Author Topic: Problem 3  (Read 15377 times)

Thomas Nutz

• Full Member
• Posts: 26
• Karma: 1
Problem 3
« on: November 12, 2012, 04:40:07 PM »
Dear Professor,

do we need to find a closed expression for the solution in problem 3, or can we express it as a sum over n?

Victor Ivrii

• Elder Member
• Posts: 2601
• Karma: 0
Re: Problem 3
« Reply #1 on: November 12, 2012, 04:45:03 PM »
Dear Professor,

do we need to find a closed expression for the solution in problem 3, or can we express it as a sum over n?

Yes, either as a series or as an integral (or present two forms)--there is no "closed" expression

Thomas Nutz

• Full Member
• Posts: 26
• Karma: 1
Re: Problem 3
« Reply #2 on: November 17, 2012, 03:21:05 PM »
Are we supposed to derive eq. 19 in lecture notes 24 (laplacian outside the disk) for problem 3b) or can we start from this formula (it given in the context of an exercise)?

Victor Ivrii

• Elder Member
• Posts: 2601
• Karma: 0
Re: Problem 3
« Reply #3 on: November 17, 2012, 04:25:27 PM »
Are we supposed to derive eq. 19 in lecture notes 24 (laplacian outside the disk) for problem 3b) or can we start from this formula (it given in the context of an exercise)?

Better forget about this formula, separate variables and use the boundary value provided and not try to derive formula at all.

Ian Kivlichan

• Sr. Member
• Posts: 51
• Karma: 17
Re: Problem 3
« Reply #4 on: November 19, 2012, 09:30:20 PM »
Hopeful solutions attached!

(parts 1,2)

Ian Kivlichan

• Sr. Member
• Posts: 51
• Karma: 17
Re: Problem 3
« Reply #5 on: November 19, 2012, 09:31:27 PM »
Hopeful solutions attached!

(part 3)

Chen Ge Qu

• Full Member
• Posts: 16
• Karma: 8
Re: Problem 3
« Reply #6 on: November 19, 2012, 09:33:01 PM »
a) Part 1 of 2

Chen Ge Qu

• Full Member
• Posts: 16
• Karma: 8
Re: Problem 3
« Reply #7 on: November 19, 2012, 09:33:48 PM »
a) Part 2 of 2

Chen Ge Qu

• Full Member
• Posts: 16
• Karma: 8
Re: Problem 3
« Reply #8 on: November 19, 2012, 09:34:25 PM »
Part b) attached

Calvin Arnott

• Sr. Member
• Posts: 43
• Karma: 17
• OK
Re: Problem 3
« Reply #9 on: November 19, 2012, 09:38:03 PM »
Problem 3

Part a. Solve:

$$\Delta u := u_{xx} + u_{yy} = 0, \phantom{\ }r < a$$
$$u \bigr|_{r=a} = f\left(\theta\right)$$
\text{in polar coordiantes } \left(r,\theta\right) \text{, with: } f(x) = \left\{\begin{aligned} &\phantom{-\ }1 &&: 0 < \theta < \pi\\ &-1 &&: \pi < \theta < 2\pi \end{aligned} \right.

If $a \le 0$, clearly the problem is not well-posed or yields only a singular point-solution, as for polar coordinates we take only $r \ge 0$. Let $a > 0$. Then this is the Dirichlet problem on the disk of radius $a$, with BC $h\left(\theta\right) = f\left(\theta\right)$. Because we have $u = f$ on the circumference of the disk, we have that $|u| \le |f| = 1$ on the disk by the maximum principle, and $|u|$ is bounded. Then, it was derived that by changing to polar coordinates $\left(x,y\right) \mapsto \left(r,\theta\right)$ with Dirichlet BC: $\{u = h\left(\theta\right) : r = a\}$, $u\left(r,\theta\right)$ has a solution of the form:

$$u\left(r,\theta\right) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^n \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)$$

$$A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\cos\left(n\phi\right)d\phi$$

$$B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\sin\left(n\phi\right)d\phi$$

Substituting in $f\left(\theta\right)$ and splitting the integral between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$:

$$\implies A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}f\left(\phi\right)\cos\left(n\phi\right)d\phi = \frac{1}{\pi a^n}\left( \int_{0}^{\pi}\cos\left(n\phi\right)d\phi - \int_{\pi}^{2\pi}\cos\left(n\phi\right)d\phi\right)$$

$$= \frac{1}{\pi a^n}\left( \frac{1}{n}\sin\left(n\phi\right) \bigr|_{0}^{\pi} - \frac{1}{n}\sin\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = 0 \text{, as } n \in \mathbb{N}$$

$$\implies B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}f\left(\phi\right)\sin\left(n\phi\right)d\phi = \frac{1}{\pi a^n}\left( \int_{0}^{\pi}\sin\left(n\phi\right)d\phi - \int_{\pi}^{2\pi}\sin\left(n\phi\right)d\phi\right)$$

$$= \frac{1}{\pi a^n}\left( -\frac{1}{n}\cos\left(n\phi\right) \bigr|_{0}^{\pi} + \frac{1}{n}\cos\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = \frac{1}{n \pi a^n}2\left(1-\left(-1\right)^n\right) \text{, as } n \in \mathbb{N}$$

$$\implies u\left(r,\theta\right) = \sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n \frac{2\left(1-\left(-1\right)^n\right)}{n \pi } \sin\left(n \theta\right) \phantom{\ } \square$$

If instead we approached the Dirichlet problem on the disk from Poisson's formula, we have that $u\left(r,\theta\right)$ has a solution of the form:

$$u\left(r,\theta\right) = \left(a^2 - r^2\right)\int_{0}^{2\pi}\frac{h\left(\phi\right)}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}$$

Substituting in $f\left(\theta\right)$ for $h\left(\theta\right)$, and splitting between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$ gives the irreducible integrals:

$$= \left(a^2 - r^2\right)\left(\int_{0}^{\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi} - \int_{\pi}^{2\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}\right) \phantom{\ } \blacksquare$$

Part b. Solve:

$$\Delta u := u_{xx} + u_{yy} = 0, \phantom{\ }r > a$$
$$u \bigr|_{r=a} = f\left(\theta\right)$$
$$\max |u| < \infty$$
\text{in polar coordiantes } \left(r,\theta\right) \text{, with: } f(x) = \left\{\begin{aligned} &\phantom{-\ }1 &&: 0 < \theta < \pi\\ &-1 &&: \pi < \theta < 2\pi \end{aligned} \right.

Answer:If $a \le 0$, clearly the problem is not well-posed. For $r < 0$ our BC $u \bigr|_{r=a} = f\left(\theta\right)$ is undefined, as $r \ge 0$, and for $r = 0$ the singular point-BC does not specify a solution. Let $a > 0$. Then this is the Dirichlet problem on the exterior of the disk of radius $a$, with BC $h\left(\theta\right) = f\left(\theta\right)$. We can view this as an equivalent problem on the interior of the disk by considering the rotationally invariant projection of the real line of the Riemann sphere to itself with the rule: $r\mapsto \frac{1}{r}$ : $\{ z : z \in \mathbb{R}_\infty , z \ge a \} \mapsto \{ \frac{1}{z} : \frac{1}{z} \in \mathbb{R}_{\infty} , \frac{1}{z} \le \frac{1}{a}\}$. This gives a mapping the Dirichlet problem on the exterior of the disk in $r$ with BC at $a$ to the Dirichlet problem on the interior of the disk in $\frac{1}{r}$ with BC at $\frac{1}{a}$. This map exists as $|u|$ bounded as $r \rightarrow \infty$ which implies that we have a removable
singularity of $u$ at our point at infinity, so $u$ can be analytically extended there, and so may the equivalent mapped point at $0$ when $\frac{1}{r} \ \rightarrow 0$ as $r \rightarrow \infty$.

We derived that the Dirichlet problem on the interior of the disk with BC: $\{u = h\left(\theta\right) : r = a\}$, $u\left(r,\theta\right)$ has a solution of the form:

$$u\left(r,\theta\right) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^n \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)$$

$$A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\cos\left(n\phi\right)d\phi$$

$$B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\sin\left(n\phi\right)d\phi$$

Substituting in $f\left(\theta\right)$, mapping $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$, and splitting the integral between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$, gives us the solution on the exterior:

$$\implies u\left(r,\theta\right) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^{-n} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)$$

$$\implies A_n = \frac{1}{\pi a^{-n}} \int_{0}^{2\pi}f\left(\phi\right)\cos\left(n\phi\right)d\phi = \frac{a^{n}}{\pi }\left( \int_{0}^{\pi}\cos\left(n\phi\right)d\phi - \int_{\pi}^{2\pi}\cos\left(n\phi\right)d\phi\right)$$

$$= \frac{a^{n}}{\pi}\left( \frac{1}{n}\sin\left(n\phi\right) \bigr|_{0}^{\pi} - \frac{1}{n}\sin\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = 0 \text{, as } n \in \mathbb{N}$$

$$\implies B_n = \frac{1}{\pi a^{-n}} \int_{0}^{2\pi}f\left(\phi\right)\sin\left(n\phi\right)d\phi = \frac{a^{n}}{\pi }\left( \int_{0}^{\pi}\sin\left(n\phi\right)d\phi - \int_{\pi}^{2\pi}\sin\left(n\phi\right)d\phi\right)$$

$$= \frac{a^{n}}{\pi }\left( -\frac{1}{n}\cos\left(n\phi\right) \bigr|_{0}^{\pi} + \frac{1}{n}\cos\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = \frac{a^n}{n \pi }2\left(1-\left(-1\right)^n\right) \text{, as } n \in \mathbb{N}$$

$$\implies u\left(r,\theta\right) = \sum_{n=1}^{\infty} \left(\frac{a}{r}\right)^n \frac{2\left(1-\left(-1\right)^n\right)}{n \pi } \sin\left(n \theta\right) \phantom{\ } \square$$

If instead we approached the Dirichlet problem on the interior disk from Poisson's formula, we have that $u\left(r,\theta\right)$ has a solution of the form:

$$u\left(r,\theta\right) = \left(a^2 - r^2\right)\int_{0}^{2\pi}\frac{h\left(\phi\right)}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}$$

Substituting in $f\left(\theta\right)$ for $h\left(\theta\right)$, taking $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$, and splitting the integral between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$, gives us the solution on the exterior in the form of two irreducible integrals:

$$= \left(a^{-2} - r^{-2}\right) \left(\int_{0}^{2\pi}\frac{f\left(\phi\right)}{a^{-2}-2 a^{-1} r^{-1} \cos\left(\theta-\phi\right)+r^{-2}}\frac{d \phi}{2\pi}\right)$$

$$= \frac{a^2 r^2}{a^2 r^2}\left(a^{-2} - r^{-2}\right)\left(\int_{0}^{2\pi}\frac{f\left(\phi\right)}{a^{-2}-2 a^{-1} r^{-1} \cos\left(\theta-\phi\right)+r^{-2}}\frac{d \phi}{2\pi}\right)$$

$$=\left(r^2 - a^2\right)\int_{0}^{2\pi}\frac{f\left(\phi\right)}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}$$

$$= \left(r^2 - a^2\right)\left(\int_{0}^{\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi} - \int_{\pi}^{2\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}\right) \phantom{\ } \blacksquare$$
« Last Edit: November 19, 2012, 09:41:00 PM by Calvin Arnott »

Ian Kivlichan

• Sr. Member
• Posts: 51
• Karma: 17
Re: Problem 3
« Reply #10 on: November 19, 2012, 09:40:30 PM »
Calvin: it's a bit simpler to write out ((-1)^(n+1) +1) as 0 or 2!

p.s. I am envious of your skills in LaTeX

Zarak Mahmud

• Sr. Member
• Posts: 51
• Karma: 9
Re: Problem 3
« Reply #11 on: November 19, 2012, 09:55:51 PM »
Calvin: it's a bit simpler to write out ((-1)^(n+1) +1) as 0 or 2!

p.s. I am envious of your skills in LaTeX

Me too. Pretty impressive.

Victor Ivrii

Yes, simpler definitely to write $n=2m+1$ with $m=0,1,\ldots$ as "even" coefficients are $0$.