I ask to solve the following PDE:
$$u_{tt}-\triangle u=0$$ $$u(x,y,0)=x$$ $$u_t(x,y,0)=0$$
I know that the general solution to this equation is
$$u(x,t)=\displaystyle\frac{2}{3\pi t^2}\int_{B(x,t)}\displaystyle \frac{tg(y)+t^2h(y)+tDg(y)\cdot(y-x)}{(t^2-|y-x|^2)^{1/2}}$$
for $t>0$. This is given from the Evans.
Then I replace the values to get
$$u(x,t)=\displaystyle\frac{2}{3\pi t^2}\int_{B(x,t)}\displaystyle\frac{ty_1+t(1,0)\cdot(y-x)}{(t^2-|x-y|^2)^{1/2}}dy$$
taken $y=(y_1,y_2)$ and $x=(x_1,x_2)$
$$u(x,t)=\displaystyle\frac{2}{3\pi t^2}\int_{B(x,t)}\displaystyle\frac{2ty_1-tx_1}{(t^2-|x-y|^2)^{1/2}}dy$$
splitting this equation in two equations
$$u(x,t)=\displaystyle\frac{4}{3\pi t}\int_{B(x,t)}\displaystyle\frac{y_1}{(t^2-|x-y|^2)^{1/2}}dy-\displaystyle\frac{2x_1}{3\pi t}\int_{B(x,t)}\displaystyle\frac{1}{(t^2-|x-y|^2)^{1/2}}dy$$
I know how to solve the second equation using spherical coordinates, but how can I solve the first one? Any advice?
Thanks!
You do not need to go through all that work. Since the initial condition depends only on $x$, the same will be true of the solution. This leads to a one-dimensional wave equation, and the solution is $$ u(x,y)=x. $$