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Set $U=R(r)\Theta (\theta )$ in the Laplace equation, multiply by $r^2/U$, and separate:Putting $x=\cos \theta$ turns the angular equation into the Legendre differential equation. For a solution smooth on the whole spherical axis, the allowed eigenvalues are $\lambda =n(n+1)$ with angular functions $P_n(\cos \theta )$. The radial equation has the two solutions $r^n,r^{-n-1}$. Superposition yields the regular axisymmetric expansionThe coefficients and convergence region are set by the domain and boundary data. Without regularity at the poles, singular Legendre functions and more general separated degrees would also be possible; they are excluded for this exterior spherical field.

The far-field matching fixes $A_1=v_0$ and all other growing coefficients to zero. The radial Neumann boundary condition at $r_0$ isIndependence of the angular modes gives $B_1=v_0{r}_0^3/2$ and all other decaying coefficients zero. Thus the exterior harmonic potential with no flux through a sphere isIts derivative vanishes on the sphere and its difference from $v_{0}r\cos \theta$ tends to zero at infinity. A spatial constant may be added if the far-field condition is intended only to specify the gradient or the leading growing term; matching the potential difference to zero fixes that constant to zero.

For a defining function $\varphi$ with $d\varphi \ne 0$, the characteristic hypersurface test is that the principal symbol vanish at $d\varphi$.

For the wave equation with speed $c>0$,Thus its characteristic hypersurfaces satisfy ${\varphi }_t^2=c^2|{\nabla }_x\varphi {|}^2$. In one space dimension the two families are $x\pm ct=\text{constant}$; cones are characteristic away from their vertices.

For the free Schrodinger equation, in normalized units,Its total-order characteristic hypersurfaces satisfy ${\nabla }_x\varphi =0$. Their normal is purely temporal, so locally they are constant-time hypersurfaces. Multiplying the equation by a nonzero constant or choosing the opposite sign convention does not change this test.

For the Laplace equation,There are no real characteristic hypersurfaces for the Laplace equation, since their normal cannot be zero. These statements concern the ordinary total-order principal symbol, not a weighted space-time grading.

The interior elliptic regularity assertion for the Laplace equation is that a harmonic function is smooth, in fact real analytic, throughout $U$. No boundary regularity of its unspecified boundary values is implied.

First let $u\in C^2(U)$ and $\Delta u=0$. On a ball compactly contained in $U$, differentiating its spherical average and applying the divergence theorem expresses that derivative as a constant factor times ${\int }_{B_r}\Delta u$, which is zero. The spherical average tends to $u$ at the center as $r\to 0$. This proves the mean value property for harmonic functions.

Choose a radially symmetric smooth mollifier ${\eta }_r$, supported in $B_r$, with integral one. By integrating the spherical mean value property,whenever $B_r(x)⋐U$. For fixed $r$ the right side is a smooth convolution, since all derivatives can be placed on ${\eta }_r$. Thus $u$ is smooth. The same argument proves the Weyl lemma for a distributionally harmonic $u\in L_{\mathrm{loc}}^1$: first mollify $u$, apply the fixed-radius identity, and let the mollification radius tend to zero in distributions to obtain the same smooth representative.

To prove real analytic regularity, differentiating the fixed-radius convolution gives the interior derivative estimate for a harmonic functionfor any harmonic $v$. All derivatives of $u$ are harmonic. On nested balls between $B_R(a)$ and $B_{R/2}(a)$, apply this estimate $k$ times, decreasing the radius by $R/(2k)$ each time. For $|\alpha |=k$,The inequality $k^k\le e^{k}k!$ follows by integrating $\log s$ below the sum defining $\log (k!)$. Apply the one-dimensional Taylor theorem along each segment, expanding directional derivatives by the multinomial formula. The remainder is bounded by $M(A{\sum }_j|h_j|{)}^k$, so it tends to zero for sufficiently small $h$. This gives a locally convergent multivariate Taylor series. A harmonic function is real analytic in the interior.

The usual inhomogeneous elliptic regularity statement also follows: if $\Delta u=f$ and $f$ is smooth, take a cutoff $\chi$ equal to one near a given point and set $w=\Phi \ast (\chi f)$, where $\Phi$ is a fundamental solution of the Laplace equation with $\Delta \Phi =\delta$. Moving every derivative to the compactly supported smooth function $\chi f$ shows $w$ is smooth. Locally $u-w$ is harmonic, so $u$ is smooth there too.

The Cauchy problem for a partial differential equation here prescribes both the value $u{|}_{\Gamma }=g$ and the normal derivative ${\partial }_{n}u{|}_{\Gamma }=h$, together with $\Delta u=0$. Only one of these traces would be boundary data for a usual elliptic boundary problem, rather than full Cauchy data.

Every real hypersurface is a non-characteristic hypersurface for the Laplace equation. In local real analytic coordinates flattening the real analytic hypersurface $\Gamma$, the coefficient of the second transverse derivative is nonzero: its principal coefficient is the squared length of the conormal. The equation can therefore be solved for that second derivative. The normal derivative data determine the transverse first derivative, because the coefficient relating them is nonzero and the tangential first derivatives are already determined by $g$.

The coefficients, flattened Cauchy data, and coordinate change are all real analytic. The Cauchy-Kovalevskaya theorem applies, giving a unique local real analytic solution around each point of $\Gamma$. This is a local existence assertion, not a claim of stable dependence in arbitrary Sobolev space norms.

The Cauchy-Kovalevskaya theorem cannot be applied to merely $C^2$, non-$C^3$ Cauchy data. It requires real analytic data.

There is also no $C^2$ solution of the Laplace equation on a neighborhood of a point where one of these prescribed traces fails to be $C^3$. By interior elliptic regularity, any such solution would be smooth and real analytic. On the real analytic hypersurface retained from the preceding part, both its restriction and its normal derivative would then be real analytic, hence $C^3$. This contradicts the prescribed trace. There is no solution on a neighborhood of all of $\Gamma$ with the stated non-$C^3$ data. This does not exclude solutions near other points where the data happen to be real analytic.

Put $\mathbf{x}=r\mathbf{\omega }$, with $|\mathbf{\omega }|=1$. The Jacobian determinant in spherical coordinates gives $dS=r^{2}d\omega =r^2\sin \theta d\theta d\phi$, soDifferentiation under this fixed-domain integral yieldsThe last step is the divergence theorem and the Laplace equation. Hence $f$ is constant on radii for which the closed ball lies inside $V$. Continuity at the origin gives ${\lim }_{r\to 0}f(r)=\varphi (0)$, proving the mean value property for harmonic functions:

The spatial factor $r^l{Y}_l^m$ is a regular solid harmonic. Besides the Laplace equation ${\nabla }^{2}F=0$, homogeneity gives $\mathbf{r}\cdot \nabla F=lF$. The divergence of the fluid displacement isDefine the scalar dilation amplitude $D_r=r{U}^{\prime }+(l+3)U+(l/r)V^{\prime }$; this is the quantity denoted $\Delta$ in the question, not the notation ${\Delta }_L$ for a Lagrangian pressure perturbation. Then mass conservation gives $\hat{\rho }=-\rho D_r$.

The equilibrium pressure gradient and homogeneity identity giveConsequently the adiabatic equation of state yieldsFor the force equations, the product rule givesEquating the coefficients of $F{\mathbf{e}}_r$ and $\nabla F$ in the self-gravitating adiabatic displacement equations givesFinally, applying the Laplacian to the gravitational perturbation givesso the Poisson equation becomesThese are the required interior equations for uniform-density stellar oscillation.

There is a radial degeneracy at $l=0$: $F$ is spatially constant and $\nabla F=0$, so the displacement is independent of $V$. The second force equation then cannot be inferred by equating independent vectors. For nonzero $\omega$ it may be imposed as an auxiliary definition of $V$, but it is not an additional physical radial equation. At zero frequency the radial equations should be used directly. This distinction matters for interpreting the zero factor in (c).

Use the Wirtinger derivative $q_z=(q_x-i{q}_y)/2$ and write $z=x+iy$. Since $q$ solves the Laplace equation, $q_z$ is a holomorphic function. The transforms in the question select solutions with sufficient decay at infinity. In this class the solution, when it exists, is unique; hence reflection of the symmetric data about $y=\ell /2$ givesFor $\gamma >0$, this uniqueness follows directly from Green's first identity: the homogeneous problem has integral $\int |\nabla q{|}^2+\gamma {\int }_{y=0,\ell }{q}^2=0$. For arbitrary real $\gamma$, the decaying-class qualifications are explained in part (c). Without a condition at infinity, symmetry of the data alone would not force symmetry of every solution.

Put $c=q(0,0)$, $f(x)=q(x,0)$, and introduce a known transformThe bottom Robin boundary condition gives $q_y=\gamma f$, while the top one gives $q_y=-\gamma f$. The bottom side is traversed from infinity to zero. Integration by parts therefore givesThe top side is traversed from $i\ell$ towards infinity, and its exponential supplies $e^{k\ell }$, givingOn the vertical side, $dz=idy$ and the prescribed Neumann boundary condition is $q_x=g$. ThusBoth the side orientations and the factor $1/2$ from the Wirtinger derivative are essential to these signs.

To recover the printed dispersion relation, interpret the surface maintained flat as an imposed flat upper boundary for the perturbation, so its normal velocity is zero. This is the rigid-lid approximation. Work with perturbation velocity potentials ${\varphi }_{-}=A{e}^{ky}{e}^{ikx+\sigma t}$ below the interface and ${\varphi }_{+}=B\cosh [k(y-h)]e^{ikx+\sigma t}$ in the upper layer. They solve the Laplace equation, decay as $y\to -\infty$, and obey ${\partial }_y{\varphi }_{+}(h)=0$.

The linear kinematic boundary conditions at the interface give $(\sigma +ikU){\eta }_0=kA$ and $\sigma {\eta }_0=-kB\sinh (kh)$. Equality of pressures, using the linear Bernoulli equation, gives $(\sigma +ikU)A=\sigma B\cosh (kh)$. The equal fluid densities cancel the interface gravitational restoring term. Eliminating $A,B,{\eta }_0$ yieldsWith $q=\tanh (kh)$, the roots areFor every $k>0$ one root has positive real part. Thus the vortex sheet has a Kelvin-Helmholtz instability at every wavelength in this model. If $kh\gg 1$, $q\to 1$ and the growing rate is $Uk/2$. If $kh\ll 1$, $q\sim kh$ and it is $Uk\sqrt{kh}$.

There is a physical qualification to the wording. A dynamically deformable free surface at finite gravity is not exactly a flat lid. Its linear conditions combine to ${\sigma }^2{\varphi }_{+}(h)+g{\partial }_y{\varphi }_{+}(h)=0$. Writing the upper potential as $B\cosh (ky)+C\sinh (ky)$ and using the same interface conditions instead givesThe stated quadratic follows in the strong-gravity flat-surface limit, or with the flat condition imposed exactly. It is not the exact dispersion relation of a freely moving upper surface at arbitrary finite $g$. This makes explicit the boundary interpretation needed for the requested formula.

Use the standard ideal-fluid assumptions of incompressible flow, inviscid flow, gravity $-g{\mathbf{e}}_y$, and no surface tension. Irrotational flow gives ${\mathbf{u}}_j=\nabla {\phi }_j$ in each layer, and incompressibility gives Laplace equationThe rigid-wall boundary conditions are ${\phi }_{1y}=0$ at $y=h_1$ and ${\phi }_{2y}=0$ at $y=-h_2$. At the moving interface, each fluid has the same normal velocity as the interface:The dynamic condition is continuity of pressure, $p_1=p_2$ at $y=\eta$. Bernoulli equation in each fluid reads ${\phi }_{jt}+\frac{1}{2}|\nabla {\phi }_j{|}^2+p_j/{\rho }_j+gy=C_j(t)$. Choosing the potential gauges to subtract a common reference pressure, the interface condition becomesTangential velocities need not agree across an inviscid interface.

Linearize about the flat interface at rest, $\eta =0$, ${\phi }_j=0$. Both potentials still satisfy Laplace equation, now on the fixed strips $0<y<h_1$ and $-h_2<y<0$, with the same rigid-wall conditions. Discarding products of perturbations and evaluating the interface conditions at $y=0$ givesFor a normal mode proportional to $e^{i(kx-\omega t)}$, the vertical profiles satisfy ${\Phi }_j^{\prime \prime }-k^2{\Phi }_j=0$, the interface velocity is $-i\omega \eta$, and the wall-normal derivatives vanish. Smallness of the amplitude relative to wavelength and layer depths justifies this linearization.

For the bounded-domain correction, let $\mathcal{S}$ be the family of Perron subfunctions for the Poisson equation $v\in C^0(\overline{\Omega })$ with $v\le \phi$ on $\partial \Omega$. The subfunction condition is ball comparison: whenever $\overline{B}\subset \Omega$ and a classical solution $H$ of $\Delta H=f$ on $B$ satisfies $v\le H$ on $\partial B$, it also satisfies $v\le H$ throughout $B$. For a $C^2$ function this is the sign condition $\Delta v\ge f$. A superfunction reverses both inequalities.

The Perron solution is the pointwise upper envelope of the subfunctions:This is the Perron method for the Dirichlet problem with the Poisson equation replacing the homogeneous Laplace equation. The power barrier for an exterior cone supplies a global member of $\mathcal{S}$ and a global superfunction, as constructed below. The permitted subfunction-superfunction comparison therefore makes this envelope finite. The standard interior Perron theorem, available in the question, gives $u\in C^2(\Omega )$ and $\Delta u=f$; neither result needs to be proved in this part.

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.

The PDF specifies decay as $y\to \infty$; the local TeX drops this limit from its displayed boundary data. Take $a=4\pi$, so the conformal map is $w=\sin (z/4)$. Its components areFor $-2\pi <x<2\pi$ and $y>0$, $\eta >0$. The bottom edge maps monotonically to $(-1,1)$, and the two vertical sides map to the real rays $(-\infty ,-1)$ and $(1,\infty )$. The derivative $\cos (z/4)/4$ never vanishes in the interior. Injectivity follows from the identities for equal sines: the alternatives $z_1/4=z_2/4+2\pi n$ and $z_1/4=\pi -z_2/4+2\pi n$ give only the identical point within this strip. Surjectivity can be seen directly: for fixed $\xi$ and $v=y/4>0$, solve $\sin (x/4)=\xi /\cosh v$; thenOn $v>\max (0,\mathrm{arcosh}|\xi |)$, with the lower endpoint read as $0$ when $|\xi |\le 1$, the right side increases strictly from zero to infinity. Thus every point of the upper half-plane has a unique preimage.

By conformal invariance of harmonicity, write the transformed harmonic function as $\Phi (\xi ,\eta )$. Its real-axis boundary data areThe inverse sine on $(-1,1)$ is its real principal branch. For bounded integrable data, take the Fourier transform convention $\hat{\Phi }(k,\eta )={\int }_{\mathbb{R}}\Phi (\xi ,\eta )e^{-ik\xi }d\xi$. The Laplace equation becomes ${\hat{\Phi }}_{\eta \eta }-k^2\hat{\Phi }=0$, with the bounded decaying solution $\hat{\Phi }=\hat{F}(k)e^{-|k|\eta }$. Inverting the Fourier transform givesConsequently the Poisson kernel for the upper half-plane yieldsThe decay is uniform across the strip: $|w{|}^2={\sin }^2(x/4)+{\sinh }^2(y/4)\ge {\sinh }^2(y/4)$ and, for $|w|>1$, $|\Phi |\le \Vert F{\Vert }_{L^1}|w|/(\pi (|w|-1{)}^2)\to 0$. This verifies the printed decay condition. Boundary limits give the prescribed values at continuity points away from the corners.

For the given sine data,Values assigned at $s=\pm 1$ do not alter the integral. There is no solution continuous on the entire closed half-strip for these data: the bottom edge has corner limits $\pm 1$, but the side edges have corner limits zero. The formula is the bounded Dirichlet problem solution with the specified limits away from those two corners. This is the usual interpretation of the printed problem; boundedness excludes additional boundary-singular harmonic functions, while decay alone without a regularity or boundedness condition does not establish uniqueness.

The general-order Cauchy-Kovalevskaya theorem applies to analytic data on a non-characteristic hypersurface. In local coordinates $(t,z)$, a real analytic system of order $m$ in an unknown vector $U$ is in normal form whenwith $F$ real analytic near the initial jet. Analytic data ${\partial }_t^{j}U(0,z)=h_j(z)$ for $0\le j<m$ determine a unique real analytic solution near each initial point. More general systems, including different orders for the components, have the same conclusion after solving for the highest normal derivatives: the corresponding coefficient matrix, or highest-derivative Jacobian for a nonlinear system, must be invertible. This is the system's non-characteristic hypersurface condition. The uniqueness asserted by the Cauchy-Kovalevskaya theorem is initially uniqueness in the real analytic class.

A real analytic coordinate change flattens a real analytic initial hypersurface. Non-characteristicity permits solving for its highest normal derivatives by the real analytic implicit function theorem. Introduce all derivatives through order $m-1$ as additional unknowns. Their normal derivatives are either another jet variable, a tangential first derivative of a jet variable, or the right-hand side of the original equation. This gives a first-order real analytic system. The derivative-compatibility identities have zero initial data and are preserved by the system; equivalently, its real analytic coefficient recurrence reproduces the derivatives of $U$. This explains both reductions, rather than assuming arbitrary real analytic hypersurfaces are already flat.

For the first example, the principal symbol of the Laplace equation at the conormal $d{x}_1$ is $1$, so the plane is a non-characteristic hypersurface. The theorem therefore applies to real analytic $u{|}_{x_1=0}$ and ${\partial }_{1}u{|}_{x_1=0}$:This does not give Hadamard well-posedness in smooth or Sobolev space norms. For $n\ge 2$, the harmonic functionshave zero value data and normal derivative data $e^{-\sqrt{N}}\cos (N{x}_2)$ tending to zero with every fixed tangential derivative. At any fixed $x_1=\delta >0$, however, $u_N(\delta ,0)$ diverges. Thus the elliptic Cauchy problem for a partial differential equation is unstable despite real analytic solvability. The exceptional case $n=1$ is the elementary affine ordinary differential equation and has no such tangential high-frequency instability.

Fix the Riemann curvature tensor conventionFor the limiting affine connection, every product of two nonzero connection coefficients is zero, because its only nonzero coefficients have spatial upper index and both lower indices $t$. Its only possibly nonzero curvature components, apart from antisymmetry in the last two indices, areThe Ricci tensor is thereforeAll other components vanish, so the limiting connection is a connection with vanishing Ricci tensor exactly when $\varphi$ satisfies the Laplace equation on spatial Euclidean space. This is Ricci-flatness, not necessarily vanishing curvature: $\varphi =x^2-y^2$ is a harmonic function but has nonzero Hessian matrix and hence a nonflat limiting connection. Reversing the curvature sign convention reverses the displayed Ricci sign but leaves the equivalence unchanged.

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.

Interpret the malformed printed domain inequality as the first quadrant $x>0,y>0$, as indicated by the two axis boundary conditions. There is also a genuine missing hypothesis: decay of the boundary data does not require the harmonic function itself to be bounded. For zero data, both $q=0$ and $q=xy$ satisfy the Laplace equation and both edge values, but the Wirtinger derivative of the latter isThus the data do not determine an arbitrary solution's derivative without a growth restriction. We derive the integral representation for the bounded solution, which is also the standard decaying solution when the continuous edge data decay. Any other admissible unbounded solution can differ by a zero-boundary harmonic function.

The conformal map $w=z^2$ sends the quadrant bijectively to the complex upper half-plane: the positive real axis maps to $s>0$, and the positive imaginary axis maps to $s<0$. Define its real-axis boundary function byThe compatibility at the corner makes $f$ continuous there. The Dirichlet Poisson integral in a quadrant follows from the upper-half-plane Poisson kernel:It is bounded by the supremum of the bounded edge data, and the kernel's approximate identity property supplies both Dirichlet boundary conditions. More explicitly, for $z^2=a+ic$, where $a=x^2-y^2$ and $c=2xy>0$, this integral isIt displays directly which part of the real-line boundary corresponds to each original axis.

To obtain its derivative, first take real data and use the Schwarz integral formula. A holomorphic function whose real part is the Poisson solution in the $w$-plane isThe regularizing subtraction is independent of $w$ and affects only the imaginary normalization; it permits bounded data without an artificial integrability assumption. Then $q=\mathrm{Re}F(z^2)$. By the chain rule and the Wirtinger derivative,The factor is $z$, not $2z$: taking the real part contributes the factor $1/2$ in the complex derivative. Differentiation under the integral sign is justified uniformly on compact interior subsets because the denominator stays away from the real boundary and decays quadratically at infinity. Splitting the real integral at zero and substituting $s=r^2$ or $s=-r^2$ gives the holomorphic derivative of a quadrant Dirichlet solution:This converges for bounded data, and in particular for the stated sufficiently decaying data. No interior pole is encountered since $\mathrm{Im}(z^2)=2xy>0$. For complex-valued data the same formula holds by applying the real-data construction separately to their real and imaginary parts and using linearity.

An equivalent form, often convenient when the data's derivatives decay, comes from integration by parts. Sincethe two endpoint terms are $-g_2(0)/z^2$ and $+g_1(0)/z^2$. They cancel by the specified compatibility, leavingThis is the harmonic quadrant solution from tangential boundary derivatives applied to the actual derivatives $g_1^{\prime },g_2^{\prime }$; the supplied $g_1,g_2$ in this question are boundary values, not derivatives. Omitting those primes would solve a different boundary problem.