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The nullclines are $x=f(y)$ and $y=g(x)$, both decreasing curves in the positive quadrant. To their left/right the horizontal velocity is respectively positive/negative; below/above the second nullcline the vertical velocity is positive/negative. Bounded positive synthesis functions give an inward-pointing sufficiently large rectangle, while the axes point into the quadrant. More directly, put $H(x)=f(g(x))-x$: $H(0)>0$, and $H(x)<0$ for large enough $x$. Continuity therefore gives a positive equilibrium.

At an equilibrium, the Jacobian matrix is $J=\left(\begin{array}{cc}-1& f^{\prime }(y_{\ast })\\ g^{\prime }(x_{\ast })& -1\end{array}\right)$, with eigenvalues $-1\pm \sqrt{f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })}$. Its logarithmic-slope product isbecause $f(y_{\ast })=x_{\ast }$ and $g(x_{\ast })=y_{\ast }$. Thus a product exceeding one makes this equilibrium a saddle and gives $H^{\prime }(x_{\ast })>0$. Immediately on its left $H<0$, and immediately on its right $H>0$. Since $H(0)>0$ and $H$ is eventually negative, there must be other roots on both sides, crossing in the attracting direction. For simple roots those have $H^{\prime }<0$, hence two negative Jacobian eigenvalues: the saddle separates at least two stable steady-state alternatives. The divergence is $-2$, so periodic attractors cannot replace these equilibria in a bounded planar region. Nongeneric flat crossings or intervals of equilibria require a nonlinear stability interpretation; the strict saddle inequality alone does not guarantee that every other root is hyperbolic. This is the usual mutual repression stability criterion behind bistability.

The conclusion also has a nonlinear trapping justification. Choose $l<u$ on the left of the saddle with $H(l)>0>H(u)$; such endpoints exist by the signs just proved. The rectangle $[l,u]\times [g(u),g(l)]$ is forward invariant: on the vertical edges, $f(y)-l\ge f(g(l))-l>0$ and $f(y)-u\le f(g(u))-u<0$, while the horizontal-edge velocities point inward or are tangent. The same construction on the right of the saddle gives a second disjoint trapping rectangle. Each contains a nonempty compact attracting invariant set, obtained by intersecting its nested forward images; strict vertical inwardness also rules out invariant boundary trajectories. If a downward crossing is isolated, endpoints can be taken arbitrarily close to it, proving nonlinear stability, and the absence of periodic orbits makes nearby trajectories converge to that equilibrium. Intervals of equilibria instead give attracting alternatives consisting of families, so bounded monotone functions without nondegeneracy do not necessarily produce two isolated hyperbolic attractors.

For an inviscid flow of constant density $\rho$, driven by a conservative body force $-\nabla \Phi$, the Euler equation isThe vector identity $(u\cdot \nabla )u=\nabla (|u{|}^2/2)-u\times (\nabla \times u)$ and irrotationality give the Unsteady Bernoulli equation. Specifically, wherever a single-valued potential $u=\nabla \varphi$ is available,On a connected potential-flow region this meansThe integration function is spatially constant but need not be constant in time; it can be removed by changing the potential's time-dependent gauge. A simply connected irrotational region provides the requisite global potential. For a barotropic fluid the pressure term is replaced by $\int dp/\rho (p)$; here the liquid density is constant.

For the tube, let $U(t)$ be its approximately uniform axial velocity. The large reservoir's surface velocity is negligible and gravity is neglected as stipulated. Entrance-region inertia contributes an effective length of order $a$, negligible against $L$. The potential difference between the outlet and the reservoir is consequently $LU$ to leading order. Subtract Bernoulli at the reservoir surface, where pressure is $P+p_a$, from its value at the outlet, where pressure is $p_a$ and speed is $U$. This givesThe inertial term cannot be dropped during startup. With $U_{\infty }=\sqrt{2P/\rho }$, separate variables:Hence the inviscid startup in a pressure-driven tube givesInitially $U\sim Pt/(\rho L)$, the acceleration of the stationary liquid column. At long times $U\to \sqrt{2P/\rho }$, the inviscid pressure-to-kinetic-energy limit. The result applies while the tube remains full and reservoir dimensions are large enough that surface motion and the neglected entrance inertia remain small.

Use units $c=1$ and signature $(+---)$. For a massive test particle, write its four-momentum as $p^i=m{u}^i$, where $u^i=d{x}^i/d\tau$, $\tau$ is Minkowski proper time, and $u^i{u}_i=1$. The stated flat-background scalar-force theory of gravity then givesContract with $p_i=m{u}_i$. The two terms cancel:Thus rest mass is conserved and the acceleration law isThere is no mass or composition parameter in this equation. Particles released at the same event with the same velocity consequently have identical trajectories. In a static weak field at small speed, its spatial part becomes $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$, since raising a spatial index introduces a minus sign. The field equation likewise reduces to ${\nabla }^2\Phi =4\pi G\rho$ for nonrelativistic matter, because $T^i{}_i\simeq \rho$ and $\square =-{\nabla }^2$ in the static limit.

Therefore yes: the model agrees with the Eötvös test of universal free fall. In the point-particle/test-body approximation, inertial and passive gravitational masses have the same universal ratio. This is the weak equivalence principle tested by the Eötvös experiment; it is not a claim about strongly self-gravitating bodies or a proof of all aspects of the Einstein equivalence principle.

Assume $V>0$ and that its derivatives do not change substantially over the field interval traversed in the estimate. With initially negligible kinetic energy, $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$. The force $-V_{,\varphi }$ changes the field velocity on the Hubble friction time. Keeping $H$ and $V_{,\varphi }$ approximately constant for that local estimate, and starting at rest, solves the field equation explicitly:After one Hubble time,Writing $A=M_{\mathrm{Pl}}^2{V}_{,\varphi }^2/V^2$, the condition $A<1$ keeps this kinetic-to-potential ratio below about $0.31$, comfortably inside the inflationary bound. Integrating once more gives the fractional potential change estimateover the same interval. Thus the characteristic time for an order-one potential change is $H^{-1}/A$, up to constants of order one. The requested local Hubble-timescale criterion isUsing a strict inequality with one as the order-one threshold is conventional; its coefficient is not an exact universal end-of-inflation bound. Much smaller $A$ gives the controlled slow-roll regime and many Hubble times of cosmic inflation if the bounds persist.

After the transient in relaxation to the slow-roll attractor, $3H\dot{\varphi }\simeq -V_{,\varphi }$. Differentiating this approximate relation givesTherefore the standard sufficient conditions for negligible acceleration throughout slow roll are ${ϵ}_V\ll 1$ andThe absolute value matters: large negative curvature also drives rapid evolution. The more precise local acceleration criterion is $|{\eta }_V-{ϵ}_V|\ll 1$, with the usual separate smallness conditions avoiding a tuned cancellation. If a field starts exactly at rest on a nonzero slope, its initial acceleration is not negligible; the displayed exponential transient describes its approach to the attractor.

These are local timescale and slow-roll conditions, rather than a global duration theorem from derivatives at a single point. A smooth potential can have a narrow flat shoulder followed by a steep drop, ending cosmic inflation soon despite small starting derivatives. The conditions must hold over the traversed interval. This is the additional smooth-evolution assumption implicit in the estimate.

Use primes for conformal-time derivatives. The cold-dark-matter Cosmological Euler equation in synchronous gauge is $v_C^{\prime }+\mathcal{H}{v}_C=0$, so multiplying by $a$ gives $(a{v}_C{)}^{\prime }=0$. ThereforeIn an expanding universe $a$ increases, damping any initial peculiar velocity as $a^{-1}$. During radiation domination $a\propto \tau$, so the velocity scales as ${\tau }^{-1}$; during matter domination $a\propto {\tau }^2$, it scales as ${\tau }^{-2}$. This is the nonrelativistic limit of the physical-momentum redshift derived in Q1. It concerns the bulk velocity in the synchronous coordinate frame, not a growing peculiar velocity sourced in another gauge. A cold-matter rest-frame choice also removes the residual synchronous time displacement relevant to that component. The decay statement assumes expanding evolution and the pressureless, linear approximation used by the equation.

With negligible cold-dark-matter velocity, the synchronous perfect-fluid density equation gives ${\delta }_C^{\prime }=-h^{\prime }/2$. Differentiate and use the trace Einstein equation, taking the cold matter sound speed to vanish and its density to dominate:In a spatially flat matter-dominated universe, $a\propto {\tau }^2$, $\mathcal{H}=2/\tau$, and the conformal Friedmann equation gives $4\pi G{a}^2{\rho }_C=3{\mathcal{H}}^2/2=6/{\tau }^2$. HencePut ${\delta }_C\propto {\tau }^s$. The indicial equation is $s(s-1)+2s-6=(s-2)(s+3)=0$. These two independent modes span the solution space, soEquivalently the growing mode is proportional to $a$ and the decaying mode to $a^{-3/2}$. The coefficients are set by initial data. Since synchronous gauge has residual coordinate modes, interpreting the density contrast physically requires fixing its residual freedom consistently, for example by the cold-matter rest frame and metric tensor integration convention; the mode equation alone is not a gauge-invariant observable.

The continuity and Cosmological Euler equation in synchronous gauge combine particularly simply. Differentiate ${\delta }_C^{\prime }=-\nabla \cdot v_C-h^{\prime }/2$ and add $\mathcal{H}{\delta }_C^{\prime }$:The velocity terms cancel by the Cosmological Euler equation in synchronous gauge, so this step does not require setting the initial velocity exactly to zero.

For subhorizon modes deep in radiation domination, take the rapidly oscillating radiation perturbation to be negligible in the averaged gravitational forcing. Also neglect the subdominant matter self-gravity to leading order in ${\rho }_C/{\rho }_r$. With $a\propto \tau$ and $\mathcal{H}=1/\tau$, the resulting equation isIntegrating twice giveswhere the arbitrary reference time ${\tau }_{\ast }$ makes the logarithm dimensionless and can be absorbed into $B$. This is logarithmic growth of matter perturbations during radiation domination.

The neglect of matter self-gravity is needed for the displayed form to be exact within the leading radiation-background approximation. Uniform radiation alone leaves a nonzero term $4\pi G{a}^2{\rho }_C{\delta }_C$ if the cold-matter density is retained. Relative to ${\mathcal{H}}^2$, its coefficient is $3{\Omega }_C(a)/2$, small deep in the radiation era but not near equality. The Mészáros equation retains that effect in a matter-plus-radiation background; its solution basis tends to a constant and a logarithm at $a/a_{\mathrm{eq}}\ll 1$, recovering the present leading result. It connects smoothly to matter-era growth rather than allowing the logarithmic approximation to be extrapolated indefinitely.

Galaxy-scale modes enter the Hubble radius before matter-radiation equality. The earlier radiation forcing and horizon-entry matching ordinarily generate a nonzero logarithmic coefficient, even though subsequent oscillatory radiation forcing is small. A Fourier mode with entry time ${\tau }_h\sim k^{-1}$ accumulates $\log ({\tau }_{\mathrm{eq}}/{\tau }_h)\sim \log (k{\tau }_{\mathrm{eq}})$ before equality. Across a large scale-factor range this factor need not be small, so treating the density contrast as exactly frozen would lose an important part of the galaxy-scale seed amplitude and its scale dependence. Growth is nevertheless much slower than ${\delta }_C\propto a$ in matter domination. This is the Mészáros effect, and the logarithmic matching contributes to the small-scale matter cold-dark-matter transfer function.

Use the prescribed-speed time parametrization of a curve. On a regular part traversed in increasing $u$, let $q(u)=\Vert F^{\prime }(u)\Vert >0$ and assume the prescribed speed $v(F(u))$ is positive. The chain rule for velocity givesIntegrate this scalar ordinary differential equation with an adaptive Runge-Kutta method, arranging output at the exact times $t_k=t_0+k\Delta t$. Evaluate $P_k=F(u(t_k))$. Internal integration steps may be shorter than $\Delta t$; output interpolation must meet the same error tolerance, so variable numerical steps do not create variable frame times.

An equivalent implementation precomputesby adaptive numerical integration. This function is strictly increasing. Bracket each $t_k$ in the table and invert using safeguarded Newton method or interval bisection; monotonicity makes each solution unique. Stop at $u=1$, with a shorter final time interval if the total travel time is not a multiple of $\Delta t$. This accounts for changes in both the curve's parameter speed and the physical speed, unlike equal increments in $u$.

If the specified speed vanishes, examine the reciprocal-speed integral at that event. It may diverge, giving asymptotic approach, or be integrable, giving arrival in finite time. A scalar speed does not determine whether motion reverses or how it resumes after stopping; such behavior needs the physical dynamics or a chosen continuation. The positive-speed formula applies between these events, with the direction changed if the traversal reverses.

Use the equal-time points to estimate velocity and acceleration with centered finite differences:At the endpoints use one-sided formulas of the desired order. A differentiated path and its time parametrization can instead supply these quantities. Let $g_k$ denote the gravitational acceleration, directed downwards. In the accelerating camera, the effective downward vector is $g_k-a_k$, so the locally felt upward vector is $U_k=a_k-g_k$. For a stationary camera this gives $-g_k$, fixing the sign.

The apparent-gravity camera frame is obtained by orthogonal projection and a cross product:When $W_k\ne 0$, $X_k,Z_k$ are perpendicular unit vectors, and $Y_k$ completes a right-handed orthonormal basis: $X_k\times Y_k=Z_k$. Also $U_k=(U_k\cdot X_k)X_k+\Vert W_k\Vert Z_k$, so the felt-up vector lies in the $XZ$ plane, with its transverse component pointing towards positive $Z$.

The stated exclusion of $a_k=g_k$ ensures $U_k\ne 0$, but it does not ensure $W_k\ne 0$. If felt-up is parallel to travel, the roll angle is undetermined rather than the construction being impossible. Every transverse orthonormal basis then has the required plane property. For continuity, project the previous $Z$ into $X_k^{\perp }$ and normalize it, using $Y_k=Z_k\times X_k$. If this projection is also too small, choose a coordinate unit vector least aligned with $X_k$ and project that vector instead. Switch to this transported choice near the parallel case to avoid numerical amplification. A stopped vehicle likewise needs a chosen limiting travel tangent vector, since its zero velocity itself has no direction. None of this requires a Frenet frame, which can fail where curvature vanishes.

Write $v=u_x$ and $D={\partial }_t+v{\partial }_x$. The Euler equations and conservation of mass give $D\rho =-\rho v_x$ and $\rho Du=-p_x{e}_x$. The internal-energy density of the ideal gas is $e=p/(\gamma -1)$. The adiabatic pressure equation therefore givesMultiply the momentum equation by $u$ and use continuity to obtain the kinetic-energy balanceAdding the two equations moves ${\partial }_x(pv)$ into the flux. Thus the entire system is ${\partial }_{t}Q+{\partial }_{x}F=0$, whereThis derives the conservative energy flux of a polytropic ideal gas. In particular one-dimensional spatial dependence does not remove the transverse velocities from the total energy. The energy flux can equally be written $\rho v(u^2/2+w)$ with specific enthalpy $w=\gamma p/[(\gamma -1)\rho ]$.

Let $U=|u_1|$ and $q=(\gamma -1)/(\gamma +1)$, the strong-shock limit of the normal velocity ratio. In the plane containing the upstream direction and the shock normal, take $u_1=U{e}_X$ and $n=\cos \beta e_X+\sin \beta e_Y$. The normal velocity is multiplied by $q$ while the tangential velocity is unchanged, henceEliminating $\beta$ gives the strong-shock polar for a perfect gas:Equivalently this is the circleBoth signs of the perpendicular component are possible by changing shock orientation. The endpoints $u_{X2}=qU,U$ lie on the upstream axis. An original sketch for $\gamma =5/3$ illustrates the locus and a tangent ray:

The deflection is the angle of the ray from the origin to the downstream velocity, rather than the shock-normal angle $\beta$. Its maximum occurs where this ray is tangent to the circle. The circle's center is at distance $\gamma U/(\gamma +1)$ and its radius is $U/(\gamma +1)$, so the tangent right triangle provesThis is the maximum deflection through a strong perfect-gas shock; for $\gamma =5/3$ it is about $36.87$ degrees.

For fixed volume define $E_B={\int }_V{B}^2/(2{\mu }_0)dV$ and $J=\nabla \times B/{\mu }_0$. Differentiate and use the ideal magnetohydrodynamic induction equation. The vector identitythen gives the general balanceThe volume term is the work done by the magnetic force on the fluid, with the opposite sign for magnetic energy. Under the force-free assumption of part (b), it vanishes, leaving the requested boundary expression. Without that assumption, or another reason for zero net Lorentz work, a boundary-only balance is not generally valid.

For an axisymmetric boundary its normal has no azimuthal component. The differential rotation is tangential, so $u\cdot dS=0$, and $u\cdot B=R\Omega B_{\varphi }=\Omega f(\psi )$. Therefore ${\dot{E}}_B={\mu }_0^{-1}{\oint }_S\Omega f(\psi )B\cdot dS$. A thin axisymmetric tube between neighboring flux labels carries flux $2\pi d\psi$. Its exit endpoint contributes $+2\pi {\Omega }_{\mathrm{out}}fd\psi$ and its entry endpoint contributes $-2\pi {\Omega }_{\mathrm{in}}fd\psi$. Pair the endpoints, counting each once, to obtainThe endpoint convention fixes the sign. Include each connected tube segment through $V$ once; a field line with no boundary crossing contributes nothing. This is magnetic-energy injection by differential boundary rotation. A common angular velocity at both ends produces no such injection.