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I would like to know how to handle the following pde.

What makes it difficult for me to solve it is the fact that both boundary conditions for $x$ aren't zero.

Here's the equation:

$$u_{tt} - u_{xx} =0, \ \ t \ge 0, \ x \in [0, \pi] $$

$$u(0,x) = 1, \ \ u_t(0,x)=1+x, \ \ x \in [0, \pi ]$$

$$u_x(t,0) = u_x(t, \pi)=0, \ t \ge 0$$

Let $u(t,x) = T(t)X(x)$

So it all comes down to solving the following: $\frac{X''(x)}{X(x)} = \frac{T''(t)}{T(t)} = \lambda <0$ (otherwise the solutions are zero)

Next I get that $X_n(x) = A_n \cos (nx), \ \ n \in \mathbb{Z} $ (because $X'(0)=0$ and $X(0)=0$ )

and $T(t)=C \cos (\sqrt{ \lambda}t) + D \sin(\sqrt{ \lambda}t) $

$T(0)=C$

and $u(0,x) = T(0)X(x) = CX(x) = 1$

How can it be?

Could you tell me where I make a mistake and help me solve this?

Thank you!

Hagrid
  • 2,581

1 Answers1

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Cosines aren't the only eigenfunctions: $X(x)=\text{constant}$ also works. It's true that the case $\lambda>0$ gives only trivial solutions, but you overlooked the possibility that $\lambda=0$.

Hans Lundmark
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  • Ok, so if $X''(x) = 0$, then $X(x)=ax+b$, but $X'(0)=0$, so $X(x)=ax+b, X'(0)=a=0$ and $X(x)=b$. With $T$ we have $T(t) = \alpha x + \beta$ and $u(0,x) = T(0)X(x) = \beta b = 1$ and $u_t(0,x) = T'(0)X(x) = \alpha X(x) = 1+x$ and this suggests that $X$ is dependent on $x$, doesn't it? – Hagrid Feb 06 '15 at 08:18
  • So there aren't any solutions for $\lambda <0$? – Hagrid Feb 06 '15 at 08:18
  • None (except zero) which satisfy the boundary conditions $X'(0)=X'(\pi)=0$. By the way, I notice now that you have written the wrong conditions; you say $X(0)=0$, but that should be $X'(\pi)=0$. – Hans Lundmark Feb 06 '15 at 11:26
  • And what you should do with the initial data $u(0,x)$ and $u_t(0,x)$ is expand them in the basis of eigenfunctions, so that you can solve the problem for each Fourier mode separately. I have absolutely no idea what you are trying to do in the second sentence of your first comment... – Hans Lundmark Feb 06 '15 at 11:29