An expansion for 2d Euler equation

Question

Let $R>0$ be a large constant, such that for any $x \in \Omega$, $\Omega \subset B_R(x)$. Consider the following problem in $\mathbb{R}^2$: $$ -\varepsilon^2 \Delta u=1_{\{u>a\}} \text { in }\, B_R(0) \quad u=0 \text { on }\, \partial B_R(0), $$ where $a>0$ is a constant. Then, the above equation has a solution $U_{\varepsilon , a}$, which can be written as $$ U_{ \varepsilon, a}(y)= \left\{\begin{aligned}& a+\frac{1}{4 \varepsilon^2}(s_{ \varepsilon}^2-|y|^2), && |y| \leq s_{ \varepsilon}, \\ & a \ln \frac{|y|}{R} / \ln \frac{s_{ \varepsilon}}{R}, && s_{ \varepsilon} \leq|y| \leq R,\end{aligned}\right. $$ where $s_{\varepsilon}$ is the constant, such that $U_{\varepsilon, a} \in C^1(B_R(0))$.

So, $s_{\varepsilon}$ satisfies $$ -\frac{s_{\varepsilon }}{2 \varepsilon ^2}=\frac{a}{s_{\varepsilon} \ln \frac{s_{\varepsilon}}{R}} . $$

From $$ \frac{s_{\varepsilon } \sqrt{\ln \frac{R}{s_{\varepsilon }}}}{\varepsilon }=\sqrt{2 a}, $$ we see that if $\varepsilon>0$ is small, the above equation is uniquely solvable for $s_{\varepsilon}>0$ small. Moreover, we have the following expansion for $s_{\varepsilon}$ : $$ s_{\varepsilon }=\frac{\sqrt{2 a} \varepsilon}{\sqrt{|\ln \varepsilon |}}\Big(1+O\Big(\frac{\ln |\ln \varepsilon |}{|\ln \varepsilon |}\Big)\Big) . $$ Above are some calculations in the paper [Planar vortex patch problem in incompressible steady flow]. But I can not find out the last estimate. Any comments are welcome.

Answer 1

Let's consider the simplified equation (which is really WLOG, since you are just absorbing constants) $$ x \sqrt{-\ln x} = y $$ for $x,y > 0$ and $x,y \ll 1$. (Here $x$ is playing the role of $s_\epsilon$ and $y$ is playing the role of $\epsilon$.) This equation has an explicit solution in terms of the Lambert W function $$ x = \exp\left( \frac12 W_{-1}(-2 y^2) \right) $$ Unfortunately this formula doesn't really help in obtaining the asymptotics.

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Instead, let's do this by hand. It is convenient to square the equation given above. Let $w = x^2$ and $z = 2y^2$ we have the above equation squares to $$ w \ln w = -z $$ Consider the ansatz $$ w_{\delta} = -\frac{z}{\ln z} (1+ \delta) $$ We can compute $$ w_\delta \ln w_\delta = - \frac{z}{\ln z} (1+\delta) \Big[ \ln z + \ln(-\frac1{\ln z}) + \ln(1+\delta) \Big] $$ so $$ w_\delta \ln w_\delta + z = - z \Big[\underbrace{\frac{1}{\ln z}\ln(-\frac{1}{\ln z}) + \delta + \frac{\delta}{\ln z} \ln( - \frac{1}{\ln z}) + \frac{1+\delta}{\ln z} \ln(1+\delta)}_{=: g(z,\delta)} \Big] $$

For $z$ small, we have that $\zeta := \frac{1}{\ln z} \ln(-\frac{1}{\ln z}) > 0$, and it converges to $0$ as $z \searrow 0$. So for all $z$ sufficiently small, you have, for $\delta \in (-1/2,0)$, that $$ g(z,\delta) - \frac{1+\delta}{\ln z} \ln(1+\delta) \in [\zeta + \frac32 \delta, \zeta + \delta ] $$ Observing next that for $\delta \in [-1/2,0]$ there is a constant $M$ such that $$ - M \delta \leq \ln(1+\delta) \leq 0 $$ we see that by taking $z$ even smaller, so $M/\ln(z)$ is small in absolute value, we have the following result

Lemma For all $\delta \in [-1/2,0]$ and $z$ sufficiently small, we have $$ g(z,\delta) \in [\zeta + 2 \delta, \zeta + \frac12 \delta] $$

Since we know that $\lim_{z\to 0} \zeta = 0$, we can restrict further to those $z$ small enough such that $\zeta < \frac15$. But now the lemma implies

$$ g(z,-1/2) \leq \frac15 - \frac14 < 0, \qquad g(z,0) = \zeta > 0 $$

so there exists $\delta \in [-1/2,0]$ such that $g(z,\delta) = 0$. Furthermore, this requires $0 \in [\zeta + 2\delta, \zeta + \frac12 \delta]$ which means that $\delta \in [-2\zeta, -\frac12 \zeta]$. As $g(z,\delta)$ is the remainder of $w_\delta \ln w_\delta + z$, we conclude

Theorem There exists a $z_0 > 0$ such that for all $z\in (0,z_0)$, there exists a $\delta \in [-2 \zeta, -\frac12 \zeta]$, where $\zeta = -\frac{1}{\ln z} \ln(-\frac{1}{\ln z})$, such that $w_\delta = - \frac{z}{\ln z}(1 + \delta)$ solves $w_\delta \ln w_\delta = - z$.

Unwrapping the change of variables from $(x,y)$ to $(w,z)$, you get

Corollary There exists a $y_0 > 0$ such that for all $y\in (0,y_0)$, the expression $x \sqrt{-\ln x} = y$ can be solved by $$ x = \frac{y}{\sqrt{|\ln y|}} (1 + \omega) $$ for some $\omega = O\left(\left| \frac{\ln|\ln y|}{\ln y} \right|\right)$.