2
$\begingroup$

I have a problem with this first order DE: let $-\infty and

$u'(x)+(\lambda+q(x))u(x)=0,\tag{1}$ where $u$ is a continuous and real valued, while $\lambda$ is a parameter not depending on $x$.

A strange non trivial boundary condition is given, namely $\alpha u(a)+\alpha'u'(a)+\beta u(b)+\beta'u'(b)=0.$

Then I have to show that this problem admits at most three eigenvalues.

What I have tried: basically to convert this problem into a Sturm Liouville problem, however I couldn't conclude anything.

Can anybody help me?

How to go through this kind of problems? thanks in advance.

-Guido-

  • 0
    Ah, it's an eigenvalue of an *operator*, not an eigenvalue of a problem. Got it. Thanks.2012-10-29

1 Answers 1

1

The equation is a linear first order differential equation, whose solution is $ u(x)=C\,e^{-\lambda x-\int_a^xq(t)dt}. $ We may take $C=1$. Then $\begin{align*} u(a)&=e^{-\lambda a},\\ u'(a)&=-(\lambda+q(a))e^{-\lambda a},\\ u(b)&=e^{-\lambda b-\int_a^bq(t)dt},\\ u'(b)&=-(\lambda+q(b))e^{-\lambda b-\int_a^bq(t)dt}. \end{align*}$ Let $k=e^{-\int_a^bq(t)dt}$. The boundary condition is then $ \alpha\,e^{-\lambda a}-\alpha'(\lambda+q(a))\,e^{-\lambda a}+\beta\,k\,^{-\lambda b}-\beta'\,k\,(\lambda+q(b))e^{-\lambda b}=0, $ which can be written as the following equation in the unknown $\lambda$: $ -\alpha'\lambda+k(\beta-\beta'q(b))e^{-\lambda(b-a)}-k\,\beta'\lambda\,e^{-\lambda(b-a)}=\alpha'q(a)-\alpha. $ You have to study the number of solutions of this equation according to the possible values of the parameters.