7.1 GREEN S FIRST IENTITY MATH 14B Solution Key HW 05 7.1 GREEN S FIRST IENTITY 1. erive the 3-dimensional maximum principle from the mean value property. SOLUTION. We aim to prove that if u is harmonic in the bounded set 3 and u is continuous on = bdy then the maximum and minimum of u occurs on bdy and nowhere else, provided that u is not constant. We proceed by proving the contrapositive. That is, if u admits an interior extremum then u must be constant. Since u is continuous on the closed, bounded set, then u must attain both its maximum, say M and its minimum, say m, somewhere in. Suppose that u attains an interior extremum at the point x 0 and let x be any other point in. Since is open and connected in 3 it follows that is polygonally path connected. 1 Let Γ be the polygonal path in joining x 0 and x and let δ be the shortest distance separating Γ and bdy : δ = min{ x y : x Γ, x bdy }. By the compactness of there exists a sequence of balls B R (x i ), for i = 0, 1,..., n, with x i on Γ, satisfying R δ, x i+1 in B R (x i ), x in B R (x n ). By the mean-value property we see that u is identically equal to M in each B R (x i ), i = 0, 1,,... n; hence, u(x ) = M. Since our choice of x was completely arbitrary, u must be identically equal to M throughout and, by continuity of u, throughout all of. Indeed, this proves that if u is not constant in, then u attains its maximum value on the boundary of and nowhere else. Similarly, this argument applied to u establishes that if u is non-constant, it can attain its minimum value on bdy and nowhere else.. Prove the uniqueness up to constant of the Neumann problem using the energy method. SOLUTION. Suppose that u 1 and u solve the Neumann problem for harmonic functions, that is, u = 0 in = g(x) on bdy. 1 This is an elementary result from point-set topology in Math 118A The existence of such a number δ > 0 follows by the continuity of the norm and the compactness of. Again, a result in math 118A. 1 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 013
7.1 GREEN S FIRST IENTITY Put w = u 1 u, then w solves the homogeneous Neumann problem w = 0 in w = 0 in bdy. An application of Green s first identity makes it plain that w w ds = w w d x + bdy Since w/ = 0 on bdy and w = 0 in, we see that 0 = w w d x = w d x. w w d x. Since the integrant w 0, it follows by the vanishing theorem that w 0 in, that is, w 0 in and we deduce that w is constant in, say w C. By our construction we see that u 1 = u + C in and we have established uniqueness of the Neumann problem up to a constant as required. 5. Prove irichlet s principle for the Neumann boundary condition. It asserts that among all real-valued functions w(x) on the quantity E[w] = 1 w d x bdy hw ds is the smallest for w = u, where u is the solution of the Neumann problem u = 0 in, = h(x) on bdy. It is required to assume that the average of the given function h(x) is zero (by Exercise 6.1.11). Notice three features of this principle (i) There is no constraint at all on the trail functions w(x). (ii) The function h(x) appears in the energy. (iii) The functional E[w] does not change if a constant is added to w(x). Hint: Follow the method in Section 7.1. Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 013
7.1 GREEN S FIRST IENTITY SOLUTION. For the sake of completeness we present the solution to Exercise 6.1.11. Suppose that u solves the Neumann problem u = f in = g on bdy. By taking v 1 in Green s 1st identity we see that ds = u d x. bdy Indeed, this imposes a necessary condition in order for the Neumann problem to be well-posed. More specifically, it must be the case that bdy g(x) ds = f (x) d x. Armed with this result we now return to our original problem and see that it must be the case that for f (x) = 0 h(x) ds = ds = 0. bdy In other words, the average of the given function h(x) is zero. By Exercise, the solution to the well-posed Neumann problem is unique up to a constant. Fix a solution, say u bdy 3 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 013
7.1 GREEN S FIRST IENTITY and let w be any other trial function. By putting v = u w and we see that E[w] = 1 = 1 w d x bdy u v d x hw ds bdy = 1 ( u v) ( u v) d x h(u v) ds bdy = 1 ( u u u v + v v) d x = 1 ( u u v + v ) d x = 1 u d x = E[u] + 1 v d x = E[u] + 1 v d x = E[u] + 1 v d x + bdy h(u v) ds bdy hu ds + 1 v d x u vd x + u vd x + v u d x, bdy bdy bdy h(u v) ds h(u v) ds hv ds v ds u vd x + bdy hv ds where we invoke Green s 1st identity for the last line. Now, since u = 0 in we obtain that E[w] E[u] = 1 Since the integrant v 0 it follows that E[w] E[u] 0, v d x. hence; and the desired conclusion follows. E[w] E[u] 6. Let A and B be two disjoint bounded spatial domains, and let be their exterior. So bdy = bdy A bdy B. Consider a harmonic function u(x) in that tends to zero at infinity, which is constant on bdy A and constant on bdy B, and which satisfies bdy A ds = Q > 0 and bdy B ds = 0. Interpretation: The harmonic function u(x) is the electrostatic potential of two conductors, A and B; Q is the charge on A, while B is uncharged. 4 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 013
7.1 GREEN S FIRST IENTITY (a) Show that the solution is unique. Hint: Use the Hopf maximum principle. (b) Show that u 0 in. Hint: If not, then u(x) has a negative minimum. Use the Hopf principle again. (c) Show that u > 0 in. SOLUTION. First, we recall the Hopf form of the maximum principle as presented in Exercise 1(a) on page 177 of our textbook. If u(x) is a non-constant harmonic function in a connected plane domain with a smooth boundary that has a maximum at x 0 (necessarily on the boundary by the strong maximum principle), then / > 0 at x 0 where n is the unit outward normal vector. (a) The Hopf principle is pis aller since its proof is outside the scope of this class. Accordingly, we present an alternative proof. Suppose that u 1 and u solve our PE of interest, namely, u = 0, in, u = a, on bdy A, u = b, on bdy B, ds = Q > 0 bdy A and bdy B u(x) 0, uniformly, as x. Put w = u 1 u, then w solves w = 0, in, w = 0, on bdy = bdy A bdy B, bdy w ds = bdy A w ds = w(x) 0, uniformly, as x. bdy B ds = 0 w ds = 0 Let x 0 be an arbitrary point in. Since w(x) 0 as x, it follows that given ɛ > 0, the radius R can be chosen sufficiently large so that x 0 < R, and w(x) < ɛ for x R. In the bounded region R defined by the intersection of and the ball x < R, the maximum principle applies. Since the boundary of R consists partly of bdy and partly of the surface of the sphere x = R, and since w 0 on bdy while w < ɛ on x = R, if follows that w(x) < ɛ throughout R. In particular, 5 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 013
7.1 GREEN S FIRST IENTITY w(x 0 ) < ɛ. Since our choice of ɛ may be arbitrarily small it follows that w(x 0 ) = 0, hence w(x 0 ) = 0. Also, since the choice of x 0 was arbitrary we see that w 0, or u 1 = u in and we have established uniqueness. (b) First note that it cannot be the case that u is constant, otherwise u 0 since u vanishes at infinity. More specifically, u 0, so n u = / 0, which in turn contradicts the hypothesis that the charge on A is positive. Second, when u is assumed to be non-constant we argue by means of a contradiction. Suppose that there exist a point x such that u(x ) < 0. By applying the Hopf principle to u we see that the minimum of u, that is, the maximum of u is attained on bdy and nowhere else, say at x 0, and /(x 0 ) > 0, or equivalently /(x 0 ) < 0. If the minimum is attained on bdy A, then the condition ds = Q > 0, bdy A makes it plain that it cannot be the case that / is negative on all of bdy A. More specifically, there must exist a point x A bdy A such that /(x A ) > 0, but then when moving outward along the normal directional derive it must be the case that the value of u is decreasing as we move away from the boundary of A into our domain, inasmuch u is constant on bdy A. But this contradicts our assumption that the minimum value of u is attained on bdy A and nowhere else. Conversely, if the minimum of u is attained on bdy B, then the condition ds = 0, bdy B along with the fact that / 0 on bdy B makes it plain that there must exist a point x B bdy B such that /(x B ) > 0, but then again we arrive at a contradiction. Therefore it follows that u 0 on. (c) We argue by means of a contradiction. Suppose that there exists a point x where u(x ) = 0. Since u cannot be constant it follows by the continuity of u there exist a small enough neighborhood along x where u is negative. By applying the argument from part (b) to a point in such a neighborhood we arrive at yet another contradiction. 7. (Rayleigh-Ritz approximation to the harmonic function u in with u = h on bdy.) Let w 0, w 1,..., w n be arbitrary functions such that w 0 = h on bdy and w 1 = = w n = 0 on bdy. The problem is to find constants c 1, c,..., c n so that w 0 + c 1 w 1 + + c n w n had the least possible energy. Show that the constants must solve the linear system ( w j, w k )c k = ( w 0, w j ) for j = 1,,..., n. k=1 6 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 013
7.1 GREEN S FIRST IENTITY SOLUTION. First, recall the inner-product notation and put ( w j, w k ) = w j w k d x = ( w k, w j ) w = w 0 + c 1 w 1 + + c n w n = w 0 + c j w j, where w 0, c j, and w j have the properties described in the statement of our problem for j = 1,,..., n. It follows that the energy is given by E[w] = 1 = 1 w d x w w d x = 1 ( w, w) = 1 ( w 0 + c j w j, w 0 + = 1 ( w 0, w 0 ) + ( w 0, = 1 ( w 0, w 0 ) + ( w 0, = 1 ( w 0, w 0 ) + = 1 ( w 0, w 0 ) + c j w j ) c j w j ) + ( c j w j, w 0 ) + ( c j w j, c j w j ) + ( c j w j, ( w 0, w j )c j + ( w 0, w j )c j + 1 c j w j ) ( w j, w j )c + j ( w j, w j )c + j k=1 j k k=1 j k c j w j ) ( w k, w j )c k c j ( w k, w j )c k c j. For the choice of coefficients c k for k = 1,,..., n which minimize E, the partial derivative of E with respect to each c k must vanish. In other words, for each k = 1,,..., n, it follows that c k E[w] = 0. More specifically, 0 = ( w 0, w k ) + ( w k, w k )c k + ( w k, w j )c j = ( w 0, w k ) + ( w k, w j )c j, j k 7 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 013
7.1 GREEN S FIRST IENTITY hence; as required. ( w k, w j )c j = ( w 0, w k ) for k = 1,,..., n, 8 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 013