Solutions to Sample Midterm 2 Math 121, Fall 2004

1. Use Fourier series to find the solution u(x, y) of the following boundary value problem for Laplace’s equation in the semi-infinite strip 0 < x < 1, y > 0:

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

u(0, y) = u(1, y) = 0, u(x, 0) = 1, u(x, y) → 0 as y → ∞.

Solution.

• The separated solutions of Laplace’s equation that satisfy the boundary conditions at x = 0, 1 and as y → ∞ are sin(nπx)e^{−nπy}, where n is a positive integer. We therefore look for a solution of the form

u(x, y) = ∑_{n=1}^{∞} b_{n} sin(nπx)e^{−nπy}.

Imposing the boundary condition at y = 0, we obtain

∑_{n=1}^{∞} b_{n} sin(nπx) = 1,

so

b_{n} = 2 ∫_{0}^{1} sin(nπx) dx
= 2 [\frac{− cos(nπx)}{nπ}]_{0}^{1}
= 2 [\frac{(−1)^{n+1}}{nπ} + \frac{1}{nπ} ]_{0}^{1}
= \{ \begin{matrix}4/(nπ) & for n odd, \\ 0 & for n even. \end{matrix}

The solution is therefore

u(x, y) = \frac{4}{π} \{sin(πx)e^{−πy} + \frac{1}{3} sin(3πx)e^{−3πy} + \frac{1}{5} sin(5πx)e^{−5πy} + …\} .

2. Use Fourier series to find the solution u(x, t) of the following initialboundary value problem for the wave equation in 0 < x < 1 and t > 0:

\frac{∂^{2}u}{∂t^{2}} − \frac{∂^{2}u}{∂x^{2}} = 0,

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

u(x, 0) = 0,

\frac{∂u}{∂t}(x, 0) = x.

Solution.

• The separated solutions of the wave equation that are zero at t = 0 and satisfy the boundary conditions at x = 0, 1 are t and cos(nπx) sin(nπt), where n = 1, 2, . . .. We therefore look for a solution of the form

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

Differentiating this series with respect to t, we find that

\frac{∂u}{∂t}(x, t) = \frac{1}{2}a_{0} + ∑_{n=1}^{∞} nπa_{n} cos(nπx) cos(nπt).

Imposing the initial condition for ∂u/∂t at t = 0, we get the Fourier cosine expansion:

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

Hence, for n ≥ 1 we have

nπa_{n} = 2 ∫_{0}^{1} x cos(nπx) dx
= 2 [\frac{x sin(nπx)}{nπ} + \frac{cos(nπx)}{(nπ)^{2}}]_{0}^{1}
= 2 [\frac{(−1)^{n}}{(nπ)^{2}} − \frac{1}{(nπ)^{2}} ]_{0}^{1}
= \{ \begin{matrix}−4/(nπ)^{2} & for n odd, \\ 0 & for n even, \end{matrix}

and

a_{n} = \{ \begin{matrix}−4/(nπ)^{3} & for n odd, \\ 0 & for n even. \end{matrix}

For n = 0, we get

a_{0} = 2 ∫_{0}^{1} x dx
= 2 [\frac{1}{2}x^{2}]_{0}^{1}
= 1.

Hence, the solution is

u(x, t) = \frac{1}{2}t − \frac{4}{π^{3}} \{cos(πx) + \frac{1}{3^{3}} cos(3πx) + \frac{1}{5^{3}} cos(5πx) + …\} .

3. Use Fourier transforms to solve the following initial value problem for u(x, t) in −∞ < x < ∞, t > 0:

\frac{∂u}{∂t} = −\frac{∂^{4}u}{∂x^{4}},

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

Write the solution for u(x, t) as a convolution, but do not compute any inverse transforms explicitly. How smooth is the solution for t > 0?

Solution.

• Let

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

be the Fourier transform of u with respect to x. Then, taking the Fourier transform of the initial value problem, we get

\frac{∂û}{∂t} = −(−ik)^{4}û,

û(k, 0) = f̂(k),

where f̂ is the Fourier transform of f. It follows that

\frac{∂û}{∂t} = −k^{4}û,

û(k, 0) = f̂(k),

which has the solution

û(k, t) = f̂(k)e^{−k^{4}t}.

According to the convolution theorem, if f, g have Fourier transforms f̂, ĝ respectively then f̂·ĝ is the Fourier transform of \frac{1}{2π}f ∗g. It follows that

u(x, t) = ∫_{−∞}^{∞} G(x − y, t)f(y) dy

where

Ĝ(k, t) = \frac{1}{2π} e^{−k^{4}t}.

The solution is smooth (infinitely differentiable with respect to x) for t > 0 since its Fourier transform decays exponentially quickly as k → ∞ (assuming, for example, that f̂(k) is a bounded function of k).

4. (a) Give the formulas for the Fourier transform f̂(k) of a function f(x) and the inverse Fourier transform.

(b) Compute the Fourier transform of e^{−|x|}.

(c) State Parseval’s theorem, and use it to evaluate

∫_{0}^{∞} \frac{1}{(1 + k^{2})^{2}} dk.

Solution.

• (a) A function f(x) and its Fourier transform f̂(k) are related by

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

• (b) If f(x) = e^{−|x|}, then using

|x| = \{ \begin{matrix}x & for x ≥ 0, \\ −x & for x ≤ 0, \end{matrix}

and changing x → −x in the integral for −∞ < x < 0, we find that

f̂(k) = \frac{1}{2π} ∫_{−∞}^{∞} e^{−|x|}e^{−ikx} dx,
= \frac{1}{2π} \{∫_{−∞}^{0} e^{(1−ik)x} dx + ∫_{0}^{∞} e^{−(1+ik)x} dx\}
= \frac{1}{2π} ∫_{0}^{∞} \{e^{−(1−ik)x} + e^{−(1+ik)x}\} dx
= −\frac{1}{2π} [\frac{e^{−(1−ik)x}}{1 − ik} + \frac{e^{−(1+ik)x}}{1 + ik} ]_{0}^{∞}
= \frac{1}{2π} [\frac{1}{1 − ik} + \frac{1}{1 + ik} ]_{0}^{∞}
= \frac{1}{π} \frac{1}{1 + k^{2}} .

• (c) Parseval’s theorem states that

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

For f(x) = e^{−|x|}, we compute that

∫_{−∞}^{∞} |f(x)|^{2} dx = 2 ∫_{0}^{∞} e^{−2x} dx
= − [e^{−2x}]_{0}^{∞}
= 1.

It follows from Parsevals theorem and (b) that

\frac{2}{π^{2}} ∫_{0}^{∞} \frac{1}{(1 + k^{2})^{2}} dk = \frac{1}{2π} · 1,

so

∫_{0}^{∞} \frac{1}{(1 + k^{2})^{2}} dk = \frac{π}{4} .

• Remark. The integral in (c) can also be evaluated directly by use of the substitution k = tan θ, which gives

∫_{0}^{∞} \frac{1}{(1 + k^{2})^{2}} dk = ∫_{0}^{π/2} \frac{1}{(1 + tan^{2} θ)^{2}} sec^{2} θ dθ
= ∫_{0}^{π/2} \frac{1}{sec^{4} θ} sec^{2} θ dθ
= ∫_{0}^{π/2} \frac{1}{sec^{2} θ} dθ
= ∫_{0}^{π/2} cos^{2} θ dθ
= \frac{π}{2} · \frac{1}{2}
= \frac{π}{4} ,

which verifies Parseval’s theorem explicitly in this case.

5. Use Laplace transforms to solve the following initial value problem: y′′ + 2y′ + 2y = 1,

y(t) = 0, y′(0) = 1.

Solution.

• Let Y (p) be the Laplace transform of y(t). Then, taking the Laplace

transform of the ODE and using the initial conditions, we get that

p^{2}Y − 1 + 2pY + 2Y = \frac{1}{p}.

Solving for Y, we get

Y (p) = \frac{1}{p^{2} + 2p + 2} + \frac{1}{p(p^{2} + 2p + 2)}.

We have p^{2} + 2p + 2 = (p + 1)^{2} + 1, so (from L13 of the table)

L^{−1} [\frac{1}{p^{2} + 2p + 2}] = e^{−t} sin t.

Also

\frac{1}{p(p^{2} + 2p + 2)} = \frac{1}{2} [\frac{1}{p} − \frac{p + 2}{p^{2} + 2p + 2}]
= \frac{1}{2} [\frac{1}{p} − \frac{p + 1}{(p + 1)^{2} + 1} − \frac{1}{(p + 1)^{2} + 1}] .

So (from L1, L13, L14) we have

L^{−1} [\frac{1}{p(p^{2} + 2p + 2)}] = \frac{1}{2} [1 − e^{−t} cos t − e^{−t} sin t]

Hence, combining these inverse transforms, we get

y(t) = \frac{1}{2} [1 − e^{−t} cos t + e^{−t} sin t] .

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

y′′ − 4y = 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′′ − 4y = f(t),
y(0) = y′(0) = 0.

Solution.

• (a) The derivative of y has a jump discontinuity of size one at t = 0. The solution is therefore

y(t) = \{ \begin{matrix}y_{+}(t) & for t ≥ 0, \\ 0 & for t < 0, \end{matrix}

where

y′′_{+} − 4y_{+} = 0 for t > 0,
y_{+}(0) = 0, y′_{+}(0) = 1.

The general solution of the ODE is y_{+}(t) = a cosh 2t + b sinh 2t, and the initial conditions imply that a = 0 and b = 1/2. Hence,

y(t) = \frac{1}{2} sinh 2t for t ≥ 0.

• (b) The solution is

y(t) = \frac{1}{2} ∫_{0}^{t} sinh 2(t − s)f(s) ds.