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Use the Minkowski metric $\mathrm{diag}(1,-1)$ and the angular field $u=\beta \varphi$. The Euler-Lagrange equation becomes $u_{tt}-u_{xx}+m^2\sin u=0$. For the dimensionless light-cone coordinatesthis is $u_{\xi \eta }=\sin u$. This normalization keeps the coupling in the relation between the physical field and its angle, rather than silently setting $\beta =1$.

A Bäcklund transformation is a system of first-order differential relations that maps a solution to another solution. One convention for the Sine-Gordon Bäcklund transformation uses a nonzero parameter $a$ and defines $v={\mathcal{B}}_a[u]$ byA compatible initial value or integration constant selects a particular transformed solution. Put $w=(v+u)/2$ and $d=(v-u)/2$. Differentiating gives $w_{\xi \eta }=\cos d\sin w$ and $d_{\xi \eta }=\cos w\sin d$. Their sum and difference yieldThus compatibility of the first-order relations contains the field equations for both fields, and the transformed physical field $v/\beta$ also solves Sine-Gordon theory.

The transformation provides a generating conservation law. On the branch close to the original field write $v=u+2d$. Its first equation isIt recursively determines a formal small-$a$ expansion of $d$ in local derivatives of $u$:At every subsequent order the new coefficient occurs linearly, so this recursion continues indefinitely. The second Bäcklund equation and the first imply the exact identityIndeed ${\partial }_{\eta }\cos d=-a\sin d\sin (u+d)$ and ${\partial }_{\xi }\cos (u+d)=-a^{-1}\sin d\sin (u+d)$. Comparing powers of $a$ therefore gives local conserved currents. If ${\partial }_{\eta }{P}_j={\partial }_{\xi }{Q}_j$, our coordinate convention givesFor localized fields approaching vacua at spatial infinity, the boundary flux vanishes and ${\mathcal{I}}_j$ is conserved. The formal expansion need not converge: each coefficient is a separately exact local conservation law.

For example $P_0=u_{\xi }^2/2$, $Q_0=1-\cos u$, a light-cone combination of energy and momentum. The next coefficient is a derivative improvement, so it contributes no independent charge under the same decay conditions. At order $a^2$, remove the improvement generated by ${\partial }_{\xi }(u_{\xi }{u}_{\xi \xi })$ and multiply by $-2$. One obtains the genuinely higher conservation lawIt can also be checked directly using $u_{\xi \eta }=\sin u$. Continuing the recursion, and using the opposite light-cone construction, produces the local conserved-charge hierarchy of sine-Gordon theory, with infinitely many nontrivial higher-spin charges after derivative improvements are removed. This is the Bäcklund generating current for sine-Gordon conserved charges. The hierarchy is the characteristic field-theory form of classical integrability; an ordinary energy conservation law alone would not supply these constraints.

The same transformation constructs solutions rather than only currents. Starting from the vacuum $u_0=0$, its two first-order equations integrate toFor $a=e^{\rho }>0$ the exponent is $m\cosh \rho x-m\sinh \rho t+{\delta }_a$. This is a Sine-Gordon kink with velocity $\tanh \rho$, center set by ${\delta }_a$ and classical rest mass $8m/{\beta }^2$. Negative parameters can supply the corresponding opposite-orientation seeds.

The allowed Bianchi permutability for sine-Gordon Bäcklund transformations gives the two-step field algebraically. For vacuum seed and $a\ne b$,Branches of the inverse tangent must be continued smoothly; its principal value alone does not specify the vacuum labels of a multi-kink field. All angles here are $u=\beta \varphi$. In the physical-field version each field difference in the superposition formula carries $\beta$; the formula printed without it implicitly uses the $\beta =1$ angular-field convention.

To display two real scattering solutions, take $0<v<1$, set $\Gamma =(1-v^2{)}^{-1/2}$, and put $X=m\Gamma x$, $Y=m\Gamma vt$. Choosing $a=e^{\rho }$, $b=e^{-\rho }$ with $v=\tanh \rho$ and zero phase constants yields the Sine-Gordon kink-antikink scattering solutionChoosing instead $b=-e^{-\rho }$ yields the Sine-Gordon two-kink solutionThe latter has net angular winding $4\pi$, while the former has zero net winding. At large positive or negative time they separate into localized kinks with velocities $\pm v$. Repeated commuting transformations give general multi-soliton fields. Continuing the kink-antikink velocity to an imaginary value also gives a real Sine-Gordon breather:up to the irrelevant overall field sign and translations.

The scattering interpretation ties the construction to the conserved hierarchy. In an exact multi-soliton sector, the incoming species and rapidities reappear after the collision, up to permutation: the solitons change positions, not their asymptotic shapes or velocities, and the collision emits no radiation. In the two-kink example the large-time centers obey $|X|=|Y|-\log v+o(1)$, so a right-moving trajectory acquires a shift $-2\log v/(m\Gamma )$. These shifts encode the interaction even though the collision is elastic. The commuting construction makes the net displacement in a many-soliton collision the sum of its pairwise displacements, independent of how the collisions are ordered; this is pairwise additivity of soliton shifts and the classical counterpart of factorized scattering.

Conservation of the entire hierarchy is much stronger than conservation of energy and momentum: its independent rapidity-weighted sums constrain the whole asymptotic soliton data and rule out particle production in this sector. Generic initial fields may also contain radiation scattering data, so this statement is about the exact soliton collisions, not a claim that every initial field is a pure soliton. The Bäcklund map unifies the conserved hierarchy, explicit soliton construction and elastic, pairwise scattering picture of classical integrability.

We prove finite propagation speed using a shrinking cone energy argument. Fix $T>0$ and $x_{\ast }$, and suppose the Cauchy data vanish on $B(x_{\ast },T)$. For $0\le s<T$, define the local wave energyThe homogeneous wave equation gives the local conservation lawDifferentiate the integral over the moving ball. Its boundary moves inward with speed one, so the divergence theorem givesHere $n$ is the outward unit normal, ${\partial }_n$ the normal derivative, and ${\nabla }_{\tan }$ the component of the gradient tangent to the boundary. Since $E(0)=0$ and $e\ge 0$, we have $E(s)=0$. Hence both ${\varphi }_t$ and $\nabla \varphi$ vanish inside the backward light cone. Integrating ${\varphi }_t(s,x_{\ast })=0$ from the zero initial displacement gives $\varphi (s,x_{\ast })=0$; continuity then gives $\varphi (T,x_{\ast })=0$.

Applying the same energy estimate to the difference of two solutions proves the domain of dependence assertion. If $\mathrm{dist}(x_{\ast },K)>T$, the initial ball misses $K$, and the preceding argument proves the stated support bound. No disturbance propagates faster than one.

The sign convention makes this a defocusing semilinear wave equation:Its wave energy includes a nonnegative potential term:For a smooth function solution, finite propagation speed preserves compact support on finite time intervals. Differentiate under the integral and use integration by parts:Equivalently the local conservation law has density $e={\varphi }_t^2/2+|\nabla \varphi {|}^2/2+{\varphi }^4/4$ and flux $-{\varphi }_t\nabla \varphi$. All three terms are nonnegative, and $E=0$ forces both Cauchy data to vanish. Hence this is a positive conserved energy, rather than the indefinite energy associated with the opposite sign.

Use downward distance $y=H-z$, with $z$ the height above the bottom, and write $W=W_s(a)>0$. This convention will also reproduce the printed negative deposit-front velocity in part (d). The ideal simple-cubic sphere packing has packing fraction $\pi /6$; leaving ${\varphi }_{\max }$ symbolic keeps the jump calculation independent of its value. In suspension, the kinematic sedimentation equation isIntegrate this conservation law across a moving discontinuity to obtain the Rankine-Hugoniot conditionThe brackets denote values on the increasing-$y$ side minus those on the other side. A sedimentation shock is a concentration jump moving at this secant slope of the particle flux. Deposited material is stationary and has zero flux.

For the upper clearing front, the states are $0$ and ${\varphi }_{\max }/8$, giving $U_1=7W/8$ downward. For the lower deposition front, the states are ${\varphi }_{\max }/8$ and ${\varphi }_{\max }$, giving $U_2=-W/8$, upward. ThusIn terms of height above the bottom, the shock paths areThey meet when $H=Wt$. Therefore complete settling occurs atThe final height also follows directly from particle volume fraction conservation, ${\varphi }_{\max }{h}_c=({\varphi }_{\max }/8)H$. The stationary deposit after the meeting carries no particle flux.

The diagram plots time vertically against height, as requested. Its shock slopes have the opposite spatial sign to $U_1,U_2$, because those velocities were defined in the downward coordinate.

For the finite-mass similarity solution above, the total disk mass isSubstitute $\xi =1/(\tau \sqrt{x})$, so $x={\tau }^{-2}{\xi }^{-2}$ and $x^{-3/2}|dx|=2\tau d\xi$. This givesThe requested linear increase of mass is therefore correct. It is supplied by the infinite-radius boundary, rather than by a source-free isolated finite disk. The inward mass accretion rate iswhich tends to zero at the origin and to $3\pi C{\bar{\nu }}_0{\Sigma }_0$ at infinity.

For Keplerian rotation, the specific angular momentum is $\ell =\sqrt{G{M}_{\star }r}$. The total disk angular momentum would beThis integral diverges at $\xi =0$, corresponding to infinite radius. There is no finite constant total angular momentum for the nonzero printed similarity solution. The seemingly time-independent last expression is an infinite quantity; it cannot prove a finite conservation law.

The problem remains visible under a cutoff. With $r\le r_{0}R$, write $J_R=J_{\ast }{E}_1(1/(\tau \sqrt{R}))$, where $J_{\ast }=4\pi C{\Sigma }_0\sqrt{G{M}_{\star }}{r}_0^{5/2}$ and $E_1$ is the exponential integral. ThenEven the finite change after subtracting a common logarithmic divergence is not zero. Direct integration of the disk diffusion equation gives the boundary identityfor this profile the outer expression has a nonzero limit, giving the same finite change rate. An inner zero torque alone does not impose zero outer angular-momentum flux.

More generally, the printed family $A+B{e}^{-\xi }$ can have finite angular momentum at both ends only if $A=0$ and $A+B=0$, hence it is trivial. Thus the request for a nonzero finite constant $J_d$ is incompatible with the stated viscosity exponent, similarity ansatz and infinite radial domain. A physical finite-domain or altered similarity problem would need different boundary data or assumptions; silently assigning a finite value to the divergent integral does not repair it.

For this chemical reaction network, the law of mass action gives the reaction rates are $k_1{s}^{2}e$, $k_{2}c$ and $k_{3}c$. Their stoichiometry givesHere the convention for $k_1$ absorbs any combinatorial factor for two identical reactants. Direct differentiation gives the two conservation lawsEliminating $e$ and $p$ leavesFor positive $s_0,e_0,k_1$, use the dimensionless variables $u=s/s_0$, $v=c/e_0$ and $\tau =k_1{e}_0{s}_{0}t$. With prime marks denoting differentiation in $\tau$, this becomeswhere

Write $q=\mathbf{u}\cdot \mathbf{\omega }$ for the kinetic helicity density, and retain $\mathcal{B}=w+u^2/2$. Since $\mathbf{\omega }$ is a curl, $\nabla \cdot \mathbf{\omega }=0$. Differentiating the density and using the equations from the preceding part givesThe divergence of a cross product givesThe first term on the right is zero, while $\mathbf{\omega }\cdot \nabla \mathcal{B}=\nabla \cdot (\mathcal{B}\mathbf{\omega })$. Thus the kinetic helicity conservation law has fluxThe last equality for the flux uses the vector triple product identity. Integrating and applying the divergence theorem gives $d{\int }_{V}qdV/dt=-{\int }_{\partial V}{\mathbf{F}}_{H_k}\cdot \mathbf{n}dS$ for a fixed volume. Hence its total kinetic helicity is conserved when the boundary flux vanishes; for example, tangency of both velocity field and vorticity to the boundary is sufficient. The local conservation law does not by itself imply that its density follows each fluid particle unchanged.

Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio $\gamma >1$, and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is $e=p/[(\gamma -1)\rho ]$, and the specific enthalpy is $w=\gamma p/[(\gamma -1)\rho ]$. Integrating the local conservation laws across a thin stationary control volume makes the mass flux $\rho u$, momentum flux $p+\rho u^2$, and energy flux $u(\rho u^2/2+\gamma p/(\gamma -1))$ continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state $p=K{\rho }^{\gamma }$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.

In the Strong-shock Rankine-Hugoniot conditions, $p_1/({\rho }_1{u}_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r={\rho }_2/{\rho }_1$. The first two flux balances give $u_2=u_1/r$ and $p_2={\rho }_1{u}_1^2(1-1/r)$ to leading order. The energy balance becomesDiscarding the unchanged-state root $r=1$ gives the compressive shock compression ratio and the required limits:

For an axisymmetric viscous gravity current on a cone, the cylindrical radius at slant distance $x$ is $x\cos \alpha$. A surface annulus therefore has area $2\pi x\cos \alpha dx$. The local downslope volume flux is the same as in part (a), but conservation of mass must account for this growing circumference. ThusThis is a leading thin-film description away from the apex, with $0<\alpha <\pi /2$ and thickness small relative to the local curvature and streamwise length scales.

The original PDF uses the long-time criterion $x_N\gg (V\csc \alpha {)}^{1/3}$; the local TeX incorrectly reads $V\cos \alpha$. To see why the PDF criterion is the relevant one, scale $h\sim V/(x_N^2\cos \alpha )$. In the bulk, the hydrostatic-gradient term relative to direct downslope gravity isHence the leading outer evolution, with $\beta =K\sin \alpha$, isFor a fixed-volume similarity solution, let $x_N\propto t^b$ and $h\propto t^a$. Conservation of mass gives $a+2b=0$, while the evolution equation gives $a-1=3a-b$. Thus $b=1/5$, $a=-2/5$. Writing $h=t^{-2/5}H(\xi )$, $\xi =x/t^{1/5}$, yieldsThe positive branch $H=c\sqrt{\xi }$ satisfies this equation when $c^2=1/(5\beta )$. It givesThe ideal outer current has zero thickness beyond its front. Its mass integral isThereforeThe reduced equation is a conservation law after multiplication by $x$. Its front obeys the Rankine-Hugoniot condition, ${\dot{x}}_N=\beta h_N^2=x_N/(5t)$, consistent with the mass-derived front position. The outer profile has a finite jump at the front; the omitted hydrostatic-gradient term resolves a narrow front region. There is also a small apex region where the outer gradient and conical curvature violate the local approximations. Neither region sets the leading conserved-volume scaling above.

For a radial function, the equation isTherefore, assuming the population density and radial flux decay at infinity and are regular at the origin,Thus $N$ is conserved; it is the total population obtained by integrating density over the plane. This is the global form of the equation's conservation law.

SetThenwhile the radial diffusion term has the same prefactor and equalsEquating them and multiplying by $x$ givesso the proposed ansatz is a similarity solution.

Regularity at $x=0$ makes the constant of integration zero. Wherever $f>0$,for some $x_0>0$. The nonnegative weak solution continues as $f=0$ once this parabola reaches zero:This is the two-dimensional Barenblatt profile for quadratic porous-medium diffusion; its degenerate diffusivity permits compact support.

Indeed, conservation requiresso $x_0=(16/\pi {)}^{1/4}$ if desired. At time $t$ the populated disk has radius $x_0(NDt{)}^{1/4}$ and hence area

At an endemic equilibrium, $I^{\ast }>0$, so $0=\dot{I}=I^{\ast }(\beta S^{\ast }-\gamma )$ forcesThe equation $\dot{R}=0$ gives $R^{\ast }=\gamma I^{\ast }/\alpha$. The conservation law $S^{\ast }+I^{\ast }+R^{\ast }=N$ therefore yieldsA positive endemic state exists exactly when $\beta N>\gamma$, equivalently when the basic reproduction number ${\mathcal{R}}_0=\beta N/\gamma$ exceeds one.

Fick's first law gives the diffusive fluxLocal conservation says $C_t+\nabla \cdot \mathbf{J}=0$, and therefore

For $D(C)=kC$, conservation of the deposited amount requiresThe dimensions satisfy $[M]=[C]L^2$ and $[k]=L^2/([C]T)$. Thus $Mkt$ has dimension $L^4$, so the similarity length is $(Mkt{)}^{1/4}$. Requiring the concentration scale to have dimension $[C]$ givesHence

In polar coordinates, substitution into
$C_t=k{r}^{-1}(rC{C}_r{)}_r$ givesMultiplying by $\xi$ and integrating once, with regularity and zero radial flux at the origin, yieldsWhere $F>0$, this reduces to $F^{\prime }=-\xi /4$, soNonnegativity and zero flux at the moving front give the compactly supported profileThe normalization givesand hence $A=1/\sqrt{2}$ and ${\xi }_0^4=32$. This is the Two-dimensional Barenblatt solution with diffusivity proportional to concentration, and its support radius isso $N=32$.

Now add the linear reaction term and writeSubstitution cancels the terms $aC$ and leavesChoosingproduces $G_{\tau }=k\nabla \cdot (G\nabla G)$, as in the linear-reaction time change for quadratic nonlinear diffusion. Therefore

For $a=0$, the front grows like $t^{1/4}$ while the central concentration decays like $t^{-1/2}$. For $a>0$, $\tau \sim e^{at}/a$, so the front grows like $e^{at/4}$ and the central concentration like $e^{at/2}$. For $a<0$, $\tau \to 1/|a|$: the front approaches a finite limiting radius while the concentration decays exponentially to zero.

Adding the equations for the free enzyme and its two complexes givesThusThis conservation law says that the enzyme is neither created nor destroyed: each enzyme molecule is either free, bound in $C_1$, or bound in $C_2$.

Under $a_{\alpha }^I\mapsto a_{\alpha }^I+{\partial }_{\alpha }{\Lambda }^I$, the Chern--Simons term changes bybecause the contraction with two commuting derivatives vanishes. The source term changes byCurrent conservation removes the last term, so

When the only nonpotential force points radially, its curl has no radial component. The radial absolute vorticity therefore obeys a local flux conservation law, although vortex stretching can change it along a fluid trajectory.

A similarity profile is the time-independent function left after expressing a evolving field in its scaled coordinate, for example $\rho (z,t)={\rho }_{0}f(z/Z(t))$. Integrating such profiles converts a global conservation law into an algebraic relation among the time-dependent scales.