Author Topic: HA8-P1  (Read 596 times)

Emily Deibert

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HA8-P1
« on: November 06, 2015, 02:55:24 PM »
Problem 1. http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.P.html

(a) We are asked to find the solutions of $\Delta u = u_{xx} + u_{yy} + u_{zz} = k^2u$ that depend only on $r$. So, we must turn this into spherical coordinates. By the methods of the textbook section 6.3.2, we arrive at:

\begin{equation}
\Delta u(r, \phi{}, \theta{}) = u_{rr} + \frac{2u_r}{r} + \frac{u_{\phi{} \phi{}}}{r^2} + \frac{\cot(\phi{})u_{\phi}}{r^2} + \frac{u_{\theta{} \theta{}}}{\sin^2(\phi{})r^2} = k^2u
\end{equation}

Note that these are defined as in the textbook: $r$ is the radius, $\phi$ is the latitude, and $\theta$ is the longitude.

Now, the problem asks for a solution that depends only on r. In this case, any derivatives with respect to the other variables should be zero. So canceling these terms, we can write the equation as:

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

Now we can make use of the hint provided in the problem, namely that we should make the substitution $u(r) = \frac{v(r)}{r}$. Then $u''(r)$ and $u'(r)$ are:

\begin{equation}
\begin{cases}
u_r = \frac{v'(r)}{r} - \frac{v(r)}{r^2} \\
u_{rr} = \frac{v''(r)}{r} - 2\frac{v'(r)}{r^2} + 2\frac{v}{r^3}
\end{cases}
\end{equation}

Now plugging these all into the problem, we have:

\begin{equation}
\frac{v''(r)}{r} - 2\frac{v'(r)}{r^2} + 2\frac{v}{r^3} + \frac{2}{r} \Big( \frac{v'(r)}{r} - \frac{v(r)}{r^2} \Big) = k^2\frac{v(r)}{r} \longrightarrow
\end{equation}
\begin{equation}
\frac{v''(r)}{r} - 2\frac{v'(r)}{r^2} + 2\frac{v}{r^3} + 2\frac{v'(r)}{r^2} - 2\frac{v(r)}{r^3} = k^2\frac{v(r)}{r} \longrightarrow
\end{equation}
\begin{equation}
\frac{v''(r)}{r} = k^2\frac{v(r)}{r} \longrightarrow
\end{equation}
\begin{equation}
v''(r) = k^2v
\end{equation}

We now have a simple ODE which can be solved by the usual methods. Recall that $k>0$ as defined in the problem. The solution to the ODE is:

\begin{equation}
v(r) = Ae^{kr} + Be^{-kr}
\end{equation}

Now we put this back in terms of $u(r)$, where we recall that $u(r) = \frac{v(r)}{r}$. So:

\begin{equation}
u(r) = Ae^{kr}r^{-1} + Be^{-kr}r^{-1}
\end{equation}
« Last Edit: November 06, 2015, 03:07:07 PM by Emily Deibert »

Emily Deibert

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Re: HA8-P1
« Reply #1 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}
« Last Edit: November 06, 2015, 03:12:23 PM by Emily Deibert »