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**HA8 / Re: HA8-P1**

« **on:**November 06, 2015, 03:10:20 PM »

(b) This part is very similar to the first, so I will omit several steps for the sake of convenience. Recall that for a solution that depends only on $r$, we can reduce the problem to:

\begin{equation}

\Delta u(r) = u_{rr} + \frac{2}{r}u_r = -k^2u

\end{equation}

By the same method of making a substitution $u(r) = \frac{v(r)}{r}$ as in part (a), we will arrive at the same derivatives as before. This will give us an equation in terms of $v(r)$:

\begin{equation}

v''(r) = -k^2v

\end{equation}

Again, this is a simple ODE. By the usual methods, this will lead to a solution:

\begin{equation}

v(r) = Ce^{ikr} + De^{-ikr}

\end{equation}

Plugging this back in terms of $u(r)$, we arrive at a solution:

\begin{equation}

u(r) = Ce^{ikr}r^{-1} + De^{-ikr}r^{-1}

\end{equation}

\begin{equation}

\Delta u(r) = u_{rr} + \frac{2}{r}u_r = -k^2u

\end{equation}

By the same method of making a substitution $u(r) = \frac{v(r)}{r}$ as in part (a), we will arrive at the same derivatives as before. This will give us an equation in terms of $v(r)$:

\begin{equation}

v''(r) = -k^2v

\end{equation}

Again, this is a simple ODE. By the usual methods, this will lead to a solution:

\begin{equation}

v(r) = Ce^{ikr} + De^{-ikr}

\end{equation}

Plugging this back in terms of $u(r)$, we arrive at a solution:

\begin{equation}

u(r) = Ce^{ikr}r^{-1} + De^{-ikr}r^{-1}

\end{equation}