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I am having trouble with this problem. Here is the problem:

I might need some tips on how to go through this problem. I have a sense on solving the cases for the separation constant, but I am having trouble on how to do it in this scenario. Here's the question:

Solve Laplace’s equation $[u_{xx}+u_{yy} = 0.]$ (This equation should be in polar form)

inside a circular annulus $R_1 < r < R_2$ with boundary conditions: $\varphi(R_1,\theta) = 0$ and $\frac{\partial \varphi}{\partial r} (R_2,θ) = f(θ)$

I understand the process of Separation Of Variables and the Laplace Equation. The point I am stuck on at the moment is evaluating the Boundary Condition and the Initial Condition.

Also, this is my first time on this website and I am trying to get familiar with it. I might need some tips on how to input in these figures correctly.

Dmoreno
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  • Could you walk us up to the point you get stuck? Do you know the circular laplacian, for example? – Danny W. Sep 29 '14 at 22:45
  • I have already edited the question. The only thing I am confused on is trying to evaluate the boundary condition and the initial condition for each case for the separation constant, K. –  Sep 29 '14 at 22:49

1 Answers1

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The usual separation of variables in polar coordinates produces

$$\phi \left( r,\theta \right) =\sum _{n=0}^{\infty } \left( A_{{n}}\sin \left( n\theta \right) +B_{{n}}\cos \left( n\theta \right) \right) \left( {r}^{n}C_{{n}}+{r}^{-n}E_{{n}} \right) +F\ln \left( r \right)$$

which is rewritten as

$$\phi \left( r,\theta \right) =\sum _{n=1}^{\infty } \left( A_{{n}}\sin \left( n\theta \right) +B_{{n}}\cos \left( n\theta \right) \right) \left( {r}^{n}C_{{n}}+{r}^{-n}E_{{n}} \right) +G+ F\ln \left( r \right)$$ Applying the boundary condition at $r=R_1$ we obtain

$$\sum _{n=1}^{\infty } \left( A_{{n}}\sin \left( n\theta \right) +B_{{n }}\cos \left( n\theta \right) \right) \left( {R_{{1}}}^{n}C_{{n}}+{R _{{1}}}^{-n}E_{{n}} \right) +F\ln \left( R_{{1}} \right) +G =0 $$

From this last equation we derive that $G=-Fln(R_1)$ and

$${R_{{1}}}^{n}C_{{n}}+{R_{{1}}}^{-n}E_{{n}}= 0$$

Then we have

$$ E_{{n}}=-{R_{{1}}}^{2\,n}C_{{n}} $$

The solution takes the form

$$\phi \left( r,\theta \right) =\sum _{n=1}^{\infty } \left( A_{{n}}\sin \left( n\theta \right) +B_{{n}}\cos \left( n\theta \right) \right) \left( {r}^{n}-{r}^{-n}{R_{{1}}}^{2\,n} \right)-Fln(R_1)+Fln(r) $$

Applying the boundary condition at $r=R_2$ we obtain

$$\sum _{n=1}^{\infty } \left( A_{{n}}\sin \left( n\theta \right) +B_{{n }}\cos \left( n\theta \right) \right) n \left( {R_{{2}}}^{n-1}+{R_{{2 }}}^{-n-1}{R_{{1}}}^{2\,n} \right)+F/R_2 =f \left( \theta \right) $$

Finally the constants $F$, $A_n$ and $B_n$ are determined using the standard expressions of Fourier series.

Juan Ospina
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  • Okay I understand this concept. Is this for the condition when the separation constant is greater than zero? –  Sep 30 '14 at 02:13
  • I guess I do not know how you got $n\theta$ for $F(\theta)$ and I do not understand how you got $r^n$ and $r^{-n}$ from the very beginning of your work. –  Sep 30 '14 at 11:26
  • Hi @Juan Ospina, I do not understand what you mean. –  Sep 30 '14 at 14:11
  • Okay here is the work I have: Use Separation of Variables so assume the solution $\phi (r,\theta)$ = $F(r)G(\theta)$. By Separation of Variables, it would form out to $\frac{r}{F}$ $\frac{\partial}{\partial r}$ $r (\frac{\partial F}{\partial r}) = \lambda$ –  Sep 30 '14 at 14:22
  • Sorry, I am trying to get familiar with Latex. It is my first time on this website. Here is what I edited: Okay here is the work I have: Use Separation of Variables so assume the solution $\phi (r,\theta)$ = $F(r)G(\theta)$. By Separation of Variables, it would form out to $\frac{r}{F}$ $\frac{\partial}{\partial r}$ $r (\frac{\partial F}{\partial r}) = -\lambda F$ For the function $G(\theta)$, I got $\frac{\partial^2 G}{\partial^2 \theta} = \lambda G$. I only made observations for the cases of $\lambda$ for $F(R)$. –  Sep 30 '14 at 14:31
  • Okay, solving your equations you obtain $$F \left( r \right) ={\it C_1},{r}^{\sqrt {\lambda}}+{\it C_2},{r}^ {-\sqrt {\lambda}} $$ $$ G \left( \theta \right) ={\it A_1},\sin \left( \sqrt {\lambda}\theta \right) +{\it A_2},\cos \left( \sqrt {\lambda}\theta \right)

    $$ Given that G must be a periodic function of $\theta$ it is necessary to make $\sqrt {\lambda}=n$ where $n$ is an integer. Then the solution is a sum for all values of $n$. Do you agree?

     – Juan Ospina Sep 30 '14 at 14:32
  • Given that you need to solve a problem inside a circular region only the case with $\lambda>0$ is admissible. Do you agree? – Juan Ospina Sep 30 '14 at 14:54
  • I got that we need to solve the case for $\lambda > 0$, but I do not understand why we set $\sqrt \lambda = n$ –  Sep 30 '14 at 15:10
  • Now pay attention to the equation for $G \left( \theta \right)$. When $\lambda >0$ you obtain trigonometric functions. You need periodic solutions respect to $\theta$ and a function of the form $sin(c\theta)$ is periodic only when $c$ is an integer. For this reason you must to make $\sqrt {\lambda}=n$. Do you agree? – Juan Ospina Sep 30 '14 at 15:16
  • Oh okay that makes more sense now. Thank you for your help and for your patience. I really appreciate it. –  Sep 30 '14 at 15:18
  • You are welcome. – Juan Ospina Sep 30 '14 at 15:20
  • One more thing, I am not sure if I am computing the constants in the Fourier Series. I have $A_0 = \frac{1}{n2\pi(R_1^{n-1} + R_1^{-n-1} R_2^{2n})} [\int_0^{2\pi} F(\theta) d\theta.]$. I have $A_n = \frac{1}{n\pi(R_1^{n-1} + R_1^{-n-1} R_2^{2n})} [\int_0^{2\pi} F(\theta) \sin(n\theta) d\theta.]$. Is this right? If it is, I should be able to compute $B_n$ –  Oct 01 '14 at 03:54
  • Hi @SSivetts, please look the little modifications in the computation above. Then $$F=\frac{1}{2} ,{\frac {\int _{0}^{2,\pi }!f \left( \theta \right) {d\theta}R _{{2}}}{\pi }} $$ Your expression for $A_n$ is correct and with a similar expression for $B_n$. All the best. – Juan Ospina Oct 01 '14 at 10:59
  • Okay that makes more sense. Thank you for all your help again. I wish you the best as well. –  Oct 01 '14 at 11:26