Author Topic: HA #4, problem 1  (Read 149 times)

Shaghayegh A

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HA #4, problem 1
« on: October 10, 2016, 01:25:42 PM »
I am having trouble with problem 1 of home assignment 1, it asks to find u(x,t) for :

$$\begin{align*}
& u_{tt}-c^2u_{xx}=0, &&t>0, x>0,  \\\
&u|_{t=0}= \phi (x),   &&x>0, \\
&u_t|_{t=0}= c\phi'(x),  &&x>0, \\
&u|_{x=0}=\chi(t), &&t>0.
\end{align*}$$

My solution:  $u=f(x+ct)+g(x-ct)$ where f and g are some functions. By the boundary conditions,
$$\begin{align*}
& f(x)+g(x)=\phi(x) \\\
& f'(x)-g'(x)=\phi ' (x) \implies f(x)-g(x)=\phi(x)\\\
\end{align*}$$ So $f(x)=\phi(x)$ and $g(x)=0$ so $f(x+ct)=\phi(x+ct)$ , but is this true for all x>0? Because it seems that t can be negative here and we must say $$f(x+ct)=\phi(x+ct),  x>ct $$
Thank you

Victor Ivrii

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Re: HA #4, problem 1
« Reply #1 on: October 10, 2016, 02:31:16 PM »
Quote
Because it seems that $t$ can be negative here
Please note $t>0$ in equation

PS Also: you posted it originally into the wrong board "Chapter 3"