two

Appendix A VectorCalculusinTwoDimensions The purpose of this appendix is to review the basics of vector calculus in the two dimensions. Most, if not all, this material should be famili ar to the student who has taken a basic course in multivariable calculus, but it is worth col lecting together the necessary results. We will assume you are familiar with the basics of pa rtial derivatives, including the equality of mixed partical (assuming they are continuou s), the chain rule, implcit differentiation. In addition, some familiarity with multip le integrals is assumed, although we will review the highlights. Proofs can be found in most vec tor calculus text, including [9,168,171]. We begin with a discussion of plane curves and domains. Many p hysical quantities, including force and velocity, are determined by vector field s, and we review the basic concepts. The key differential operators in planar vector ca lculus are the gradient and divergence operations, along with the Jacobian matrix for m aps from R2to itself. There are three basic types of line integrals: integrals with resp ect to arc length, for computing lengths of curves, masses of wires, center of mass, etc., ord inary line integrals of vector fields for computing work and fluid circulation, and flux line i ntegrals for computing flux of fluids and forces. Next, we review the basics of double inte grals of scalar functions over plane domains. Line and double integrals are connected by the justly famous Green’s theorem, which isthetwo-dimensional versionofthefundam ental theorem ofcalculus. The integration by parts argument required to characterize the adjoint of a partial differential operator rests on the closely allied Green’s formula. A.1. Plane Curves. We begin our review by collecting together the basic facts co ncerning geometry of plane curves. A curveC⊂R2is parametrized by a pair of continuous functions x(t) =/parenleftbigg x(t) y(t)/parenrightbigg ∈R2, (A.1) where the scalar parameter tvaries over an (open or closed) interval I⊂R. When it exists, the tangent vector to the curve at the point xis described by the derivative, dx dt=/squaresmallsolidx=/parenleftbigg/squaresmallsolidx /squaresmallsolidy/parenrightbigg . (A.2) We shall often use Newton’s dot notation to abbreviate deriv atives with respect to the parametert. 12/11/12 1203 c/ci∇cleco√y∇t2012 Peter J. Olver Cusped Curve Circle Figure Eight Figure A.1. Planar Curves. Physically, we can think of a curve as the trajectory describ ed by a particle moving in the plane. The parameter tis identified with the time, and so x(t) gives the position of the particle at time t. The tangent vector/squaresmallsolidx(t) measures the velocity of the particle at time t; its magnitude†/ba∇dbl/squaresmallsolidx/ba∇dbl=/radicalbig/squaresmallsolidx2+/squaresmallsolidy2is the speed, while its orientation (assuming the velocity is nonzero) indicates the instantaneous direction of motio n of the particle as it moves along the curve. Thus, by the orientation of a curve, we mean the direction of motion or parametrization, as indicated by the tangent vector. Rever sing the orientation amounts to moving backwards along the curve, with the individual tan gent vectors pointing in the opposite direction. The curve parametrized by x(t) is called smoothprovided its tangent vector is con- tinuous and everywhere nonzero:/squaresmallsolidx/\e}atio\slash=0. This is because curves with vanishing derivative may have corners or cusps; a simple example is the first curve p lotted in Figure A.1, which has parametrization x(t) =/parenleftbigg t2 t3/parenrightbigg ,/squaresmallsolidx(t) =/parenleftbigg 2t 3t2/parenrightbigg , and has a cusp at the origin when t= 0 and/squaresmallsolidx(0) = 0. Physically, a particle trajectory remains smooth as long as the speed of the particle is never ze ro, which effectively prevents theparticlefrominstantaneously changing itsdirectiono fmotion. Aclosedcurveis smooth if, in addition to satisfying/squaresmallsolidx(t)/\e}atio\slash=0at all points a≤t≤b, the tangents at the endpoints match up:/squaresmallsolidx(a) =/squaresmallsolidx(b). A curve is called piecewise smooth if its derivative is piecewise continuous and nonzero everywhere. The corners in a piecewi se smooth curve have well- defined right and left tangents. For example, polygons, such as triangles and rectangles, are piecewise smooth curves. In this book, all curves are ass umed to be at least piecewise smooth. Acurve is simpleifithasno self-intersections: x(t)/\e}atio\slash=x(s)whenever t/\e}atio\slash=s. Physically, this means that the particle is never in the same position twi ce. A curve is closedifx(t) is defined for a≤t≤band its endpoints coincide: x(a) =x(b), so that the particle ends †Throughout this chapter, we always use the standard Euclidean inner p roduct and norm. Withsomecare, all ofthe concepts canbeadapted to otherchoices of inner product. In differential geometry and relativity, one even allows the inner product and norm to vary from point to point, [60]. 12/11/12 1204 c/ci∇cleco√y∇t2012 Peter J. Olver up where it began. For example, the unit circle x(t) = (cost,sint)Tfor 0 ≤t≤2π, is closed and simple†, while the curve x(t) = (cost,sin2t)Tfor 0 ≤t≤2π, is not simple since it describes a figure eight that intersect s itself at the origin. Both curves are illustrated in Figure A.1. Assuming the tangent vector/squaresmallsolidx(t)/\e}atio\slash=0, then the normalvector to the curve at the pointx(t) is the orthogonal or perpendicular vector /squaresmallsolidx⊥=/parenleftbigg/squaresmallsolidy −/squaresmallsolidx/parenrightbigg (A.3) of the same length /ba∇dbl/squaresmallsolidx⊥/ba∇dbl=/ba∇dbl/squaresmallsolidx/ba∇dbl. Actually, there are two such normal vectors, the other being the negative −/squaresmallsolidx⊥. We will always make the “right-handed” choice (A.3) of norm al, meaning that as we traverse the curve, the normal always poin ts to our right. If a simple closed curve Cis oriented so that it is traversed in a counterclockwise dir ection — the standard mathematical orientation — then (A.3) describes t he outwards-pointing normal. If we reverse the orientation of the curve, then both the tang ent vector and normal vector change directions; thus (A.3) would give the inwards-point ing normal for a simple closed curve traversed in the clockwise direction. The same curve Ccan be parametrized in many different ways. In physical terms , a particle can move along a prescribed trajectory at a variety of different speeds, and these correspond to different ways of parametrizing the curve. Con version from one parame- trization x(t) to another/tildewidex(τ) is effected by a change of parameter , which is a smooth, invertible function t=g(τ); the reparametrized curve is then /tildewidex(τ) =x(g(τ)). We require thatdt/dτ=g′(τ)>0 everywhere. This ensures that each tcorresponds to a unique value ofτ, and, moreover, the curve remains smooth and is traversed in the same overall direction under the reparametrization. On the other hand, i fg′(τ)<0 everywhere, then the orientation of the curve is reversed under the reparamet rization. We shall use the notation −Cto indicate the curve having the same shape as C, but with the reversed orientation. Example A.1. The function x(t) = (cost,sint)Tfor 0< t < π parametrizes a semi-circle of radius 1 centered at the origin. If we set‡τ=−cottthen we obtain the less evident parametrization /tildewidex(τ) =/parenleftbigg1√ 1+τ2,−τ√ 1+τ2/parenrightbiggT for−∞<τ <∞ †For a closed curve to be simple, we require x(t)/negationslash=x(s) whenever t/negationslash=sexceptat the ends, wherex(a) =x(b) is required for the ends to close up. ‡The minus sign is to ensure that dτ/dt > 0. 12/11/12 1205 c/ci∇cleco√y∇t2012 Peter J. Olver Interior Point Bounded Domain A Simple Closed Curve Figure A.2. Topology of Planar Domains. of thesamesemi-circle, in the samedirection. In the familiar parametrization, the velocity vector has unit length, /ba∇dbl/squaresmallsolidx/ba∇dbl ≡1, and so the particle moves around the semicircle in the counterclockwise direction with unit speed. In the second p arametrization, the particle slows down near the endpoints, and, in fact, takes an infinite amount of time to traverse the semicircle from right to left. A.2. Planar Domains. A plate or other two-dimensional body occupies a region in th e plane, known as a domain. The simplest example is an open circular disk Dr(a) =/braceleftbig x∈R2/vextendsingle/vextendsingle/ba∇dblx−a/ba∇dbl<r/bracerightbig (A.4) of radiusrcentered at a point a∈R2. In order to properly formulate the mathematical tools needed to understand boundary value problems and dyna mical equations for such bodies, we first need to review basic terminology from point s et topology of planar sets. Many of the concepts carry over as stated to subsets of any hig her dimensional Euclidean spaceRn. Let Ω⊂R2be any subset. A point a∈Ω is called an interior point if some small disk centered at ais entirely contained within the set: Dε(a)⊂Ω for some ε >0; see Figure A.2. The set Ω is openif every point is an interior point. A set Kisclosedif and only if its complement Ω = R2\K={x/\e}atio\slash∈K}is open. Example A.2. Iff(x,y) is any continuous real-valued function, then the subset {f(x,y)>0}wherefis strictly positive is open, while the subset {f(x,y)≥0}wheref is non-negative is closed. One can, of course, replace 0 by an y other constant, and also reverse the direction of the inequalities, without affectin g the conclusions. In particular, the set Dr={x2+y2<r2} (A.5) consisting of all points of (Euclidean) norm strictly less t hanr, defines an open disk of radiusrcentered at the origin. On the other hand, Kr={x2+y2≤r2} (A.6) 12/11/12 1206 c/ci∇cleco√y∇t2012 Peter J. Olver Figure A.3. Open Sets Defined by a Hyperbola. is theclosed disk of radiusr, which includes the bounding circle Cr={x2+y2=r2}. (A.7) A point x⋆is alimit point of a set Ω if there exists a sequence of points x(n)∈Ω converging to it, so that†x(n)→x⋆asn→ ∞. Every point x∈Ω is a limit point (just take allx(n)=x) but the converse is not necessarily valid. For example, the points on the circle (A.7) are all limit points for the open disk (A.5). The closureof a set Ω, written Ω, is defined as the set of all limit points of Ω. In particular, a s etKis closed if and only if it contains all its limit points, and so K=K. Theboundary∂Ω of a subset Ω consists of all limit points which are not interior points. If Ω is open, t hen its closure is the disjoint union of the set and its boundary Ω = Ω∪∂Ω. Thus, the closure of the open disk Dris the closed disk Dr=Dr∪Cr; the circle Cr=∂Dr=∂Drforms their common boundary. An open subset that can be written as the union, Ω = Ω1∪Ω2, of two disjoint, nonempty, open subsets, so Ω1∩Ω2=∅, is called disconnected . For example, the open set Ω ={x2−y2>1} (A.8) is disconnected, consisting of two disjoint “sectors” boun ded by the two branches of the hyperbolax2−y2= 1; see Figure A.3. On the other hand, the complementary open set /hatwideΩ ={x2−y2<1} (A.9) isconnected , and consists of all points between the two hyperbolas. A subset is called bounded if it is contained inside a (possibly large) disk, i.e., Ω ⊂Dr for somer>0, as in the second picture in Figure A.2. Thus, both the close d and the open disks (A.5), (A.6) are bounded, whereas the two hyperbolic s ectors (A.8), (A.9) are both unbounded. The class of subsets for which the boundary value problems fo r the partial differential equations of equilibrium mechanics are properly prescribe d can now be defined. †See Section 12.5 for more details on convergence. 12/11/12 1207 c/ci∇cleco√y∇t2012 Peter J. Olver Rectangle Annulus Wedge Figure A.4. Planar Domains. Definition A.3. Aplanar domain isa connected, open subset Ω ⊂R2whose bound- ary∂Ω consists of one or more piecewise smooth, simple curves, su ch that Ω lies entirely on one side of each of its boundary curve(s). The last condition is to avoid dealing with pathologies. For example, the subset Ω \C obtained by cutting out a curve Cfrom the interior of an open set Ω would not be an allowable domain. Example A.4. The open rectangle R={a<x<b,c<y<d }is an open, con- nected and bounded domain. Its boundary is a piecewise smoot h curve, since there are corners where the tangent does not change continuously. Theannulus r2<x2+y2<R2,for fixed 0 <r<R, (A.10) is an open, connected, bounded domain whose boundary consis ts of two disjoint concentric circles. The degenerate case of a punctured disk , whenr= 0, isnota domain since its boundary consists of a circle and a single point — the origin. Another well-studied example is the wedge-shaped domain W={α<θ<β }con- sisting of all points whose angular coordinate θ= tan−1y/xlies between two prescribed values. If 0 < β−α <2π, then the wedge is a domain whose boundary consists of two connected rays. However, if β=α+2π, then the wedge is obtained by cutting the plane along a single ray at angle α. The latter case does not comply with our definition of a domain since the wedge now lies on both sides of its boundary r ay. Any connected domain is automatically pathwise connected meaning that any two points can be connected by (i.e., are the endpoints of) a curv e lying entirely within the domain. If the domain is bounded, which is the most important case for boundary value problems, then its boundary consists of one or more piecewis e smooth, simple, closed curves. A bounded domain Ω is called simply connected if it has just one such boundary curve; this means that Ω is connected and has no holes, and so i ts boundary ∂Ω =Cis a simple closed curve that contains Ω in its interior. For ins tance, an open disk and a rectangle are both simply connected, whereas an annulus is n ot. The Jordan Curve Theorem states the intuitively obvious, bu t actually quite deep, result that any simple closed curve divides the plane R2into two disjoint, connected, open 12/11/12 1208 c/ci∇cleco√y∇t2012 Peter J. Olver Figure A.5. Vector Fields. domains — its interior, which is bounded and simply connected, and its exterior, which is unbounded and not simply connected. This result is illust rated in the final figure in Figure A.2; the interior of the indicated simple closed curv e is shaded in gray while the exterior is in white. Note that the each subdomain lies entir ely on one side of the curve, which forms their common boundary. The following result is often used to characterize the simpl e connectivity of more general planar subsets, including unbounded domains. Lemma A.5. A planar domain Ω⊂R2issimply connected if it is connected and, moreover, if the interior of any simple closed curve C⊂Ωis also contained in Ω. For example, an annulus (A.10) is not simply connected becau se the interior of a circle going around the hole is not entirely contained within the an nulus. On the other hand, the unbounded domain (A.9) lying between two branches of a hyper bola is simply connected, even though its boundary consists of two disjoint, unbounde d curves. A.3. Vector Fields. A vector-valued function v(x,y) =/parenleftbigg v1(x,y) v2(x,y)/parenrightbigg is known as a (planar) vector field . A vector field assigns a vector v(x,y) to each point ( x,y)Tin its domain of definition, and hence defines a (in general nonlinear) function v:Ω→R2. The vector field can be conveniently depicted by drawing an arrow representing the vectorv=v(x,y) starting at its point of definition ( x,y)T. See Figure A.5 for some representative sketches. Example A.6. Vector fields arise very naturally in physics and engineerin g appli- cations from physical forces: gravitational, electrostat ic, centrifugal, etc. A force field f(x,y) = (f1(x,y),f2(x,y))Tdescribes the direction and magnitude of the force experi- enced by a particle at position ( x,y). In a planar universe, the gravitational force field exerted by a point mass concentrated at the origin has, accor ding to Newtonian gravita- tional theory, magnitude proportional to†1/r, wherer=/ba∇dblx/ba∇dblis the distance to the origin, †In three-dimensional Newtonian gravity, 1 /ris replaced by 1 /r2. 12/11/12 1209 c/ci∇cleco√y∇t2012 Peter J. Olver and is directed towards the origin. Thus, the vector field des cribing gravitational force has the form f=−γx /ba∇dblx/ba∇dbl=/parenleftBigg −γx/radicalbig x2+y2,−γy/radicalbig x2+y2/parenrightBiggT , (A.11) whereγ >0 denotes the constant of proportionality, namely the produ ct of the two masses times the universal gravitational constant. The same force law applies to the attraction, γ >0, and repulsion, γ <0, of electrically charged particles. Newton’s Laws of planetary motion produce the second order s ystem of differential equations md2x dt2=f. The solutions x(t) describe the trajectories of planets subject to a central g ravitational force, e.g., the sun. They also govern the motion of electric ally charged particles under a central electric charge, e.g., classical (i.e., not quantu m) electrons revolving around a cen- tral nucleus. In three-dimensional Newtonianmechanics, p lanets move along conic sections — ellipses in the case of planets, and parabolas and hyperbol as in the case of non-recurrent objects like some comets. Interestingly (and not as well-kn own), the corresponding two- dimensional theory is not as neatly described — the typical o rbit of a planet around a planar sun does not form a simple closed curve, [ 57]! Example A.7. Another important example is the velocity vector field vof a steady- statefluidflow. Thevector v(x,y)measurestheinstantaneousvelocityofthefluidparticles (molecules or atoms) as they pass through the point ( x,y). “Steady-state” means that the velocity at a point ( x,y) does not vary in time — even though the individual fluid particles are in motion. If a fluid particle moves along the cu rvex(t) = (x(t),y(t))T, then its velocity at time tis the derivative v=/squaresmallsolidxof its position with respect to t. Thus, for a time-independent velocity vector field v(x,y) = (v1(x,y),v2(x,y))T, the fluid particles will move in accordance with an autonomous, first order syste m of ordinary differential equations dx dt=v1(x,y),dy dt=v2(x,y). (A.12) According to the basic theory†of systems of ordinary differential equations , an individua l particle’s motion x(t) will be uniquely determined solely by its initial position x(0) =x0. In fluid mechanics, the trajectories of particles are known a s thestreamlines of the flow. The velocity vector vis everywhere tangent to the streamlines. When the flow is ste ady, the streamlines do not change in time. Individual fluid parti cles experience the same motion as they successively pass through a given point in the domain occupied by the fluid. As a specific example, consider the vector field v(x,y) =/parenleftbigg −ωy ωx/parenrightbigg , (A.13) †See Section 20.2 for details. 12/11/12 1210 c/ci∇cleco√y∇t2012 Peter J. Olver Rotational Flow Source Spiral Sink Figure A.6. Steady State Fluid Flows. for fixedω>0, which is plotted in the first figure in Figure A.5. The corres ponding fluid trajectories are found by solving the associated first order system of ordinary differential equations /squaresmallsolidx=−ωy,/squaresmallsolidy=ωx, with initial conditions x(0) =x0,y(0) =y0. This is a linear system, and can be solved by the eigenvalue and eigenvector techniques presented in Cha pter 9. The resulting flow x(t) =x0cosωt−y0sinωt, y (t) =x0sinωt+y0cosωt, corresponds to a fluid that is uniformly rotating around the o rigin. The streamlines are concentric circles, and the fluid particles rotate around th e circles in a counterclockwise direction with angular velocity ω, as illustrated in Figure A.6. Note that the fluid velocity vis everywhere tangent to the circles. The origin is a stagnat ion point, since the velocity fieldv=0vanishes there, and the particle at the origin does not move. As another example, the radial vector field v(x,y) =αx=/parenleftbigg αx αy/parenrightbigg (A.14) corresponds to a fluid source, α >0, or sink,α <0, at the origin, and is plotted in the second figure in Figure A.5. The solution to the first order sys tem of ordinary differential equations/squaresmallsolidx=αxwithinitial conditions x(0) =x0givesthe radial flow x(t) =eαtx0. The streamlines are the rays emanating from the origin, and the m otion is outwards (source) or inwards (sink) depending on the sign of α. As in the rotational flow, the origin is a stagnation point. Combining the radial and circular flow vector fields, v(x,y) =/parenleftbigg αx−ωy ωx+αy/parenrightbigg (A.15) leads toa swirling source or sink —think of thewater drainin g out of your bathtub. Again, the flow is found by integrating a linear system of ordinary di fferential equations /squaresmallsolidx=αx−ωy,/squaresmallsolidy=ωx+αy. 12/11/12 1211 c/ci∇cleco√y∇t2012 Peter J. Olver Solving as in Chapter 9, we find that the fluid particles follow the spiral streamlines x(t) =eαt/parenleftbig x0cosωt−y0sinωt/parenrightbig , y (t) =eαt/parenleftbig x0sinωt+y0cosωt/parenrightbig , again illustrated in Figure A.6. Remark: Allofthephaseportraitsfor linearsystemsoffirst order o rdinarydifferential equations in two variables presented in Section 9.3 can be re interpreted as streamline plots for steady state fluid flows. Additional, nonlinear examples , along with numerical solution techniques, can be found in Chapter 20. Remark: Of course, physical fluidmotionoccurs inthree-dimension al space. However, any planar flow can also be viewed as a particular type of three -dimensional fluid motion that does not depend upon the vertical coordinate. The motio n on every horizontal plane is the same, and so the planar flow represents a cross-section of the full three-dimensional motion. Forexample,slicingasteadyflowpastaverticalcyl inderbyatransversehorizontal plane results in a planar flow around a circle; see Figure fcyl . A.4. Gradient and Curl. In the same vein, a scalar-valued function u(x,y) is often referred to as a scalar field, since it assigns a scalar to each point ( x,y)Tin its domain of definition. Typical physical examples of scalar fields include temperature, defl ection of a membrane, height of a topographic map, density of a plate, and so on. Thegradient operator ∇maps a scalar field u(x,y) to the vector field ∇u= gradu=/parenleftbigg ∂u/∂x ∂u/∂y/parenrightbigg (A.16) consisting of its two first order partial derivatives. The sc alar fielduis often referred to as apotential function for its gradient vector field ∇u. For example, the gradient of the potential function u(x,y) =x2+y2is the radial vector field ∇u= (2x,2y)T. Similarly, the gradient of the logarithmic potential function u(x,y) =−γlogr=−1 2γlog(x2+y2) is the gravitational force (A.11) exerted by a point mass con centrated at the origin. Addi- tional physical examples include the velocity potential of certain fluid velocity vector fields and the electromagnetic potential whose gradient describe s the electromagnetic force field. Not every vector field admits a potential because not every ve ctor field lies in the range of the gradient operator ∇. Indeed, if u(x,y) has continuous second order partial derivatives, and/parenleftbigg v1 v2/parenrightbigg =v=∇u=/parenleftbigg ux uy/parenrightbigg , then, by the equality of mixed partials, ∂v1 ∂y=∂ ∂y/parenleftbigg∂u ∂x/parenrightbigg =∂ ∂x/parenleftbigg∂u ∂y/parenrightbigg =∂v2 ∂x. 12/11/12 1212 c/ci∇cleco√y∇t2012 Peter J. Olver The resulting equation ∂v1 ∂y=∂v2 ∂x(A.17) constitutes one of the necessary conditions that a vector fie ld must satisfy in order to be a gradient. Thus, for example, the rotational vector field (A .13) does not satisfy (A.17), and hence is nota gradient. There is nopotential function for such circulating flows. The difference between the two terms in (A.17) is known as the curlof the planar vector field v= (v1,v2), and denoted by† ∇∧v= curlv=∂v2 ∂x−∂v1 ∂y. (A.18) Notice that the curl of a planar vector field is a scalar field. ( In contrast, in three dimen- sions, the curl of a vector field is a vector field — see (B.79).) Thus, a necessary condition for a vector field to be a gradient is that its curl vanish ident ically:∇∧v≡0. Even if the vector field has zero curl, it still may not be a grad ient. Interestingly, the general criterion depends only upon the topology of the doma in of definition, as clarified in the following theorem. Theorem A.8. Letvbe a smooth vector field defined on a domain Ω⊂R2. If v=∇ufor some scalar function u, then∇ ∧v≡0. IfΩis simply connected, then the converse holds :if∇∧v≡0thenv=∇ufor some potential function udefined on Ω. As we shall see, this result is a direct consequence of Green’ s Theorem A.26. Example A.9. The vector field v=/parenleftbigg−y x2+y2,x x2+y2/parenrightbiggT (A.19) satisfies∇ ∧v≡0. However, there is no potential function defined for all ( x,y)/\e}atio\slash= (0,0) such that ∇u=v. As the reader can check, the angular coordinate u=θ= tan−1y x(A.20) satisfies∇θ=v, butθis not well-defined on the entire domain since it experiences a jump discontinuity of magnitude 2 πas we go around the origin. Indeed, Ω = {x/\e}atio\slash=0}isnot simply connected, and so Theorem A.8 does not apply. On the ot her hand, if we restrict vto any simply connected subdomain /hatwideΩ⊂Ω that does not encircle the origin, then the angular coordinate (A.20) can be unambiguously and smoothl y defined on /hatwideΩ, and does serve as a single-valued potential function for v; see Exercise A.5.30 †In this text, we adopt the more modern wedge notation ∧for what is often denoted by a cross×. 12/11/12 1213 c/ci∇cleco√y∇t2012 Peter J. Olver In fluid mechanics, the curl of a vector field measures the loca l circulation in the associated steady state fluid flow. If we place a small paddle w heel in the fluid, then its rate of spinning will be in proportion to ∇ ∧v. (An explanation of this fact will appear below.) The fluid flow is called irrotational if its velocity vector field has zero curl, and hence, assuming Ω is simply connected, is a gradient: v=∇u. In this case, the paddle wheel will not spin. The scalar function u(x,y) is known as the velocity potential for the fluid motion. Similarly, a force field that is given by a gra dientf=∇ϕis called a conservative force field , and the function ϕdefines the force potential. Supposeu(x) =u(x,y)isascalarfield. Givenaparametrizedcurve x(t) = (x(t),y(t))T, the composition f(t) =u(x(t)) =u(x(t),y(t)) indicates the behavior as we move along the curve. For example, if u(x,y) represents the elevation of a mountain range at position (x,y), andx(t) represents our positionat time t, thenf(t) =u(x(t)) is our altitude at time t. Similarly, if u(x,y) represents the temperature at ( x,y), thenf(t) =u(x(t)) measures our temperature at time t. The rate of change of the composite function is found through the chain rule df dt=d dtu(x(t),y(t)) =∂u ∂xdx dt+∂u ∂ydy dt=∇u·/squaresmallsolidx, (A.21) and hence equals the dot product between the gradient ∇u(x(t)) and the tangent vector /squaresmallsolidx(t) to the curve at the point x(t). For instance, our rate of ascent or descent as we travel through the mountains is given by the dot product of ou r velocity vector with the gradient of theelevationfunction. The dot product between the gradient and afixed vector a= (a,b)Tis known as the directional derivative of the scalar field u(x,y) in the direction a, and denoted by ∂u ∂a=a·∇u=aux+buy. (A.22) Thus, the rate of change of ualong a curve x(t) is given by its directional derivative ∂u/∂/squaresmallsolidx=∇u·/squaresmallsolidx, as in (A.21), in the tangent direction. This leads us to one i mportant interpretation of the gradient vector. Proposition A.10. The gradient ∇uof a scalar field points in the direction of steepest increase of u. The negative gradient, −∇u, which pointsin the opposite direction, indicates the direction of steepest decrease of u. For example, if u(x,y) represents the elevation of a mountain range at position ( x,y) on a map, then ∇utells us the direction that is steepest uphill, while −∇upoints directly downhill — the direction water will flow. Similarly, if u(x,y) represents the temperature of a two-dimensional body, then ∇utells us the direction in which it gets hottest the fastest. Heat energy (like water) will flow in the opposite direction, namely in the direction of the vector−∇u. This basic fact underlies the derivation of the multi-dime nsional heat and diffusion equations. You need to be careful in how you interpret Theorem 19.38. Cle arly, the faster you move along a curve, the faster the function u(x,y) will vary, and one needs to take this into account when comparing the rates of change along differe nt curves. The easiest way to normalize is to assume that the tangent vector a=/squaresmallsolidxhas norm 1, so /ba∇dbla/ba∇dbl= 1 and we 12/11/12 1214 c/ci∇cleco√y∇t2012 Peter J. Olver are going through xwith unit speed. Once this is done, Theorem 19.38 is an immedi ate consequence of the Cauchy–Schwarz inequality (3.18). Inde ed, /vextendsingle/vextendsingle/vextendsingle/vextendsingle∂u ∂a/vextendsingle/vextendsingle/vextendsingle/vextendsingle=|a·∇u| ≤ /ba∇dbla/ba∇dbl /ba∇dbl∇u/ba∇dbl=/ba∇dbl∇u/ba∇dbl,when /ba∇dbla/ba∇dbl= 1, withequalityifandonlyif a=c∇upointsinthesamedirectionasthegradient. Therefore, the maximum rate of change is when a=∇u//ba∇dbl∇u/ba∇dblis the unit vector in the direction of the gradient, while the minimum is achieved when a=−∇u//ba∇dbl∇u/ba∇dblpoints in the opposite direction. As a result, Theorem 19.38 tells us how to move if w e wish to minimize a scalar function as rapidly as possible. Theorem A.11. A curve x(t)will realize the steepest decrease in the scalar field u(x)if and only if it satisfies the gradient flow equation /squaresmallsolidx=−∇u, ordx dt=−∂u ∂x(x,y), dy dt=−∂u ∂y(x,y).(A.23) The only points at which the gradient does not tell us about th e directions of in- crease/decrease are the critical points , which are, by definition, points where the gradient vanishes: ∇u=0. These include local maxima or minima of the function, i.e., mountain peaks or bottoms of valleys, as well as other types of critica l points like saddle points that represent mountain passes. In such cases, we must look at the second or higher order derivatives to tell the directions of increase/decrease; s ee Section 19.3 for details. Remark: Theorem A.11 forms the basis of gradient descent methods fo r numerically approximating the maxima and minima of functions. One begin s with a guess ( x0,y0) for the minimum and then follows the gradient flow in to the min imum by numerically integratingthesystemofordinarydifferentialequations( A.23). Thisideawillbedeveloped in detail in Chapter 19. Example A.12. Consider the function u(x,y) =x2+2y2. Its gradient vector field is∇u= (2x,4y)T, and hence the gradient flow equations (A.23) take the form /squaresmallsolidx=−2x,/squaresmallsolidy=−4y. The solution that starts out at initial position ( x0,y0)Tis x(t) =x0e−2t, y (t) =y0e−4t. (A.24) Note that the origin is a stable fixed point for this linear dyn amical system, and the solutions x(t)→0converge exponentially fast to the minimum of the function u(x,y). If we start out not on either of the coordinate axes, so x0/\e}atio\slash= 0 andy0/\e}atio\slash= 0, then the trajectory (A.24) is a semi-parabola of the form y=cx2, wherec=y0/x2 0. These curves, along with the four coordinate semi-axes, are the paths to follow to get to the minimum 0the fastest. 12/11/12 1215 c/ci∇cleco√y∇t2012 Peter J. Olver Figure A.7. Level Sets and Gradient. Level Sets Letu(x,y) be a scalar field. The curves defined by the implicit equation u(x,y) =c (A.25) holding the function u(x,y) constant are known as its level sets . For instance, if u(x,y) represents the elevation of a mountain range, then its level sets are the usual contour lines on a topographic map. The Implicit Function Theorem tells us that, away from critical points, the level sets of a planar function are simple, thoug h not necessarily closed, curves. Theorem A.13. Ifu(x,y)has continuous partial derivatives, and, at a point, ∇u(x0,y0)/\e}atio\slash=0, then the level set passing through the point (x0,y0)Tis a smooth curve near the point in question. Critical points, where ∇u=0, are either isolated points, or points of intersection of level sets. For example the level sets of the function u(x,y) = 3x2−2x3+y2are plotted in Figure ls . The function has critical points at (0 ,0)Tand (1,0)T. The former is a local minimum, and forms an isolated level point, while the l atter is a saddle point, and is the point of intersection of the level curves {u= 1}. If we parametrize an individual level set by x(t) = (x(t),y(t))T, then (A.25) tells us that the composite function u(x(t),y(t)) =cis constant along the curve and hence its derivative d dtu(x(t),y(t))=∇u·/squaresmallsolidx= 0 vanishes. We conclude that the tangent vector/squaresmallsolidxto the level set is orthogonal to the gradient direction ∇uat each point. In this manner, we have established the follow ing additional important interpretation of the gradient, whic h is illustrated in Figure A.7. Theorem A.14. The gradient ∇uof a scalar field uis everywhere orthogonal to its level sets {u=c}. 12/11/12 1216 c/ci∇cleco√y∇t2012 Peter J. Olver Figure A.8. Orthogonal System of Ellipses and Parabolas. Comparing Theorems A.11 and A.14, we conclude that the curve s of steepest descent are always orthogonal (perpendicular) to the level sets of t he function. Thus, if we want to hike uphill the fastest, we should keep our direction of tr avel always perpendicular to the contour lines. Similarly, if u(x,y) represents temperature in a planar body at position (x,y) then the level sets are the curves of constant temperature, known as the isotherms . Heat energy will flow in the negative gradient direction, and hence orthogonally to the isotherms. Example A.15. Consider again the function u(x,y) =x2+2y2from Example A.12. Its level sets u(x,y) =x2+ 2y2=cform a system of concentric ellipses centered at the origin. Theorem A.14 implies that the parabolic traject ories (A.24) followed by the solutions to the gradient flow equations form an orthogonal s ystem of curves to the ellipses. This is evident in Figure A.8, showing that the ellipses and p arabolas intersect everywhere at right angles. A.5. Integrals on Curves. As you know, integrals of scalar functions,/integraldisplayb af(t)dt, are taken along real intervals [a,b]⊂R. In higher dimensional calculus, there are a variety of poss ible types of integrals. The closest in spirit to one-dimensional integration are “l ine†integrals”, in which one integrates along a curve. In planar calculus, line integral s come in three flavors. The most basicarethe integralsofscalar functions withrespect to a rclength. Such integralsareused to compute lengths of curves, and masses of one-dimensional objects like strings and wires. Thesecond andthirdvarietiesareused tointegrateavector fieldalongacurve. Integrating the tangential component of the vector field is used, for inst ance, to compute work and measure circulation along the curve. The last type integrat es the normal component of the †A more accurate term would be “curve integral”, but the terminology is stan dard and will not be altered in this text. 12/11/12 1217 c/ci∇cleco√y∇t2012 Peter J. Olver vector field along the curve, and represents flux (fluid, heat, electromagnetic, etc.) along the curve. Arc Length Thelengthof the plane curve x(t) over the parameter range a≤t≤bis computed by integrating the (Euclidean) norm†of its tangent vector: L(C) =/integraldisplayb a/vextenddouble/vextenddouble/vextenddouble/vextenddoubledx dt/vextenddouble/vextenddouble/vextenddouble/vextenddoubledt=/integraldisplayb a/radicalbig/squaresmallsolidx2+/squaresmallsolidy2dt. (A.26) The formula is justified by taking the limit of sums of lengths of small approximating line segments, [ 9]. For example, if the curve is given as the graph of a function y=f(x) for a≤x≤b, then its length is computed by the familiar calculus formul a L(C) =/integraldisplayb a/radicalBigg 1+/parenleftbiggdy dx/parenrightbigg2 dx. (A.27) It is important to verify that the length of a curve does not de pend upon any particular parametrization (or even direction of traversal) of the cur ve. Example A.16. The length of a circle x(t) =/parenleftbigg rcost rsint/parenrightbigg , 0≤t≤2π, of radiusris given by L(C) =/integraldisplay2π 0/vextenddouble/vextenddouble/vextenddouble/vextenddoubledx dt/vextenddouble/vextenddouble/vextenddouble/vextenddoubledt=/integraldisplay2π 0rdt= 2πr, verifying the well-known formula for its circumference. On the other hand, the curve x(t) =/parenleftbigg acost bsint/parenrightbigg , 0≤t≤2π, (A.28) parametrizes an ellipse with semi-axes a,b. Its arc length is given by the integral s=/integraldisplay2π 0/radicalbig a2sin2t+b2cos2t dt. Unfortunately, this integral cannot be expressed in terms o f elementary functions. It is, in fact, anelliptic integral , [145], so named for this very reason! A curve is said to be parametrized by arc length , written x(s) = (x(s),y(s))T, if one traverses it with constant, unit speed, which means that /vextenddouble/vextenddouble/vextenddouble/vextenddoubledx ds/vextenddouble/vextenddouble/vextenddouble/vextenddouble= 1 (A .29) †Alternative norms lead to alternative notions of curve length, of import ance in the study of curved spaces in differential geometry. In Einstein’s theory of rel ativity, one allows the norm to vary from point to point, and hence length will vary over space. 12/11/12 1218 c/ci∇cleco√y∇t2012 Peter J. Olver tn Figure A.9. The Moving Frame for an Ellipse. at all points. In other words, the length of that part of the cu rve between arc length parameter values s=s0ands=s1is exactly equal to s1−s0. To convert from a more general parameter tto arc length s=σ(t), we must compute s=σ(t) =/integraldisplayt a/vextenddouble/vextenddouble/vextenddouble/vextenddoubledx dt/vextenddouble/vextenddouble/vextenddouble/vextenddoubledt and so ds=/ba∇dbl/squaresmallsolidx/ba∇dbldt=/radicalbig/squaresmallsolidx2+/squaresmallsolidy2dt.(A.30) Theunit tangent to the curve at each point is obtained by differentiating with respect to the arc length parameter: t=dx ds=/squaresmallsolidx /ba∇dbl/squaresmallsolidx/ba∇dbl=/parenleftBigg/squaresmallsolidx/radicalbig/squaresmallsolidx2+/squaresmallsolidy2,/squaresmallsolidy/radicalbig/squaresmallsolidx2+/squaresmallsolidy2/parenrightBigg ,so that /ba∇dblt/ba∇dbl= 1.(A.31) (As always, we require/squaresmallsolidx/\e}atio\slash=0.) Theunit normal to the curve is orthogonal to the unit tangent, n=t⊥=/parenleftbiggdy ds,−dx ds/parenrightbigg =/parenleftBigg/squaresmallsolidy/radicalbig/squaresmallsolidx2+/squaresmallsolidy2,−/squaresmallsolidx/radicalbig/squaresmallsolidx2+/squaresmallsolidy2/parenrightBigg ,so that/ba∇dbln/ba∇dbl= 1, n·t= 0.(A.32) At each point on the curve, the vectors t,nform an orthonormal basis of R2known as the moving frame along the curve. For example, for the ellipse (A.28) with sem i-axesa,b, the unit tangent and normal are given by t=1 a2+b2/parenleftbigg −asint bcost/parenrightbigg , n=1 a2+b2/parenleftbigg bcost asint/parenrightbigg , and graphed in Figure A.9. Actually, a curve has two unit norm als at each point — one points to our right side and the other to our left side as we mov e along the curve. The normalnin (A.32) is the right-handed normal, and is the traditional one to choose; the opposite, left-handed normal is its negative −n. If we traverse a simple closed curve in a counterclockwise direction, then the right-handed norma lnis the unit outward normal, pointing to the curve’s exterior. 12/11/12 1219 c/ci∇cleco√y∇t2012 Peter J. Olver Arc Length Integrals We now explain how to integrate scalar functions along curve s. Suppose first that C is a (piecewise) smooth curve that is parametrized by arc len gth,x(s) = (x(s),y(s)) for 0≤s≤ℓ, whereℓis the total length of C. Ifu(x) =u(x,y) is any scalar field, we define itsarc length integral along the curve Cto be /integraldisplay Cuds=/integraldisplayℓ 0u(x(s),y(s))ds. (A.33) For example, if ρ(x,y) represents the density at position ( x,y) of a wire bent in the shape of a curveC, then the arc length integral/integraldisplay Cρ(x,y)dscomputes the total mass of the wire. In particular, the length of the curve is (tautologously) gi ven by L(C) =/integraldisplay Cds=/integraldisplayℓ 0ds=ℓ. (A.34) If we use an alternative parametrization x(t), witha≤t≤b, then the arc length integral is computed using the change of parameter formula (A.30), an d so /integraldisplay Cuds=/integraldisplayb au(x(t))/vextenddouble/vextenddouble/vextenddouble/vextenddoubledx dt/vextenddouble/vextenddouble/vextenddouble/vextenddoubledt=/integraldisplayb au(x(t),y(t))/radicalBigg/parenleftbiggdx dt/parenrightbigg2 +/parenleftbiggdy dt/parenrightbigg2 dt.(A.35) Changing the orientation of the curve does notalter the value of this type of line integral. Moreover, if we break up the curve into two nonoverlapping pi eces, then the arc length integral decomposes into a sum: /integraldisplay Cuds=/integraldisplay −Cuds,/integraldisplay Cuds=/integraldisplay C1uds+/integraldisplay C2uds, C =C1∪C2.(A.36) Example A.17. A circular wire radius 1 has density proportional to the dist ance of the point from the xaxis. The mass of the wire is computed by the arc length integr al /contintegraldisplay C|y|ds=/integraldisplay2π 0|sint|dt= 4. The arc length integral was evaluated using the parametriza tionx(t) = (acost,asint)T for 0≤t≤2π, wherebyds=/ba∇dbl/squaresmallsolidx/ba∇dbldt=dt. Line Integrals of Vector Fields There are two intrinsic ways of integrating a vector field alo ng a curve. In the first version, we integrate its tangential component v·t, wheret=dx/dsis the unit tangent vector, with respect to arc length. 12/11/12 1220 c/ci∇cleco√y∇t2012 Peter J. Olver Definition A.18. Theline integral of a vector field valong a parametrized curve x(t) is given by /integraldisplay Cv·dx=/integraldisplay Cv1(x,y)dx+v2(x,y)dy=/integraldisplay Cv·tds. (A.37) To evaluate the line integral, we parametrize the curve by x(t) fora≤t≤b, and then /integraldisplay Cv·dx=/integraldisplayb av(x(t))·dx dtdt=/integraldisplayb a/bracketleftbigg v1(x(t),y(t))dx dt+v2(x(t),y(t))dy dt/bracketrightbigg dt.(A.38) This result follows from the formulae (A.30,31) for the arc l ength and unit tangent vector. In general, line integrals are independent of how the curve i s parametrized — as long as it is traversed in the same direction. Reversing the directi on of parameterization, i.e., changing the orientation of the curve, changes the sign of th e line integral — because it reverses the direction of the unit tangent. As before, line i ntegrals can be decomposed into sums over components: /integraldisplay −Cv·dx=−/integraldisplay Cv·dx,/integraldisplay Cv·dx=/integraldisplay C1v·dx+/integraldisplay C2v·dx, C=C1∪C2. (A.39) In the second formula, one must take care to orient the two par tsC1,C2in the same direction as C. Example A.19. LetCdenote the circle of radius rcentered at the origin. We will compute the line integral of the rotational vector field (A.1 9), namely /contintegraldisplay Cv·dx=/contintegraldisplay Cydx−xdy x2+y2. The circle on the integral sign serves to remind us that we are integrating around a closed curve. We parameterize the circle by x(t) =rcost, y (t) =rsint, 0≤t≤2π. Applying the basic line integral formula (A.38), we find /contintegraldisplay Cydx−xdy x2+y2=/integraldisplay2π 0−r2sin2t−r2cos2t r2dt=−2π, independent of the circle’s radius. Note that the parametri zation goes around the circle once in the counterclockwise direction. If we go around once in the clockwise direction, e.g., by using the parametrization x(t) = (rsint,rcost), then the resulting line integral equals +2π. Ifvrepresents the velocity vector field of a steady state fluid, t hen the line integral (A.37) represents the circulation of the fluid around the curve. Indeed, v·tis proportional to the force exerted by the fluid in the direction of the curve, and so the circulation integral measures the average of the tangential fluid forces around th e curve. Thus, for example, 12/11/12 1221 c/ci∇cleco√y∇t2012 Peter J. Olver the rotational vector field (A.19) has a net circulation of −2πaround any circle centered at the origin. The minus sign tells us that the fluid is circula ting in the clockwise direction — opposite to the direction in which we went around the circle . Afluidflow is irrotational ifthecirculationiszero forallclosedcurves. Anirrotati onal flow will not cause a paddle wheel to rotate — there will be just as much fluid pushing in one direction as in the opposite, and the net tangential forc es will cancel each other out. The connection between circulation and the curl of the veloc ity vector field will be made evident shortly. If the vector field happens to be the gradient of a scalar field, then we can readily evaluate its line integral. Theorem A.20. Ifv=∇uis a gradient vector field, then its line integral /integraldisplay C∇u·dx=u(b)−u(a) (A .40) equals the difference between the potential function’s valu es at the endpoints a=x(a) andb=x(b)of the curve C. Corollary A.21. LetΩbe a connected domain. A scalar field has zero gradient, ∇u(x,y) =0, for all(x,y)T∈Ω, if and only if u(x,y)≡cis constant on Ω. Proof: Indeed,, if a,bare any two points in Ω, then, by connectivity, we can find a curve connecting them. Then (A.40) implies that u(b) =u(a), which shows that uis constant. Q.E.D. Thus, the line integral of a gradient field is independent of path ; its value does not depend on how you get from point ato point b. In particular, if Cis a closed curve, then /contintegraldisplay C∇u·dx= 0, since the endpoints coincide: a=b. In fact, independence of path is both necessary and sufficient for the vector field to be a gradient. Theorem A.22. Letvbe a vector field defined on a domain Ω. Then the following are equivalent: (a)The line integral/integraldisplay Cv·dxis independent of path. (b)/contintegraldisplay Cv·dx= 0for every closed curve C. (c)v=∇uis the gradient of some potential function defined on Ω. In such cases, on any connected component, a potential funct ion can be computed by integrating the vector field u(x) =/integraldisplayx av·dx. (A.41) Hereais any fixed point (which defines the zero potential level), an d we evaluate the line integral over any curve that connects atox; path-independence says that it does not matter which curve we use to get from atox. The proof that ∇u=vis left as an exercise. 12/11/12 1222 c/ci∇cleco√y∇t2012 Peter J. Olver Example A.23. The line integral /integraldisplay Cv·dx=/integraldisplay C(x2−3y)dx+(2−3x)dy of the vector field v=/parenleftbig x2−3y,2−3x/parenrightbigTis independent of path. Indeed, parametrizing a curveCby (x(t),y(t)),a≤t≤b, leads to /integraldisplay C(x2−3y)dx+(2−3x)dy=/integraldisplayb a/bracketleftbigg (x2−3y)dx dt+(2−3x)dy dt/bracketrightbigg dt =/integraldisplayb ad dt/parenleftbig x3−3xy+2y/parenrightbig dt=/parenleftbig x3−3xy+2y/parenrightbig/vextendsingle/vextendsingle/vextendsingle/vextendsingleb t=a. The result only depends on the endpoints a= (x(a),y(a))T,b= (x(b),y(b))T, and not on the detailed shape of the curve. Integrating from a=0tob= (x,y) produces the potential function u(x,y) =x3−3xy+2y. As guaranteed by (A.41), ∇u=v. On the other hand, the line integral /integraldisplay Cv·dx=/integraldisplay C(x3−2y)dx+x2dy of the vector field v=/parenleftbig x3−2y,x2/parenrightbigTis not path-independent, and so vdoes not admit a potential function. Indeed, integrating from (0 ,0) to (1,1) along the straight line segment {(t,t)|0≤t≤1}, produces /integraldisplay C(x3−2y)dx+x2dy=/integraldisplay1 0/parenleftbig t3−2t+t2/parenrightbig dt=−5 12. On the other hand, integrating along the parabola {(t,t2)|0≤t≤1}, yields a different value/integraldisplay C(x3−2y)dx+x2dy=/integraldisplay1 0/parenleftbig t3−2t2+2t3/parenrightbig dt=1 12. Ifvrepresents a force field, then the line integral (A.37) repre sents the amount of workrequired to move along the given curve. Work is defined as forc e, or, more correctly, the tangential component of the force in the direction of mot ion, times distance. The line integral effectively totals up the infinitesimal contri butions, the sum total representing the total amount of work expended in moving along the curve. N ote that the work is independent of the parametrization of the curve. In other wo rds (and, perhaps, counter- intuitively†), the amount of work expended doesn’t depend upon how fast yo u move along the curve. †Thereason thisdoesn’tagreewithourintuitionaboutworkisthatwearen ottakingfrictional effects into account, and these are typically velocity-dependent . 12/11/12 1223 c/ci∇cleco√y∇t2012 Peter J. Olver According to Theorem A.22, the work does not depend on the rou te you use to get from one point to the other if and only if the force field admits a potential function: v=∇u. Then, by (A.40), the work is just the difference in potential at the two points. In particular, for a gradient vector field there is no net work re quired to go around a closed path. Flux The second type of line integral is found by integrating the n ormal component of the vector field along the curve:/integraldisplay Cv·nds. (A.42) Using the formula (A.32) for the unit normal, we find that the i nner product can be rewritten in the alternative form v·n=v1dy ds−v2dx ds=v⊥·t, wheret=dx/dsis the unit tangent, while v⊥= (−v2,v1)T(A.43) is a vector field that is everywhere orthogonal to the velocit y vector field v= (v1,v2)T. Thus, the normal line integral (A.42) can be rewritten as a ta ngential line integral /integraldisplay Cv·nds=/integraldisplay Cv1dy−v2dx=/integraldisplay Cv∧dx=/integraldisplay Cv⊥·dx=/integraldisplay Cv⊥·tds.(A.44) Ifvrepresents the velocity vector field for a steady-state fluid flow, then the inner product v·nwith the unit normal measures the fluxof the fluid flow across the curve at the given point. The flux is positive if the fluid is moving in th e normal direction nand negative if it is moving in the opposite direction. If the vec tor field admits a potential, v=∇u, then the flux v·n=∇u·n=∂u ∂n(A.45) equals its normal derivative , i.e., the directional derivative of the potential functio nuin the normal direction to the curve. The line integral/integraldisplay Cv·ndssums up the individual fluxes, and so represents the total flux across the curve, mean ing the total volume of fluid that passes across the curve per unit time — in the direction a ssigned by the unit normal n. In particular, if Cis a simple closed curve and nis the outward normal, then the flux integral (A.42) measures the net outflow of fluid across C; if negative, it represents an inflow. The total flux is zero if and only if the total amount of fl uid contained within the curve does not change. Thus, in the absence of sources or sink s, an incompressible fluid, such as water, will have zero net flux around any closed curve s ince the total amount of fluid within any given region cannot change. 12/11/12 1224 c/ci∇cleco√y∇t2012 Peter J. Olver Figure A.10. Double Integration Domain. Example A.24. For the radial vector field v=x=/parenleftbigg x y/parenrightbigg ,we have v⊥=/parenleftbigg −y x/parenrightbigg . As we saw in Example A.7, vrepresents the fluid flow due to a source at the origin. Thus, the resulting fluid flux across a circle Cof radiusris computed using the line integral /contintegraldisplay Cv·nds=/contintegraldisplay Cxdy−ydx=/integraldisplay2π 0r2sin2t+r2cos2tdt= 2πr2. Therefore, the source fluid flow has a net outflow of 2 πr2units across a circle of radius r. This is not an incompressible flow! A.6. Double Integrals. We assume that the student is familiar with the foundations o f multiple integration, and merely review a few of the highlights in this section. Giv en a scalar function u(x,y) defined on a domain Ω, its double integral /integraldisplay/integraldisplay Ωu(x,y)dxdy=/integraldisplay/integraldisplay Ωu(x)dx (A.46) is equal to the volume of the solid lying underneath the graph ofuover Ω. If u(x,y) represents the density at position ( x,y)Tin a plate having the shape of the domain Ω, then the double integral (A.46) measures the total mass of th e plate. In particular, area Ω =/integraldisplay/integraldisplay Ωdxdy is equal to the areaof the domain Ω. In the particular case when Ω =/braceleftbig ϕ(x)<y<ψ(x), a<x<b/bracerightbig (A.47) is given as the region lying between the graphs of two functio ns, as in Figure A.10, then we can evaluate the double integral by repeated scalar integ ration, /integraldisplay/integraldisplay Ωu(x,y)dxdy=/integraldisplayb a/parenleftBigg/integraldisplayψ(x) ϕ(x)u(x,y)dy/parenrightBigg dx, (A.48) 12/11/12 1225 c/ci∇cleco√y∇t2012 Peter J. Olver inthetwocoordinatedirections. Fubini’sTheoremstatest hatonecanequallywellevaluate the integral in the reverse order /integraldisplay/integraldisplay Ωu(x,y)dxdy=/integraldisplayd c/parenleftBigg/integraldisplayβ(y) α(y)u(x,y)dx/parenrightBigg dy (A.49) in the case Ω =/braceleftbig α(y)<x<β(y), c<y<d/bracerightbig (A.50) lies between the graphs of two functions of y. Example A.25. Compute the volume of the solid lying under the positive part of the paraboloid z= 1−x2−y2. Note that z >0 if and only if x2+y2<1, and hence we should evaluate the double integral /integraldisplay/integraldisplay Ω(1−x2−y2)dxdy over the unit disk Ω =/braceleftbig x2+y2<1/bracerightbig . We may represent the disk in the form (A.47), so that Ω =/braceleftbig −/radicalbig 1−x2<y</radicalbig 1−x2,−1<x<1/bracerightbig . Therefore, we evaluate the volume by repeated integration /integraldisplay/integraldisplay Ω[1−x2−y2]dxdy=/integraldisplay1 −1/bracketleftBigg/integraldisplay√ 1−x2 −√ 1−x2(1−x2−y2)dy/bracketrightBigg dx =/integraldisplay1 −1/bracketleftbigg (y−x2y−1 3y3)/vextendsingle/vextendsingle/vextendsingle√ 1−x2 y=−√ 1−x2/bracketrightbigg dx=/integraldisplay1 −14 3(1−x2)3/2dx=1 2π. The final integral is most easily effected via a trigonometric substitution. Alternatively, and much easier, one can use polar coordinat es to evaluate the integral. The unit disk takes the form D={0≤r<1,0≤θ<2π}, and so /integraldisplay/integraldisplay D(1−x2−y2)dxdy=/integraldisplay/integraldisplay D(1−r2)rdrdθ=/integraldisplay1 0/parenleftbigg/integraldisplay2π 0(r−r3)dθ/parenrightbigg dr=1 2π. We are using the standard formula dxdy=rdrdθ (A.51) for the area element in polar coordinates, [ 9]. Thepolarintegrationformula(A.51)isaconsequenceofthe generalchangeofvariables formula for double integrals. If x=x(s,t), y =y(s,t), is an invertible change of variables that maps ( s,t)T∈Dto (x,y)T∈Ω, then /integraldisplay/integraldisplay Ωu(x,y)dxdy=/integraldisplay/integraldisplay DstU(s,t)/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂(x,y) ∂(s,t)/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (A.52) 12/11/12 1226 c/ci∇cleco√y∇t2012 Peter J. Olver HereU(s,t) =u(x(s,t),y(s,t)) denotes the function when rewritten in the new variables, while ∂(x,y) ∂(s,t)= det/parenleftbigg xsxt ysyt/parenrightbigg =∂x ∂s∂y ∂t−∂x ∂t∂y ∂s(A.53) is theJacobian determinant of the functions x,ywith respect to the variables s,t, which measures the local change in area under the map. In the event that the domain of integration is more complicat ed than either (A.47) or (A.50), then one performs “surgery” by chopping up the domai n Ω = Ω1∪Ω2∪···∪Ωk into smaller pieces. The pieces Ωiare not allowed to overlap, and so have at most their boundary curves in common. The double integral /integraldisplay/integraldisplay Ωu(x,y)dxdy=/integraldisplay/integraldisplay Ω1u(x,y)dxdy+···+/integraldisplay/integraldisplay Ωku(x,y)dxdy (A.54) can then be evaluated as a sum of the double integrals over the individual pieces. A.7. Green’s Theorem. For double integrals, the role of the Fundamental Theorem of Calculus is played by Green’s Theorem . The Fundamental Theorem relates an integral over an interv alI= [a,b] to an evaluation at the boundary ∂I={a,b}, which consists of the two endpoints of the interval. In a similar manner, Green’s Theorem relates cert ain double integrals over a planar domain Ω to line integrals around its boundary curve( s)∂Ω. Theorem A.26. Letv(x)be a smooth vector field defined on a bounded domain Ω⊂R2. Then the line integral of varound the boundary ∂Ωequals the double integral of the curl of vover the domain. This result can be written in either of the eq uivalent forms /integraldisplay/integraldisplay Ω∇∧vdx=/contintegraldisplay ∂Ωv·dx,/integraldisplay/integraldisplay Ω/parenleftbigg∂v2 ∂x−∂v1 ∂y/parenrightbigg dxdy=/contintegraldisplay ∂Ωv1dx+v2dy. (A.55) An outline of the proof appears in Exercise A.7.8. Green’s Th eorem was first formu- lated in 1828 by the English mathematician and miller George Green, and, contempora- neously, by the Russian mathematician Mikhail Ostrogradsk i. Example A.27. Let us apply Green’s Theorem A.26 to the particular vector fie ld v= (0,x)T. Since∇∧v≡1, we find /contintegraldisplay ∂Ωxdy=/integraldisplay/integraldisplay Ωdxdy= area Ω. (A.56) This means that we can compute the area of a planar domain by co mputing the indicated line integral around its boundary! For example, to compute t he area of a disk Drof radius r, we parametrize its bounding circle Crby (rcost,rsint)Tfor 0≤t≤2π, and compute areaDr=/contintegraldisplay Crxdy=/integraldisplay2π 0r2cos2tdt=πr2. 12/11/12 1227 c/ci∇cleco√y∇t2012 Peter J. Olver If we interpret vas the velocity vector field associated with a steady state flu id flow, then the right hand side of formula (A.55) represents the cir culation of the fluid around the boundary of the domain Ω. Green’s Theorem implies that th e double integral of the curl of the velocity vector must equal this circulation line integral. According to Exercise , if we divide the double integral in (A.55) by the area of the domain, 1 area Ω/integraldisplay/integraldisplay Ω∇∧vdx= MΩ[∇∧v], we obtain the mean of the curl ∇ ∧vof the vector field over the domain. In particular, if the domain Ω is very small, then ∇ ∧vdoes not vary much, and so its value at any point in the domain is more or less equal to the mean. On the oth er hand, the right hand side of (A.55) represents the circulation around the bounda ry∂Ω. Thus, we conclude that the curl ∇∧vof the velocity vector field represents an “infinitesimal cir culation” at the point it is evaluated. In particular, the fluid is irrotation al, with no net circulation around any curve, if and only if ∇∧v≡0 everywhere. Under the assumption that its domain of definition is simply connected, Theorem A.22 tell us that thi s is equivalent to the existence of a velocity potential uwith∇u=v. Theorem A.28. A vector field vdefined on a simply connected domain Ω⊂R2 admits a potential, v=∇ϕfor someϕ:Ω→Rif and only if ∇∧v≡0. We can also apply Green’s Theorem A.26 to flux line integrals o f the form (A.42). Using the identification (A.44) followed by (A.55), we find th at /contintegraldisplay ∂Ωv·nds=/contintegraldisplay ∂Ωv⊥·dx=/integraldisplay/integraldisplay Ω∇∧v⊥dxdy. However, note that the curl of the orthogonal vector field (A. 43), namely ∇∧v⊥=∂v1 ∂x+∂v2 ∂y=∇·v, (A.57) coincides with the divergence of the original velocity field. Combining these together, we have proved the divergence or flux form of Green’s Theorem: /integraldisplay/integraldisplay Ω∇·vdxdy=/contintegraldisplay ∂Ωv·nds. (A.58) As before, Ω is a bounded domain, and nis the unit outward normal to its boundary ∂Ω. In the fluid flow interpretation, the right hand side of (A.58) represents the net fluid flux out of the region Ω. Thus, the double integral of the diver gence of the flow vector must equal this net change in area. Thus, in the absence of sou rces or sinks, the divergence of the velocity vector field, ∇·vwill represent the local change in area of the fluid at each point. In particular, if the fluid is incompressible if and on ly if∇·v≡0 everywhere. An ideal fluid flow is both incompressible, ∇ ·v= 0, and irrotational, ∇ ∧v=0. Assuming its domain is simply connected, we introduce veloc ity potential u(x,y), so∇u= v. Therefore, 0 =∇·v=∇·∇u=uxx+uyy. (A.59) 12/11/12 1228 c/ci∇cleco√y∇t2012 Peter J. Olver Therefore, the velocity potential for an incompressible, i rrotational fluid flow is a harmonic function, i.e., it satisfies the Laplace equation! Water wav es are typically modeled in this manner, and so many problems in fluid mechanics rely on th e solution to Laplace’s equation. 12/11/12 1229 c/ci∇cleco√y∇t2012 Peter J. Olver