Author Topic: Problem 4  (Read 8325 times)

Ian Kivlichan

  • Sr. Member
  • ****
  • Posts: 51
  • Karma: 17
    • View Profile
Problem 4
« on: November 19, 2012, 03:29:55 AM »
For Problem 4, it seems the solution can only defined up to a constant - is that alright?

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Problem 4
« Reply #1 on: November 19, 2012, 05:11:26 AM »
For Problem 4, it seems the solution can only defined up to a constant - is that alright?

This was discussed in the lecture 24, section 1
http://www.math.toronto.edu/courses/apm346h1/20129/L24.html#sect-24.1

Ian Kivlichan

  • Sr. Member
  • ****
  • Posts: 51
  • Karma: 17
    • View Profile
Re: Problem 4
« Reply #2 on: November 19, 2012, 11:19:38 AM »
Oh - oops! I hadn't read that far yet. Thank you!

Ian Kivlichan

  • Sr. Member
  • ****
  • Posts: 51
  • Karma: 17
    • View Profile
Re: Problem 4
« Reply #3 on: November 19, 2012, 09:30:33 PM »
Hopeful solutions attached! :)

Note that for 4.b), we require that as r goes to infinity, u goes to 0, which forces $A_0 = 0$.
« Last Edit: November 19, 2012, 09:37:39 PM by Ian Kivlichan »

Chen Ge Qu

  • Full Member
  • ***
  • Posts: 16
  • Karma: 8
    • View Profile
Re: Problem 4
« Reply #4 on: November 19, 2012, 09:34:58 PM »
Both parts attached
« Last Edit: November 20, 2012, 01:46:32 AM by Chen Ge Qu »

Calvin Arnott

  • Sr. Member
  • ****
  • Posts: 43
  • Karma: 17
  • OK
    • View Profile
Re: Problem 4
« Reply #5 on: November 19, 2012, 09:38:35 PM »
Problem 4

Part a.

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

Answer:
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 Neumann problem on the disk of radius $ a $, with BC $h\left(\theta\right) = f\left(\theta\right)$. It was derived that by changing to polar coordinates $\left(x,y\right) \mapsto \left(r,\theta\right)$ has a solution of the form:

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

Differentiating this equation for $u\left(r,\theta\right)$ with respect to $r$ gives us our boundary condition at $r=a$, $f\left(\theta\right)$.

$$ u_{r}\left(r,\theta\right) = \sum_{n=1}^{\infty} n r^{n-1} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)  $$

$$ u_{r}\left(a,\theta\right) = \sum_{n=1}^{\infty} n a^{n-1} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right) = f\left(\theta\right) $$

$$ \text{Notice that: }\int_{0}^{2\pi}f\left(\theta\right) d \theta = \int_{0}^{\pi} d \theta - \int_{\pi}^{2\pi} d \theta = \pi - 2 \pi + \pi = 0 $$

So our series representation for $f$ has a free coefficient and we may find a unique solution up to a constant $C \in \mathbb{R}$, despite losing a degree of freedom to differentiation. Our Fourier coefficients at $r = a$ are then given by:

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

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

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

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

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

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

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

$$ \implies u\left(r,\theta\right) = \sum_{n=1}^{\infty} \left(\frac{r^n}{a^{n-1}}\right) \frac{2\left(1-\left(-1\right)^n\right)}{ \pi n^2 } \sin\left(n \theta\right) + C, \phantom{\ } C \in \mathbb{R} \phantom{\ } \blacksquare $$



Part b. Solve:

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

Answer:
If $a \le 0$, clearly the problem is not well-posed. For $r < 0$ our BC $ u_r \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 Neumann 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 Neumann problem on the exterior of the disk in $r$ with BC at $a$ to the Neumann 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 $r \rightarrow \infty$.

Now- because for our BC again we have: $ \int_{0}^{2\pi}f\left(\theta\right) d \theta $ so our BC has a free coefficient, and the Neumann problem on the interior of the disk with BC: $\{u = h\left(\theta\right) : r = a\}$, $u\left(r,\theta\right)$ has a unique solution up to a constant $C \in \mathbb{R}$ of the form:

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

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

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

Mapping $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$ gives us the equivalent solution on the exterior:

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

Because we get a negative term when differentiating we have to be a bit more careful with the signs of our coefficients. Our BC at $\frac{1}{r} = \frac{1}{a}$, $u_r = f\left(\theta\right)$ gives us:

$$u_{r} = \sum_{n=1}^{\infty} \left(-n\right) r^{-n-1} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right) $$

$$u_{r} \bigr|_{\frac{1}{a}} = \sum_{n=1}^{\infty} \left(-n\right) a^{-n-1} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)= f\left(\theta\right) $$

So our Fourier coefficients are given by:

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

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

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


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

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

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

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

$$ \implies u\left(r,\theta\right) = \sum_{n=1}^{\infty} \left(\frac{a^{n+1}}{r^n}\right) \frac{2\left(\left(-1\right)^n - 1\right)}{ \pi n^2 } \sin\left(n \theta\right) + C, \phantom{\ } C \in \mathbb{R} \phantom{\ } \blacksquare $$
« Last Edit: November 19, 2012, 09:40:20 PM by Calvin Arnott »

Ian Kivlichan

  • Sr. Member
  • ****
  • Posts: 51
  • Karma: 17
    • View Profile
Re: Problem 4
« Reply #6 on: November 19, 2012, 09:41:48 PM »
Calvin: I think you should set your constants to zero, as in Lecture 24.

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Problem 4
« Reply #7 on: November 20, 2012, 06:03:07 AM »
Calvin: I think you should set your constants to zero, as in Lecture 24.

Not really: lecture suggests that we can do it imposing an extra condition $\int _\Omega u \, dS=0$.

A better question: why solution exists? Related: Does solution of the same equation but with b.c. $u_r|_{r=a}=1$ exist?

Ian Kivlichan

  • Sr. Member
  • ****
  • Posts: 51
  • Karma: 17
    • View Profile
Re: Problem 4
« Reply #8 on: November 20, 2012, 12:04:28 PM »
Calvin: I think you should set your constants to zero, as in Lecture 24.

Not really: lecture suggests that we can do it imposing an extra condition $\int _\Omega u \, dS=0$.

A better question: why solution exists? Related: Does solution of the same equation but with b.c. $u_r|_{r=a}=1$ exist?
I don't think we can have solutions with BC $u_r|_{r=a}=1$ since all our coefficients would become 0 (since $ \int_{0}^{2\pi}f\left(\theta\right) \sin(n\theta) d \theta  = \int_{0}^{2\pi}f\left(\theta\right) \cos(n\theta) d \theta = 0$ for $n \ge 1$). I guess that the condition there should be that $u_r|_{r=a}$ be Fourier-decomposable, without a constant ($A_0$).
« Last Edit: November 20, 2012, 12:07:09 PM by Ian Kivlichan »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Problem 4
« Reply #9 on: November 29, 2012, 04:54:15 AM »
Calvin: I think you should set your constants to zero, as in Lecture 24.

Not really: lecture suggests that we can do it imposing an extra condition $\int _\Omega u \, dS=0$.

A better question: why solution exists? Related: Does solution of the same equation but with b.c. $u_r|_{r=a}=1$ exist?
I don't think we can have solutions with BC $u_r|_{r=a}=1$ since all our coefficients would become 0 (since $ \int_{0}^{2\pi}f\left(\theta\right) \sin(n\theta) d \theta  = \int_{0}^{2\pi}f\left(\theta\right) \cos(n\theta) d \theta = 0$ for $n \ge 1$). I guess that the condition there should be that $u_r|_{r=a}$ be Fourier-decomposable, without a constant ($A_0$).

Yes, it is true--but it is exactly what happens with this $f$! And this is what my question was about