Wave equation and weird notation

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SUMMARY

The discussion revolves around solving the wave equation $u_{tt}=u_{xx}$ with initial conditions $u(x,0)=0$ and $u_t(x,0)=\chi_{[-1,1]}(x)$. The characteristic function $\chi_{[-1,1]}(x)$ acts as an indicator function, defining the initial velocity profile. The solution approach involves using the D'Alembert formula, where the initial condition is transformed using Heaviside step functions. The final expression for the solution is $u(x,t)=\frac{1}{2}\int_{x-t}^{x+t}(H(s+1)-H(s-1))\,ds$.

PREREQUISITES
  • Understanding of wave equations, specifically $u_{tt}=u_{xx}$.
  • Familiarity with initial conditions in partial differential equations.
  • Knowledge of characteristic functions and Heaviside step functions.
  • Proficiency in applying the D'Alembert solution method for wave equations.
NEXT STEPS
  • Study the D'Alembert solution for wave equations in detail.
  • Learn about the properties and applications of Heaviside step functions.
  • Explore alternative methods for solving wave equations without Heaviside functions.
  • Investigate the implications of initial conditions on the behavior of wave solutions.
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Students and researchers in applied mathematics, particularly those focusing on partial differential equations and wave phenomena, as well as educators preparing for examinations in mathematical physics.

Markov2
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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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