Advanced Analysis Math 121, Fall 2004 Solutions, Midterm 2

1. [20%] Use Laplace transforms to find the solution y(t) of the following initial value problem

y′′ + 9y = f(t),
y(t) = 0, y′(0) = 0,

where f(t) is an arbitrary function. Express your answer as a convolution.

Solution.

• Let Y (p) = L[y(t)] and F (p) = L[f(t)], where L denotes the Laplace

transform. Then

p^{2}Y + 9Y = F (p),

so

Y (p) = \frac{F (p)}{p^{2} + 9}.

From the tables,

L^{−1} [\frac{1}{p^{2} + 9}] = \frac{1}{3} sin(3t).

Therefore, using the convolution theorem, we have

y(t) = \frac{1}{3} ∫_{0}^{t} sin[3(t − s)]f(s) ds.

2. [20%] (a) Say what jump conditions the solution y(t) of the following initial value problem satisfies at t = 0, and find the solution directly (do not use Laplace transforms):

y′ + 2y = δ(t),
y(t) = 0 for t < 0.

(b) Write the solution of the following initial value problem, where f(t) is an arbitrary function, as a convolution (you don’t need to derive your answer):

y′ + 2y = f(t),
y(0) = 0.

Solution.

• (a) The solution y(t) has a jump discontinuity of size one at t = 0, so that y(t) for t > 0 satisfies the IVP

y′ + 2y = 0,
y(0) = 1.

The solution of the ODE is

y(t) = ce^{−2t}

where c is a constant of integration, and the initial condition implies that c = 1. Hence,

y(t) = \{ \begin{matrix}e^{−2t} & for t > 0, \\ 0 & for t < 0. \end{matrix}

• (b) The solution is given by

y(t) = ∫_{0}^{t} e^{−2(t−s)}f(s) ds.

3. [20%] Use Fourier series to find the solution u(x, y) of the following boundary value problem for Laplace’s equation in the strip 0 < x < 1, y > 0:

\frac{∂^{2}u}{∂x^{2}} + \frac{∂^{2}u}{∂y^{2}} = 0,

\frac{∂u}{∂x}(0, y) = \frac{∂u}{∂x}(1, y) = 0,

u(x, 0) = x, u(x, y) is bounded as y → ∞.

What does the solution u(x, y) approach as y → ∞?

Solution.

• The appropriate linear combination of separated solutions of Laplace’s equation that satisfy the boundary conditions at x = 0, 1 and as y → ∞ is

u(x, y) = \frac{1}{2}a_{0} + ∑_{n=1}^{∞} a_{n} cos(nπx)e^{−nπy},

where the a_{n} are constants. Imposing the boundary condition at y = 0, we obtain that

x = \frac{1}{2}a_{0} + ∑_{n=1}^{∞} a_{n} cos(nπx),

so

a_{n} = 2 ∫_{0}^{1} x cos(nπx) dx.

Evaluating this integral, we find that a_{0} = 1, and for n ≥ 1

a_{n} = \{ \begin{matrix}−4/(nπ)^{2} & for n odd, \\ 0 & for n even. \end{matrix}

Hence the solution is

u(x, y) = \frac{1}{2} − \frac{4}{π^{2}} \{cos(πx)e^{−πy} + \frac{1}{3^{2}} cos(3πx)e^{−3πy}

+\frac{1}{5^{2}} cos(5πx)e^{−5πy} + …\} .

It follows that u(x, y) → 1/2 as y → ∞. Thus, the temperature approaches a constant value equal to the mean value of the boundary data at y = 0.

4. [20%] Reduce the boundary conditions to homogeneous boundary conditions and use Fourier series to find the solution u(x, t) of the following initial-boundary value problem for the heat equation in 0 < x < 1 and t > 0:

\frac{∂u}{∂t} = \frac{∂^{2}u}{∂x^{2}},

u(0, t) = 0, u(1, t) = 1, u(x, 0) = 0.

What does the solution u(x, t) approach as t → ∞?

Solution.

• The steady-state solution u = x satisfies the non-homogeneous boundary conditions and the PDE. We write

u(x, t) = x + v(x, t).

Then v(x, t) satisfies

\frac{∂v}{∂t} = \frac{∂^{2}v}{∂x^{2}} ,

v(0, t) = 0, v(1, t) = 0, v(x, 0) = −x.

We look for a solution for v of the form

v(x, t) = ∑_{n=1}^{∞} b_{n} sin(nπx)e^{−n^{2}π^{2}t}.

This function satisfies the PDE and the boundary conditions at x = 0, 1 for any choice of the constants b_{n}. Imposing the initial condition at t = 0 we get that

−x = ∑_{n=1}^{∞} b_{n} sin(nπx).

Using the expression for the Fourier sine coefficients, we get

b_{n} = −2 ∫_{0}^{1} x sin(nπx) dx
= \frac{2(−1)^{n}}{nπ} .

The solution of the original problem is therefore

u(x, t) = x + \frac{2}{π} ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} sin(nπx)e^{−n^{2}π^{2}t}.

As t → ∞, we have u(x, t) → x, so the temperature approaches the steady-state temperature distribution associated with the given nonhomogeneous boundary conditions.

5. [20%] (a) Consider the following initial value problem for u(x, t)

\frac{∂u}{∂t} = \frac{∂^{3}u}{∂x^{3}},

u(x, 0) = f(x),

where −∞ < x < ∞, t > 0. Solve for the Fourier transform of u(x, t) with respect to x, and write the solution for u(x, t) as a Fourier integral, but do not attempt to invert the transform explicitly.

(b) State Parseval’s theorem for the Fourier transform. Use your solution from (a) to show that

∫_{−∞}^{∞} u^{2}(x, t) dx = ∫_{−∞}^{∞} f^{2}(x) dx

for all times t. (This result expresses ‘conservation of energy’.)

Solution.

• (a) Let

û(k, t) = \frac{1}{2π} ∫_{−∞}^{∞} u(x, t)e^{−ikx} dx

be the Fourier transform of u(x, t) with respect to x. Then, Fourier tranforming the PDE, we find that

\frac{∂û}{∂t} = (ik)^{3}û,

û(k, 0) = f̂(k),

where f̂ is the Fourier transform of f. Solving this ODE, we get

û(k, t) = f̂(k)e^{(ik)^{3}t}.

Using this expression in the Fourier inversion formula,

u(x, t) = ∫_{−∞}^{∞} û(k, t)e^{ikx} dk,

and writing (ik)^{3} = −ik^{3}, we obtain the solution

u(x, t) = ∫_{−∞}^{∞} f̂(k)e^{ikx−ik^{3}t} dk.

• (b) Parseval’s theorem states that

\frac{1}{2π} ∫_{−∞}^{∞} |f(x)|^{2} dx = ∫_{−∞}^{∞} |f̂(k)|^{2} dk.

Since |e^{iθ}| = 1 for any real number θ, we have that

|e^{−ik^{3}t}| = 1.

Hence, for any t > 0, we have

|û(k, t)| = |f̂(k)e^{−ik^{3}t}| = |f̂(k)| .

It then follows from Parseval’s theorem and this equation that

∫_{−∞}^{∞} u^{2}(x, t) dx = 2π ∫_{−∞}^{∞} |û(k, t)|^{2} dk
= 2π ∫_{−∞}^{∞} |f̂(k)|^{2} dk
= ∫_{−∞}^{∞} f^{2}(x) dx.