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This is a MathOverflow cross-post (currently no answer there) and a generalization of a previous MathOverflow question, "Reducing system of equations involving Erf, Error Function".

Consider the system of equations: $$1/2 + {\rm Erf}(x) - \alpha {\rm Erf}(\frac{x+y}{2})=0$$ $$-1/2 + {\rm Erf}(y) - \alpha {\rm Erf}(\frac{x+y}{2})=0,$$ where $x \le y$ and ${\rm Erf}$ is the Error Function. By ansatz it is clear that one solution is $(x,y)=(-{\rm Erf}^{-1}(1/2),{\rm Erf}^{-1}(1/2))$, independent of $\alpha$. The question is to prove the uniqueness of this solution, for any $\alpha$.

(Update: Robert's visual below demonstrates how there are clearly multiple solutions for a robust interval of $\alpha \in (0.5, \approx 0.8)$, and so the general $\alpha$ formulation was not helpful. I am revising my question to specifically focus on uniqueness in the case of $\alpha =1/2$.)

I am particularly interested in a proof for $\alpha = 1/2$, but I hope that by phrasing the problem in terms of a general $\alpha$, someone may see an elegant elementary proof.

Observe that proving that $x+y=0$ is sufficient. My previous MO question includes a proof of uniqueness in this way for the case where $\alpha=1$, with two proofs (one by Noam Elkies, one by myself). Unfortunately these proofs do not appear to generalize to other values of $\alpha$ (Noam's proposition is false for $\alpha = 1/2$, my approach does not appear to generalize), even though I strongly suspect the general statement.

Numerics: I have embarked on some numerical exploration. If one considers the equation $${\rm Erf}(x) + {\rm Erf}(y) = 2\alpha {\rm Erf}(\frac{x+y}{2}),$$ For $\alpha=1$ we know that $y=-x$ and $y=x$ are solutions. Numerically, it appears as through $y=-x$ is always a solution, while the second solution is $\alpha$-dependent, and not always so nice. For $\alpha=1/2$ the second equation appears to be ${\rm Erf}(c_{1/2} y) - {\rm Erf}(c_{1/2}x) =1$ where $c_{1/2} \approx 0.57285884$. For other values of $\alpha$ the equation appears to be of the form ${\rm Erf}(c_\alpha y - d_\alpha) - {\rm Erf}(c_\alpha x +d_\alpha) =1$. Any insight on what the value of $c_{1/2}$ might be analytically or why $d_{1/2}$ might be zero would be much appreciated.

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The solution is not always unique. For $1/2 < \alpha < .79654774210531568819$ approximately, there are two other solutions. For example with $\alpha = 0.6$ the other solutions are approximately $x = -1.910219091, y = .006118254868$ and $ x = -.006118254868, y = 1.910219091$. This can be seen graphically: here are the curves given by your two equations (the first in red, second in blue) for different values of $\alpha$ from $0.3$ to $1$.

The value near 0.8 is actually the value of $\exp{-(\text{erf}^{-1}(1/2))^2}$. This is the value of $\alpha$ for which the two curves have the same slope at the point $(-\text{erf}^{-1}(1/2), \text{erf}^{-1}(1/2))$.

enter image description here

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  • $\begingroup$ Ah, yes indeed, and thanks for the really helpful visual. I should have seen this, thanks. I'm going to revise my question for specifically $\alpha = 1/2$, since the general $\alpha$ case wasn't a helpful (or correct generalization). Really I'm only interested in $\alpha = 1/2$ (and would like to find an analytic proof). $\endgroup$ – Johan Apr 26 '12 at 2:06

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