TITLE: Solution of an equation with Jacobi theta function
QUESTION [6 upvotes]: 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 $x=1$).

REPLY [6 votes]: 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.