I must solve the following problem with initial boundary value:
\begin{cases} u_t = c^2 u_{xx} \hspace{3cm} 0<x<L, \hspace{0.3cm} t>0, \hspace{0.3cm}c>0\\ u(x,0) = f(x) \hspace{2.3cm} 0 \leq x \leq L\\ hu(0,t)- u_{x}(0,t) = 0 \hspace{0,79cm} t \geq 0, \hspace{0.3cm} h>0\\ hu(L,t)+ u_{x}(L,t) = 0 \hspace{0.68cm} t \geq 0, \hspace{0.3cm} h>0 \end{cases}
My attempt:
To do this, I proceeded by the method of separation of variables assuming a solution of the form $u(x,t) = X(x)T(t)$.
Then, $u_{t} = XT'$ and $u_{xx} = X''T$, so $\frac{T'}{c^{2}T} = \frac{X''}{X} = -\lambda$, where $\lambda$ is a constant.
We obtain the ODE $X'' + \lambda X = 0$
The boundary conditions give:
$$hu(L,t) + u_{x}(L,t) = hX(L)T(t) + X'(L)T(t) = 0 \Rightarrow hX(L) + X'(L) = 0$$ $$hu(0,t) - u_{x}(0,t) = hX(0)T(t) - X'(0)T(t) = 0 \Rightarrow hX(0) - X'(0) = 0$$
We mount the following problem:
\begin{cases} X'' + \lambda X = 0\\ hX(0)-X'(0) = 0 \hspace{0.65cm} (1)\\ hX(L) + X'(L) = 0 \hspace{0.5cm} (2) \end{cases}
if $\lambda > 0$, say $\lambda = k^2$ with $k> 0$, then
$$X(x) = a \cos(kx) + b \sin(kx)$$
By equation (1) we have: $ha-kb = 0$
By equation (2) we have: $(ha +kb)\cos(kL) + (hb-ka)\sin(kL) = 0$
As $ha = kb$, then $(2kb)\cos(kL) + (hb-ka)\sin(kL) = 0$
Now my question is: How can I continue with this? Is my reasoning correct? I haven't really found a way to be able to continue with this problem. I need to find $X$ and its coefficients $a$ and $b$. Then get the eigenvalues $\lambda_n$ and then find $T$. I appreciate any help you can give me.
Note: I had recently posted this problem, but I have now made a more detailed attempt.