I have the following PDE:

\begin{cases} u_t = u_{xx}, \ x\in (0, 2), \ t>0 \\[6pt] u(x, 0) = \sin\frac{\pi x}{2}, \ x \in [0, 2] &(i)\\[6pt] u(0, t) = 1, \ t>0 &(ii)\\[6pt] u(2, t) = 0, \ t>0 &(iii)\\[6pt] \end{cases}

I know how to solve boundary problems by using the method of separation of variables (and Fourier Series), but in this case I have trouble with the last condition.

Given that I have obtained the solution of the first ODE in the form of $X(x) = c_1\cos(px) + c_2\sin(px)$ (where $p$ is $\lambda = -p^2$), from the first boundary condition I will get that $c_1=1$.

By plugging it in the second boundary condition I obtain the following: $$\cos(2p) + c_2\sin(2p) = 0$$ I am looking for a hint on how to proceed from here. Do I need to express $p$ in terms of $c_2$ and continue from there? And if so, how can I proceed with finding the coefficients? Thanks.

  • $\begingroup$ Any reason you choose $\lambda = -p^2$ vs $\lambda = p^2$? $\endgroup$ – Cuhrazatee Feb 20 '19 at 20:24
  • 1
    $\begingroup$ @Cuhrazatee, because when $\lambda$ is positive, the only solution is the trivial one, if I'm not mistaken $\endgroup$ – aniri Feb 20 '19 at 20:29
  • $\begingroup$ Yeah, that's right. I didn't see that at first sight. $\endgroup$ – Cuhrazatee Feb 20 '19 at 22:25

You can't immediately proceed with separation of variables since the boundary conditions aren't homogeneous. To "fix" this, break up the solution into $u(x,t) = w(x) + v(x,t)$. Here, $w(x)$ is the steady-state solution that exactly matches the boundary conditions:

\begin{cases} w_{xx} = 0 \\ w(0) = 1 \\ w(2) = 0 \end{cases}

Solving the above gives $w(x) = \frac12(2-x)$. Then, it remains to solve

\begin{cases} v_t = v_{xx} \\ v(0,t) = v(2,t) = 0 \\ v(x,0) = \sin \dfrac{\pi x}{2} - \frac12(2-x) \end{cases}

which can be done with separation of variables. The general solution is

$$ v(x,t) = \sum_{n=1}^\infty c_n\sin\left(\frac{n\pi x}{2}\right)e^{-n^2\pi^2 t/4} $$

You can take it from here.

| cite | improve this answer | |
  • $\begingroup$ thank you for the help! $\endgroup$ – aniri Feb 21 '19 at 14:41

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.