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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-

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    I don't know what is meant by an eigenvalue of a problem.2012-10-29
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    Is it $u'$ or $u''$ in the equation?2012-10-29
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    @JuliánAguirre it is $u'$, it is a first order DE2012-10-29
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    @GerryMyerson I'm following the language introduced in the Birkhoff Rota Ordinary differential equation book and you can define the operator $L[u]=u'+q(x)u$ so for $\lambda$ to be an eigenvalue it means that $L[u]=\lambda u$ has a non trivial solution $u$.2012-10-29
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    Ah, it's an eigenvalue of an *operator*, not an eigenvalue of a problem. Got it. Thanks.2012-10-29

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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.