I am trying to solve the wave equation:

$u_{tt}$ = $u_{xx}$

With initial values:

$u(x,0) =\begin{cases} x^3 - x, &\text{for }|x|\le 1,\ \\0, &\text{for }|x|\ge1\end{cases}$

$u_t(x,0) =\begin{cases} 1 - x^2, \text{for } |x|\le 1, \\ 0, \text{for } |x|\ge1 \end{cases}$

In the domain $D= \{(x,t) | - \infty < x<\infty, \space\space t>0\} $

I am using D'Alembert's Formula and after inputting the values in I have:

$u(x,t) = \frac{1}{2}[\frac{2}{3}(x+t)^3 + \frac{4}{3}(x-t)^3 - 2(x-t)] $

I am not sure where to go from here.

I think because the initial conditions are not differentiable at some points then there will be points in the solution that are not differentiable

So I'd need to consider the different cases of $|x+t|\le 1$?

I could be going down the complete wrong path so any help is appreciated.


  • $\begingroup$ The tag (differential-equations) is intended for questions about ordinary differential equations, there is a separate tag for pdes; see the tag-wiki and the tag-excerpt. (The tag-excerpt is also shown when you are adding a tag to a question.) $\endgroup$ Mar 11, 2014 at 19:53
  • $\begingroup$ just to point out you may of made a mistake with D'Alembert's formula, as I get $$\frac{1}{2}\left[\frac{4}{3}(x-t)^{3} + \frac{2}{3}(x+t)^{3} - 2(x-t)\right]$$? Though it could be me with the mistake :). $\endgroup$
    – Chinny84
    Mar 11, 2014 at 22:21
  • $\begingroup$ Thank you @Chinny84 I've corrected the mistake, do you know where I go from here? $\endgroup$
    – JohnVa
    Mar 12, 2014 at 9:11

1 Answer 1


In fact the real result should be $u(x,t)=\dfrac{f(x+t)+f(x-t)+g(x+t)-g(x-t)}{2}$ , where $f(x)=\begin{cases}x^3-x&\text{for}~|x|\leq1\\0&\text{for}~|x|\geq1\end{cases}$ and $g(x)=\begin{cases}-\dfrac{2}{3}&\text{for}~x\leq-1\\x-\dfrac{x^3}{3}&\text{for}~|x|\leq1\\\dfrac{2}{3}&\text{for}~x\geq1\end{cases}$


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