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I am looking for asymptotic solutions to the equation $$\alpha^{-1}x+\sqrt{\pi}\frac{\sqrt{x}}{2}\text{erf}\left(\frac{\sqrt{x}}{2}\right)=\beta^{-1}e^{-x/4},\qquad \alpha\ll1,\beta\gg1.$$ When $\alpha$ is large and the first term is negligible, this is easy to do, but I don't know how to proceed with the opposite case.

What I've tried for now is the following: For $\alpha,\beta^{-1}=0$, which is the limiting case, I get $x=0$, hence I have to introduce a scaling $x=\epsilon\hat x$, where $\epsilon=\epsilon(\alpha,\beta)\ll1$. Introducing this into the equation above allows me to simplify terms and reduce the equation (if I'm not wrong) to $$\epsilon(1\color{red}{+}\alpha/2)\hat x=\alpha\beta^{-1},$$ therefore I can balance the equation by choosing $\epsilon=\alpha\beta^{-1}$ and finally $$\hat x\approx\color{red}{2/(2+\alpha)}\qquad\Rightarrow\qquad x\approx\alpha\beta^{-1}.$$

Is this correct? Any hints or help on this?

Thanks in advance!

$\color{red}{\text{Edit: The leading order term had a mistake, I have corrected it.}}$

Marc
  • 846

1 Answers1

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Let $\tilde \beta = 1/\beta$. Multiplying by $\alpha$ and getting rid of the square roots, we can rewrite the equation as $$x - \alpha \tilde \beta e^{-x/4} + \alpha x \int_0^{1/2} e^{-x t^2} d t = 0.$$ Now we can look for $x$ in the form $\sum c_{i,j} \alpha^i \tilde \beta {}^j$ by substituting the sum into the equation and taking the bivariate Taylor expansion around $\alpha = 0, \,\tilde \beta = 0$.

Taking $x = \alpha \tilde \beta + \sum_{i = 0}^3 c_{i, 3 - i} \alpha^i \tilde \beta {}^{3 - i}$ gives $$c_{3, 0} \alpha^3 + \left( c_{2, 1} + \frac 1 2 \right) \alpha^2 \tilde \beta + c_{1, 2} \alpha \tilde \beta {}^2 + c_{0, 3} \tilde \beta {}^3 = 0,$$ therefore we get one third-order term $-\alpha^2 \tilde \beta/2$.

On the next step we get two fourth-order terms, which gives the approximation $$x \approx \frac \alpha \beta -\frac {\alpha^2} {2 \beta} + \frac {\alpha^3} {4 \beta} - \frac {\alpha^2} {4 \beta^2}.$$

Maxim
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