Author Topic: HA3-P6  (Read 1452 times)

Victor Ivrii

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Chi Ma

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Re: HA3-P6
« Reply #1 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 $R(x,t)=\{ (x',t'):\, 0< x'-ct' < x-ct,\, 0< x'+ct' < x+ct\}$.
« Last Edit: October 08, 2015, 10:53:50 AM by Chi Ma »

Bruce Wu

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Re: HA3-P6
« Reply #2 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?

Victor Ivrii

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Re: HA3-P6
« Reply #3 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"

Chi Ma

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Re: HA3-P6
« Reply #4 on: October 08, 2015, 09:16:21 AM »
What happened to the Jacobian? Do we need to multiply the integral by $2c$?

Victor Ivrii

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Re: HA3-P6
« Reply #5 on: October 08, 2015, 10:21:40 AM »
What happened to the Jacobian? Do we need to multiply the integral by $2c$?
Yes!

Chi Ma

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Re: HA3-P6
« Reply #6 on: October 08, 2015, 10:54:57 AM »
Thanks Prof. I have revised the solution.

Andrew Lee Chung

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Re: HA3-P6
« Reply #7 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?
« Last Edit: October 08, 2015, 03:23:23 PM by Andrew Lee Chung »

Victor Ivrii

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Re: HA3-P6
« Reply #8 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?