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MATH 3150-1, PRACTICE MIDTERM EXAM 2
OCTOBER 24 2008
Total points: 100/100. Problem 1 (30 pts) The goal of this problem is to solve the Heat Equation with mixed boundary conditions
ut = 3uxx for 0 < x < 1 and t > 0 ux(0, t) = 0 for t > 0 u(1, t) = 0 for t > 0 u(x, 0) = f (x) for 0 < x < 1 (a) Use separation of variables to show that a general solution to (1) is
u(x, t) =
∑^ ∞
n=
an cos (λnx) exp[− 3 λ^2 nt], where λn =
2 n + 1 2
π.
(b) Consider the inner product (u, v) =
0 u(x)v(x)dx. Given the orthogonality relations valid for n = 0, 1 , 2 ,... and m = 0, 1 , 2 ,...
(cos(λnx), cos(λmx)) =
1 2 if^ n^ =^ m 0 if n 6 = m, show that an = 2
0
cos(λnx)f (x)dx, for n = 0, 1 , 2 ,...
(c) Solve problem (1) with f (x) = cos(3πx/2) + 2 cos(7πx/2). Problem 2 (30 pts) Consider the 2D Laplace equation below, which models the steady state temperature distribution of a square plate where the right and left sides are kept in an ice bath and the bottom and top sides have prescribed temperatures f 1 (x) and f 2 (x) respectively.
uxx + uyy = 0, for 0 < x < 1 and 0 < y < 1 u(0, y) = u(1, y) = 0, for 0 < y < 1 u(x, 0) = f 1 (x), for 0 < x < 1 u(x, 1) = f 2 (x), for 0 < x < 1. (a) Explain why it is possible to decompose (2) into the two subproblems below (the x and y below are implicitly in (0, 1)).
(P1)
vxx + vyy = 0, v(0, y) = v(1, y) = 0, v(x, 0) = f 1 (x), v(x, 1) = 0
(P2)
wxx + wyy = 0, w(0, y) = w(1, y) = 0, w(x, 0) = 0, w(x, 1) = f 2 (x)
(b) Show that if we assume that the solution to (P2) is w(x, y) = X(x)Y (y), then separa- tion of variables gives X′′^ + kX = 0, X(0) = 0, X(1) = 0 Y ′′^ − kY = 0, Y (0) = 0
(c) Assuming k = μ^2 > 0, obtain the product solutions to (P2) wn(x, y) = Bn sin(nπx) sinh(nπy)
(d) Write down the general form of a solution to (P2), and use the formulas at the end of the exam to express Bn in terms of f 2 (x). (e) In a similar way it is possible to obtain the product solutions to (P1), vn(x, y) = An sin(nπx) sinh(nπ(1 − y)). Write down the general form of a solution to (P1) and give an expression for An in terms of f 1 (x). (f) Solve (2) with f 1 (x) = 100 and f 2 (x) = 100x(1 − x). You may use the identity below (valid for n = 1, 2 ,.. .): ∫ (^1)
0
x(1 − x) sin(nπx)dx =
2((−1)n^ − 1) π^3 n^3
Problem 3 (30 pts) Consider a circular plate of radius 1 with initial temperature distri- bution of the form f (r, θ) = g(r) cos 2θ and where the outer rim of the plate is kept in an ice bath. The temperature distribution u(r, θ, t) satisfies the 2D Heat equation
ut = ∆u for 0 < r < 1, 0 ≤ θ ≤ 2 π and t > 0 u(r, θ, 0) = f (r, θ) for 0 < r < 1 and 0 ≤ θ ≤ 2 π u(1, θ, t) = 0 for 0 ≤ θ ≤ 2 π and t > 0 Because the initial temperature distribution is a multiple of cos 2θ, the solution can be shown to be
u(r, θ, t) =
∑^ ∞
n=
a 2 nJ 2 (α 2 nr) cos 2θ exp[−α^22 nt].
where α 2 n denotes the n−th zero of the Bessel function of the first kind of order 2, and
a 2 n =
πJ2+1^2 (α 2 n)
0
∫ (^2) π
0
f (r, θ)J 2 (α 2 nr) cos 2θ dθ rdr for n = 1, 2 ,...
(a) Solve (3) with the initial temperatures f 1 (r, θ) = J 2 (α 2 , 1 r) cos 2θ and f 2 (r, θ) = J 2 (α 2 , 2 r) cos 2θ.
(b) The steady state temperature distribution is u = 0. Of the initial temperatures f 1 (r, θ) and f 2 (r, θ), which decays faster to the steady state? Justify your answer.
