### Author Topic: HA8-P4  (Read 4337 times)

#### Victor Ivrii

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##### HA8-P4
« on: November 07, 2015, 07:19:31 AM »
« Last Edit: November 08, 2015, 09:33:41 AM by Victor Ivrii »

#### Emily Deibert

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##### Re: HA8-P4
« Reply #1 on: November 08, 2015, 11:51:00 AM »
As in problem three, we have a general solution (before imposing the boundary condition) given by http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.10. Now we must differentiate this general solution in order to make use of the boundary conditions. This yields:

u_r(r. \theta{}) = \sum_{n=1}^{\infty}nr^{n-1}(A_n\cos(n\theta{}) + C_n\sin(n\theta{}))

Evaluating this at a, as per the boundary condition:

u_r(a, \theta{}) = \sum_{n=1}^{\infty}na^{n-1}(A_n\cos(n\theta{}) + C_n\sin(n\theta{})) = f(\theta)

Using the Fourier method, we find the forms of the coefficients to be:

A_n = \frac{1}{\pi{}na^{n-1}}\int_0^{2\pi{}}f(\theta{}')\cos(n\theta{}')d\theta{}' = \frac{1}{\pi{}na^{n-1}}\Big[\int_0^{\pi{}}\cos(n\theta{}')d\theta{}' - \int_{\pi}^{2\pi{}}\cos(n\theta{}')d\theta{}'\Big]  = 0

and

C_n = \frac{1}{\pi{}na^{n-1}}\int_0^{2\pi{}}f(\theta{}')\sin(n\theta{}')d\theta{}' = \frac{1}{\pi{}na^{n-1}}\Big[\int_0^{\pi{}}\sin(n\theta{}')d\theta{}' - \int_{\pi}^{2\pi{}}\sin(n\theta{}')d\theta{}'\Big] \longrightarrow $$= \frac{1}{\pi{}na^{n-1}}\Big[\frac{1-\cos(n\pi)}{n}-\frac{\cos(n\pi)-\cos(2n\pi)}{n}\Big]= \frac{1}{\pi{}na^{n-1}} \frac{2(1-\cos(n\pi{}))}{n}$$

So:
C_n = \frac{2(1-\cos(n\pi{}))}{\pi{}n^{2}a^{n-1}}

As in the previous problem, only terms with odd n survive. This results in:

C_n = \frac{4}{\pi{}n^{2}a^{n-1}}

Now, plugging these coefficients back into the general solution, we arrive at the solution:

u(r, \theta{}) = \frac{4}{\pi{}} \sum_{n \geq 1, odd} \frac{r^n\sin(n\theta{})}{n^{2}a^{n-1}}
« Last Edit: November 08, 2015, 11:52:37 AM by Emily Deibert »

#### Emily Deibert

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##### Re: HA8-P4
« Reply #2 on: November 08, 2015, 12:07:45 PM »
(b) Part b is similar, but we consider equations outside the disk. This time the general solution is given by  http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.11. Much of the solution is the same as part (a) so I will omit several steps. Note that the derivative in this case is:

u_r(r. \theta{}) = \sum_{n=1}^{\infty}-nr^{-n-1}(B_n\cos(n\theta{}) + D_n\sin(n\theta{}))

As with the $A_n$ terms in part (a), the $B_n$ terms will go to zero. The $D_n$ terms will become:

D_n = \frac{-a^{n+1}2(1-\cos(n\pi{}))}{\pi{}n^{2}}

Only odd-n terms survive; so:

D_n = \frac{-4a^{n+1}}{\pi{}n^2}

Plugging into the general solution, we get a final answer:

u(r, \theta{}) = -\frac{4}{\pi{}} \sum_{n \geq 1, odd} \frac{a^{n+1} \sin(n \theta{})}{n^2 r^n}

#### Emily Deibert

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##### Re: HA8-P4
« Reply #3 on: November 11, 2015, 02:23:09 PM »
By the way, we can only satisfy the problem if $h(\theta)$ is such that $\int_0^{2\pi}h(\theta)d\theta = 0$. This is clearly satisfied in this case.