In this chapter we study elliptical PDEs. That is, PDEs of the form. 2 u = lots,

Transcription

1 Chapter 8 Elliptic PDEs In this chapter we study elliptical PDEs. That is, PDEs of the form 2 u = lots, where lots means lower-order terms (u x, u y,..., u, f). Here are some ways to think about the physical applications of elliptic PDEs. The steady-state of the heat equation solves Laplace s equation u t κ 2 u = 0, x, t > 0 u(x, 0) = f(x), x, u(x, t) = g(x), x, t > 0, κ 2 u = 0, x, u(x) = g(x), x. If we think of ρ(x) as a charge density, the potential, u, associated with the generated electric field satisfies Poisson s Equation 2 V = ρ(x) ǫ 0, x, V = f(x), x, where ǫ 0 is the permitivity of free space and the Electric field is given by E = V. Suppose Γ is a closed, non intersecting curve in R 3. The minimal surface with boundary Γ is the minimum of the functional J(u) := 1 + u 2 x + u2 y dxdy, 133

2 Elliptic PDEs where is the projection of the surface onto the xy plane and u is the height of the surface above the xy plane. Homework (8.1) will show the minimum satisfies Plateau s Equation (1 + u 2 y)u xx + (1 + u 2 x)u yy 2u x u y u xy = 0, u(x) = f(x) x. If u x and u y are small compared to one, Laplace s equation approximates Plateau s equation Bounded Domains Our spectral methods work for elliptic PDEs as well. Consider the PDE (approximating a soap film) on the square = {(x, y) 0 < x < π, 0 < y < π}: 2 u = 0, on u(0, y) = f(y), 0 < y < π, u(π, y) = g(y), 0 < y < π, u(x, 0) = 0, 0 < x < π, u(x, π) = 0, 0 < x < π. We find conditions on f and g so that the solution satisfies u C, () C(). Since we have homogeneous boundary conditions on the top can bottom of the square, we attempt to find a solution of the form u(x, y) = C n (x) sin(ny). i=1 The last two boundary conditions are satisfied. Putting this sum (formally) in the PDE, we get C n(x) n 2 C n sin(ny) = 0. i=1 Orthogonality implies C n n2 C n = 0 or C n (x) = c 1 e nx + c 2 e nx. Recall sinh(x) = 1 2 (enx e nx ) and cosh(x) = 1 2 (enx + e nx ). We may express the solution to C n n2 C n = 0 as C n (x) = c 1 e nx + c 2 e nx = c 1 sinh(nx) + c 2 cosh(nx) = c 1 sinh(nx) + c 2 sinh(n(π x)). The last version is the easiest to work with. Thus, so far, we have u(x, y) = a n sinh(nx) + b n sinh(n(π x) sin(ny). At x = 0 we need i=1 f(y) = b n sinh(nπ) sin(ny). i=1

3 8.2. The Fundamental Solution 135 Using orthogonality again, we find b n = Similarly, we find at x = π, a n = 2 π sinh(nπ) 2 π sinh(nπ) π 0 π 0 f(y) sin(πy) dy. g(y) sin(πy) dy. To get convergence we first apply Dirichlet s theorem. We need f, g to be continuously differentiable and their odd extension to be continuous. This requires f(0) = g(0) = f(π) = g(π) = 0. Next we note that, for large n, 1/ sinh(nπ) e nπ. When we put a n and b n back in the sum, we get terms like sinh(nx)/ sinh(nπ). For large n and 0 < x < π we see sinh(nx) sinh(nπ) en(x π). As in the parabolic case we see the Fourier coefficients converge exponentially fast, as long as x is away from the boundary, and so, u C, (). To get u(x, y) continuous on the closure of, we need both a n and b n to decay like 1/n 2 since the exponential decay goes away on the boundary. For this will will have to integrate both f and g by parts twice. This will require, in addition to the other conditions, f, g C 2 (0, π). Suppose we wanted to solve on the square = {(x, y) 0 < x < π, 0 < y < π}: 2 u = 0, on u(0, y) = f 1 (y), 0 < y < π, u(π, y) = f 2 (y), 0 < y < π, u(x, 0) = g 1 (x), 0 < x < π, u(x, π) = g 2 (x), 0 < x < π. We could do this by first solving, as above, the problems 2 u 1 = 0, 2 u 2 = 0, u 1 (0, y) = f 1 (y), 0 < y < π u 2 (0, y) = 0, 0 < y < π u 1 (π, y) = f 2 (y), 0 < y < π, u 2 (π, y) = 0, 0 < y < π, u 1 (x, 0) = 0, 0 < x < π, u 2 (x, 0) = g 1 (x), 0 < x < π, u 1 (x, π) = 0, 0 < x < π u 2 (x, π) = g 2 (x), 0 < x < π. The solution to our problem is then u = u 1 + u The Fundamental Solution We want to solve Poisson s equation 2 u = f with a variety of boundary conditions and domains - both bounded and unbounded. We follow our usual procedure for constructing a Green s function. Set Lu = 2 u. We need to compute the adjoint. For the moment let us suppose the domain

4 Elliptic PDEs is bounded subset of R N and normal (meaning the divergence theorem applies). Integrating by parts twice, we find v 2 u d = v u n dσ u v n dσ + u 2 v d, where dσ is the surface element on in R n 1 and single integral represents a multiple integral in R N. Rearranging we get Green s Second Identity: (8.1) (v 2 u u 2 v) d = v u n dσ u v n dσ. This shows L v = 2 v, and L is formally self adjoint. As before we will need to solve L v = δ(x x ). We have to know the boundary conditions for our problem before we can prescribe the boundary conditions on v. In general the solution will have the form, according to (8.1), with G(x, x ) = v(x, x ) u(x) = G(x, x )f(x ) d + u(x ) x G(x, x ) n dσ G(x, x ) x u(x ) n dσ, where x means the gradient with respect to the primed variables. Since 2 yg(x, y) = 0, x = y, we might expect the solution to be radially symmetric. This suggest trying to solve 2 u = 0 in polar coordinates. To do this, set r 2 = x x2 2 + x2 N, and u(x 1, x 2,... x N ) = U(r). Then u xi = U r xi r. Here xi r = xi r. Thus and We can easily solve u xix i = U rr x i r 2 u = = x i r + U r x2 i r r r 2, N u xix i = U rr + N 1 r i=1 1 N 1 u r N 1 r. r r 1 r N 1 r r N 1 u = 0. r This ODE is called a Cauchy-Euler equation. Written differently the equation is r 2 U rr + (N 1)rU r = 0. Its form suggests U = r α might work: we find (α(α 1) + (N 1)α)r α = 0. For N > 2, we get α = 0 and α = 2 N. Thus the general solution is c 1 + c 2 r 2 N. If N = 2, the Cauchy-Euler equation becomes 1 r (ru ) = 0. U r

5 8.3. Poisson s equation with Dirichlet and Neumann Boundaries 137 We can directly integrate this to find U = c 1 + c 2 ln r. We therefore have that a solution to 2 Φ(x) = 0 is c2 ln r N = 2 Φ(x) = c 2 r N 3, N 2 where we have dropped the trivial solution c 1. To find the constants we, formally, use Green s second identity, (8.1), and, for convenience, we set x = 0 and integrate over a ball or radius r. We find 1 = = = B r(0) B r(0) B r(0) = c 2 2π. 2 x Φ(x, x ) dx Φ(x, x ) 2 x 1 dx + x Φ(x, x ) n = B r(0) 2π 0 (1 x Φ(x, x ) Φ(x, x ) x 1) n dl c 2 r rdθ Thus c 2 = 1/2π. A similar calculation shows c 2 = 1/4π in three dimensions. In Cartesian coordinates we define the fundamental solution to be ln( x 2 +y 2 ) 2π N = 2 Φ(x) =, 1 N = 3 4π(x 2 +y 2 +z 2 ) 1/2 (check Φ solves the PDE!!). When we write Φ(x, x ) we mean Φ(x x ) as Φ is only a function of the distance between its arguments. In the next section we deal with boundaries Poisson s equation with Dirichlet and Neumann Boundaries Next we consider Poisson s equation with either Dirichlet boundary conditions. Let be an open bounded set. Poisson s equation is 2 u = f, u = g(x). An examination of Green s second identity, (8.1), we need to require G(x, x ) = 0 for x since u n is not known on the boundary. We expect the solution to be u(x) = G(x, x )f(x ) d + u(x ) x G(x, x ) n dl The only Green s function so far is the fundamental solution Φ. Let us set G(x, x ) = Φ(x, x ) + R(x, x ) and require R(x, x ) to satisfy 2 R = 0 for all x, x. Moreover, we require R(x, x ) = Φ(x, x ) for all x. To summarize

6 Elliptic PDEs Definition 8.1. The Green s function, G(x, x ), for the Dirichlet problem satisfies a) 2 x G(x, x ) = 0 for x = x. b) G(x, x ) = 0 for all x. c) The Green s function has the decomposition G(x, x ) = Φ(x, x ) + R(x, x ) where 2 x R(x, x ) = 0 for all x, x and R(x, x ) = Φ(x, x ) for all x, and Φ(x, x) is the fundamental solution. Theorem 8.2. Suppose u C 2 () solves 2 u = f, u = g(x), and G is a Green s function for the Dirichlet problem. Then u may be written u(x) = G(x, x )f(x ) dx + g(x ) x G(x, x ) n dl. Proof. We can make a rigorous proof by following Theorem 8.6 below. Next we consider 2 u = f, u n = g(x). In general this PDE will not have a solution (see homework problem (8.2)), and if a solution does exist, it is not unique since a constant may be added to it. We ignore these issues for now. Definition 8.3. The Green s function, G(x, x ), for the Neumann problem satisfies a) 2 x G(x, x ) = 0 for x = x. b) x G(x, x ) n = 0 for all x. c) The Green s function has the decomposition G(x, x ) = Φ(x, x ) + R(x, x ) where 2 x R(x, x ) = 0 for all x, x and x R(x, x ) n = x Φ(x, x ) n for all x. Theorem 8.4. Suppose u C 2 () solves 2 u = f, u n = g(x), and G is a Green s function for the Neumann problem. Then u may be written u(x) = G(x, x )f(x ) dx G(x, x )g(x ) dl. Proof. This identical to the proof above for Dirichlet problem.

7 8.3. Poisson s equation with Dirichlet and Neumann Boundaries 139 Theorem 8.5. The Green s function is symmetric. That is, G(x, x ) = G(x, x) for all x, x. Proof. This is a repeat of Theorem 6.7 in Chapter 4. Here, L = L so the proof is easier. Set u = G(x 1, x ) and v = G(x 2, x ) in Green s second identity. We find, formally, [G(x 1, x ) 2 x G(x 2, x ) G(x 2, x ) 2 x G(x 1, x )] d (8.2) = G(x 1, x ) x G(x 2, x ) n dl G(x 2, x ) x G(x 1, x ) n dl. The area integral on the left is equal to G(x 2, x 1 ) G(x 1, x 2 ) while the integrals on the right are both zero with either Dirichlet or Neumann boundary conditions. Let us prove at least one theorem rigorously. Theorem 8.6. (Representation Theorem) Suppose R 2 is a normal domain and that u C 2 (). Then for all x, u(x) can be writen as a double, single, and volume potential. That is, u(x) = u(y) y Φ(x, y) n dl y Φ(x, y) u(y) n dl y + Φ(x, y) 2 u(y) dy. Proof. The key is again Green s second identity. If u and v are smooth functions, then (u 2 v v 2 u) d = u v n dl v u n dl. We wish to apply this identity to u and Φ. However, Φ has a singularity when x = y. To avoid the singularity we remove a disk from around x. Let x be fixed and let ǫ < dist(x, ). Let B ǫ (x) be the disk of radius ǫ centered at x, and ǫ = /B ǫ (x) (all points in not in B ǫ (x)). Applying Green s second identity to u, v = Φ on ǫ. Since 2 Φ = 0 on ǫ, we get Φ(x, y) 2 u(y) dy = (8.3) ǫ u(y) y Φ(x, y) n dl y Φ(x, y) u(y) n dl y. ǫ ǫ We need to take the limit as ǫ 0. We will evaluate the limits in the order they appear in (8.3). Note that Φ(x, y) 2 u(y) dy ǫ B ǫ(x) Φ(x, y) 2 u(y) dy.

8 Elliptic PDEs Since u C 2 (), 2 u has a maximum on. That is, 2 u(y) M for some M and all y. Thus Φ(x, y) 2 ln( x y ) u(y) dy 2 u(y) dy 2π B ǫ(x) B ǫ(x) ln( x y ) M dy 2π Thus B ǫ(x) = M = M B ǫ(0) 2π ǫ lim Φ(x, y) 2 u(y) dy = ǫ 0 ǫ 0 ln(r) r dr dθ 2π 0 ln(r) 2π r dr dθ Mǫ2. Φ(x, y) 2 u(y) dy. The next two boundary integrals in (8.3) are over two surfaces: and B ǫ (x). We are only interested in the latter surface. The first boundary integral in (8.3) is u(y) 1 2π y ln x y n dl y. Here B ǫ(x) n = x y r x1 y 1 = r, x 2 y 2, r with r 2 = (x 1 y 1 ) 2 + (x 2 y 2 ) 2. Moreover, x1 y 1 y ln x y = r 2, x 2 y 2 r 2, and y ln x y n = r2 r 3. We have u(y) 1 B ǫ(x) 2π y ln x y n dl y = 1 2π 2π 0 u(x + ǫˆθ) 1 ǫ ǫdθ, where ˆθ = (cos θ, sin θ) (the integration is clockwise here). Since u is continuous, lim u(y) 1 ǫ 0 ǫ 2π y ln x y n dl y = u(y) 1 2π y ln x y n dl y = lim u(x + ǫˆθ) 1 2π 0 ǫ ǫdθ u(y) 1 2π y ln x y n dl y u(x). ǫ 0 1 2π

9 8.4. Examples 141 We leave it as an exercise to show as ǫ 0. B ǫ(x) Φ(x, y) u(y) n 0 Finally, taking the limit as ǫ 0 on both sides of (8.3) we get the representation for u stated in the theorem. Remark 8.7. If we believe 2 Φ(x, y) = δ(x y) and blindly apply Green s second identity, we get the representation for u stated in Theorem 8.6. Thus we can use delta functions as a quick way of correctly calculating improper integrals. Corollary 8.8. The Green s function in Definitions 8.1 and 8.3 satisfies u(x) = u(y) y G(x, y) n dl y G(x, y) u(y) n dl y + G(x, y) 2 u(y) dy. Proof. Since G(x, y) = Φ(x, y) + R(x, y) and R is harmonic on, R satisfies 0 = u(y) y R(x, y) n dl y R(x, y) u(y) n dl y + R(x, y) 2 u(y) dy. It therefore can be added to Φ without changing the statement in the theorem Examples Example 8.9. Let = {(x, y) x R, y > 0}, and consider 2 u = 0, u = g(x). We implement Definition 8.1. We look for a Green s function of the form G(x, x ) = 1 2π ln x x + R(x, x ) with R = Φ when x. Set x = (x, y) and x = (x, y ). We need R to be harmonic ( 2 R = 0) and R = Φ on the boundary. The only non trivial harmonic function we have is ln r. Let r 1 = x x. We will put a phoney point outside of the domain which we call x. Set r 2 = x x. Then the Green s function has the form G(x, x ) = 1 2π ln r 1 1 2π ln r 2. We need to chose x so that r 1 = r 2 when x is on the boundary (y = 0 here) and r 2 = 0 for all x. To accomplish this we set x = (x, y) (sketch a picture of the three points and imagine x on the boundary). Thus G(x, y, x, y ) = 1 4π ln (x x ) 2 + (y y ) 2 1 4π ln (x x ) 2 + (y + y ) 2 = 1 4π ln (x x ) 2 + (y y ) 2 (x x ) 2 + (y + y ) 2 It is easy to check this G satisfies the properties listed in Definition 8.1.

10 Elliptic PDEs By Theorem 8.3 or Equation (8.2) (here n = (0, 1)) u(x, y) = g(x ) x G(x, x ) n dl = = y π G y y =0 g(x )dx g(x ) (x x ) 2 + y 2 dx. To be consistent we should check our solution directly. For simplicity we take the space L () to mean the set of bounded continuous functions on. For a function g L we denote the bound on g to be g. Theorem Suppose g C(R) L (R). Then (i) u C () L (), (ii) 2 u = 0 in, (iii) lim u(x, y) = g(x 0). (x,y) (x 0,0) (iv) u C(). Proof. The integral describing u is challenging to work with since it is not only an improper integral because of the domain, but also the integrand is unbounded when y = 0. Set K(x, y, x y ) = π[(x x ) 2 + y 2 ]. Our solution is u(x, y) = K(x, y, x )g(x ) dx. For y > 0 we make the substitutions z = x x and z = y tan θ. We compute It follows K(x, y, x ) dx = u(x, y) = = = = 1. π/2 π/2 π/2 π/2 y π[(x x ) 2 + y 2 ] dx y π[z 2 + y 2 ] dz y 2 sec 2 θ πy 2 (1 + tan 2 θ) dθ dθ π K(x, y, x )g(x ) dx g K(x, y, x ) dx = g.

11 8.4. Examples 143 This shows u L (). Next we show u C(). Let x, y be arbitrary with y > 0. Set for some L, y 0 and L y. Set L = {(x, y) L x L, 0 < y 0 y L y } u n (x, y) = n n g(x )y π((x x ) 2 + y 2 ) dx. Then u n C( L ). We need to show u n converges uniformly to u on L. Then Theorem C, Theorem (1.41), applies and u C( L ). Since x, y is arbitrary, it follows u C(). Clearly u n u pointwise as n. We have, after setting Δu = u n (x, y) u(x, y) and z = x x, n y Δu g π((x x ) 2 + y 2 ) dx + = g n x = g n+l 2g y π 2g y π 2g L y π(n L). n L n L y π(z 2 + y 2 ) dz + y π(z 2 + y 2 ) dz + dz z 2 + y 2 dz z 2 n x n L n y π((x x ) 2 + y 2 ) dx y π(z 2 + y 2 ) dz y π(z 2 + y 2 ) dz This shows u n (x, y) converges uniformly to u(x, y) on L. It follows u C(). The proof u C () is similar only we repeatedly apply Theorem D, Theorem We outline one iteration. Let x, y be given and let L be defined as before. We need to check the conditions of Theorem D. Since x K(x, y, x ) and K(x, y, x ) are continuous on L, Theorem 1.8 applies and x u n (x, y) = n n x K(x, y, x )g(x ) dx is continuous on L. We have u n (x, y) converges at one point (all points) in L. To show x u n converges uniformly on L we show the sequence { x u n } is uniformly Cauchy. Note that x K(x, y, x ) = 2y π x x [(x x ) 2 + y 2 ] 2.

12 Elliptic PDEs Following the same arguments as above we have, assuming n > m x u n x u m m n 2y π 4y π g x x [(x x ) 2 + y 2 ] 2 dx + n+l m L 4L y g 2π(m L) 2 z (z 2 + y 2 ) 2 dz n m 2y π x x [(x x ) 2 + y 2 ] 2 dx for n > m. This shows the convergence in uniform and Theorem D applies. Thus lim n x u n = x lim n u n. That is, u x (x, y) = The other derivatives are found in a similar way. x K(x, y, x )g(x ) dx. To establish (iii) in the theorem fix x 0 R. Since g C(R), for all ǫ > 0 there exists a δ > 0 such that g(x) g(x 0 ) < ǫ/2 for all x such that x x 0 < δ. Thus, since K(x, y, x ) dx = 1, u(x, y) u(x 0, y) y g(x ) g(x 0 ) π (x x ) 2 + y 2 dx For the first integral we have x0+δ + x 0 δ = I + II. y g(x ) g(x 0 ) π (x x ) 2 + y 2 dx R\B δ (x 0) y g(x ) g(x 0 ) π (x x ) 2 + y 2 dx I < x0+δ x 0 δ x0+δ x 0 δ ǫ/2. K(x, y, x g(x ) g(x 0 ) dx K(x, y, x )ǫ dx To estimate the second integral we need a lower bound on x x. Let us require x to satisfy x x 0 < δ/2. The integrand in II requires requires x < x 0 δ or x > x 0 + δ, and so x x 0 δ. Thus That is, x x 0 x x + x x 0 x x + δ/2 x x x x 0. x x 1 2 x x 0.

13 8.4. Examples 145 This gives as a bound on the second integral. Indeed y g(x ) g(x 0 ) II = R\B δ (x 0) π (x x ) 2 + y 2 dx 2g y R\B δ (x 0) π (x x 0) dx 4g δπ y. Putting the two together we find u(x, y) g(x 0 ) < ǫ/2 + 4g δπ y for all x x 0 < δ/2. Thus by restricting y, for every ǫ > 0 there is a neighborhood of (x 0, 0) such that u(x, y) g(x 0 ) < ǫ, and Part (iii) in the theorem follows. Standard continuity arguments now show u C(). Example Let = {(x, y) x 2 + y 2 = a 2, a > 0}, and consider 2 u = 0, u = g(x). Again we implement Definition 8.1. We look for a Green s function of the form G(x, x ) = 1 2π ln x x + R(x, x ). Following the insight we gained in the previous example, we place a phoney point, x, somewhere in the plane so that G(x, x ) = 1 2π ln x x 1 2π ln x x = 0 when x. Moreover, the symmetry of the problem suggest putting x on the line formed by x and the origin. and Let x = r, x = ρ, x = R and φ the the angle between x and x. Then x x 2 = x x + x x 2x x = r 2 + ρ 2 2rρ cos φ, x x 2 = x x + x x 2x x = ρ 2 + R 2 2ρR cos φ. If we set x x = x x when x (when ρ = a), we would conclude R = r. This is impossible since we need x outside of the domain. Instead of looking for the ratio x x / x x = 1, we try x x / x x = γ for x. We expect γ = γ(a, r). If such a γ can be found, our Green s function is G(x, x ) = 1 2π ln x x 1 2π ln(γ x x ).

14 Elliptic PDEs Setting x x 2 = γ 2 x x 2 when ρ = a, and equating the cos φ terms, we find r = Rγ 2. Equating the remaining terms shows γ = a/r. This in turn implies γ = r a. Thus the Green s function is G(x, x ) = 1 4π ln x x 2 1 = 1 4π ln a 2 r 2 4π ln(γ2 x x 2 ) = 1 r 2 + ρ 2 2rρ cos φ ρ 2 + R 2 2ρR cos φ x x 4π ln 2 γ 2 x x 2 If x = (r cos θ, r sin θ) and x = (ρ cos θ, ρ sin θ ), then φ = θ θ. Moreover, since R = a 2 /r, G(x, x ) = 1 a 2 4π ln (r 2 + ρ 2 2rρ cos(θ θ )) r 2 ρ 2 + a 4 2ra 2 ρ cos(θ θ ) By Theorem 8.3 or Equation 8.2 the solution may be written as u(x, y) = g(x ) x G(x, x ) n dl. Here and dl = a dθ. Thus, x G(x, x ) n = G ρ (8.4) u(r, θ) = 1 2π π π = a ρ=a 2π Equation (8.4) is called Poisson s formula. 1 (r/a) 2 r 2 + a 2 2ar cos(θ θ ), a 2 r 2 r 2 + a 2 2ar cos(θ θ ) g(θ ) dθ. Theorem Equation (8.4) satisfies 2 u = 0 in and u C (). Proof. We recall Theorem 1.34 from Chapter 1. Namely, if with F and F x continuous, then f(x) = f (x) = b a b a F (x, y) dy, F (x, y) x Now we observe that, if r < a, the dominator in (8.4) is never zero and we may move as many derivatives inside the integral as we wish. That is, u C (). We leave it as an exercise to check 2 u = Maximum Principle Based on the physical problems Laplace s equation describes, we expect the maximum and minimum of a harmonic function ( 2 u = 0) to occur on the boundary. This is easy to show. dy.

15 8.5. Maximum Principle 147 Theorem Let be a open bounded set in R n. Suppose u C 2 () C() satisfies 2 u 0. Then max u = max u. Proof. Suppose for the moment 2 u < 0 in. Since u is continuous on a closed bounded set, u has a maximum. Suppose the maximum is in. By the second derivative test 2 u 0. This implies 2 u 0 contrary to our assumption. Thus the maximum occurs on the boundary. Now suppose 2 u 0 in. Set, for ǫ > 0, v = u + ǫ x2 2n. Then 2 v = 2 u ǫ < 0. By the result in the previous paragraph max where d = max u max Since, we have and the result follows. v = max v = max u + ǫ x2 2n x 2. Since this holds for all ǫ > 0, 2n max u max u. max u max u, max u + ǫd, We have not shown a solution to an elliptic PDE depends continuously on the data yet. Theorem Consider on a bounded open set. Set N = sup Proof. Set v = u + Nx2 2n. Then, in, By the previous theorem u max u max 2 u = f, u = g(x) f and M = sup g. Then, for all x, u(x) M + Nd. 2 v = 2 u N 0. v = max v = max u + Nx2 M + Nd. 2n Similarly, set v = u + Nx2 2n. Then, in, 2 v = 2 u N 0. Thus u v max v = max u v = max + Nx2 M + Nd. 2n That is, u (M + Nd). The result follows.

16 Elliptic PDEs Corollary The solution to depends continuously on f and g. 2 u = f, u = g(x) Proof. Let u i, for i = 1, 2 solve Then the difference w = u i u 2 solves 2 u i = f i, u i = g i (x). 2 w = Δf, w = Δg, where Δf = f 1 f 2 and Δg = g 1 g 2. Thus w(x) = u 1 (x) u 2 (x) sup Δg + sup Δf d, and we see small changes in f and g lead to small changes in u.

17 8.6. Homework Homework Exercise 8.1. Suppose is a normal domain. Suppose a function u is specified on the boundary,, of and we want to find the minimal surface over subject our constraint u at the boundary. That is, we want to minimize the functional (minimal surface area) J(u) := 1 + u 2 x + u2 y dxdy. Minimize J by considering an arbitrary smooth function ϕ which is zero on the boundary of. Then consider J(u + ǫϕ) := 1 + (u + ǫϕ) 2 x + (u + ǫϕ)2 y dxdy. Now compute and require J(u+ǫϕ) ǫ ǫ=0 = 0 to derive Plateau s equation (1 + u 2 y )u xx + (1 + u 2 x )u yy 2u x u y u xy = 0. Exercise 8.2. Let be a bounded, normal domain. Consider the elliptic PDE 2 u = f, x u n = g. (a) Prove no solution exists if f d = g dσ. What is the physical significance of this condition? (b) If the condition is satisfied, is the solution unique? Provide a proof either way. (c) Find a condition which will make the solution unique. Exercise 8.3. Consider the elliptic PDE on the square = {(x, y) 0 < x < π, 0 < y < π}: 2 u = 0, on (a) Find the formal solution. u x (0, y) = f(y), 0 < y < π, u x (π, y) = g(y), 0 < y < π, u(x, 0) = 0, 0 < x < π, u(x, π) = 0, 0 < x < π. (b) Find conditions on f and g so that your formal solution satisfies u C, () C().

18 Elliptic PDEs Exercise 8.4. Consider the Laplace s equation on the channel = {(x, y) 0 < x <, 0 < y < π}: 2 u = 0, on (a) Find the formal solution. u(0, y) = f(y), 0 < y < π, u(x, 0) = 0, 0 < x <, u(x, π) = 0, 0 < x <. (b) Find conditions on f so that your formal solution satisfies u C, () C(). Exercise 8.5. Let = {(x, y) R 2 : 0 < y, 0 < x}. Solve, by constructing a Green s function, 2 u = 0 0 < x, 0 < y, u(x, 0) = f(x), 0 < x, u(0, y) = 0, 0 < y, u(x, y) is bounded for large x, y and f(x) is continuous and bounded. The answer is u(x, y) = y 1 π (x x ) 2 + y 2 1 (x + x ) 2 + y 2 f(x )dx. 0 Exercise 8.6. Let = {(x, y) R 2 : 0 < y, 0 < x}. Solve, by constructing a Green s function, 2 u = 0, 0 < x, 0 < y, u n = f(x), 0 < x, u(0, y) = 0, 0 < y. (x,0) Exercise 8.7. Let be the semicircle (0 < r < a, 0 < θ < π). Find the Green s function, G(x, x ) satisfying 2 G = δ( x x ), with G(x, x ) = 0 when x is on the boundary. Exercise 8.8. Consider the PDE with Robin boundary conditions on an open set 2 u = f, u n + α(x)u = g(x) x, where α(x) > 0 on. What properties does a Green s function have for this PDE. Formulate a definition like Definition 1.4 or 1.5 in the notes. Find, formally, a representation, using the Green s function, of the solution - as in Theorem 1.7 or 1.8 in the notes. Exercise 8.9. Let be an open, normal, bounded subset of R 2. Prove that the Green s function to Poisson s equation with Dirichlet boundary conditions is unique. Hint: suppose there were two Green s functions. Set u(y) = G 1 (x, y) G 2 (x, y) for a fixed x. Show 2 u = 0 and u = 0 on. 1. Show Φ(x, y) u(y) n 0 B ǫ(x) as ǫ 0 to complete the proof of Theorem 8.3

19 8.6. Homework Make the proof the Green s function is symmetric rigorous by applying Green s second identity on ǫ and take the ǫ 0 as was done in the proof of Theorem 8.3 Exercise If is not bounded, show that the solution to 2 u = 0 may not have a maximum. Exercise Consider the Helmholtz equation on 2 u u = h u = f. (a) Suppose is normal and bounded. Use energy methods to show the solution is unique. STATE the assumptions you need on u. (b) Suppose the boundary of is fractal (Not smooth so energy methods do not apply). By using ideas from elementary calculus, develop a maximum principle to show the solution is unique. (You will have to home cook your own maximum principle.) Exercise Consider Laplace s equation on the square 0 x 1, 0 y 1 u with nonhomogeneous Neumann boundary conditions: ν = 0 on the sides {y = 0, 0 x 1}, {x = 0, 0 y 1}, {x = 1, 0 y 1}, and u ν = f(x) on the side {y = 1, 0 x 1}. (a) Show that no solution exists if the condition (b) Is the solution unique? Prove or disprove. 1 0 f(x) dx = 0 is not satisfied. Exercise Let be an open, normal, bounded subset of R n. Let u C 2 () C(). Moreover, suppose that u(x) satisfies 2 u(x) = 0, x. (a) Prove max u = max u. x x (b) Prove the solution to Poisson s equation is unique using two different methods. Exercise Suppose that u, v are C 2 () C() and solve 2 u = f 1 2 v = f 2 u = g 1 v = g 2. Moreover, let f 1 f 2 0 for all x, and g 1 g 2 0 for all x. Prove that u v 0 for all x.