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APM346-2016F => APM346--Tests => FE => Topic started by: Victor Ivrii on December 13, 2016, 07:49:01 PM

Title: FE1
Post by: Victor Ivrii on December 13, 2016, 07:49:01 PM
Solve by the method of characteristics the BVP for a wave equation
&u_{tt}-  9 u_{xx}=0,\qquad 0<x<\infty , \; t>0\label{1-1}\\[2pt]
& u(x,0)=f(x),\label{1-2}\\[2pt]
& u_t(x,0)=g(x),\label{1-3}\\[2pt]
& u_x (0,t)= h(t)\label{1-4}
with $f(x)=4\cos(x)$,    $g(x)=6\sin (x) $ and  $h(t)=\sin (3t)$. You need to find a continuous solution.
Title: Re: FE1
Post by: Sajjan Heerah on December 14, 2016, 08:43:29 AM
My solution attempt for question 1
Title: Re: FE1
Post by: Shentao YANG on December 14, 2016, 10:39:10 AM
you forget $c$ when you are doing the integration in the second part
Title: Re: FE1
Post by: Victor Ivrii on December 14, 2016, 11:08:16 AM

This is incorrect solution. Obviously it does not satisfy the boundary condition.

I finished to grade this problem. One third of those who wrote got  it right, and another third made a single error (forgot $C$ in the definition of $\psi$, which is selected to make it continuous). There is no point to post a wrong solution, especially the scan of poor handwriting.

If nobody posts a correct solution, I will post one, Dec 19 or 20
Title: Re: FE1
Post by: Luyu CEN on December 14, 2016, 12:11:48 PM
The general solution for u is
\begin{equation*}u = \phi(x+3t) + \psi(x-3t)\end{equation*}
Now impose the initial conditions (2) and (3)
&\phi(x) + \psi(x) = f(x)\\
&\phi'(x) + \psi'(x) = g(x)\end{align}
Solve the above to give
&\phi(x) = \frac{1}{2}f(x) + \frac{1}{6}\int_{0}^{x} g(x') dx'  \\
&\psi(x)  = \frac{1}{2}f(x) - \frac{1}{6}\int_{0}^{x} g(x') dx'\end{align}
Therefore, when x > 3t,
\begin{equation}u(x) = \phi(x+3t) + \psi(x-3t) =  \frac{1}{2}[f(x+3t) + f(x-3t)] + \frac{1}{6}\int_{x-3t}^{x+3t} g(x') dx'\end{equation}
To find what u is when x<3t, impose initial condition (4)
\begin{equation*}\phi'(3t) + \psi'(-3t) = h(t)\end{equation*}
Which implies
\begin{equation*}\phi(3t) - \psi(-3t) = 3\int_{0}^{t} h(t') dt' + C\end{equation*}
Let $x =-3t, x<0, t = -\frac{x}{3}$
\begin{equation*}\phi(-x) - \psi(x) = 3\int_{0}^{-x/3} h(t')dt' + C\\
\psi(x) = \phi(-x) - 3\int_{0}^{-x/3} h(t')dt' + C\end{equation*}
Then for u to be continuous, $\psi(x)$ must be continuous at 0,
\begin{equation} \psi(0_{+}) = \frac{1}{2}f(0) = \phi(0) + C =\frac{1}{2}f(0)+C=\psi(0_{-})\end{equation}
We get C = 0
Therefore when x< 3t,
\begin{equation} u = \frac{1}{2}[f(x+3t)+f(3t-x)] + \frac{1}{6}[\int_{0}^{x+3t} g(x') dx' + \int_{0}^{3t-x} g(x') dx' ]- 3\int_{0}^{t-x/3} h(t')dt'\end{equation}
Now plug in $f, g, h$, we get
&u = \cos (x+3t) + 3\cos (x-3t), x>3t\\[2pt]
&u = \cos (x+3t) + 2\cos (3t-x) + 1, 0<x<3t

Title: Re: FE1
Post by: Victor Ivrii on December 14, 2016, 01:08:42 PM
Luyu CEN

PS. In LaTeX we write \cos to get $\cos$ (upright, with a proper space after), and so on