So I was solving the 1D heat equation with time-dependent boundary conditions and zero initial condition:
PDE: $\frac{\partial u}{\partial t} = k\frac{\partial^{2} u}{\partial x^{2}}$
BC: $u(0,t)=A(t)$, $u(L,t)=B(t)$
IC: $u(x,0)=0$
So I went about solving this with separation of variables
$u(x,t) = \phi (x)h(t)$
to get the ODE's
$\frac{1}{kh}\frac{\partial h}{\partial t} = \frac{1}{\phi}\frac{\partial^{2} \phi}{\partial x^{2}} = -\lambda$
solving for $\phi(x)$ I get
$\phi(x) = c_{1}\sin x\sqrt{\lambda} + c_{2}\cos x\sqrt{\lambda}$
and using $\phi(0)=A(t)$ and $\phi(L)=B(t)$ I solved for the constants:
$c_{2} = A(t)$ and $c_{1} = \frac{B(t)-A(t)\cos L\sqrt{\lambda}}{\sin L\sqrt{\lambda}}$
solving for $h(t)$ is fairly simple, and I get $h(t) = c_{h}\exp{-\lambda kt}$ then I put together $u(x,t) = \phi (x)h(t)$ and get
$u(x,t) = c_{h}\exp{-\lambda kt}[\frac{B(t)-A(t)\cos L\sqrt{\lambda}}{\sin L\sqrt{\lambda}}\sin x\sqrt{\lambda} + A(t)\cos x\sqrt{\lambda}]$
using $u(x,0) = 0 $ I should be able to solve for $c_{h}$, but that gets me that $c_{h} = 0$ and then $u(x,t) = 0$. Is the function $u(x,t)$ actually zero in this case, because of the initial condition, or did I just do something wrong while solving for $u(x,t)$? I've only ever done this problem when the initial condition was along the lines of $u(x,0)=f(x)$, which gives a non-zero solution.