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Solve $\nabla^{2}u=0$ in the region $\rho>a$, $0<\theta<\pi$ such that $u(\rho,0)=u(\rho,\pi)=0$ and $u(a, \theta) = u_{0}$ and $u\to0$ as $\rho\to\infty$.

We have the general solution $$u(\rho,\theta)=A_{0} + B_{0}\ln{\rho} + \sum^{\infty}_{n=1}(A_{n}\rho^{n}+B_{n}\rho^{-n})(C_{n}\cos{n\theta}+D_{n}\sin{n\theta}).$$

Now my lecturer says applying $u=0$ at $\theta=0, \pi$ means that $A_{0}=B_{0}=0$ and $C_{n}=0$.

But I can't really see this.

\begin{align}u(\rho,0) &= A_{0}+B_{0}\ln{\rho} + \sum_{n=1}^{\infty}(A_{n}\rho^{n}+B_{n}\rho^{-n})C_{n} = 0, & (*)\\ u(\rho,\pi)&= A_{0}+B_{0}\ln{\rho}+\sum^{\infty}_{n=1}(A_{n}\rho^{n}+B_{n}\rho^{-n})C_{n}(-1)^{n}=0.& (**)\end{align}

I don't see how in any way this implies the above? I notice we can do $(*)-(**)$: $$\sum_{n=1}^{\infty}(A_{n}\rho^{n}+B_{n}\rho^{-n})C_{n}(1-(-1)^{n})=0$$ Then the summand is zero whenever $n$ is even, but this isn't stated above.

user2850514
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  • Note that the equations $f(\rho) = 0$ should hold for all $\rho > a$. The result you need to establish $C_n = 0$ is the following: if $g(x) = \sum_{n=0}^\infty a_n x^n = 0$ for all $x$ (or all $x$ in some compact interval) then $a_n\equiv 0$. – Winther Jan 09 '15 at 03:27
  • Oh that makes perfect sense thanks. How about for $A_{0}$ and $B_{0}$ though? – user2850514 Jan 09 '15 at 03:33
  • If $C_n = 0$ we get $A_0 + B_0\log\rho = 0$ for all $\rho$ so $B_0=0$ and therefore $A_0=0$. – Winther Jan 09 '15 at 03:34

1 Answers1

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I noticed you received help in the comments, but from the BC, we can determine some of the constants right off the bat. Since $u\to 0$ as $\rho\to\infty$, $B_0 = 0$; otherwise, we have $\ln(\rho)\to\infty$ as $\rho\to\infty$ so $u$ wouldn't be bounded. Now, $B_n = 0$, because again, the solution wouldn't be bounded since $\rho^n\to\infty$ as $\rho\to\infty$. Thus, we can write the general solution as $$ u(\rho,\theta) = A_0 + \sum_{n=1}^{\infty}\frac{1}{\rho^n}\bigl[C_n\cos(n\theta) + D_n\sin(n\theta)\bigr] $$ Using the BC, we have \begin{alignat}{2} u(\rho,0)&= A_0+ \sum_{n=1}^{\infty}\frac{1}{\rho^n}C_n &&={}0\\ u(\rho,\pi)&= A_0+ \sum_{n=1}^{\infty}\frac{1}{\rho^n}C_n(-1)^n &&={}0 \end{alignat} Therefore, $$ \sum_{n=1}^{\infty}\frac{C_n}{\rho^n}\bigl[1-(-1)^n\bigr] = 0 $$ which is easier to deal with then the equation you derived.

dustin
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