I have the following pde and boundary conditions I am trying to solve by separation of variables. I am doing something wrong with my formulation because my boundaries are incorrect when I plot my solution.
The Problem
$u$ is a function of space ($x$) and time ($t$).
$$\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}$$ $$s.t.$$ $$u(0,t) = f(t)$$ $$\frac{\partial}{\partial x}u(L,t) = 0$$ $$u(x,0) = C = constant$$
My Attempt
Using separation of variables, I arrive at:
$$u(x,t) = Ae^{-\lambda Kt}\cdot(c_1\cos(\sqrt{\lambda}x)+c_2\sin(\sqrt{\lambda}x))$$
Now satisfying the first boundary condition: $$u(0,t) = Ae^{-\lambda Kt}\cdot(c_1\cos(0)+c_2\sin(0))= Ae^{-\lambda Kt}\cdot c_1 = f(t)$$
$$c_1 = \frac{f(t)}{Ae^{-\lambda Kt}}$$
Now the second BC:
$$\frac{\partial}{\partial x}u(L,t) = Ae^{-\lambda Kt}\cdot \sqrt{\lambda}(-c_1\sin(\sqrt{\lambda}L)+c_2\cos(\sqrt{\lambda}L)) = 0$$
This is satisfied only when $$-c_1\sin(\sqrt{\lambda}L)+c_2\cos(\sqrt{\lambda}L) = 0$$
$$c_2 = \frac{f(t)}{Ae^{-\lambda Kt}}\tan(\sqrt{\lambda}L)$$
Final boundary condition:
$$u(x,0) = c_1\cos(\sqrt{\lambda}x)+c_2\sin(\sqrt{\lambda}x) = \frac{f(t)}{Ae^{-\lambda Kt}}\cos(\sqrt{\lambda}x)+\frac{f(t)}{Ae^{-\lambda Kt}}\tan(\sqrt{\lambda}L)\sin(\sqrt{\lambda}x)= C$$
$$\frac{f(t)}{Ae^{-\lambda Kt}}(\cos(\sqrt{\lambda}x)+\tan(\sqrt{\lambda}L)\sin(\sqrt{\lambda}x)) = C$$ $$A = \frac{f(t)}{Ce^{-\lambda Kt}}(\cos(\sqrt{\lambda}x)+\tan(\sqrt{\lambda}L)\sin(\sqrt{\lambda}x))$$
I am pretty stuck up to here. For one, I do not know what the eigenvalue $\lambda$ sould be ($= \frac{n\pi}{L}$?) and I do not know how if I did this right. If so, I would start by computing $A$ because it appears that $c_1$ and $c_2$ depend on $A$. Thanks in advance for any advice.