# [SOLVED]eigenfunction spatial comoponent

#### dwsmith

##### Well-known member
$$\varphi\psi'' = -c^2\varphi^{(4)}\psi\Rightarrow \frac{\psi''}{c^2\psi} = -\frac{\varphi^{(4)}}{\varphi} = -\beta^4.$$
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
$$X_n(x)\sim\sin\alpha_nx - \sinh\alpha_nx - \frac{\sin\alpha_nL - \sinh\alpha_nL}{\cos\alpha_nL - \cosh\alpha_nL}(\cos\alpha_nx - \cosh\alpha_nx)$$