Problem with satisfying Boundary conditions for 1D heat PDE

Question

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.

Answer 1

I've noticed two mistakes. First one is that you have three constants, $A$, $c_1$, and $c_2$, plus $\lambda$, and only three boundary conditions. When you multiply two constants (say $Ac_1$) you get a new constant. Let's call this $C_1$. Then your general solution is $$u(x,t) = e^{-\lambda Kt}\cdot(C_1\cos(\sqrt{\lambda}x)+C_2\sin(\sqrt{\lambda}x))$$ Now using the three equations for boundary conditions you can get $C_1$, $C_2$, and $\lambda$.

Also, I've noticed that for the final boundary condition you used $t=0$, but you still have $t$ in your equation. You should not repeat the same mistake after you solved your first issue.

Answer 2

The point is that $u(x,t) = Ae^{-\lambda Kt}\cdot(c_1cos(\sqrt{\lambda}x)+c_2sin(\sqrt{\lambda}x))$ is a solution if and only if $A$, $c_1$ and $c_2$ are constant. When you impose that $u(0,t) = f(t)$ you find a $c_1$ which depends on t and this is not possible. If $c_1$ depends on t,$u(x,t) = Ae^{-\lambda Kt}\cdot(c_1cos(\sqrt{\lambda}x)+c_2sin(\sqrt{\lambda}x))$ is not a solution of your differential equation. You should look at a solution in the form

$u(x,t)=c_1(x) e^{t/k} (c_2(t) e^{x \sqrt{k}}+c_3(t) e^{-x \sqrt{k}})$.