Author Topic: TT1-P3  (Read 1751 times)

Victor Ivrii

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TT1-P3
« on: October 21, 2015, 08:49:38 PM »
Find  solution to
\begin{align}
&u_{tt}-4u_{xx}=0, && t>0, \ x>0,\label{eq-3-1}\\[2pt]
&u|_{t=0}=e^{-x}, && x>0,\label{eq-3-2}\\[2pt]
&u_t|_{t=0}=6e^{-x}, && x>0,\label{eq-3-3}\\[2pt]
&u|_{x=0}= e^{-2t}, &&t>0.\label{eq-3-4}
\end{align}

Zaihao Zhou

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Re: TT1-P3
« Reply #1 on: October 21, 2015, 09:23:59 PM »
\begin{equation}   u(x,t) = \phi (x+2t) + \psi (x-2t)       \end{equation}
where $\phi (x) = -e^{-x}$ and $\psi (x) = 2e^{-x} $ when $ x > 2t$. When $ 0< x < 2t$, $ \psi (x) = 2e^x$
So for $ x > 2t$: \begin{equation}   u(x,t) = -e^{-x-2t} + 2e^{2t-x}       \end{equation}
For $ 0 < x < 2t$: \begin{equation}   u(x,t) = -e^{-x-2t} + 2e^{x-2t}       \end{equation}

Bruce Wu

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Re: TT1-P3
« Reply #2 on: October 21, 2015, 09:50:24 PM »
I believe that for $0<x<2t$ the solution is actually \begin{equation}\large
u(x,t)=e^{x-2t}-e^{-x-2t}+e^{-2t+x}\end{equation}
« Last Edit: October 21, 2015, 09:54:19 PM by Fei Fan Wu »

Zaihao Zhou

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Re: TT1-P3
« Reply #3 on: October 21, 2015, 10:08:44 PM »
For $ 0 < x < 2t$:
\begin{equation} \phi (2t) + \psi (-2t) = e^{-2t}   \end{equation}
Let $ x = -2t$
\begin{equation} \phi (-x) + \psi (x) = e^{x}   \end{equation}
\begin{equation}  \psi (x) = -\phi (-x) + e^{x}   \end{equation}
Here $ \phi(-x) = -e^x$. So
\begin{equation}  \psi (x) = e^x + e^{x} = 2e^x  \ \ \ \  \      \   x < 0 \end{equation}
\begin{equation}  u(x,t) = \phi (x+2t) + \psi (x-2t) = -e^{-x-2t} + 2e^{x-2t} \end{equation}

Please correct me if I'm wrong

Bruce Wu

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Re: TT1-P3
« Reply #4 on: October 21, 2015, 10:23:52 PM »
Here we have an inhomogeneous Dirichlet boundary condition. Using http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.html#mjx-eqn-eq-2.6.10 I obtained the solution I gave above

Bruce Wu

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Re: TT1-P3
« Reply #5 on: October 21, 2015, 10:43:19 PM »
Ah, I see. Our expressions are actually equivalent. I have just forgotten to collect the $e^{x-2t}$ terms. My apologies