Consider one of the solutions to the Laplace equation $$u(x,t)=\sum_{n=1}^{\infty} C_{n}e^{-\alpha _{n}^2kt}\left [ \frac{-\alpha_{n} }{h}\cos(\alpha _{n}x) + B_{n}\sin(\alpha _{n}x) \right ],$$ where $\alpha _{n}^2$ is a positive eigenvalue. An initial condition is given as $$u(x,0)=\sum_{n=1}^{\infty}C_{n}\left [ \frac{-\alpha_{n} }{h}\cos(\alpha _{n}x) + B_{n}\sin(\alpha _{n}x) \right ].$$
I am not showing my work that got me to this solution, because that would be a bit of time waster, both for me and whoever is reading.
My question is:
From this information, what would be the constants(usually given as an expression with an integral) of $C_{n}$ and $B_{n}$?
I usually do not have trouble with this, but the $C_{n}$ seems to be where I am confused. I understand that this would be a full Fourier seires, but I am not too sure how to continue. Could anyone provide me some feedback?
Thanks in advance!
Edit: The boundary conditions are: $$u_{x}(0,t) +hu(0,t)=0 \ \ \ \ \text{and} \ \ \ \ u_{x}(1,t)=0, t>0.$$ Also, I am just showing the solution for the positive eigenvalues, since I do not need help finding the constants for the negative one.
Let's start off with the eigenfunction in $x$
$$ X_n(x) = -\frac{\alpha_n}{h}\cos(\alpha_n x) + B_n\sin(\alpha_n x) $$
The first B.C. gives
$$ X_n'(0) + hX_n(0) = \alpha_n B_n - \alpha_n = 0 \implies B_n = 1 $$
With that out of the way, the second B.C. gives
$$ X_n'(1) = \alpha_n\left[\frac{\alpha_n}{h}\sin \alpha_n + \cos\alpha_n\right] = 0 $$
$$ \implies \alpha_n\tan\alpha_n = -h $$
This is a necessary condition for $\alpha_n$. There is no closed form, but you can solve it numerically given any $h$.
To show the eigenfunctions are orthogonal, note that for $n\ne m$ we have
$$ \int_0^1 {X_m}''X_n = {X_m}'X_n\Bigg|_0^1 - X_n{X_m}'\Bigg|_0^1 + \int_0^1 X_m{X_n}'' $$
However
\begin{align} {X_m}'X_n\Bigg|_0^1 - X_n{X_m}'\Bigg|_0^1 &= -{X_m}'(0)X_n(0) + X_m(0){X_n}'(0) \\ &= hX_m(0)\cdot X_n(0) - X_m(0)\cdot hX_n(0) = 0 \end{align}
Therefore
$$ \int_0^1 {X_m}''X_n - \int_0^1 X_m{X_n}'' = 0 \implies -(\alpha_m^2-\alpha_n^2)\int_0^1 X_mX_n = 0 $$
But $\alpha_m \ne \alpha_n$, so the integral must be $0$
The remaining condition is
$$ u(x,0) = f(x) = \sum_n C_n X_n(x) $$
where $f(x)$ is a known function. Using orthogonality
$$ \int_0^1 f(x)X_m(x) dx = \sum_n C_n \int_0^1 X_n(x) X_m(x) dx = C_m \int_0^1 \big[X_m(x)\big]^2\ dx $$
$$ \implies C_n = \frac{\int_0^1 f(x) X_n(x)\ dx}{\int_0^1 \big[X_n(x)\big]^2\ dx} $$
where the integrals depend on $\alpha_n$. Again, numerical computation is useful here
If you follow the same method for the other case, i.e. $X'' = \mu^2 X$, you'll find
$$ X_n(x) = \frac{\mu_n}{h}\cosh(\mu_n x) + \sinh(\mu_n x) $$
and $ \mu_n\tanh\mu_n = h $. There is one solution if $h > 0$.
Note that the formula for $C_n$ works the same whether the eigenvalue is positive, negative, or zero.