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Integrate ${\nabla }^{2}u=0$ over a large half-ball and apply the divergence theorem. The hemispherical contribution vanishes by the assumed decay. On $z=0$, the outward normal is $-{\mathbf{e}}_z$, so $\partial u/\partial n=-p$. Hence compatibility requiresThis is the usual solvability condition for the Neumann problem.

Choose $0<r<\mathrm{dist}(x,\partial \Omega )$. Differentiating the ball mean-value formula with respect to its centre and then using the divergence theorem givesConsequently the interior derivative estimate for a harmonic function yieldsThe constant may depend on $x$ and its distance from the boundary, but it is independent of $u$.

Fix $x\in \Omega$ and $R<\mathrm{dist}(x,\partial \Omega )$. The mean value property for harmonic functions isTo prove the spherical identity, translate $x$ to the origin and put $F(r)=|S^{n-1}{|}^{-1}{\int }_{S^{n-1}}u(r\theta )d{S}_{\theta }$. The divergence theorem givesHence $F(r)={\lim }_{s\downarrow 0}F(s)=u(0)$. Integrating the spherical identity in polar coordinates gives the ball identity.

Let $K={\partial }_t$ and let the future unit normal to ${\Sigma }_{\tau }$ be $n=f^{-1}{\partial }_t$. The conserved Killing energy is the fluxSubstitution of the Klein-Gordon scalar stress-energy tensor gives exactlyApply the divergence theorem to the slab $[{\tau }_1,{\tau }_2]\times {\mathbb{R}}^3$. The spatial-boundary flux vanishes because $\psi$ decays, and ${\nabla }_a{J}^a=0$ makes the two time-slice fluxes equal. Thus this conserved scalar-field energy in a static spacetime is independent of $\tau$.

Dulac theorem states the following. Let $D$ be a simply connected region for the planar system $\dot{z}=f(z)$, and suppose there is a continuously differentiable function $B:D\to \mathbb{R}$ for whichhas one strict sign throughout $D$. Then there is no periodic orbit contained in $D$.

To prove it, suppose a periodic orbit $\Gamma$ existed and let $A$ be its interior. Simple connectedness ensures $A\subset D$. By the planar divergence theorem,The vector field $f$ is tangent to its trajectory $\Gamma$, so its scalar product with the outward normal $n$ vanishes. The right-hand side is therefore zero. The left-hand side cannot be zero because its integrand has one strict sign, a contradiction.

The Poincaré-Bendixson theorem states that if a forward trajectory of a smooth planar system remains in a compact set and its nonempty omega-limit set contains no equilibrium point, then that omega-limit set is a periodic orbit.

The Navier-Stokes equation for constant mass density $\rho$, dynamic viscosity $\mu$, and body force per unit mass $\mathbf{f}$ iswhere the Newtonian fluid stress tensor isTaking the dot product with $\mathbf{u}$ givesIncompressibility impliesand converts the left side into a local time derivative plus the divergence of kinetic-energy flux. The divergence theorem therefore gives the kinetic-energy balance for an incompressible Newtonian fluidThe four terms are respectively outward advective transport of kinetic energy, mechanical power supplied by surface traction, power supplied by the body force, and irreversible viscous dissipation into heat.

Because ${\epsilon }_{ij}$ and ${\mu }_{ij}$ are symmetric and time-independent, the linear anisotropic dielectric relations giveand similarlyIntegrate the identity from part (b) over $V$ and apply the divergence theorem to obtainWhen $\mathbf{J}=0$, this is conservation of electromagnetic energy. The Poynting theorem in a linear anisotropic medium identifiesas the energy density andas the energy flux, or Poynting vector.

Close the surface with the unit disk $D$ in the plane $z=0$. Its outward normal is $-e_z$, but $F\cdot (-e_z)=-z^4=0$ there. AlsoThe $x$ and $y$ terms integrate to zero by symmetry. The divergence theorem therefore gives

Suppose ${\varphi }_1,{\varphi }_2$ solve the Dirichlet problem and put $u={\varphi }_1-{\varphi }_2$. Then ${\nabla }^{2}u=0$ in $V$ and $u=0$ on $S$. The divergence theorem applied to $u\nabla u$ gives Green's identityThus $\nabla u=0$, so $u$ is constant; its zero boundary value makes it zero. The Dirichlet solution is therefore unique.

For homogeneous Neumann data the same calculation again shows that $u$ is constant, but the boundary condition does not determine that constant. Hence, whenever a Neumann solution exists, adding any constant produces another solution. This is the standard uniqueness of Poisson equation distinction.

The integrand defining $\epsilon (t)$ is nonnegative by the dominant energy condition, henceApply the spacetime divergence theorem to $J^a=T^{ab}{V}_b$ on the slab $S$. The flux through the cylinder at $r\to \infty$ vanishes by the stated decay. With the Lorentzian boundary orientation, the two spacelike boundary contributions givePart (ii) makes the right-hand side nonnegative, so for $t_1\ge t_0$,Therefore $\epsilon$ is a nonnegative monotone decreasing energy functional.

The Roche lobe of star 1 is the volume around it bounded by the critical closed equipotential passing through the inner Lagrange point $L_1$. Material on a lower potential surface remains confined to star 1; at the critical surface a path opens toward star 2.

Let $\mathbf{g}=-\nabla \Psi$ and take $d\mathbf{S}$ outward. For a nearly spherical interior surface, the divergence theorem givesThe self-gravity term contributes $-4\pi G{M}_1$. The companion lies outside $S$, so its potential is harmonic inside and contributes zero net flux. For the centrifugal term, ${\nabla }^2(-{\Omega }^2{s}^2/2)=-2{\Omega }^2$, so its acceleration has divergence $2{\Omega }^2$ and contributes $2{\Omega }^{2}V$. Using $V=4\pi r^3/3$ and $S=4\pi r^2$,The sign is the outward-normal component; the dominant self-gravity is inward and therefore negative.

For two body-force-free Stokes flows $({\mathbf{u}}^{(1)},{\mathbf{\sigma }}^{(1)})$ and $({\mathbf{u}}^{(2)},{\mathbf{\sigma }}^{(2)})$ in the same domain, the Lorentz reciprocal theorem for Stokes flow isIndeed, the difference of the two volume integrands isbecause both flows are incompressible and the Newtonian fluid stress tensor is symmetric. The divergence theorem proves the boundary identity.

On a rigid body, ${\mathbf{u}}^{(r)}={\mathbf{U}}^{(r)}+{\mathbf{\Omega }}^{(r)}\times \mathbf{x}$. The reciprocal theorem becomesWriting $(\mathbf{F},\mathbf{G}{)}^T=\mathsf{R}(\mathbf{U},\mathbf{\Omega }{)}^T$ and choosing arbitrary pairs of rigid velocities shows thatThus the hydrodynamic resistance matrix is symmetric.

The divergence theorem is ${\iint }_{\partial V}F\cdot ndS={\iiint }_V\nabla \cdot FdV$. Here $\nabla \cdot F=3z+2yz$. The $y$ term integrates to zero by symmetry, while the ellipse has area $\pi /\sqrt{ab}$. Hence the flux is

Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives

Letbe the scalar moment of inertia and total kinetic energy. Sincesubstitution of the Euler momentum equation reduces the stress contribution, by the divergence theorem and isotropic pressure $\mathbf{P}=P1$, toFor self-gravity, $\int \rho \mathbf{r}\cdot \mathbf{F}dV=\Omega$, where $\Omega <0$ is the gravitational potential energy. Thus the stellar virial theorem is

Apply hydrostatic equilibrium to the isothermal core, taking its boundary pressure to be $P_c$. Its ideal-gas pressure integral scales as $\int P,dV\propto M_c{T}_c$, its volume as $R_c^3$, and its gravitational energy as $-G{M}_c^2/R_c$. The virial theorem therefore givesor, after absorbing fixed dimensional and structural factors into positive constants,Hydrogen-burning reactions are extremely temperature-sensitive, so expansion cools and suppresses burning while contraction heats and enhances it. This stellar thermostat keeps the shell and adjoining isothermal core near an approximately fixed $T_c$.

At fixed $M_c,T_c$, differentiating $P_c(R_c)$ gives

For a homologous envelope, hydrostatic balance gives $P_c\propto G{M}^2/R^4$ and the ideal-gas temperature scale gives $T_c\propto GM/R$. Eliminating $R$ yieldsup to composition and gravitational constants common to the sequence. A matching core exists only if this required pressure does not exceed $P_{c,\max }\propto T_c^4/M_c^2$. ThereforeThis maximum fractional isothermal-core mass is the mechanism behind the Schönberg-Chandrasekhar limit.

Divergence theorem says ${\int }_V\nabla \cdot FdV={\int }_{\partial V}F\cdot ndS$ for the outward normal. Here $\nabla \cdot (x{e}^{-r^2})=(3-2r^2)e^{-r^2}$, so radial integration gives $4\pi a^3{e}^{-a^2}$. On the sphere $F\cdot n=a{e}^{-a^2}$, whose integral is the same $4\pi a^3{e}^{-a^2}$.

Differentiate the ball mean-value formula and use the divergence theorem:ConsequentlyThus one may take $C(n)=n$.

Integrating the Euler-Lagrange equation and applying the divergence theorem on the compact boundaryless manifold givesEquivalently, if this integral were nonzero, replacing $u$ by $u+C$ would leave the gradient term unchanged and make the energy unbounded below in one direction, so no minimizer could exist.

Let ${\Sigma }_t$ be a Cauchy hypersurface with induced metric $h_{ij}$, lapse $N$, shift $N^i$, and future unit normal $n^a$. For the normalization of the action in the question, the canonical momentum density isThe equal-time canonical commutation relations areandWith the conventional extra factor $1/2$ in the action, $\Pi$ loses the factor two.

For complex classical solutions, the Klein-Gordon inner product isThe integrand is a conserved current because both fields obey the Klein-Gordon equation. Applying the divergence theorem between two Cauchy hypersurfaces shows that the value is independent of the foliation, provided there is no boundary flux.