# Solution of the wave equation $u_{tt}-u_{xx}=0$.

Consider the wave equation $u_{tt}-u_{xx}=0$ for $(x,t)\in \mathbb{R}\times(0,\infty)$ and $u(x,0)=f(x)$, $u_t(x,0)=g(x)$, $x\in \mathbb{R}$.

Let $u_i$ be the solution of the above problem with $f=f_i$ and $g=g_i$ for $i=1,2$ where $f_i:\mathbb{R}\to \mathbb{R}$ and $g_i:\mathbb{R}\to \mathbb{R}$ are given $C^2$ functions satisfying $f_1(x)=f_2(x)$ and $g_1(x)=g_2(x)$ for every $x\in [-1,1]$. Which of the following are necessarily true?

1. $u_1(0,1)=u_2(0,1)$
2. $u_1(1,1)=u_2(1,1)$
3. $u_1(\frac{1}{2},\frac{1}{2})=u_2(\frac{1}{2},\frac{1}{2})$
4. $u_1(0,2)=u_2(0,2)$

Since this is a general wave equation problem, it's solution is given by $$u(x,t)=\frac{1}{2}\{f(x+t)+f(x-t)\}+\frac{1}{2}\int_{x-t}^{x+t}g(\tau)d\tau$$ Now $$u_i(x,t)=\frac{1}{2}\{f_i(x+t)+f_i(x-t)\}+\frac{1}{2}\int_{x-t}^{x+t}g_i(\tau)d\tau$$ for $i=1,2$. So it looks if I put $i=1$, and using the relations of $f_1=f_2$ etc, all the options turn out to be correct. But answer says only option 1 and option 3 are correct. Where is the mistake? How to solve this problem? Any hint or help will be great. Thanks.

See this $$u_i(x,t)=\frac{1}{2}\{f_i(x+t)+f_i(x-t)\}+\frac{1}{2}\int_{x-t}^{x+t}g_i(\tau)d\tau$$ for $i=1,2$
Test for each options just by putting values of $x$ and $t$. Remember that $f_1(x)=f_2(x)$ and $g_1(x)=g_2(x)$ only for all values in $[-1,1]$.