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.
Best Answer
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.