### Author Topic: HA #4, problem 1  (Read 1501 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

Because it seems that $t$ can be negative here
Please note $t>0$ in equation