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Wave equation and weird notation

Markov

Member
Feb 1, 2012
149
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!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,183
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!
$\chi$ is called the "characteristic function". It's an indicator function, defined as follows:

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

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.
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
As Ackbach said (via his link to wolfram) convert the IC using a pair of Heaviside step functions

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

then use the D'Alembert solution

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

Markov

Member
Feb 1, 2012
149
Okay, I'm almost there, in this case, we have $f=0,$ the solution is just $u(x,t)=\dfrac12\displaystyle\int_{x-c}^{x+c}g(s)\,ds,$ and $g(s)$ should be expressed as the Heaviside step functions as you mentioned, but I don't know how to make it work with the D'lembert formula.
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Hint: $\displaystyle \int H(x)dx = x H(x) + c$.
 

Markov

Member
Feb 1, 2012
149
I still don't get it very well, how to do it with $H(x+1)$ for example?

Thanks a lot!
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
$\displaystyle \int H(x+a)dx = (x+a) H(x+a) + c$
 

Markov

Member
Feb 1, 2012
149
Okay I get that, since I have $u(x,t)=\dfrac12\displaystyle\int_{x-c}^{x+c}g(s)\,ds,$ so the solution equals $u(x,t)=\displaystyle\frac{1}{2}\int_{x-t}^{x+t}{\left( H(s+1)-H(s-1) \right)\,ds},$ is that what you mean?
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Yep, that it!
 

Hurry

New member
May 31, 2012
3
Hi, I'm currently studying this because I have a test tomorrow but I don't get very well the solution, is there a way to solve it without using the Heaviside step function?