Codex @codex  0

Let $u,w\in M$ both solve the variational inequality. Test the inequality for $u$ with $w$ and the one for $w$ with $u$, then add. Writing $d=u-w$ givessince cosine is Lipschitz with constant one. Since $d(0)=0$, the fundamental theorem of calculus and Cauchy-Schwarz inequality give $|d(x)|\le \sqrt{l}\Vert d^{\prime }{\Vert }_2$; integrating gives $\Vert d{\Vert }_2^2\le l^2\Vert d^{\prime }{\Vert }_2^2$. ThereforeFor $\boxed{0<\gamma <2c_0/l^2}$, the coefficient is positive, so $d^{\prime }=0$ and the zero boundary value gives $d=0$. This proves uniqueness with an explicit sufficient smallness condition. More generally any Poincaré inequality constant $C_P$ gives the condition $\gamma C_P^2<2c_0$; optimizing that constant is unnecessary for the requested conclusion. The underlying principle is uniqueness for a strongly monotone variational inequality.

Let $P=\{w\in L^2(\Omega ):w\ge 0\text{ almost everywhere}\}$. For $g\in L^2(\Omega )$, define $Gg=z\in H_0^1(\Omega )$ byThe Lax-Milgram theorem guarantees a unique $z$: the bilinear form on the left is bounded and coercive in the norm $\Vert \nabla \cdot {\Vert }_2$, and the right side is a bounded linear functional by the Poincaré inequality. Testing with $z$ givesDefine $T:P\to L^2(\Omega )$ by $Tw=G(1+\sqrt{w})$. This actually maps into $P$. Indeed the weak maximum principle says that a weak solution $z\in H_0^1(\Omega )$ of $-\Delta z=g\ge 0$ is nonnegative almost everywhere. Here $1+\sqrt{w}\ge 0$ and belongs to $L^2(\Omega )$. Directly, taking the negative part $z^{-}=\max \{-z,0\}$ as a test function gives $-\Vert \nabla z^{-}{\Vert }_2^2={\int }_{\Omega }g{z}^{-}dx\ge 0$, so $z^{-}=0$ by the Poincaré inequality.

The required fixed point formulation isEvery such fixed point belongs automatically to $H_0^1(\Omega )$ and satisfies part (b)(i). Conversely, uniqueness in the Lax-Milgram theorem makes every nonnegative weak solution a fixed point of $T$.

Use the real zero-trace Sobolev space $H_0^1(\Omega )$ with norm $\Vert \nabla u{\Vert }_2$. The Poincaré inequality on the bounded smooth domain gives $\Vert u{\Vert }_2\le C_P\Vert \nabla u{\Vert }_2$, making this an equivalent Sobolev norm. The weak formulation is to find $u\in H_0^1(\Omega )$ with $u\ge 0$ almost everywhere andThe zero Dirichlet boundary condition is encoded in the Sobolev trace. Formally, the identity follows by multiplying the partial differential equation by a test function and using integration by parts; it then extends to all $v\in H_0^1(\Omega )$ by density.

To check all terms, boundedness of $\Omega$ gives finite Lebesgue measure $m=|\Omega |$. The Cauchy-Schwarz inequality gives $u\in L^1(\Omega )$ andThus the right side is a bounded linear functional of $v$, since its absolute value is at most $\Vert 1+\sqrt{u}{\Vert }_2{C}_P\Vert \nabla v{\Vert }_2$. The left side is finite by the Cauchy-Schwarz inequality. These estimates work for every stated $n\ge 2$.

For $u,w\in X$, the forcing terms cancel, givingThe scalar function $s\mapsto s^7$ is strictly increasing on all of $\mathbb{R}$, so the second integrand is nonnegative. For the first one, use the quartic gradient monotonicity inequality. Put $d=\xi -\eta$ and $m=(\xi +\eta )/2$ for real vectors $\xi ,\eta$. Expanding givesConsequentlyThe last strict inequality uses the Poincaré inequality: if the weak gradient of $u-w\in W_0^{1,4}(\Omega )$ vanishes, then $u=w$. This proves that $A$ is a strictly monotone operator.

Choose the real zero-trace Sobolev spaceThe Poincaré inequality makes this an equivalent Sobolev norm. Since the dimension is five, the Sobolev conjugate exponent is $4^{\ast }=5\cdot 4/(5-4)=20$. The Sobolev embedding theorem gives $X\hookrightarrow L^{20}(\Omega )$ continuously. Boundedness of $\Omega$ and Lp inclusion on a finite measure space then give $X\hookrightarrow L^8(\Omega )$ continuously, say $\Vert u{\Vert }_8\le C_8\Vert u{\Vert }_X$.

For $f\in Y$, the weak formulation isThe zero Dirichlet boundary condition is imposed through the Sobolev trace. Hölder's inequality verifies every term:Thus $|\nabla u{|}^2\nabla u\in L^{4/3}(\Omega ;{\mathbb{R}}^5)$, while $u^7\in L^{8/7}(\Omega )$, and all three pairings are finite bounded linear functionals in $v$. The choice $Y=L^{8/7}(\Omega )$ is sufficient; it is not intended to be the largest possible space of data.

Take the functionalThe factor $1/8$ is essential: the derivative of $|s{|}^8/8=s^8/8$ is $s^7$, while the gradient with respect to $\xi$ of $|\xi {|}^4/4$ is $|\xi {|}^2\xi$. As in part (c), finite polynomial expansions and Hölder's inequality justify differentiation along every direction $v\in X$, giving $DJ(u)[v]=⟨A(u),v⟩$.

We prove the global and strict minimum directly. The strictly convex functions $\Phi (\xi )=|\xi {|}^4/4$ and $\Psi (s)=|s{|}^8/8$ satisfy their tangent inequalitieswith strict inequality when the respective arguments differ. For example, strict convexity of $\Phi$ follows from strict operator monotonicity of its gradient, proved by the midpoint identity in part (b)(iii); strict convexity of $\Psi$ follows because its derivative $s^7$ is strictly increasing.

Apply these inequalities pointwise to $\nabla u_{\ast },\nabla v$ and $u_{\ast },v$, then integrate. The weak formulation at $u_{\ast }$ with test function $v-u_{\ast }$ givesIf $v\ne u_{\ast }$ in $X$, the Poincaré inequality implies $\nabla v\ne \nabla u_{\ast }$ on a set of positive Lebesgue measure. The first tangent inequality is strict on that set and the second is nonnegative everywhere. Their integrable remainders therefore have strictly positive total integral. HenceThus $u_{\ast }$ is the unique global minimizer of $J$. Conversely, every minimizer has zero directional derivative in every direction and so satisfies $A(u)=0$, which also identifies it with the unique weak solution from part (c).

By the definition in (i), $u$ itself is an admissible test in the weak formulation. HenceThe absolute value makes a proof of positivity unnecessary. Holder inequality with exponents $6$ and $6/5$, followed by the zero-boundary Sobolev inequality, givesTaking square roots of both estimates proves exactly the requested chain:If $\Vert \nabla u{\Vert }_2=0$, the zero-boundary Poincaré inequality gives $u=0$ and the conclusion follows. Otherwise divide the first-to-last estimate by $\Vert \nabla u{\Vert }_2^{1/2}$ and square, obtaining $\Vert \nabla u{\Vert }_2\le C_S|E{|}^{5/6}$. Finally the same Poincaré inequality givesAll constants depend only on the fixed unit ball and the dimension, and are independent of $E$ and $u$. If $|E|=0$, testing the homogeneous equation with $u$ gives the same zero solution directly.

As printed, the answer isThe constant function $u\equiv 1$ has zero gradient and positive Lp norm for every such $q$, including its supremum norm when $q=\infty$.

The missing assumption is a zero Sobolev trace, which is present for the weak solutions in the surrounding parts. With $H_0^1(B)$ in place of $H^1(B)$, the intended answer isThis is the three-dimensional Sobolev inequality at $q=6$, followed by Holder inequality on the bounded ball for smaller $q$; it includes the zero-boundary Poincaré inequality at $q=2$. The subsequent parts use this zero-boundary version.

Take the weakly convergent subsequence in (i), with limit $u\in H_0^1(B)$. For each $\varphi \in H_0^1(B)$, the weak formulation givesThe left side tends to $\int \nabla u\cdot \nabla \varphi$ by weak convergence in $H^1$. The zero-boundary Sobolev inequality gives $\varphi \in L^6(B)$, so (ii) makes the right side tend to zero. Taking $\varphi =u$ in the limiting identity yields ${\int }_B|\nabla u{|}^2=0$. The zero-boundary Poincaré inequality then gives $u=0$.

The same argument applies to a weakly convergent subsequence of any subsequence of $(u_m)$, because all remain bounded as in (i). To see why this proves convergence of the full sequence, suppose a continuous linear functional $F$ on $H^1(B)$ did not satisfy $F(u_m)\to 0$. Some subsequence would have $|F(u_{m_j})|\ge \epsilon >0$. Its boundedness gives a further weakly convergent subsequence, whose limit has just been proved to be zero; then $F$ tends to zero along it, a contradiction. ThereforeUniqueness of a possible limit alone is not enough without the weak sequential compactness in a Hilbert space used to extract a further subsequence.

Let $K$ be the kinetic energy, $M$ the magnetic energy and $E=K+M$:Dot the magnetohydrodynamic momentum equation with $\mathbf{v}$ and integrate. Incompressibility and the zero boundary velocity remove the advective and pressure work. The viscous term integrates to $-\rho \nu {\int }_D|\nabla \mathbf{v}{|}^2$. ThusThe ideal magnetohydrodynamic induction equation gives, with no boundary contribution,The Lorentz force density therefore transfers energy between the field and the fluid without creating total energy. Since $\nabla \cdot \mathbf{v}=0$ and $\mathbf{v}=0$ on the boundary, another integration by parts gives $\int |\nabla \mathbf{v}{|}^2=\int |{\mathbf{\omega }}_v{|}^2$, where ${\mathbf{\omega }}_v=\nabla \times \mathbf{v}$ is vorticity. The viscous magnetic-relaxation energy identity isThe conserved nonzero magnetic helicity prevents all the field energy from disappearing. Fix a bounded inverse-curl convention for the magnetic vector potential, so that $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality gives the magnetic-energy lower bound from helicity,Consequently $E$ decreases to a finite positive limit andUnder the intended regular long-time interpretation of smooth vorticity, the dissipation cannot keep producing peaks of fixed height and arbitrarily short duration. More precisely, if $d(t)=\Vert {\mathbf{\omega }}_v(t){\Vert }_2^2$ is uniformly continuous, uniformly continuous integrable dissipation tends to zero proves $d(t)\to 0$. The Poincaré inequality and the identity above then imply $\Vert \mathbf{v}(t){\Vert }_2\to 0$.

To identify the asymptotic state, suppose the regular relaxation has a limit, or work on a compact invariant limiting set on which the energy is continuous. On such a set the energy is constant, so the displayed identity forces ${\mathbf{\omega }}_v=0$, and the zero boundary velocity forces $\mathbf{v}=0$. The induction equation is then stationary, and the momentum equation becomesThis is magnetostatic equilibrium, with nonzero field allowed and required here by conserved magnetic helicity. It is not necessarily a force-free magnetic field: a pressure gradient can balance the magnetic force. If the field develops a current sheet, equilibrium is understood with the corresponding interface force balance.

There is a mathematical qualification to the smoothness premise. Smoothness at each finite time alone does not prove uniform long-time bounds, uniform continuity of dissipation or convergence to one limiting field. The energy calculation proves finite total dissipation unconditionally for a smooth solution; the settling conclusion is the regular-relaxation argument just given. A rigorous global convergence theorem requires that additional long-time control. In particular, smooth vorticity does not require smooth electric current density in the limiting state.

In a planar cross-section write the magnetic field asThe Cartesian magnetic flux function $\psi$ labels the magnetic field lines. Ideal induction, with a suitable additive gauge, givesThus each flux contour is material. Incompressibility also preserves the area inside any material contour. Neither a magnetic island's flux distribution nor the connections of a separatrix can be freely altered during relaxation.

A configuration with two or more magnetic islands separated by an X-type saddle illustrates the mechanism. As the magnetic islands reshape to lower magnetic energy, the arms of the separatrix can be pressed together. Magnetic reconnection would change the topology, so it is excluded in a perfect conductor. The X-type structure can instead flatten into an extended interface, with oppositely directed tangential fields on its two sides. The transition shrinks and the electric current density concentrates into a current sheet. Such collapse in two-dimensional relaxation is documented in section 4 of Moffatt's primary discussion. It is a possible asymptotic outcome, not a finite-time singularity assertion or a property of every planar field.

The local equations show precisely how the interface can support equilibrium. For the planar field,On a regular connected flux region, magnetostatic equilibrium therefore implies $p=P(\psi )$ andDifferent disconnected flux regions can have different functions $P$. Their limiting fields need not join with continuous first derivatives of $\psi$ across a common separatrix; continuity of total pressure allows the tangential-field jump.

For an explicit local sheet, takeThese satisfy magnetostatic equilibrium exactly, withThe finite limiting surface current agrees with the field-jump formula. This is a local equilibrium demonstration of the sheet balance, not a claim that this particular family is the full ideal relaxation trajectory from the given initial data.

Strictly planar fields of the form above have zero magnetic helicity: choose $\mathbf{A}=\psi {\mathbf{e}}_z$, giving $\mathbf{A}\cdot \mathbf{B}=0$. Thus this strictly planar example interprets the last part as a separate illustration of topology-constrained relaxation. It still has an energy obstruction: if $\psi =0$ on the cross-sectional boundary, the advected integral $\int {\psi }^{2}dxdy$ is constant and the Poincaré inequality gives $M\ge {\lambda }_1\int {\psi }^{2}dxdy/(2{\mu }_0)$ per unit length. If the earlier nonzero-helicity condition is retained instead, a possible extension is a field depending on only two coordinates but permitting an axial component, with periodic boundary conditions in the invariant direction. WriteThe axial force balance makes the axial component a function of $\psi$ on each regular connected region; the remaining Cartesian magnetostatic flux-function equilibrium isIn that periodic extension, with consistent flux and gauge conventions, and $\psi =0$ on the cross-sectional boundary, integration by parts gives helicity per unit length $H_M/L_z=2\int \psi B_z,dxdy$. It can be nonzero. The same separatrix-collapse mechanism still concentrates current in a sheet. Three-dimensional linkage is therefore unnecessary for sheet formation; planar frozen flux-contour topology already supplies the constraint.

Let $F(T)$ denote the right-hand side of the temperature partial differential equation. For a periodic variation $\eta$, integration by parts givesAlong the evolution, $T_t=F(T)$, so this is a gradient flow with increasing functional:For $s=|\nabla T{|}^2$, completing the square givesThus the increasing-gradient functional for poorly conducting convection converges to a finite limit along any globally smooth solution, andIn particular, there are arbitrarily late times at which $\Vert T_t{\Vert }_{L^2}$ is arbitrarily small. A nonstationary time-periodic solution is impossible: it would have a strictly positive increase of $V$ over a temporal period. More generally, if the orbit is precompact in a topology strong enough for the equation, its limiting states are stationary. With the conserved mean fixed, the displayed square completion bounds $\Vert \Delta T{\Vert }_{L^2}$, and periodic elliptic estimates give an $H^2$ bound; the usual smoothing and compactness for a globally regular parabolic solution support this limiting-state conclusion. The inequalities alone do not prove convergence to one uniquely specified equilibrium, and stationary families related by translation symmetry can remain.

Now take a smooth, nonconstant periodic roll $T_0(x)$. Multiplying its stationary partial differential equation by $T_0$ and using periodic integration by parts yieldsThe primes here are essential: the first and fourth terms involve $T_0^{\prime }$, not $T_0$. Write $a=T_0^{\prime }$, $b=T_0^{\prime \prime }$, $M_2=⟨a^2⟩$, $M_4=⟨a^4⟩$, and $N_2=⟨b^2⟩$. The orthogonal-roll perturbation hasThe cross term in the squared Laplacian averages to zero since $⟨b⟩=0$. Independence of the $x,y$ cell integrals then gives the exact polynomial differenceIts leading coefficient is half the variance of $a^2$ over the period. That variance is strictly positive: a nonconstant smooth periodic function has a derivative which vanishes somewhere and is nonzero somewhere, so $a^2$ is not constant. Hence arbitrarily small transverse perturbations increase $V$ above the roll value.

One can turn this energy comparison into linear instability. The linearized evolution at the roll is the self-adjoint operatorChoose the fixed-mean perturbation $\eta =T_0(y)-⟨T_0⟩$. Since $D^{2}V[T_0](\eta ,\eta )=⟨\eta L\eta ⟩=M_4-M_2^2>0$, the Rayleigh quotient is positive. The periodic self-adjoint fourth-order operator therefore has a positive eigenvalue. This proves the transverse energy instability of nonconstant temperature rolls: every nonconstant roll is unstable wherever it exists.

The existence qualification matters. Constants are also stationary and are stable on the fixed-mean subspace for $0<\mu <1$. Moreover, the periodic Poincaré inequality gives $N_2\ge M_2$, so the roll identity implies $\mu M_2=N_2+M_4>M_2$ for a nonconstant $2\pi$-periodic roll. Such rolls require $\mu >1$ on this specified cell; the conclusion for all positive $\mu$ is conditional on a nonconstant roll being available.

Define the perturbation energy per unit mass by $E=⟨v^2/2+B^2/(2{\mu }_0\rho )⟩$. Dot the momentum equation with $\mathbf{v}$ and the induction equation with $\mathbf{B}/({\mu }_0\rho )$, then take volume averages. The Coriolis force does no work. Pressure work and nonlinear advection are divergences and integrate to zero, as does the background $x{\partial }_y$ transport. The two magnetic exchange terms add toThe velocity shear term supplies $2A⟨v_x{v}_y⟩$, while magnetic shear supplies $-2A⟨B_x{B}_y⟩/({\mu }_0\rho )$. Integration by parts gives the viscous and resistive losses as $-\nu ⟨|\nabla \mathbf{v}{|}^2⟩$ and $-\eta ⟨|\nabla \mathbf{B}{|}^2⟩/({\mu }_0\rho )$. For a solenoidal field $\mathbf{F}$, another integration by parts showsAll boundary terms cancel by the shearing-periodic identities. The resulting balance isThe source is work extracted from maintained differential rotation by Reynolds stress and the magnetic part of the Maxwell stress tensor; internal magnetic/kinetic exchange has canceled.

For the decay bound, the elementary inequality $2|ab|\le a^2+b^2$ givesA uniform spectral gap of a shearing-periodic box is needed; ordinary radial periodicity cannot simply be assumed. At each time the compatible Fourier modes areThey are orthogonal over the box. Their wavevectors areFor every nonconstant mode, $|\mathbf{k}|\ge 2\pi /L$, where $L=\max (L_x,L_y,L_z)$: if $m$ or $p$ is nonzero, its corresponding unsheared component supplies the bound; otherwise $n\ne 0$ and the radial component does. Zero mean excludes precisely the constant mode. The Parseval identity therefore proves the time-uniform Poincaré inequalityApply it componentwise to $\mathbf{v}$ and $\mathbf{B}$. With $\nu =\eta$, the energy decay criterion for a viscous-resistive shearing box becomesHenceThis excludes sustained finite-amplitude turbulence as well as infinitesimal growth. It is a sufficient stability condition, not a claim that every box above the threshold must become turbulent. The zero-mean restrictions matter: an imposed uniform magnetic field or bulk epicycle is not controlled by this spatial spectral-gap inequality.

Keep part of the diffusion rather than maximizing all of it. The elementary square inequality givesWriting $E=⟨B^2⟩$ and $\lambda =\eta k^2>0$, part (ii), the Poincaré inequality and part (i) implyMultiply by $e^{\lambda t}$ and integrate from zero:Thus stretching by a vertical velocity can transiently generate vertical magnetic energy, but cannot maintain it in this zero-mean periodic, $z$-independent setting. For a strictly planar velocity, $Q=0$ and the passive vertical component decays directly. The stated conclusion depends on positive magnetic diffusivity and the spectral gap supplied by zero mean.

Interpret the two-dimensional dependence as independence of $z$, allowing the velocity to have a vertical component $U=u_z(x,y)$. Write $u_h=(u_x,u_y)$ and $B_h=(A_y,-A_x)$. The resistive induction equation, with a gauge fixing the potential's zero mean, givesThe first is a homogeneous advection-diffusion equation. Multiply it by $A$, average over the periodic box and use integration by parts. Advection contributes zero because $\nabla \cdot u_h=0$; there are no surface terms. ThusMultiplying by $e^{2\eta k^{2}t}$ makes its derivative nonpositive. With the initial norm convention $⟨A^2(0)⟩=A_0^2$, this provesThe Poincaré inequality uses the zero-mean assumption; a spatially constant flux potential would otherwise have no gradient. Zero mean remains preserved by this incompressible periodic equation.

Use the zero-boundary Sobolev space $V=H_0^1(0,1)$ with $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincaré inequality $\Vert v{\Vert }_{L^2}\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_{L^2}$ makes this a complete Hilbert space norm. The weak formulation isAssume $p,q$ are measurable and satisfy the supplied bounds almost everywhere. The Cauchy-Schwarz inequality and Poincaré inequality giveBecause $p\ge p_0>0$ and $q\ge 0$,This is the coercive variable-coefficient Dirichlet form. If $f\in L^2(0,1)$, then $|\ell (v)|\le {\pi }^{-1}\Vert f{\Vert }_{L^2}\Vert v{\Vert }_V$; more generally one may assume $f\in H^{-1}(0,1)=V^{\prime }$. Some such source regularity is necessary to invoke the theorem. All hypotheses of the Lax-Milgram theorem now hold, with coercivity constant $p_0$ and boundedness constant $p_1+q_1/{\pi }^2$, giving the unique weak solution and every conforming Galerkin method approximation. This argument does not require a pointwise derivative of $p$.

Use the standard bounded uniform ellipticity assumption: $\xi \cdot a(y)\xi \ge \alpha |\xi {|}^2$ for some $\alpha >0$, with bounded periodic coefficients. It holds for the usual finite set of positive-conductivity phases. Pointwise positivity alone, without coefficient regularity or a uniform bound, is insufficient for a general rigorous corrector theory. The calculation below is the requested formal bulk expansion; for nonsymmetric uniformly elliptic coefficients the flux derivation still works.

Set $y=x/ϵ$. In a two-scale expansion for periodic conductivity, differentiation becomesThe leading, order ${ϵ}^{-2}$, equation isFor fixed $x$, multiply by $u_0$ minus its cell mean and integrate over the periodic cell. Opposite-face terms cancel, givingHence $u_0$ is independent of $y$.

The order ${ϵ}^{-1}$ equation is nowFor each coordinate direction, choose a periodic zero-mean corrector $w_j$ satisfying the periodic conductivity cell problemThe right side has zero cell mean. In weak form,for every periodic test function $\varphi$. On the space of zero-mean periodic functions, the Poincaré inequality and uniform ellipticity give coercivity, so the corrector exists and is unique by the Lax-Milgram theorem. Discontinuous phase coefficients are handled by this weak formulation.

The order ${ϵ}^{-1}$ solution iswhere the last term is independent of $y$ and does not affect leading flux. Thus the leading microscopic flux isAt order one the equation readsAverage over $Y$. The second term integrates to zero by periodicity, and $|Y|=1$. ThereforeThe effective tensor is constant, so the last equation is exactly $-a_{ij}^{\ast }{\partial }_{x_i}{\partial }_{x_j}{u}_0=f$. The macroscopic boundary condition is $u_0=0$ on $\partial \Omega$. Higher bulk correctors usually do not individually vanish on the physical boundary; a boundary corrector in periodic homogenization handles this without changing the leading homogenized equation.

For symmetric $a$, a useful check on the effective conductivity isThe extra term involving $\nabla w_i$ vanishes by the weak cell equation. This formula proves symmetry. With $w_{\xi }={\xi }_j{w}_j$, it also gives $\xi \cdot a^{\ast }\xi \ge \alpha {\int }_Y|\xi +\nabla w_{\xi }{|}^2\ge \alpha |\xi {|}^2$, confirming that the effective tensor is positive-definite.

A hyperbolic Riemann surface in potential theory is a noncompact Riemann surface admitting a positive Green function on a Riemann surface. With pole $p$, this is a positive function harmonic off $p$, having local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and obtained as the minimal such positive function by exhaustion. This is a noncircular definition for the present proof. For a simply connected surface it is equivalent to disk conformal type; for arbitrary surfaces potential-theoretic hyperbolicity and disk universal-cover type should be distinguished.

The unit disk is an example. At pole zero its function is $G(z,0)=-\log |z|$: it is positive in the punctured disk, harmonic there, has the correct logarithmic singularity, and tends to zero at the boundary. If another positive function has the same singularity and is harmonic elsewhere, subtract $G$ and apply the maximum principle for harmonic functions on disks of radius $r<1$; the boundary lower bound is $\log r$. Letting $r\uparrow 1$ shows that the other function dominates $G$. Thus $G$ is indeed the positive Green function on a Riemann surface. At a general pole it is

Here is a Green-function exhaustion proof of disk uniformization. The main analytic idea is to build conformal maps on bordered disks from their Green functions and pass to an extremal limit. The positive global Green function prevents that limit from becoming constant.

First exhaust the noncompact simply connected surface $R$ by relatively compact domains $R_n$ containing $p$, each with a smooth Jordan boundary. This topological step does not use conformal uniformization: take regular neighborhoods of finitely many coordinate disks covering successive compact sets and fill bounded complementary components. Simple connectedness and the Jordan curve theorem make the filled neighborhoods topological disks; enlarge them to be nested.

Each $R_n$ has a zero-boundary Green function on a Riemann surface $G_n(\cdot ,p)$. One analytic construction, independent of conformal uniformization, is as follows. Equip the compact bordered domain with a smooth conformal metric. Multiply $-\log |\zeta |$ by a smooth cutoff function equal to one near $p$ and zero near the boundary; call the result $s$. Distributionallywhere $k$ is smooth and supported away from $p$. Solve $\Delta v=k$ with zero boundary data. This auxiliary Dirichlet problem can be obtained by minimizing $\frac{1}{2}\int |\nabla v{|}^2+\int kv$ over functions with zero boundary trace: the Poincaré inequality gives coercivity, minimization in a Hilbert space gives a weak solution, and interior and smooth-boundary elliptic regularity give a smooth solution. Then $G_n=s+v$ is harmonic away from $p$, has the logarithmic pole, and vanishes at the boundary. It is positive by the maximum principle for harmonic functions, applied after removing a sufficiently small circle around its positive pole. This construction uses ordinary linear elliptic existence, not the desired uniformization theorem. Equivalently, the same bordered-domain Green functions can be constructed by the Perron method for the Dirichlet problem using local logarithmic barriers.