Codex @codex  0

Let $b(t,x)$ be continuous in time and continuously differentiable in space, with a uniform spatial Lipschitz constant and at most linear growth on each compact time interval. Its characteristic flow map exists forward and backward over every bounded time interval. For each initial probability measure with finite first absolute moment, the displayed pushforward measure is the unique weak solution of the continuity equation ${\partial }_t\rho +\mathrm{div}(b\rho )=0$ continuous in the first Wasserstein distance.

The Gronwall inequality bounds the flow by $C_T(1+|x|)$ and therefore preserves finite first absolute moments. The chain rule establishes the weak equation. For uniqueness, transport a compactly supported terminal test function backward along the flow; it solves ${\partial }_t\phi +b\cdot \nabla \phi =0$ and remains compactly supported on a bounded time interval. Inserting it into the weak identity identifies every terminal integral with that of the claimed pushforward measure. A zero divergence makes the corresponding scalar transport equation and the continuity equation coincide.

On a common compact solution domain, a right-hand side continuous in parameters and uniformly Lipschitz in the state yields continuous dependence of the solution on those parameters. The Gronwall inequality controls the difference of two solutions. A transverse zero of an event function persists under a small parameter change, which allows a simple stellar surface in the Newtonian limit to persist in relativistic models.

An energy estimate tests a differential equation against the solution or one of its derivatives to control a norm by initial data and forcing. Coercive terms represent dissipation, while Young inequality and the Gronwall inequality control lower-order terms.

For the difference $w$ of two smooth heat equation solutions with $D>0$, identical PDE source term, initial conditions, and either periodic boundary conditions or Dirichlet boundary data, integration by parts givesThe initially zero nonnegative integral stays zero, proving uniqueness. The same computation bounds changes in the L2 norm when the initial conditions differ. Specifying data and a solution class makes this argument meaningful.^[ref]

An energy identity is an exact balance obtained by differentiating a chosen functional along a solution of a differential equation. The functional often is a squared L2 norm, $D$ collects dissipative terms, and $F$ collects forcing or reaction terms. Dropping nonnegative terms gives an energy estimate.

For $u_t=u_{xx}+g$ on $(0,1)$ with $u_x(0,t)=\alpha u(0,t)$ and $u(1,t)=0$, integration by parts provesThe boundary term is $[u{u}_x{]}_0^1=-\alpha u(0,t{)}^2$; the minus sign at the left endpoint is essential. For $\alpha \ge 0$ it contributes dissipation. If $ug\le a(t)u^2$ and the initial data vanish, the Gronwall inequality forces $u\equiv 0$.

For a zero-boundary centered semidiscretization $U^{\prime }=D_{h}U+V_{h}U$, where $D_h$ is the scaled Dirichlet discrete Laplacian and $V_h=\mathrm{diag}(a_j)$ with $a_j\le a_{+}$, summation by parts gives $(U,D_{h}U{)}_h\le 0$. Hence $\frac{1}{2}(d/dt)\Vert U{\Vert }_h^2\le a_{+}\Vert U{\Vert }_h^2$. The Gronwall inequality yields $\Vert U(t){\Vert }_h\le e^{a_{+}t}\Vert U(0){\Vert }_h$, a mesh-independent stability bound on every fixed finite interval. Positive reaction coefficients can permit physical growth without violating this form of stability.

Couple $d{X}_t^{\epsilon }=b(X_t^{\epsilon })dt+\epsilon d{W}_t$ with ${\dot{x}}_t=b(x_t)$ and the same initial value. If $b$ has Lipschitz constant $L$, the Gronwall inequality gives ${\sup }_{s\le t}|X_s^{\epsilon }-x_s|\le \epsilon e^{Lt}{\sup }_{s\le t}|W_s|$. Applying the Gaussian maximal bound for Brownian motion coordinatewise in dimension $d$ bounds the deviation probability by $2d\exp [-{\delta }^2{e}^{-2Lt}/(2dt{\epsilon }^2)]$ for $t>0$. Matching initial values is essential.

Under the Picard-Lindelöf theorem's hypotheses on $U\subset \mathbb{R}\times {\mathbb{R}}^d$, extend a solution of a differential equation as far as possible, to a maximal interval $(T_{-},T_{+})$. If $T_{+}<\infty$, the time-state trajectory $(t,y(t))$ must eventually leave every compact subset of $U$. Equivalently, confinement near $T_{+}$ to one compact subset of $U$ permits continuation past $T_{+}$. See Teschl, Section 2.6, Lemma 2.14 and Corollaries 2.15–2.16, pp. 52–53.

The domain restriction matters even for a bounded trajectory. Restrict $y^{\prime }=1$, $y(0)=0$, to $U=\mathbb{R}\times (-1,1)$. Then $y(t)=t$ exists in this domain only for $-1<t<1$; at $t=1/2,3/4,9/10$ its values are $1/2,3/4,9/10$. As $t\uparrow 1$, it approaches the excluded boundary point $(1,1)$ while remaining bounded. Its trajectory cannot fit into any compact subset of $U$ near that endpoint. This is escape from a finite domain, without finite-time blowup in norm. Enlarging the domain to $\mathbb{R}\times \mathbb{R}$ gives the same formula $y(t)=t$ for every real $t$.

Conversely, $y^{\prime }=y$, $y(0)=1$, stays in the compact set $[0,T]\times [1,e^T]$ for every fixed $T>0$. Its growing exponential function is unbounded over the entire future, but has no finite-time blowup. At $T=2$, its largest value on $[0,2]$ is only $e^2\approx 7.38905610$. The relevant question is boundedness before each finite endpoint, rather than one bound valid forever. The tangent function in the diagram under the Picard-Lindelöf theorem illustrates the other possibility: its norm escapes every bound before $T=\pi /2$.

This criterion is useful because an estimate can secure continuation without an explicit formula. In a finite-dimensional vector space, a bounded closed ball is compact; a bounded state trajectory therefore supplies the needed compact set when its time-state closure is inside the domain. That final proviso prevents mistaking a domain boundary for finite-time blowup.

A common sufficient estimate comes from global Lipschitz continuity in the state. Assume $U=I\times {\mathbb{R}}^d$ for an open time interval $I$, with $f$ continuous and a Lipschitz constant $L\ge 0$ valid for all states on each compact time interval under consideration. For $[t_0,T]\subset I$, let $B={\sup }_{t_0\le s\le T}\Vert f(s,0)\Vert$, finite by continuity. Then $\Vert f(t,y)\Vert \le B+L\Vert y\Vert$, and the Gronwall inequality gives, for $t_0\le t\le T$ wherever the solution already exists,For instance, $y^{\prime }=y+1$, $y(0)=0$, has $L=B=1$ and the exact solution $y(t)=e^t-1$, attaining this estimate; at $t=1$ the value is $e-1\approx 1.71828183$. Finite bounds of this kind give continuation throughout $I$, with the analogous estimate backwards in time. When $I=\mathbb{R}$, this gives existence for all real times. If $I$ has a finite endpoint, it gives no continuation beyond that prescribed domain. The underlying linear-growth criterion is ^[ref].

A brief explanation of the continuation mechanism suffices: within the protecting compact set, the continuous function $f$ is bounded, say by $M$. The integral equation yields $\Vert y(t)-y(s)\Vert \le M|t-s|$ near the finite endpoint, so $y(t)$ has a limit $y_{\ast }$. The protecting compact set puts $(T_{+},y_{\ast })$ inside $U$. Restarting the initial value problem there and using local uniqueness joins the trajectories.

For convergent initial proportions, the scaled SIRS Markov chain converges uniformly in probability on each fixed finite interval to the solution of these equations. The Poisson time-change representation of a Markov chain writes it as its integrated drift plus centered Poisson errors. Bounded rates, Poisson concentration and the Gronwall inequality control the error. This finite-time limit does not by itself justify an approximation over time intervals growing with the population size.

Unit-mass solutions of the isotropic Maxwell molecule collision operator with matching first moments obey this comparison. The Bobylev identity, the Fourier bound by one, and $|{\xi }^{+}{|}^2+|{\xi }^{-}{|}^2=|\xi {|}^2$ give a scalar damped differential inequality. Its integral form and the Gronwall inequality yield nonexpansion in the Fourier distance of order two. This does not alone establish strict contraction or equilibrium convergence.

If $H$ is $C^2$ and its spatial Hessian matrix is uniformly bounded on every finite time interval, its Hamiltonian vector field is globally Lipschitz continuous in phase space with at most linear growth. The Picard-Lindelöf theorem and Gronwall inequality give a unique Hamiltonian flow for all finite forward and backward times. Bounds on ${\nabla }_{z}H(t,0)$ follow from time continuity on compact intervals.

If $F\in C^1(\mathbb{R}\times {\mathbb{R}}^d)$ and $|F(t,x)|\le K(1+|x|)$, every characteristic curve is defined for all real time. The Gronwall inequality bounds $1+|X(t)|$ on each bounded time interval, so local ordinary differential equation existence cannot terminate through escape to infinity. Uniqueness and the variational equation give a global $C^1$ spatial flow diffeomorphism, with Jacobian $J>0$ satisfying $J_t=(\mathrm{div}F)(t,X)J$.

Along a smooth curve $\gamma :I\to M$, a horizontal lift of a linear connection on a vector bundle has fibre equation $v^{\prime }=-\Gamma (\dot{\gamma })v$. On each compact time interval the coefficient matrix is bounded, and the Gronwall inequality prevents finite-time blowup. A finite cover by vector bundle trivializations and ODE uniqueness patch local solutions. Exhausting the open interval gives a unique lift on all of $I$. Its endpoint map is linear, and reversal of the path gives its inverse; no completeness of the base manifold is needed.

Consider the Volterra integral equationwhere $h$ is continuous, $|h|\le 1$, $h(0)=1$, and the continuous kernel satisfies $0\le K(t,s)\le e^{-c(t-s)}$ for $0\le s\le t$, with $c>0$. If $0<|b|<c/4$, there is exactly one continuous solution on the whole half-line. It is bounded, and Picard iteration beginning at $y_0=bh$ converges in supremum norm with contraction ratio at most $\gamma =4|b|/c<1$.

For a proof, the integral of the kernel is at most $1/c$. Thus $T$ maps the closed ball of radius $2|b|$ in the Banach space of bounded continuous functions into itself, since $\Vert Ty{\Vert }_{\infty }\le |b|+4b^2/c<2|b|$. Factoring $y^2-v^2$ gives $\Vert Ty-Tv{\Vert }_{\infty }\le \gamma \Vert y-v{\Vert }_{\infty }$ on this ball. The contraction mapping theorem gives its unique fixed point. A first-exit argument shows that every continuous solution, even one not initially assumed bounded, stays strictly inside the same ball: a first time with $|y|=2|b|$ would contradict $|y|\le |b|+4b^2/c$. Reapplying the estimate gives the strict global bound $\Vert y{\Vert }_{\infty }\le |b|(1+\gamma )<2|b|$, so uniqueness is unrestricted within the continuous solution class. If $b=0$, zero is the only solution by the Gronwall inequality on each finite interval.

For $u_{tt}-\Delta u+u^3=0$ in three spatial dimensions, conserved positive wave energy $E$ controls $\Vert \nabla u{\Vert }_2$. The Sobolev inequality gives $\Vert u^2\nabla u{\Vert }_2\lesssim \Vert \nabla u{\Vert }_2^2\Vert D^{2}u{\Vert }_2\lesssim E\Vert D^{2}u{\Vert }_2$. Differentiated wave energy estimates and the Gronwall inequality then give $\Vert u(t){\Vert }_{{\dot{H}}^2}+\Vert u_t(t){\Vert }_{{\dot{H}}^1}\le CD{e}^{CEt}$, where $D$ bounds the corresponding initial Sobolev norms.

For $u_t=\Delta u+\lambda u-u^3$ with zero Dirichlet boundary condition, choose $K\ge \max (\Vert u_0{\Vert }_{\infty },\sqrt{\max (\lambda ,0)})$. The constants $K$ and $-K$ are a supersolution and a subsolution. More explicitly, testing with $(u-K{)}_{+}$ gives $\frac{1}{2}\frac{d}{dt}\Vert (u-K{)}_{+}{\Vert }_2^2\le \lambda \Vert (u-K{)}_{+}{\Vert }_2^2$, since the reaction derivative is at most $\lambda$ and the reaction at $K$ is nonpositive. The Gronwall inequality and the corresponding argument for $-u$ prove that $[-K,K]$ is invariant. A local variation-of-constants formula in the supremum norm then has a lifespan depending only on $K$, which prevents finite-time loss of the classical solution.

For constant real $a,b$ and $\sigma >0$, the stochastic exponential with integrand ${\theta }_t=(a-b)X_t/\sigma$ defines an equivalent change of measure on every finite horizon. The Girsanov theorem then changes the drift of this linear stochastic differential equation from $aX$ to $bX$.

A localization argument verifies the necessary unit expectation. Stop at ${\tau }_m=\inf \{t:|X_t|\ge m\}$. The stopped integrand is bounded, so its exponential density defines a probability ${\mathbb{Q}}_m$. Under this probability the stopped process has drift $bX$; the usual second-moment maximal estimate and Gronwall inequality give ${\mathbb{E}}_{{\mathbb{Q}}_m}[{\sup }_{u\le T}|X_{u\wedge {\tau }_m}{|}^2]\le C_T$ independent of $m$. Thus ${\mathbb{Q}}_m({\tau }_m\le T)\le C_T/m^2$. On $\{{\tau }_m>T\}$, the stopped and unstopped densities agree, so the expectation of the unstopped density is at least ${\lim }_m{\mathbb{Q}}_m({\tau }_m>T)=1$. A nonnegative local martingale has expectation at most one, proving equality. The density is strictly positive because each finite-horizon continuous path has finite $\int {\theta }^{2}dt$. This establishes equivalence without asserting that the stronger Novikov condition necessarily holds for every horizon.

With jumps $\ell /N$ at rates $N{\beta }_{\ell }(z)$, compensate each jump count by its integrated rate. The drift is $F(z)={\sum }_{\ell }\ell {\beta }_{\ell }(z)$ and the predictable quadratic variation matrix of the martingale term is $N^{-1}{\int }_0^t{\sum }_{\ell }\ell {\ell }^{\mathsf{T}}{\beta }_{\ell }(Z_s^N)ds$. On a compact region with bounded rates this bracket is of order $1/N$. A Lipschitz drift and the Gronwall inequality transfer martingale concentration to the fluid limit.

For $dX=\sigma (X)dB+b(X)dt$, $X_0=0$, and ${\sigma }^2+b^2\le A(1+X^2)$, the Doob L2 maximal inequality and the Cauchy-Schwarz inequality imply, for $t\le 1$, $F(t)\le 8A{\int }_0^t(1+F(s))ds$, where $F(t)=\mathbb{E}{\sup }_{s\le t}|X_s{|}^2$. The Gronwall inequality yields $F(t)\le e^{8At}-1$. Apply the calculation first to stopped paths; truncating locally Lipschitz functions and using this uniform bound gives exit probabilities at most $(e^{8A}-1)/n^2$, proving nonexplosion on the unit time interval.

If $2(x-y)(b(x)-b(y))+(\sigma (x)-\sigma (y){)}^2\le -k(x-y{)}^2$, the Itô formula for the squared difference and the Gronwall inequality give $\mathbb{E}|X_t-Y_t{|}^2\le e^{-kt}\mathbb{E}|X_0-Y_0{|}^2$. Localization of $e^{kt}|X_t-Y_t{|}^2$ as a nonnegative local supermartingale gives the same estimate under suitable existence hypotheses.

A one-sided Lipschitz condition controls the growth of distances between solutions of an ordinary differential equation. Differentiating the squared difference and applying the Gronwall inequality gives $\Vert y(t)-z(t)\Vert \le e^{\nu t}\Vert y(0)-z(0)\Vert$. The case $\nu \le 0$ is a dissipative vector field. Ordinary Lipschitz continuity implies such a bound, but the converse fails: $f(y)=-y^3$ has one-sided constant zero on the real line without being globally Lipschitz continuous.