8
$\begingroup$

I tried to proceed as expected: set $u = X(x)Y(y)$, then you get $$\frac{X''}{X} = -\frac{Y''}{Y} = -\lambda$$. Assume $\lambda>0$, and $\lambda = z^2$ then you get $$X = C_1e^{-zx}+C_2e^{zx}$$, $$Y = C_3\cos(zy)+C_4\sin(zy)$$. So then we get the from the first two conditions $X(0) = C_1+C_2 = 0$, and $Y(0) = C_3 = 0$. Then we get $$X(x) = C_2\sinh(zx)C_4\sinh(z) = \sin(x)$$ and $$C_2\sinh(z)C_4\sin(zy) = y^2$$ from the second two conditions. I have no idea where to go from there.

$\endgroup$
  • $\begingroup$ How about the other two cases $\lambda < 0$ and $\lambda = 0.$ It looks like the latter will give you a solution that satisfies the boundary condition. $\endgroup$ – dezdichado Jul 19 '17 at 19:14
  • $\begingroup$ @dezdichado does that mean I can just ignore the $\lambda \neq 0$ cases? $\endgroup$ – J.J. Jul 19 '17 at 19:38
  • $\begingroup$ Not necessarily. You have to consider all cases, but it looks like $\lambda\neq 0$ cases will not have a solution. Assuming your proof above was correct, you need to do the same for the negative case. $\endgroup$ – dezdichado Jul 19 '17 at 20:45
  • $\begingroup$ @dezdichado But what I wrote above was not really a proof. $\endgroup$ – J.J. Jul 19 '17 at 20:56
  • $\begingroup$ it basically is proving that the positive case does not give a solution. The last equality you obtained is impossible, so the solution does not exist. $\endgroup$ – dezdichado Jul 19 '17 at 21:55
1
$\begingroup$

The boundary conditions at point $(1,1)$ introduce a singularity $$\begin{cases} u(x,1)=\sin(x) \quad\to\quad u(1,1)=\sin(1)\\ u(1,y)=y^2 \quad\to\quad u(1,1)=1 \end{cases}$$

enter image description here

This can be overcome with the solution in term of Fourier series:

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.