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I would like to find solutions of the following differential equation:

$ \sum_{1}^{\infty} a_n f(nx) + f''(x)+ x^2 f(x)=\lambda f(x)$

For example in space of function from $\mathbb R^*$ to $\mathbb C$

If we modify the sum in the differential equation by posing $g(t)=f(e^{t})$, and make a Fourier transform like advise here (for an equation with infinite sum but without derivation) it seems it does not work.

Which method would you suggest ?

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If the numbers $b_k=\sum_{n=1}^\infty a_nn^k$ are all fnite, you can find a power series solution: plug $$f(x)=\sum_{k=0}^\infty c_k x^k,$$ and you obtain a recurrency which determines $c_k$: $$c_{k+2}=\frac{1}{(k+1)(k+2)}\left((\lambda-b_k)c_k-c_{k-1}\right).$$ So you can set $c_0$ and $c_1$ arbitrarily, and then determine all $c_k$. To obtain a convergent series you need further assumptions on $a_n$ which will imply a growth estimate of $b_k$.

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  • $\begingroup$ Yes, if the $a_n$ decrease extremely rapidely it can provide a solution, but in the case the $a_n$ do not decrease faster than $n^{-k}$ ? for example if for all n we have $a_n=1$ ? Do you think there is a way to obtain a solution using convolution or Fourier transform? $\endgroup$ – Bertrand Dec 20 '14 at 17:56
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    $\begingroup$ You need some assumption on $a_n$, other wise you have to explain what the first term of your equation means. $\endgroup$ – Alexandre Eremenko Dec 20 '14 at 19:20
  • $\begingroup$ Alexander, first I would like to thank you for your help! Condition on $a_n$ I am interested in can be all $a_n$ are lower than a constant. In this case if there is a solution f then asymptotic of f at infinity is at least in $x^a$ with $a>1$. in fact if it is not possible to find an explicit solution showing there is a solution and finding an asymptotic a infinity will be already a good result for me. Do you know a method to find an asymptotic ? $\endgroup$ – Bertrand Dec 21 '14 at 8:54
  • $\begingroup$ @Bertrand If $a_n$ are just bounded and $f$ grows, how do you understand the sum? (formally the series diverges in this case). $\endgroup$ – fedja Dec 25 '14 at 11:58
  • $\begingroup$ Fedja, Mistake from my side, of course I was meaning f is $x^{-a}$ with $a>1$. $\endgroup$ – Bertrand Dec 28 '14 at 9:28

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