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Chapter 21 TheCalculusofVariations We have already had ample opportunity to exploit Nature’s pr opensity to minimize. Minimization principles form one of the most wide-ranging m eans of formulating mathe- matical models governing the equilibrium configurations of physical systems. Moreover, many popular numerical integration schemes such as the powe rful finite element method are also founded upon a minimization paradigm. In this chapt er, we will develop the basic mathematicalanalysisofnonlinearminimizationprincipl esoninfinite-dimensionalfunction spaces — a subject known as the “calculus of variations”, for reasons that will be explained as soon as we present the basic ideas. Classical solutions to minimization problems in the calculus of variations are prescribed by boundary value pro blems involving certain types of differential equations, known as the associated Euler–La grange equations. The math- ematical techniques that have been developed to handle such optimization problems are fundamental in many areas of mathematics, physics, enginee ring, and other applications. In this chapter, we will only have room to scratch the surface of this wide ranging and lively area of both classical and contemporary research. The history of the calculus of variations is tightly interwo ven with the history of mathematics. The field has drawn the attention of a remarkabl e range of mathematical luminaries, beginning with Newton, then initiated as a subj ect in its own right by the Bernoulli family. The first major developments appeared in t he work of Euler, Lagrange and Laplace. In the nineteenth century, Hamilton, Dirichle t and Hilbert are but a few of the outstanding contributors. In modern times, the calculu s of variations has continued to occupy center stage, witnessing major theoretical advan ces, along with wide-ranging applications in physics, engineering and all branches of ma thematics. Minimizationproblemsthatcanbeanalyzedbythecalculuso fvariationsservetochar- acterize the equilibrium configurations of almost all conti nuous physical systems, ranging through elasticity, solid and fluid mechanics, electro-mag netism, gravitation, quantum me- chanics, string theory, and many, many others. Many geometr ical configurations, such as minimal surfaces, can be conveniently formulated as optimi zationproblems. Moreover, nu- merical approximations to the equilibrium solutions of suc h boundary value problems are based on a nonlinear finite element approach that reduced the infinite-dimensional mini- mization problem to a finite-dimensional problem, to which w e can apply the optimization techniques learned in Section 19.3; however, we will not pur sue this direction here. Wehave, infact, alreadytreatedthesimplestproblemsinth ecalculusofvariations. As we learned in Chapters 11 and 15, minimization of a quadratic functional requires solving an associated linear boundary value problem. Just as the van ishing of the gradient of a function of several variables singles out the critical poin ts, among which are the minima, 12/11/12 1143 c/ci∇cleco√y∇t2012 Peter J. Olver Figure 21.1. The Shortest Path is a Straight Line. both local and global, so a similar “functional gradient” wi ll distinguish the candidate functions that might be minimizers of the functional. The fin ite-dimensional calculus leads to a system of algebraic equations for the critical poi nts; the infinite-dimensional functional analog results a boundary value problem for a non linear ordinary or partial differential equation whose solutions are the critical func tions for the variational problem. So, the passage from finite to infinite dimensional nonlinear systems mirrors the transition from linear algebraic systems to boundary value problems. 21.1. Examples of Variational Problems. The best way to appreciate the calculus of variations is by in troducing a few concrete examples of both mathematical and practical importance. So me of these minimization problems played a key role in the historical development of t he subject. And they still serve as an excellent means of learning its basic constructi ons. Minimal Curves, Optics, and Geodesics Theminimal curve problem istofindtheshortestpathbetweentwospecifiedlocations. In its simplest manifestation, we are given two distinct poi nts a= (a,α) and b= (b,β) in the plane R2, (21 .1) and our task is to find the curve of shortest length connecting them. “Obviously”, as you learn in childhood, the shortest route between two points is a straight line; see Figure 21.1. Mathematically, then, the minimizing curve should be the gr aph of the particular affine function† y=cx+d=β−α b−a(x−a)+α (21.2) that passes through or interpolates the two points. However , this commonly accepted “fact” — that (21.2) is the solution to the minimization prob lem — is, upon closer inspec- tion, perhaps not so immediately obvious from a rigorous mat hematical standpoint. †We assume that a/negationslash=b, i.e., the points a,bdo not lie on a common vertical line. 12/11/12 1144 c/ci∇cleco√y∇t2012 Peter J. Olver Let us see how we might formulate the minimal curve problem in a mathematically precise way. For simplicity, we assume that the minimal curv e is given as the graph of a smooth function y=u(x). Then, according to (A.27), the length of the curve is given by the standard arc length integral J[u] =/integraldisplayb a/radicalbig 1+u′(x)2dx, (21.3) where we abbreviate u′=du/dx. The function u(x) is required to satisfy the boundary conditions u(a) =α, u (b) =β, (21.4) in order that its graph pass through the two prescribed point s (21.1). The minimal curve problem asks us to find the function y=u(x) that minimizes the arc length functional (21.3) among all “reasonable” functions satisfying the pre scribed boundary conditions. The reader might pause to meditate on whether it is analytica lly obvious that the affine function (21.2) is the one that minimizes the arc length inte gral (21.3) subject to the given boundary conditions. One of the motivating tasks of the calc ulus of variations, then, is to rigorously prove that our everyday intuition is indeed corr ect. Indeed, the word “reasonable” isimportant. For the arc length functional (21.3) to be defined, the function u(x) should be at least piecewise C1, i.e., continuous with a piecewise continuous derivative. Indeed, if we were to all ow discontinuous functions, then the straight line (21.2) does not, in most cases, give th e minimizer; see Exercise . Moreover, continuous functions which are not piecewise C1need not have a well-defined arclength. The moreseriously one thinksabout these issues , theless evident the“obvious” solution becomes. But before you get too worried, rest assur ed that the straight line (21.2) is indeed the true minimizer. However, a fully rigorous proo f of this fact requires a careful development of the mathematical machinery of the calculus o f variations. A closely related problem arises in geometrical optics. The underlying physical prin- ciple, first formulated by the seventeenth century French ma thematician Pierre de Fermat, is that, when a light ray moves through an optical medium, it t ravels along a path that minimizes the travel time. As always, Nature seeks the most e conomical†solution. In an inhomogeneous planar‡optical medium, the speed of light, c(x,y), varies from point to point, depending on the optical properties. Speed equals th e time derivative of distance traveled, namely, the arc length of the curve y=u(x) traced by the light ray. Thus, c(x,u(x)) =ds dt=/radicalbig 1+u′(x)2dx dt. Integrating from start to finish, we conclude that the total t ravel time along the curve is equal to T[u] =/integraldisplayT 0dt=/integraldisplayb adt dxdx=/integraldisplayb a/radicalbig 1+u′(x)2 c(x,u(x))dx. (21.5) †Assuming time = money! ‡For simplicity, we only treat the two-dimensional case in the text. S ee Exercise for the three-dimensional version. 12/11/12 1145 c/ci∇cleco√y∇t2012 Peter J. Olver a b Figure 21.2. Geodesics on a Cylinder. Fermat’s Principle states that, to get from point a= (a.α) to point b= (b,β), the light ray follows the curve y=u(x) that minimizes this functional subject to the boundary conditions u(a) =α, u (b) =β, If the medium is homogeneous, e.g., a vacuum†, thenc(x,y)≡cis constant, and T[u] is a multiple of the arc length functional (21.3), whose minimiz ers are the “obvious” straight lines traced by the light rays. In an inhomogeneous medium, t he path taken by the light ray is no longer evident, and we are in need of a systemat ic method for solving the minimization problem. Indeed, all of the known laws of geome tric optics, lens design, focusing, refraction, aberrations, etc., will be conseque nces of the geometric and analytic properties of solutions to Fermat’s minimization principl e, [22]. Another minimizationproblem ofa similarilkisto construc t thegeodesics ona curved surface, meaning the curves of minimal length. Given two poi ntsa,blying on a surface S⊂R3, we seek the curve C⊂Sthat joins them and has the minimal possible length. For example, if Sis a circular cylinder, then there are three possible types o f geodesic curves: straight line segments parallel to the center line; arcs of circles orthogonal to the center line; and spiral helices, the latter illustrated in F igure 21.2. Similarly, the geodesics on a sphere are arcs of great circles. In aeronautics, to mini mize distance flown, airplanes follow geodesic circumpolar paths around the globe. Howeve r, both of these claims are in need of mathematical justification. In order to mathematically formulate the geodesic minimiza tion problem, we suppose, for simplicity, that our surface S⊂R3is realized as the graph‡of a function z=F(x,y). We seek the geodesic curve C⊂Sthat joins the given points a= (a,α,F(a,α)),andb= (b,β,F(b,β)),lying on the surface S. †In the absence of gravitational effects due to general relativity. ‡Cylinders are not graphs, but can be placed within this framework by pas sing to cylindrical coordinates. Similarly, spherical surfaces are best treated in sph erical coordinates. In differential geometry, [ 60], one extends these constructions to arbitrary parametrized surfac es and higher dimensional manifolds. 12/11/12 1146 c/ci∇cleco√y∇t2012 Peter J. Olver Let us assume that Ccan be parametrized by the xcoordinate, in the form y=u(x), z =v(x) =F(x,u(x)), where the last equation ensures that it lies in the surface S. In particular, this requires a/ne}ationslash=b. The length of the curve is supplied by the standard three-di mensional arc length integral (B.18). Thus, to find the geodesics, we must minimiz e the functional J[u] =/integraldisplayb a/radicalBigg 1+/parenleftbiggdy dx/parenrightbigg2 +/parenleftbiggdz dx/parenrightbigg2 dx =/integraldisplayb a/radicalBigg 1+/parenleftbiggdu dx/parenrightbigg2 +/parenleftbigg∂F ∂x(x,u(x))+∂F ∂u(x,u(x))du dx/parenrightbigg2 dx,(21.6) subject to the boundary conditions u(a) =α, u(b) =β.For example, geodesics on the paraboloid z=1 2x2+1 2y2(21.7) can be found by minimizing the functional J[u] =/integraldisplayb a/radicalbig 1+(u′)2+(x+uu′)2dx. (21.8) Minimal Surfaces The minimal surface problem is a natural generalization of t he minimal curve or geodesicproblem. Initssimplest manifestation, wearegiv enasimpleclosedcurve C⊂R3. The problem is to find the surface of least total area among all those whose boundary is the curve C. Thus, we seek to minimize the surface area integral areaS=/integraldisplay/integraldisplay SdS over all possible surfaces S⊂R3with the prescribed boundary curve ∂S=C. Such an area–minimizing surface is known as a minimal surface for short. For example, if Cis a closed plane curve, e.g., a circle, then the minimal surface will just be the planar region it encloses. But, if the curve Ctwists into the third dimension, then the shape of the minimizing surface is by no means evident. Physically, if we bend a wire in the shape of the curve Cand then dip it into soapy water, the surface tension forces in the resulting soap film w ill cause it to minimize surface area, and hence be a minimal surface†. Soap films and bubbles have been the source of much fascination, physical, æsthetical and mathematical, over the centuries. The minimal surface problem is also known as Plateau’s Problem , named after the nineteenth century †More accurately, the soap film will realize a local but not necessarily gl obal minimum for the surface area functional. Nonuniqueness of local minimizers can b e realized in the physical experiment — the same wire may support more than one stable soap film. 12/11/12 1147 c/ci∇cleco√y∇t2012 Peter J. Olver ∂ΩC Figure 21.3. Minimal Surface. French physicist Joseph Plateauwho conducted systematice xperiments onsuch soapfilms. A satisfactory mathematical solution to even the simplest v ersion of the minimal surface problem was only achieved in the mid twentieth century, [ 135,140]. Minimal surfaces and related variational problems remain an active area of conte mporary research, and are of importance in engineering design, architecture, and biolo gy, including foams, domes, cell membranes, and so on. Let us mathematically formulate the search for a minimal sur face as a problem in the calculus of variations. For simplicity, we shall assume that the bounding curve C projects down to a simple closed curve Γ = ∂Ω that bounds an open domain Ω ⊂R2in the (x,y) plane, as in Figure 21.3. The space curve C⊂R3is then given by z=g(x,y) for (x,y)∈Γ =∂Ω. For “reasonable” boundary curves C, we expect that the minimal surface Swill be described as the graph of a function z=u(x,y) parametrized by ( x,y)∈Ω. According to the basic calculus formula (B.41), the surface area of such a graph is given by the double integral J[u] =/integraldisplay/integraldisplay Ω/radicalBigg 1+/parenleftbigg∂u ∂x/parenrightbigg2 +/parenleftbigg∂u ∂y/parenrightbigg2 dxdy. (21.9) To find the minimal surface, then, we seek the function z=u(x,y) that minimizes the surface area integral (21.9) when subject to the Dirichlet b oundary conditions u(x,y) =g(x,y) for ( x,y)∈∂Ω. (21.10) As we will see, (21.57), the solutions to this minimization p roblem satisfy a complicated nonlinear second order partial differential equation. A simple version of the minimal surface problem, that still c ontains some interesting features, is to find minimal surfaces with rotational symmet ry. Asurface of revolution is obtained by revolving a plane curve about an axis, which, for definiteness, we take to be thexaxis. Thus, given two points a= (a,α),b= (b,β)∈R2, the goal is to find the curve y=u(x) joining them such that the surface of revolution obtained b y revolving the curve around the x-axis has the least surface area. Each cross-section of the r esulting surface is a circle centered on the xaxis; see Figure srev . According to Exercise B.5.4, the area of 12/11/12 1148 c/ci∇cleco√y∇t2012 Peter J. Olver such a surface of revolution is given by J[u] =/integraldisplayb a2π|u|/radicalbig 1+(u′)2dx. (21.11) We seek a minimizer of this integral among all functions u(x) that satisfy the fixed bound- ary conditions u(a) =α, u(b) =β.The minimal surface of revolution can be physically realized by stretching a soap film between two circular wires , of respective radius αandβ, that are held a distance b−aapart. Symmetry considerations will require the minimizin g surface to be rotationally symmetric. Interestingly, the r evolutionary surface area func- tional (21.11) is exactly the same as the optical functional (21.5) when the light speed at a pointisinverselyproportionaltoitsdistancefromthehor izontalaxis: c(x,y) = 1/(2π|y|). Isoperimetric Problems and Constraints The simplest isoperimetric problem is to construct the simple closed plane curve of a fixed length ℓthatencloses thedomain oflargest area. Inother words, wes eek to maximize area Ω =/integraldisplay/integraldisplay Ωsubject to the constraint length ∂Ω =/contintegraldisplay ∂Ω=ℓ, over all possible domains Ω ⊂R2. Of course, the “obvious” solution to this problem is that the curve must be a circle whose perimeter is ℓ, whence the name “isoperimetric”. Note that the problem, as stated, does not have a unique solut ion, since if Ω is a maximing domain, any translated or rotated version of Ω will also maxi mize area subject to the length constraint. To makeprogress on theisoperimetric problem, let us assume that the boundary curve is parametrized by its arc length, so x(s) = (x(s),y(s))Twith 0≤s≤ℓ, subject to the requirement that/parenleftbiggdx ds/parenrightbigg2 +/parenleftbiggdy ds/parenrightbigg2 = 1. (21.12) According to (A.58), we can compute the area of the domain by a line integral around its boundary, area Ω =/contintegraldisplay ∂Ωxdy=/integraldisplayℓ 0xdy dsds, (21.13) and thus we seek to maximize the latter integral subject to th e arc length constraint (21.12). We also impose periodic boundary conditions x(0) =x(ℓ), y (0) =y(ℓ), (21.14) that guarantee that the curve x(s) closes up. (Technically, we should also make sure that x(s)/ne}ationslash=x(s′) for any 0 ≤s < s′< ℓ, ensuring that the curve does not cross itself.) A simpler isoperimetric problem, but one with a less evident solution, is the following. Among all curves of length ℓin the upper half plane that connect two points ( −a,0) and (a,0), find the one that, along with the interval [ −a,a], encloses the region having the largest area. Of course, we must take ℓ≥2a, as otherwise the curve will be too short 12/11/12 1149 c/ci∇cleco√y∇t2012 Peter J. Olver to connect the points. In this case, we assume the curve is rep resented by the graph of a non-negative function y=u(x), and we seek to maximize the functional /integraldisplaya −audx subject to the constraint/integraldisplaya −a/radicalbig 1+u′2dx=ℓ. (21.15) In the previous formulation (21.12), the arc length constra int was imposed at every point, whereas here it is manifested as an integral constraint. Bot h types of contraints, pointwise andintegral, appearinawiderangeofappliedandgeometric alproblems. Such constrained variational problems can profitably be viewed as function sp ace versions of constrained optimization problems. Thus, not surprisingly, their anal ytical solution will require the introduction of suitable Lagrange multipliers. 21.2. The Euler–Lagrange Equation. Even the preceding limited collection of examples of variat ional problems should al- ready convince the reader of the tremendous practical utili ty of the calculus of variations. Let us now discuss the most basic analytical techniques for s olving such minimization problems. We will exclusively deal with classical techniqu es, leaving more modern direct methods — the function space equivalent of gradient descent and related methods — to a more in–depth treatment of the subject, [ cvar]. Let us concentrate on thesimplest classof variationalprob lems, inwhich the unknown is a continuously differentiable scalar function, and the fu nctional to be minimized depends upon at most its first derivative. The basic minimization pro blem, then, is to determine a suitable function y=u(x)∈C1[a,b] that minimizes the objective functional J[u] =/integraldisplayb aL(x,u,u′)dx. (21.16) The integrand is known as the Lagrangian for the variational problem, in honor of La- grange, one of the main founders of the subject. We usually as sume that the Lagrangian L(x,u,p) is a reasonably smooth function of all three of its (scalar) arguments x,u, and p, which represents the derivative u′. For example, the arc length functional (21.3) has Lagrangian function L(x,u,p) =/radicalbig 1+p2, whereas in the surface of revolution problem (21.11), we have L(x,u,p) = 2π|u|/radicalbig 1+p2. (In the latter case, the points where u= 0 are slightly problematic, since Lis not continuously differentiable there.) In order to uniquely specify a minimizing function, we must i mpose suitable boundary conditions. All of the usual suspects — Dirichlet (fixed), Ne umann (free), as well as mixed and periodic boundary conditions — that arose in Chapter 11 a re also relevant here. In the interests of brevity, we shall concentrate on the Dirich let boundary conditions u(a) =α, u (b) =β, (21.17) although some of the exercises will investigate other types of boundary conditions. The First Variation According to Section 19.3, the (local) minimizers of a (suffic iently nice) objective function defined on a finite-dimensional vector space are ini tially characterized as critical 12/11/12 1150 c/ci∇cleco√y∇t2012 Peter J. Olver points, wheretheobjectivefunction’s gradientvanishes. Ananalogousconstructionapplies in the infinite-dimensional context treated by the calculus of variations. Every sufficiently nice minimizer of a sufficiently nice functional J[u] is a “critical function”, meaning that its functional gradient vanishes: ∇J[u] = 0. Indeed, the mathematical justification of this fact outlined in Section 19.3 continues to apply here; s ee, in particular, the proof of Theorem 19.40. Of course, not every critical point turns o ut to be a minimum — maxima, saddles, and many degenerate points are also critic al. The characterization of nondegenerate critical points as local minima or maxima rel ies on the second derivative test, whose functional version, known as the second variati on, will be is the topic of the following Section 21.3. But we are getting ahead of ourselves. The first order of busin ess is to learn how to compute the gradient of a functional defined on an infinite-di mensional function space. As noted in the general Definition 19.37 of the gradient, we must first impose an inner product on the underlying space. The gradient ∇J[u] of the functional (21.16) will be defined by the same basic directional derivative formula: /an}b∇acketle{t∇J[u];v/an}b∇acket∇i}ht=d dtJ[u+tv]/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0. (21.18) Herev(x) is a function that prescribes the “direction” in which the d erivative is computed. Classically, vis known as a variation in the function u, sometimes written v=δu, whence the term “calculus of variations”. Similarly, the gradient operator on functionals is often referred to as the variational derivative , and often written δJ. The inner product used in (21.18) is usually taken (again for simplicity) to be the sta ndard L2inner product /an}b∇acketle{tf;g/an}b∇acket∇i}ht=/integraldisplayb af(x)g(x)dx (21.19) on function space. Indeed, while the formula for the gradien t will depend upon the under- lying inner product, cf. Exercise , the characterization of critical points does not, and so the choice of inner product is not significant here. Now, starting with (21.16), for each fixed uandv, we must compute the derivative of the function h(t) =J[u+tv] =/integraldisplayb aL(x,u+tv,u′+tv′)dx. (21.20) Assuming sufficient smoothness of the integrand allows us to b ring the derivative inside the integral and so, by the chain rule, h′(t) =d dtJ[u+tv] =/integraldisplayb ad dtL(x,u+tv,u′+tv′)dx =/integraldisplayb a/bracketleftbigg v∂L ∂u(x,u+tv,u′+tv′)+v′∂L ∂p(x,u+tv,u′+tv′)/bracketrightbigg dx. Therefore, setting t= 0 in order to evaluate (21.18), we find /an}b∇acketle{t∇J[u];v/an}b∇acket∇i}ht=/integraldisplayb a/bracketleftbigg v∂L ∂u(x,u,u′)+v′∂L ∂p(x,u,u′)/bracketrightbigg dx. (21.21) 12/11/12 1151 c/ci∇cleco√y∇t2012 Peter J. Olver The resulting integral often referred to as the first variation of the functional J[u]. The condition /an}b∇acketle{t∇J[u];v/an}b∇acket∇i}ht= 0 for a minimizer is known as the weak form of the variational principle. To obtain an explicit formula for ∇J[u], the right hand side of (21.21) needs to be written as an inner product, /an}b∇acketle{t∇J[u];v/an}b∇acket∇i}ht=/integraldisplayb a∇J[u]v dx=/integraldisplayb ahvdx between some function h(x) =∇J[u] and the variation v. The first summand has this form, but the derivative v′appearing in the second summand is problematic. However, as the reader of Chapter 11 already knows, the secret to moving a round derivatives inside an integral is integration by parts. If we set r(x)≡∂L ∂p(x,u(x),u′(x)), we can rewrite the offending term as /integraldisplayb ar(x)v′(x)dx=/bracketleftbig r(b)v(b)−r(a)v(a)/bracketrightbig −/integraldisplayb ar′(x)v(x)dx, (21.22) where, again by the chain rule, r′(x) =d dx/parenleftbigg∂L ∂p(x,u,u′)/parenrightbigg =∂2L ∂x∂p(x,u,u′)+u′∂2L ∂u∂p(x,u,u′)+u′′∂2L ∂p2(x,u,u′). (21.23) So far we have not imposed any conditions on our variation v(x). We are only com- paring the values of J[u] among functions that satisfy the prescribed boundary cond itions, namely u(a) =α, u (b) =β. Therefore, we must make sure that the varied function /hatwideu(x) =u(x)+tv(x) remains within this set of functions, and so /hatwideu(a) =u(a)+tv(a) =α,/hatwideu(b) =u(b)+tv(b) =β. For this to hold, the variation v(x) must satisfy the corresponding homogeneous boundary conditions v(a) = 0, v (b) = 0. (21.24) As a result, both boundary terms in our integration by parts f ormula (21.22) vanish, and we can write (21.21) as /an}b∇acketle{t∇J[u];v/an}b∇acket∇i}ht=/integraldisplayb a∇J[u]v dx=/integraldisplayb av/bracketleftbigg∂L ∂u(x,u,u′)−d dx/parenleftbigg∂L ∂p(x,u,u′)/parenrightbigg/bracketrightbigg dx. 12/11/12 1152 c/ci∇cleco√y∇t2012 Peter J. Olver Since this holds for all variations v(x), we conclude that† ∇J[u] =∂L ∂u(x,u,u′)−d dx/parenleftbigg∂L ∂p(x,u,u′)/parenrightbigg . (21.25) This is our explicit formula for the functional gradient or v ariational derivative of the func- tional (21.16) with Lagrangian L(x,u,p). Observe that the gradient ∇J[u] of a functional is afunction. Thecritical functions u(x) are, by definition, those for which the functional gradient vanishes: satisfy ∇J[u] =∂L ∂u(x,u,u′)−d dx∂L ∂p(x,u,u′) = 0. (21.26) In view of (21.23), the critical equation (21.26) is, in fact , a second order ordinary differ- ential equation, E(x,u,u′,u′′) =∂L ∂u(x,u,u′)−∂2L ∂x∂p(x,u,u′)−u′∂2L ∂u∂p(x,u,u′)−u′′∂2L ∂p2(x,u,u′) = 0, (21.27) known as the Euler–Lagrange equation associated with the variational problem (21.16), in honor of two of the most important contributors to the subjec t. Any solution to the Euler– Lagrangeequationthatissubjecttotheassumedboundaryco nditionsformsacriticalpoint for the functional, and hence is a potential candidate for th e desired minimizing function. And, in many cases, the Euler–Lagrange equation suffices to ch aracterize the minimizer without further ado. Theorem 21.1. Suppose the Lagrangian function is at least twice continuou sly dif- ferentiable :L(x,u,p)∈C2. Then any C2minimizer u(x)to the corresponding functional J[u] =/integraldisplayb aL(x,u,u′)dx, subject to the selected boundary conditions, must satisfy the associated Euler–Lagrange equation (21.26). Let us now investigate what the Euler–Lagrange equation tel ls us about the examples of variational problems presented at the beginning of this s ection. One word of caution: there do exist seemingly reasonable functionals whose mini mizers are not, in fact, C2, and hence do not solve the Euler–Lagrange equation in the cla ssical sense; see [ 12] for examples. Fortunately, in most variational problems that a rise in real-world applications, such pathologies do not appear. Curves of Shortest Length — Planar Geodesics Let us return to the most elementary problem in the calculus o f variations: finding the curve of shortest length connecting two points a= (a,α),b= (b,β)∈R2in the plane. As we noted in Section 21.2, such planar geodesics min imize the arc length integral J[u] =/integraldisplayb a/radicalbig 1+(u′)2dxwith Lagrangian L(x,u,p) =/radicalbig 1+p2, †See Exercise for a complete justification. 12/11/12 1153 c/ci∇cleco√y∇t2012 Peter J. Olver subject to the boundary conditions u(a) =α, u (b) =β. Since∂L ∂u= 0,∂L ∂p=p/radicalbig 1+p2, the Euler–Lagrange equation (21.26) in this case takes the f orm 0 =−d dxu′ /radicalbig 1+(u′)2=−u′′ (1+(u′)2)3/2. Since the denominator does not vanish, this is the same as the simplest second order ordinary differential equation u′′= 0. (21.28) Therefore, the solutions to the Euler–Lagrange equation ar e all affine functions, u=cx+ d, whose graphs are straight lines. Since our solution must al so satisfy the boundary conditions, the only critical function — and hence the sole c andidate for a minimizer — is the straight line y=β−α b−a(x−a)+α (21.29) passing through the two points. Thus, the Euler–Lagrange eq uation helps to reconfirm our intuition that straight lines minimize distance. Be that as it may, the fact that a function satisfies the Euler– Lagrange equation and the boundary conditions merely confirms its status as a cr itical function, and does not guarantee that it is the minimizer. Indeed, any critical function is also a candidate formaximizing the variational problem, too. The nature of a critical funct ion will be elucidated by the second derivative test, and requires some further work. Of course, for the minimum distance problem, we “know” that a straight line cannot maximize distance, and must be the minimizer. Nevertheless, the reader should h ave a small nagging doubt that we have completely solved the problem at hand ... Minimal Surface of Revolution Consider next the problem of finding the curve connecting two points that generates a surface of revolution of minimal surface area. For simplic ity, we assume that the curve is given by the graph of a non-negative function y=u(x)≥0. According to (21.11), the required curve will minimize the functional J[u] =/integraldisplayb au/radicalbig 1+(u′)2dx,with Lagrangian L(x,u,p) =u/radicalbig 1+p2,(21.30) where we have omitted an irrelevant factor of 2 πand used positivity to delete the absolute value on uin the integrand. Since ∂L ∂u=/radicalbig 1+p2,∂L ∂p=up/radicalbig 1+p2, 12/11/12 1154 c/ci∇cleco√y∇t2012 Peter J. Olver the Euler–Lagrange equation (21.26) is /radicalbig 1+(u′)2−d dxuu′ /radicalbig 1+(u′)2=1+(u′)2−uu′′ (1+(u′)2)3/2= 0. (21.31) Therefore, to find the critical functions, we need to solve a n onlinear second order ordinary differential equation — and not one in a familiar form. Fortunately, there is a little trick†we can use to find the solution. If we multiply the equation by u′, we can then rewrite the result as an exact derivative u′/parenleftbigg1+(u′)2−uu′′ (1+(u′)2)3/2/parenrightbigg =d dxu/radicalbig 1+(u′)2= 0. We conclude that the quantityu/radicalbig 1+(u′)2=c, (21.32) is constant, and so the left hand side is a first integral for the differential equation, as per Definition 20.26. Solving for‡ du dx=u′=√ u2−c2 c results in an autonomous first order ordinary differential eq uation, which we can immedi- ately solve:/integraldisplayc du√ u2−c2=x+δ, whereδis a constant of integration. The most useful form of the left hand integral is in terms of the inverse to the hyperbolic cosine function cosh z=1 2(ez+e−z), whereby cosh−1u c=x+δ, and hence u=ccosh/parenleftbiggx+δ c/parenrightbigg . (21.33) In this manner, we have produced the general solution to the E uler–Lagrange equation (21.31). Anysolutionthatalsosatisfiestheboundarycondi tionsprovidesacriticalfunction for the surface area functional (21.30), and hence is a candi date for the minimizer. The curve prescribed by the graph of a hyperbolic cosine func tion (21.33) is known as acatenary . It isnota parabola, even though to the untrained eye it looks similar . Interestingly,thecatenaryisthesameprofileasahangingc hain. Owingtotheirminimizing properties, catenaries are quitecommon in engineering des ign — for instance a chain hangs in the shape of a catenary, while the arch in St. Louis is an inv erted catenary. †Actually, as withmany tricks, this isreallyanindicationthat someth ingprofound is goingon. Noether’s Theorem, a result of fundamental importancein modern ph ysics that relates symmetries and conservation laws, [ 78,146], underlies the integration method; see Exercise for further details. ‡The square root is real since, by (21.32), |u| ≤ |c|. 12/11/12 1155 c/ci∇cleco√y∇t2012 Peter J. Olver So far, we have not taken into account the boundary condition s It turns out that there are three distinct possibilities, depending upon the config uration of the boundary points: (a) There is precisely one value of the two integration constan tsc,δthat satisfies the two boundary conditions. In this case, it can be proved that this catenary is the unique curve that minimizes the area of its associated surface of re volution. (b) There are two different possible values of c,δthat satisfy the boundary conditions. In this case, one of these is the minimizer, and the other is a spu rious solution — one that corresponds to a saddle point for the surface area funct ional. (c) There are novalues of c,δthat allow (21.33) to satisfy the two boundary conditions. This occurs when the two boundary points a,bare relatively far apart. In this configuration, the physical soap film spanning the two circul ar wires breaks apart into two circular disks, and this defines the minimizer for th e problem, i.e., unlike cases (a) and (b), there is nosurface of revolution that has a smaller surface area than the two disks. However, the “function”†that minimizes this configuration consists of two vertical lines from the boundary points to th exaxis, along with that segment on the axis lying between them. More precisely, we can approximate this function by a sequence of genuine functions that givepr ogressively smaller and smallervaluestothesurface area functional (21.11),but t heactual minimum isnot attained among the class of (smooth) functions; this is illu strated in Figure cats . Further details are relegated to the exercises. Thus, even in such a reasonably simple example, a number of th e subtle complications can already be seen. Lack of space precludes a more detailed d evelopment of the subject, and we refer the interested reader to more specialized books on the calculus of variations, including [ 47,78]. The Brachistochrone Problem The most famous classical variational principle is the so-c alledbrachistochrone prob- lem. The compound Greek word “brachistochrone” means “minimal time”. An experi- menter lets a bead slide down a wire that connects two fixed poi nts. The goal is to shape the wire in such a way that, starting from rest, the bead slide s from one end to the other in minimal time. Na¨ ıve guesses for the wire’s optimal shape , including a straight line, a parabola, a circular arc, or even a catenary are wrong. One c an do better through a careful analysis of the associated variational problem. Th e brachistochrone problem was originally posed by the Swiss mathematician Johann Bernoul li in 1696, and served as an inspiration for much of the subsequent development of the su bject. We take, without loss of generality, the starting point of th e bead to be at the origin: a= (0,0). The wire will bend downwards, and so, to avoid distractin g minus signs in the subsequent formulae, we take the vertical yaxis to point downwards. The shape of the wire will be given by the graph of a function y=u(x)≥0. The end point b= (b,β) is †Here “function” must be taken in a verybroad sense, as this one does not even correspond to a generalized function! 12/11/12 1156 c/ci∇cleco√y∇t2012 Peter J. Olver Figure 21.4. The Brachistrochrone Problem. assumed to lie below and to the right, and so b >0 andβ >0. The set-up is sketched in Figure 21.4. To mathematically formulate the problem, the first step is to find the formula for the transit time of the bead sliding along the wire. Arguing as in our derivation of the optics functional (21.5), if v(x) denotes the instantaneous speed of descent of the bead when it reaches position ( x,u(x)), then the total travel time is T[u] =/integraldisplayℓ 0ds v=/integraldisplayb 0/radicalbig 1+(u′)2 vdx, (21.34) whereℓis the overall length of the wire. We shall use conservation of energy to determine a formula fo r the speed vas a function of the position along the wire. The kinetic energy o f the bead is1 2mv2, where mis its mass. On the other hand, due to our sign convention, the potential energy of the bead when it is at height y=u(x) is−mgu(x), where gthe gravitational force, and we take the initial height as the zero potential energy level. T he bead is initially at rest, with 0 kinetic energy and 0 potential energy. Assuming that frict ional forces are negligible, conservation of energy implies that the total energy must re main equal to 0, and hence 0 =1 2mv2−mgu. We can solve this equation to determine the bead’s speed as a f unction of its height: v=/radicalbig 2gu. (21.35) Substituting this expression into (21.34), we conclude tha t the shape y=u(x) of the wire is obtained by minimizing the functional T[u] =/integraldisplayb 0/radicalBigg 1+(u′)2 2gudx, (21.36) subjet to the boundary conditions u(0) = 0, u (b) =β. (21.37) 12/11/12 1157 c/ci∇cleco√y∇t2012 Peter J. Olver The associated Lagrangian is L(x,u,p) =/radicalbigg 1+p2 u, where we omit an irrelevant factor of√2g(or adopt physical units in which g=1 2). We compute ∂L ∂u=−/radicalbig 1+p2 2u3/2,∂L ∂p=p/radicalbig u(1+p2). Therefore, the Euler–Lagrange equation for the brachistoc hrone functional is −/radicalbig 1+(u′)2 2u3/2−d dxu′ /radicalbig u(1+(u′)2)=−2uu′′+(u′)2+1 2/radicalbig u(1+(u′)2)= 0.(21.38) Thus, the minimizing functions solve the nonlinear second o rder ordinary differential equa- tion 2uu′′+(u′)2+1 = 0. Rather thantrytosolvethisdifferentialequationdirectly , wenotethattheLagrangian does not depend upon x, and therefore we can use the result of Exercise , that states that the Hamiltonian function H(x,u,p) =L−p∂L ∂p=1/radicalbig u(1+p2) defines a first integral. Thus, H(x,u,u′) =1/radicalbig u(1+(u′)2)=k,which we rewrite as u(1+(u′)2) =c, wherec= 1/k2is a constant. (This can be checked by directly calculating dH/dx≡0.) Solving for the derivative u′results in the first order autonomous ordinary differential equation du dx=/radicalbigg c−u u. As in (20.7), this equation can be explicitly solved by separ ation of variables, and so /integraldisplay/radicalbiggu c−udu=x+k. The left hand integration relies on the trigonometric subst itution u=1 2c(1−cosθ), whereby x+k=1 2c/integraldisplay/radicalbigg 1−cosθ 1+cosθsinθdθ=1 2c/integraldisplay (1−cosθ)dθ=1 2c(θ−sinθ). 12/11/12 1158 c/ci∇cleco√y∇t2012 Peter J. Olver Figure 21.5. A Cycloid. The left hand boundary condition implies k= 0, and so the solution to the Euler–Lagrange equation are curves parametrized by x=r(θ−sinθ), u =r(1−cosθ). (21.39) With a little more work, it can be proved that the parameter r=1 2cis uniquely prescribed bytherighthandboundarycondition, andmoreover, theresu ltingcurvesuppliestheglobal minimizer of the brachistochrone functional, [ cvar]. The miniming curve is known as a cycloid, which, according to Exercise , can be visualized as the curve traced by a point sitting on the edge of a rolling wheel of radius r, as plotted in Figure 21.5. Interestingly, in certain configurations, namely if β <2b/π, the cycloid that solves the brachistrochrone problem dips below the right hand endpoint b= (b,β), and so the bead is moving upwards when it reaches the end of the wire. 21.3. The Second Variation. The solutionstotheEuler–Lagrangeboundary valueproblem arethe criticalfunctions for the variational principle, meaning that they cause the f unctional gradient to vanish. For finite-dimensional optimization problems, being a crit ical point is only a necessary condition for minimality. One must impose additional condi tions, based on the second derivative of the objective function at the critical point, in order to guarantee that it is a minimum and not a maximum or saddle point. Similarly, in the calculus of variations, the solutions to the Euler–Lagrange equation may also inclu de (local) maxima, as well as other non-extremal critical functions. To distinguish bet ween the possibilties, we need to formulate a second derivativetest for the objectivefuncti onal. In thecalculus of variations, the second derivative of a functional is known as its second variation , and the goal of this section is to construct and analyze it in its simplest manife station. For a finite-dimensional objective function F(u1,...,un), the second derivative test was based on the positive definiteness of its Hessian matrix. The justification was based on the second order Taylor expansion of the objective functi on at the critical point. In an analogous fashion, we expand an objective functional J[u] near the critical function. Consider the scalar function h(t) =J[u+tv], where the function v(x) represents a variation. The second order Taylor expansion ofh(t) takes the form h(t) =J[u+tv] =J[u]+tK[u;v]+1 2t2Q[u;v]+···. 12/11/12 1159 c/ci∇cleco√y∇t2012 Peter J. Olver The first order terms are linear in the variation v, and, according to our earlier calculation, given by an inner product h′(0) =K[u;v] =/an}b∇acketle{t∇J[u];v/an}b∇acket∇i}ht between the variation and the functional gradient. In parti cular, ifuis a critical function, then the first order terms vanish, K[u;v] =/an}b∇acketle{t∇J[u];v/an}b∇acket∇i}ht= 0 for all allowable variations v, meaning those that satisfy the homogeneous boundary con- ditions. Therefore, the nature of the critical function u— minimum, maximum, or neither — will, in most cases, be determined by the second derivative terms h′′(0) =Q[u;v]. As argued in our proof of the finite-dimensional Theorem 19.4 3, ifuis a minimizer, then Q[u;v]≥0. Conversely, if Q[u;v]>0 for all v/ne}ationslash≡0, i.e., the second variation is positive definite, then the critical function uwill be a strict local minimizer. This forms the crux of the second derivative test. Let us explicitly evaluate the second variational for the si mplest functional (21.16). Consider the scalar function h(t) =J[u+tv] =/integraldisplayb aL(x,u+tv,u′+tv′)dx, whose first derivative h′(0) was already determined in (21.21); here we require the se cond variation Q[u;v] =h′′(0) =/integraldisplayb a/bracketleftbig Av2+2Bvv′+C(v′)2/bracketrightbig dx, (21.40) where the coefficient functions A(x) =∂2L ∂u2(x,u,u′), B(x) =∂2L ∂u∂p(x,u,u′), C(x) =∂2L ∂p2(x,u,u′),(21.41) are found by evaluating certain second order derivatives of the Lagrangian at the critical function u(x). In contrast to the first variation, integration by parts wi ll not eliminate all of the derivatives on vin the quadratic functional (21.40), which causes significa nt complications in the ensuing analysis. The second derivative test for a minimizer relies on the posi tivity of the second varia- tion. To formulate conditions that the critical function be a minimizer for the functional, So, we need to establish criteria guaranteeing the positive definiteness of such a quad- ratic functional is positive definite, meaning that Q[u;v]>0 for all non-zero allowable variations v(x)/ne}ationslash≡0. Clearly, if the integrand is positive definite at each poin t, so A(x)v2+2B(x)vv′+C(x)(v′)2>0 whenever a < x < b, andv(x)/ne}ationslash≡0,(21.42) thenQ[u;v]>0 is also positive definite. 12/11/12 1160 c/ci∇cleco√y∇t2012 Peter J. Olver Example 21.2. For the arc length minimization functional (21.3), the Lagr angian isL(x,u,p) =/radicalbig 1+p2. To analyze the second variation, we first compute ∂2L ∂u2= 0,∂2L ∂u∂p= 0,∂2L ∂p2=1 (1+p2)3/2. For the critical straight line function u(x) =β−α b−a(x−a)+α,with p=u′(x) =β−α b−a, we find A(x) =∂2L ∂u2= 0, B(x) =∂2L ∂u∂p= 0, C(x) =∂2L ∂p2=(b−a)3 /bracketleftbig (b−a)2+(β−α)2/bracketrightbig3/2≡k. Therefore, the second variation functional (21.40) is Q[u;v] =/integraldisplayb ak(v′)2dx, wherek >0 is a positive constant. Thus, Q[u;v] = 0 vanishes if and only if v(x) is a constant function. But the variation v(x) is required to satisfy the homogeneous boundary conditions v(a) =v(b) = 0, and hence Q[u;v]>0 for all allowable nonzero variations. Therefore, we conclude that the straight line is, indeed, a ( local) minimizer for the arc length functional. We have at last justified our intuitionth at the shortest distance between two points is a straight line! In general, as the following example demonstrates, the poin twise positivity condition (21.42) is overly restrictive. Example 21.3. Consider the quadratic functional Q[v] =/integraldisplay1 0/bracketleftbig (v′)2−v2/bracketrightbig dx. (21.43) We claim that Q[v]>0 for all nonzero v/ne}ationslash≡0 subject to homogeneous Dirichlet boundary conditions v(0) = 0 = v(1). This result is not trivial! Indeed, the boundary condit ions play an essential role, since choosing v(x)≡c/ne}ationslash= 0 to be any constant function will produce a negative value for the functional: Q[v] =−c2. To prove the claim, consider the quadratic functional /tildewideQ[v] =/integraldisplay1 0(v′+vtanx)2dx≥0, which is clearly non-negative, since the integrand is every where≥0. Moreover, by continu- ity, the integral vanishes if and only if vsatisfies the first order linear ordinary differential equation v′+vtanx= 0,for all 0 ≤x≤1. 12/11/12 1161 c/ci∇cleco√y∇t2012 Peter J. Olver The only solution that also satisfies boundary condition v(0) = 0 is the trivial one v≡0. We conclude that /tildewideQ[v] = 0 if and only if v≡0, and hence /tildewideQ[v] is a positive definite quadratic functional on the space of allowable variations. Let us expand the latter functional, /tildewideQ[v] =/integraldisplay1 0/bracketleftbig (v′)2+2vv′tanx+v2tan2x/bracketrightbig dx =/integraldisplay1 0/bracketleftbig (v′)2−v2(tanx)′+v2tan2x/bracketrightbig dx=/integraldisplay1 0/bracketleftbig (v′)2−v2/bracketrightbig dx=Q[v]. In the second equality, we integrated the middle term by part s, using ( v2)′= 2vv′, and noting that the boundary terms vanish owing to our imposed bo undary conditions. Since /tildewideQ[v] is positive definite, so is Q[v], justifying the previous claim. To appreciate how subtle this result is, consider the almost identical quadratic func- tional /hatwideQ[v] =/integraldisplay4 0/bracketleftbig (v′)2−v2/bracketrightbig dx, (21.44) the only difference being the upper limit of the integral. A qu ick computation shows that the function v(x) =x(4−x) satisfies the boundary conditions v(0) = 0 = v(4), but /hatwideQ[v] =/integraldisplay4 0/bracketleftbig (4−2x)2−x2(4−x)2/bracketrightbig dx=−64 5<0. Therefore,/hatwideQ[v] isnotpositive definite. Our preceding analysis does not apply be- cause the function tan xbecomes singular at x=1 2π, and so the auxiliary integral/integraldisplay4 0(v′+vtanx)2dxdoes not converge. The complete analysis of positive definiteness of quadratic functionals is quite subtle. Indeed, the strange appearance of tan xin the preceding example turns out to be an important clue! In the interests of brevity, let us just stat e without proof a fundamental theorem, and refer the interested reader to [ 78] for full details. Theorem 21.4. LetA(x),B(x),C(x)∈C0[a,b]be continuous functions. The quad- ratic functional Q[v] =/integraldisplayb a/bracketleftbig Av2+2Bvv′+C(v′)2/bracketrightbig dx ispositivedefinite, so Q[v]>0for allv/ne}ationslash≡0satisfyingthehomogeneousDirichlet boundary conditions v(a) =v(b) = 0, provided (a)C(x)>0for alla≤x≤b, and (b)For any a < c≤b, the only solution to its linear Euler–Lagrange boundary va lue problem −(Cv′)′+(A−B′)v= 0, v (a) = 0 =v(c), (21.45) is the trivial function v(x)≡0. 12/11/12 1162 c/ci∇cleco√y∇t2012 Peter J. Olver Remark: A value cfor which (21.45) has a nontrivial solution is known as a conjugate pointtoa. Thus, condition ( b) can be restated that the variational problem has no conjugate points in the interval ( a,b]. Example 21.5. The quadratic functional Q[v] =/integraldisplayb 0/bracketleftbig (v′)2−v2/bracketrightbig dx (21.46) has Euler–Lagrange equation −v′′−v= 0. The solutions v(x) =ksinxsatisfy the boundary condition v(0) = 0. The first conjugate point occurs at c=πwherev(π) = 0. Therefore, Theorem 21.4 implies that the quadratic functional (21.46) is positive definite provided the upper integration limit b < π. This explains why the original quadratic functional (21.43) is p ositive definite, since there are no conjugate points on the interval [0 ,1], while the modified version (21.44) is not, because the first conjugate point πlies on the interval (0 ,4]. In the case when the quadratic functional arises as the secon d variation of a functional (21.16),thenthecoefficientfunctions A,B,CaregivenintermsoftheLagrangian L(x,u,p) by formulae (21.41). In this case, the first condition in Theo rem 21.4 requires ∂2L ∂p2(x,u,u′)>0 (21 .47) for the minimizer u(x). This is known as the Legendre condition . The second, conjugate point condition requires that the so-called linear variational equation −d dx/parenleftbigg∂2L ∂p2(x,u,u′)dv dx/parenrightbigg +/parenleftbigg∂2L ∂u2(x,u,u′)−d dx∂2L ∂u∂p(x,u,u′)/parenrightbigg v= 0 (21 .48) has no nontrivial solutions v(x)/ne}ationslash≡0 that satisfy v(a) = 0 and v(c) = 0 for a < c≤b. In this way, we have arrived at a rigorous form of the second deri vative test for the simplest functional in the calculus of variations. Theorem 21.6. If the function u(x)satisfies the Euler–Lagrange equation (21.26), and, in addition, the Legendre condition (21.47)and there are no conjugate points on the interval, then u(x)is a strict local minimum for the functional. 21.4. Multi-dimensional Variational Problems. The calculus of variations encompasses a very broad range of mathematical applica- tions. The methods of variational analysis can be applied to an enormous variety of phys- ical systems, whose equilibrium configurations inevitably minimize a suitable functional, which, typically, represents the potential energy of the sy stem. Minimizing configurations appear as critical functions at which the functional gradie nt vanishes. Following similar computational procedures as in the one-dimensional calcul us of variations, we find that the critical functions are characterized as solutions to a syst em of partial differential equations, 12/11/12 1163 c/ci∇cleco√y∇t2012 Peter J. Olver known as the Euler–Lagrange equations associated with the v ariational principle. Each solution to the boundary value problem specified by the Euler –Lagrange equations sub- ject to appropraite boundary conditions is, thus, a candida te minimizer for the variational problem. In many applications, the Euler–Lagrange boundar y value problem suffices to single out the physically relevant solutions, and one need n ot press on to the considerably more difficult second variation. Implementation of the variational calculus for functional s in higher dimensions will be illustrated by looking at a specific example — a first order var iational problem involving a single scalar function of two variables. Once this is fully u nderstood, generalizations and extensions to higher dimensions and higher order Lagrangia ns are readily apparent. Thus, we consider an objective functional J[u] =/integraldisplay/integraldisplay ΩL(x,y,u,ux,uy)dxdy, (21.49) having the form of a double integral over a prescribed domain Ω⊂R2. TheLagrangian L(x,y,u,p,q ) is assumed to be a sufficiently smooth function of its five argu ments. Our goal is to find the function(s) u=f(x,y) that minimize the value of J[u] when subject to a set of prescribed boundary conditions on ∂Ω, the most important being our usual Dirichlet, Neumann, or mixed boundary conditions. For simp licity, we concentrate on the Dirichlet boundary value problem, and require that the mini mizer satisfy u(x,y) =g(x,y) for ( x,y)∈∂Ω. (21.50) The First Variation The basic necessary condition for an extremum (minimum or ma ximum) is obtained in precisely the same manner as in the one-dimensional frame work. Consider the scalar function h(t)≡J[u+tv] =/integraldisplay/integraldisplay ΩL(x,y,u+tv,ux+tvx,uy+tvy)dxdy depending on t∈R. Thevariation v(x,y) is assumed to satisfy homogeneous Dirichlet boundary conditions v(x,y) = 0 for ( x,y)∈∂Ω, (21.51) to ensure that u+tvsatisfies the same boundary conditions (21.50) as uitself. Under these conditions, if uis a minimizer, then the scalar function h(t) will have a minimum at t= 0, and hence h′(0) = 0. When computing h′(t), we assume that the functions involved are sufficiently smoo th so as to allow us to bring the derivative inside the integral, an d then apply the chain rule. At t= 0, the result is h′(0) =d dtJ[u+tv]/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=/integraldisplay/integraldisplay Ω/parenleftbigg v∂L ∂u+vx∂L ∂p+vy∂L ∂q/parenrightbigg dxdy, (21.52) 12/11/12 1164 c/ci∇cleco√y∇t2012 Peter J. Olver where the derivatives of Lare all evaluated at x,y,u,ux,uy. To identify the functional gradient, we need to rewrite this integral in the form of an in ner product: h′(0) =/an}b∇acketle{t∇J[u];v/an}b∇acket∇i}ht=/integraldisplay/integraldisplay Ωh(x,y)v(x,y)dxdy, where h=∇J[u]. To convert (21.52) into this form, we need to remove the offend ing derivatives from v. In two dimensions, the requisite integration by parts formu la is based on Green’s Theo- rem A.26, and the required formula can be found in (15.89), na mely, /integraldisplay/integraldisplay Ω∂v ∂xw1+∂v ∂yw2dxdy=/contintegraldisplay ∂Ωv(−w2dx+w1dy)−/integraldisplay/integraldisplay Ωv/parenleftbigg∂w1 ∂x+∂w2 ∂y/parenrightbigg dxdy, (21.53) in which w1,w2are arbitrary smooth functions. Setting w1=∂L ∂p, w2=∂L ∂q, we find /integraldisplay/integraldisplay Ω/parenleftbigg vx∂L ∂p+vy∂L ∂q/parenrightbigg dxdy=−/integraldisplay/integraldisplay Ωv/bracketleftbigg∂ ∂x/parenleftbigg∂L ∂p/parenrightbigg +∂ ∂y/parenleftbigg∂L ∂q/parenrightbigg/bracketrightbigg dxdy, where the boundary integral vanishes owing to the boundary c onditions (21.51) that we impose on the allowed variations. Substituting this result back into (21.52), we conclude that h′(0) =/integraldisplay/integraldisplay Ωv/bracketleftbigg∂L ∂u−∂ ∂x/parenleftbigg∂L ∂p/parenrightbigg −∂ ∂y/parenleftbigg∂L ∂q/parenrightbigg/bracketrightbigg dxdy=/an}b∇acketle{t∇J[u];v/an}b∇acket∇i}ht,(21.54) where ∇J[u] =∂L ∂u−∂ ∂x/parenleftbigg∂L ∂p/parenrightbigg −∂ ∂y/parenleftbigg∂L ∂q/parenrightbigg is the desired first variation or functional gradient. Since the gradient vanishes at a critical function, we conclude that the minimizer u(x,y) must satisfy the Euler–Lagrange equation ∂L ∂u(x,y,u,ux,uy)−∂ ∂x/parenleftbigg∂L ∂p(x,y,u,ux,uy)/parenrightbigg −∂ ∂y/parenleftbigg∂L ∂q(x,y,u,ux,uy)/parenrightbigg = 0. (21.55) Once we explicitly evaluate the derivatives, the net result is a second order partial differ- ential equation, which you are asked to write out in full deta il in Exercise . Example 21.7. As a first elementary example, consider the Dirichlet minimi zation problem J[u] =/integraldisplay/integraldisplay Ω1 2/parenleftbig u2 x+u2 y/parenrightbig dxdy (21.56) that we first encountered in our analysis of the Laplace equat ion (15.103). In this case, the associated Lagrangian is L=1 2(p2+q2),with∂L ∂u= 0,∂L ∂p=p=ux,∂L ∂q=q=uy. 12/11/12 1165 c/ci∇cleco√y∇t2012 Peter J. Olver Therefore, the Euler–Lagrange equation (21.55) becomes −∂ ∂x(ux)−∂ ∂y(uy) =−uxx−uyy=−∆u= 0, which is the two-dimensional Laplace equation. Subject to t he selected boundary condi- tions, the solutions, i.e., the harmonic functions, are cri tical functions for the Dirichlet variational principle. This reconfirms the Dirichlet chara cterization of harmonic functions as minimizers of the variational principle, as stated in The orem 15.14. However, the calculus of variations approach, as developed so far, leads to a much weaker result since it only singles out the harmonic functio ns ascandidates for minimizing the Dirichlet integral; they could just as easily be maximin g functions or saddle points. When dealing with a quadratic variational problem, the dire ct algebraic approach is, when applicable, the more powerful, since it assures us that the s olutions to the Laplace equation really do minimize the integral among the space of functions satisfying the appropriate boundary conditions. However, the direct method is restric ted to quadratic variational problems, whose Euler–Lagrange equations are linear parti al differential equations. In nonlinear cases, one really does need to utilize the full pow er of the variational machinery. Example 21.8. Let us derive the Euler–Lagrange equation for the minimal su rface problem. From (21.9), the surface area integral J[u] =/integraldisplay/integraldisplay Ω/radicalbig 1+u2 x+u2 ydxdy has Lagrangian L=/radicalbig 1+p2+q2. Note that ∂L ∂u= 0,∂L ∂p=p/radicalbig 1+p2+q2,∂L ∂q=q/radicalbig 1+p2+q2. Therefore, replacing p→uxandq→uyand then evaluating the derivatives, the Euler– Lagrange equation (21.55) becomes −∂ ∂xux/radicalbig 1+u2 x+u2 y−∂ ∂yuy/radicalbig 1+u2 x+u2 y=−(1+u2 y)uxx+2uxuyuxy−(1+u2 x)uyy (1+u2 x+u2 y)3/2= 0. Thus, a surface described by the graph of a function u=f(x,y) is a critical function, and hence a candidate for minimizing surface area, provided it s atisfies the minimal surface equation (1+u2 y)uxx−2uxuyuxy+(1+u2 x)uyy= 0. (21.57) Weareconfronted withacomplicated, nonlinear, second ord er partialdifferential equation, which has been the focus of some of the most sophisticated and deep analysis over the past two centuries, with significant progress on understanding i ts solution only within the past 70 years. In this book, we have not developed the sophisticat ed analytical, geometrical, and numerical techniques that are required to have anything of substance to say about its solutions. We refer the interested reader to the advance d texts [ 135,140] for further developments in this fascinating problem. 12/11/12 1166 c/ci∇cleco√y∇t2012 Peter J. Olver Example 21.9. The small deformations of an elastic body Ω ⊂Rnare described by thedisplacement field,u:Ω→Rn. Each material point x∈Ω in the undeformed body will move to a new position y=x+u(x) in the deformed body /tildewideΩ ={y=x+u(x)|x∈Ω}. The one-dimensional case governs bars, beams and rods, two- dimensional bodies include thin plates and shells, while n= 3 for fully three-dimensional solid bodies. See [ 8,87] for details and physical derivations. For small deformations, we can use a linear theory to approxi mate the much more complicated equations of nonlinear elasticity. The simple st case is that of a homogeneous and isotropic planar body Ω ⊂R2. The equilibrium mechanics are described by the defor- mation function u(x) = (u(x,y),v(x,y))T. A detailed physical analysisof the constitutive assumptions leads to a minimization principle based on the f ollowing functional: J[u,v] =/integraldisplay/integraldisplay Ω/bracketleftbig1 2µ/ba∇dbl∇u/ba∇dbl2+1 2(λ+µ)(∇·u)2/bracketrightbig dxdy =/integraldisplay/integraldisplay Ω/bracketleftbig/parenleftbig1 2λ+µ/parenrightbig (u2 x+v2 y)+1 2µ(u2 y+v2 x)+(λ+µ)uxvy/bracketrightbig dxdy.(21.58) The parameters λ,µare known as the Lam´ e moduli of the material, and govern its intrinsic elastic properties. They are measured by performing suitab le experiments on a sample of the material. Physically, (21.58) represents the stored (o r potential) energy in the body under the prescribed displacement. Nature, as always, seek s the displacement that will minimize the total energy. To compute the Euler–Lagrange equations, we consider the fu nctional variation h(t) =J[u+tf,v+tg], in which the individual variations f,gare arbitrary functions subject only to the given homogeneous boundary conditions. If u,vminimize J, thenh(t) has a minimum at t= 0, and so we are led to compute h′(0) =/an}b∇acketle{t∇J;f/an}b∇acket∇i}ht=/integraldisplay/integraldisplay Ω(f∇uJ+g∇vJ)dxdy, which we write as an inner product (using the standard L2inner product between vector fields) between the variation fand the functional gradient ∇J= (∇uJ,∇vJ)T. For the particular functional (21.58), we find h′(0) =/integraldisplay/integraldisplay Ω/bracketleftbig/parenleftbig λ+2µ/parenrightbig (uxfx+vygy)+µ(uyfy+vxgx)+(λ+µ)(uxgy+vyfx)/bracketrightbig dxdy. We use the integration by parts formula (21.53) to remove the derivatives from the vari- ationsf,g. Discarding the boundary integrals, which are used to presc ribe the allowable boundary conditions, we find h′(0) =−/integraldisplay/integraldisplay Ω/parenleftBigg/bracketleftbig (λ+2µ)uxx+µuyy+(λ+µ)vxy/bracketrightbig f+ +/bracketleftbig (λ+µ)uxy+µvxx+(λ+2µ)vyy/bracketrightbig g/parenrightBigg dxdy. 12/11/12 1167 c/ci∇cleco√y∇t2012 Peter J. Olver The two terms in brackets give the two components of the funct ional gradient. Setting them equal to zero, we derive the second order linear system o f Euler–Lagrange equations (λ+2µ)uxx+µuyy+(λ+µ)vxy= 0,(λ+µ)uxy+µvxx+(λ+2µ)vyy= 0, (21.59) known as Navier’s equations , which can be compactly written as µ∆u+(µ+λ)∇(∇·u) =0 (21.60) for the displacement vector u= (u,v)T. The solutions to are the critical displacements that, under appropraite boundary conditions, minimize the potential energy functional. Since we are dealing with a quadratic functional, a more deta iled algebraic analy- sis will demonstrate that the solutions to Navier’s equatio ns are the minimizers for the variational principle (21.58). Although only valid in a lim ited range of physical and kine- maticalconditions, the solutions to theplanar Navier’seq uations and itsthree-dimensional counterpart are successfully used to model a wide class of el astic materials. In general, the solutions to the Euler–Lagrange boundary va lue problem are critical functions for the variational problem, and hence include al l (smooth) local and global min- imizers. Determination of which solutions are geniune mini ma requires a further analysis of the positivity properties of the second variation, which is beyond the scope of our intro- ductory treatment. Indeed, a complete analysis of the posit ive definiteness of the second variation of multi-dimensional variational problems is qu ite complicated, and still awaits a completely satisfactory resolution! 12/11/12 1168 c/ci∇cleco√y∇t2012 Peter J. Olver