4.11 Exercises

These are homework exercises to accompany Libl's "Differential Equations for Engineering" Textmap. This is a textbook targeted for a one semester first course on differential equations, aimed at engineering students. Prerequisite for the course is the basic calculus sequence.

4.1: Boundary value problems

Hint for the following exercises: Note that when $λ greater than 0$ , then $\cos (\sqrt{λ} (t - a))$ and $\sin (\sqrt{λ} (t - a))$ are also solutions of the homogeneous equation.

Your Turn

(challenging)

Suppose $x^{‴} + λ x = 0$ and $x (0) = 0, x^{'} (0) = 0, x (1) = 0$ . Suppose that $λ greater than 0$ . Find an equation that all such eigenvalues must satisfy. Hint: Note that $- \sqrt[3]{λ}$ is a root of $r^{3} + λ = 0$ .

Answer

$\frac{\sqrt{3}}{3} e^{\frac{- 3}{2} \sqrt[3]{λ}} - \frac{\sqrt{3}}{3} \cos (\frac{\sqrt{3} \sqrt[3]{λ}}{2}) + \sin (\frac{\sqrt{3} \sqrt[3]{λ}}{2}) = 0$

4.2: The Trigonometric Series

Your Turn

Suppose $f (t)$ is defined on $open parenthesis minus π, π ]$ as

$\begin{matrix} - 1 i f - π < t \leq 0, \\ 1 i f 0 < t \leq π . \end{matrix}$

Extend periodically and compute the Fourier series of $f (t)$ .

There is another form of the Fourier series using complex exponentials $e^{n t}$ for $n = \dots, - 2, - 1, 0, 1, 2, \dots$ instead of $\cos (n t)$ and $\sin (n t)$ for positive $n$ . This form may be easier to work with sometimes. It is certainly more compact to write, and there is only one formula for the coefficients. On the downside, the coefficients are complex numbers.

Your Turn

Let

$f (t) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (n t) + b_{n} \sin (n t) .$

Use Euler’s formula $e^{i θ} = \cos (θ) + i \sin (θ)$ to show that there exist complex numbers $c_{m}$ such that

$f (t) = \sum_{m = - \infty}^{\infty} c_{m} e^{i m t} .$

Note that the sum now ranges over all the integers including negative ones. Do not worry about convergence in this calculation. Hint: It may be better to start from the complex exponential form and write the series as

$c_{0} + \sum_{m = 1}^{\infty} c_{m} e^{i m t} + c_{- m} e^{- i m t} .$

Your Turn

Suppose $f (t)$ is defined on $open parenthesis minus π, π ]$ as $f (t) = \sin (π t)$ . Extend periodically and compute the Fourier series.

Answer

$\sum_{n = 1}^{\infty} \frac{(π - n) \sin (π n + π^{2}) + (π + n) \sin (π n - π^{2})}{π n^{2} - π^{3}} \sin (n t)$

Your Turn

Suppose $f (t)$ is defined on $open parenthesis minus π, π ]$ as $f (t) = t^{4}$ . Extend periodically and compute the Fourier series.

Answer

$\frac{π^{4}}{5} + \sum_{n = 1}^{\infty} \frac{(- 1)^{n} (8 π^{2} n^{2} - 48)}{n^{4}} \cos (n t)$

4.3: More on the Fourier series

Your Turn

Let

$0 if minus 3 less than t less than or equal to 0,; t if 0 less than t less than or equal to 3,$

extended periodically. Suppose

F (t)

is the function given by the Fourier series of

f

. Without computing the Fourier series evaluate.

$F open parenthesis 2 close parenthesis$
$F open parenthesis minus 2 close parenthesis$
$F open parenthesis 4 close parenthesis$
$F open parenthesis minus 4 close parenthesis$
$F open parenthesis 3 close parenthesis$
$F open parenthesis minus 9 close parenthesis$

Your Turn

Let

$f (t) = t^{2} f o r - 2 < t \leq 2$

extended periodically.

Compute the Fourier series for $f (t)$ .
Write out the series explicitly up to the $3 to the power rd$ harmonic.

Answer

$\frac{8}{6} + \sum_{n = 1}^{\infty} \frac{16 (- 1)^{n}}{π^{2} n^{2}} \cos (\frac{n π}{2} t)$
$\frac{8}{6} - \frac{16}{π^{2}} \cos (\frac{π}{2} t) + \frac{4}{π^{2}} \cos (π t) - \frac{16}{9 π^{2}} \cos (\frac{3 π}{2} t) + \dots$

Your Turn

Let

$f (t) = t f o r λ < t \leq λ (f o r s o m e λ)$

extended periodically.

Compute the Fourier series for $f (t)$ .
Write out the series explicitly up to the $3 to the power rd$ harmonic.

Answer

$\sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1} 2 λ}{n π} \sin (\frac{n π}{λ} t)$
$\frac{2 λ}{π} \sin (\frac{π}{λ} t) - \frac{λ}{π} \sin (\frac{2 π}{λ} t) + \frac{2 λ}{3 π} \sin (\frac{3 π}{λ} t) - \dots$

Your Turn

Let

$f (t) = \frac{1}{2} + \sum_{n = 1}^{\infty} \frac{1}{n (n^{2} + 1)} \sin (n π t) .$

Compute

f^{'} (t)

Answer

$f^{'} (t) = \sum_{n = 1}^{\infty} \frac{π}{n^{2} + 1} \cos (n π t)$

Your Turn

Let

$f (t) = \frac{1}{2} + \sum_{n = 1}^{\infty} \frac{1}{n^{3})} \cos (n t) .$

Find the antiderivative.
Is the antiderivative periodic?

Answer

$F (t) = \frac{t}{2} + C + \sum_{n = 1}^{\infty} \frac{1}{n^{4}} \sin (n t)$
no

Your Turn

Let

$f (t) = \frac{t}{2} f o r - π < t \leq π$

extended periodically.

Compute the Fourier series for $f (t)$ .
Plug in $t = \frac{π}{2}$ to find a series representation for $\frac{π}{4}$ .
Using the first 4 terms of the result from part b) approximate $\frac{π}{4}$ .

Answer

$\sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n} \sin (n t)$
$f$ is continuous at $t = \frac{π}{2}$ so the Fourier series converges to $f (\frac{π}{2}) = \frac{π}{4}$ . Obtain $\frac{π}{4} = \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{2 n - 1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots$ .
Using the first $4$ terms get $76 / 105 approximately equals 0.72$ (quite a bad approximation, you would have to take about $50$ terms to start to get to within $0.01$ of $\frac{π}{4}$ ).

Your Turn

Let

$0 if minus 2 less than t less than or equal to 0,; 2 if 0 less than t less than or equal to 2,$

extended periodically. Suppose

F (t)

is the function given by the Fourier series of

f

. Without computing the Fourier series evaluate.

$F open parenthesis 0 close parenthesis$
$F open parenthesis minus 1 close parenthesis$
$F open parenthesis 1 close parenthesis$
$F open parenthesis minus 2 close parenthesis$
$F open parenthesis 4 close parenthesis$
$F open parenthesis minus 8 close parenthesis$

Answer

$F open parenthesis 0 close parenthesis equals 1$
$F (- 1) = 0$
$F open parenthesis 1 close parenthesis equals 2$
$F (- 2) = 1$
$F open parenthesis 4 close parenthesis equals 1$
$F (- 9) = 0$

4.4: Sine and Cosine Series

Your Turn

Let $f (t) = \frac{t}{3}$ on $0 less than or equal to t less than 3$ .

Find the Fourier series of the even periodic extension.
Find the Fourier series of the odd periodic extension.

Answer

$\frac{1}{2} + \sum_{\overset{n = 1}{n odd}}^{\infty} \frac{- 4}{π^{2} n^{2}} \cos (\frac{n π}{3} t)$
$\sum_{n = 1}^{\infty} \frac{2 {(- 1)}^{n + 1}}{π n} \sin (\frac{n π}{3} t)$

Your Turn

Let $f open parenthesis t close parenthesis equals cos open parenthesis 2 t close parenthesis$ on $0 less than or equal to t less than π$ .

Find the Fourier series of the even periodic extension.
Find the Fourier series of the odd periodic extension.

Answer

$cos open parenthesis 2 t close parenthesis$
$\sum_{\overset{n = 1}{n odd}}^{\infty} \frac{- 4 n}{π n^{2} - 4 π} \sin (n t)$

Your Turn

Let $f (t) = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \sin (n t)$ . Solve $x^{″} - x = f (t)$ for the Dirichlet conditions $x open parenthesis 0 close parenthesis equals 0$ and $x open parenthesis π close parenthesis equals 0$ .

Answer

$\sum_{n = 1}^{\infty} \frac{- 1}{n^{2} (1 + n^{2})} \sin (n t)$

Your Turn

(challenging)

Let $f (t) = t + \sum_{n = 1}^{\infty} \frac{1}{2^{n}} \sin (n t)$ . Solve $x^{″} + π x = f (t)$ for the Dirichlet conditions $x open parenthesis 0 close parenthesis equals 0$ and $x open parenthesis π close parenthesis equals 1$ . Hint: Note that $\frac{t}{π}$ satisfies the given Dirichlet conditions.

Answer

$\frac{t}{π} + \sum_{n = 1}^{\infty} \frac{1}{2^{n} (π - n^{2})} \sin (n t)$

4.5: Applications of Fourier series

Your Turn

Let $F (t) = \sin (2 π t) + 0.1 \cos (10 π t)$ . Find the steady periodic solution to $x^{″} + \sqrt{2} x = F (t)$ . Express your solution as a Fourier series.

Answer

$x = \frac{1}{\sqrt{2} - 4 π^{2}} \sin (2 π t) + \frac{0.1}{\sqrt{2} - 100 π^{2}} \cos (10 π t)$

Your Turn

Let $F (t) = \sum_{n = 1}^{\infty} e^{- n} \cos (2 n t)$ . Find the steady periodic solution to $x^{″} + 3 x = F (t)$ . Express your solution as a Fourier series.

Answer

$x = \sum_{n = 1}^{\infty} \frac{e^{- n}}{3 - (2 n)^{2}} \cos (2 n t)$

Your Turn

Let $F open parenthesis t close parenthesis equals | t |$ for $minus 1 less than or equal to t less than or equal to 1$ extended periodically. Find the steady periodic solution to $x^{″} + \sqrt{3} x = F (t)$ . Express your solution as a series.

Answer

$x = \frac{1}{2 \sqrt{3}} + \sum_{\overset{n = 1}{n odd}}^{\infty} \frac{- 4}{n^{2} π^{2} (\sqrt{3} - n^{2} π^{2})} \cos (n π t)$

Your Turn

Let $F open parenthesis t close parenthesis equals | t |$ for $minus 1 less than or equal to t less than or equal to 1$ extended periodically. Find the steady periodic solution to $x^{″} + π^{2} x = F (t)$ . Express your solution as a series.

Answer

$x = \frac{1}{2 \sqrt{3}} - \frac{2}{π^{3}} t \sin (π t) + \sum_{\overset{n = 3}{n odd}}^{\infty} \frac{- 4}{n^{2} π^{4} (1 - n^{2})} \cos (n π t)$

4.6: PDEs, Separation of Variables, and the Heat Equation

Your Turn

Find a series solution of

$\begin{matrix} u_{t} = u_{x x}, \\ u (0, t) = u (1, t) = 0, \\ u (x, 0) = 100 f o r 0 < x < 1 . \end{matrix}$

Your Turn

Find a series solution of

$\begin{matrix} u_{t} = u_{x x}, \\ u_{x} (0, t) = u_{x} (π, t) = 0, \\ u (x, 0) = 3 \cos (x) + \cos (3 x) f o r 0 < x < π . \end{matrix}$

Your Turn

Find a series solution of

$\begin{matrix} u_{t} = \frac{1}{3} u_{x x}, \\ u_{x} (0, t) = u_{x} (π, t) = 0, \\ u (x, 0) = \frac{10 x}{π} f o r 0 < x < π . \end{matrix}$

Your Turn

Find a series solution of

$\begin{matrix} u_{t} = u_{x x}, \\ u (0, t) = 0, u (1, t) = 100, \\ u (x, 0) = \sin (π x) f o r 0 < x < 1 . \end{matrix}$

Hint: Use the fact that $u open parenthesis x, t close parenthesis equals 100 x$ is a solution satisfying $u_{t} = u_{x x}, u (0, t) = 0, u (1, t) = 100$ . Then usesuperposition.

Your Turn

(challenging)

Suppose that one end of the wire is insulated (say at $x equals 0$ ) and the other end is kept at zero temperature. That is, find a series solution of

$\begin{matrix} u_{t} = k u_{x x}, \\ u_{x} (0, t) = u (L, t) = 0, \\ u (x, 0) = f (x) f o r 0 < x < L . \end{matrix}$

Express any coefficients in the series by integrals of $f (x)$ .

Your Turn

(challenging)

Suppose that the wire is circular and insulated, so there are no ends. You can think of this as simply connecting the two ends and making sure the solution matches up at the ends. That is, find a series solution of

$\begin{matrix} u_{t} = k u_{x x}, \\ u (0, t) = u (L, t), u_{x} (0, t) = u_{x} (L, t) \\ u (x, 0) = f (x) f o r 0 < x < L . \end{matrix}$

Express any coefficients in the series by integrals of $f (x)$ .

Your Turn

Find a series solution of

$\begin{matrix} u_{t} = 3 u_{x x}, \\ u (0, t) = u (π, t) = 0, \\ u (x, 0) = 5 \sin (x) + 2 \sin (5 x) f o r 0 < x < π . \end{matrix}$

Answer

$u (x, t) = 5 \sin (x) e^{3 t} + 2 \sin (5 x) e^{- 75 t}$

Your Turn

Find a series solution of

$\begin{matrix} u_{t} = 0.1 u_{x x}, \\ u_{x} (0, t) = u_{x} (π, t) = 0, \\ u (x, 0) = 1 + 2 \cos (x) f o r 0 < x < π . \end{matrix}$

Answer

$u (x, t) = 1 + 2 \cos (x) e^{- 0.1 t}$

Your Turn

Suppose that the temperature on the wire is fixed at $0$ at the ends, $L equals 1$ , $k equals 1$ , and $u open parenthesis x, 0 close parenthesis equals 100 sin open parenthesis 2 π x close parenthesis$ .

What is the temperature at $x = \frac{1}{2}$ at any time.
What is the maximum and the minimum temperature on the wire at $t equals 0$ .
At what time is the maximum temperature on the wire exactly one half of the initial maximum at $t equals 0$ .

Answer

$0$
$minimum minus 100$ , $maximum 100$
$t = \frac{\ln 2}{4 π^{2}}$

4.7: One dimensional wave equation

Your Turn

Solve

$\begin{matrix} y_{t t} = 9 y_{x x}, \\ y (0, t) = y (1, t) = 0, \\ y (x, 0) = \sin (3 π x) + \frac{1}{4} \sin (6 π x) f o r 0 < x < 1, \\ y_{t} (x, 0) = 0 f o r 0 < x < 1 . \end{matrix}$

Your Turn

Solve

$\begin{matrix} y_{t t} = 4 y_{x x}, \\ y (0, t) = y (1, t) = 0, \\ y (x, 0) = \sin (3 π x) + \frac{1}{4} \sin (6 π x) f o r 0 < x < 1, \\ y_{t} (x, 0) = \sin (9 π x) f o r 0 < x < 1 . \end{matrix}$

Your Turn

(challenging)

Suppose that you have a vibrating string and that there is air resistance proportional to the velocity. That is, you have

$\begin{matrix} y_{t t} = a^{2} y_{x x} - k y_{t}, \\ y (0, t) = y (1, t) = 0, \\ y (x, 0) = f (x) f o r 0 < x < 1, \\ y_{t} (x, 0) = 0 f o r 0 < x < 1 . \end{matrix}$

Suppose that $0 less than k less than 2 π a$ . Derive a series solution to the problem. Any coefficients in the series should be expressed as integrals of $f (x)$ .

Your Turn

Solve

$\begin{matrix} y_{t t} = y_{x x}, \\ y (0, t) = y (π, t) = 0, \\ y (x, 0) = \sin (x) f o r 0 < x < π, \\ y_{t} (x, 0) = \sin (x) f o r 0 < x < π . \end{matrix}$

Answer

$y (x, t) = \sin (x) (\sin (t) + \cos (t))$

Your Turn

Solve

$\begin{matrix} y_{t t} = 25 y_{x x}, \\ y (0, t) = y (2, t) = 0, \\ y (x, 0) = 0 f o r 0 < x < 2, \\ y_{t} (x, 0) = \sin (π x) + 0.1 \sin (2 π t) f o r 0 < x < 2 . \end{matrix}$

Answer

$y (x, t) = \frac{1}{5 π} \sin (π x) \sin (5 π t) + \frac{1}{100 π} \sin (2 π x) \sin (10 π t)$

Your Turn

Solve

$\begin{matrix} y_{t t} = 2 y_{x x}, \\ y (0, t) = y (π, t) = 0, \\ y (x, 0) = x f o r 0 < x < π, \\ y_{t} (x, 0) = 0 f o r 0 < x < π . \end{matrix}$

Answer

$y (x, t) = \sum_{n = 1}^{\infty} \frac{2 (- 1)^{n + 1}}{n} \sin (n x) \cos (n \sqrt{2} t)$

4.8: D’Alembert solution of the wave equation

Your Turn

The d’Alembert solution still works if there are no boundary conditions and the initial condition is defined on the whole real line. Suppose that $y_{t t} = y_{x x}$ (for all $x$ on the real line and $t greater than or equal to 0$ ), $y open parenthesis x, 0 close parenthesis equals f open parenthesis x close parenthesis$ , and $y sub t open parenthesis x, 0 close parenthesis$ , where

$\begin{matrix} 0 i f x < - 1, \\ x + 1 i f - 1 \leq x < 0, \\ - x + 1 i f 0 \leq x < 1 \\ 0 i f x > 1 . \end{matrix}$

Solve using the d’Alembert solution. That is, write down a piecewise definition for the solution. Then sketch the solution for $t = 0, t = 1 / 2, t = 1$ , and $t equals 2$ .

Your Turn

Using the d’Alembert solution solve $y sub t t equals 9 y sub x x$ , $0 < x < 1, t > 0, y (0, t) = y (1, t) = 0, y (x, 0) = \sin (2 π x)$ , and $y sub t open parenthesis x, 0 close parenthesis equals sin open parenthesis 3 π x close parenthesis$ .

Answer

$y (x, t) = \frac{\sin (2 π (x - 3 t)) + \sin (2 π (3 t + x))}{2} + \frac{\cos (3 π (x - 3 t)) - \cos (3 π (3 t + x))}{18 π}$

Your Turn

Take $y sub t t equals 4 y sub x x$ , $0 < x < 1, t > 0, y (0, t) = y (1, t) = 0, y (x, 0) = x - x^{2}$ , and $y sub t open parenthesis x, 0 close parenthesis equals 0$ . Using the D’Alembert solution find the solution at

$t equals 0.1$ ,
$t equals 1 / 2$ ,
$t equals 1$ .

You may have to split your answer up by cases.

Answer

$\begin{matrix} x - x^{2} - 0.04 i f 0.2 \leq x \leq 0.8 \\ 0.6 x i f x \leq 0.2 \\ 0.6 - 0.6 x i f x \geq 0.8 \end{matrix}$
$y (x, \frac{1}{2}) = - x + x^{2}$
$y (x, 1) = x - x^{2}$

Your Turn

Take $y sub t t equals 100 y sub x x$ , $0 < x < 4, t > 0, y (0, t) = y (4, t) = 0, y (x, 0) = F (x)$ , and $y sub t open parenthesis x, 0 close parenthesis equals 0$ . Suppose that $F (0) = 0, F (1) = 2, F (2) = 3, F (3) = 1$ . Using the D’Alembert solution find

$y open parenthesis 1, 1 close parenthesis$ ,
$y open parenthesis 4, 3 close parenthesis$ ,
$y open parenthesis 3, 9 close parenthesis$ .

Answer

$y (1, 1) = - \frac{1}{2}$
$y open parenthesis 4, 3 close parenthesis equals 0$
$y (3, 9) = \frac{1}{2}$

4.9: Steady state temperature and the Laplacian

Your Turn

Let $R$ be the region described by $0 less than x less than π$ and $0 less than y less than π$ . Solve the problem

$Δ u = 0, u (x, 0) = \sin x, u (x, π) = 0, u (0, y) = 0, u (π, y) = 0 .$

Your Turn

Let $R$ be the region described by $0 less than x less than 1$ and $0 less than y less than 1$ . Solve the problem

$\begin{matrix} u_{x x} + u_{y y} = 0, \\ u (x, 0) = \sin (π x) - \sin (2 π x), u (x, 1) = 0, \\ u (0, y) = 0, u (1, y) = 0 . \end{matrix}$

Your Turn

Let $R$ be the region described by $0 less than x less than 1$ and $0 less than y less than 1$ . Solve the problem

$\begin{matrix} u_{x x} + u_{y y} = 0, \\ u (x, 0) = u (x, 1) = u (0, y) = u (1, y) = C . \end{matrix}$

for some constant $C$ . Hint: Guess, then check your intuition.

Your Turn

Let $R$ be the region described by $0 less than x less than π$ and $0 less than y less than π$ . Solve

$Δ u = 0, u (x, 0) = 0, u (x, π) = π, u (0, y) = y, u (π, y) = y .$

Hint: Try a solution of the form $u (x, y) = X (x) + Y (y)$ (different separation of variables).

Your Turn

Use the solution of Exercise $4$ to solve

$Δ u = 0, u (x, 0) = \sin x, u (x, π) = π, u (0, y) = y, u (π, y) = y .$

Hint: Use superposition.

Your Turn

Let $R$ be the region described by $0 less than x less than w$ and $0 less than y less than h$ . Solve the problem

$\begin{matrix} u_{x x} + u_{y y} = 0, \\ u (x, 0) = 0, u (x, h) = f (x), \\ u (0, y) = 0, u (w, y) = 0 . \end{matrix}$

The solution should be in series form using the Fourier series coefficients of $f (x)$ .

Your Turn

Let $R$ be the region described by $0 less than x less than w$ and $0 less than y less than h$ . Solve the problem

$\begin{matrix} u_{x x} + u_{y y} = 0, \\ u (x, 0) = 0, u (x, h) = 0, \\ u (0, y) = f (y), u (w, y) = 0 . \end{matrix}$

The solution should be in series form using the Fourier series coefficients of $f (y)$ .

Your Turn

Let $R$ be the region described by $0 less than x less than w$ and $0 less than y less than h$ . Solve the problem

$\begin{matrix} u_{x x} + u_{y y} = 0, \\ u (x, 0) = 0, u (x, h) = 0, \\ u (0, y) = 0, u (w, y) = f (y) . \end{matrix}$

The solution should be in series form using the Fourier series coefficients of $f (y)$ .

Your Turn

Let $R$ be the region described by $0 less than x less than 1$ and $0 less than y less than 1$ . Solve the problem

$\begin{matrix} u_{x x} + u_{y y} = 0, \\ u (x, 0) = \sin (9 π x), u (x, 1) = \sin (2 π x), \\ u (0, y) = 0, u (1, y) = 0 . \end{matrix}$

Hint: Use superposition.

Your Turn

Let $R$ be the region described by $0 less than x less than 1$ and $0 less than y less than 1$ . Solve the problem

$\begin{matrix} u_{x x} + u_{y y} = 0, \\ u (x, 0) = \sin (π x), u (x, 1) = \sin (π x), \\ u (0, y) = \sin (π y), u (1, y) = \sin (π y) . \end{matrix}$

Hint: Use superposition.

Your Turn

Let $R$ be the region described by $0 less than x less than 1$ and $0 less than y less than 1$ . Solve the problem

$Δ u = 0, u (x, 0) = \sum_{n = 1}^{\infty} \sin (n π x), u (x, 1) = 0, u (0, y) = 0, u (1, y) = 0 .$

Answer

$u (x, y) = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \sin (n π x) (\frac{\sinh (n π (1 - y))}{\sinh (n π)})$

Your Turn

Let $R$ be the region described by $0 less than x less than 1$ and $0 less than y less than 2$ . Solve the problem

$Δ u = 0, u (x, 0) = 0.1 \sin (π x), u (x, 2) = 0, u (0, y) = 0, u (1, y) = 0 .$

Answer

$u (x, y) = 0.1 \sin (π x) (\frac{\sinh (π (2 - y))}{\sinh (2 π)})$

4.10: Dirichlet problem in the circle and the Poisson kernel

Your Turn

Using series solve $Δ u = 0, u (1, θ) = g (θ)$ for the following data. Hint: trig identities.

$g (θ) = 1 / 2 + 3 \sin (θ) + \cos (3 θ)$
$g (θ) = \cos (3 θ) + 3 \sin (3 θ) + \sin (9 θ)$
$g (θ) = 2 \cos (θ + 1)$
$g (θ) = \sin^{2} (θ)$

Your Turn

Try and guess a solution to $Δ u = - 1, u (1, θ) = 0$ . Hint: try a solution that only depends on $r$ . Also first, don’t worry about the boundary condition.
Now solve $Δ u = - 1, u (1, θ) = \sin (2 θ)$ using superposition.

Answer

$u = \frac{- 1}{4} r^{2} + \frac{1}{4}$
$u = \frac{- 1}{4} r^{2} + \frac{1}{4} + r^{2} \sin (2 θ)$

Your Turn

(challenging)

Derive the Poisson kernel solution if the region is a circle of radius $ρ$ rather than $1$ . That is, solve $Δ u = 0, u (ρ, θ) = g (θ)$ .

Answer

$u (r, θ) = \frac{1}{2 π} \int_{- π}^{π} \frac{ρ^{2} - r^{2}}{ρ - 2 r ρ \cos (θ - α) + r^{2}} g (α) d α$

Adapted from Differential Equations for Engineers by Jiří Lebl (Oklahoma State University), hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.