I have a PDE:
$$u_{xx} - u_{yy} = 0$$
And my boundary conditions are:
$u = -\sin(2 \pi x)$ on $x+y = -1$
$u = \sin(2 \pi x)$ on $x-y=1$
Now I can find the characteristic variables $\phi = x+y$ and $\psi = x-y$, and after reducing the equation to standard form I find the general solution to be:
$$u = F(x+y) + G(x-y)$$
Now when I try to impose the boundary conditions:
$F(-1) + G(2x+1) = -\sin(2 \pi x)$
$F(2x-1) + G(1) = \sin(2 \pi x)$
My first difficulty is solving that system for $F$ and $G$, I cant quite see how to solve that.
My second issue is showing that this solution is uniquely determined for the region bound by $y < 0$ and the two lines where the data is given. For this I can see that we have determined $F$ and $G$ on the lines, but how do we know this also gives us a unique solution inside the region?
Any help is very much appreciated,
Thanks
First we can rewrite $\sin(2\pi x)$ as: $$\sin(2\pi x)=\sin\left(\pi(x+y)+\pi (x-y)\right)$$ $$=\sin(\pi(x+y))\cos(\pi(x-y))+\cos(\pi(x+y))\sin(\pi(x-y))......(1)$$
On $x+y=-1$, we have $$u = F(-1) + G(x-y) = -\sin(2\pi x)=-\sin(-\pi)\cos(\pi(x-y))-\sin(\pi(x-y))\cos(-\pi)\implies F(-1) + G(x-y) =\sin(\pi(x-y))\implies G(\eta) =\sin(\pi\eta)-F(-1) ......(2)$$
Similarly:
On $x-y=1$, we have $$u = F(x+y) + G(1) = \sin(2\pi x)=\sin(\pi(x+y))\cos(-\pi)+\cos(\pi(x+y))\sin(-\pi)\implies F(x+y) + G(1) =-\sin(\pi(x+y))\implies F(\xi) =-\sin(\pi\xi)- G(1)......(3)$$
Thus $$F(-1)=-\sin(-\pi)- G(1)=-G(1)=c=\text{const}$$
So $$u(x,y)=F(x+y)+G(x-y)=-\sin(\pi(x+y))+c+\sin(\pi (x-y))-c$$ $$=-\sin(\pi(x+y))+\sin(\pi (x-y))$$