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- Jan 26, 2012

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$\chi$ is called the "characteristic function". It's an indicator function, defined as follows:I have $u_{tt}=u_{xx},$ $x\in\mathbb R,$ $t>0,$ $u(x,0)=0$ and $u_t(x,0)=\chi_{[-1,1]}(x).$

What does mean the last condition? In such case, how to solve the equation then?

Thanks!

$$\chi_{A}(x)=\begin{cases}1,\quad x\in A\\ 0,\quad x\not\in A\end{cases}.$$

In your case, the function $\chi_{[-1,1]}(x)$ looks like a box function. It comes in from negative infinity at zero, bumps up to $1$ at $x=-1$, stays $1$ until $x=1$, and then drops back down to zero and stays there for the rest of the positive real axis. It looks like this.

I am not competent enough to help you solve your problem, however. Jester would be the man.

- Jan 26, 2012

- 183

$u_t(0,x) = g(x) = H(x+1) - H(x-1)$

then use the D'Alembert solution

$\displaystyle u = \frac{f(x+t)+f(x-t)}{2} + \frac{1}{2}\int_{x-t}^{x+t} g(s)ds$.

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- Jan 26, 2012

- 183

Hint: $\displaystyle \int H(x)dx = x H(x) + c$.

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- #6

- Jan 26, 2012

- 183

$\displaystyle \int H(x+a)dx = (x+a) H(x+a) + c$

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- #8

- Jan 26, 2012

- 183

Yep, that it!