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For $s\ge 0$ and $\xi \in \mathbb{R}$, define the characteristic curve $X(t;s,\xi )$ byThe bounded derivative $F_x$ makes $F(t,\cdot )$ globally Lipschitz, uniformly in $t$. On each finite time interval, $|F(t,x)|\le |F(t,0)|+L|x|$, so Gronwall inequality prevents finite-time escape. The Picard-Lindelöf theorem therefore gives a unique trajectory for every finite $t\ge 0$. Differentiation in $\xi$ givesso the characteristic flow map is a $C^1$ increasing diffeomorphism.

Along a characteristic, the chain rule changes the equation intoTracing $(t,x)$ backward by the flow therefore givesThe regularity of the flow makes this a $C^1$ classical solution. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.

Let $X$ and $Y$ be two solutions with the same initial value and Brownian motion, and put $Z=X-Y$. Itô formula givesStop when either process or the stochastic integral becomes large. Taking expectations, using the assumed one-sided Lipschitz bound, and then removing the localization givesThe Gronwall inequality yields $\mathbb{E}{Z}_t^2=0$. Thus $X_t=Y_t$ almost surely at every rational time, and path continuity makes the two processes indistinguishable. This proves pathwise uniqueness.

Let $u$ and $v$ be two weak solutions with the same initial value, and put $w=u-v$. Bilinearity givesPair with $w$. The first transport term vanishes because $P_{N}u$ is divergence free. Part a and the Young inequality giveThereforeThe coefficient is integrable because $v\in L^2(0,T;V)$. Since $w(0)=0$, the Gronwall inequality gives $w=0$ on $[0,T]$. The global weak solution is unique.

For $h=f_1-f_2$, $h\prime =\sqrt{1+x}(f_1+f_2)h$ and $h(0)=0$. On every compact subinterval the coefficient is continuous, so Gronwall inequality gives $h=0$; hence equality holds throughout $[0,a)$.

Take the $L^2$ inner product of the temperature equation with ${\theta }_m$. The spectral projection disappears against ${\theta }_m\in {\tilde{H}}_m$, and part (a)(iii) cancels transport. ThusThe periodic Poincaré inequality, part (i), and the Cauchy-Schwarz inequality implyThe Gronwall inequality therefore gives, for $0\le t\le T$,This defines a bound $K_0(T)$ independent of $m$, and part (i) then givesIntegrating the energy identity and using the same bound on its right-hand side gives

It remains to estimate the time derivative. For $\phi \in H_{\mathrm{per}}^1$, the Fourier projection ${\Pi }_m$ is a contraction in $H^1$, and the skew identity from part (a) givesThe sobolev embedding theorem and the periodic elliptic estimate for the Stokes operator bound $\Vert u_m{\Vert }_{\infty }$ by $C|A{u}_m|$. Moreover,The already obtained bounds therefore implywith $K_0^{\prime }(T)$ independent of $m$.

Let $(u_1,{\theta }_1)$ and $(u_2,{\theta }_2)$ be two weak solutions with the same initial data, and set $U=u_1-u_2$ and $\Theta ={\theta }_1-{\theta }_2$. The diagnostic Stokes equation givesSubtracting the temperature equations yieldsPair this equation with $\Theta$. The term transported by $u_1$ vanishes by the skew-symmetry of incompressible transport, while the other nonlinear term satisfiesThe forcing difference is at most $C\Vert \Theta {\Vert }_2^2$. ConsequentlyThe coefficient is integrable on $[0,T]$ because ${\theta }_2\in L^2(0,T;H^1)$. Since $\Theta (0)=0$, the Gronwall inequality gives $\Theta =0$, and the Stokes equation then gives $U=0$. The weak solution is unique.

SetSymmetry here means $W(-z)=W(z)$, so $\nabla W(0)=0$ and $K(0)=0$. The bounded Hessian matrix and the fact that $J$ is an orthogonal matrix giveThus the particle system is ${\dot{X}}^i=N^{-1}{\sum }_{j}K(X^i-X^j)$. On the configuration space with norm $\Vert X{\Vert }_{\max }={\max }_i|X^i|$, its vector field has Lipschitz constant at most $2L$:The Picard-Lindelöf theorem gives local existence and uniqueness. Linear growth and the Gronwall inequality give $\Vert X(t){\Vert }_{\max }\le e^{2Lt}\Vert X(0){\Vert }_{\max }$, which rules out escape to infinity in finite time and extends the solution to every $t\ge 0$.

The associated empirical measure isHere each ${\delta }_{X^i(t)}$ is a Dirac measure. The finite first absolute moment follows from $\int |x|d{\mu }_t^N=N^{-1}{\sum }_i|X^i(t)|<\infty$. Pairing equal particle labels defines a coupling, soConsequently

A finite first moment suffices for the velocity, but may not suffice for the printed potential. For example, $W(z)=|z{|}^2/2$ has all the stated properties. If $\mu$ is the law of $(S,0)$ with density $2s^{-3}$ on $s\ge 1$, then $\mathbb{E}S=2$ but $\mathbb{E}{S}^2=\infty$, and $(W\ast \mu )(x)=\infty$ for every $x$. Thus the literal definition of $V$ needs a finite second absolute moment, or another hypothesis ensuring integrability of $W$.

For the specified class ${\mathcal{P}}_1$, define the velocity fields directly by $b_{\mu }=K\ast \mu$ and $b_{\hat{\mu }}=K\ast \hat{\mu }$. Equivalently, use the renormalized interaction potentialThe bounded Hessian matrix and $\nabla W(0)=0$ implyThus $V_{\mathrm{ren}}$ is finite, is twice continuously differentiable in space, and has $\nabla V_{\mathrm{ren}}=\nabla W\ast \mu$, by dominated differentiation under the integral sign. Its rotated negative gradient is exactly $b_{\mu }$. Whenever $W\ast \mu$ is finite, the two potentials differ only by an $x$-independent constant and give the same transport equation.

We check the hypotheses for characteristic representation of a measure-valued continuity equation. For every $x,y$,For any coupling $\gamma \in \Gamma ({\mu }_t,{\hat{\mu }}_t)$, the Lipschitz condition on $K$ also givesTaking the infimum over couplings proves the uniform boundApplying the same estimate at two times of one curve proves time continuity of $b_{\mu }$, uniformly in $x$. Also $|m_1({\mu }_t)-m_1({\mu }_s)|\le d_1({\mu }_t,{\mu }_s)$, so $m_1({\mu }_t)$ is bounded on compact time intervals. Spatial partial derivatives of $b_{\mu }$ are continuous and bounded by $L$, using dominated convergence in space and weak convergence of probability measures in time for the bounded continuous kernel $DK$. All corresponding facts hold for $\hat{\mu }$.

The characteristic representation of a measure-valued continuity equation says that a continuous velocity field, continuously differentiable and globally Lipschitz in space and with locally uniform linear growth, has a global characteristic flow map $X_{\mu }(t;s,a)$ solvingFor each $f\in {\mathcal{P}}_1$, its pushforward measure is the unique weak solution in $C([0,\infty ),{\mathcal{P}}_1)$ of the associated continuity equation. The hypotheses have all been verified above, and the fields have zero divergence, so these are also the two required transport equations. Explicitly,For completeness, the Gronwall inequality givesso these pushforward measures have finite first absolute moments. Integrating $|X_{\mu }(t;0,a)-X_{\mu }(s;0,a)|$ against $f$ proves continuity in the Wasserstein distance. The chain rule proves the weak identity. For uniqueness at a fixed terminal time $t$, use the backward test function $\phi (s,x)=\psi (X_{\mu }(t;s,x))$, where $\psi$ is continuously differentiable with compact support. It satisfies ${\partial }_s\phi +b_{\mu }\cdot \nabla \phi =0$ and has compact spatial support uniformly for $0\le s\le t$, by the backward linear-growth bound. The weak identity with temporal cutoffs therefore gives $\int \psi d{\rho }_t=\int \psi (X_{\mu }(t;0,a))df(a)$, identifying every weak solution with the displayed pushforward measure.

Finally compare the two characteristic curves with the same initial point $a$. Put ${\Delta }_t=X_{\mu }(t;0,a)-X_{\hat{\mu }}(t;0,a)$. The spatial Lipschitz condition and the boxed velocity estimate givefor almost every time; the integral form gives the same conclusion at zeros of $\Delta$. The Gronwall inequality impliesThe pushforward measure $(X_{\mu }(t;0,\cdot ),X_{\hat{\mu }}(t;0,\cdot ){)}_{\#}f$ is a coupling of ${\rho }_t,{\hat{\rho }}_t$. Integrating the last bound against the probability measure $f$ provesThe same constant $L$ depends only on the uniform bound for the Hessian matrix of $W$, and is independent of the prescribed curves and their common initial data.

The nonnegative quadrant is invariant because both right-hand sides have the corresponding population as a factor. There $u_2^{\prime }\le \rho \gamma u_2$. The differential inequality, or the Gronwall inequality, givesBiologically $\gamma <0$ means $\delta >r_2$: per-capita removal exceeds even the grey squirrels' maximum uncrowded growth rate. Competition can only strengthen the decline.

The existence assertion is false under the hypotheses printed in the source. No measurability of $\theta$ is specified. Even requiring $\theta$ to be a Borel measurable function does not suffice: a coefficient can be positive at every point while approaching zero too rapidly near the starting point. This is the obstruction described by pointwise positivity does not ensure weak existence.

Here is a counterexample in dimension one, with $\Theta =1$:Suppose that a solution to the martingale problem with domain $C_c^{\infty }(\mathbb{R})$ existed. The cutoff argument from part (a)(ii), now with $\mathcal{L}=\theta \Delta$, makes $X$ a continuous local martingale withThe criterion for a continuous local martingale with bounded quadratic variation makes it a square-integrable martingale, and the Itô isometry gives $\mathbb{E}{X}_t^2=\mathbb{E}[X{]}_t$.

Apply the occupation-times formula with the indicator of the singleton $\{0\}$. Since that singleton has zero Lebesgue measure,Thus $X$ spends zero ordinary time at zero. Away from zero, $\theta (x)\le x^2$, soThe Gronwall inequality forces $\mathbb{E}{X}_t^2=0$ for every $t$. Continuity then forces $X\equiv 0$ almost surely, contradicting either the zero occupation time just proved or $[X{]}_t=2t$ for that constant path. Consequently

For completeness, the intended construction works if $\theta$ is a Borel measurable function and bounded away from zero on every compact set. In particular, it works for continuous strictly positive $\theta$. Let $B$ be standard $d$-dimensional Brownian motion, and define a clock and its inverse byThe Brownian path on each finite interval has compact range, so the added hypothesis makes $A_u$ finite. The clock is continuous and strictly increasing, and $A_u\ge u/(2\Theta )$, so $A_u\to \infty$. Its inverse is continuous and satisfies ${\tau }_t\le 2\Theta t$. Each ${\tau }_t$ is a stopping time for the Brownian filtration.

For $f\in C_c^{\infty }({\mathbb{R}}^d)$, the optional sampling theorem applied at these bounded random times shows thatis a true martingale in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}^B$. The inverse-clock substitution $d{\tau }_t=2\theta (X_t)dt$, valid almost everywhere, rewrites it asThis proves the Brownian time change for a variable diffusion speed:The factor $2$ in the clock is necessary because standard Brownian motion has infinitesimal generator $\Delta /2$.

There need not be a weak solution of a stochastic differential equation $d{X}_t=\sqrt{2\theta (X_t)}d{B}_t$, $X_0=0$, even when $\theta$ is a bounded Borel measurable function strictly positive everywhere. For the displayed example, any solution would be a continuous local martingale with $[X{]}_t=2{\int }_0^t\theta (X_s)ds\le 2t$, hence a square-integrable martingale. The occupation-times formula applied to $1_{\{0\}}$ gives $0={\int }_0^t{1}_{\{X_s=0\}}d[X{]}_s=2{\int }_0^t{1}_{\{X_s=0\}}ds$. Since $\theta (x)\le x^2$ away from zero, $\mathbb{E}{X}_t^2\le 2{\int }_0^t\mathbb{E}{X}_s^{2}ds$. The Gronwall inequality makes $X\equiv 0$, a contradiction. Thus a diffusion amplitude positive at a single problematic point cannot repair arbitrarily slow neighbouring motion. A local positive lower bound, as in the Brownian time change for a variable diffusion speed, rules out this example.

If the rate functions are bounded and Lipschitz on an invariant compact state space, the Poisson time-change representation of a Markov chain has drift $F={\sum }_j{v}_j{\beta }_j$. Each centered Poisson term is uniformly small after division by $n$. The Gronwall inequality then bounds the distance to $\dot{z}=F(z)$ by the initial error and the centered counting error, multiplied by $e^{LT}$. A Poisson maximal concentration bound gives exponentially decreasing error probabilities for a fixed horizon and fixed error tolerance.

Before the difference $Z$ hits zero, its diffusion coefficient has magnitude at most $KZ$ and its drift is at least $-LZ$. Itô's formula bounds the drift of $f_{ϵ}(Z)$ by $(L+K^2)f_{ϵ}(Z)$. Its stochastic integrand is bounded by $K/(4ϵ)$, making expectation legitimate on finite horizons. The Gronwall inequality and the value $f_{ϵ}(0)=1/ϵ$ then exclude finite-time contact from a strictly positive starting difference.

The stochastic SI model with susceptible fraction $X_n$, initial limit $a$, and total infection rate $n\lambda X_n(1-X_n)$ has deterministic limitThe Poisson time-change representation of a Markov chain writes $X_n$ as its integrated drift minus a centered Poisson error. On $[0,T]$ its clock is at most $n\lambda T/4$. The Poisson maximal concentration bound makes every fixed error tolerance have summable probabilities in $n$, so the Borel-Cantelli lemma gives uniform error decay almost surely. The Gronwall inequality transfers that decay to $X_n-x$, because the drift is Lipschitz continuous. The assertion is for fixed finite intervals and does not claim approximation over times growing with $n$.

For the weak energy solution of a variable-coefficient wave equation, uniqueness can be proved without using $u_t$ as a spatial $H_0^1$ test function. For a difference with zero data, fix $s$ and test by $v(t)={\int }_t^{s}u(r)dr$ for $t<s$, extended by zero afterwards. Put $z(t)={\int }_0^{t}u(r)dr$, so $v(t)=z(s)-z(t)$. Integration by parts in time for the symmetric principal form, and in space for the first-order terms, giveFor short enough intervals the last term is absorbed. The Gronwall inequality forces $u=0$, and repetition proves uniqueness on the full interval. This antiderivative test is valid at the stated space-time $H^1$ regularity, unlike an unqualified direct energy test by $u_t$.

For globally Lipschitz drift $b$, couple $d{X}^{\epsilon }=b(X^{\epsilon })dt+\epsilon dW$ and $\dot{x}=b(x)$ from the same point. The Gronwall inequality gives ${\sup }_{s\le t}|X_s^{\epsilon }-x_s|\le \epsilon e^{Lt}{\sup }_{s\le t}|W_s|$. Thus convergence is uniform on each finite horizon. The Feynman-Kac formula for generator ${\epsilon }^2\Delta /2+b\cdot \nabla +c$ has weight $\exp \int c$, with a positive sign. Bounded continuous $c$ is uniformly continuous on a compact neighbourhood of the deterministic trajectory, and bounded continuous $f$ supplies a deterministic integrable bound. The dominated convergence theorem proves the displayed limit without requiring global uniform continuity of $c$.