MHB Wave equation and weird notation

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The discussion revolves around solving the wave equation $u_{tt}=u_{xx}$ with initial conditions $u(x,0)=0$ and $u_t(x,0)=\chi_{[-1,1]}(x)$, where $\chi_{[-1,1]}(x)$ is the characteristic function indicating a box function between -1 and 1. Participants clarify that the last condition represents an initial velocity profile, which can be expressed using Heaviside step functions for easier integration. The D'Alembert solution is suggested for solving the equation, leading to a specific formulation involving integrals of the characteristic function. There is a request for alternative methods to solve the problem without Heaviside functions, indicating a need for further clarification on the solution process. The conversation highlights the complexities of applying initial conditions in wave equations.
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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!
 
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Markov said:
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.
 
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

$\displaystyle u = \frac{f(x+t)+f(x-t)}{2} + \frac{1}{2}\int_{x-t}^{x+t} g(s)ds$.
 
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.
 
Hint: $\displaystyle \int H(x)dx = x H(x) + c$.
 
I still don't get it very well, how to do it with $H(x+1)$ for example?

Thanks a lot!
 
$\displaystyle \int H(x+a)dx = (x+a) H(x+a) + c$
 
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?
 
Yep, that it!
 
  • #10
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?
 

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