whence limy!x y2 u(y) =f(x): Consequently, u2C( ). Uniqueness of the solution follows from Corollary 5.3. Example 5.19 Let<0, leta;b> 0, let =fx= (x1;x2)2R2jx1j<a;jx2j<bg and letf2C(@ ) be dened by f(x) =c with a constant c>0. Then there is a unique solution u2C( )\C1( ) of
u(x) +u(x) = 0; x2 ; uj@ =f: For,w(x) =cis a supersolution with wj@ =f, cf. Corollary 5.15. To construct a subsolution vwithvj@ =f, note that v1(x1;x2) =c ep a+ep a ep x1+ep x1 66 is a solution, hence a subsolution with v1(x1;x2)( =c;jx1j=a;jx2jb c;jx1ja1;jx2j=b: Also v2(x1;x2) =c ep b+ep b ep x2+ep x2 is a solution, hence a subsolution with v2(x1;x2)( c;jx1j=a;jx2jb =c;jx1ja1;jx2j=b: Consequently v(x) = max(v1(x);v2(x)) is a subsolution with vj@ =f. Corollary 5.18 thus implies that the boundary value problem has a unique solution. For an arbitrary domain and for arbitrarily given boundary data it is dicult to nd sub- and supersolutions with boundary values equal to the given data. In the following we show that it suces to nd sub- and supersolutions, which satisfy the boundary condition locally. Denition 5.20 Let R2be a bounded open set. We call x02@ a regular boundary point, if to every 2R;  > 0 andr >jjthere is a supersolution w2C( ) and a subsolution v2C( ) satisfying w(x0) =v(x0) =and w(x)( ; x2 \B(x0); r; x2 nB(x0); v(x)( ; x2 \B(x0); r; x2 nB(x0): Theorem 5.21 Let R2be a bounded open domain, let 0and let f2C(@ ;R). Ifx02@ is a regular point, then the solution uof
u(x) +u(x) = 0; x2 constructed by Perron's method is continuous at x0and satises u(x0) =f(x0): 67 Proof: Let">0. Then there is >0 such that jf(x)f(x0)j<" for allx2@ withjxx0j<. By assumption there is a supersolution wand a subsolution vsatisfyingw(x0) =f(x0) +";v(x0) =f(x0)"and w(x)w(x0) =f(x0) +"; forjxx0j<; w(x)sup y2@ jf(y)j; forjxx0j; v(x)v(x0) =f(x0)"; forjxx0j<; v(x) sup y2@ jf(y)j; forjxx0j: These conditions imply w(x)f(x0) +"f(x); x2@ \B(x0); w(x) sup y2@ jf(y)j f(x); x2@ nB(x0); v(x)f(x0)"f(x); x2@ \B(x0); v(x) sup y2@ jf(y)j f(x); x2@ nB(x0); hencevj@ fwj@ . Therefore we have v2Sf, henceSf6=;. This implies that the solution uexists and satises v(x)u(x)w(x); for allx2 . Thus, lim sup x!x0 x2 u(x)lim sup x!x0 x2 w(x) = limx!x0 x2 w(x) =f(x0) +" lim infx!x0 x2 u(x)lim infx!x0 x2 v(x) = limx!x0 x2 v(x) =f(x0)": Since">0 was arbitrary, we infer that lim sup x!x0 x2 u(x) = lim infx!x0 x2 u(x) =f(x0); which implies limx!x0 x2 u(x) =f(x0): Thereforeuis continuous at x0. 68 x0 aB(x0) B=2(a) BR(a) Ball with center aintersecting @ atx0 Corollary 5.22 Let0, let R2be a bounded open set and let f2 C(@ ;R). If every point of @ is regular, then there is a unique solution u2 C( ;R)of
u(x) +u(x) = 0; x2 uj@ =f: This result follows immediately from the preceding theorem. It remains to nd a criterion for regular boundary points. Theorem 5.23 Let R2be a bounded open set and let x02@ . Assume that0. If there is an open ball BR2n such that B\ =fx0g; thenx0is a regular boundary point. Proof: Let2R,r >jjand >0 be given. We must construct suitable super- and subsolutions. We rst assume that 0. LetB=BR(a) be the ball with BR(a)\ =fx0g: Without restriction of generality we can assume that R < 2. Otherwise we shrinkBuntil this estimate holds. This estimate for Rand the equation BR(a)\ =fx0gimply BR(a)B=2(a)B(x0): For simplicity we write =B=2(a)nBR(a): I.) To construct a supersolution we let u2C()\C1() be the solution of
u(x) +u(x) = 0; x2; u(x) =r; x2@B=2(a); u(x) =; x2@BR(a): 69 This solution is given by u(x) =( C1J0(p jxaj) +C2N0(p jxaj);for<0; C1+C2lnjxaj; for= 0; with suitable constants C1andC2, which can be determined from the boundary conditions. We can use this formula to extend uto the region R2nBR(a), which contains as a subset. The extended function satises the Helmholtz equation in the whole domain of denition. Therefore, for y2R2withjyaj> = 2 the radial symmetry of uand the maximum principle applied to the region y=Bjyaj(a)nBR(a) yield for all x2ythat min 0;;u(y)
= min(0;min @yu)u(x)max(0;max @yu) = max 0;;u(y)
: (5.13) We can insert x2@B=2(a)yinto this inequality. Since uhas the value r on@B=2(a), the second inequality in (5.13) can only hold if u(y)r;for ally2R2n@B=2(a): (5.14) This implies min 0;;u(y)
=0, whence, the rst inequality in (5.13) yields u(x);for allx2R2nBR(a): (5.15) Now dene w=uj :Thenwis a solution of the Helmholtz equation, hence it is a supersolution. Moreover, x02@BR(a) implies w(x0) =u(x0) =: If we note that R2nBR(a) and nB(x0)R2nB=2(a), we obtain from (5.14) and (5.15) that won ;andwron nB(x0): Thereforewsatises all conditions required from the supersolution in Deni- tion 5.20. II.) To construct a subsolution vletu2C()\C1() be the solution of
u(x) = 0; x2; u(x) =r; x2@B=2(a); u(x) =; x2@BR(a): This solution has the form u(x) =C1+C2lnjxaj 70 with constants C1;C2uniquely determined by the boundary conditions. We use this formula to extend uto the region R2nBR(a). The extended function satises the potential equation in the whole domain of denition. Choose y2R2 withjyaj>=2. Sinceuis radially symmetric with respect to a, it has the constant value u(y) on the circle @Bjyaj(a), whence Corollary 5.8(ii) implies for allx2ythat min(;u(y)) = min @yu<u (x)<max @yu= max(;u(y)): (5.16) This inequality must hold for x2@B=2(a)y. For suchxwe haveu(x) =r. The rst inequality in (5.16) can therefore only hold if u(y)<r. This implies max(;u(y)) =, so the second inequality in (5.16) yields that u(x)<for all x2y. Sinceywas an arbitrary point outside of the ball B=2(a), we obtain uonR2nBR(a); ur;onR2nB=2(a): Since0 by assumption, it follows that u0. Consequently v=uj is a non-positive solution of the potential equation, hence vis a subsolution, by Theorem 5.14. Since x02@BR(a), we have v(x0) =. From the inequalities above it is immediately seen that vsatises all conditions required in Deni- tion 5.20 from the subsolution. III.) It remains to construct a supersolution and a subsolution in the case  >0. To this end let ^ wand ^vbe the super- and subsolution to the value con- structed in the preceding part of the proof. Since the negative of a supersolution is a subsolution and the negative of a subsolution is a supersolution, it follows that w=^v; v =^w are a supersolution and a subsolution, respectively, satisfying the estimates required for a regular point. Consequently x0is a regular point. Example 5.24 Let R2be a bounded, open and convex set. Then to every pointx2@ there is a ball Bsuch that \B=fxg, hence every boundary point is regular. Therefore the Dirichlet boundary value problem
u(x) +u(x) = 0; x2 uj@ =f; with0 has a unique solution u2C( )\C1( ) to every f2C(@ ). Example 5.25 Let @ be a nite subset such that @ is two times contin- uously dierentiable at every point of @ n. Assume that through every point yof a straight line `is passing such that is locally on one side of `aty. Then to every point x2@ there is a ball BwithB\ =fxg, hence every point of@ is regular, and the Dirichlet problem can be uniquely solved. In particular, the Dirichlet problem can be uniquely solved if @ 2C2: 71 A domain satisfying the conditions of Example 5.25 72 6 Fundamental solution, Green's function 6.1 Convolution integrals Theorem 6.1 Let1p <1; '2L1(Rn;C)andf2Lp(Rn;C). Then for almost allx2Rnthe integral F(x) =Z Rn'(xy)f(y)dy=Z Rn'(y)f(xy)dy exists. The function Fdened by this integral belongs to Lp(Rn;C)and satises kFkLp(Rn)k'kL1(Rn)kfkLp(Rn): Here kukLp(Rn)=Z Rnju(x)jpdx1=p denotes the norm in the Banach space Lp(Rn;C). Proof: We haveZ RnZ Rnj'(y)jjf(xyjpdxdy =Z Rnj'(y)jZ Rnjf(xy)jpdxdy =Z Rnj'(y)jdyZ Rnjf(x)jpdx=k'kL1(Rn)kfkp Lp(Rn): (6.1) Thus, Tonelli's theorem yields (x;y)7!j'(y)jjf(xy)jp
2L1(RnRn); whence Fubini's theorem implies thatR Rnj'(y)jjf(xy)jpdyexists for almost allx2Rn, consequently y7!j'(y)j1=pjf(xy)j2Lp(Rn) for almost all x. For p= 1 we therefore get from (6.1) Z RnjF(x)jdxZ RnZ Rnj'(y)jjf(xy)jdydxk'kL1(Rn)kfkL1(Rn): This completes the proof for p= 1. For 1 < p <1let 1< q <1satisfy 1 p+1 q= 1. Then H older's inequality yields Z Rnj'(xy)f(y)jdyp =Z Rnj'(y)f(xy)jdyp Z Rnj'(y)j1 qj'(y)j1 pjf(xy)jdyp k'kp q 1Z Rnj'(y)jjf(xy)jpdy : The right hand side is bounded for almost all x. It thus follows for these x thaty!'(xy)f(y)2L1(Rn). Furthermore, the last inequality and (6.1) together imply Z RnF(x)jpdx=Z RnZ Rn'(xy)f(y)dyp dx  k'kp q L1(Rn)k'kL1(Rn)kfkp Lp(Rn)= (k'kL1(Rn)kfkLp(Rn))p: 73 Remark: Often one uses the notation F(x) =Z Rn'(xy)f(y)dy= ('f)(x): The operatoris called convolution. With this notation the inequality just proved is k'fkLp(Rn)k'kL1(Rn)kfkLp(Rn): Consequently f7!'f:Lp(Rn)!Lp(Rn) is a linear and continuous mapping with norm not greater than k'kL1(Rn) 6.2 Fundamental solution Denition 6.2 Let2C. The fundamental solution Fof the Helmholtz equation
u+u= 0 in Rn,n= 2;3, is dened as follows: (i) Letn= 3. Forx2R3withx6= 0 set F(x) =eip jxj 4jxj: (ii) Letn= 2. Forx2R2withx6= 0 set F(x) =8 >< >:1 4N0(p jxj); 6= 0; 1 2lnjxj;  = 0; whereN0is Neumann's function of order 0. For the square root we take the branch satisfying 0 argp < . In the following we mainly study the fundamental solution in R3. Analogous results hold for the fundamental solution in R2. Lemma 6.3 The three-dimensional fundamental solution Fis innitely dier- entiable in R3nf0gand satises
F(x) +F(x) = 0; x6= 0: 74 Proof: It is obvious that Fis innitely dierentiable in R3nf0g. To show that the Helmholtz equation is satised let r=r(x) =jxj. Then (
+)F(x) = (
+)eip r(x) 4r(x) = 3X i=1"@r @xi2@2 @r2+@2r @x2 i@ @r# +! eip r 4r = 3X i=1"xi jxj2@2 @r2+1 jxjx2 i jxj3@ @r# +! eip r 4r =@2 @r2+2 r@ @r+eip r 4r= 0: Since@ @xiF(x) = (ip 1 jxj)eip jxj 4jxjxi jxj, it follows that to every R> 0 there exist constantsC1;C2such that jF(x)jC1 jxj;@ @xiF(x)C2 jxj2 for allxwith 0<jxjR, hence F;@ @xiF2L1(BR(0);C); whereBR(0) =fx2R3jxj<Rg. For2Cn[0;1) it follows that Reip <0; whencejF(x)jand@ @xiF(x)decay exponentially for jxj!1 . This implies that F;@ @xiF2L1(R3;C): Theorem 6.4 Assume that 2Cn[0;1). Forf2L2(R3;C)set u(x) =Z R3F(xy)f(y)dy=(Ff)(x); x2R3: Thenu2H1(R3;C)is a weak solution of the equation
u+u=f: (6.2) Proof: To prove that u=Ff2H1(R3) note that since F2L1(R3), @ @xiF2L1(R3) andf2L2(R3), we conclude from Theorem 6.1 that u=Ff2L2(R3); vi=@ @xiF
f2L2(R3);fori= 1;2;3: 75 Consequently, to prove that u2H1(R3) it suces to show that viis the weak derivative of u. To verify this let '2 C1(R3). Then (u;'xi) =Z R3Z R3F(xy)f(y)dy@ @xi'(x)dx =Z R3Z R3F(xy)@ @xi'(x)dxf(y)dy =Z R3lim r!0Z R3nBr(y)F(xy)@ @xi'(x)dxf(y)dy =Z R3lim r!0Z R3nBr(y)@ @xiF(xy)'(x)dxZ @Br(y)ni(x)F(xy)'(x)dSx f(y)dy =Z R3Z R3@ @xiF(xy)'(x)dxf(y)dy =Z R3Z R3@ @xiF(xy)f(y)dy'(x)dx=(vi;'): This proves that vi=@ @xiu; i= 1;2;3, henceu2H1(R3). In the computation above we used Gau' theorem. n(x) = (n1(x);n2(x);n3(x)) denotes the interior unit normal vector to @Br(y). We also used that lim r!0Z @Br(y)ni(x)F(xy)'(x)dSx= lim r!0Z @Br(y)ni(x)eip r 4r'(x)dSx lim r!0sup R3j'j1 4Z @Br(y)1 rdSx= sup R3j'jlim r!0r= 0: To prove that uis a weak solution of (6.2) in R3we must by Denition 3.31 show that (ru;r') +(u;') = (f;') (6.3) holds for all '2 C1(R3;C). To prove this we use the rst Green's formula and proceed similarly. We compute (ru;r') +(u;') =Z R3Z R3 rxF(xy)
rx'(x)F(xy)'(x)
dxf(y)dy =Z R3lim r!0Z R3nBr(y) rxF(xy)
rx'(x)F(xy)'(x) dxf(y)dy =Z R3lim r!0Z R3nBr(y)(
x)F(xy)'(x)dx +Z @Br(y)@ @nxF(xy)'(x)dSx f(y)dy =Z R3lim r!0Z @Br(y)@ @reip r 4r'(x)dSxf(y)dy: (6.4) 76 Now lim r!0Z jxyj=r@ @reip r 4r'(x)dSx (6.5) =lim r!0eip r 4r ip 1 r
Z jxyj=r'(y) + ('(x)'(y))dSx='(y); since lim r!0eip r 4r ip 1 r
Z jxyj=r'(x)'(y)dSx lim r!0sup x2@Br(y)j'(x)'(y)j1 4r(jp j+1 r)4r2= 0: (6.6) Here we used the continuity of '. We combine (6.5) with (6.4) and obtain (6.3). Consequently, uis a weak solution. Remark 6.5 Iffis more regular then u(x) =Z R3F(xy)f(y)dy is not only a weak solution, but also a classical solution of
u+u= 0. For, a weak solution uis a classical solution if u2C2(R3). This regularity of uis obtained for example if f2C1(R3) and jf(x)j;j@ @x1f(x)j;:::;j@ @x3f(x)jC for allx2R3, with a suitable constant C. To see this, suppose that fsatises these conditions and that Re ip <0. Then @ @xiu(x) =Z R3@ @xiF(xy)f(y)dy =Z R3@ @yiF(xy)f(y)dy=Z R3F(xy)@ @yif(y)dy and @2 @xj@xiu(x) =Z R3@ @xjF(xy)@ @yif(y)dy; and some technical considerations show that these derivatives exist in the clas- sical sense. 77 6.3 Green's function Convolution with the fundamental solution yields a solution of the Helmholtz equation in the whole space R3, but it does not yield the solution of a boundary value problem, since the boundary condition will not be satised in general. To nd a replacement for the fundamental solution in case of a boundary value problem assume that 2Cand that the boundary value problem
u(x) +u(x) =f(x); x2 ; u(x) =g(x); x2@ ; has a solution uin the domain R3. Assuming that the solution is regular enough such that Green's formula can be applied we obtain with the fundamen- tal solution F Z F(xy)f(y)dy=Z F(xy)(
y+)u(y)dy = lim r!0Z nBr(x)F(xy)(
y+)u(y)dy = lim r!0Z nBr(x)(
y+)F(xy)u(y)dy (6.7) +Z @ F(xy)@ @nyu(y)@ @nyF(xy)u(y)dSy +Z @Br(x)F(xy)@ @nyu(y)@ @nyF(xy)u(y)dSy =Z @ F(xy)@ @nyu(y)@ @nyF(xy)u(y)dSyu(x); since (
y+)F(xy) = 0 in the domain nBr(x), and since lim r!0Z @Br(x)F(xy)@ @nyu(y)dSy= 0 lim r!0Z @Br(x)@ @nyF(xy)u(y)dSy=u(x); which is proved as in (6.5), (6.6). From (6.7) we thus obtain u(x) =Z F(xy)f(y)dy+Z @ F(xy)@ @nyu(y)@ @nyF(xy)u(y)dSy: This is a representation formula for the solution in terms of the boundary values uj@ and@u @nof the solution. Sinceuj@ is equal to g, we can insert g(y) foru(y) in the boundary inte- gral on the right hand side of the representation formula. However, the normal 78 derivative@u=@n appearing in the boundary integral is unknown and can only be determined by solving the boundary value problem. Yet, one gets a repre- sentation formula which does not contain the unknown normal derivative if one replaces the fundamental solution Fby the Green's function for the Dirichlet boundary value problem G(x;y) =F(xy) +w(x;y); wherew:  !Cis dened as follows: For x2 letv: !Cbe the solution of
v(y) +v(y) = 0;fory2 v(y) =F(xy);fory2@ : Then set w(x;y) =v(y): The function Gsatises for every '2C( ) andx2 (i)G(x;y) = 0; y2@ ; (ii) (
y+)G(x;y) = 0; y2 ; x6=y, (iii) lim r!0Z @Br(x)'(y)G(x;y)dSy= 0; (iv) lim r!0Z @Br(x)'(y)@ @nyG(x;y)dSy = lim r!0Z @Br(x)'(y)@ @nyF(xy)dSy+ lim r!0Z @Br(x)'(y)@ @nyw(x;y)dSy ='(x): (nyis the interior normal vector to @Br(x).) (ii) { (iv) are precisely the properties needed in the computation (6.7). Therefore the same computation yields for Ginstead ofFand for the solution of the Dirichlet boundary value problem, using that G(x;y) = 0 fory2@ , u(x) =Z G(x;y)f(y)dyZ @ @ @nyG(x;y)g(y)dSy: (6.8) Of course, the determination of Grequires to solve the Dirichlet boundary value problem in . Therefore Gcan only be constructed if it is known in advance that the Dirichlet boundary value problem has a solution. The Green's function cannot be used to answer the questions of existence and uniqueness of boundary value problems. But if Gcan be determined explicitly it oers a means to represent, to compute and to study properties of the solution. 79 It is also possible to dene the Green's function for the Neumann boundary value problem
u(x) +u(x) =f(x); x2 ; @ @nu(x) =g(x); x2@ ; where2Cis a given constant. The Green's function for this problem is G(x;y) =F(xy) +w(x;y); wherew:  !Cis dened as follows: For x2 letv: !Cbe the solution of
v(y) +v(y) = 0;fory2 @ @nyv(y) =@ @nyF(xy);fory2@ : Then set w(x;y) =v(y): The function Gsatises @ @nyG(x;y) = 0; y2@ ; and we obtain the representation formula for the solution of the Neumann boundary value problem u(x) =Z G(x;y)f(y)dy+Z @ G(x;y)g(y)dSy: 6.4 The Green's function for the potential equation in a ball. Pois- son's representation formula. It is possible to determine the Green's function explicitly in some cases. Here we derive the Green's function for a ball BR(0) in Rnwithn= 2 andn= 3. To this end let R> 0 and consider the Kelvin transformation K:Rnnf0g! Rnnf0gdened by K(x) =R jxj2 x: (Re ection at the sphere with radius R.) Lemma 6.6 Forx;y2Rnwith 0<jxj<R,jyj=Rwe have jyK(x)j=R jxjjyxj: 80 Proof: We have jyK(x)j2=jK(x)j2+jyj22y
K(x) = =R jxj4 jxj2+R22R jxj2 x
y =R jxj2 R2+R jxj2 jxj22R jxj2 x
y =R jxj2 (jyj2+jxj22x
y) =R jxj2 jyxj2: Theorem 6.7 The Green's function to the Dirichlet problem for the potential equation in the ball BR(0)R3is G(x;y) =1 4jxyj+w(x;y) with w(x;y) =8 >>< >>:R 4jxj1 jyR jxj
2xj; 0<jxj<R 1 4R; x = 0: Proof: For 0<jxj<R we haveR jxj2 x=RR jxj>R. Thus, for all x2BR(0) y7!w(x;y)2C1 BR(0)
; and
yw(x;y) = 0: Also, for 0<jxj<R andy2@BR(0) we have w(x;y) =1 4jxj RjyK(x)j=1 4jxyj=F(xy): Clearly, for x= 0 andy2@BR(0) w(x;y) =1 4R=1 4jyj=F(xy): Consequently, wsatises
yw(x;y) = 0; (x;y)2BR(0)BR(0) w(x;y) =F(xy);(x;y)2BR(0)@BR(0); hence G(x;y) =F(xy) +w(x;y) is the Green's function. 81 Corollary 6.8 LetBR(0)R3and letu2C1 BR(0)
\C2 BR(0)
be a solu- tion of the Dirichlet problem
u(x) = 0; x2BR(0); u(x) =f(x); x2@BR(0): Then this solution is given by the Poisson representation formula u(x) =1 4RZ jyj=RR2jxj2 jxyj3f(y)dSy: Proof: Ifu2C1 BR(0)
\C2 BR(0)
, then the derivation of the representation formula (6.8) is valid, hence u(x) =Z @BR(0)@ @nyG(x;y)f(y)dSy: (6.9) Now for 0<jxj<R andjyj=R @ @nyG(x;y) =y jyj
ry1 4jxyjR 4jxj1 jyR jxj
2xj =1 4jxyj2(yx) jxyj
y jyj+R 4jxj1 jyR jxj
2xj2yR jxj
2x jyR jxj
2xj
y jyj =1 4jxyj3(yx)
y jyj+jxj2 4R21 jxyj3 yR jxj
2x
y jyj =1 4jxyj3 (yx)
y jyj+jxj R
2 yR jxj
2x

y jyj =1 4jxyj3 jyj+jxj2 R2jyj
=1 4RR2jxj2 jxyj3: Insertion into (6.9) yields the formula claimed in the lemma. For R2the Green's function to the Dirichlet problem for the potential equation is dened by G(x;y) =1 2lnjxyj+w(x;y); where for every x2
yw(x;y) = 0; y2 w(x;y) =1 2lnjxyj; y2@ : By the same method as in the three-dimensional case one obtains: 82 Theorem 6.9 (i)The Green's function for the Dirichlet problem to the poten- tial equation in the circle BR(0)R2is G(x;y) =1 2lnRjxyj jxjjyR jxj
2xj: (ii)Letu2C1(BR(0))\C2(BR(0)) be a solution of
u(x) = 0; x2BR(0); (6.10) u(x) =f(x); x2@BR(0): (6.11) Thenuis given by the Poisson representation formula u(x) =1 2RZ jyj=RR2jxj2 jxyj2f(y)dsy =1 2Z2 0(R22) R22Rcos('#) +2f(#)d#; wherex= (;'), in polar coordinates. Remark 6.10 Under the assumptions of Theorem 6.9 the boundary data fare continuously dierentiable. However, by Theorem 5.5 we know that the Dirich- let problem (6.10), (6.11) in R2has a unique solution for every continuous functionf. In fact, using Theorem 5.4 it is not dicult to prove by approxima- tion of a given continuous function fby continuously dierentiable functions that the Poisson representation formula also holds if fis only continuous. Up to now we have not shown that the Dirichlet problem for the potential equation in a ball BR(0)R3has a solution. Of course, the integral u(x) =1 4RZ jyj=RR2jxj2 jxyj3f(y)dSy in the Poisson representation formula in R3exists for every function f2 C @BR(0)
. Therefore one surmises that as in R2the solution of the Dirichlet problem in a three-dimensional ball exists and is given by this integral formula if the boundary data are continuous. This is true and can be proved directly by showing that the function ugiven by the formula is twice continuously dif- ferentiable in BR(0) and satises
u(x) = 0; x2BR(0; limx!z x2BR(0)u(x) =f(z);for allz2@BR(0): The proof of the rst assertion is obvious, but the second assertion is dicult to verify, since the denominator of the integrand in the Poisson representation formula tends to zero if xconverges to a boundary point. We do not analyse this boundary behavior here, but investigate a similar integral in Section 7, where we prove a general existence result for the Dirichlet problem in bounded domains in R3. 83 6.5 Green's function for the half space As last example we determine the Green's function for the Dirichlet problem in the half space H=fx= (x1;x2;x3)2R3jx3>0g. We have @H=f(x1;x2;0)j(x1;x2)2R2g=R2: Tof2 C(R2) and2Cone wants to nd a solution u2C2(H)\C(H) of
u(x) +u(x) = 0; x2H u(x0;0) =f(x0); x0= (x1;x2)2R2: Lemma 6.11 The Green's function for the Dirichlet problem in the half space His given by G(x;y) =1 4eip jxyj jxyj1 4eip j^xyj j^xyj wherex= (x1;x2;x3)2H,^x= (x1;x2;x3); y2Hwithx6=y. Proof: Obviously we have (
y+)1 4eip j^xyj j^xyj= 0; x;y2H; x6=y: Fory= (y1;y2;0)2@Handx2Hwe obtain j^xyj=p (x1y1)2+ (x2y2)2+ (x3)2=jxyj; hence G(x;y) =1 4eip jxyj jxyj1 4eip jxyj jxyj= 0: Corollary 6.12 Let2Cn[0;1)and letu2C1(H)\C2(H)be a solution of
u(x) +u(x) = 0; x2H uj@H=f; ju(x)j;jru(x)j C; x2H: Then forx2H u(x) =1 2Z @H@ @nyeip jxyj jxyjf(y)dy= +1 2Z @H@ @y3eip jxyj jxyjf(y)dy: Remark 6.13ju(x)j;jru(x)jCis a condition for the behavior of u(x) and ru(x) ifjxj!1 . 84 Proof: Letx2Handy2Hwithjyj2jxj. Then j^xyjjxyjjyjjxj1 2jyj: Thus, since Re ip <0, forx;y2Hwithjyjmax(1;2jxj) jG(x;y)j2 42 jyje1 2Reip jyjC1ecjyj; with a suitable constant C1>0 andc=1 2Reip  > 0. Similarly, forjyj max(1;2jxj) jryG(x;y)jC2ecjyj: Let R=fx2Hjjxj<Rg:Thenuhas the representation u(x) =Z @ R@ @nyG(x;y)u(y)G(x;y)@ @nyu(y)dSy =Z y2@H jyj<R@ @nyG(x;y)u(y)dSy Z jyj=R y3>0@ @nyG(x;y)u(y)G(x;y)@ @nyu(y)dSy: This formula holds for all R> 0. Thus, u(x) = lim R!1u(x) =lim R!1Z y2@H jyj<R@ @nyG(x;y)u(y)dSy lim R!1Z jyj=R y3>0@ @nyG(x;y)u(y)G(x;y)@ @nyu(y)dSy =Z y2@H@ @nyG(x;y)f(y)dSy: (6.12) Here we use that lim R!1Z jyj=R y3>0@ @nyG(x;y)u(y)G(x;y)@ @nyu(y)dSy lim R!1Z jyj=R y3>0(C2ecRC+C1ecRC)dSy= 0: Now, fory= (y1;y2;0)2@H @ @nyj^xyj=@ @y3j^xyj=y3+x3 j^xyj y3=0 =x3 jxyj=@ @y3jxyj=@ @nyjxyj: 85 Consequently, for y2@H @ @nyG(x;y) =@ @nyeip jxyj 4jxyj@ @nyeip j^xyj 4j^xyj =@ @nyeip jxyj 4jxyj+@ @nyeip jxyj 4jxyj =1 2@ @nyeip jxyj 4jxyj: Insertion of this expression into the representation formula (6.12) yields the statement of the corollary. 86 7 Integral equation method 7.1 The boundary integral equations Let R3be a bounded open set. In this section we study the Dirichlet and Neumann boundary value problems for the Helmholtz equation in and in the complement R3n and show that all four problems can be solved uniquely, if the boundary @ is suciently smooth. The Green's function method cannot be used for this, since for such general domains the Green's functions cannot be determined explicitly. Instead, we use the method of boundary integral equations. To explain the method consider the Dirichlet problem
u(x) +u(x) = 0; x2 (7.1) u(x) =f(x); x2@ : (7.2) If the boundary @ is smooth, then in suciently small neighborhoods of every pointx02@ the boundary will be almost planar. This suggets to represent the solution of (7.1), (7.2) by using the representation formula derived in Corol- lary 6.12 for the Dirichlet problem in the half space, for which the boundary is a plane. We thus try to nd the solution uin the form u(x) =1 2Z @ @ @nyeip jxyj jxyjv(y)dSy: (7.3) with a suitable function v2C(@ ;C). In this equation uis called double layer potential ,vis called the boundary layer . As will be seen, the double layer potentialusatises the Helmholtz equation
u(x) +u(x) = 0 for everyx2R3n@ . To see how vmust be chosen, note that if is equal to the half space Hand ifuis the solution of the Dirichlet problem in Hto the boundary condition uj@H=f, then we know from Corollary 6.12 that uis given by (7.3) with the choice v=f. This means that for every boundary point x0 we have limx!x0 x2 u(x) =f(x0) =v(x0): (7.4) One cannot expect that ugiven by (7.3) satises this simple limit relation also for curved boundaries. Instead, we shall show that for general boundaries @ a correction term appears on the right hand side of (7.4). This correction term is given by the jump relation limx!x0 x2 u(x) =v(x0) +1 2Z @ @ @nyeip jx0yj jx0yjv(y)dSy; x 02@ : (7.5) 87 From this jump relation we see that uis a solution of the Dirichlet boundary value problem (7.1), (7.2), if the boundary layer vsatises v(x) +1 2Z @ @ @nyeip jxyj jxyjv(y)dSy=f(x); (7.6) for allx2@ . This is an integral equation for the unknown function v2 C(@ ;C) with the given right hand side f2C(@ ;C). If a solution vof this integral equation can be determined to a given function f, then the double layer potential udened in (7.3) with vas the boundary layer is a solution of the Dirichlet boundary value problem to the boundary data f. Thus, if this integral equation is solvable for every f2C(@ ;C), the Dirichlet boundary value problem is solvable for all continuous boundary data. Therefore we must study under what conditions the boundary integral equation is solvable. Forx2@ we write (Kv)(x) =1 2Z @ @ @nyeip jxyj jxyjv(y)dSy: (7.7) With this notation the integral equation (7.6) can be written in the short form (I+K)v=f; (7.8) whereIdenotes the identity operator. The solution of the Dirichlet boundary value problem for the Helmholzu equation is thus reduced to the determination of the inverse of the linear operator I+K. To solve the Neumann boundary value problem
u(x) +u(x) = 0; x2 ; (7.9) @ @nu(x) =f(x); x2@ ; (7.10) we represent the solution by a single layer potential u(x) =1 2Z @ eip jxyj jxyjv(y)dSy; x2 : (7.11) The single layer potential satises the Helmholtz equation
u(x) +u(x) = 0 for allx2R3n@ and is continuous across the boundary @ , but the normal derivative satises a jump relation. To state this relation, we dene for x2@ @ @nu(x+) = lim s!0 s>0@ @su(x+snx);@ @nu(x) = lim s!0 s<0@ @su(x+snx): (7.12) 88 The jump relation is @ @nu(x) =v(x) +1 2Z @ @ @nxeip jxyj jxyjv(y)dSy; x2@ ; (7.13) where@ @nxdenotes the derivative with respect to the variable xin the direction of the exterior normal vector nxatx2@ . Note that@ @nu(x) is the limit of the normal derivative from the interior of . Therefore the Neumann boundary condition (7.10) is satised if the boundary layer vsatises the integral equation v+K0v=f; with the operator K0dened by (K0v)(x) =1 2Z @ @ @nxeip jxyj jxyjv(y)dSy: (7.14) Therefore the Neumann problem is solvable if the operator ( I+K0) : C(@ ;C)!C(@ ;C) is invertible. To determine whether the inverse ( I+K)1exists we must rst study the operator K. It is not obvious that the operator Kis well dened, since the denominator of the integral kernel in the expression on the right hand side of (7.7) vanishes at y=x; the integrand might thus have a non-integrable singularity. In Section 7.2 we therefore study this integral and show that it exists for everyv2C(@ ;C); we also show that the mapping x7!(Kv)(x) :@ !C is continuous, which implies that K:C(@ ;C)!C(@ ;C) is a linear operator. At the end of Section 7.2 we prove the jump relation (7.5). In Section 7.3 we study the single layer potential and verify the jump relation (7.13). In Section7.4 we shortly review the functional analytic theory of compact operators. With this theory we study in Sections 7.5 and 7.6 the invertibility of the mappings ( I+K0) and (I+K) and use the results to solve the Neumann and Dirichlet boundary value problems. 7.2 Properties of the double layer potential In this section we investigate the behavior of the double layer potential when x lies in R3n but is close to @ and when xbelongs to@ . In these investigations we use a technical lemma, which we prove rst. Throughout Section 7 we assume that R3is a bounded domain with @ 2C2and thatv2C(@ ;C). We need some notations and denitions: Let x02@ be an arbitrarily chosen point. We can choose the x1;x2;x3{coordinate system such that the x1;x2{plane is tangential to @ atx0and such that the exterior normal vector 89 nx0points into the direction of the negative x3-axis. Forx= (x1;x2;x3)2R3 we writex0= (x1;x2). By this choice of the coordinate system we have x= (x0;x3); x 0= (x0 0;0): Since@ 2C2, a parametrization of a neighborhood VR=VR(x0)@ ofx0 in@ is given by y07! y0;'(y0)
:B0 RR2!VR; whereB0 R=fy02R2jy0x0 0j<Rg, and where '2C2(B0 R;R) satises j'(y0)jMjy0x0 0j2;jr'(y0)jM0jy0x0 0j; with suitable constants M;M0>0. Fory2VRwe thus have y= y0;'(y0)
; y2VR: The exterior unit normal vector at y2VRis given by ny=1p 1 +jr'(y0)j2r'(y0) 1 2R3: The constants R,MandM0depend on x0, but since @ is of class C2, they can be chosen independently of x02@ . After these preparations we can formulate the technical lemma: Lemma 7.1 Let2C. There is a function g:f(x;y)jx2R3; x0=x0 0; y2 VRg!Cand constants C;c0, which can be chosen independent of x02@ , such that jg(x;y)jCec0jxyjjxyj2 and such that for every v2C VR;C
and for every x2R3withx0=x0 0 Z VR@ @nyeip jxyj jxyjv(y)dSy=Z B0 Rx3h(jxyj) jxyj3v(y)dy0+Z VRg(x;y) jxyj3v(y)dSy; whereh(r) =eip r(1ip r). Proof: We have at y= y0;'(y0)
2VR p 1 +jr'j2@ @nyeip jxyj jxyj =eip jxyj jxyj 1 jxyj+ip yx jyxj
r' 1 =eip jxyj(ip jxyj1) jxyj3 r'
(y0x0)('(y0)x3)
=h(jxyj)x3+r'
(y0x0)'(y0) jxyj3 =x3h(jxyj) +g1(x;y) jxyj3; 90 with g1(x;y) =h(jxyj) r'(y0)
(y0x0)'(y0)
: (7.15) Thus, Z VR@ @nyeip jxyj jxyjv(y)dSy =Z B0 R@ @nyeip jxyj jxyjv(y)p 1 +jr'(y0)j2dy0 =Z B0 Rx3h(jxyj) jxyj3v(y)dy0+Z B0 Rg1(x;y) jxyj3v(y)dy0 =Z B0 Rx3h(jxyj) jxyj3v(y)dy0+Z VRg(x;y) jxyj3v(y)dy; withg(x;y) =g1(x;y)p 1+jr'(y0)j2. From (7.15) we obtain jg(x;y)jjg1(x;y)jjh(jxyj)j M0jy0x0j2+Mjy0x0j2
Cec0jxyjjxyj2: If we now choose x=x02@ , hencex3= 0, we obtain from Lemma 7.1 that (Kv)(x) =1 2Z @ @ @nyeip jxyj jxyjv(y)dSy =1 2Z VRg(x;y) jxyj3v(y)dSy+1 2Z @ nVR@ @nyeip jxyj jxyjv(y)dSy: Since g(x;y) jxyj3Cec0jxyjjxyj2 jxyj3Cec0jxyj jxyj; it follows that the integral in the double layer potential exists for every x2@ , hence it denes a function Kv:@ !C. The next theorem shows that this function is H older continuous. Theorem 7.2 (i) (H older continuity) There is a constant Msuch that for all v2C(@ ;C)andx(1);x(2)2@ j(Kv)(x(1))(Kv)(x(2))jMjx(1)x(2)j1 4kvk1: Hence, the function Kvdened by the double layer potential is H older continu- ous with exponent 1=4. Here we use the norm kvk1= supx2@ jv(x)j: (ii) The linear operator v7!Kv:C(@ )!C(@ )is bounded. 91 Proof: (i) With2jx(1)x(2)jlet V=fx2@ jxx(1)j<g: Thenx(2)2Vwith dist (x(2);@ nV) = inf y2@ nVjx(2)yj 2: (7.16) By Lemma 7.1 we have (Kv)(x(1))(Kv)(x(2)) =1 2Z @ nV@ @nyeip jx(1)yj jx(1)yj@ @nyeip jx(2)yj jx(2)yj v(y)dy +1 2Z Vg(x(1);y) jx(1)yj3v(y)dSy1 2Z Vg(x(2);y) jx(2)yj3v(y)dSy =I1+I21+I22: (7.17) Since the constants C;c0in Lemma 7.1 are independent of x02@ , we have that g(x(i);y) jx(i)yj3Cec0jx(i)yj jx(i)yjC0 jx(i)yj; for allx(i);y2@ , whence jI2ij 1 2Z VC0 jx(i)yjkvk1dSy 1 2C0kvk1Z V1 jx(i)yjdSyC1kvk1; (7.18) with a suitable constant C1only depending on @ . The mean value theorem yields fory2 nVthat @ @nyeip jx(1)yj jx(1)yj@ @nyeip jx(2)yj jx(2)yj (x(1)x(2))
rx@ @nyeip jxyj jxyj x=x jx(1)x(2)jC2 jxyj323C2 3jx(1)x(2)j; withxon the line segment connecting x(1)andx(2)and with a constant C2only depending on @ . In the last step we used that (7.16) implies jyxj=2 for ally2@ nV. Thus jI1j1 2Z @ 23C2 3jx(1)x(2)jkvk1dSyC3kvk11 3jx(1)x(2)j;(7.19) 92 with a constant C3only depending on @ . Choose =jx(1)x(2)j1=4. Then =jx(1)x(2)j1=4=1 jx(1)x(2)j3=4jx(1)x(2)j2jx(1)x(2)j; forjx(1)x(2)j1 2
3=4. For such x(1);x(2)the relations (7.17) { (7.19) are valid. Together these relations yield j(Kv)(x(1))(Kv)(x(2))j  (2C1+C31 3jx(1)x(2)j)kvk1 = (2C1+C3)jx(1)x(2)j1=4kvk1: This proves (i). To prove (ii) note that for x2@ j(Kv)(x)j=1 2Z @ @ @nyeip jxyj jxyjv(y)dSy1 2Z @ @ @nyeip jxyj jxyjdSykvk1; which implieskKvk1Ckvk1withC=1 2R @ @ @nyeip jxyj jxyjdSy. Next we prove the jump relation (7.5): Theorem 7.3 (Jump relations) Let R3be a bounded domain with @ 2 C2, and letny2R3be the exterior unit normal vector to @ aty2@ . Assume that2Candv2C(@ ;C). Forx2R3n@ set w(x) =1 2Z @ @ @nyeip jxyj jxyjv(y)dSy: (i)Thenw2C1(R3n@ )satises
w(x) +w(x) = 0; x2R3n@ : (ii)Forx02@ we have the jump relations limx!x0 x2 w(x) =v(x0) +1 2Z @ @ @nyeip jx0yj jx0yjv(y)dSy limx!x0 x2R3n w(x) =v(x0) +1 2Z @ @ @nyeip jx0yj jx0yjv(y)dSy: Proof: (i) Since for all 2N3 0  (x;y)7!@jj @x@ @nyeip jxyj jxyj 2C (R3n@ )@ ;C
; 93 it follows as usual that w2C1(R3n@ ;C) with @jj @xw(x) =1 2Z @ @ @ny@jj @xeip jxyj jxyjv(y)dSy forx2R3n@ . In particular, this implies (
+)w(x) =1 2Z @ @ @ny(
x+)eip jxyj jxyjv(y)dSy= 0; since1 4eip jxyj jxyjis the fundamental solution of the Helmholtz equation. This proves (i). To prove (ii) assume that x2 is a point on the line normal to @ atx0, hence x= (x0;x3),x3>0,x0= (x0 0;0) andx0=x0 0. Lemma 7.1 yields w(x) +v(x0)1 2Z @ @ @nyeip jx0yj jx0yjv(y)dSy=I1+I2+I3; (7.20) with I1=v(x0)1 2Z B0 Rx3 jxyj3h(jxyj)v(y)dz; I2=1 2Z VRg(x;y) jxyj3g(x0;y) jx0yj3 v(y)dz; I3=1 2Z @ nVR@ @nyeip jxyj jxyj@ @nyeip jx0yj jx0yj v(y)dSy: In the following lemmas we derive estimates for jI1j;jI2j;jI3j. Lemma 7.4 To every">0there is1=1(R)>0such that jI3j<" for all 0<x 3<1. Proof: Forx3!0 it follows that x!x0, hence the integrand of I3tends to zero uniformly in @ nVR. This yields the statement. Lemma 7.5 To" > 0there isR1>0such that for all RR1and all 1x3>0 jI2j<": 94 Proof: By Lemma 7.1 we have jI2jkvk1 2Z VRCec0jxyj jxyj+Cec0jx0yj jx0yjdSyC0kvk1 Z VR1 jx0yjdSy<"; forRR1withR1suciently small. Lemma 7.6 Fory= z;'(z)
withz2BRwe have 1 jxyj31 jx(z;0)j33M(1 +MR)3 1 jx(z;0)j2 3M(1 +MR)3(R2+x2 3)1 2 jx(z;0)j3: Proof: Setr1=jx(z;0)j,r2=jxyj,r=jx0zj. Since xy=x(z;0)(0;'(z)); the inverse triangle inequality yields jr1r2jj(0;'(z)jMr2; and the triangle inequality implies r2 r1r1+j'(z)j r11 +j'(z)j r1 +Mr; r r2r1 r2r2+j'(z)j r21 +j'(z)j r1 +Mr: Thus, 1 r3 21 r3 1(r1r2)(r2 1+r1r2+r2 2) r3 1r3 2Mr2 r2 2r1 r2+ 1 +r2 r11 r2 1 M(1 +Mr)2(3 + 2Mr)1 r2 13M(1 +MR)31 r2 1: The statement follows from this inequality and from r2 1=r2+x2 3R2+x2 3. Lemma 7.7 1 2Z B0 Rx3 jx(z;0)j3dz= 1x3p x2 3+R21: Proof: We use polar coordinates ( r;) with the origin at x0 0=x0and note again thatjx(z;0)j2=r2 1=x2 3+r2to conclude 1 2Z B0 Rx3 jx(z;0)j3dz=1 2Z2 0ZR 0x3p x2 3+r23rdrd =x31p x2 3+r2r=R r=0= 1x3p x2 3+R2: 95 Lemma 7.8 To">0there is a constant R2>0such that for all 0<RR2 there is2=2(R)>0such that jI1j3" holds for all x3<2. Proof: Note thath(0) = 1, which means that the continuous function (z;x 3)!v(x0)h(j(x0;x3)(z;'(z))j)v(z;'(z)) =v(x0)h(jxyj)v(y) has a zero at ( z;x 3) = (x0;0) =x0. Consequently there are R2>0;2>0 such that forjx0zj<R 2and 0<x 3<2 v(x0)h(jxyj)v(y)<": ForRR2andx3<2we thus obtain from Lemma 7.6 and 7.7 that jI1j=v(x0) 1 2Z B0 Rx3 jx(z;0)j3dz+x3p x2 3+R2! 1 2Z B0 Rx3 jxyj3h(jxyj)v(y)dz =1 2Z B0 Rx3 jx(z;0)j3 v(x0)h(jxyj)v(y)
dz +1 2Z B0 Rx3 jx(z;0)j3x3 jxyj3 h(jxyj)v(y)dz+v(x0)x3p x2 3+R2 "1 2Z B0 Rx3 jx(z;0)j3dz +khvk13M(1 +MR)3(R2+x2 3)1 21 2Z B0 Rx3 jx(z;0)j3dz +jv(x0)jx3p x2 3+R2(7.21) "+khvk13M(1 +MR)3Rr 1 +2 R2 +jv(x0)j2 R; where we used the notation khvk1= sup y2@ x2 jh(jxyj)v(y)j: Now choose R2>0 small enough such that khvk13M(1+MR 2)3R2(1+"2)1 2< ". Subsequently, to 0 < R < R 2choose2=2(R) small enough such that 2 R(1 +jv(x0)j)< ". From (7.21) we then obtain jI1j<3"forRR2and 0<x 3<2. 96 End of the proof of the Theorem 7.3: To">0 set R= minfR1;R2g;  = minf1(R);2(R)g; (7.22) whereR1;R2;1;2are the numbers given in Lemmas 7.4, 7.5 and 7.8. From these lemmas and from (7.20) we then obtain for all 0 <x 3<and for ^v(x0) =v(x0) +1 2Z @ @ @nyeip jx0yj jx0yjv(y)dSy that jw(x)^v(x0)jjI1j+jI2j+jI3j<5"; whence lim x3!0 x3>0w(x0;x3) = lim s!0 s>0w(x0snx0) = ^v(x0): (7.23) It remains to show that this limit also holds if xapproaches x0not along the line normal to the boundary. To this end we note that the limit (7.23) is uniform with respect to x02@ , since the numbers R1,R2,1,2in (7.22) can be chosen independently of x0, which is seen by examination of the proof. To " >0 we can therefore choose 3>0 such thatjw(zsnz)^v(z)j< ", for allz2@ and all 0< s <  3. By Theorem 7.2 we have that ^ v2C(@ ). Consequently there is4>0 such thatj^v(z)^v(x0)j< "for allz2@ withjzx0j<  4. By these estimates we obtain for all points zsnzfrom the neighborhood U(x0) =fzsnzjz2@ ;jzx0j<4;0<s< 3g\ ofx0in that jw(zsnz)^v(x0)jjw(zsnz)^v(z)j+j^v(z)^v(x0)j<2": This means that limx!x0 x2 w(x) = ^v(x0); which proves the rst jump relation in statement (ii). The second jump relation is proved analogously. 7.3 Properties of the single layer potential We next study the single layer potential (7.11). We assume that R3is a bounded open set with @ 2C2and that2C. Just as in the previous section it can be shown that to v2C(@ ;C) there are constants C,c0such that for all x;y2@ the estimate @ @nxeip jxyj jxyjCec0jxyj jxyj holds. It follows that for every v2C(@ ;C) and allx2@ the integral in (7.14) exists. (7.14) thus denes a function ( K0v) :@ !C. 97 Theorem 7.9 (i) There is a constant Msuch that for all v2C(@ ;C)and x(1);x(2)2@ j(K0v)(x(1))(K0v)(x(2))jMjx(1)x(2)j1 4kvk1: (ii) The linear operator v7!K0v:C(@ )!C(@ )is bounded. Theproof is similar to the proof of Theorem 7.2. Therefore we omit it. Theorem 7.10 Assume that v2C(@ ;C). Forx2R3n@ set !(x) =1 2Z @ eip jxyj jxyjv(y)dSy: Then (i)!belongs toC(R3;C)\C1(R3n@ ;C)and satises (
+)!(x) = 0; x2R3n@ : (ii)Atx2@ the one sided derivatives@! @n(x)dened in (7.12) exist and satisfy @! @n(x) =v(x) +1 2Z @ @ @nxeip jxyj jxyjv(y)dSy: To prove this theorem we need a lemma. Lemma 7.11 Forx2@ ands2Rletxs=x+snx. Then lim s!0Z @ @ @seip jxsyj jxsyj+@ @nyeip jxsyj jxsyj v(y)dSy =Z @ @ @nxeip jxyj jxyj+@ @nyeip jxyj jxyj v(y)dSy: (7.24) Proof: Since lim s!0@ @seip jxsyj jxsyj=@ @nxeip jxyj jxyj; for ally2@ nfxg, it suces to show that the limit can be interchanged with the integral. To verify this we constuct a majorant for the integrand on the left hand side of (7.24), which is independent of s. The result is then implied by Lebesgue's integration theorem. Note rst that to every r0there is a constant c1such thatj@ @reip r rjc1 r2for all 0<rr0. Together with this estimate it thus follows for all s2Rand all y2@ withy6=xthat @ @seip jx+snxyj jx+snxyj+@ @nyeip jxsyj jxsyj =@ @reip r r r=jxsyj(xsy)
(nxny) jxsyjc1 jxsyj2jnxnyj: (7.25) 98 Since@ is bounded and of class C2, there is a constant c2such that for all y2@ jnxnyjc2jxyj: (7.26) We choose s0>0 small enough such that the line segment fx+snxjjsjs0g interescts@ only in the point x. It then follows by standard considerations that there is a constant c3>0 such that for all jsjs0we have jxyjc3jxsyj: From this estimate and from (7.25), (7.26) we conclude that @ @seip jx+snxyj jx+snxyj+@ @nyeip jxsyj jxsyjc1c2jxyj c1 3jxyj2=C jxyj; with the constant C=c1c2c3independent of s. The functionC jxyjis integrable over the two-dimensional manifold @ , hence it is a majorant for the integrand on the left hand side of (7.24). Proof of Theorem 7.10: The proof of (i) is standard and we omit it. In the proof of (ii) we restrict ourselves to the verication of the formula for@! @n(x). The other formula is proved in the same way. From Lemma 7.11 and from the jump relations in Theorem 7.3(ii) we con- clude that @! @n(x) = lim s!0 s<01 2Z @ @ @seip jxsyj jxsyjv(y)dSy = lim s!0 s<01 2Z @ @ @seip jxsyj jxsyj+@ @nyeip jxsyj jxsyj v(y)dSy lim s!0 s<01 2Z @ @ @nyeip jxsyj jxsyjv(y)dSy =1 2Z @ @ @nxeip jxyj jxyj+@ @nyeip jxyj jxyj v(y)dSy +v(x)1 2Z @ @ @nyeip jxyj jxyjv(y)dSy =v(x) +1 2Z @ @ @nxeip jxyj jxyjv(y)dSy: 7.4 Compact operators on a Banach space To study the solvability of the boundary integral equations we use a result from functional analysis, which we present here. In the following Xdenotes a Banach space with normk
k. 99 Denition 7.12 A linear operator T:X!Xis called bounded if there is a constantCsuch that kTxkCkxk for allx2X. Theorem 7.13 A linear operator T:X!Xis bounded if and only if it is continuous. Proof: IfTis continuous at 0 it follows that there is >0 such that kTxk1 for allx2Xwithkxk. Since for every y2X,y6= 0 we have ky kykk=kyk kyk=; it follows that kTyk=kTkyk y kyk
k=kyk kT y kyk
k 1 kyk: This proves that Tis bounded. On the other hand, assume that Tsatises kTxkCkxk for allx2X. Lety2X," >0, and set=" C. Then for all z2Xwith kzykit follows kT(z)T(y)k=kT(zy)kCkzyk"; henceTis continuous at y. Sinceywas arbitrary, Tis continuous on X. Denition 7.14 A linear operator T:X!Xis called compact if to every bounded sequence fxngnXthe sequence of images fTxngnhas a subsequence, which converges in X. Lemma 7.15 A compact operator is bounded. Proof: If the compact operator Twould not be bounded then there would exist a sequencefxngnXwithkxnk= 1 andkTxnkn, for alln2N. The sequencefTxngnwould not have a convergent subsequence, hence Tis not compact. Remember that for a linear operator T:X!Xa number2Cwith the property that there is x2X,x6= 0 satisfying Txx= 0 is called eigenvalue ofT. The element xis called eigenvector. The set E=fx2XjTxx= 0g is a linear subspace of Xcalled eigenspace of the eigenvalue . The dimension ofEis called the multiplicity of . 100 Denition 7.16 LetT:X!Xbe a bounded operator. The resolvent set (T) ofTconsists of all points 2C, which are not eigenvalues and for which the operator ( TI) :X!Xis surjective. Here Iis the identity. The complement ( T) =Cn(T) is called spectrum of T. Clearly,2(T) if and only if TIis injective and surjective. Hence  belongs to the resolvent set if and only if ( TI)1exists. Theorem 7.17 LetT:X!Xbe compact. (T)is a countable set with no accumulation point dierent from zero. Each nonzero 2(T)is an eigenvalue ofTwith nite multiplicity. If Xhas innite dimension, then 0belongs to (T). I only give part of the proof. The complete proof can be found for example in the book of Alt, pp. 363. Proof: I.) First I show that the eigenvalues of Tdo not accumulate at a point 6= 0. Otherwise there would exist a sequence fngnof distinct eigenvalues of Twith eigenvectors xnsuch that 06=n!6= 0. LetMnbe the subspace spanned by the nvectorsx1;:::;xn. The space Mnis invariant under T; for if x2Mnthenx=c1x1+:::+cnxn, hence Tx=T(c1x1+:::+cnxn) =1c1x1+:::+ncnxn2Mn; thusT(Mn)Mn. Since eigenvectors to distinct eigenvalues are linearly independent, the vec- torsx1;x2;:::are linearly independent. Therefore Mn1is a proper subspace ofMnand there is yn2Mnsuch thatkynk= 1 and dist( yn;Mn1) = 1. This holds since Mnis isomorphic to Rn. With the sequence fyngnthus dened I show thatf1 nTyngncontains no Cauchy sequence, contradicting the assump- tion thatTis compact. (Note that f1 nyngnis a bounded sequence.) We have form<n 1 nTyn1 mTym=yn 1 mTym1 n(Tn)yn
where the second term on the right belongs to Mn1becauseym2Mn1,Mn1 is invariant under Tand (Tn)yn2Mn1. Since dist( yn;Mn1) = 1, it follows that each element of the sequence f1 nTyngnhas distance1 from any other one, showing that no subsequence of this sequence can be convergent. II.) If there would be an eigenvalue 6= 0 of innite multiplicity we could derive a contradiction by exactly the same arguments, dening nbyn=for alln and choosing forfxngna sequence of linearly independent eigenvectors to . It remains to show that if 6= 0 is not an eigenvalue it belongs to (T), hence the rangeR(TI) is equal to X. To this end it is shown that R(TI) is closed and that there is no nontrivial complementary space. For the details I refer to the book of Alt. 101 This theorem shows that if Tis a compact operator, then for given h2Xthe equation (I+T)x=his solvable if1 is not an eigenvalue of T. 7.5 Solution of the Neumann problem I use Theorem 7.17 to show that the integral equations ( I+K0)v=fand (I+K0)v=fcan be solved, which implies that the interior and exterior Neumann boundary value problems have solutions. To this end I show that K0 is a compact operator on the Banach space C(@ ;C). The norm on this Banach space is kvk1= sup x2@ jv(x)j: Denition 7.18 Let Rm. A sequencefvngnof functions vn: !Cis called uniformly equicontinuous if to every ">0 there is>0 such that jvn(x)vn(y)j<" for alln2Nand allx;y2 withjxyj<. Theorem 7.19 (Arzela-Ascoli)2Letfvngnbe a bounded, uniformly equicon- tinuous sequence of functions on . Then there is a uniformly convergent sub- sequencefvn`g`. Corollary 7.20 The operator K0:C(@ ;C)!C(@ ;C)is compact. Proof: Letfvngnbe a bounded sequence in C(@ ;C). By Theorem 7.9(ii) the operator K0is bounded, which implies that also the sequence fK0vngnis bounded in C(@ ;C). Let" >0, setC= supkvnk1and letM > 0 be the constant from Theorem 7.9(i). This theorem implies for all x1;x22@ with jx1x2j<=" CM4 that j(K0vn)(x1)(K0vn)(x2)jMjx1x2j1=4kvnk1MC" CM =": Thus,fK0vngnis a uniformly equicontinuous sequence. Therefore all the as- sumptions of Theorem 7.19 are satised for this sequence, from which we con- clude that it has a subsequence converging with respect to the norm k
k1of C(@ ;C). This means that K0is compact. Lemma 7.21 Let2Cn[0;1). Then 1and1are no eigenvalues of K0. 2Cesare Arzel a (1847{1912), Giulio Ascoli (1843{1896). 102 Proof: We prove rst that 1 is not an eigenvalue. To this end it suces to show ifv2C(@ ;C) satises (I+K0)v= 0; (7.27) thenv= 0, since this implies that the kernel of K0(1)Iis equal tof0g. To verify that vvanishes, we dene for x2R3the single layer potential u(x) =1 2Z @ eip jxyj jxyjv(y)dSy: (7.28) Equation (7.27) and the jump relations from Theorem 7.10(ii) imply that u solves the boundary value problem
u(x) +u(x) = 0; x2 ; @ @nu(x) = 0; x2@ : The rst Green's formula yields 0 =Z @ @ @nu(x)u(x)dS=Z
u(x)u(x) +ru(x)
ru(x)dS =Z (ju(x)j2+jru(x)j2)dx =iImZ ju(x)j2dx+Z jru(x)j2Reju(x)j2dx: Since Im6= 0 or Re<0 it follows from this equation that u0 in . Since the single layer potential uis continuous on R3it thus follows that uis also a solution of the boundary value problem
u(x) +u(x) = 0; x2R3n u(x) = 0; x2@(R3n ): To apply the Green's formula in R3n note that for 2Cn[0;1) we have by our choice of the square root that Re ip <0, whence ju(x)j 1 2Z @ eReip jxyj jxyjjv(y)jdSyeReip dist(x;@ ) dist(x;@ )1 2Z @ jv(y)jdSy: Thereforeju(x)jdecreases exponentially for jxj!1 . Because ru(x) =1 2Z @ rxeip jxyj jxyjv(y)dSy; 103 it follows in the same way that also jru(x)jdecreases exponentially for jxj!1 . Therefore the rst Green's formula yields Z R3n (ju(x)j2+jru(x)j2)dx = lim R!1Z R3n jxj<R ju(x)j2+jru(x)j2
dx = lim R!1Z @ @ @nu(x)u(x)dS+Z jxj=R@ @nu(x)u(x)dS = 0: As above it follows from this equation that u= 0 in R3n , whence u0 in R3. Again using the jump relations from Theorem 7.10, we now conclude for allx2@ that 0 =@u @nx(x+)@u @nx(x) =v(x) +1 2Z @ @ @nxeip jxyj jxyjv(y)dSy v(x)1 2Z @ @ @nxeip jxyj jxyjv(y)dSy=2v(x); hencev= 0:Therefore1 is not an eigenvalue. To prove that 1 is not an eigenvalue we assume that v2C(@ ;C) satises (I+K0)v= 0: We insertvinto (7.28). The jump relations for single layer potentials then imply that usolves the boundary value problem
u(x) +u(x) = 0; x2R3n ; @ @nu(x) = 0; x2@(R3n ): Proceeding as above we conclude from this that u= 0 in R3n and in , from which we infer by the jump relations that v= 0. Therefore 1 is not an eigenvalue. Corollary 7.22 Let R3be a bounded open set with @ 2C2. Suppose that 2Cn[0;1). Then the interior Neumann boundary value problem
u(x) +u(x) = 0; x2 ; @ @nu(x) =f(x); x2@ 104 and the exterior Neumann boundary value problem
u(x) +u(x) = 0; x2R3n ; @ @nu(x+) =f(x); x2@(R3n ) ju(x)j;jru(x)j=O(eReip jxj);jxj!1 (7.29) have unique solutions for all f2C(@ ;C) =C(@(R3n );C). The solutions are given by the single layer potentials u(x) =1 2Z @ eip jxyj jxyjv(y)dSy; x2 ; (7.30) wherevsatises the integral equation (I+K0)v=ffor the interior problem and(I+K0)v=ffor the exterior problem. Proof: Since by Corollary 7.20 the operator K0is compact and since by Lemma 7.21 the number 1 is not an eigenvalue of this operator, it follows from Theorem 7.17 that 1 belongs to the resolvent set of K0, whence the map- ping (I+K0) :C(@ ;C)!C(@ ;Cis invertible. Consequently, the boundary integral equation ( I+K0)v=fhas a unique solution v2C(@ ;C) . With thisvas boundary layer the single layer potential ufrom (7.30) is a solution of the interior Neumann boundary value problem. To prove that the solution is unique let ^ube another solution of the same problem. Then w=u^usatises
w(x) +w(x) = 0; x2 @ @nw(x) = 0; x2@ ; hence the rst Green's formula yields 0 =Z @ @ @nw(x)w(x)dS=Z
w(x)w(x) +jrw(x)j2dx =Z jw(x)j2+jrw(x)j2
dx =iImZ jw(x)j2dx+Z jrw(x)j2Rejw(x)j2dx: This implies w= 0, hence u= ^u. Therefore the solution is unique. A solutionuof the exterior Neumann boundary value problem is obtained if we insert the unique solution vof the boundary integral equation ( I+K0)v=f into (7.30). As in the proof of Lemma 7.21 we see that udened in this way satises the radiation condition (7.29). To prove uniqueness of the solution suppose that ^ uis a second solution. We apply the rst Green's formula to w= u^uin the exterior domain R3n as in the proof of Lemma 7.21, noting that w(x) decreases exponentially for jxj!1 because of the radiation condition (7.29) satised by uand ^u, and conclude as above that w= 0. 105 7.6 Solution of the Dirichlet problem To solve the interior and exterior Dirichlet problems we must show that the boundary integral equations ( I+K)v=fand (I+K)v=fare solvable. Theorem 7.23 The operator K:C(@ ;C)!C(@ ;C)is compact. This theorem is proved in the same way as Corollary 7.20 using Theorem 7.2 instead of Theorem 7.9. Lemma 7.24 Let2Cn[0;1). Then 1and1are no eigenvalues of K. Proof: Foru;v2C(@ ) we write hu;vi@ =Z @ u(x)v(x)dSx: By interchanging the order of integration we obtain hK0u;vi@ =Z @ 1 2Z @ @ @nxeip jxyj jxyju(y)dSyv(x)dSx =Z @ u(y)1 2Z @ @ @nxeip jxyj jxyjv(x)dSxdSy =hu;Kvi@ : (7.31) Now let= 1 or=1 and assume that v2C(@ ) satises (I+K)v= 0: From (7.31) we conclude for all u2C(@ ) that h(I+K0)u;vi@ =hu;(I+K)vi@ = 0: Since by Lemma 7.21 neither 1 nor 1 is an eigenvalue of the compact operator K0, the mapping (I+K0) :C(@ )!C(@ ) is surjective. Therefore there isu2C(@ ) such that (I+K0)u=v. Thus, Z @ jv(x)j2dSx=hv;vi@ =h(I+K0)u;vi@ = 0: Consequently vmust be equal to zero and cannot be an eigenfunction. This implies that is not an eigenvalue of K. Corollary 7.25 Let R3be a bounded open set with @ 2C2. Suppose that 2Cn[0;1). Then the interior Dirichlet boundary value problem
u(x) +u(x) = 0; x2 u(x) =f(x); x2@ ; 106 and the exterior Dirichlet boundary value problem
u(x) +u(x) = 0; x2R3n u(x) =f(x); x2@(R3n ); ju(x)j;jru(x)j=O(eReip jxj);jxj!1 have unique solutions for all f2C(@ ;C). The solutions are given by the double layer potentials u(x) =1 2Z @ @ @nyeip jxyj jxyjv(y)dSy; wherevsatises the integral equation (I+K)v=ffor the interior problem and(I+K)v=ffor the exterior problem. This corollary is proved as the corresponding result for the Neumann problem. 107 8 Hilbert space methods 8.1 Elliptic dierential operators, weak solutions Let Rnbe an open set and let Lu(x) =X jj1 jj1D a(x)Du(x)
; x2 be a linear dierential operator of second order with coecient functions : !C; ;2Nn 0;jj;jj1: The sum X j+j=2a(x)D+u(x) is called principle part of this operator. Denition 8.1 (i) The operator Lis called elliptic if for all 2Rn,6= 0 and allx2 X j+j=2a(x)+6= 0: (ii)Lis called strongly elliptic if to every x2 there is>0 such that for all 2Rn ReX j+j=2a(x)+ jj2: is the ellipticity constant. (iii)Lis uniformly strongly elliptic if Lis strongly elliptic with an ellipticity constant which can be chosen independent of x2 . Example: Choose a(x) =( 1; if=;jj= 1 0; otherwise. ThenX jj1 jj1D a(x)Du(x)
=nX i=1@2 @x2 iu(x) =
u(x): For this operator we have X jj=jj=1a(x)+=nX i=12 i=jj2; 108 consequently
is uniformly strongly elliptic with ellipticity constant = 1. In the following I assume that a: !Ris measurable and bounded for all multi-indices ;2Nn 0withjj;jj1. The operator Lu=X jj1 jj1D(aDu) is in divergence form. For such operators the Denition 3.31 of weak solutions for the Helmholtz equation can be generalized immediately: Denition 8.2 Letf2L2( ;C) and2C. (i) The function u2H1( ;C) is called weak solution of the partial dierential equationX jj1 jj1D a(x)Du(x)
+u(x) =f(x) in , if for all '2 C1( ;C) the equation X jj1 jj1(1)jjZ a(x)Du(x)D'(x)dx+Z u(x)'(x)dx=Z f(x)'(x)dx holds. (ii) Letg2H1( ;C). Thenu2H1( ;C) is called weak solution of the Dirichlet boundary value problem X jj1 jj1D a(x)Du(x)
+u(x) =f(x); x2 uj@ =gj@ ; ifuis a weak solution of the partial dierential equation and if ug2 H1( ;C). In the following I write for u;v2H1( ;C) B(u;v) =Z X jj1 jj1(1)jja(x)Du(x)Dv(x)dx: (8.1) With this denition it follows that u2H1( ;C) is a weak solution of X jj1 jj1D(aDu) +u=f 109 if and only if B(u;') +(u;') = (f;') (8.2) for all'2 C1( ). For the Laplace operator L=
we have B(u;v) =(ru;rv) : Insertion of this expression into (8.2) shows that for the Helmholtz equation Denition 3.31 of weak solutions coincides with Denition 8.2. 8.2 Coercivity of sesquilinear forms to elliptic operators Denition 8.3 LetXbe a vector space over Cwith normkuk, and let (u;v)7! [u;v] :XX!Cbe a mapping. This mapping is called 1. a sesquilinear form, if [u+v;w ] =[u;w] +[v;w]; [u;v +w] =[u;v] +[u;w]; 2. symmetric, if [ u;v] =[v;u], 3. bounded, ifj[u;v]jKkukkvk, 4. strictly coercive, if there is c>0 such that [ u;u]ckuk2for allu2X. The simplest example of a symmetric, bounded, strictly coercive sesquilinear form is the scalar product ( u;v) on a Hilbert space. The mapping ( u;v)7! B(u;v) :H1( ;C)H1( ;C)!Cdened in (8.1) is linear in the rst argument and antilinear in the second argument, hence Band of course also Bare sesquilinear forms. In this section we study the coercivity of B, which is a slightly weaker property than strict coercivity. In the formulation of the respective result we write for u2H1( ;C) juj1; = (ru;ru)1=2 : With this notation one has kuk2 1; =kuk2 +juj2 1; : Theorem 8.4 Leta: !Cbe bounded measurable with a(x) = (1)j+ja(x); ifjj+jj1; a(x) =a(x)2R; ifjj=jj= 1; and assume that Lu=X jj;jj1D(aDu) 110 is uniformly strongly elliptic with ellipticity constant  > 0. Then ^B(u;v) = B(u;v)is a symmetric and bounded sesquilinear form on H1( ;C), which satises ^B(u;u)c1juj2 1; c2kuk2 ;for allu2H1( ;C); (8.3) where c1= 2; c 2=K2 2+K; K =X j+j1kak1: Denition 8.5 A sesquilinear form ^Bsatisfying (8.3) with suitable constants c1>0 andc20 is called coercive on H1( ;C). Remark 8.6 Sincea2Rforjj=jj= 1 the condition of strong ellipticity isX jj=jj=1a+jj2: Proof of the theorem: Foru;v2H1( ;C) we have B(u;v) =X jj;jj1(1)jj(aDu;Dv) =X jj;jj1(1)jj+jj+jj(aDu;Dv) =X jj;jj1(1)jj(Du;aDv) =B(v;u): Thus,Bis symmetric. Also we have jB(u;v)jX jj;jj1kak1j(Du;Dv)jCkuk1; kvk1; : ThusBis bounded. To see thatBis coercive dene B0(u;v) =X jj=jj=1(aDu;Dv): The above calculation shows that B0is symmetric. Since a(x)2Rforjj= jj= 1 we thus obtain for real valued functions u;vthat B0(u;v) =B0(v;u) =B0(v;u): 111 Thus, ifu2H1( ;C) andu1= Reu,u2= Imu, it follows B0(u;u) =B0(u1+iu2;u1+iu2) =B0(u1;u1)B0(u2;u2) iB0(u2;u1) +iB0(u1;u2) =B0(u1;u1)B0(u2;u2) =Z X jj=1 jj=1a(x) Du1(x)Du1(x) +Du2(x)Du2(x)
dx =Z X jj=1 jj=1a(x) ru1(x)
++X jj=1 jj=1a(x) ru2(x)
+ dx Z jru1(x)j2+jru2(x)j2dx=juj2 1; : This together with ab" 2a2+1 2"b2yields B(u;u) =B0(u;u)X j+j=1(1)jj(aDu;Du)(a00u;u) juj2 1; X j+j=1kak1juj1; kuk0; ka00k1kuk2 0; juj2 1; " 2juj2 1; 1 2"K2kuk2 0; Kkuk2 0; : Choosing"=shows that ^B=Bis coercive with the constant c2given in the theorem. The proof is complete. 8.3 Existence of weak solutions to elliptic equations The coercivity of the sesquilinear form Ballows to prove that boundary value problems to elliptic operators have weak solutions. To show this we reformulate Denition 8.2 of weak solutions slightly. Assume that the operator L=P jj1 jj1D(aD) satises the assumption of Theorem 8.4, let 2Rand letf2L2( ;C). By Denition 8.2 b.) the functionu2 H1( ) is a weak solution of the homogeneous Dirichlet boundary value problem Lu+u =fin ; (8.4) uj@ = 0; (8.5) if for allv2 C1( ;C) B(u;v) +(u;v) = (f;v) (8.6) 112 holds. The sesquilinear form Bis bounded on H1( ;C), hence it is continuous in both arguments. Therefore, since C1( ;C) is dense in H1( ;C), equation (8.6) holds for all v2 C1( ;C) if and only if it holds for all v2 H1( ;C). Using that is real, we conclude that u2 H1( ;C) is a weak solution of the homogeneous Dirichlet boundary value problem (8.4), (8.5) if and only if B(v;u) +(v;u) = (v;f) (8.7) holds for all v2 H1( ;C). We have thus reduced the problem of the existence of weak solutions to an abstract problem for symmetric sesquilinear forms B on the Hilbert space H1( ). Accordingly, the existence proof is based on the coercivity of Band on the following easy result: Lemma 8.7 Let[u;v]be a symmetric, bounded, strictly coercive sequilinear form on a Banach space XoverC. Then [u;v]is a scalar product on X. The associated norm u= [u;u]1=2is equivalent to the norm kuk. The space Xis complete with respect to the norm u, whenceXis a Hilbert space with the scalar product [u;v]. Proof: Obviously every symmetric, strictly coercive sesquilinear form is a scaler product. From the boundedness and the strict coercivity we obtain ckuk2[u;u] =u2Kkuk2; (8.8) which means that k
k and
are equivalent norms. If fung1 n=1is a Cauchy sequence with respect to the norm
, then (8.8) implies that fung1 n=1is also a Cauchy sequence with respect to the norm k
k. SinceXis complete with respect to this norm, there is a limit element u2Xof this Cauchy sequence. From (8.8) we obtain lim n!1uunlim n!1K1 2kuunk= 0; henceuis also the limit of fung1 n=1with respect to the norm
. Therefore X is complete with respect to this norm. Corollary 8.8 Let Rnbe an open set, let L=X jj;jj1D(aD) satisfy the assumptions of Theorem 8.4, and let <c2with c2=K2 2+K; K =X j+j1kak1; 113 where > 0is the ellipticity constant of L. Then the homogeneous Dirichlet boundary value problem (8.4) ,(8.5) has a unique weak solution u2 H1( ;C) for allf2L2( ;C). This solution satises kuk1; max2 ;1 c2
kfk : Proof: Dene the sesquilinear form [ u;v] on H1( ;C) H1( ;C) by [u;v] =B(u;v)(u;v) : (8.9) Theorem 8.4 implies that this sesquilinear form is symmetric, bounded and satises for u2 H1( ;C) [u;u] =B(u;u)(u;u)  2juj2 1; (c2+)kuk2 ckuk2 1; ;(8.10) withc= min( 2;c2)>0. Thus, [u;v] is strictly coercive. Consequently, by Lemma 8.7 this sesquilinear form is a scalar product on H1( ;C) with norm u2= [u;u] =B(u;u)(u;u) : Moreover, the linear form h: H1( )!Cdened by h(v) =(v;f) is bounded because (8.10) yields jh(v)jkfk kvk kfk kvk1; kfk c1 2v: The Riesz representation theorem (Corollary 3.7) thus implies that there is a unique function u2 H1( ;C) satisfying [v;u] =h(v) for allv2 H1( ;C). By denition of [ u;v] this equation is equivalent to (8.7). Consequently, uis the unique weak solution of the boundary value problem. This solution satises ckuk2 1; [u;u] =h(u)kfk kuk1; ; hencekuk1; 1 ckfk . This proves the corollary. Letf2L2( ) andg2H1( ). By denition, u2H1( ) is a weak solution of the inhomogeneous Dirichlet boundary value problem Lu+u=f; (8.11) uj@ =gj@ ; (8.12) 114 ifw=ug2 H1( ) and B(v;u) +(v;u) = (v;f) for allv2 H1( ). This implies that wsatises B(v;w) +(v;w) = (v;fg) B(v;g) for allv2 H1( ). On the other hand, if w2 H1( ) satises this equation for allv2 H1( ), thenu=w+gis a weak solution of the problem (8.11), (8.12). Corollary 8.9 Let the assumptions of Corollary 8.8 be satised. Then for all  <c2, allf2L2( ) andg2H1( ) there is a unique weak solution u2 H1( )of the inhomogeneous Dirichlet boundary value problem (8.11) ,(8.12) . Proof: Let the linear form h: H1( )!Cbe dened by h(v) = (v;gf) +B(v;g): The function u=g+wis a weak solution of the Dirichlet boundary value problem if and only if w2 H1( ) satises [v;w] =h(v) for allv2 H( ), where [ v;w] is the sesquilinear form dened in (8.9). From the boundedness of Bwe have jh(v)j  kgfk kvk +Kkgk1; kvk1;  kgfk +Kkgk1;
kvk1;  kgfk +Kkgk1;
c1 2u; where in the last step we used (8.10). Therefore his a bounded linear form on the Hilbert space H1( ) equipped with the scalar product [ u;v]. Consequently, by Corollary 3.7 applied to this Hilbert space there is a unique solution w2  H1( ). Example 8.10 The operator L=
does not have lower order terms, hence c2= 0. Therefore there is a unique weak solution of
u+u=f; uj@ =gj@ for all<0,f2L2( ),g2H1( ). 115 9 Eigenvalue problems, spectral theory 9.1 The Friedrichs' extension of the operator L LetL=P jj;jj1DaDwitha: !Cbounded measurable, and let f2L2( ). By denition u2H1( ;C) is a weak solution of Lu=fif and only if B(u;') = (f;') for all'2 C1( ). In this point of view L=PDaDis merely a symbolic expression. Yet, we can attach a precise meaning to Land dene it as an operator on the Hilbert space L2( ;C) as follows: The domain of denition D(L) ofLis given by D(L) =n u2H1( )9 f2L2( )8 '2 C1( ):B(u;') = (f;') o : It is immediately seen that D(L) is a linear subspace of H1( ). Note that if u2D(L) then the function f2L2( ) satisfying B(u;') = (f;') for all '2 C1( ), which exists by denition, is unique. For, if g2L2( ) is a second such function then (f;') =B(u;') = (g;') ; whence (fg;') = 0 for all'2 C1( ). Since C1( ) is dense in L2( ), this equation implies f=g. Therefore for u2D(L) we can dene Lu:=f: This denes a linear operator L:D(L)L2( )!L2( ). For this operator the equation Lu=fholds if and only if uis a weak solution of this equation in the above sense. We obtain an operator LD\adapted" to the homogeneous Dirichlet problem if we restrict this operator to the set H1( )\D(L) : LD=Lj H1( )\D(L): This operator has the following property: For f2L2( ) the equation LDu=f holds if and only if uis a solution of X jj1 jj1D(aDu) =f; uj@ = 0 116 in the weak sense. If in particular a2C1( ) for alljj;jj1, then for u2 C1( ) the expressionP jj;jj1D(aDu) can be computed in the classical sense. By partial integration it thus follows for u;'2 C1( ) that B(u;') =X jj;jj1D(aDu);' : Sincef= D(aDu)2L2( ) and since C1( ) H1( )H1( ), it follows by denition of LandLDthat Lu=LDu=X jj;jj1D(aD)u foru2 C1( ). Consequently, on C1( ) the operators LandLDcoincide with the classical dierential operatorP jj;jj1D(aD), both are extensions of this operator. LDis called the Friedrichs' extension ofP jj;jj1D(aD) on  C1( ). Corollary 8.8 implies that for  <c2andf2L2( ) there is a unique solutionuof LDuu=f; and this solution satises kuk kuk1; max2 ;1 c2 kfk : This means that the inverse operator (LD)1:L2( )!D(L)L2( ) exists, and that this operator satises for all f2L2( ) kuk =k(LD)1fk max2 ;1 c2 kfk : Therefore this operator is bounded. Consequently (1;c2)(LD): Thus, the spectrum ( LD) belongs to the complement of ( 1;c2) inC. In the following the spectrum will be determined precisely. 117 9.2 Existence of eigenvalues in bounded domains The results of this section are based on the following fundamental result: Theorem 9.1 (Rellich selection theorem.) Let Rnbe a bounded open subset. Every bounded sequence in H1( )has a subsequence, which converges in the norm of L2( ). We omit the proof. It can be found for example in the books of Alt and Leis. In this section we always assume that a: !Care bounded measurable with a(x) = (1)j+ja(x); ifjj+jj1; a(x) =a2R; ifjj=jj= 1; and that Lu(x) =X jj;jj1D a(x)Du(x)
; is a uniformly elliptic operator with ellipticity constant . Let2Cbe an eigenvalue and u2D(LD) be an eigenfunction of LD, hence (LD)u= 0: By denition this holds if and only if u2 H1( ) satises B(u;') +(u;') = 0 for all'2 C1( ). This is equivalent to B(u;v) +(u;v) = 0 for allv2 H1( ). Lemma 9.2 Every eigenvalue of LDis real. Eigenfunctions u1andu2to distinct eigenvalues are orthogonal: (u1;u2) =B(u1;u2) = 0: Proof: Letu2 H1( ) be an eigenfunction to the eigenvalue . Then B(u;u) +(u;u) = 0: The symmetry of BimpliesB(u;u)2R, hence =B(u;u) kuk2 2R: 118 Ifu1andu2are eigenfunctions to the distinct eigenvalues 1;2, then 1(u1;u2) =B(u1;u2) =B(u2;u1) =2(u2;u1) =2(u1;u2) ; hence (12)(u1;u2) = 0. Since 126= 0, this yields ( u1;u2) = 0, whence B(u1;u2) =1(u1;u2) = 0: The proof is complete. LetMbe a nite dimensional linear subspace of H1( ) spanned by eigenfunc- tions ofLD. We denote by M?the linear space of all functions in H1( ), which are orthogonal to Mwith respect to the scalar product ( u;v). Since j(u;v)jkuk kvk kuk1kvk1, the scalar product ( u;v) is continuous with respect to the norm kuk1, whenceM?is closed. We allow M=;, in which caseM?= H1( ). Theorem 9.3 Ifu2M?withkuk = 1 exists such that B(u;u) = min v2M? kvk =1(B(v;v)); then=B(u;u)is an eigenvalue of LDanduis an eigenfunction to this eigenvalue. Proof: Note thatkv kvk k = 1 for allv6= 0, hence min v2M? kvk =1(B(v;v)) = min v2M? Bv kvk ;v kvk
 = min v2M? B(v;v) kvk2
: It follows thatB(u;u) kuk2 =B(u;u) B(v;v) kvk2 for allv, hence7! B(u+v;u+v) ku+vk2 :R!Rhas the minimum at = 0 for allv2M?. Thus 0 =d dB(u+v;u +v) ku+vk2 j=0 =d dB(u;u) +2 ReB(u;v) +2B(v;v) kuk2 +2 Re (u;v) +2kvk2 j=0 =2 ReB(u;v)kuk2 B(u;u)2 Re (u;v) kuk2 = 2 Re (B(u;v) +(u;v)): SinceM?is a linear space it follows that iv2M?ifv2M?. Thus Im B(u;v) +(u;v)
= Re (iB(u;v)i(u;v)
= Re B(u;iv) +(u;iv)
= 0; 119 hence B(u;v) +(u;v) = 0 (9.1) for allv2M?. Letwbe one of the nitely many eigenfunctions which span M, and letbe the eigenvalue to w. Then B(w;v) +(w;v) = 0 for allv2 H1( ). Since u2M?we have (w;u) = 0, thusB(w;u) = 0, whenceB(u;v) = (u;v) = 0 for all v2M. Together with (9.1) this implies B(u;v) +(u;v) = 0 for all v2 H1( ), whence is an eigenvalue and uis an eigenfunction. Theorem 9.4 Let Rnbe a bounded open set. The function B(v;v) assumes a minimum on the set fv2M?jkvk = 1g, which is not smaller thanc2, wherec20is the constant from the coercivity estimate proved in Theorem 8.4. (The minimum is not unique.) Proof: It has been shown in Theorem 8.4 that Bis coercive. This implies for allu2H1( ) withkuk = 1 that B(u;u) 2juj2 1; c2kuk2 c2; consequently the inmum = inf v2M? kvk =1 B(v;v)
c2 exists, and we can select a sequence fukgkfv2M?jkvk = 1gsatisfying lim k!1B(uk;uk) =: The coercivity implies  2jukj2 1; B(uk;uk) +c2!+c2; which yields kukk1= kukk2 +jukj2 1;
1=2c with a suitable constant C. Hencefukgkis bounded in H1( ). In general, the sequencefukgkdoes not converge. However, we can select a convergent subsequence: Let u2=B(u;u). SinceB(u;v) is a sesquilinear form, the parallelogram equality holds: u+v2+uv2= 2u2+ 2v2: 120 Thus u`uk2= 2u`2+ 2uk2u`+uk2(9.2) =u`2+ 2uk2ku`+ukk2 u`+uk ku`+ukk 2 2u`2+ 2uk22ku`+ukk2 : Here we used thatu`+uk ku`+ukk 2fv2M?jkvk = 1g;whenceu`+uk ku`+ukk . Sincefukgkis bounded in H1( ) and since is bounded, there is a sub- sequencefuksgsconverging in L2( ), by the Rellich selection theorem. Let u2L2( ) be the limit function. Denoting the subsequence by fu0 kgk, for simplicity, we obtain from the continuity of the norm that kuk = 1 and ku0 k+u0 `k !k2uk = 2, fork;`!1 . The inequality (9.2) together with the coercivity of Bthus yields for k;`!1 that  2ju0 `u0 kj2 1; u0 `u0 k2+c2ku0 `u0 kk2 !22+ 2242= 0: Consequentlyfu0 kgkconverges in H1( ) with limit function u, since the limits inL2( ) and H1( ) coincide. From the continuity of Bon H1( ) H1( ) we thus conclude B(u;u) = lim k!1B(u0 k;u0 k) == inf v2M? kvk =1B(v;v): From the closedness of M?we conclude that u2fv2M?jkvk = 1g, hence uis a minimum ofB(v;v) on this set. Corollary 9.5 Let Rnbe a bounded open set and let M H1( ) be a nite dimensional linear space spanned by eigenfunctions of LD, or letM=;. Then there is an eigenvalue ofLDand an eigenfunction u2M?to, which satisfykuk = 1 and =B(u;u) = min v2M? kvk =1B(v;v)c2: Proof: Combination of the preceding two theorems. 9.3 Spectral theorem and resolvent set Also in this section we assume that Rnis a bounded open set and that Lu(x) =X jj;jj1D a(x)Du(x)
121 is a uniformly elliptic operator with bounded, measurable coecient functions a: !Csatisfying a(x) = (1)j+ja(x);j+j1 a(x) =a(x)2R;jj=jj= 1: Theorem 9.6 (Spectral theorem for LD)There is a countably innite sequencefmgmRof eigenvalues ofLDsatisfying c212:::m:::!1; m!1; where the eigenvalues are repeated according to multiplicity. Moreover, there is a sequencefumgm H1( ) of corresponding eigenfunctions, which form a complete orthonormal system in L2( ). Proof: We construct the sequences fmgmandfumgmby induction: If 1;:::;mandu1;:::;umare already constructed, let Mmbe the space spanned byu1;:::;um. Denem+1andum+12M? mto be the eigenvalue and eigen- function ofLDsatisfyingkum+1k = 1 and m+1=B(um+1;um+1) = min v2M? m kvk =1B(v;v); which exist according to Corollary 9.5. The corollary also yields mc2. SinceM? m+1M? mit follows that m+1= min v2M? m kvk =1B(v;v)min v2M? m+1 kvk =1B(v;v) =m+2: Moreover,m!1 form!1 . Otherwise there would exist C > 0 with mCfor allm. The coercivity of Byields  2jumj2 1;  B(um;um) +c2kumk2 =m(um;um) +c2 =m+c2C+c2; whencefumgmis bounded in H1( ). By the Rellich selection theorem we could select a subsequence converging in L2( ). However, such a subsequence does not exist, since ( u`;um) = 0 implies ku`umk2 =ku`k2 +kumk2 = 2; whenever`6=m. Therefore m!1 . By construction, fumgmis an orthonormal system in L2( ). If it is not complete there is f2L2( ) dierent from zero such that (um;f) = 0 122 for allm. In Theorem 8.4 we proved that the sesquilinear from B(u;v)+(c2+1)(u;v) is strictly coercive, which implies that there is w2 H1( ),w6= 0, such that B(v;w) + (c2+ 1)(v;w) = (v;f) for allv2 H1( ). For the eigenfunctions umwe thus obtain (m+c2+ 1)(um;w) =B(um;w) + (c2+ 1)(um;w) = (um;f) = 0: Becausem+c2+ 11, it follows that ( um;w) = 0 for all m, thusw2M? k for allk. Settingw0=w kwk , we obtain for all k k+1= min v2M? k kvk =1B(v;v)B(w0;w0): This contradicts k!1 fork!1 . Consequently the orthonormal system fumgmis complete in L2( ). Corollary 9.7 Letfmgmbe the eigenvalues constructed in the preceding the- orem and letfumgmbe the complete orthonormal system of eigenfunctions. (a)u2L2( )belongs to H1( )if and only ifP1 m=1jmjj(u;um)j2<1. In this case lim `!1ku`X m=1(u;um) umk1; = 0; 1X m=1mj(u;um) j2B(u;u): (b)u2L2( )belongs toD(LD)if and only ifP1 m=1jmj2j(u;um)j2<1. In this case LDu=1X m=1m(u;um)um: Proof: (a) Foru2L2( ) andk`set uk`=`X m=k(u;um)um2 H1( ): 123 Choose <c2. ThenB(v;w)(v;w) is a strictly coercive sesquilinear form on H1( ), hence ckuk;`k2 1;  B(uk`;uk`)(uk`;uk`) =`X m;s=k(u;um)(u;us) B(um;us)(um;us)
=`X m;s=k(u;um)(u;us) (m) (um;us) =`X m=k(m)j(u;um)j2: From this inequality we conclude that ifP1 m=1jmjj(u;um)j2<1, then fP` m=1(u;um)umg`converges in H1( ). Since the limit in H1( ) coincides with the limit uinL2( ), we obtain u2 H1( ). On the other hand, if u2 H1( ) we compute similarly 0 B(uu1`;uu1`)(uu1`;uu1`) =B(u;u)(u;u)kX m=1(m)j(u;um)j2; hence 1X m=1mj(u;um)j2B(u;u) + (u;u)1X m=1j(u;um)j2 =B(u;u): (b) By denition, u2D(LD) if and only if u2 H1( ) and there is f2L2( ) such that B(v;u) = (v;f); (9.3) for allv2 H1( ). In this case we have LDu=f. SinceP1 m=1(v;um)um converges to v2 H1( ) in H1( ), we obtain from the continuity of Bthat B(v;u) =1X m=1(v;um)B(um;u) =1X m=1(v;um)m(um;u) and (v;f) =1X m=1(v;um) (um;f): Thus (9.3) holds if and only if 1X m=1 m(um;u)(um;f)
(v;um) = 0 124 for allv2 H1( ). Setting v=ukshows that (9.3) holds if and only if k(uk;u) = (uk;f);for allk2N: (9.4) Thus,u2D(LD) if and only if u2 H1( ) and there is f2L2( ) satisfying (9.4). Ifu2D(LD) we conclude from (9.4) that 1X m=1jmj2j(u;um)j2=1X m=1j(f;um)j2<1 and LDu=f=1X m=1(f;um)um=1X m=1m(u;um)um: On the other hand, ifP1 m=1jmj2j(u;um)j2<1we conclude from the above thatu2 H1( ). Dene a function f2L2( ) byf=P1 m=1m(u;um)um. Since this function satises (9.4), we infer that u2D(LD). Corollary 9.8 To every2Cnfmgmand everyf2L2( ;C)there is a unique solution uof LDuu=f given by u=1X m=1(f;um) mum: Consequently (LD) =Cnfmgm,(LD) =fmgm. Remark 9.9 This result means, of course, that the Dirichlet problem X jj;jj1D(aDu) +u=f uj@ = 0 has a unique weak solution for all 6=mand allf2L2( ). Proof: Fromu=P1 m=1(f;um) mumit follows (u;um) =(f;um) m; hence 1X m=1jmj2j(u;um)j2=1X m=1m m2 j(f;um)j2C1X m=1j(f;um)j2<1: 125 Corollary 9.7 thus shows that u2D(LD) and LDuu=1X m=1(m)(u;um)um=1X m=1(f;um)um=f: The solution is unique since LDis injective. For, 0 =LDvv=1X m=1(m)(v;um)um yields together with m6= 0 that (v;um) = 0 for all m, hencev= 0. 126 10 Linear hyperbolic equations of second order 10.1 Hyperbolic dierential operators The wave equation@2 @t2u(x;t) =c
xu(x;t) is a hyperbolic equation. We now show that the spectral theorem from Section 9 can be used to prove existence of solutions for the wave equation and other hyperbolic equations. LetLu(x) =P jj1 jj1D a(x)Du(x)
be a linear dierential operator of second order. In the remainder I always assume that the coecients of the principal part L0u(x) =X jj=1 jj=1a(x)D+u(x) are real valued functions: a(x)2R;forjj=jj= 1: One uses the set of zeros of the \principal symbol" p(x;) =X jj=jj=1a(x)+; 2Rn; of the dierential operator Lto classify the operator. An operator, whose set of zeros only consists of 0 2Rnis elliptic. A subsetMofRnis called conic with vertex at 0 if 2Mimplies2M for all0. Sincep(x;) is homogeneous of order 2 with respect to , it follows that if is a zero of p(x;), thenis a zero for all 2R, hence the set of zeros ofp(x;) is a conic subset of Rnsymmetric with respect to the vertex 0. The operator Lis called hyperbolic if the set of zeros of the principal symbol pis a double cone. This is made precise in the following Denition 10.1 The operator Lis hyperbolic at x2Rn, if there is a vector 6= 0 such that every line in Rnparallel to, not passing through the origin, intersects the set fjp(x;) = 0gin precisely two distinct points. Example 10.2 LetP jj;jj1D x a(x)D x
,x= (x1;:::;xn)2Rn, be an elliptic operator satisfying X jj=jj=1a(x)+>0; 2Rn; 6= 0: Then Lu(x;t) =@2 @t2u(x;t)X jj1 jj1D x a(x)D xu(x;t)
127 is a hyperbolic operator. To see this note that with x;2Rn,t;2Rthe principal symbol is p(x;t;; ) =2X jj=1 jj=1a(x)+: Set= (0;:::; 0;1)2Rn+1. Every line in Rn+1parallel toand not passing through the origin is of the form 7!(;) with2Rn,6= 0. For such the equation p(x;t;; ) =2X jj=jj=1a(x)+= 0 has the two distinct solutions =qP jj=jj=1a(x)+:In particular, the d'Alembert operator @2 tc
x=@2 @t2cnX i=1@2 @x2 i is hyperbolic for every constant c>0. Therefore the wave equation @2 @t2u(x;t) =c
xu(x;t) is a hyperbolic equation. 10.2 Energy estimate for the wave equation, uniqueness of solutions Let Rnbe an open set, let f: [0;1)!Cbe a bounded continuous function. Theorem 10.3 Letu:C2( [0;1);C)\C( [0;1);C)be a solution of the initial-boundary value problem @2 @t2u(x;t) =c
xu(x;t) +f(x;t); (x;t)2 (0;1) u(x;t) = 0; (x;t)2@ [0;1) u(x;0) =u(0)(x); ut(x;0) =u(1)(x); x2 : with a constant c>0. Thenusatises the energy estimate E(u;t)1=2E(u;0)1=2+Zt 0kf(t)k dt; where the energy E(u;t)is dened by E(u;t) =Z 1 2jut(x;t)j2+c 2jrxu(x;t)j2dx: 128 Proof: Sinceuis two times continuously dierentiable we have d dtE(u;t) =d dtZ 1 2jut(x;t)j2+c 2jrxu(x;t)j2dx =Z Re utt(x;t)ut(x;t) +crxu(x;t)
rxut(x;t) dx =Z Re utt(x;t)c
xu(x;t)
ut(x;t) dx = ReZ f(x;t)ut(x;t)dxkf(t)k kut(t)k 2kf(t)k E(u;t)1=2: Nowd dtE(u;t) =d dt E(u;t)1=2
2= 2E(u;t)1=2d dtE(u;t)1=2: Combination of these relations yields d dtE(u;t)1=2kf(t)k : Integration yields the stated estimate. Corollary 10.4 The initial-boundary value problem @2 @t2u(x;t) =c
xu(x;t) +f(x;t) u(x;t) = 0;(x;t)2@ [0;1) u(x;0) =u(0)(x); ut(x;0) =u(1)(x); x2 has at most one solution u2C2( [0;1);C)\C( [0;1);C). Proof: Letuandvbe two solutions. Then the dierence w=uvsatises @2 @t2w(x;t) =c
xw(x;t) w(x;t) = 0; (x;t)2@ [0;1) w(x;0) =wt(x;0) = 0; x2 : Form the energy estimate it thus follows Z 1 2jwt(x;t)j2+c 2jrxw(x;t)j2dxE(w;0) = 0: Consequently wt(x;t) = 0 for all ( x;t)2 [0;1) andw(x;0) = 0 for all x2 , whence w(x;t) =Zt 0wt(x;)d= 0; for all (x;t)2 [0;1). Thus,u=v. 129 10.3 Existence of weak solutions of initial-boundary value problems to hyperbolic equations In the following I consider hyperbolic dierential operators of the form@2 @t2L, where L=X jj1 jj1D x a(x)D x
is a uniformly strongly elliptic dierential operator with bounded measurable coecient functions a: !Csatisfying a(x) = (1)j+ja(x);j+j1 a(x) =a(x)2R;jj=jj= 1: Rnis a bounded open set, for T >0 ZT= (0;T): denotes a cylindric subset of Rn+1, and foru:ZT!Cand 0< t < T the functionu(t) : !Cis dened by u(t)
(x) =u(x;t): The goal of this section is to show that the initial-boundary value problem @2 @t2u(x;t) =Lu(x;t) +f(x;t); (x;t)2ZT u(x;t) = 0; (x;t)2@ (0;1) u(x;0) =u(0)(x); ut(x;0) =u(1)(x); x2 ; has a weak solution. In order to give the denition of weak solutions inhomo- geneous Sobolev spaces must be introduced: Denition 10.5 ForT >0,m2Nlet H(t) m(ZT;C) =fu2L2(ZT;C)@k @tku2L2(ZT;C);kmg: H(t) m(ZT) is a Hilbert space with the scalar product (u;v)(t) m=mX k=0@k @tku;@k @tkv ZT and the normkuk(t) m= (u;u)(t) m
1=2. 130 Theorem 10.6 (Sobolev embedding theorem.) LetT >0and0T. Then there is a unique continuous linear mapping B:H(t) 1(ZT;C)!L2( ;C) satisfying (Bu)(x) =u(x;); for allu2C [0;T]
\H(t) 1(ZT;C). A proof can be found in the book of Alt, p.249. Denition 10.7 The function Buis called the trace of the mapping u2 H(t) 1(ZT) on fgand is denoted by uj fg. As in Section 9 let LD:D(LD)L2( )!L2( ) denote the Friedrichs' extension of the operatorP jj;jj1D(aD) in . Denition 10.8 LetT > 0,f2L2(ZT;C),u(0)2D(LD),u(1)2 H1( ;C). A functionu:ZT!Cis a weak solution of the Dirichlet initial-boundary value problem @2 tu(x;t) =Lu(x;t) +f(x;t); (x;t)2ZT; u(x;t) = 0; (x;t)2@ (0;T); u(x;0) =u(0)(x); ut(x;0) =u(1)(x); x2 :(10.1) if 1.u2H(t) 2(ZT), 2.u(t)2D(LD), for almost all t2(0;T), 3.@2 tu(t) =LDu(t) +f(t), for almost all t2(0;T), 4.uj f0g=u(0); utj f0g=u(1). Theorem 10.9 Let Rnbe an open bounded set. To every f2H1(ZT;C), u(0)2D(LD)andu(1)2 H1( ;C)there is a weak solution of the Dirichlet initial-boundary value problem (10.1) . This solution is given by u(x;t) =1X m=1m(t)um(x); wherefumg1 m=1is the complete orthonormal system of eigenfunctions of the operatorLD, and where m: [0;1)!Cis the solution of the initial value problem @2 @t2m(t) +mm(t) = f(t);um
m(0) = u(0);um
@ @tm(0) = u(1);um
: 131 Heremis the eigenvalue to um. Clearly, this implies for m>0 m(t) = cos(p mt)(u(0);um) +1pmsin(p mt)(u(1);um) +Zt 01pmsinp m(t)
f();um
d: Form= 0 we obtain m(t) = (u(0);um) + (u(1);um) +Zt 0(t) f();um
d; and form<0 m(t) = cosh(p mt)(u(0);um) +1pmsinh(p mt)(u(1);um) +Zt 01pmsinhp m(t)
f();um
d: Sincec21:::m:::!1 form!1 there are only nitely many eigenvalues m0. Proof: From the explicit expression for mgiven above we obtain for s= 0;1;2 j@s tm(t)j C(t)p 1 +jmjs j(u(0);um) j +1p 1 +jmjj(u(1);um) j+1p 1 +jmjZt 0j f();um
jd +s2j f(t);um
j; (10.2) where s2=( 0; s6= 2 1; s= 2; and C(t) =8 >< >:^Cep1t;if1<0 ^C(1 +t);if1= 0 ^C; if1>0; with a suitable constant ^C. Using the Cauchy-Schwarz inequality, which yields ja+b+c+dj24(a2+b2+c2+d2) 132 andZt 0j(f();um)jd2 tZt 0j(f();um)j2d; we obtain from (10.2) j@s tm(t)j24C(t)2(1 +jmj)s j(u(0);um) j2 +1 1 +jmjj(u(1);um) j2+t 1 +jmjZt 0j f();um
j2d + 4s2j f(t);um
j2: (10.3) The assumptions u(0)2D(LD) andu(1)2 H1( ) imply 1X m=1 1 +jmj2
j(u(0);um) j2<1; 1X m=1 1 +jmj
j(u(1);um) j2<1; cf. Corollary 9.7. Moreover, from f2H1(ZT) it follows by the theorem of Fubini that f(t)2H1( ) for almost all t>0. Thus, Corollary 9.7 (a) yields 1X m=1(1 +jmj)Zt 0j f();um
j2d Zt 0K1 B f();f()
+kf()k2  d K2Zt 0kf()k2 1; dKkfk2 1;Zt<1; with suitable constants K1;K2>0. From these estimates and from (10.3) we obtain 1X m=1jmj2jm(t)j2C1C(t)2(1 +t); (10.4) 1X m=1j@s tm(t)j2C1C(t)2(1 +t) + 4s21X m=1j f(t);um
j2 =C1C(t)2(1 +t) + 4s2kf(t)k2 : Thus, fors= 0;1;2 ZT 01X m=1j@s tm(t)j2dtZT 0C1C(t)2(1 +t) + 4s2kf(t)k2 dt C1C(T)2(1 +T)T+ 4s2kfk2 ZT; (10.5) 133 where we used that C(t) is an increasing function. Since u`=P` m=1mum satises for k` k@s t(u`uk)k2 ZT= (@s t(u`uk);@s t(u`uk))ZT=`X m;j=k(@s tmum;@s tjuj)ZT =`X m;j=kZT 0@s tm@s tjdt(um;uj) =`X m=kZT 0j@s tm(t)j2dt; we obtain from (10.5) that f@s tu`g`is a Cauchy sequence in L2(ZT) with lim `!1@s `u`=1X m=1@s tmum; fors= 0;1;2 . This means that fu`g`converges in the space H(t) 2(ZT;C) and that the limit function u2H(t) 2(ZT;C) satises @s tu=1X m=1@s tmum: (10.6) Also, since u(t) =P1 m=1m(t)umimplies u(t);um
=m(t), we infer from (10.4) that 1X m=1jmj2j u(t);um
j2<1; whenceu(t)2D(LD) for allt0 and LDu(t) =1X m=1m u(t);um
um; (10.7) by Corollary 9.7. Summing up, we conclude from (10.6) and (10.7) that @2 tu(t) =@2 t1X m=1m(t)um=1X m=1@2 tm(t)um =1X m=1 mm(t) + f(t);um
 um =1X m=1m u(t);um
um+1X m=1 f(t);um
um =LDu(t) +f(t): 134 Thus, the rst three conditions of Denition 10.8 are satised. To verify the last condition note that m(0) = u(0);um
and@ @tm(0) = (u(1);um) yield uj f0g=1X m=1m(0)um=1X m=1(u(0);um) um=u(0); @tuj f0g=1X m=1@t(0)um=1X m=1(u(1);um) um=u(1): This completes the proof. 135 A Appendix: Bessel and Neumann functions Letmbe a nonnegative integer. The Neumann function of order mis Nm(x) =(1 2x)m m1X k=0(mk1)! k!(1 4x2)k +2 ln(1 2x)Jm(x) (1 2x)m 1X k=0[ (k+ 1) + (m+k+ 1)](1 4x2)k k!(m+k)!;(A.1) whereJmis the Bessel function and where the {function is dened by (1) = ; (m) = +m1X k=1k1; m2: Here = lim m!1 1 +1 2+1 3+:::+1 mln(m) = 0:5772156649 ::: denotes the Euler constant. In particular, for m= 0 we obtain N0(x) =2  ln(1 2x) +
J0(x) +2 1 4x2 (1!)2(1 +1 2)(1 4x2)2 (2!)2+ (1 +1 2+1 3)(1 4x2)3 (3!)2::: :(A.2) To determine the asymptotic behavior of Nmat 0 we use the representation Jm(x) =1X k=0(1)k k!(m+k)!x 22k+m of the Bessel function, which implies J0(0) = 1; Jm(x) =O(xm); x!0; m1: From this relation and from (A.1) and (A.2) we thus obtain for x!0 that N0(x) =2 ln(x) +O(1); N1(x) =2 x1+O(xln(x)); Nm(x) =2m(m1)! xm+O(xm+2); m2: The following lemma is needed in the proof of Theorem 5.4. 136 Lemma A.1 Letm2N. Then we have for Rr0that jJm(ir)j jJm(iR)jeRR r(m2 s2+1)1=2dsr Rm : Proof: Forr= 0 the statement holds since Jm(0) = 0. To prove the statement forr>0 setum(r) =Jm(ir). This function satises u00 m(r) +1 ru0 m(r) + (1m2 r2)um(r) = 0 for positive r. Sinceru00 m(r) +u0 m(r) = (ru0 m(r))0, this yields (ru0 m(r))0(1 +m2 r2)rum(r) = 0: (A.3) For any complex valued function vwe have (1 2jvj2)0= (1 2vv)0= Rev0v. Multi- plication of (A.3) with1 2rum0(r) therefore results in 1 2jru0 m(r)j20 (1 +m2 r2)1 2jrum(r)j20 = 0; hence d dr1 2jru0 m(r)j2(1 +m2 r2)1 2jrum(r)j2 =m2 r3jrum(r)j20: (A.4) Form2Nwe haveum(0) = 0 and ru0 m(r)jr=0= 0. Using this, we obtain by integration of (A.4) over the interval [0 ;r] that jru0 m(r)j2(1 +m2 r2)jrum(r)j20: Sinceumdoes not have a zero on the interval (0 ;1), as we showed in Theo- rem 4.6, this implies that ju0 m(r)j jum(r)jr m2 r2+ 1: The series representation of Jmshows thatu0 m(r) um(r)=J0 m(ir) Jm(ir)is real for real r. This impliesu0 m(r) um(r)=u0 m(r) um(r)andjum(r)j0 jum(r)j= Reu0 m(r) um(r)=u0 m(r) um(r). From the last inequality we therefore conclude that jum(r)j0 jum(r)j=u0 m(r) um(r)r m2 r2+ 1 holds for all r>0 or that jum(r)j0 jum(r)j=u0 m(r) um(r)r m2 r2+ 1 137 holds for all r > 0. The second inequality cannot be true, since it implies jum(r)j0<0, which in view of um(0) = 0 is impossible. The rst inequality is equivalent to (lnjum(r)j)0r m2 r2+ 1: Integration of this inequality over the interval [ r;R] yields jum(r)j jum(R)jeRR r(m2 s2+1)1=2ds: The lemma follows from eRR r(m2 s2+1)1=2dseRR rm sds=emln(R=r)=r R
m. 138