Finding the eigenvalues and eigenfunctions to this sturm-liouville operator

Question

I am asked to find the eigenvalues and eigenfunctions to the operator $A$, defined by $$Au = -\frac{d^2u}{dx^2}-6\frac{du}{dx}$$ and also $0<x<L, \quad u(0) = u(L) = 0$.

Rewriting the operator to sturm-liouville form gives $$Au = -\frac{1}{e^{6x}}(e^{6x}u')'.$$

To find the eigenvalues, eigenvectors I am solving $$Au = \lambda u \iff -u''-6u'-\lambda u = 0$$ where the characteristic equation $x^2 + 6x + \lambda = 0$ has solution $x=-3\pm \sqrt{9-\lambda}.$

What can I say about $\lambda$ to further solve this? How do I proceed? And how do I incorporate the restrictions on $u$?

The solution is supposed to be $$\lambda_k = 9 + (\frac{k\pi}{L})^2$$ $$u_k(x) = e^{-3x}\sin{\frac{k\pi}{L}x}, \ k=1,2,.....$$

Answer 1

The key is to notice that $\lambda>9$ and the characteristic equation has two complex roots that are conjugates of one another.

The general solution is $u(x)=Ce^{q_1 x} + De^{q_2 x}$ where

$q_1=-3+i\sqrt{\lambda-9}$ and $q_2=-3-i\sqrt{\lambda-9}$ are the complex roots of the characteristic equation

This is equivalent to $u=e^{-3x}(Acos(\sqrt{\lambda-9}x)+Bsin(\sqrt{\lambda-9} x))$

The boundary conditions require $B=0$ and $\sqrt{\lambda-9}L=k\pi$, which gives the desired form for u. Substituting this in the original differential equation gives

$\lambda=9+(\frac{k\pi}{L})^2$

Answer 2

$$ Af=\lambda f \iff -f''-6f'=\lambda f $$ To normalize the solutions, set $f(0)=0$ and $f'(0)=1$. Assume that $f=e^{\alpha x}$. Then $$ -\alpha^2-6\alpha=\lambda \\ \alpha^2+6\alpha=-\lambda \\ (\alpha+3)^2=9-\lambda\\ \alpha = -3 \pm \sqrt{9-\lambda} $$ The solutions so far are $$ f(x)=e^{-3x}\frac{\sin(\sqrt{\lambda-9}x)}{\sqrt{\lambda-9}} $$ Now the condition $f(L)=0$ determines the eigenvalues $\lambda$: $$ \frac{ \sin(\sqrt{\lambda-9}L)}{\sqrt{\lambda-9}}=0. $$ $\lambda=9$ is not a solution because the limit of the above as $\lambda\rightarrow 9$ is non-zero. The solutions are $$ \lambda = 9+n^2\pi^2/L^2,\;\;\; n=1,2,3,\cdots. $$ The resulting eigenfunctions are $$ f_n = e^{-3x}\frac{\sin(n\pi x/L)}{n\pi/L},\;\;\; n=1,2,3,\cdots. $$