Author Topic: FE-1  (Read 3111 times)

Victor Ivrii

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FE-1
« on: December 18, 2015, 07:39:28 PM »
Solve by the method of characteristics the BVP for a wave equation
\begin{align}
&4 u_{tt}-  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 (0,t)= h(t)\label{1-4}
\end{align}
with $f(x)=2xe^{-2x^2}$,   $g(x)=0$ and  $h(t)=te^{-t^2/2}$.

Yiqi Shi

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Re: FE-1
« Reply #1 on: December 18, 2015, 10:37:02 PM »
\begin{equation}
u_{tt}-\frac{1}{2}u_{xx}=0
\end{equation}
\begin{equation}
u(x,t)=\phi(x+\frac{1}{2}t)+\psi(x-\frac{1}{2}t)
\end{equation}
by(2) and (3)
\begin{equation}
\phi(x)+\psi(x)=f(x)=2xe^{-2x^2}\\
\phi(x)-\psi(x)=g(x)=0
\end{equation}
Then
\begin{equation}
\phi(x)=xe^{-2x^2}\\
\psi(x)=xe^{-2x^2}
\end{equation}
Thus for $x>\frac{1}{2}t$
\begin{equation}
\mu(x,t)=(x+\frac{1}{2}t)e^{-2(x+\frac{1}{2}t)^2}+(x-\frac{1}{2}t)e^{-2(x-\frac{1}{2}t)^2}
\end{equation}
Now consider  $x<\frac{1}{2}t$
by (4)
\begin{equation}
\phi(\frac{1}{2}t)+\psi(-\frac{1}{2}t)=te^{-t^2/2}
\end{equation}
let $x=-\frac{1}{2}t$, then $t=-2x$
\begin{equation}
\phi(-x)+\psi(x)=-2xe^{-(-2x)^2/2}\\
\psi(x)=-2xe^{-2x^2}-\phi(-x)\\
\psi(x)=-2xe^{-2x^2}+xe^{-2x^2}\\
\psi(x)=-xe^{-2x^2}
\end{equation}
Thus for $x<\frac{1}{2}t$
\begin{equation}
\mu(x,t)=(x+\frac{1}{2}t)e^{-2(x+\frac{1}{2}t)^2}-(x-\frac{1}{2}t)e^{-2(x-\frac{1}{2}t)^2}
\end{equation}
« Last Edit: December 18, 2015, 10:48:59 PM by Yiqi Shi »

Vivian Tan

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Re: FE-1
« Reply #2 on: December 18, 2015, 10:43:19 PM »
For this I used D'Alembert formulas:

for $ x > \frac{t}{2}$:
\begin{equation}
u(x, t) = \frac{1}{2}[2(x + \frac{t}{2})e^{-2(x+\frac{t}{2})^2} + 2(x - \frac{t}{2})e^{-2(x-\frac{t}{2})^2}] + \int^{x+\frac{t}{2}}_{x-\frac{t}{2}}0 dx'
\end{equation}

\begin{equation}
u(x, t) =  \frac{1}{2}[2(x + \frac{t}{2})e^{-2(x+\frac{t}{2})^2} + 2(x - \frac{t}{2})e^{-2(x-\frac{t}{2})^2}]
\end{equation}

for $ 0 < x < \frac{t}{2}$:
\begin{equation}
u(x, t) = \frac{1}{2}[2(x + \frac{t}{2})e^{-2(x+\frac{t}{2})^2} + 2(\frac{t}{2} - x)e^{-2(\frac{t}{2} - x)^2}]  + (t - 2x)e^{\frac{(t - 2x)^{2}}{2}}
\end{equation}

Victor Ivrii

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Re: FE-1
« Reply #3 on: December 19, 2015, 07:18:13 AM »
For $x>t/2$ both solutions coincide (and are correct). However for $0<x<t/2$ Vivian made an error (here only, not in the paper).


I just graded this problem. Almost half solved it either correctly or with minor (computational) errors (as $0<x<t/2$) getting 15 or 13 pts respectively. About third forgot about region $0<x<t/2$ but got a correct answer as $x>t/2$ getting around 9 pts  (depending on some specifics of solution). Few had no clue what they are doing.

So, a correct answer (by Yiqi (whose love to letter $\mu$ was already noticed :D)
\begin{equation*}
u(x, t) = \left\{\begin{aligned}
& \frac{1}{2}\bigl(x + \frac{t}{2}\bigr)e^{-2(x+\frac{t}{2})^2} + \bigl(x - \frac{t}{2}\bigr)e^{-2(x-\frac{t}{2})^2} && x>\frac{t}{2}>0,\\
& \frac{1}{2}\bigl( x + \frac{t}{2}\bigr)e^{-2(x+\frac{t}{2})^2} - \bigl(x - \frac{t}{2}\bigr)e^{-2(x-\frac{t}{2})^2} && 0< x<\frac{t}{2}.
\end{aligned}\right.
\end{equation*}
« Last Edit: December 19, 2015, 10:27:58 AM by Victor Ivrii »