Taubes disk barriers and monotone Dirichlet iteration

First, $\log \mathrm{\mid }z{\mathrm{\mid }}^2$ is locally integrable and harmonic off zero. For a smooth compactly supported test function $\phi$, apply [F14] on an annulus excluding $\mathrm{\mid }z\mathrm{\mid }\le \epsilon$ to the one-form $\log \mathrm{\mid }z{\mathrm{\mid }}^2\ast d\phi -\phi \ast d\log \mathrm{\mid }z{\mathrm{\mid }}^2$. Its inner-boundary contribution is $-\epsilon \log ({\epsilon }^2){\int }_0^{2\pi }{\mathrm{\partial }}_r\phi (\epsilon ,\theta )d\theta +2{\int }_0^{2\pi }\phi (\epsilon ,\theta )d\theta$, which tends to $4\pi \phi (0)$; the outer contribution is zero. Thus $\mathrm{\Delta }\log \mathrm{\mid }z-z_j{\mathrm{\mid }}^2=4\pi {\delta }_{z_j}$ by translation. For $z\mathrm{\ne }{z}_j$, put $t_j=a\mathrm{/}(\mathrm{\mid }z-z_j{\mathrm{\mid }}^2+a)$. Then $0<t_j<1$, $e^{u_{-}}={\prod }_j(1-t_j{)}^{m_j}$, and direct radial differentiation gives $\mathrm{\Delta }{u}_{-}=-\frac{4}{a}{\sum }_j{m}_j{t}_j^2+4\pi {\sum }_j{m}_j{\delta }_{z_j}$ distributionally. If $N>0$, let $t_{\ast }:={\max }_j{t}_j$; then $1-e^{u_{-}}\ge t_{\ast }$ and ${\sum }_j{m}_j{t}_j^2\le N{t}_{\ast }$. Hence $\mathrm{\Delta }{u}_{-}+\kappa (1-e^{u_{-}})-4\pi D\ge (\kappa -4N\mathrm{/}a)t_{\ast }\ge 0$. For $N=0$, $u_{-}=0$ and the subsolution inequality is equality.

The denominator $R^2-{\bar{z}}_{j}z$ has no zeros on ${\bar{D}}_R$ because $\mathrm{\mid }{z}_j\mathrm{\mid }<R$. The logarithmic fundamental-solution identity and the boundary equality $R^2\mathrm{\mid }z-z_j{\mathrm{\mid }}^2=\mathrm{\mid }{R}^2-{\bar{z}}_{j}z{\mathrm{\mid }}^2$ for $\mathrm{\mid }z\mathrm{\mid }=R$ give $\mathrm{\Delta }{u}_{s,R}=4\pi D$ and $u_{s,R}=0$ on $\mathrm{\partial }{D}_R$. Consequently $A_R=e^{u_{s,R}}$ is smooth and nonnegative on ${\bar{D}}_R$, vanishes at the centers, and equals one on the boundary. Since $\mathrm{\Delta }{\psi }_R=-\kappa$, the supersolution residual away from the Dirac masses is $\mathrm{\Delta }{\psi }_R+\kappa (1-A_R{e}^{{\psi }_R})=-\kappa A_R{e}^{{\psi }_R}\le 0$.

By [F10], $H_0^1(D_R;\mathbb{R})$ is a real Hilbert space with the full $W^{1,2}$ inner product; [F2] gives its norm. For any $L_R>0$, [F11] and the pointwise Euclidean inequality $\mathrm{\mid }\mathrm{\nabla }v\cdot \mathrm{\nabla }w\mathrm{\mid }\le \mathrm{\mid }\mathrm{\nabla }v\mathrm{\mid }\mathrm{\mid }\mathrm{\nabla }w\mathrm{\mid }$ give $\mathrm{\mid }a(v,w)\mathrm{\mid }\le (1+L_R)\mathrm{\parallel }v{\mathrm{\parallel }}_{W^{1,2}}\mathrm{\parallel }w{\mathrm{\parallel }}_{W^{1,2}}$, so the form is bounded. Also $a(v,v)\ge \min \{1,L_R\}\mathrm{\parallel }v{\mathrm{\parallel }}_{W^{1,2}}^2$, so it is coercive. For $f\in L^2(D_R)$, [F11] gives $\int \mathrm{\mid }fw\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }w{\mathrm{\parallel }}_2$, and [F12] then gives $\mathrm{\mid }\int fw\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }w{\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }w{\mathrm{\parallel }}_{W^{1,2}}$. Thus [F3] gives a unique zero-Dirichlet weak solution of $(-\mathrm{\Delta }+L_R)u=f$. The resolver is linear by uniqueness and positivity preserving: if $f\ge 0$, then $-u$ is a weak subsolution of $(-\mathrm{\Delta }+L_R)(-u)=-f\le 0$ with zero trace, so [F4] gives $u\ge 0$.

The lower and upper barriers have the same logarithmic singular part, so $w=u_{-}-u_{s,R}-{\psi }_R$ extends continuously across every center. Their subsolution and supersolution inequalities imply $\mathrm{\Delta }w-c_{R}w\ge 0$, where $c_R=\kappa {\int }_0^1\exp ((1-t)(u_{s,R}+{\psi }_R)+t{u}_{-})dt\ge 0$, extended continuously at the centers. On $\mathrm{\partial }{D}_R$, $w=u_{-}<0$ (and $w=0$ when $D$ is empty). Applying [F4] to $(-\mathrm{\Delta }+c_R)w\le 0$ with nonpositive boundary trace gives $w\le 0$, hence $v_0=u_{-}-u_{s,R}\le {\psi }_R$.

Set $G_R(v)=\kappa +L_{R}v-\kappa A_R{e}^v$ and choose $L_R\ge \kappa {\max }_{{\bar{D}}_R}(A_R{e}^{{\psi }_R})$. For $v\le {\psi }_R$, $G_R^{\mathrm{\prime }}(v)=L_R-\kappa A_R{e}^v\ge 0$, so the iteration map $T_R=(-\mathrm{\Delta }+L_R{)}^{-1}{G}_R$ is order preserving. To prove the lower comparison, put $z=v_0-T_R{v}_0$. The logarithmic singularities in $v_0=u_{-}-u_{s,R}$ cancel, so $v_0\in H^1(D_R)$; also $T_R{v}_0\in H_0^1(D_R)$. The subsolution inequality of step 1.1 and $\mathrm{\Delta }{u}_{s,R}=4\pi D$ from step 1.2 cancel the Dirac terms and give, distributionally, $\mathrm{\Delta }{v}_0+\kappa (1-e^{u_{-}})\ge 0$. Since $u_{-}=u_{s,R}+v_0$ and $e^{u_{s,R}}=A_R$, this is $(-\mathrm{\Delta }+L_R)v_0\le G_R(v_0)$. The resolvent equation for $T_R{v}_0$ therefore makes $z$ a weak subsolution of $(-\mathrm{\Delta }+L_R)z\le 0$. Its boundary trace is that of $v_0$, namely $u_{-}{\mathrm{\mid }}_{\mathrm{\partial }{D}_R}\le 0$ because $u_{s,R}=0$ there; the inequality is strict for nonempty $D$ and is equality for the empty divisor. The weak maximum principle [F4], with $L_R>0$, gives $z\le 0$, hence $v_0\le T_R{v}_0$. For the upper comparison, $\mathrm{\Delta }{\psi }_R=-\kappa$ and the supersolution residual in step 1.2 give $(-\mathrm{\Delta }+L_R){\psi }_R\ge G_R({\psi }_R)$; both ${\psi }_R$ and $T_R{\psi }_R$ have zero trace, so positivity of the resolver from step 1.3 gives $T_R{\psi }_R\le {\psi }_R$. Thus $v_0\le v_1\le \cdots \le {\psi }_R$; every iterate after $v_0$ has zero Dirichlet trace. Each has a smooth representative: starting from smooth $v_0$ and $A_R$, rewrite its shifted resolvent equation as a Laplace equation, use boundedness and [F5] for $W^{2,p}$, then [F6]–[F7] and local bootstrapping; this also identifies continuous representatives up to the boundary. These representatives realize the order pointwise.

The monotone iterates have a bounded pointwise limit $v_R$ on the finite disk. If $C\ge {\sup }_{D_R}({\psi }_R-v_0{)}^2$, then $0\le C-(v_R-v_k{)}^2\uparrow C$. Applying [F13] and subtracting the finite integral of C gives ${\int }_{D_R}(v_R-v_k{)}^2\to 0$, so $v_k\to v_R$ in $L^2(D_R)$. For functions $v,w$ with $v_0\le v,w\le {\psi }_R$, the pointwise mean-value formula gives $\mathrm{\mid }{e}^v-e^w\mathrm{\mid }\le e^{{\psi }_R}\mathrm{\mid }v-w\mathrm{\mid }$; hence $\mathrm{\parallel }{G}_R(v)-G_R(w){\mathrm{\parallel }}_2\le (L_R+\kappa \mathrm{\parallel }{A}_R{\mathrm{\parallel }}_{\mathrm{\infty }}{e}^{\max {\psi }_R})\mathrm{\parallel }v-w{\mathrm{\parallel }}_2$. Subtract the two resolvent equations for $T_{R}v,T_{R}w$ and test with $u=T_{R}v-T_{R}w$. The form in step 1.3 satisfies $a(u,u)\ge \min \{1,L_R\}\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,2}}^2$, while [F11] and [F12] bound the right side by $\mathrm{\parallel }{G}_R(v)-G_R(w){\mathrm{\parallel }}_2\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,2}}$; division when $u\mathrm{\ne }0$ (and trivially when $u=0$) gives $\mathrm{\parallel }{T}_{R}v-T_{R}w{\mathrm{\parallel }}_{W^{1,2}}\le \min \{1,L_R{\}}^{-1}\mathrm{\parallel }{G}_R(v)-G_R(w){\mathrm{\parallel }}_2$. Therefore $T_R{v}_k\to T_R{v}_R$ in $H_0^1$; the shifted sequence $v_{k+1}=T_R{v}_k$ has the same $L^2$ limit as $v_k$, so uniqueness of an $L^2$ limit gives $v_R=T_R{v}_R$. It is a weak solution of $-\mathrm{\Delta }{v}_R=\kappa (1-A_R{e}^{v_R})$ with zero trace.

Since $v_0\le v_R\le {\psi }_R$ and $A_R$ is bounded on ${\bar{D}}_R$, the right-hand side in step 3.1 lies in $L^p(D_R)$ for every finite $p$. For a prescribed $0<\alpha <1$, choose $p>2\mathrm{/}(1-\alpha )$. The disk is bounded and smooth, so [F5] gives $v_R\in W^{2,p}$; [F6] gives a $C^{1,\beta }$ representative with $\beta >\alpha$ and zero boundary values. The nonlinear right-hand side is then $C^{0,\alpha }$, and [F7] upgrades the weak solution to $C^{2,\alpha }({\bar{D}}_R)$. Repeated local Schauder bootstrapping, using [F9], makes $v_R$ smooth in $D_R$. Setting $u_R=u_{s,R}+v_R$ proves its distributional equation and logarithmic center behavior.

If $u_R$ were positive somewhere, its boundary value is zero and it tends to $-\mathrm{\infty }$ at each center, so it would attain a positive interior maximum away from the centers. There $\mathrm{\Delta }{u}_R=\kappa (e^{u_R}-1)>0$ by its equation, contradicting the second-derivative test at a maximum. Thus $u_R\le 0$; together with the lower iteration bound in step 2.2 and its limit in step 3.1 this gives $u_{-}\le u_R\le 0$.

Fix a disk $U=B_S(0)$ containing the centers in its interior and let $s=2{\sum }_j{m}_j\log \mathrm{\mid }z-z_j\mathrm{\mid }$. For every sufficiently large $R$, $w_R=u_R-s$ is smooth on $\bar{U}$. On $\mathrm{\partial }U$, steps 1.1 and 5.1 give the uniform bounds $u_{-}-s\le w_R\le -s$, whose endpoints are finite there. On $U$, $\mathrm{\Delta }{w}_R=\kappa (e^s{e}^{w_R}-1)=\kappa (e^{u_R}-1)\in [-\kappa ,0]$. A quadratic barrier $Q(z)=M+\kappa (S^2-\mathrm{\mid }z{\mathrm{\mid }}^2)\mathrm{/}4$, with $M$ larger than the uniform boundary bound, satisfies $\mathrm{\Delta }(w_R-Q)\ge 0$ and $\mathrm{\Delta }(-w_R-Q)\ge 0$, and both differences are nonpositive on $\mathrm{\partial }U$; the disk maximum principle [F4], with zero lower-order coefficient, yields $\mathrm{\mid }{w}_R\mathrm{\mid }\le Q$.

On any smaller compact disk $K⋐U$, [F8] applied on finitely many nested balls and the uniform $L^{\mathrm{\infty }}$ bound of step 6.1 yields uniform $W^{2,p}$ bounds for $w_R$ for every finite $p>2$, since its Laplacian is bounded. For a prescribed $0<\alpha <1$, choose $p>2\mathrm{/}(1-\alpha )$ and then $\beta$ with $\alpha <\beta <1-2\mathrm{/}p$; [F6] gives a uniform $C^{1,\beta }$ bound on a slightly larger compact disk. As $e^s$ is smooth and $\mathrm{\Delta }{w}_R=\kappa (e^s{e}^{w_R}-1)$, the right-hand side has a uniform $C^{0,\alpha }$ bound there; [F9] then yields a uniform $C^{2,\alpha }$ bound on $K$. These bounds are independent of $R$ and are the local compactness estimates used for disk exhaustion.