MATH 124B Solution Key HW 03

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1 6.1 LAPLACE S EQUATION MATH 124B Solution Key HW LAPLACE S EQUATION 4. Solve u x x + u y y + u zz = 0 in the spherical shell 0 < a < r < b with the boundary conditions u = A on r = a and u = B on r = b, where A and B are constants. Hint: Look for a solution depending only on r. SOLUTION. Since n = 3, the radial Laplacian takes the form u r r + n 1 u r = u r r + 2 r r u r = 0. Put v = u r then our 2nd order OE becomes the 1st order OE v r + 2 r v = 0. Recognizing the product rule we rewrite our OE as hence; Integrating in r we obtain (r 2 v) r = 0, v = C 1 r 2 = u r. u = C 1 r 1 + C 2. The boundary conditions u(a) = A and u(b) = B reveal that A = C 2 C 1 a B = C 2 C 1 b. In other words, C 1 = ab a b (A B) and C aa bb 2 = a b. Putting all of this together we see that u(r) = ab a b (A aa bb B)r 1 + a b = B + A B r 1 b 1. a 1 b 1 5. Solve u x x + u y y = 1 in r < a with u(x, y) vanishing on r = a. 1 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 124B, Winter 2013

2 6.1 LAPLACE S EQUATION SOLUTION. Since n = 2, the radial Laplacian takes the form By putting v = u r we recast our OE as By the product rule we see that Hence, More specifically, Integrating in r makes it plain that u r r + n 1 u r = u r r + 1 r r u r = 1. r v r + v = r. (r v) r = r v r + v = r. r v = r2 2 + C 1. v = r 2 + C 1 r = u r. u = r2 4 + C 1 log r + C 2. The term log r introduces a singularity at r = 0 so it must be the case that C 1 = 0. Hence; u(r) = r2 4 + C 2. Since u(x, y) vanishes at r = a, we get u(a) = 0 and obtain that u(r) = r2 4 a Solve u x x + u y y + u zz = 1 in the spherical shell a < r < b with u(x, y, z) vanishing on both the inner and outer boundaries. SOLUTION. Since n = 3, the radial Laplacian takes the form By putting v = u r we recast our OE as u r r + n 1 u r = u r r + 2 r r u r = 1. v r + 2 r v = 1. 2 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 124B, Winter 2013

3 6.1 LAPLACE S EQUATION Multiplying both sides by a factor of r 2 and recognizing the product rule we obtain Integrating in r we get that Or equivalently, Integrating in r we see that (r 2 v) r = r 2 v r + 2r v = r 2. r 2 v = r3 3 + C 1. v = r 3 + C 1 r 2 = u r. u = r2 6 C 1 r + C 2. Since u vanishes on the inner and outer boundaries we have u(a) = u(b) = 0, that is, a 2 6 C 1 a + C 2 = 0 More specifically, b 2 6 C 1 b + C 2 = 0. ab(b a) C 1 = and C 2 = a2 + ab + b Putting all of this together we now see that u(r) = r2 a 2 6 ab(a + b)(a 1 r 1 ) A spherical shell with inner radius 1 and outer radius 2 has a steady-state temperature distribution. Its inner boundary is held at 100 C. Its outer boundary satisfies u/ r = γ < 0, where γ is a constant. (a) Find the temperature. Hint: The temperature depends only on the radius. (b) What are the hottest and coldest temperatures? (c) Can you choose γ so that the temperature on its outer boundary is 20 C? SOLUTION. 3 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 124B, Winter 2013

4 6.1 LAPLACE S EQUATION (a) After making a reduction to the radial variable, again the boundary conditions are symmetric, we aim to find the solution u to the OE u r r + 2 r u r = 0, u(1) = 100, u r (2) = γ. The general solution to this radial OE is given by u(r) = C 1 r + C 2, The 2nd boundary condition makes it plain that C 1 = 4γ, thus, u(r) = 4γ r + C for some constant C. The first (inner) boundary condition then given us that C = 100 4γ. Therefore, the desired solution is precisely 4γ(1 r) u(r) = r (b) In this scenario, the temperature goes up as r decreases and vice-versa. Thus, the hottest temperature is observed when r = 1 and is 100 C. The coldest temperature is observed when r = 2 and is (100 2γ) C. Indeed, this agrees with our physical intuition because there is a net flux out of the shell at the outer boundary as u r = γ < 0 there. Thus there must be a net heat flux into the inner boundary, and therefore, the temperature is hotter there. (c) Yes, simply choose γ = 40 in the equation 100 2γ for the temperature on the outer boundary. 10. Prove the uniqueness of the irichlet problem u = f in, u = g on bdy by the energy method. That is, after subtracting two solutions w = u v, multiply the Laplace equation for w by w itself and use the divergence theorem. SOLUTION. Suppose that u and v both solve our PE of interest. That is, u = f in v = f in and u = g on bdy. v = g on bdy. By putting w = u v we see that w satisfies the PE w = 0 in w = 0 on bdy. 4 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 124B, Winter 2013

5 6.2 RECTANGLES AN CUBES Now, we integrate the integral of w 2 over the region. This integral must be positive since its integrant is positive. More specifically, we shall show that it must, in fact, be zero. First, we notice that w 2 dv = Second, we use the familiar vector calculus identity (w w) = w w + w w ( w w) dv. (10.1) to rewrite the right hand side (10.1) in a form suitable for the divergence theorem. Thus, (10.1) becomes [ (w w) w w] dv. The 2nd term is automatically zero since w = 0 in. An application of the divergence theorem now makes it plain that (w w)dv = bdy (w w) n ds = bdy w w n ds, where n denotes the unit vector normal to the surface bdy enclosing. The boundary condition w = 0 on bdy makes it plain that our integrant is zero, hence the integral is zero as well. i.e., w 2 dv = (w w)dv = 0. Finally, since the original integrant w 2, is positive, it follows by the vanishing theorem on page 416 in our textbook that w 2 is identically zero. Hence, w must be constant. However, the condition on the boundary, bdy tells us that the constant must in fact be zero. Hence; w 0 u = v, and we ve established uniqueness. 6.2 RECTANGLES AN CUBES 1. Solve u x x + u y y = 0 in the rectangle 0 < x < a, 0 < y < b with the following boundary conditions: u x = a on x = 0, u x = 0 on x = a u y = b on y = 0, u y = 0 on y = b. Hint: Note that the necessary condition of Exercise is satisfied. A shortcut is to guess that the solution might be a quadratic polynomial in x and y. 5 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 124B, Winter 2013

6 6.2 RECTANGLES AN CUBES SOLUTION. Our hint suggests that the solution is a quadratic polynomial in x and y, i.e., u(x, y) = Ax 2 + Bx y + C y 2 + x + E y + F. We have the partial derivatives u x (x, y) = 2Ax + B y + u y (x, y) = Bx + 2C y + E. We now use the boundary conditions to solve for the missing constants. Since a = u x (0, y) = B y +, must hold for all values of y, it follows that B = 0 and = a. Next, using 0 = u x (a, y) = 2Aa + B y + = 2Aa a, we find that A = 1/2. Similarly, b = u y (x, 0) = E. Solving 0 = u y (x, b) = 2C b + b gives B = 1/2. Hence, u(x, y) = 1 2 x y 2 ax + b y + F solves our PE of interest for any constant F. 4. Find the harmonic function in the square {0 < x < 1, 0 < y < 1} with the boundary conditions u(x, 0) = x, u(x, 1) = 0, u x (0, y) = 0, u x (1, y) = y 2. SOLUTION. The standard approach of separating variables is doomed to fail since it is impossible to calculate eigenfunctions in the absence of homogeneous boundary conditions in either variable. Instead, we solve two subproblems in which each have homogenous boundary conditions in one variable. Notice that if v is a harmonic function on the unit square with boundary conditions v(x, 0) = x, v x (0, y) = 0 v(x, 1) = 0, v x (1, y) = 0, and w is a harmonic function on the unit square with boundary conditions then u = v + w solves our original problem. w(x, 0) = 0, w x (0, y) = 0 w(x, 1) = 0, w x (1, y) = y 2, Applying the technique of separating variables to v(x, y) = X (x)y (y) gives X /X + Y /Y. Since v has homogeneous boundary conditions in the x-variable, we choose the sign of our separation constant λ so that we obtain the OE s X + λx = 0 and Y λy = 0. 6 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 124B, Winter 2013

7 6.2 RECTANGLES AN CUBES A routine calculation makes it plain the the eigenvalues for X are λ n = (nπ) 2 with corresponding eigenfunctions X n (x) = cos nπx. Next, when λ > 0, then Y = 0 has the solution Y (y) = C y +. Using the boundary condition 0 = Y (1) = C +, we find that C = and hence, Y (y) = (1 y). If λ > 0, then Y λy = 0 has solution of the form Y (y) = Acosh nπ y + B sinh nπ y. The boundary condition implies that 0 = Y (1) = Acosh nπ + B sinh nπ, and so giving the solution Putting these together, B = Acoth nπ, Y (y) = A cosh nπ y coth nπ sinh nπ y. v(x, y) = 1 2 A(1 y) + A n cosh nπ y coth nπ sinh nπ y cos nπx. (6.2) n=1 Using the remaining boundary condition, x = v(x, 0) = 1 2 A 0 + A n cos nπx, n=1 which is just the Fourier cosine series of x. Hence, and for n = 1, 2, 3,..., A n = A 0 = x d x = 1, x cos nπx d x = 2 [( 1)n 1] (nπ) 2. Next, we solve the boundary problem for w. Again, we separate variables and choose the sign of λ so that Y + λy = 0, and X λx = 0. Solving the homogeneous problem for Y, if λ = 0 we have the solution Y (y) = C y +, 7 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 124B, Winter 2013

8 6.2 RECTANGLES AN CUBES but the boundary conditions imply that C = = 0, hence 0 is not an eigenvalue. For λ > 0, we write ω 2 = λ and obtain a solution of the form The boundary condition Y (y) = Acos ω y + B sin ω y. 0 = Y (0) = Acos 0 + B sin 0 = A, implies that A = 0. The second boundary condition 0 = Y (1) = B sin ω implies that ω = π, 2π, 3π,..., Hence; we have eigenvalues λ n = (nπ) 2 with corresponding eigenfunctions Solving the OE X λ n X = 0 gives The boundary condition implies that B = 0. Hence; Y n (y) = sin nπ y, for n = 1, 2, 3,... X (x) = Acosh nπx + B sinh nπx. 0 = X (0) = Anπ sinh 0 + Bnπ cosh 0 = Bnπ X (x) = Acosh nπx. Putting X and Y together along with the super positioning principle gives w(x, y) = B n cosh nπx sin nπ y. (6.3) n=1 The final boundary condition gives the equation y 2 = w x (1, y) = B n nπ sinh nπ sin nπ y. Using the familiar formula for the Fourier sine series coefficients we obtain: 2 B n = nπ sinh nπ 1 0 n=1 2 y 2 sin nπ y d y = nπ sinh nπ ( 1)n nπ + 2(( 1)n 1. (nπ) 3 Taking u = v + w by adding (6.2) and (6.3) with the coefficients A n and B n as above completes our problem. 8 Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 124B, Winter 2013