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I have been struggling with this equation for some time and I do not seem to find any conclusive answer (it's from my research, not a homework).

It has to do with the real solutions $x$ to the following equation

$$ x + x f(x) = 1 + f(1),$$ where $$ f(x) = 2\sum_{n=1}^\infty \mathrm{e}^{(-ax^2-b) n^2} $$ with $a$ and $b$ strictly positive.

I know that $x=1$ solves the equation trivially; from simulations, I cannot find a contradiction to the fact that it should be the only solution. However, I cannot prove nor disprove that $x=1$ is the only solution.

I have tried using the upper bound $$ f(x) \leq \frac{ax^2+b+1}{ax^2+b}\mathrm{e}^{-ax^2 -b},$$ and i have tried relating $f(x)$ with the Elliptic theta function $$ f(x) = -1 + \theta_3(0,\mathrm{e}^{-ax^2 -b});$$ I have also tried to prove that $x = -x f(x) + 1 +f(1)$ is a contractive mapping; however, I have only found (quite restrictive) sufficient conditions on $a$ and $b$ for it to be true.

If someone manages to solve it or help me find a counterexample, I will gladly acknowledge their contribution in the paper i am writing.

EDIT: $x = -x f(x) +1 +f(1)$ is not a contractive mapping in general. I have found counterexamples where it is not (but the equation still only has one solution $x=1$).

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  • $\begingroup$ You just need consider the monotonicity of the function $g(x)=x+f(x)$ with $x>-b/a$. It is seems that $g^{''}(x)<0$ for all $x>-b/a$ and thus $g(x)-g(1)=0$ has at most two real solution. $\endgroup$
    – Zhou
    May 22, 2018 at 7:34
  • $\begingroup$ Hi @Zhou, thank you for your input! While it makes a lot of sense, I had a typo in the problem: the function I am looking at is $g(x) = x + x f(x)$. $\endgroup$
    – Riccardo Sven Risuleo
    May 22, 2018 at 7:51

2 Answers 2

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Let $\, g(x) := \theta_3(0,\mathrm{e}^{-ax^2 -b}).\,$ Your question about solutions to $\, x + x f(x) = 1 + f(1) \,$ is now about $\, x g(x) = g(1).\,$ Now $\,g(x)\,$ is a bell shaped curve with $\, g(x) > 0 \,$ and $\, g(-x) = g(x).\,$ If we can prove that $\,xg(x)\,$ is monotone increasing we are done. If it holds for $\,a=1, b=0\,$ then it holds in general.

In that special case, using Jacobi theta identities (mentioned by Mikhail Skopenkov in another answer), $\,xg(x) = g(\pi/x)\sqrt{\pi}$ and since $\,g(x)\,$ is monotone decreasing for $\,x>0,\,$ then $\,g(\pi/x)\,$ must be monotone increasing and we are done.

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  • $\begingroup$ Hi! Thank you for your input; could you give me a pointer as to how I can prove that $xg(x)$ is monotone increasing? I get that $$\frac{\mathrm d}{\mathrm d x} \{x g(x)\} = g(x) - 4 a x^2 \sum_{n=1}^\infty \mathrm{e}^{-a x^2 -b}n^2$$ which is not obviously positive. Or is there some clever trick? $\endgroup$
    – Riccardo Sven Risuleo
    May 23, 2018 at 7:03
  • $\begingroup$ I cannot thank you enough. Please, send me an email with your full name and affiliation if you want an acknowledgment in my paper. Otherwise, thank you very much! $\endgroup$
    – Riccardo Sven Risuleo
    May 25, 2018 at 9:37
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Have you tried Jacobi's identities for theta-functions? At least for $a=1, b=0$ the identity $$xg(x)=\sqrt{\pi}\,g(\pi/x)$$ implies that the function $xg(x):=xf(x)+x$ is monotone increasing.

Edit: the identity corrected, thanks to Somos.

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  • $\begingroup$ @Somos: Surely, many thanx $\endgroup$
    – Mikhail Skopenkov
    May 24, 2018 at 17:55
  • $\begingroup$ I cannot credit you both for the answer; if you would like and acknowledgment in my paper, please send me an email with your full name and affiliation. Otherwise, thank you very much! $\endgroup$
    – Riccardo Sven Risuleo
    May 25, 2018 at 9:36
  • $\begingroup$ That was just a standard trick with teta-functions estimates $\endgroup$
    – Mikhail Skopenkov
    May 25, 2018 at 18:58

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