Math 257/316 Assignment 10 Solutions 1. Find the eigenvalues and eigenfunctions of the following Sturm-Liouville problem on [0, 1] X 00 (x) + λX(x) = 0, X 0 (0) = 0, X(1) = 0, and use them to solve the following heat equation with mixed BCs and a source term: ut = uxx + cos(5πx/2)t, ux (0, t) = 0, 0 < x < 1, t > 0, u(1, t) = 0 u(x, 0) = cos(7πx/2) There are three cases to consider: • Case 1: λ := −µ2 < 0. Then the general solution is X(x) = Aeµx + Be−µx . The first BC is 0 = X 0 (0) = µ(A − B) so B = A, and then the second BC is 0 = X(1) = A(eµ + e−µ ) which implies A = 0. So no non-zero solutions. • Case 2: λ = 0. Then the general solution is X(x) = Ax + B. The first BC is 0 = X 0 (0) = A so A = 0, and then the second BC is 0 = X(1) = B. So, again, no non-zero solutions. • Case 3: λ := µ2 > 0. Then X(x) = A sin(µx) + B cos(µx). The first BC is 0 = X 0 (0) = µA, so A = 0, and then the second BC is 0 = X(1) = B cos(µ), so µ = (1/2 + k)π, k = 0, 1, 2, 3, . . .. So the eigenvalues and eigenfunctions are: λk = (1/2 + k)2 π 2 , Xk (x) = cos((1/2 + k)πx), k = 0, 1, 2, 3, . . . To solve the heat problem, we use an eigenfunction expansion u(x, t) = ∞ X ck (t)Xk (x) = k=0 ∞ X k=0 1 ck (t) cos((k + 1/2)πx) (which has the correct BCs built in). Plugging into the equation, we find 0 = ut − uxx − t cos(5πx/2) ∞ ∞ X X = c0k (t) cos((k + 1/2)πx) − ck (t)[−(k + 1/2)2 π 2 ] cos((k + 1/2)πx) − t cos(5πx/2) k=0 = = ∞ X k=0 c0k (t) + (k + 1/2)2 π 2 ck (t) cos((k + 1/2)πx) − t cos(5πx/2) k=0 ∞ X 0 ck (t) + (k + 1/2)2 π 2 ck (t) cos((k + 1/2)πx) k=0,k6=2 + c02 (t) + (5/2)2 π 2 c2 (t) − t cos(5πx/2) So for k 6= 2, we require c0k (t) = −(k + 1/2)2 π 2 ck (t) =⇒ ck (t) = bk e−(k+1/2) 2 π2 t , while for k = 2 we need c02 (t) + (5π/2)2 c2 (t) = t. h i0 2 2 2 The integrating factor is e(5π/2) t giving e(5π/2) t c2 (t) = te(5π/2) t , and so integration by parts yields 2 2 2 e(5π/2) t c2 (t) = (5π/2)−2 e(5π/2) t (t−(5π/2)−2 )+b2 =⇒ c2 (t) = (5π/2)−2 (t−(5π/2)−2 )+b2 e−(5π/2) t . It remains to find the constants bk , using the initial condition: ∞ X cos(7πx/2) = u(x, 0) = bk cos((k + 1/2)πx) + −(5π/2)−4 + b2 cos(5πx/2), k=0,k6=2 and inspection shows that b2 = (5π/2)−4 , b3 = 1, bk = 0 for k 6= 2, 3, and so h i 2 2 u(x, t) = (5π/2)−2 t + (5π/2)−4 (e−(5π/2) t − 1) cos(5πx/2) + e−(7π/2) t cos(7πx/2). 2. Consider the problem of steady heat conduction in a semi-circular annulus ∆u = 1 ∂2u ∂ 2 u 1 ∂u + + 2 2 = 0, 2 ∂r r ∂r r ∂θ 2 1 < r < 2, 0 < θ < π, u(1, θ) = u(2, θ) = 0, u(r, 0) = 0, u(r, π) = 1. Separate variables, u(r, θ) = R(r)Θ(θ), and find the Euler equation to be solved for R(r). Show that it can also be converted to the Sturm-Liouville form 1 (rR0 )0 + λ R = 0, r R(1) = R(2) = 0. Find the eigenfunctions Rn and eigenvalues λn , and check the orthogonality condition Z 2 1 0 m 6= n Rn (r)Rm (r) dr = , C m=n r 1 where you should determine the constant C. Hence, find the solution u(r, θ) by eigenfunction expansion. Hint: remember to include the weighting function 1/r when calculating the generalized Fourier coefficients. Substituting u(r, θ) = R(r)Θ(θ) into the equation gives r2 R0 Θ00 R00 +r =− = −λ. R R Θ The R problem is the Euler ODE 0 = r2 R00 (r) + rR0 (r) + λR(r) = 0, R(1) = 0 = R(2), (for convenience we have written the separation constant with a minus sign, since it is the R problem which has the zero BCs at both ends), which we can re-write in Sturm-Liouville form: 1 (rR0 )0 + λ R = 0, r R(1) = 0 = R(2). Solutions of the Euler ODE are of the form rµ where µ solves the indicial equation √ 0 = µ2 + λ =⇒ µ = ±i λ, and so the general solution is √ √ R(r) = A sin( λ ln x) + B cos( λ ln x). The BCs give 0 = R(1) = B and √ 0 = R(2) = A sin( λ ln 2) √ =⇒ R(r) = A sin( λ ln x) √ =⇒ 3 λ ln 2 = nπ, n = 1, 2, 3, . . . , so λn = nπ n2 π 2 , R (r) = sin ln r . n ln 2 ln2 (2) Compute, by changing variables y = ln r/ ln 2 (so dy = (1/ ln 2)(dr/r)), Z 1 2 1 Rn (r)Rm (r) dr = r 2 Z sin(nπ ln r/ ln 2) sin(mπ ln r/ ln 2)dr/r Z 1 0 m 6= n = ln 2 sin(nπy) sin(mπy)dy = . ln 2 m=n 0 2 1 Now the Θ problem, Θ00 = λn Θ, Θ(0) = 0 has solutions Θn (θ) = sinh(nπθ/ ln 2), so u(r, θ) = ∞ X cn sinh(nπθ/ ln 2) sin(nπ ln r/ ln 2). n=1 It remains to satisfy the last BC: 1 = u(r, π) = ∞ X cn sinh(nπ 2 / ln 2) sin(nπ ln r/ ln 2). n=1 Using the eigenfunction expansion coefficient formula, the coefficients are given by R2 Z 1 1 2 2 1 sin(nπ ln r/ ln 2) r dr sinh(nπ / ln 2)cn = R 2 2 = ln 2 sin(nπy)dy 1 ln 2 0 1 sin (nπ ln r/ ln 2) r dr 4 2 n odd nπ (1 − cos(nπ)) = = . 0 n even nπ So 4 u(r, θ) = π ∞ X n=1,n odd sinh(nπθ/ ln 2) ln r sin nπ . n sinh(nπ 2 / ln 2) ln 2 4