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I would like to know if we can find a real function $v(x)$ and a complex function $f(x)$, such that they solve the following differential equation (with $\alpha$ a complex, $0<Re(\alpha)<1$):

$$(x^2 f(x)')'+v(x)f(x) = \alpha f(x)$$

with the constraint that the imaginary part of the following integral is zero, i.e.,

$$Im (\int_0^{\infty} \ln(x) \Big(\mathcal{F}_c \circ\phi (v(x) f(x))\Big) \Big(\mathcal{F}_c \circ\phi (\overline{f(x)})\Big) dx)=0 $$

denoting $\phi(g(x))=\frac{1}{x} g(\frac{1}{x})$ and $\mathcal{F}_c$ to be the cosine transform $\mathcal{F}_c (f(x))=\int_0^{\infty} f(t) \cos(xt) dt $.

If there was no $\ln(x)$ in the integral above, the result would be direct as all couples $(v(x),f(x))$ that are solutions of the differential equation would verify the second condition (using a property of cosine transform, providing the terms are well defined). However, due to the term $\ln(x)$ in the integral I did not manage to find the couples $(v(x),f(x))$ which fulfill the conditions above.

Any idea on a way to treat problems of this type?

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  • $\begingroup$ --I would like to know if we can find $v(x)$ a real function and $f(x)$ a complex function-- ### --I did not managed to characterise the couples (v(x),f(x)) fullfilling the conditions above--. So, which verb is the correct one? $\endgroup$
    – fedja
    Commented Dec 18, 2017 at 5:30
  • $\begingroup$ Correct verb is "find" ! I corrected to be clear. $\endgroup$
    – Bertrand
    Commented Dec 20, 2017 at 18:00
  • $\begingroup$ Some parentheses are lacking in the integrand, which makes unclear the argument of $\phi$. Also, the use of $\circ$ is ambiguous or wrong. What's being transformed by $\mathcal{F}_c$? Why not just writing a double integral? $\endgroup$
    – Pietro Majer
    Commented Dec 20, 2017 at 18:58
  • $\begingroup$ I think we could take $f(x)=0$ and $\nu(x)=0$ for all values of $x$, to give one solution. So the answer is "yes". $\endgroup$
    – Ben McKay
    Commented Dec 20, 2017 at 20:33

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