Solutions to Math. 401 (Fulling) Test C: Heat Equation and Laplace's Equation - Prof. Step, Exams of Mathematics

Download Solutions to Math. 401 (Fulling) Test C: Heat Equation and Laplace's Equation - Prof. Step and more Exams Mathematics in PDF only on Docsity! Math. 401 (Fulling) 1 December 2003 Test C – Solutions 1. (40 pts.) Consider the heat equation ∂u ∂t = ∂2u ∂x2 (0 < x <∞, 0 < t <∞) with the boundary condition ∂u ∂x (t, 0) = 0 and the initial data u(0, x) = f(x). (a) Solve the problem by separation of variables or an equivalent transform method. Transform method: A Fourier cosine transform in x is called for, because of the Neumann condition at 0 and the infinite interval. Let’s agree that a tilde indicates the Fourier cosine transform, without a factor of √ 2/π. Take the transform of the PDE and the initial condition: ∂ũ ∂t = −ω2ũ, ũ(0, ω) = f̃(ω) = ∫ ∞ 0 f(x) cos(ωx) dx. It follows that ũ(t, ω) = ũ(0, ω)e−ω 2t = f̃(ω)e−ω 2t. Therefore, u(t, x) = 2 π ∫ ∞ 0 ũ(t, ω) cos(ωx) dω = 2 π ∫ ∞ 0 f̃(ω) cos(ωx)e−ω 2t dω, where f̃(ω) = ∫ ∞ 0 f(x̃) cos(ωx̃) dx̃. Separation of variables: Look for separated solutions u = T (t)X(x) and conclude that T ′ T = −ω2 = − X ′′ X with X ′(0) = 0 and no other homogeneous boundary conditions except that X be well behaved (bounded) as x → +∞. Therefore, X(x) = cos(ωx) with no restriction on ω except ω ≥ 0. Thus u(t, x) = ∫ ∞ 0 A(ω) cos(ωx)e−ω 2t dω with f(x) = ∫ ∞ 0 A(ω) cos(ωx) dω. Clearly A(ω) is the same thing we called 2π f̃(ω) in the first method, and we can finish up the problem the same way we did there. 401C-F03 Page 2 (b) Find a Green function, G, such that u(t, x) = ∫ ∞ 0 G(t, x, x̃)f(x̃) dx̃. Substitute the formula for f̃ into the formula for u and interchange the order of integration: u(t, x) = ∫ ∞ 0 dx̃ ∫ ∞ 0 dω 2 π cos(ωx) cos(ωx̃)e−ω 2tf(x̃). Thus G(t, x, x̃) = 2 π ∫ ∞ 0 cos(ωx) cos(ωx̃)e−ω 2t dω. Can we evaluate this integral? Yes, indeed! G(t, x, x̃) = 1 2π ∫ ∞ 0 ( eiωx + e−iωx )( eiωx̃ + e−iωx̃ ) e−ω 2t dω = 1 2π ∫ ∞ 0 ( eiω(x+x̃) + eiω(x−x̃) + eiω(−x+x̃) + e−iω(x+x̃) ) e−ω 2t dω = 1 2π ∫ ∞ −∞ ( eiω(x+x̃) + eiω(x−x̃) ) e−ω 2t dω = 1√ 4πt ( e−(x−x̃) 2/4t + e−(x+x̃) 2/4t ) . (At the last step a useful formula at the end of the test has been used.) Thus G equals the Green function for the heat equation on the whole real line, plus an “image” term representing a fictitious temperature source at the point x = −x̃ in the unphysical region. 2. (30 pts.) Solve ∂2u ∂x2 + ∂2u ∂y2 = 0 (0 < x <∞, 0 < y < L) with the boundary data ∂u ∂y (x, 0) = 0 = ∂u ∂y (x, L) (0 < x <∞) and u(0, y) = g(y) (0 < y < L). Separation of variables: As usual for Laplace’s equation, we get X′′ X = −Y ′′ Y = λ.