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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

Wooster
  • 3,775

1 Answers1

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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 $$u(x,y)=F(x+y)+G(x-y)=-\sin(\pi(x+y))+c+\sin(\pi (x-y))-c$$ $$=-\sin(\pi(x+y))+\sin(\pi (x-y))$$

mike
  • 5,604
  • Thanks for the answer, I'm a little lost though. Would you mind elaborating on how you make the final deduction that $F$ and $G$ are given by (2) and (3) respectively? – Wooster Jun 17 '14 at 20:48
  • We know the solution is $u(x,y)=F(x+y)+G(x-y)$. We just don't know what the explicit forms $F(x+y)$ and $G(x-y)$ are. After we expressed $\sin(2\pi x)=u(x+y)+v(x-y)$, we used the boundary conditions to nail down the forms of $F(x+y)$ and $G(x-y)$. – mike Jun 17 '14 at 20:57
  • Do you mean that G(x - y) is given by (2) (setting x-y = 0, we see that F(-1) = 0), and F(x + y) is given by (3) (setting x + y = 0, we see that G(1) = 0, or just look at 2 to see that G(1) = 0)? Nice solution though, I like it. – breeden Jun 17 '14 at 21:05
  • There was an error in (1) which leads errors in (2) and (3). I think that I have corrected all errors in (1)-(3). Please take a look at the answer again. – mike Jun 17 '14 at 23:57