Green functions and Sturm-liouville eigenfunctions

Question

I have the differential operator, $L$ that is defined by $$L=-y''.$$ on the interval $(0,1)$ along with the boundary conditions $$y'(0)=y(0) \qquad\text{and}\qquad y'(1)+y(1)=0.$$ The Sturm-Liouville problem is defined by $$L[y]=\lambda \,y''.$$ I can not find the eigenvalues ​or the eigenfunctions. How do I find the Green's function?

Answer 1

The differential equation you seek is

$$y''+\lambda y = 0$$

which had the general solution

$$y(x) = A \cos{\sqrt{\lambda} x} + B \sin{\sqrt{\lambda} x}$$

The boundary value given by $y'(0)=y(0)$ implies that $A = B \sqrt{\lambda}$. The boundary value given by $y'(1) + y(1) = 0$ implies that

$$-A \sqrt{\lambda} \sin{\sqrt{\lambda}} + B \sqrt{\lambda} \cos{\sqrt{\lambda}} + A \cos{\sqrt{\lambda}} + B \sin{\sqrt{\lambda}}=0$$

Using the relation derived from the first boundary condition, we get

$$-\lambda \sin{\sqrt{\lambda}} + \sqrt{\lambda} \cos{\sqrt{\lambda}} + \sqrt{\lambda} \cos{\sqrt{\lambda}} + \sin{\sqrt{\lambda}}=0$$

or, after a little algebra,

$$\tan{\sqrt{\lambda}} = \frac{2 \sqrt{\lambda}}{\lambda-1}$$

Here is a plot of the LHS and RHS so you can see the intersections and hence, the square roots of the eigenvalues. Clearly, for the first few, they must be computed numerically. For the $k$th eigenvalue, where $k$ is large,

$$\lambda_k \sim k^2 \pi^2$$

The eigenfunction (unnormalized) corresponding to that eigenvalue is

$$y_k(x) = \sqrt{\lambda_k} \cos{\sqrt{\lambda_k} x} + \sin{\sqrt{\lambda_k} x}$$

The Green's function may then be found from the eigenvalues/eigenfunctions as follows:

$$G(x,x') = \sum_{k=1}^{\infty} \frac{y_k(x) y_k(x')}{\lambda-\lambda_k}$$

where $G$ satisfies

$$\frac{d^2}{dx^2} G(x,x') + \lambda G(x,x') = \delta(x-x')$$