Lemma 10.24.18F2C09635072E

Proof. Barriers and the radial inverse. Set $L=1-\log h=1+\log (1+1/r)$ and $w=-2\log L$. Differentiation gives

The function $Q$ is positive, since $L\geq 1$ and $c>2$. Its limits are $2/c$ at $0$ and $1-2/c$ at infinity, so $0<q_-\leq Q\leq q_+<\infty $ for some constants $q_\pm $. Choose $a_0\leq \min (0,\log q_-)$ and $b_0\geq \max (0,\log q_+)$. Then $l=w+a_0$ and $H=w+b_0$ satisfy

Exhaust $\mathbb P^1\setminus \{ 0\} $ by the disks $|\zeta |<R$ in the coordinate $\zeta =1/z$, and solve the equation on them with boundary value $w$. For $q=|\zeta |^2$, the equation becomes

where $\Delta $ is the Euclidean real Laplacian.

For $k>0$, the regular radial solution $v=V(q)$ of $(-\Delta +k)v=F(q)$ satisfies

The $j$-fold iterate of the integral kernel applied to the constant $1$ is $q^ j/(4^ j(j!)^2)$, so successive approximation converges uniformly on every finite interval. The coefficient of $A_0$ in the solution is $\sum _{j\geq 0}(kq/4)^ j/(j!)^2>0$, so exactly one value of $A_0$ gives any prescribed value at $q=R^2$. Moreover,

and this formula gives smoothness at $q=0$ by induction. The maximum principle for $-\Delta +k$ shows that this radial inverse is unique and preserves order.

Choose $k\geq \sup f\mathrm{e}^{H}$ on the fixed disk, and let $T(v)$ be the radial solution of $(-\Delta +k)T(v)=kv-f\mathrm{e}^{v}+g$ with boundary value $w$; the right side is nondecreasing in $v$ on $[l,H]$. The barrier inequalities give $l\leq T(l)$ and $T(H)\leq H$, so the iterates of $T$ starting from $l$ increase and stay below $H$. The derivative formula gives a uniform bound for their first derivatives on the fixed interval. Compactness and monotonicity give uniform convergence; passing to the limit in (10.24.20) proves the equation, and the averaged derivative formula bootstraps the limit to smoothness.

The same barriers bound the solutions on every fixed disk in $\zeta $ as $R\to \infty $. For an exact solution $U(q)$ of the nonlinear equation,

and induction gives bounds for all derivatives on compact intervals. A diagonal limit is therefore smooth on the $\zeta $-plane, including the point $z=\infty $, and satisfies $u=-2\log (-\log r)+O(1)$ at $0$.

The asymptotics, mass and energy. In the variable $t=\log r$, the equation reads

Put $v(t)=u(t)+c\log (1+\mathrm{e}^ t)$ and $V(s)=v(-s)$. Then

Convexity and this sublinear decay imply $V'<0$ and $V'\to 0$ at infinity. The function $J=(V')^2-2k_0\mathrm{e}^{V}$ satisfies $J'=-2k_0'\mathrm{e}^{V}$ and tends to zero. Since $k_0'=O(\mathrm{e}^{-s})$ and $\mathrm{e}^{V}\asymp s^{-2}$,

Thus $(\mathrm{e}^{-V/2})'=\sqrt{c/2}(1+O(\mathrm{e}^{-s}))$, and integration gives $\mathrm{e}^{-V/2}=\sqrt{c/2}(s+A)+O(\mathrm{e}^{-s})$. This identity and the differential equation give all three asymptotic formulas.

The local potential $u+c\log (1+r)$ extends subharmonically across $0$ by lemma 5.24.7. Its radial slope tends to zero at the puncture, so the extension adds no point mass. The density of $g_1$ is asymptotic to

Stokes' theorem for currents gives mass $c$. The energy integral at the cusp is bounded by $C\int ^\infty (\log s)/s^2\, \mathrm{d}s$, and the integral against the smooth form $\alpha $ is finite as well. $\square$