I am stuck in a part of a proof perhaps somebody can help me.
I've got the following eigenvalue problem:
$$u'' + \lambda u = 0 \quad 0 <x < L$$ With boundary conditions: $$ -u'(0) + \sigma_1 u(0) = 0, \quad u'(L) + \sigma_2 u(L) = 0$$ Where both $ \sigma_1$ and $ \sigma_2$ are constants.
By writing $\lambda = k^2$, for real $k$, the general solution is : $$u(x) = A \cos kx + B \sin kx$$ and imposing the boundary conditions: $u(0) = A$, $u(L) =A \cos kL + B \sin kL$, $u'(x)= -kA \sin kx +kB \cos kx$
Now substituting the given boundary values: $u'(0) = kB$, and $u'(L) = -kA \sin kL +kB \cos kL$
Which gives:
$$-kB + \sigma_1A = 0 $$ $$-kA \sin kL +kB \cos kL + \sigma_2(A \cos kL + B \sin kL)=0$$
The last equation can be re-arranged to: $$(-kA+\sigma_2B) \sin KL + (kB + \sigma_2A)\cos kL = 0 $$ Giving: $$\frac{(-kA+\sigma_2B)}{(kB + \sigma_2A)}\tan kL = -1$$ Now solving for $\tan kL$ gives:
$$\tan kL = -\frac {(kB + \sigma_2A)}{(-kA+\sigma_2B)}$$
By using the first BC we obtain:
$$\tan kL =\frac {(\sigma_1 + \sigma_2)k}{(k^2-\sigma_1 \sigma_2)}$$
But how do I get to the corresponding eigen function:
$$u_n(x) = u(x;k_n) \quad u(x;k) = K(k) \sin(kx + θ(k))$$ where:
$$\tan(θ(k))= \frac {k}{\sigma_1}$$
For some undetermined normalisation constant $K(k)$
Hint:
Recall that: $$ A\cos x + B\sin x= K\sin(x+\phi) $$ with $$ K=\sqrt{A^2+B^2},\quad \phi=\arctan\frac AB. $$