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In the proof of Theorem 2 in this paper here on arxiv on page 10 for $k=2$ it is claimed that if the Wronskian of two solutions $y_1,y_2$ to the differential equation

$$-y''_i(x) + q(x) y_i(x) = \lambda_i y_i(x)$$ is zero at some position $x_0$ (so $W(y_1,y_2)(x_0)=0$) then we also have that $W'(y_1,y_2)(x_0)=0$. I first thought that this is a trivial consequence of Abel's identity, but then I noticed that the two $y_i$'s satisfy different ODE's, in the sense that $\lambda_1 \neq \lambda_2$. Thus, Abel's identity no longer holds and I could not see why this is then the case.

Does anybody here have an idea, where this comes from?

Please write a comment, if anything is unclear.

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  • $\begingroup$ Solutions to your equations are exponential functions, so, assuming that the claim holds, you can easily compute everything explicitly! $\endgroup$
    – Alex Degtyarev
    Commented Dec 29, 2014 at 13:18
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    $\begingroup$ @AlexDegtyarev you are wrong $-y''(x) +q(x) y(x) = \lambda y(x)$ does not have exponential solutions in general. I added the variable $x$ in order to make this more clear $\endgroup$
    – Withoutsugar
    Commented Dec 29, 2014 at 13:21
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    $\begingroup$ Of course, this won't hold for general functions. In the paper, the claim is being made for specific (trigonometric) functions. $\endgroup$
    – Christian Remling
    Commented Dec 29, 2014 at 18:25
  • $\begingroup$ If $q(x) \equiv 0$, then for $\lambda_{1} = \frac{\pi^{2}}{4}$ and $\lambda_{2} = \frac{\pi^{2}}{9}$ we have that $W\left(\sin(\sqrt{a}x),\sin(\sqrt{b}x)\right) = 0$ for $x=0$, but $W\left(\sin(\sqrt{a}x),\sin(\sqrt{b}x)\right) = \frac{\pi}{6}$ for $x=1$. $\endgroup$
    – Tadashi
    Commented Dec 29, 2014 at 19:14

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