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 $x=1$).



Solution of an equation with Jacobi theta function
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.


