Robin boundary conditions


### Eigenvalues for Robin boundary conditions

#### Case of Robin-Robin boundary condition

For boundary value problem \begin{align} &X''+\lambda X =0\\[3pt] &(X'-\alpha X)|_{x=0}=0,\qquad (X'+\beta X)(l)=0 \end{align} let us find positive eigenvalues $\lambda=k^2$ ($k>0$) first. We have $$X= A\cos(k x) +B\sin (k x).$$ Then \begin{array}{rcr} -\alpha A& +& k B=0,\\ (\beta-k\tan (k l)) A& +& (k+\beta \tan(kl))B=0 \end{array} and calculating determinant and taking it equal to $0$ we get $$\tan (kl)=\frac{(\alpha+\beta)k}{k^2-\alpha\beta}.$$

Solve it graphically:

$\alpha=\beta$, varies

$\alpha=-.5$, $\beta$ varies

We see that there is an infinite number of roots, tending to $+\infty$.

Let us find negative eigenvalues $\lambda=-k^2$ ($k>0$). We have \begin{array}{rcr} -\alpha A& +& k B=0,\\ (\beta+k\tanh (k l)) A& +& (k+\beta \tanh(kl))B=0 \end{array} and calculating determinant and taking it equal to $0$ we get $$\tanh (kl)=-\frac{(\alpha+\beta)k}{k^2+\alpha\beta}.$$

Solve it graphically:

$\alpha=\beta$, varies

$\alpha=-1$, $\beta$ varies

$\alpha=-2$, $\beta$ varies

$\alpha=-3$, $\beta$ varies

We see that there is just none, or one, or two positive roots.