Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the second-order ODE

$-(py')'+(q-\lambda$$w)y=wf$ $(1)$

with $y$ an $L^2$ complex-valued function on $[a,b]$ subject to the boundary conditions:

$\alpha_1y(a)-\alpha_2y'(a)=0$ and $\beta_1y(b)-\beta_2y'(b)=0$

where $f$ is also an arbitrary $L^2$ complex-valued function on $[a,b]$, $p$,$q$ and $w$ are real-valued $C^1$ functions on $[a,b]$ and $\lambda$ is a complex constant.

Suppose that we are able to construct a function which, when evaluated for a given $f$, returns a solution $y$ that meets the above constraints. Is this $y$ then a unique solution to $(1)$?

[Alternative characterisation: I have avoided functional analysis terminology above so as not to make the problem too generic and thus eliminate important information, but if we take $(1)$ to define an operator $D$ from a subset of $L^2$ onto $L^2$, I am wondering whether a second operator $E$ which determines a $y$ for any given $f$ is in fact the inverse. P.S. The methodology for constructing $E$ given $D$ and the circumstances under which this is possible are already clear to me.]

share|improve this question
Does your solution $y$ turns out to be $C^2$? If $y$ is merely in $L^2$ (only defined almost everywhere), what do you mean by the boundary conditions? –  GWu Apr 21 '11 at 19:21
I assume you are referring to the fact that {a} and {b} are both of Lebesgue measure zero so all $f$ in $L^2$ would satisfy the boundary conditions (in which case is $C^1$ sufficient to make the boundary conditions meaningful)? Anyway, yes, the strong version is that $y$ and $y'$ are continuously differentiable. A weaker requirement - which I am not properly studying but am thinking about all the same - would be that $y$ and $py'$ are absolutely continuous, ie differentiable in a weak sense. –  Josef K. Apr 21 '11 at 23:07
Under the following circumstances I think you can prove uniqueness: introduce a new variable x=y' and examine the homogeneous version of your equation, then you have a first order system, and the result should be unique if for the resulting system (x',y')=F(x,y) the function F is locally Lipschitz continuous. If you can prove bounds for the norms on x and y, I think then that continuity in F suffices. I could be wrong about all of this though. –  Tim Seguine Nov 13 '11 at 15:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.