# Toronto Math Forum

## APM346-2015F => APM346--Home Assignments => HA3 => Topic started by: Victor Ivrii on September 28, 2015, 01:06:14 PM

Title: HA3-P6
Post by: Victor Ivrii on September 28, 2015, 01:06:14 PM
Problem 2 here
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.4.P.html#problem-2.4.P.2 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.4.P.html#problem-2.4.P.2)
Title: Re: HA3-P6
Post by: Chi Ma on October 07, 2015, 12:19:57 AM
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{x-ct}{2c} \right) - h(0)
\end{align}
where $R(x,t)=\{ (x',t'):\, 0< x'-ct' < x-ct,\, 0< x'+ct' < x+ct\}$.
Title: Re: HA3-P6
Post by: Bruce Wu on October 08, 2015, 12:03:01 AM
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?
Title: Re: HA3-P6
Post by: Victor Ivrii on October 08, 2015, 08:04:22 AM
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"
Title: Re: HA3-P6
Post by: Chi Ma on October 08, 2015, 09:16:21 AM
What happened to the Jacobian? Do we need to multiply the integral by $2c$?
Title: Re: HA3-P6
Post by: Victor Ivrii on October 08, 2015, 10:21:40 AM
What happened to the Jacobian? Do we need to multiply the integral by $2c$?
Yes!
Title: Re: HA3-P6
Post by: Chi Ma on October 08, 2015, 10:54:57 AM
Thanks Prof. I have revised the solution.
Title: Re: HA3-P6
Post by: Andrew Lee Chung on October 08, 2015, 03:19:59 PM
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?
Title: Re: HA3-P6
Post by: Victor Ivrii on October 08, 2015, 05:13:06 PM
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?