I'm working through chapter 4 of Analysis of Transport Phenomena by Deen (ISBN 9780195084948) which uses Finite Fourier Transforms to solve PDEs by separation of variables. I'm looking at example 4.5-2, with the entire text for the example shown here:
By the way I think that's a typo on Figure 4-7, the boundary condition at $x=0$ should be $\Theta=0$.
I can work through the first part of the derivation. I start with the separation of variables assumption:
$$\Theta (x,y) = \Phi_n(x) \Theta_n(y)$$
Substitute this into the differential equation:
$$\nabla^2 \Theta = 0$$
$$\frac{\partial^2 \Theta}{\partial x^2}+\frac{\partial^2 \Theta}{\partial y^2}=0$$
$$\frac{d^2 \Phi_n}{dx^2}\Theta_n + \Phi_n \frac{d^2 \Theta_n}{dy^2}=0$$
Rearrange and set equal to a constant:
$$\frac{1}{\Phi_n}\frac{d^2 \Phi_n}{dx^2}=-\frac{1}{\Theta}\frac{d^2 \Theta_n}{dy^2}=-\lambda_n^2$$
First looking at the $\Phi_n(x)$ and applying the boundary conditions at $x=0$ and $x=1$, I get Eq. 4.5-17. Taking the solution to be of the form:
$$\Phi_n(x) = a_n \sin \lambda_n x + b_n \cos \lambda_n x$$
Applying the B.C. at $x=0$ gives $b_n=0$, so that leaves
$$\Phi_n = a_n \sin \lambda_n x,\;\;\;\;n=1,2,…,$$
as the Eigenfunction (4.5-18 in the text), and applying the B.C. at $x=1$ gives
$$\lambda_n=-\text{Bi} \tan \lambda_n$$
as the Eigenvalues (4.5-19 in the text).
My problem is with 4.5-20, where it gives the value of $a_n$ as
$$a_n=\left( \frac{2 \lambda_n}{\lambda_n – \cos \lambda_n \sin \lambda_n} \right) ^{1/2}$$
All it says is that this is the 'normalization condition', but gives no further elaboration here or elsewhere in the text. The earlier example 4.3-3 that the example references doesn't help either, as there they just left $a_n$ as an undetermined constant.
So how is $a_n$ being determined here? I know that in general the normalization condition is where a constant is chosen just to make the math easier: for example if we have two Diriclet boundary conditions in the domain $[0,\ell]$ then we choose $a_n=\sqrt{2/\ell}$ for $\Phi_n(x) = a_n \sin n \pi x / \ell$ so that when we use the orthogonality of the basis functions we get
$$\left< \Phi_n,\Phi_m \right>=\int_0^\ell \Phi_n (x) \Phi_m (x) dx = \begin{cases} 1,\;\;\;\; n=m\\ 0,\;\;\;\; n \neq m \end{cases}$$
Instead of some other constant we have to carry through. Is this the same here, just that $a_n$ is a lot more complex because of the more complicated Robin boundary condition at $x=1$?
Best Answer
Having the eigenfunctions be orthogonal for this problem requires that the following be true:
$$\left< \Phi_n , \Phi_m \right> = \int_0^1 \Phi_n(x) \Phi_m(x) = C_{nm} \delta_{nm} = \begin{cases} C_{nn} & \text{if}\;n=m \\ 0 & \text{if}\; n \neq m \end{cases}$$
However for convenience we want them to be orthonormal, which means that
$$\left< \Phi_n , \Phi_m \right> = \int_0^1 \Phi_n(x) \Phi_m(x) = \delta_{nm} = \begin{cases} 1 & \text{if}\;n=m \\ 0 & \text{if}\; n \neq m \end{cases}$$
Or in other words $C_{nm}=1$.
For this problem we then solve the following:
$$\left< \Phi_n , \Phi_m \right> = \int_0^1 \Phi_n(x) \Phi_m(x) = \int_0^1 a_n \sin (\lambda_n x) a_m \sin (\lambda_m x) dx$$
This is also orthogonal, so we have
$$\int_0^1 a_n \sin (\lambda_n x) a_m \sin (\lambda_m x) dx = \begin{cases} \frac{{a_n}^2}{2} \left( \sin^2 \lambda_n + \cos^2 \lambda_n - \frac{1}{\lambda_n} \sin \lambda_n \cos \lambda_n \right) & \text{if $\,\,n = m$} \\ 0 & \text{if $\,\,n \neq m$} \end{cases} $$
We then set the result for the $n=m$ case equal to 1 and then solve for the only remaining unknown $a_n$:
$$\frac{{a_n}^2}{2} \left( \sin^2 \lambda_n + \cos^2 \lambda_n - \frac{1}{\lambda_n} \sin \lambda_n \cos \lambda_n \right)=1$$
$$a_n = \left( \frac{2 \lambda_n}{\lambda_n - \sin \lambda_n \cos \lambda_n} \right)^{1/2}$$