Toronto Math Forum
APM3462016F => APM346Tests => FE => Topic started by: Victor Ivrii on December 13, 2016, 07:49:01 PM

Solve by the method of characteristics the BVP for a wave equation
\begin{align}
&u_{tt} 9 u_{xx}=0,\qquad 0<x<\infty , \; t>0\label{11}\\[2pt]
& u(x,0)=f(x),\label{12}\\[2pt]
& u_t(x,0)=g(x),\label{13}\\[2pt]
& u_x (0,t)= h(t)\label{14}
\end{align}
with $f(x)=4\cos(x)$, $g(x)=6\sin (x) $ and $h(t)=\sin (3t)$. You need to find a continuous solution.

My solution attempt for question 1

you forget $c$ when you are doing the integration in the second part

Sajjan
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

The general solution for u is
\begin{equation*}u = \phi(x+3t) + \psi(x3t)\end{equation*}
Now impose the initial conditions (2) and (3)
\begin{align}
&\phi(x) + \psi(x) = f(x)\\
&\phi'(x) + \psi'(x) = g(x)\end{align}
Solve the above to give
\begin{align}
&\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(x3t) = \frac{1}{2}[f(x+3t) + f(x3t)] + \frac{1}{6}\int_{x3t}^{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(3tx)] + \frac{1}{6}[\int_{0}^{x+3t} g(x') dx' + \int_{0}^{3tx} g(x') dx' ] 3\int_{0}^{tx/3} h(t')dt'\end{equation}
Now plug in $f, g, h$, we get
\begin{align}
&u = \cos (x+3t) + 3\cos (x3t), x>3t\\[2pt]
&u = \cos (x+3t) + 2\cos (3tx) + 1, 0<x<3t
\end{align}

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