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A constant-energy phase-space trajectory of the harmonic oscillator is the ellipseIts area is $\pi \sqrt{2E/(m{\omega }^2)}\sqrt{2mE}=2\pi E/\omega$. Thus the action and slow energy change areThe action variable is an adiabatic invariant when $|\dot{\omega }|\ll {\omega }^2$, with $\omega$ staying positive. Averaging over many short oscillations therefore keeps $I$ approximately fixed; an increase in frequency increases the energy in the same proportion.

The Hamiltonian is $\boxed{H(q,p)=p^2/(2m)+V(q)}$. On a regular family of closed phase-space trajectories, action-angle variables are canonical coordinates $(\theta ,I)$ in which $\theta$ is periodic modulo $2\pi$ and $H=H(I)$. Hamilton's equations then give $\dot{I}=0$ and $\dot{\theta }=dH/dI$.

The action variable isThe orientation is that of the motion, so the integral is the positive enclosed phase-space area divided by $2\pi$.

For the Fourier mode $e^{i(i\theta +j\varphi )}$, the forward Euler method has amplification factorIt is real and ranges from $1-4\mu (1+c)$ to $1$. The two-dimensional Fourier stability analysis therefore requires its minimum to be at least $-1$. Thus the exact stable range isThe endpoint is included; the highest-frequency mode then has amplification $-1$ and bounded norm.

Use ${\hat{u}}^n(\theta )={\sum }_m{u}_m^n{e}^{-im\theta }$, initially for finite-support sequences and then by Plancherel theorem for ${\ell }^2$. A shift by $k$ multiplies the transform by $e^{ik\theta }$. Hence the recurrence givesHere the last equality is understood with time-zero data, and the scheme is assumed well-defined, so the denominator is nonzero or any removable singularity is filled continuously.

If $|H|\le 1$, Plancherel gives $\Vert u^n{\Vert }_2\le \Vert u^0{\Vert }_2$ for every $n$. Conversely, if the continuous amplification factor has $|H({\theta }_0)|>1$, it exceeds $1+\delta$ on an interval of positive length. Choose a nonzero initial Fourier transform supported in that interval. Then $\Vert u^n{\Vert }_2\ge (1+\delta {)}^n\Vert u^0{\Vert }_2$, contradicting a uniform stability bound. Therefore uniform power stability holds exactly whenWithout continuity the exact condition is an essential-supremum bound, since isolated Fourier frequencies have zero measure. The fixed recurrence here uses a bound uniform over all time steps; mesh-dependent finite-time growth bounds are a different stability convention.

Fix the scaling constants as $U=A/z$, $\delta =Bz$, with $A=F/(\rho \nu )$ and $B=\nu \sqrt{\rho /F}$, so $A{B}^2=\nu$. The Stokes streamfunction is then $\psi =\nu zf(\eta )$. Write $g=f^{\prime }/\eta$. Differentiation givesSubstitution in the boundary-layer equation, cancelling $A^2/z^3$, gives $-(f/\eta )g^{\prime }-g^2=g^{\prime \prime }+g^{\prime }/\eta$. Multiplication by ${\eta }^3$ therefore gives the required similarity equationAxis regularity requires zero radial velocity and a finite smooth axial velocity: choose the streamfunction gauge $f(0)=0$ and require $f(\eta )=c{\eta }^2+O({\eta }^4)$. Equivalently $f^{\prime }(0)=0$, $f^{\prime }/\eta$ has a finite limit, and $(f^{\prime }/\eta {)}^{\prime }\to 0$ at the axis. The constant $c$ is not specified in advance. In the quiescent ambient fluid require $f^{\prime }/\eta \to 0$ and $f^{\prime }-f/\eta \to 0$ as $\eta \to \infty$, so both velocity components vanish. The localized physical jet has finite axial volume flux, corresponding to finite $f(\infty )$.

Finally the prescribed force fixes the amplitude through the finite momentum-flux normalizationThe axis conditions express regularity rather than independently prescribed arbitrary derivatives; the far-field conditions express matching to the resting fluid, and the integral condition supplies the jet strength.

The incompressibility condition turns the left side of the boundary-layer equation into conservative form:Multiply by $2\pi \rho r$ and integrate from the axis to infinity. Regularity at the axis and decay in the ambient fluid eliminate the radial boundary terms. Thus the momentum flux is constant,Matching the applied force sets this constant to $F$. Scaling convection against radial viscosity gives $U^2/z\sim \nu U/{\delta }^2$, while the flux gives $F\sim \rho U^2{\delta }^2$. Combining them yieldsThe slenderness requirement is $\delta /z\sim \nu \sqrt{\rho /F}\ll 1$, equivalent to a large force-based Reynolds number $\sqrt{F/\rho }/\nu$.

In a steady axisymmetric flow without swirl, time derivatives and azimuthal derivatives vanish. For a slender high-Reynolds number jet of width $\delta \ll z$, radial viscous derivatives dominate axial ones by $(z/\delta {)}^2$. The transverse momentum balance makes pressure nearly constant across the jet, matching the constant ambient pressure, so the leading axial pressure gradient vanishes. Retaining axial convection and radial diffusion in the Navier-Stokes equation gives the boundary-layer equationThe radial velocity is small but its convective term is of the same order as axial convection because radial gradients are larger.

The preceding condition is symmetric in $b,d$, so ${\ell }_{[b;d]}=0$. On a patch where the potential representation supplied in the question holds, write ${\ell }_a={\partial }_a\varphi$. DefineThen $k_{a;b}=e^{\varphi }({\ell }_{a;b}+{\ell }_a{\ell }_b)$. Since $e^{\varphi }\ne 0$, the condition on $\ell$ is exactly $\boxed{k_{a;b}=0}$, so $k$ is a parallel covector field. Closed covectors have this scalar-potential representation locally by the Poincaré lemma; a global potential additionally requires exactness, for example on a simply connected domain.

Substitute $T^a{}_{bc}={\ell }^a{g}_{bc}$ into the preceding identity, using metric compatibility. Set $M_{bd}={\ell }_{b;d}+{\ell }_b{\ell }_d$ and $S=g^{bd}{M}_{bd}={\nabla }_a{\ell }^a+{\ell }_a{\ell }^a$. Contract the first and third curvature indices, in the stated curvature convention, to obtainIf the two Ricci tensors agree, then $M_{bd}=S{g}_{bd}$. Taking the trace in dimension $n>1$ gives $S=nS$, so $S=0$ and $M=0$. Conversely $M=0$ makes the Ricci difference vanish. Thus $\boxed{{\tilde{R}}_{bd}=R_{bd}⟺{\ell }_{b;d}+{\ell }_b{\ell }_d=0}$.

At any chosen point take normal coordinates for $g$, so $\Gamma =0$ there. Substitute $\tilde{\Gamma }=\Gamma +T$ into the stated Riemann curvature tensor formula and subtract the original curvature. At that point the mixed products with $\Gamma$ vanish and ordinary derivatives of $T$ equal its covariant derivatives. The result isIn unit-weight antisymmetrization notation this is $\boxed{2T^a{}_{b[d;c]}-2T^a{}_{e[d}{T}^e{}_{c]b}}$. Every term is a tensor, so the identity established in normal coordinates holds in every coordinate system.

The coordinate transformation law for a affine connection has a homogeneous term with one upper and two lower index factors, and an inhomogeneous term involving second coordinate derivatives. The latter is the same for both Levi-Civita connections, so it cancels in their difference. Consequentlywhich is exactly the tensor component transformation law. Equivalently, $(\tilde{\nabla }-\nabla {)}_{X}Y$ is linear over functions in both $X$ and $Y$.

Use the mostly-plus metric, so $\square$ reduces to ${\nabla }^2$ for static fields. Seek a Green function satisfying $({\nabla }^2-{\lambda }^2)G={\delta }^{(3)}$. Convolving it with $-{\mu }_0\mathbf{J}$ gives the required solution.

For $R>0$, write $G=f(R)/R$. The homogeneous equation is $f^{\prime \prime }-{\lambda }^{2}f=0$. Decay discards the growing exponential, while the normalization at the origin follows from ${\nabla }^2(1/R)=-4\pi {\delta }^{(3)}$. Therefore for every $\lambda \ge 0$,The $1/R$ singularity fixes the delta coefficient; the extra factor is smooth to leading order at zero. For a conserved localized static current, integration by parts also gives $\nabla \cdot \mathbf{A}=0$, as required by the constraint.

At $\lambda =0$ the far field has algebraic multipole decay; for an ordinary localized steady current the vector potential starts at dipole order. At $\lambda >0$ the Yukawa potential gives exponential suppression on length scale ${\lambda }^{-1}$, with an overall bound of order $e^{-\lambda |\mathbf{x}|}/|\mathbf{x}|$ for bounded source support.

Vary the four-potential with compactly supported variation. Antisymmetry of the electromagnetic field tensor gives $\delta (F_{\mu \nu }{F}^{\mu \nu })=4F^{\mu \nu }{\partial }_{\mu }\delta A_{\nu }$. Integration by parts therefore givesThus the Euler-Lagrange field equation is $\boxed{{\partial }_{\mu }{F}^{\mu \nu }-{\lambda }^2{A}^{\nu }=-{\mu }_0{J}^{\nu }}$.

A gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. Commuting derivatives leave $F_{\mu \nu }$ unchanged. At $\lambda =0$, the source term changes by $c^{-1}\int {\partial }_{\mu }\chi J^{\mu }$, a boundary term because the current is conserved. With suitable boundary conditions the action is gauge invariant. For $\lambda >0$, the mass term changes by terms involving $2A^{\mu }{\partial }_{\mu }\chi +(\partial \chi {)}^2$, so the same arbitrary gauge transformations are not a symmetry.

Take the divergence of the field equation. Antisymmetry gives ${\partial }_{\nu }{\partial }_{\mu }{F}^{\mu \nu }=0$, and conservation gives ${\partial }_{\nu }{J}^{\nu }=0$. Thus ${\lambda }^2{\partial }_{\nu }{A}^{\nu }=0$. For $\lambda >0$ this is the Lorenz constraint in Proca theory, and expanding $F$ givesFor $\lambda =0$ the same reduced equation holds when the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$ is imposed. More generally it is enough that the divergence be spacetime constant, but the conventional vanishing boundary conditions set that constant to zero.

Let $m_0={\min }_{x}u(x,0)>0$, which exists by periodicity and continuity. Fix any finite interval $[0,T]$ on which the classical solution is smooth. Choose $ϵ>0$ with $ϵ(1+T)<m_0/2$ and consider the strictly positive decreasing barrier $b(t)=m_0-ϵ(1+t)$.

Initially $u>b$. If this strict inequality first failed, compactness of the spatial circle would give a first contact $(x_{\ast },t_{\ast })$ with $u=b>0$. At that contact, the spatial minimum of $u-b$ has $u_x=0$, $u_{xx}\ge 0$, and the first-contact condition gives $(u-b{)}_t\le 0$. But the equation yieldsa contradiction. Therefore $u>b$ throughout the interval. Letting $ϵ\downarrow 0$ gives $\boxed{u(x,t)\ge m_0>0}$ on $[0,T]$. Since $T$ was arbitrary, positivity persists for every time for which the assumed smooth solution exists. This argument is a parabolic comparison principle proved directly and does not assume uniform parabolicity before positivity has been established.

Use a Fourier transform convention for which $-\Delta$ has symbol $|\xi {|}^2$. The distributional equation gives $\hat{u}=\hat{f}/(|\xi {|}^2+m^2)$ for real nonzero $m$, since this smooth positive multiplier has no zeros. Consequentlywhere $\boxed{C_m=\max (1,m^{-2})}$. This bound is independent of $s$. For $f\in H^s$, the same multiplier defines $u\in H^{s+2}$; alternatively approximate $f$ by Schwartz functions and pass to the limit using the estimate. This is the massive Laplacian isomorphism on Sobolev spaces.

For $a=1+|\xi {|}^2$, the inequality $a\le ϵa^2+1/(4ϵ)$, integrated against $|\hat{v}{|}^2$, gives the interpolation estimateFor a finite global bound in the last request, assume $\Vert u{\Vert }_{L^2}<\infty$ as well as boundedness. This assumption is necessary: the smooth bounded solution $u\equiv m$ has infinite $H^2$ norm on ${\mathbb{R}}^n$. Put $M=\Vert u{\Vert }_{\infty }$, $U=\Vert u{\Vert }_2$. Then $u^3\in L^2$ with norm at most $M^{2}U$. Incorporate the constant drift on the left:The constant-drift massive elliptic estimate follows because the modulus of this symbol is at least $|\xi {|}^2+m^2$, so the same multiplier argument first establishes $u\in H^2$ and then gives the stronger a priori boundIt is allowed to depend on $\alpha$ but in fact does not need to. Once regularity has been established, the earlier estimate and interpolation also give the conventional absorption proof: $\Vert u{\Vert }_{H^2}\le C_m(M^{2}U+|\alpha |\Vert u{\Vert }_{H^1})$, and choosing $ϵ$ small absorbs the $H^2$ term.