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For the corrected bounded-domain problem, the interior theorem for the Perron method for the Dirichlet problem gives a classical solution of the Poisson equation, and the power barrier for an exterior cone argument gives its continuous extension with the required Dirichlet boundary condition. Hence existence holds in $C^2(\Omega )\cap C^0(\overline{\Omega })$.

If $u_1,u_2$ are two such classical solutions, then $w=u_1-u_2$ is a harmonic function with zero boundary trace. The weak maximum principle for elliptic operators, applied to $w$ and $-w$ on the bounded domain, gives $w\le 0$ and $-w\le 0$. ConsequentlyThere is exactly one solution after adding boundedness of $\Omega$. For the printed unbounded-domain version, the half-plane counterexample in part (i) disproves uniqueness; an additional condition at infinity or a suitable restricted solution class would instead have to be specified.

Choose a smooth cutoff function $\eta$ supported in $B_2$, equal to one on $B_1$, with $0\le \eta \le 1$ and $|D\eta |\le C$. For $i<n$, put $v_i=D_{i}u$. The zero Dirichlet boundary condition implies that the tangential boundary derivative $v_i$ vanishes on the flat face. Differentiating the Poisson equation gives $\Delta v_i=D_{i}f$. Test this equation with ${\eta }^2{v}_i$. The boundary terms vanish because $v_i=0$ on the flat face and $\eta =0$ near the curved face. Integration by parts in the tangential direction moves $D_i$ off $f$, yieldingThis is the essential Caccioppoli inequality step: it involves $f$, rather than $Df$. Expanding $D_i({\eta }^2{v}_i)$ and using the Cauchy-Schwarz inequality and the Young inequality givesThe term $f\eta (D_i\eta )v_i$ is bounded by $C{f}^2+C|D\eta {|}^2{v}_i^2$. Absorbing the first term and summing over $i<n$ proves the hinted estimate:All mixed second partial derivatives are now controlled, since $D_{ni}u=D_{in}u$. The remaining normal derivative is recovered from the equation:The Cauchy-Schwarz inequality bounds its squared L2 norm by $n$ times the sum of the squared L2 norms on the right. Including the already controlled zeroth and first partial derivatives, and taking square roots, gives the second-derivative estimate at a flat Dirichlet boundary:The full second-order Sobolev norm is controlled without any norm of $Df$.

For the self-gravitating incompressible slab, let the fluid occupy $-H<z<H$, with mass density ${\rho }_0=\Sigma /(2H)$ and a vacuum exterior. Poisson equation and hydrostatic equilibrium giveA constant ambient pressure could be added without changing the result. Choose perturbations proportional to $e^{ikx-i\omega t}$, using rotational symmetry in the horizontal plane to put the wavevector along $x$.

For the fluid displacement, the linearized momentum equation inside the uniform slab is $-{\omega }^2\mathbf{\xi }=-\nabla (p^{\prime }/{\rho }_0+{\varphi }^{\prime })$. For $\omega \ne 0$ it follows that $\mathbf{\xi }=\nabla \chi$, with $p^{\prime }/{\rho }_0+{\varphi }^{\prime }={\omega }^2\chi$. Incompressibility gives $\nabla \cdot \mathbf{\xi }=0$, henceNeutral limits of these surface modes follow by continuity. Stationary vorticity perturbations form a separate zero-frequency sector; the nonzero-frequency surface modes are irrotational by the momentum equation itself, without an extra irrotational-flow assumption.

The Eulerian mass density perturbation vanishes in the bulk because the equilibrium mass density is uniform and the displacement is divergence free. If the two vertical surface displacements are ${\eta }_{+}={\xi }_z(H)$ and ${\eta }_{-}={\xi }_z(-H)$, the surface density perturbation of a displaced uniform interface givesThus Poisson equation reduces to Laplace equation for ${\varphi }^{\prime }$ inside and outside. The potential is continuous, while integrating through each surface givesThe perturbing potential must decay at vertical infinity. The kinematic conditions are ${\eta }_{\pm }={\partial }_z\chi (\pm H)$. The free surface pressure condition is the vanishing Lagrangian pressure perturbation, $p^{\prime }+{\eta }_{\pm }{p}_0^{\prime }=0$; at the top this gives $p^{\prime }(H)/{\rho }_0=g_s{\eta }_{+}$.

Put $q=kH>0$. For even scalar-potential symmetry, chooseThen ${\eta }_{-}=-{\eta }_{+}$: this is a thickness-changing mode, not a vertical bending mode. Write $\eta ={\eta }_{+}$; the top derivative jump gives $-kC(\cosh q+\sinh q)=4\pi G{\rho }_0\eta$. Since $\cosh q+\sinh q=e^q$,Substitute these values into ${\omega }^2\chi (H)=g_s\eta +{\varphi }^{\prime }(H)$. Using $4\pi G{\rho }_0=g_s/H$ gives the even-mode dispersion relation for the surface modes of a self-gravitating incompressible slab:The positive term comes from the restoring hydrostatic free-surface pressure in the background gravitational field. The negative term is the attraction due to the perturbed gravitational potential of the displaced surfaces, which tends to reinforce the disturbance. Both effects ultimately involve the slab's gravity, but enter the boundary condition through different terms.

For odd scalar-potential symmetry, instead takeNow ${\eta }_{-}={\eta }_{+}$, so the surfaces bend in the same direction. The jump again gives $C=-4\pi G{\rho }_0\eta /(k{e}^q)$, but $\chi (H)=\eta \tanh q/k$ and ${\varphi }^{\prime }(H)=-4\pi G{\rho }_0\eta /[k(1+\coth q)]$. The same pressure condition yieldsThe bottom kinematic, jump and pressure conditions follow with the indicated parity and give the same dispersion relation, so no boundary condition has been discarded.

To determine the gravitational instability of an incompressible slab, factor the two dimensionless expressions asFor the even mode the bracket is strictly increasing, with derivative $1+e^{-2q}>0$, is negative at zero and tends to positive infinity. It therefore has a unique positive zeroThe even mode is unstable for $0<kH<q_c$, neutral at $q_c$, and oscillatory for $kH>q_c$. Negative ${\omega }_e^2$ gives a growing mode with $\omega =i\sqrt{-{\omega }_e^2}$ under the stated time convention. For the odd mode, $1-e^{-2q}<2q$ for every $q>0$, so the bracket is always positive: all odd modes with positive wavenumber are stable.

At long wavelengths $F_e=-q+2q^2+O(q^3)$ and $F_o=q-2q^2/3+O(q^3)$: a thickness or column-density disturbance has a gravitational instability, whereas a bending disturbance is restored. At short wavelengths both approach $q-1/2$, so both surface branches are stable. The slab equilibrium is therefore linearly unstable overall, because it admits the long-wavelength even modes despite the stability of its odd sector.

Let $L=\Delta +c$ with real $c$, and letThe Fredholm alternative for an elliptic Dirichlet problem says that $N$ is finite-dimensional. If $N=0$, all the prescribed data admit a unique solution. In general the exact boundary compatibility in the self-adjoint Fredholm alternative isThese conditions are necessary and sufficient; when they hold, all solutions form $u_0+N$. Thus in the second branch some data are incompatible, while compatible data have nonunique solutions. The outward normal derivative fixes the displayed sign.

To prove the alternative, put $X=\{w\in C^{2,\alpha }(\overline{\Omega }):w{|}_{\partial \Omega }=0\}$ and $Y=C^{0,\alpha }(\overline{\Omega })$. The standard zero-boundary Poisson equation theorem makes $\Delta :X\to Y$ an isomorphism. Multiplication by $c$, followed by ${\Delta }^{-1}$, is a compact operator $K:X\to X$: the embedding $C^{2,\alpha }\hookrightarrow C^{0,\alpha }$ is compact, and multiplication is bounded. Writing $u=\psi +w$ givesThe Fredholm alternative for a compact operator implies finite kernel and cokernel of equal dimension, and invertibility exactly when the kernel vanishes.

For the explicit compatibility conditions, use the self-adjoint Dirichlet realization of $L$ on $L^2$. It has compact resolvent: a sufficiently large positive shift of $-L$ is coercive, its inverse gains two derivatives, and the Rellich-Kondrachov compactness theorem makes that inverse compact. The Fredholm solvability condition for a self-adjoint operator is $F\perp N$. Standard boundary elliptic regularity upgrades the resulting weak solution to $C^{2,\alpha }$ for these data and boundary. Thus it gives the same kernel and solvability as the preceding Hölder-space formulation.

Finally Green second identity, with $z=0$ on the boundary and $Lz=0$, givesConsequently $\int zF=\int zf+\int \psi {\partial }_{\nu }z$, proving both necessity and sufficiency with the stated sign. If using complex data and real $c$, replace $z$ by $\overline{z}$ in the pairings.

Let the free surface be at $r=R_1$ and take the Newtonian gravitational potential to vanish at infinity. Mass conservation fixes $R_1=(3M_1/(4\pi {\rho }_1){)}^{1/3}$. The enclosed mass is $m(r)=4\pi {\rho }_1{r}^3/3$, and hydrostatic equilibrium requires $d{p}_0/dr=-{\rho }_{1}Gm(r)/r^2$. With vacuum outside, $p_0(R_1)=0$, givingThe pressure is zero outside. A prescribed constant external pressure would simply add that constant to $p_0$.

The bulk Eulerian and Lagrangian fluid perturbations obey $\delta \rho =0$ because the mass density is uniform and the Lagrangian displacement has zero divergence. The linearized Euler momentum equation in the interior isTaking the curl gives ${\omega }^2\nabla \times \mathbf{\xi }=0$. Thus every nonzero-frequency normal mode has an irrotational flow displacement; the simply connected space formed by the interior admits $\mathbf{\xi }=\nabla U$ and incompressibility gives the Laplace equation ${\nabla }^{2}U=0$. For a regular center, one spherical harmonic component is $U=A_l{r}^l{Y}_l^m$, with $|m|\le l$.

It is essential to retain the surface gravity perturbation of a uniform-density sphere. Although the bulk mass density perturbation is zero, the displaced density discontinuity givesThe Poisson equation implies that $\delta \Phi$ is continuous and its outward radial derivative has the jumpUsing regularity at the center and decay at infinity, writeThe derivative jump is $-(2l+1)C_l{R}_1^{l-1}{Y}_l^m$, so $C_l=-4\pi G{\rho }_{1}l{A}_l/(2l+1)$.

Integrating the bulk Euler momentum equation gives $\delta p/{\rho }_1={\omega }^{2}U-\delta \Phi$ for $l\ge 1$. At the free surface the Lagrangian pressure perturbation vanishes:Combining these relations yields the Kelvin stellar mode frequencyThe $2l+1$ choices of $m$ are degenerate because of spherical symmetry. For $l=1$, $r{Y}_1^m$ is a linear combination of coordinates in a Cartesian coordinate system, so $\nabla U$ is a constant displacement: the zero frequency is rigid translation of the whole isolated body. Such a translation has no restoring force. The constant $l=0$ potential generates no displacement; a radial breathing motion is excluded by incompressibility and a regular center.

The printed irrotational assertion needs the nonzero-frequency qualification, or restriction to the potential incompressible stellar surface modes. It is false for all neutral displacements: $\mathbf{\xi }=\mathbf{\Omega }\times \mathbf{r}$ has zero divergence, zero normal displacement at the surface, and $\delta p=\delta \Phi =0$, so it is a zero-frequency displacement, but $\nabla \times \mathbf{\xi }=2\mathbf{\Omega }\ne 0$. This does not change the Kelvin stellar mode spectrum just derived.

For a time-independent current, the Maxwell equations reduce to magnetostatics:Write $\mathbf{B}=\nabla \times \mathbf{A}$ using a vector potential and choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. The vector identity for the curl of a curl gives the Poisson equationThe free-space Green function therefore givesTaking the curl with respect to $\mathbf{r}$, moving it under the integral, and using $\nabla (1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ gives the Biot-Savart law

As printed, the hypotheses are insufficient. They order the barriers only on the boundary; the intended monotone iteration for a semilinear elliptic equation needs an ordered subsolution and supersolution throughout the domain. On $B_1$, take a positive first Dirichlet Laplacian eigenfunction $e_1$ with $-\Delta e_1={\lambda }_1{e}_1$, and setBoth barriers satisfy $Q{\phi }^{\pm }=0$ and vanish on the boundary, but ${\phi }^{-}>{\phi }^{+}$ inside. No function lies between them. This also disproves the unconditional conclusions in (b) and (d).

With the intended extra hypothesis ${\phi }^{-}\le {\phi }^{+}$ on $\overline{\Omega }$, solvability of the Poisson equation and the Schauder estimate give a unique $u_1\in C^{2,\alpha }$ with forcing $V({\phi }^{-})$ and boundary data $\psi$. Smooth $V$ composed with the $C^{0,\alpha }$ barrier has the required forcing regularity. The barrier inequalities giveBoth differences have nonnegative boundary values. The weak maximum principle for elliptic operators applied to their negatives gives $\boxed{{\phi }^{-}\le u_1\le {\phi }^{+}}$ under the corrected hypothesis.

Use the interior ordered subsolution and supersolution correction from part (a). Recursively solve the Poisson equation $\Delta u_{k+1}=V(u_k)$ with boundary value $\psi$, starting from $u_0={\phi }^{-}$. Part (a) gives $u_0\le u_1\le {\phi }^{+}$. If $u_{k-1}\le u_k\le {\phi }^{+}$, thenThe first difference has zero boundary values, the second nonnegative values. The weak maximum principle for elliptic operators applied to their negatives gives $u_k\le u_{k+1}\le {\phi }^{+}$. Induction proves the bounded monotone sequenceWithout interior ordering, the counterexample in part (a) disproves the claim.

The free-space Green function of the Laplacian for the Poisson equation gives the electric potentialFor $r=|\mathbf{x}|\gg R$, its electric multipole expansion begins withDefine the total electric charge and electric dipole moment byThen

The analogous solution for the magnetic vector potential isIts apparent $r^{-1}$ coefficient is ${\mu }_0(4\pi {)}^{-1}\int \mathbf{J}{d}^3{x}^{\prime }$. For each component, the supplied identity and the divergence-free vector field condition giveThe divergence theorem turns this into a surface integral outside the compact support of the current density, where $\mathbf{J}=0$. Hence $\int \mathbf{J}{d}^3{x}^{\prime }=0$, so the $1/r$ term vanishes.

For $w\in X_{\delta }$, hypothesis (2) gives $F(w)\in C^{0,\alpha }(\overline{\Omega })$. Thus $f-F(w)$ is Hölder continuous and the classical Dirichlet problem for the Poisson equation has a unique solution $u=T(w)\in C^{2,\alpha }(\overline{\Omega })$. The global Schauder estimate givesThe maximum estimate for the Poisson problem also givesThese estimates prove that $T$ is well defined and give the requested norm control.

In the Newtonian limit of general relativity, time derivatives are negligible and $T_{00}=\rho$ is the only leading component. With signature $(-,+,+,+)$,The $00$ component of the trace-reversed equation becomesIdentifying $h_{00}=-2\Phi$ therefore gives the Poisson equation of Newtonian gravity:

The nonrelativistic fluid assumptions giveAssume again a stationary field and asymptotic flatness. The harmonic-gauge Linearized Einstein equations becomewhile the other trace-reversed components are smaller. Trace reversal then gives $h_{00}={\bar{h}}_{00}/2$ to leading order, soUsing $h_{00}=-2\Phi$ yieldsin units $G=1$, which is the Poisson equation of Newtonian gravity.

In a static spherical star, hydrostatic equilibrium and the Poisson equation giveor equivalently $r^{-2}d(r^{2}g)/dr=4\pi G\rho$.

Let $\mathbf{\xi }$ be the fluid displacement field, so the velocity perturbation is $\delta \mathbf{u}={\partial }_t\mathbf{\xi }$. Linearizing the ideal-fluid momentum equation about the static state and cancelling the background hydrostatic terms givesConservation of mass says that the Lagrangian density perturbation is ${\Delta }_L\rho =-\rho \nabla \cdot \mathbf{\xi }$. The relation between Eulerian and Lagrangian fluid perturbations and adiabatic compression gives, with $\Delta =\nabla \cdot \mathbf{\xi }$,while linearized self-gravity gives $\boxed{{\nabla }^2\delta \Phi =4\pi G\delta \rho }$.

For the stated spherical harmonic displacement, the radial divergence is $r^{-2}d(r^2{\tilde{\xi }}_r)/dr$, while $\nabla {\tilde{\xi }}_h$ is radial and orthogonal to the angular gradient of $Y_l^m$. HenceEquating the radial and horizontal coefficients of $Y_l^m$ and $\nabla Y_l^m$, and applying the separated Laplacian in spherical coordinates to $\widetilde{\delta \Phi }(r)Y_l^m$, gives

Define the stellar buoyancy frequency byEliminating $\tilde{\Delta }$ between the density and pressure perturbations givesSubstitution in the radial equation, followed by use of $dp/dr=-\rho g$, yieldsRegular spherical profiles near the center have $\rho ={\rho }_c+O(r^2)$ and $p=p_c+O(r^2)$, while $m(r)=4\pi {\rho }_c{r}^3/3+O(r^5)$ and hence $g=O(r)$. Both logarithmic gradients in the definition of $N^2$ are $O(r)$, so $N^2=A{r}^2+O(r^4)$. A Sun-like radiative central stratification is stable, making $A>0$.

Write physical position as $\mathbf{r}=a(t)\mathbf{x}$ and physical velocity asSubtract the homogeneous expanding background from the pressureless Euler equation. To linear order in peculiar velocity and density contrast, the convective term is negligible andorHere $\nabla$ differentiates with respect to comoving position and $\Phi$ is the peculiar gravitational potential.

The linear continuity and Poisson equations areFor the growing mode $\delta (\mathbf{x},t)=D(a){\delta }_i(\mathbf{x})$, matter conservation gives ${\bar{\rho }}_m\propto a^{-3}$, so the potential scales asafter absorbing the fiducial normalization into ${\Phi }_i$. Integrating the linear Euler equation with a negligible decaying mode gives

Taking the divergence of Euler and using continuity and Poisson yields the linear growth equationIn the fiducial normalization used in the question this can be rearranged asThe cosmological Lagrangian displacement obeys $\dot{\mathbf{\psi }}=\mathbf{v}/a$. Combining the last two relations and choosing the growing displacement gives the Zeldovich approximation

DefineThen $\mathbf{x}={\mathbf{x}}_i-b\nabla {\Phi }_i$ and $\mathbf{v}=-a\dot{b}\nabla {\Phi }_i$. Keeping the lowest nonvanishing order in displacement in the halo angular momentum givesThe angular momentum vanishes for a spherical Lagrangian patch, and also whenever the patch's inertia principal axes align with the local tidal-field principal axes. It likewise vanishes in a locally isotropic tidal field.

Taylor-expandThe constant-gradient term vanishes by the barycentre definition. Withone obtains the tidal torque theory result$I$ is the protohalo inertia tensor and $T$ is the local tidal, or gravitational Hessian, tensor. Their eigenframe misalignment produces the torque.

In an Einstein-de Sitter universe, $D\propto a\propto t^{2/3}$. Hence $\dot{b}\propto \dot{a}\propto t^{-1/3}$ andThis linear estimate is normally stopped near turnaround. Simulated haloes subsequently gain angular momentum through nonlinear torques, anisotropic accretion, and mergers, including the orbital angular momentum of infalling subhaloes, so the early linear estimate underpredicts the final value.

In the homogeneous background, $\mathbf{v}=0$ and the physical velocity is the Hubble flowThe comoving coordinate $\mathbf{x}$ is held fixed by the time derivative in the fluid equation, soMoreover, spatial homogeneity gives $\nabla \bar{P}=0$. The Euler equation therefore reduces toTaking the divergence in three spatial dimensions givesConsistency with the Poisson equation,then requiresThis is the pressureless, zero-cosmological-constant Raychaudhuri equation, obtained by the Newtonian fluid derivation of the Raychaudhuri equation.

The elementary Newtonian derivation of the Friedmann equation follows a test particle on the boundary of a finite expanding sphere. It therefore introduces a distinguished centre and can misleadingly suggest that all matter expands away from one origin.

The fluid derivation is local: the Euler equation and Poisson equation must be compatible in every comoving neighbourhood. It consequently describes a genuinely homogeneous infinite universe, in which every comoving observer sees the same Hubble flow, without selecting a physical centre. This is the improvement captured by the Newtonian fluid derivation of the Raychaudhuri equation.

Freeze the lower-order coefficients on a ball and apply the interior Schauder estimate for the Poisson equation to a cutoff of $u$. The terms $b_i{D}_{i}u+cu$ are controlled by the Hölder interpolation inequalityChoosing $\eta$ in terms of the prescribed $\delta$ yields

The standard iteration lemma for nested balls absorbs the first term. The remaining $C^2$ norm is bounded by the interior derivative estimate and interpolation, producingEquivalently, a contradiction-and-rescaling proof would produce a globally Hölder harmonic limit forbidden by the polynomial-growth Liouville theorem for harmonic functions.

Spherical symmetry reduces the Poisson equation for gravity toAfter one integration,The omitted integration constant would represent a point mass and is absent for the dark-matter configuration alone. Choosing the additive constant so that the potential tends to zero at large radius gives

Write $\mu =GM$ and $s=\sqrt{b^2+r^2}$. The gravitational potential of the spherical isochrone model obeysThe spherical Poisson equation therefore givesIts small-radius asymptotic expansion isso the model has a finite-density core. At large radius,The $r^{-4}$ envelope has finite total mass $M$.