Eigenvalues Sturm-Liouville system

[jegues]

Homework Statement

See figure attached

Homework Equations

The Attempt at a Solution

$$\lambda > 1,$$

$$y^{''} + 2y^{'} + \alpha^{2}y = 0, \quad \alpha > 0$$

Into auxillary equation,

$$m^{2} + 2m + \alpha^{2} = 0$$

I'm stuck as to how to solve this auxillary equation.

Any ideas?

Thanks again!

 

[LCKurtz]

Either complete the square on m or use the quadratic formula.

 

[jegues]

Either complete the square on m or use the quadratic formula.

Okay so completing the square,

$$m^{2} + 2m + 1 = - \alpha^{2} +1$$

$$(m+1)^{2} = - \alpha^{2} + 1$$

$$m= \sqrt{- \alpha^{2} + 1} - 1$$

$$\forall \quad \alpha > 0, \quad \sqrt{- \alpha^{2} + 1} \quad \stackrel{\leftrightarrow}{ iff } \quad i \sqrt{\alpha^{2} - 1}$$

So,

$$m = i \sqrt{\alpha^{2} - 1} -1$$

Then,

$$y = e^{-x}\left( C_{1}cos(x\sqrt{\alpha^{2} - 1}) + C_{2}sin(x\sqrt{\alpha^{2} - 1}) \right)$$

Using the boundary conditions,

$$C_{1} = 0,$$

$$C_{2}e^{-L}sin(L\sqrt{\alpha^{2} - 1}) = 0, \quad C_{2} \neq 0$$

$$\alpha^{2} = \frac{n^{2} \pi^{2}}{L^{2}} + 1 = \lambda_{n}, \quad n \in Z, n \geq 1$$

So,

$$y_{n}(x) = e^{-x}sin(\frac{n\pi x}{L})$$