# Toronto Math Forum

## APM346-2015F => APM346--Tests => Test 1 => Topic started by: Victor Ivrii on October 21, 2015, 08:49:38 PM

Title: TT1-P3
Post by: Victor Ivrii 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}
Title: Re: TT1-P3
Post by: Zaihao Zhou on October 21, 2015, 09:23:59 PM
$$u(x,t) = \phi (x+2t) + \psi (x-2t)$$
where $\phi (x) = -e^{-x}$ and $\psi (x) = 2e^{-x}$ when $x > 2t$. When $0< x < 2t$, $\psi (x) = 2e^x$
So for $x > 2t$: $$u(x,t) = -e^{-x-2t} + 2e^{2t-x}$$
For $0 < x < 2t$: $$u(x,t) = -e^{-x-2t} + 2e^{x-2t}$$
Title: Re: TT1-P3
Post by: Bruce Wu on October 21, 2015, 09:50:24 PM
I believe that for $0<x<2t$ the solution is actually \large
u(x,t)=e^{x-2t}-e^{-x-2t}+e^{-2t+x}
Title: Re: TT1-P3
Post by: Zaihao Zhou on October 21, 2015, 10:08:44 PM
For $0 < x < 2t$:
$$\phi (2t) + \psi (-2t) = e^{-2t}$$
Let $x = -2t$
$$\phi (-x) + \psi (x) = e^{x}$$
$$\psi (x) = -\phi (-x) + e^{x}$$
Here $\phi(-x) = -e^x$. So
$$\psi (x) = e^x + e^{x} = 2e^x \ \ \ \ \ \ x < 0$$
$$u(x,t) = \phi (x+2t) + \psi (x-2t) = -e^{-x-2t} + 2e^{x-2t}$$

Please correct me if I'm wrong
Title: Re: TT1-P3
Post by: Bruce Wu 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
Title: Re: TT1-P3
Post by: Bruce Wu 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