Wave equation and weird notation

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Discussion Overview

The discussion revolves around solving the wave equation $u_{tt}=u_{xx}$ with specific initial conditions, particularly focusing on the interpretation of the initial condition involving the characteristic function $\chi_{[-1,1]}(x)$. Participants explore various methods to approach the solution, including the use of Heaviside step functions and the D'Alembert solution.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks clarification on the meaning of the initial condition $u_t(x,0)=\chi_{[-1,1]}(x)$ and how to solve the wave equation under these conditions.
  • Another participant explains that $\chi_{[-1,1]}(x)$ is the characteristic function, describing it as a box function that takes the value 1 between -1 and 1 and 0 elsewhere.
  • A suggestion is made to convert the initial condition using Heaviside step functions, leading to the expression $g(x) = H(x+1) - H(x-1)$.
  • Participants discuss the D'Alembert solution and its application, noting that in this case, $f=0$ simplifies the solution to $u(x,t)=\frac{1}{2}\int_{x-t}^{x+t}g(s)ds$.
  • There is a hint provided regarding the integration of Heaviside functions, specifically $\int H(x)dx = x H(x) + c$.
  • Further clarification is sought on how to apply this integration to $H(x+1)$, indicating some uncertainty in the process.
  • One participant confirms the formulation of the solution using the Heaviside functions, expressing it as $u(x,t)=\frac{1}{2}\int_{x-t}^{x+t}(H(s+1)-H(s-1))\,ds$.
  • A later participant expresses a desire to find a solution without using Heaviside step functions, indicating ongoing uncertainty and exploration of alternative methods.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to solve the wave equation without using Heaviside step functions, and there remains uncertainty regarding the integration process and its implications for the solution.

Contextual Notes

Participants express varying levels of understanding regarding the application of Heaviside functions and the D'Alembert solution, indicating potential gaps in foundational knowledge or assumptions about the methods discussed.

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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