# PDE Separation Of Variables: Laplace Equation Problem

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.

• 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. – user179766 Sep 29 '14 at 22:49

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.

• Okay I understand this concept. Is this for the condition when the separation constant is greater than zero? – user179766 Sep 30 '14 at 2:13
• 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. – user179766 Sep 30 '14 at 11:26
• Hi @Juan Ospina, I do not understand what you mean. – user179766 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 $\frac{r}{F}$ $\frac{\partial}{\partial r}$ $r (\frac{\partial F}{\partial r}) = \lambda$ – user179766 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 $F(R)$. – user179766 Sep 30 '14 at 14:31