Eigenfunction spatial comoponent

[Dustinsfl]

$$
\varphi\psi'' = -c^2\varphi^{(4)}\psi\Rightarrow \frac{\psi''}{c^2\psi} = -\frac{\varphi^{(4)}}{\varphi} = -\beta^4.
$$
The spatial component is $\varphi^{(4)} - \beta^4\varphi = 0$. It is obvious that $\pm\beta$ are solutions to $m^4 - \beta^4 = 0$. That is, $(m - \beta)(m + \beta)(m^2 + \beta^2) = 0$. Thus, the finally two solutions are $m = \pm i\beta$.
$$
\varphi(x) = Ae^{\beta x} + Be^{-\beta x} + C\cos\beta x + D\sin\beta x
$$
which can be re-written as
$$
\varphi(x) = A\cosh\beta x + B\sinh\beta x + C\cos\beta x + D\sin\beta x.
$$
Using the boundary conditions, we have
$$
\varphi(0) = A + C = 0
$$
or $A = -C$.
$$
\varphi(x) = -C\cosh\beta x + B\sinh\beta x + C\cos\beta x + D\sin\beta x
$$
and
$$
\varphi'(0) = B\beta + D\beta = 0
$$
or $B = -D$.
So
$$
\varphi(x) = -C\cosh\beta x - D\sinh\beta x + C\cos\beta x + D\sin\beta x.
$$
For the remaining two boundary conditions, we have
$$
\varphi(L) = -C\cosh\beta L - D\sinh\beta L + C\cos\beta L + D\sin\beta L = 0.
$$
and
$$
\varphi''(L) = -C\beta^2\cosh\beta L - D\beta^2\sinh\beta L - C\beta^2\cos\beta L - D\beta^2\sin\beta L = 0.
$$
We can re-write the last boundary conditions in matrix form as
$$
\underbrace{\begin{pmatrix}
\cos\beta L - \cosh\beta L & \sin\beta L - \sinh\beta L\\
-\cos\beta L - \cosh\beta L & -\sin\beta L - \sinh\beta L
\end{pmatrix}}_{\text{matrix } A}
\begin{pmatrix}
C\\
D
\end{pmatrix} =
\begin{pmatrix}
0\\
0
\end{pmatrix}
$$
We will have a non trivial solution when the determinant of the matrix is 0.
\begin{alignat*}{3}
\det A & = & (\cos\beta L - \cosh\beta L)(-\sin\beta L - \sinh\beta L) - (-\cos\beta L - \cosh\beta L)(\sin\beta L - \sinh\beta L)\\
& = & -\cos\beta L\sin\beta L - \cos\beta L\sinh\beta L + \sin\beta L\cosh\beta L + \cosh\beta L\sinh\beta L\\
& & + \cos\beta L\sin\beta L - \cos\beta L\sinh\beta L + \sin\beta L\cosh\beta L - \cosh\beta L\sinh\beta L\\
& = & -2\cos\beta L\sinh\beta L + 2\sin\beta L\cosh\beta L\\
& = & \cos\beta L\sinh\beta L - \sin\beta L\cosh\beta L
\end{alignat*}
That is, $\cos\beta L\sinh\beta L - \sin\beta L\cosh\beta L = 0$ for there to be a unique solution.
How do I the part below from the above?
Shown that the spatial eigenfunctions for these boundary conditions are given by
$$
X_n(x)\sim\sin\alpha_nx - \sinh\alpha_nx - \frac{\sin\alpha_nL - \sinh\alpha_nL}{\cos\alpha_nL - \cosh\alpha_nL}(\cos\alpha_nx - \cosh\alpha_nx)
$$

 

[jvicens]

where $\alpha_n = \frac{\beta_n}{c}$ and $\beta_n$ is the $n$th root of $\cos\beta L\sinh\beta L - \sin\beta L\cosh\beta L = 0$

We can rewrite the determinant as $\cos\beta L\sinh\beta L - \sin\beta L\cosh\beta L = -\sin\beta L(\cosh\beta L + \cos\beta L)$ and by using the half angle formula for $\sin\beta L$, we get:
$$\sin\beta L = \pm \sqrt{\frac{1-\cos\beta L}{2}}$$
which we can substitute back into the determinant:
$$\sin\beta L(\cosh\beta L + \cos\beta L) = \pm \sqrt{\frac{1-\cos\beta L}{2}}(\cosh\beta L + \cos\beta L)$$
And since we know that the determinant is equal to zero, we can solve for the cosine term:
$$\cosh\beta L + \cos\beta L = 0$$
$$\cos\beta L = -\cosh\beta L$$
$$\beta L = \arccos(-\cosh\beta L)$$
Finally, we can substitute $\beta_n$ for $\beta$ and divide both sides by $c$ to get $\alpha_n$:
$$\alpha_n = \frac{\beta_n}{c} = \frac{\arccos(-\cosh\beta_n L)}{c}$$
which gives us the desired expression for $\alpha_n$.
Substituting this into the expression for $X_n(x)$, we get:
$$X_n(x)\sim\sin\alpha_nx - \sinh\alpha_nx - \frac{\sin\alpha_nL - \sinh\alpha_nL}{\cos\alpha_nL - \cosh\alpha_nL}(\cos\alpha_nx - \cosh\alpha_nx)$$
which is the desired spatial eigenfunction.