what is the value of $u(1/2,1/2)$?

Question

Let $u(x,t)$ be the solution of $u_{tt}-u_{xx}=1$, $x\in \mathbb{R}$, $t>0$ with $u(x,0)=0$ and $u_t(x,0)=0$, $x\in \mathbb{R}$. Then what is the value of $u(1/2,1/2)$?

I know the formula for $u_{tt}-u_{xx}=0$, $u(x,0)=f(x), u_t(x,0)=g(x)$. More precisely in general setting we have,

$$u(x,t)={f(x+ct)+f(x-ct)\over 2}+\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)$$

But how do I solve it for non homogeneous case? In my case $f=0, g=0$

Thanks for help.

Answer 1

Given an inhomogeneous wave equation on the whole line \begin{align} > &u_{tt}-c^2u_{xx}=f(x,t),\\ &u(x,0)=0,\\ &u_t(x,0)=0. \end{align} The solution formula is \begin{equation} > u(x,t)=\frac{1}{2c}\int_{0}^{t}\int_{x-c(t-s)}^{x+c(t-s)}f(y,s)\,dy\,ds. > \end{equation}

Derivation of inhomogeneous wave equation

Thanks Rafa Budria for the link.

$$u(x,t)=\frac{1}{2c}\int_{0}^{t}\int_{x-c(t-s)}^{x+c(t-s)}f(y,s)\,dy\,ds. $$

$$u(x,t)=\frac{1}{2c}\int_{0}^{t}\int_{x-c(t-s)}^{x+c(t-s)}1\,dy\,ds. $$

$$u(x,t)=\frac{1}{2c}\int_{0}^{t}2(t-s)\,ds. $$

$$u(x,t)=\frac{1}{c}\left[ts-s^2/2\right]^{t}_{0} $$

$$u(x,t)=\left[t^2-t^2/2\right] $$

$$u(x,t)=t^2/2 $$

Therefore $u(1/2,1/2)=1/8$

P.S. Note that value of $c$ is $1$ here.