Math 342 Partial Differential Equations «Viktor Grigoryan

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1 Math 342 Partial ifferential Equations «Viktor Grigoryan 3 Green s first identity Having studied Laplace s equation in regions with simple geometry, we now start developing some tools, which will lead to representation formulas for harmonic functions in general regions. The fundamental principle that we will use throughout is the ivergence theorem, which states that divfdx= F nds () for a vector field F defined in the closure of an open solid region. In the above identity stands for the boundary of the region, n is the outward normal vector to this boundary, and the integral on the right hand side is taken with respect to the area element of the boundary. One can think of the divergence theorem as the generalization of the Fundamental Theorem of Calculus to higher dimensions. Indeed, it states the equality of the integral of the derivative of F inside the region and the values of F on the boundary. We will routinely write F for the divergence. We consider two functions, u,v, both of two or three (or more) variables. From the product rule, we have (vu x ) x =v x u x +vu xx, (vu y ) y =v y u y +vu yy, and similarly for the other partial derivatives. Adding up these identities gives (v u)= u v+v u. Integrating both sides of the above identity over the region, and using the divergence theorem for the integral on the left hand side, we get v ds = u vdx+ v udx, (2) since u n=/. This is Green s first identity. Rewriting (2) as v udx= v ds u vdx, we can think of this identity as the generalization of integration by parts, in the sense that one derivative is transferred from the function u to the function v under the integral, which results in a switched sign and a boundary term, much like the integration by parts for functions of single variable. In particular, we can take v in (2), which will give ds = udx. (3) The last identity implies a necessary condition for the Neumann problem to be well posed. Indeed, applying (3) to the solution of the Neumann problem { u=f(x) in, =h(x) on, (4) we have that necessarily h(x)ds = f(x)dx. (5) This means that unless h and f satisfy (5), the Neumann problem for the Poisson s equation cannot have a solution. For nice enough functions f and h, one can show that the solution to the Neumann problem exists, if (5) is satisfied. However uniqueness of the solution to the Neumann problem holds only up to a constant, since adding a constant to any solution of (4) will give another solution. 3. Mean value property We previously used Poisson s formula for the solution of the irichlet problem on the circle to prove the mean value property of harmonic functions in two dimensions. A similar property holds in three and higher dimensions, and we next prove this property in three dimensions using Green s first identity (2). The mean value property in three dimensions states that the average value of a harmonic function over any sphere is equal to its value at the center. Without loss of generality, we can assume that the sphere is centered at the origin, since we can use translation invariance of Laplace s equation to translate the coordinate system so that the center of the sphere coincides

2 with the origin. Let ={ x =a} be a sphere centered at the origin with radius a. The outward normal to the sphere has the direction of x, and hence, n= x x = x r, and = u n= u x r = r, since r= x 2 +y 2 +z 2 and the chain rule imply that x = x r r, y = y r r, z = z r r, and subsequently u x r = x x r + y y r + z z r = x 2 +y 2 +z 2 = r r 2 r. Using the identity /=/ r, and the fact that u is harmonic, (3) can be written as ds =0. r Converting to spherical coordinates, the above integral over the sphere r =a takes the form ˆ 2π ˆ π 0 0 u r (a,θ,φ)a 2 sinθdθdφ=0. We divide both sides of the above identity by the surface area of the sphere, 4πa 2, and notice that, since the integration is with respect to φ and θ, we can pull out the derivative with respect to the radial variable r outside of the integral, ( [ ˆ 2π ˆ π u(r,θ,φ)a sinθdθdφ]) 2 =0. r 4πa r=a Since the radius a is arbitrary, the above [ derivative ˆ must vanish for every value of a, that is, 2π ˆ π ] u(r,θ,φ)r 2 sinθdθdφ =0 r 4πr for all values of r>0. But the expression in the square brackets is exactly the average of the harmonic function u over the sphere { x =r}, so it is constant with respect to the radius r, and hence the average of u over concentric circles centered at the origin is independent of the radii of the circles. In particular passing to the limit r 0 gives: Average of u over the sphere S ={ x =r}= ˆ 2π ˆ π u(r,θ,φ)r 2 sinθdθdφ 4πr 2 =lim r 0 4π ˆ 2π ˆ π 0 0 ˆ 2π ˆ π u(r,θ,φ)sinθdθdφ= u(0)sinθdθdφ=u(0), 0 0 4π 0 0 since ˆ 2π ˆ π sinθdθdφ= 4π 0 0 as the average of the constant function over a unit sphere. The mean value property works in higher dimensions as well, and can be proved by exactly the same method, making use of the Green s first identity in higher dimensions. Just as we showed in two dimensions, the mean value property implies the maximum principle for harmonic functions, i.e. that the maximum of a harmonic function u is assumed on the boundary of open connected domains, and cannot be assumed inside the domain, unless u is everywhere constant. The idea is again based on the fact that, if the harmonic function assumes a maximum at an inner point, then it must be constant in any disk centered at this point, which lies entirely inside, due to the mean value property. 3.2 Uniqueness of irichlet s and Neumann s problems Previously we saw that the maximum principle implies uniqueness of solutions to the irichlet problem for Laplace s, and more generally Poisson s equations. Today we will see that an energy method, similar to ones used for the wave and heat equations, also implies the uniqueness for the irichlet, as well as the Neumann problem for Poisson s equation. 2

3 Let us consider Poisson s equation with either irichlet or Neumann boundary conditions, u=f(x) in u =h, or =g. Assuming that the above problem has two solutions, u and u 2,their difference, w=u u 2, will solve Laplace s equation, i.e. will be harmonic, in, and will have vanishing irichlet (Neumann) data on the boundary. But then multiplying both sides of the equation w=0 by w, and using Green s first identity (2), we get w wdx= w w ds w 2 dx=0. The integral over the boundary is zero due to the vanishing irichlet or Neumann boundary data of w, and consequently, w 2 dx=0. The function w 2 is nonnegative, and the only way the above integral can be zero, is if w 0 in. Hence, w constant in. In the case of the irichlet condition, we have w 0 on, and hence u u 2 =w 0 in as well. In the case of Neumann condition, all we can assert is that u u 2 =w constant in, that is, the solution to the Neumann problem is unique up to a constant. 3.3 irichlet s principle The quantity encountered in the above energy method has a physical meaning of potential energy. More specifically, we define the energy of a function w defined in to be E[w]= w 2 dx. (6) 2 Recall that Laplace s equation describes steady states of oscillatory (wave) or heat conduction processes, and hence we expect that the solutions of Laplace s equation will have the smallest possible energy as the preferred equilibrium states. Mathematically, this can be stated as the following. irichlet s principle. Among all functions w(x) defined in that satisfy the boundary condition w = h on, the function that has the least energy is the unique harmonic function with irichlet data h on. That is, if u is a harmonic function, then E[w] E[u] for any function w that has the same values as u on the boundary, w=u=h on. The proof of irichlet s principle follows from Green s first identity. Let u be harmonic in, and w=u=h on, then the function v=u w vanishes on the boundary. Writing w=u v, we have E[w]= 2 (u v) 2 dx= ( u v) ( u v)dx 2 = ( u 2 2 u v+ v 2 )dx=e[u] u vdx+e[v]. 2 From (2), we see that u vdx= v udx+ v ds =0, since the first integral on the right vanishes due to u being harmonic, and the second integral vanishes due to v vanishing on the boundary. But then E[w]=E[u]+E[v] E[u], since E[v] 0 as an integral of a nonnegative function. One can prove that the converse of the irichlet s principle is true as well, i.e. that the function minimizing the energy must be harmonic. In this sense harmonic functions are solutions to the variational problem associated with the energy (6), and Laplace s equation is exactly the Euler-Lagrange equation for this Lagrangian. 3

4 3.4 Conclusion Starting from the divergence theorem we derived Green s first identity (2), which can be thought of as integration by parts in higher dimensions. Using this identity, we proved several properties of harmonic functions in higher dimensions, namely, the mean value property, which implies the maximum principle; uniqueness of the irichlet and Neumann problems for Poisson s equation; and irichlet s principle. The mean value property obtained in this way generalized the mean value property in two dimensions, previously proved by Poisson s formula for the solution of irichlet s problem for the circle. Green s first identity will be used next time to derive Green s second identity, which will lead to representation formulas for harmonic functions. 4

5 32 Green s second identity, Green s functions Last time we derived Green s first identity for the pair of functions (u,v), which in three dimensions can be written as v udx= v ds u vdx. (7) Interchanging u and v, we can also write the Green s first identity for the pair (v,u), u vdx= u v ds v udx. (8) Notice that the last terms of (7) and (8) are identical, thus subtracting the first identity from the second one, we obtain ( (u v v u)dx= u v ) v ds. (9) This is Green s second identity for the pair of functions (u,v). Similar to the notion of symmetric boundary conditions for the heat and wave equations, one can define symmetric boundary conditions for Laplace s equation, by requiring that the right hand side of (9) vanish for any functions u,v satisfying the boundary conditions. It is not hard to see that (homogeneous) irichlet, Neumann and Robin boundary conditions are all symmetric. 32. Representation formula Green s second identity (9) leads to the following representation formula for the solution of the irichlet problem in a domain. If u=0 in, then for [ any point x 0, u(x 0 )= u(x) ( ) ] + ds. (0) 4π x x 0 4π x x 0 Indeed, the above formula is essentially a direct application of (9) to the pair of functions u(x) and v(x)= 4π x x 0, () the last function being a multiple of the radial solution x translated by the vector x 0. The particular choice of the factor /(4π) will be apparent from the proof of (0). The only issue with the application of (9) to the pair u and v arises from the singularity of v(x) at the point x 0. To circumvent this, we will apply (9) in the region with a ball centered at x 0 of radius ɛ cut out, where ɛ will be made to approach zero in the end. efining B ɛ = {x: x x 0 < ɛ} to be a ball of radius ɛ centered at x 0, we observe that for ɛ small enough, B ɛ will entirely lie inside. In the region ɛ =\B ɛ, where \ stands for the difference of sets, both u(x) and v(x) [ are well-defined and harmonic, so (9) implies that u(x) v ] ɛ v(x) ds =0, where ɛ is the boundary of ɛ. But ɛ consists of two pieces:, and B ɛ, which have opposite orientations, in the sense that the outward normals of and ɛ coincide on their common boundary, while those of ɛ and B ɛ have opposite [ directions. Thus, 0= u(x) v ] [ ɛ v(x) ds = u(x) v ] [ v(x) ds u(x) v ] B ɛ v(x) ds. The first term on the right hand side is exactly the right [ hand side of (0), so to prove (0) it suffices to show that u(x 0 )=lim u(x) v ] ɛ 0 B ɛ v(x) ds. (2) enoting r= x x 0 = (x x 0 ) 2 +(y y 0 ) 2 +(z z 0 ) 2, we can rewrite v(x)= /(4πr), and B ɛ ={r=ɛ}. Also notice that the outward normal to the boundary of the ball B ɛ has the direction of the vector r=x x 0, hence n=r/r. But then, /= / r, [ and the integral under the limit on the right hand side of (2) can be rewritten as u(x) v ] [ v(x) ds = u ( ) + ] r 4πr r ds. 4πr B ɛ r=ɛ 5

6 On {r=ɛ}, we have ( ) = r r r = 2 ɛ 2, and r = ɛ, which after substitution [ into the previous identity gives u(x) v ] B ɛ v(x) ds = uds+ ds =u+ɛ 4πɛ 2 r=ɛ 4πɛ r=ɛ r r, where u and / r denote the average values of respectively the functions u(x) and / on the sphere x x 0 =r=ɛ. But as ɛ 0, the points [ on the sphere converge to its center, x x 0, and we have lim u(x) v ] ɛ 0 B ɛ v(x) ds =u(x 0 )+0 (x 0), where (/)(x 0 )=lim ɛ 0 /, and as such is bounded. Hence, (2) follows, which also finishes the proof of (0). Notice that for the function v(x) given by (), we have v=0 in( \B ɛ. Then (9) gives w vdx= w vdx= v w+ w v ) B ɛ B ɛ B ɛ v w ds, for any function w. But then the above arguments will imply ( that w vdx=lim w vdx=lim ɛ ɛ 0 2 w+w+ɛ ) w =w(x 0 ), B ɛ 0 ɛ v where w is bounded by some constant times the maximum of w(x) in the closed ball B ɛ, and w and w/ are the averages of respectively w(x) and w/ over the sphere B ɛ. This shows that for any smooth function w, w vdx=w(x 0 ), and hence for the function v(x) defined by () we have v=δ(x x 0 ), (3) with δ being the irac delta function. It is easy to see that the representation formula (0) follows directly from Green s second identity and (3). In general dimensions, the (distributional) solution of the equation u=δ(x) is called a fundamental solution of Laplace s equation. Comparing this to (3), we can see that in three dimensions the radial function /(4π x ) is a fundamental solution, and v(x) is an x 0 translation of it. Observe that for the representation formula (0) the fundamental solution of Laplace s equation plays a similar role to that of the heat kernel in the context of the heat equation. In two dimensions the fundamental radial solution of Laplace s equation is v(x)= 2π log x, and the corresponding representation ˆ [ formula for the solution of Laplace s equation 2 u=0 is u(x 0 )= u(x) ( ) 2π log x x 0 2π log x x 0 ] ds. (4) The above integral is a line integral over the bounding curve of a two-dimensional region, and ds denotes the arc-length element of this boundary Green s functions The representation formulas (0) and (4) are not very useful for solving boundary value problems for the Laplace equation, since they require data of both u and / on the boundary. So by themselves these formulas will not give the solution to irichlet or Neumann problems, since we would have the data for either u or /, but not both. Notice that to derive the representation formula we used Green s second identity, and relied on the fact that the function v(x) given by () has a particular kind of singularity at the point x 0, and is harmonic everywhere else. This is encompassed in equation (3), from which the representation formula follows directly. So if we 6

7 could find another function with these properties, for which in addition either the first or the second term under the integral in (0) vanishes, then we would have solution formulas for the irichlet and Neumann problems. efinition 32. (Green s functions). The function G(x) is called a Green s function for the operator in the three dimensional domain at the point x 0, if it satisfies the following properties. (i) G(x) has continuous second derivatives and is harmonic in \{x 0 }. (ii) G(x)=0 on the boundary of. (iii) G(x)+ 4π x x 0 is finite at x 0 and is harmonic in all of. In two dimensions condition (iii) must be replaced by the requirement that G(x) log x x 2π 0 is harmonic in all of. We will use the notation G(x,x 0 ) to indicate that G is the Green s function at the point x 0. It is not hard to see that the definition of the Green s function is equivalent to requiring that G(x,x 0 ) solve the following irichlet problem { G(x,x0 )=δ(x x 0 ) in, (5) G(x,x 0 )=0 on. Indeed, it is clear from (3) that G(x,x 0 ) v(x) solves Laplace s equation, and is hence harmonic in all of. It can be shown that a Green s function exists, and must be unique as the solution to the irichlet problem (5). Using Green s function, we can show the following. Theorem If G(x,x 0 ) is a Green s function in the domain, then the solution to irichlet s problem for Laplace s equation in is given by u(x 0 )= u(x) G(x,x 0) ds. (6) The proof of this theorem is a straightforward application of Green s second identity (9) to the pair (u,g). Indeed, from (9) we have ( (u G G u)dx= u G ) G ds. In the above identity the second term on the left hand side vanishes due to u being harmonic, and the second term in the right hand side integral vanishes, since G(x,x 0 )=0 on. Also, u Gdx= u(x)δ(x x 0 )dx=u(x 0 ), and (6) follows. Equipped with Theorem 32.2 we can find the solution to the irichlet problem on a domain, provided we have a Green s function in. In practice, however, it is quite difficult to find an explicit Green s function for general domains. Next time we will see some examples of Green s functions for domains with simple geometry. One can use Green s functions to solve Poisson s equation as well. Theorem If G(x,x 0 ) is a Green s function in the domain, then the solution to the irichlet s problem for Poisson s equation u=f(x) is given by u(x 0 )= u(x) G(x,x 0) ds+ f(x)g(x,x 0 )dx. To find a solution formula for the Neumann problem, condition (ii) in the definition of a Green s function must be replaced by (ii N ) G(x) =c on the boundary of for a suitable constant c. Such a Green s function would solve { the Neumann problem G(x,x0 )=δ(x x 0 ) in, G(x,x 0 ) (7) =c on. 7

8 The divergence theorem then implies that G(x,x 0 )dx= But from (7) we have G(x,x 0 )dx= δ(x x 0 )dx=, and G ds. G ds = cds =c Area( ). Hence, the constant c is given by c=/area( ). The solution formula for the Neumann problem is then u(x 0 )=C G(x,x 0 ) (x) dx, where the Green s function G(x,x 0 ) satisfies the condition (ii N ), and C is an arbitrary constant. The proof is left as an exercise for the curious student Conclusion We derived Green s second identity from Green s first identity, which was subsequently applied to the pair of a harmonic function and the fundamental radial solution of Laplace s equation to arrive at the representation formula (0). To find a harmonic function with this representation formula one needs the boundary values of both the harmonic function and its derivative with respect to the outer normal on the boundary, which is not useful for solving the classical boundary value problems (irichlet, Neumann or Robin). To circumvent this, we defined Green s function as a fundamental solution of Laplace s equation that has vanishing irichlet data on the boundary of the domain, which leads to a solution formula for the irichlet problem. A similar Green s function can be defined to solve Neumann s problem as well. Finding a Green s function for an arbitrary domain can be quite cumbersome, however. But for domains with symmetry Green s functions can be found by geometric considerations. 8