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For two solutions of the Poisson equation with the same Dirichlet boundary condition, let $w=u_1-u_2$. Then ${\nabla }^{2}w=0$ in $V$ and $w=0$ on its boundary. By Green's first identity,Thus $w$ is constant on each connected component, and the zero boundary values make each constant zero. The Dirichlet solution is unique whenever it exists. This argument requires the regularity needed for the displayed integration by parts; it does not assert existence for arbitrary rough data.

With a Neumann boundary condition, integrating the Poisson equation and using the divergence theorem instead gives the necessary compatibility conditionOn a disconnected region this condition must hold on each connected component separately. If two solutions have the same normal derivative, their difference is harmonic with ${\partial }_{n}w=0$. The same Green's first identity again forces $\nabla w=0$. Thus Neumann solutions are unique only up to an additive constant on each connected component. Adding a constant preserves both the equation and normal derivative, so uniqueness without a normalization cannot hold. On a connected region, prescribing the mean value removes this one-dimensional freedom.

For an axisymmetric Newtonian gravitational potential, the vacuum Laplace equation in cylindrical coordinates is $R^{-1}{\partial }_R(R{\partial }_R\Phi )+{\partial }_z^2\Phi =0$. Use separation of variables, $\Phi =X(R)Z(z)$. Dividing by $XZ$ and setting the separated constant to $k^2$, with $k>0$, givesThe first equation has solutions $e^{\pm kz}$; with $u=kR$, the second is the order-zero Bessel differential equation. Regularity at the axis selects the Bessel function of the first kind $J_0(kR)$, rather than the singular Bessel function of the second kind $Y_0$. ThusThere are also zero-separation-constant modes, and other boundary conditions can select other separated families. For a localized reflection-symmetric thin disk, decay away from the disk selects $e^{-k|z|}$ on the respective half-spaces.

Apply the divergence theorem to the Poisson equation for Newtonian gravity across a narrow pillbox enclosing disk area. Writing the volume mass density as $\Sigma (R)\delta (z)$ givesReflection symmetry makes ${\Phi }_z(R,0^{+})=2\pi G\Sigma (R)$. A superposition of the regular vacuum modes therefore satisfiesThe order-zero Hankel transform and its inversion now determine the Hankel representation of thin-disk gravity:The sign is fixed by the positive derivative of an attractive Newtonian gravitational potential above the disk. For ordinary finite disks the representation uses the isolated boundary condition, fixing any extra vacuum field or additive constant. Infinite scale-free disks require suitable regularization of the potential itself.

Equatorial circular orbit balance is $v_c^2=R{\Phi }_R(R,0)$. Using $J_0^{\prime }(u)=-J_1(u)$ in the Hankel representation of thin-disk gravity givesDifferentiation under the integral is justified by convergence for regular models, or by first working off the disk and taking the limiting radial field.

Write $C={\Sigma }_0{R}_0$. For the Mestel disk, the supplied Bessel function integral gives $S(k)=-2\pi GC/k$ for $k>0$. The circular speed is thenMeanwhile the interior mass is $M(R)=2\pi {\int }_0^R(C/s)sds=2\pi CR$. ConsequentlyThe Mestel disk has a flat galaxy rotation curve, despite its nonspherical geometry.

Its infinite total mass prevents setting the Newtonian gravitational potential to zero at infinity. Indeed the integral for $\Phi$ diverges at $k=0$ because $S(k)\sim 1/k$. Subtracting the potential at a reference point removes this additive divergence. As an explicit check, the Mestel disk potential-density pair isIt is harmonic for $z\ne 0$, its vertical derivative jump is $4\pi GC/R$, and $R{\Phi }_R(R,0)=2\pi GC$. Equivalently, an Abel regularization of an oscillatory integral with damping factor $e^{-ak}$ in the radial-field integral gives $R{\int }_0^{\infty }{e}^{-ak}{J}_1(kR)dk=1-a/\sqrt{a^2+R^2}$, which tends to one as $a\downarrow 0$. The force is well-defined although the unreferenced potential is not.

The equality with $GM(R)/R$ is special to the Mestel disk, not a thin-disk shell theorem. The spherical shell theorem normally makes that formula possible, but does not apply to an astrophysical disk. Exterior annuli exert outward radial gravitational acceleration inside their radii, while interior annuli do not act exactly as point masses at the origin. For this particular scale-free surface density, their combined contributions happen to produce the same result as the interior mass expression. Changing the radial surface density or truncating the disk generally destroys the equality: enclosed mass does not determine a disc rotation curve.

Under the usual bounded-domain hypotheses, with piecewise smooth boundary and a sufficiently smooth vector field, the divergence theorem stateswhere $\mathbf{n}$ is outward. Apply it to $g\mathbf{F}$ and use the product rule $\nabla \cdot (g\mathbf{F})=\mathbf{F}\cdot \nabla g+g\nabla \cdot \mathbf{F}$ to obtainFor uniqueness, let $w=u_1-u_2$ for two classical solutions with the same Dirichlet boundary data. Then $\Delta w=0$ and $w=0$ on the boundary. Taking $\mathbf{F}=\nabla w$, $g=w$ gives Green's first identity in the formThe continuous nonnegative integrand must vanish, so $w$ is constant on each connected component. The zero boundary data make every such constant zero. Thus the solution of the Dirichlet problem for the Laplace equation is unique, if it exists. This establishes Uniqueness of the Dirichlet problem; existence is assumed. Boundedness, or suitable conditions at infinity, is needed: on an unbounded half-space the harmonic function $w=z$ has zero boundary values but is not identically zero.

For an incompressible flow, write the Newtonian fluid stress tensor as $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the rate-of-strain tensor. The Stokes equation gives ${\partial }_j{\sigma }_{ij}=0$. Symmetry of the Cauchy stress tensor therefore givesIntegrating and applying the divergence theorem expresses the viscous dissipation as boundary power:At a moving rigid body, the surface velocity is $U+\Omega \times (x-x_c)$. Its boundary-power contribution is $U\cdot F+\Omega \cdot G$, with the force and torque evaluated using the fluid's outward normal vector. Both resultants vanish for a force-free, torque-free inclusion. Thus the new viscous dissipation comes entirely from the unchanged outer boundary velocity. Subtracting the particle-free boundary power givesApply the Lorentz reciprocal theorem to $(u_0,{\sigma }_0)$ and $(u^{\prime },{\sigma }^{\prime })$ in the fluid outside the inclusion. On the outer boundary $u^{\prime }=0$, soWriting $n_p=-n$ for the particle's outward normal vector gives the required extra dissipation due to a rigid inclusion:This subtraction already accounts for the fluid volume displaced by the inclusion; it is not just the integral of the disturbance's local viscous dissipation.

For the sphere, the ambient rate-of-strain tensor $E$ is symmetric and trace free. The sphere in a uniform straining Stokes flow has zero translational and rotational velocity: inversion symmetry eliminates its force, and symmetry of $E$ eliminates its torque. To derive the disturbance, put $r=|x|$, $Q=x\cdot Ex$ and seekThe incompressible flow condition and Stokes equation reduce toA decaying family satisfying these equations is $f=A{r}^{-5}$, $g=B{r}^{-5}-5A{r}^{-7}/2$, $j=2B{r}^{-5}$. The no-slip boundary condition requires $f(a)=-1$ and $g(a)=0$, so $A=-a^5$ and $B=-5a^3/2$. ConsequentlyThe total velocity is zero at $r=a$, and the disturbance decays as $r^{-2}$. The pressure constant has been set to zero. These boundary and far-field conditions, together with Uniqueness of Stokes flow, establish the solution. There is no rotational background in a pure strain flow.

On the sphere $u^{\prime }=-u_0=-aE{n}_p$. Contracting the supplied Newtonian fluid stress tensor with $n_p$ makes its two terms proportional to $(n_p\cdot E{n}_p)n_p$ cancel, leaving $\sigma n_p=5\mu E{n}_p$. Hence the extra dissipation due to a rigid inclusion isHere ${\int }_{S^2}{n}_j{n}_{k}d\Omega =4\pi {\delta }_{jk}/3$. Taking the large outer boundary limit only after using the fixed-boundary power identity avoids replacing that prescribed boundary by an uncontrolled boundary at infinity.

If $N_v$ denotes the number of spheres per unit volume, $N_v(4\pi a^3/3)=\varphi$. ThusAdd this to the particle-free viscous dissipation density $2\mu E:E$. Comparing with $2{\mu }_{\mathrm{eff}}E:E$ gives the Einstein viscosity formula for a dilute suspension:to first order in $\varphi$. The hydrodynamic interactions neglected here contribute beyond that dilute order.

For two Stokes flows $(u^{(1)},{\sigma }^{(1)})$ and $(u^{(2)},{\sigma }^{(2)})$ of the same dynamic viscosity in the same region, with no volume force, the Lorentz reciprocal theorem statesIndeed, the divergence of their cross-work difference isThe pressure terms vanish by incompressible flow, and the derivatives of the Cauchy stress tensors vanish by the Stokes equation. The divergence theorem proves the result.

Use the convention that $F,G$ are the force and torque exerted by the body on the fluid. By Linearity of Stokes flow, $(F,G)=\mathsf{R}(U,\Omega )$. Applying the Lorentz reciprocal theorem to two independent rigid motions gives $q_1^T\mathsf{R}{q}_2=q_2^T\mathsf{R}{q}_1$, so the hydrodynamic resistance matrix is a symmetric matrix. The boundary-power identity givesThis is strictly positive for nonzero rigid motion: equality would imply $e=0$, hence a rigid motion throughout the connected fluid, which must vanish since the fluid is at rest at infinity. The no-slip boundary condition would then force $q=0$. Therefore $\mathsf{R}$ is a positive-definite matrix. Reversing to the fluid-on-body force changes the sign of the force law, not the positive resistance coefficients.

For the two rods, take the torque about $O$ and use body axes. On the $x$-rod, $X=(s,0,0)$, $0\le s\le 2L$, and its slender-body force density isOn the $y$-rod it isIntegrate $f$ and $X\times f$ along both rods, using $\int ds=2L$, $\int sds=2L^2$ and $\int s^{2}ds=8L^3/3$. A convenient dimensionally uniform statement of the complete right-angle two-rod resistance matrix isIn physical coordinates this means translation–translation entries scale as $CL$, the two translation–rotation blocks as $C{L}^2$, and rotation–rotation entries as $C{L}^3$. The planar block has inverseThis verifies the printed hydrodynamic mobility matrix with its third velocity component $L{\Omega }_z$; the TeX aid's ${\Omega }_x$ is an OCR error.

Take laboratory vertical velocity positive upwards. The body axes are $e_1=(\cos \theta ,\sin \theta )$ and $e_2=(-\sin \theta ,\cos \theta )$. In quasistatic sedimentation the force and torque exerted on the fluid equal the gravitational resultants on the body:The out-of-plane block is unforced, so its positive hydrodynamic resistance matrix gives $U_z={\Omega }_x={\Omega }_y=0$. Substitution in the planar hydrodynamic mobility matrix givesandFor $0\le \lambda <1$, this ordinary differential equation has positive right side on $0\le \theta <\pi /4$, with a stable zero at $\pi /4$. Uniqueness prevents crossing that equilibrium in finite time. For $\lambda >1$ the initial angular velocity is negative, and the stable equilibrium reached from zero is $-3\pi /4$. The heavier-end body turns clockwise through $3\pi /4$; it does not settle at $\pi /4$. More explicitly, with $K=(1-\lambda )mg/(4C{L}^2)$,where the continuous branch has $\theta -\pi /4\in (-\pi ,0)$. At $\lambda =1$, $\theta$ remains zero.

Transforming the translational velocity back to laboratory axes givesFor $\lambda \ne 1$, eliminate time between $U_h$ and the angular velocity:At either limiting orientation $\sin \theta =\cos \theta$. Thus both cases have the same net horizontal displacement,For $\lambda <1$ the drift is monotonically rightwards. For $\lambda >1$ it first moves left, reaching $x_O-x_O(0)=(2L/3)(1-\sqrt{2})$ at $\theta =-\pi /4$, then reverses. The total horizontal path length in this second case is $(2L/3)(2\sqrt{2}-1)$, whereas its net displacement is $2L/3$ rightwards. At $\lambda =1$ it falls vertically without rotation or drift; taking the infinite-time limit before $\lambda \to 1$ is consequently singular.

For the requested sedimentation drift of a weighted two-rod body sketch, a full parametric trajectory follows by also integrating $U_v/\dot{\theta }$. Put $\psi =\theta -\pi /4$ and set the initial height to zero:It tends to $-\infty$ in either case while $x_O$ tends to $2L/3$. At $\lambda <1$ both rods end pointing upwards from $O$ symmetrically; at $\lambda >1$ they end pointing downwards symmetrically. The asymptotic downward speed is $(1+2\lambda )mg/(3CL)$.

The electric field of this point charge is a translated and scaled version of the radial field in part (ii). If $\mathbf{a}\notin V$ and $\mathbf{a}\notin \partial V$, its divergence vanishes throughout $V$, so the divergence theorem gives zero flux integral.

If $\mathbf{a}$ lies inside $V$, remove a small ball centered at $\mathbf{a}$. On the remaining volume the divergence is zero. Its boundary is $\partial V$ together with the small sphere, whose outward normal for the punctured volume points towards $\mathbf{a}$. That inner sphere contributes $-q/{ϵ}_0$, by the $4\pi$ sphere flux computed in part (ii). Thus the exterior flux must be $q/{ϵ}_0$:This is Gauss's law for a point charge. The assumption that the charge is not on the boundary is essential to these alternatives.

On the outward-oriented sphere of radius $R$, $\mathbf{n}=\mathbf{x}/R$, so $\mathbf{x}\cdot d\mathbf{S}=RdS$. HenceAway from the origin, the divergence and curl of a radial vector field with $f(r)=r^{-3}$ giveIf the enclosed volume excludes the origin and its boundary avoids it, this field is smooth throughout the volume. The divergence theorem then gives $I=0$. The nonzero sphere flux is compatible with zero divergence away from the origin because the enclosed singularity prevents applying that smooth-volume theorem directly to the full ball.

For $\mathbf{F}=\nabla \varphi$, the definition of the Laplacian gives $\nabla \cdot \mathbf{F}={\nabla }^2\varphi =f$. Integrating over $V$ and applying the divergence theorem yieldsThis is the flux form of Poisson equation, for an outward-oriented piecewise smooth boundary and the usual smoothness of $\varphi$ on a neighborhood of the volume. In particular, a singularity cannot be ignored when applying this identity.

Close the truncated two-sheet double circular cone by the lateral circular cylinder $r=1$, $-1\le z\le 1$. Together these form the boundary of the solid in part (c); no planar caps are required because the cone meets the circular cylinder along both end circles. On the circular cylinder the outward unit normal is $(\cos \theta ,\sin \theta ,0)$, so $\mathbf{F}\cdot \mathbf{n}=1$, and $dS=d\theta dz$. Its flux integral isThe divergence theorem now gives ${\int }_S\mathbf{F}\cdot d\mathbf{S}+4\pi =4\pi$, using the volume integral from part (c). Henceagreeing with tangency. The isolated vertex can be handled by excising a ball of radius $\epsilon$ and passing to the limit: since $|\mathbf{F}|=O(\epsilon )$ there and the added area is $O({\epsilon }^2)$, its extra flux is $O({\epsilon }^3)$ and tends to zero.

For a conductor in electrostatic equilibrium, the interior field is zero and the surface is equipotential. The tangential exterior field is zero. A thin Gaussian pillbox gives the normal jump $E_{\mathrm{out}}\cdot \mathbf{n}-E_{\mathrm{in}}\cdot \mathbf{n}=\sigma /{ϵ}_0$. Thus, in vacuum and with $\mathbf{n}$ pointing out of the conductor,These are the electrostatic boundary conditions at a conductor.

To obtain electrostatic energy in field form, integrate over the space outside the conductors, cutting it off first at a large sphere. From the Maxwell equations, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ and $\mathbf{E}=-\nabla \varphi$, soOn a conductor boundary, the outward normal of the integration region points into the conductor, so its flux contribution is $-{\varphi }_i{Q}_i/{ϵ}_0$. At the remote outer boundary the flux integral tends to zero: for a localized charge distribution, $\varphi =O(1/r)$ and $\mathbf{E}=O(1/r^2)$. The divergence theorem therefore givesSince the field vanishes inside conductors,The field integral is over all space, although the charges occupy a finite volume. For a finite outer integration boundary there is additionally the term $({ϵ}_0/2)\oint \varphi \mathbf{E}\cdot d\mathbf{S}$.

For the electrostatic energy of alternating-charge concentric shells, the total enclosed charge after shell $n$ isBy spherical symmetry and Gauss's law, the field is zero for $r<r_1$, and in every subsequent gap, including outside the last shell,Although its direction alternates, its squared magnitude is the same in every gap. ConsequentlyThe coefficient of $1/r_1$ is $q^2/(8\pi {ϵ}_0)$, independent of the other radii and of the number of shells.

Use compatible orientations for the loop and spanning surface. The relevant Maxwell equations areThe magnetic flux transport for a moving circuit separates the change of the field from the motion of its boundary. To see the boundary term, freeze $\mathbf{B}$ at time $t$. Orient the closed swept surface as $S(t+\delta t)-S(t)+S^{\prime }$. Its side element isto first order. Hence its flux is $-\delta t{\oint }_C\mathbf{B}\cdot (\mathbf{v}\times d\mathbf{r})$. By the divergence theorem and $\nabla \cdot \mathbf{B}=0$, the sum of the three fluxes is zero, giving the purely geometric changeAdding the temporal field change and applying Stokes theorem yieldsThusThe boundary contribution is the motional electromotive force; the field contribution is Faraday's law.

Apply mass conservation to a fixed horizontal region $D$. The layer mass is $\rho {\int }_{D}hdxdy$ and the outward horizontal flux is $\rho {\int }_{\partial D}h(u,v)\cdot \mathbf{n}ds$. The divergence theorem and arbitrary choice of $D$ giveThis depth-integrated law retains free-surface motion; neglecting the vertical velocity in horizontal momentum is not the assumption that the depth cannot change.

Take the free-surface pressure to be constant. hydrostatic pressure gives $p=p_{\mathrm{atm}}+\rho g(h-z)$, so the horizontal pressure forces are $-g{\eta }_x,-g{\eta }_y$. Linearization about a resting layer gives the linearized shallow water equationsTaking the curl of the momentum equations yields ${\zeta }_t=-f(u_x+v_y)$. Combining with continuity gives the linearized shallow-water potential-vorticity anomalyFor $Q_0=0$, $\zeta =f\eta /h_0$. Taking the divergence of momentum gives ${\partial }_t(u_x+v_y)-f\zeta =-g{\nabla }_h^2\eta$. Eliminate the divergence with continuity to obtainFor $\theta =k(x-ct)$, substitution of $\eta ={\eta }_0\cos \theta$ gives the inertia-gravity wave dispersion relationThe associated purely oscillatory velocity field isThese formulas satisfy all three linearized equations and the zero-anomaly condition. A separately added spatially uniform inertial oscillation is not part of this monochromatic wave.

To first order in amplitude evaluate the wave at a particle's equilibrium coordinate $x_{\ast }$, so ${\theta }_{\ast }=k{x}_{\ast }-\omega t$. Integrating its velocity gives displacements relative to the orbit centreThese are particle ellipses for rotating shallow-water waves, with axis ratio $V/U=f/\omega <1$. For the stipulated positive signs, the velocity at phases $0,\pi /2,\pi ,3\pi /2$ points respectively right, up, left, down, with magnitudes $U,V,U,V$. The particle orbits are clockwise as time advances, since the phase decreases.

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.

For $\omega \ne 0$, put $\mu =\hat{\mu }{\omega }^2$, $C=\omega z$ and $\tau =\omega T$, where $z=\hat{C}$. Then $C_T={\omega }^2{z}_{\tau }$ and division by ${\omega }^2$ givesWe may take $\omega >0$ without loss of the dynamics by conjugating the original equation if necessary; otherwise $\tau$ reverses the time orientation. The scaling is singular at $\omega =0$ and is not a transformation for that exactly zero-frequency case.

The Hamiltonian limit of three-to-one forcing drops the terms proportional to $\omega$. For $z=x+iy$ the remaining real system isThe proposed first integral isIndeed $x_{\tau }=H_y/2$ and $y_{\tau }=-H_x/2$, hence $H_{\tau }=0$. The unperturbed equation is a planar Hamiltonian system, with a center equilibrium at the origin and three saddle equilibria atAll saddle equilibria have $H=1/3$. The factorizationshows that their central separatrix consists of the three sides of an equilateral triangle. Each level $0<H<1/3$ inside this triangle is a closed periodic orbit. Indeed, inside the triangle $0<r<1$ and ${\partial }_{r}H=2r(1+r\sin 3\theta )>0$. Each ray from the origin therefore meets each such level once, giving a compact simple closed contour with no equilibrium point on it. The nonzero vector field traverses this contour periodically; the period grows without bound as the separatrix is approached. This supplies an infinite family, not a claim that every level outside the central region is closed.

Restore the small radial perturbation. Its exact effect on the first integral isConsequently the continuum of Hamiltonian system orbits generally does not persist. The origin becomes a weak attracting focus for $\hat{\mu }<0$ or a repelling focus for $\hat{\mu }>0$, and the three hyperbolic saddle equilibria persist with perturbed stable manifold and unstable manifold. For a small positive $\hat{\mu }$, outward drift on very small orbits balances cubic damping on somewhat larger ones, selecting a stable limit cycle rather than an arbitrary energy level. Near the center equilibrium $H\simeq r^2$, so its leading radius is $r\simeq \sqrt{\hat{\mu }}$ when $\hat{\mu }$ is also small.

For a more general closed unperturbed orbit ${\Gamma }_h$, the averaged area criterion for perturbed Hamiltonian cycles says that persistence requires the averaged energy drift ${\oint }_{{\Gamma }_h}{H}_{\tau }d\tau$ to vanish. Since the unperturbed speed is $|\nabla H|/2$, the planar divergence theorem converts this leading drift towhere ${\mathcal{D}}_h$ is the enclosed region. Isolated zeros select candidate periodic orbits; a drift changing from positive inside to negative outside gives an attracting limit cycle. The separatrix triangle has mean $r^2=1/4$, so its leading flux changes sign at $\hat{\mu }=1/2$. This marks the leading possible heteroclinic transition, with higher-order corrections needed to locate it precisely.

As a cycle approaches the saddle equilibria, long residence times and splitting of the heteroclinic cycle become important. Orbits can instead drift inward to the equilibrium point at the origin or leave the periodic island and approach one of the stable states with phase locking of the full canonical equation. Those upper-branch threefold phase-locked equilibria have $C=O(1)$, so they lie outside the local $C=O(\omega )$ scaling. Thus the small perturbation gives energy selection, attracting or repelling oscillations, and possible switching/locking transitions; it does not preserve a conserved $H$ or an infinite family of neutral periodic solutions. This qualitative picture does not assume all global parameter values have the same attractor.

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:

Apply the divergence theorem to $\mathbf{F}=\psi \nabla \varphi$, for functions smooth enough on the closure of the bounded piecewise smooth volume. The product rule givesThus Green's first theorem, also called Green's first identity, isThe normal is outward. This identifies the boundary term and its sign directly from the divergence theorem.