Say I have the PDE

$$u_{tt}=4u_{xx}, u(x,0)=f(x),u_t(x,0)=g(x), 0<x<L$$

How does the solution change if I am given the boundary conditions

$$u(0,t)=u(1,t)=0$$ versus $$u_x(0,t)=u_x(1,t)=0$$?

In particular, how does the solution change if it is of the form $$u(x,t)=\sum_{n=1}^\infty\left(A_n\sin{\frac{cn\pi t}{L}}+B_n\cos{\frac{cn\pi t}{L}}\right)\sin{\frac{n\pi x}{L}}$$

  • $\begingroup$ Have you worked out the problem to see the difference? $\endgroup$ – Mhenni Benghorbal Apr 25 '14 at 19:27

Compute the initial conditions for the series and we see:

$u(0,t)=u(1,t)=0$ (due to the $\sin(\frac{n\pi}{L}x)$)

And $u_x(x,t)=\sum_{n=1}^\infty(A_n\sin(\frac{cn\pi t}{L})+B_n\cos(\frac{cn\pi t}{L}))\frac{n\pi}{L}\cos(\frac{n\pi x}{L})$

Here $u_x(0,t)=\sum_{n=1}^\infty(A_n\sin(\frac{cn\pi t}{L})+B_n\cos(\frac{cn\pi t}{L}))\frac{n\pi}{L}=0$

And $u_x(1,t)=\sum_{n=1}^\infty(-1)^n(A_n\sin(\frac{cn\pi t}{L})+B_n\cos(\frac{cn\pi t}{L}))\frac{n\pi}{L}=0$

This would impose on the values of $A_n,B_n$, the problem is that the change of initial/ boundary conditions changes the calculation of the form of the solution.

The general solution is found by solving:

$X''=-\lambda X$ for different choices of $\lambda$, i.e. $\lambda\lt 0,\lambda = 0\lambda,\gt 0$. With respect to the boundary conditions, so if those change, then the form of the (series) solution changes.

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  • $\begingroup$ for the Dirichlet and Neumann boundary conditions $\lambda\ge 0$... $\endgroup$ – DVD Jun 12 '14 at 20:14
  • $\begingroup$ @Daved, indeed that is true $\endgroup$ – Ellya Jun 12 '14 at 20:23

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