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I have the following homogeneous 1-dimensional wave equation with $c = 1$:

$$u_{tt} - u_{xx} = 0$$

With initial data

$$ u(x,0) = \phi(x) = \begin{cases} 1 & |x| \leq 1 \\ 0 & |x| > 1 \\ \end{cases}$$

and $$u_{t}(x, 0) = 0$$

I am asked to find the general solution for any $x \in \mathbb{R}, t \geq 0$ in 3 cases:

$0 < t < 1$, $t = 1$, and $t > 1$.

So far, I've come up with the general solution using D'Alembert's formula: $$u(x, t) = \frac{1}{2} [\phi(x - t) + \phi(x + t)]$$

However, I'm having trouble evaluating this solution in each of the three cases. Any help would be much appreciated.

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  • $\begingroup$ Are you sure you don't want $u_{tt} - u_{xx} = 0$ as your wave equation? The equation $u_{tt} + u_{xx} = 0$ is elliptic, and I doubt D]Alembert's applies to it. Cheers! $\endgroup$ – Robert Lewis Apr 7 '19 at 16:57
  • $\begingroup$ My apologies, yes you're right. Thanks for pointing that out! $\endgroup$ – bandar339 Apr 7 '19 at 17:00
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$$u_{tt} - u_{xx} = 0$$

The general solution (without boundary condition) is : $$u(x,y)=f(x+t)+g(x+t)$$ The functions $f$ and $g$ are not necessarily the same.

The condition $u(x,0) = \phi(x)=f(x)+g(x)$ implies $\quad\begin{cases} f(x)=\frac12\phi(x)+h(x) \\ g(x)=\frac12\phi(x)-h(x) \end{cases}$

$$ u(x,t)=\frac12\phi(x+t)+h(x+t)+\frac12\phi(x-t)-h(x-t)$$

$h(x)$ is an arbitrary function to be determined by the condition $u_t(x,0)=0$

$$ u_t(x,t)=\frac12\phi'(x+t)+h'(x+t)-\frac12\phi'(x-t)+h'(x-t)$$ $$ u_t(x,0)=\frac12\phi'(x)+h'(x)-\frac12\phi'(x)+h'(x)$$ $$u_t(x,0)=0=2h'(x) \quad\implies\quad h(x)=C$$ Your solution is confirmed : $$ u(x,t)=\frac12\phi(x+t)+\frac12\phi(x-t)$$ The function $\phi$ is a given piecewise function : $$\phi(x) = \begin{cases} 1 & |x| \leq 1 \\ 0 & |x| > 1 \\ \end{cases}$$

$$ u(x,t)=\frac12\begin{cases} 1 & x+t \leq 1 \\ 1 & -x-t \leq 1 \\ 0 & x+t > 1 \\ 0 & -x-t > 1 \\ \end{cases}+ \frac12\begin{cases} 1 & x-t \leq 1 \\ 1 & -x+t \leq 1 \\ 0 & x-t > 1 \\ 0 & -x+t > 1 \\ \end{cases}$$

$$ u(x,t)=\frac12\begin{cases} 1 & x \leq 1-t \\ 1 & x \geq -1-t \\ 0 & x > 1-t \\ 0 & x < -1-t \\ \end{cases}+ \frac12\begin{cases} 1 & x \leq 1+t \\ 1 & x \geq -1+t \\ 0 & x > 1+t \\ 0 & x < -1+t \\ \end{cases}$$

We have to consider several regions limited by

$x=1-t \quad;\quad x=1+t \quad;\quad x=-1-t \quad;\quad x=-1+t\quad$

For $\quad \boxed{0<t<1}$ :

Case $\quad x>1+t \:: \quad u=\frac12(0)+\frac12(0)=0$.

Case $\quad 1-t<x\leq 1+t \: : \quad u=\frac12(0)+\frac12(1)=\frac12$.

Case $\quad -1+t<x\leq 1-t \: : \quad u=\frac12(1)+\frac12(1)=1$.

Case $\quad -1-t\leq x< -1+t \: : \quad u=\frac12(1)+\frac12(0)=\frac12$.

Case $\quad x\leq -1-t \: : \quad u=\frac12(0)+\frac12(0)=0$.

$$u(x,t)=\begin{cases} 0 && x>1+t \\ \frac12 && 1-t<x\leq 1+t \\ 1 && -1+t<x\leq 1-t\\ \frac12 && -1-t\leq x< -1+t\\ 0 && x\leq -1-t \end{cases}$$

The above formulas are not valid for $t>1$ which is outside the domain of study specified in the wording of the question.

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IN ADDITION :

For $\quad \boxed{t>1}$ :

$$u(x,t)=\begin{cases} 0 && x>1+t \\ \frac12 && -1+t<x\leq 1+t \\ 0 && 1-t<x\leq -1+t\\ \frac12 && -1-t\leq x< 1-t\\ 0 && x\leq -1-t \end{cases}$$

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