Toronto Math Forum
APM3462015F => APM346Home Assignments => HA3 => Topic started by: Victor Ivrii on September 28, 2015, 01:06:14 PM

Problem 2 here
http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter2/S2.4.P.html#problem2.4.P.2 (http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter2/S2.4.P.html#problem2.4.P.2)

Transform the problem into the first quadrant of the characteristic coordinates $(\xi,\eta)$.
\begin{align}
4c^2\tilde{u}_{\xi \eta} &= \tilde{f}(\xi, \eta) && \xi > 0, \eta > 0 \label{a} \\
\tilde{u}_{\xi=0} &= \tilde{g} \left( \frac{\eta}{2c} \right) && \eta > 0 \\
\tilde{u}_{\eta=0} &= \tilde{h} \left( \frac{\xi}{2c} \right) && \xi > 0
\end{align}
The solution to $(\ref{a})$ is as follows.
\begin{align}
\tilde{u}(\xi,\eta)= \frac{1}{4c^2} \int_0 ^\xi
\int_0 ^\eta \tilde{f}(\xi',\eta' )\,d\eta' d\xi' + \psi(\eta) + \phi(\xi)
\end{align}
The domain of dependence is a rectangle defined as $\tilde{R}(\xi,\eta) = \{ (\xi',\eta') \vert 0< \xi' < \xi,\, 0< \eta' < \eta\}$.
Assume $\psi(0) = \phi(0) = \frac{1}{2}g(0) = \frac{1}{2}h(0)$.
\begin{align}
\phi(\xi) &= h \left( \frac{\xi}{2c} \right)  \frac{1}{2}h(0)\\
\psi(\eta) &= g \left( \frac{\eta}{2c} \right)  \frac{1}{2}g(0)
\end{align}
Translate back to the $(x,t)$ coordinates. Use the hint provided and the fact that the Jacobian is equal to $2c$.
\begin{align}
u(x,t) = \frac{1}{2c}\iint _{R(x,t)} f(x',t')\,dx'dt' + h \left( \frac{x+ct}{2c} \right) + g \left( \frac{xct}{2c} \right)  h(0)
\end{align}
where $R(x,t)=\{ (x',t'):\, 0< x'ct' < xct,\, 0< x'+ct' < x+ct\}$.

Since we have already solved the homogeneous Goursat problem in section 2.3 problem 5, and the contribution from the right hand expression is given as a hint, would it be ok to just add the two known contributions together yielding the final solution without the intermediate steps?

Since we have already solved the homogeneous Goursat problem in section 2.3 problem 5, and the contribution from the right hand expression is given as a hint, would it be ok to just add the two known contributions together yielding the final solution without the intermediate steps?
Yes, due to linearity it would be correct. But if this problem goes to Quiz you cannot refer to "the other problem"

What happened to the Jacobian? Do we need to multiply the integral by $2c$?

What happened to the Jacobian? Do we need to multiply the integral by $2c$?
Yes!

Thanks Prof. I have revised the solution.

Why do we have a rectangle as domain of dependence as compared to the one in the book?
How do we set the limits for the double integral?

Why do we have a rectangle as domain of dependence as compared to the one in the book?
How do we set the limits for the double integral?
Different problem.
You solve $u_{\xi\eta}= k(\xi,\eta)$ as $\xi>0$, $\eta>0$; $u(0,\eta)=0$, $u(\xi,0)=0$. What do we get?