Robin boundary conditions


### Eigenvalues for Robin boundary condition

#### Case of Dirichlet-Robin boundary condition

For boundary value problem \begin{align} &X''+\lambda X =0\\[3pt] &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 $A=0$ and $$\tan (k l)=-\frac{k}{\beta}.$$

Solve it graphically:

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 $$X= A\cosh(k x) +B\sinh (k x).$$ Then $A=0$ and $$\tan (k l)=-\frac{k}{\beta}.$$ $$\tanh (kl)=-\frac{k}{\beta}.$$

Solve it graphically:

We see that there is just one positive root as $\beta <-1/l$, and no positive roots at all as $\beta\ge -1/l$.

#### Case of Neumann-Robin boundary condition

For boundary value problem \begin{align} &X''+\lambda X =0\\[3pt] &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 $B=0$ and $$\tan (k l)=\frac{\beta}{k}.$$

Solve it graphically:

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 $$X= A\cosh(k x) +B\sinh (k x).$$ Then $B=0$ and $$\tanh (kl)=-\frac{\beta}{k}.$$

Solve it graphically:

We see that there is just one positive root as $\beta <0$, and no positive roots at all as $\beta\ge 0$.