maximize a product of error function

Question

I want to maximize the value of product of two error functions. Given a function \begin{align} f(a) = erf\left(\frac{c}{a}\right) erf\left(\frac{c}{\sqrt{1-a^2}}\right) , \end{align} where $c\gg 1$, and $0\le a \le1$. I do some numerical experiments, and find that maximum of $f(a)$ is obtained at $a = \sqrt{\frac{1}{2}}$. How can we prove it theoretically?

Answer 1

$$\ln(f(a))=\ln \left(\text{Erf}\left(\frac{c}{a} \right)\right)+\ln \left(\text{Erf}\left(\frac{c}{\sqrt{1-a^2}} \right)\right)$$ Let's define the function $g(x)$ for $x\in(0,1)$, $c\gg 1$ $$g(x) := \ln \left(\text{Erf}\left(\frac{c}{\sqrt{x}} \right)\right)$$

We will prove this function is concave. According to WolframAlpha, for $x>0$

$$(g(x))'' = \frac{c e^{-\frac{2c^2}{x}} \left((-2c \sqrt{x} + e^{\frac{c^2}{x}} \sqrt{\pi} (-2c^2 + 3 x) \text{Erf} \left(\frac{c}{\sqrt x}\right)\right)}{2 \pi x^{7/2} \text{Erf}^2\left(\frac{1}{\sqrt x} \right)} \tag{1}$$

Because $x \in(0,1)$, $\frac{c}{\sqrt{x}} \in \left(c,+\infty \right)$. And as $c \gg1$, the denominator of $(1)$ is asymptotically equal to $$c e^{-\frac{2c^2}{x}} \left(( e^{\frac{c^2}{x}} \sqrt{\pi} (-2c^2 ) \text{Erf} \left(\frac{c}{\sqrt x}\right)\right)$$ which is negative for all $x\in(0,1)$. We can conclude that the function $g(x)$ is concave.

Apply the Jensen's inequality: $$\ln(f(a)) \le 2\cdot\ln \left(\text{Erf}\left( \left(\frac{c}{\sqrt{\frac{a^2 +(1-a^2)}{2}}} \right)\right)\right) = 2\cdot\ln \left(\text{Erf}\left( \sqrt 2c\right)\right) $$

The equality occurs if and only if $a^2 = 1-a^2$ or $a =\frac{1}{\sqrt{2}}$