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The cubic amplitude saturation equation $R^{\prime }=R(Q^2-R^2)$ and the Picard-Lindelöf theorem imply that positive solutions cannot cross the equilibrium $Q$. If $0<R_0<Q$, then $R$ stays below $Q$ and increases. If $R_0>Q$, then it stays above $Q$ and decreases. The explicit cubic amplitude saturation solution tends to $Q$ in both cases, soThe same relation makes the distance a decreasing monotone function in any spatial norm for which the plane wave has finite norm, for example on a periodic domain. At large time, $R-Q\sim -\frac{Q}{2}(Q^2/R_0^2-1)e^{-2Q^{2}t}$.

The printed monotonicity wording must be interpreted as monotone amplitude relaxation, or monotone decay of the distance between the waves. Complex numbers have no ordinary monotone ordering, and the real part at a fixed point can oscillate. For instance, $k=U=1$, $\mu =2$ and $R_0=1/2$ give $\mathrm{Re}{\psi }_R(0,t)=R(t)\cos t$, which is not monotone despite the monotone amplitude. This result concerns the common-phase single-mode family, not stability against arbitrary spatial perturbations.

A local form of the Picard-Lindelöf theorem is as follows. Let $F(t,y)$ be continuous on a rectangle about $(t_0,y_0)$ and uniformly Lipschitz in $y$ there. Then some interval about $t_0$ admits a unique continuously differentiable solution ofThe solution is the unique fixed point of the associated integral operator. If $F$ is continuous on $[a,b]\times {\mathbb{R}}^n$ and globally Lipschitz in $y$ with one constant, the solution exists uniquely throughout $[a,b]$.

Because $\Vert b^{\prime }{\Vert }_{\infty }<\infty$, the function $b$ is Lipschitz continuous. the Picard-Lindelöf theorem, proved by iteration onconverges uniformly on every compact interval. The usual factorial estimate proves convergence for arbitrary interval length, and the Gronwall inequality proves uniqueness. Thus there is a unique global continuous solution.

For $w=W$, induction shows that $X_t^{(n)}$ is ${\mathcal{F}}_t$-measurable: its value uses only $(W_s{)}_{s\le t}$ and earlier iterates up to time $t$. The pointwise limit $X_t$ is therefore ${\mathcal{F}}_t$-measurable. Hence $X$ is adapted to $({\mathcal{F}}_t)$.

Write the scalar field energy as $E=T+V$, whereUnder the Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$, a change of variables givesA finite-energy solution of the Euler-Lagrange equation is stationary under this admissible variation. The Derrick virial identity is therefore

The static field equation is ${\varphi }^{\prime \prime }=U^{\prime }(\varphi )$, so integration givesThe polynomial before $C$ is nonnegative and vanishes, so requiring the minimum to be zero fixes $C=0$. Hence the vacuum manifold iswhich has three elements. A finite-energy scalar-field kink can join only adjacent vacua: a solution cannot cross the intermediate vacuum at finite $x$ because its first integral would have $\varphi ={\varphi }^{\prime }=0$ there and the Picard-Lindelöf theorem would make it constant. There are therefore four oriented topological sectors,comprising two increasing kinks and their two antikinks. Symmetry under $\varphi \mapsto -\varphi$ and spatial reflection generates all four from one profile.

For the $0\to 1$ sector, completing the square gives the Bogomolny boundEquality holds for the Bogomolny equationWith $y={\varphi }^2$, this becomes the logistic differential equation $y^{\prime }=2y(1-y)$. Translation invariance supplies an arbitrary center $x_0$, and the explicit kink in a phi-six model isIt tends to $0$ and $1$ at the two spatial ends and saturates the bound. Its mass is consequently

The potential factorizes asIt is even and nonnegative, with three degenerate vacua at $\varphi =-2,0,2$. Its two intervening maxima occur at $\varphi =\pm 2/\sqrt{3}$ and have height $256/27$.

A finite-energy static solution in one dimension satisfiesMultiplication by ${\varphi }^{\prime }$ and use of the vacuum boundary conditions gives the first integralAny path from a negative vacuum to a positive vacuum must pass through the intermediate vacuum $\varphi =0$. There both $U$ and ${\varphi }^{\prime }$ vanish. The Picard-Lindelöf theorem then forces a solution reaching $(\varphi ,{\varphi }^{\prime })=(0,0)$ at finite $x$ to remain there. Equivalently, the first-order orbit approaches $\varphi =0$ only as $|x|\to \infty$. This intermediate-vacuum obstruction to a kink means that a single kink cannot connect $-2$ to $2$; it splits into two elementary kinks at infinite separation.

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.

On the finite-dimensional space $H_m$, the Galerkin equation is an autonomous system of ordinary differential equations whose right-hand side is a quadratic polynomial in the coefficients of $u_m$. It is locally Lipschitz, so the Picard-Lindelöf theorem gives a unique maximal local solution.

Taking the $H$ inner product with $u_m$ givesThe advecting field $P_N{u}_m$ is divergence free. Periodicity and the skew-symmetry of incompressible transport therefore make the nonlinear term zero. HenceA finite-dimensional solution can cease to exist only if its norm diverges. This uniform bound prevents such blow-up, so the solution extends uniquely through every interval $[0,T]$.

The first equation determines $u_m$ linearly from ${\theta }_m$:Substitution into the second equation gives an ordinary differential equation on the finite-dimensional space ${\tilde{H}}_m$. Its right-hand side is polynomial, and therefore locally Lipschitz continuous. The Picard-Lindelöf theorem supplies a unique local solution. The energy estimate in part (iii) bounds ${\theta }_m$ on every finite time interval, so the finite-dimensional continuation criterion rules out finite-time escape. The solution is consequently unique on every interval $[0,T]$.

Put $\lambda ={\omega }^2-g/a>0$. The zero value of the first integral in b givesAt the endpoint $r=a$, this gives ${\dot{r}}^2=\lambda a^2/2$. The absolute speed includes radial, azimuthal and vertical components. Since $\dot{z}=\dot{r}$ there,Hence the limiting exit speed isThere is a small initial condition qualification in the wording. The exact initial conditions $r=\dot{r}=0$ give the stationary solution forever, by Picard-Lindelöf theorem. The displayed zero-first integral moving branch instead approaches that unstable equilibrium as $t\to -\infty$.

For a precise small disturbance, release the bead with zero relative velocity at $r=ϵ>0$. Its first integral is $J=-\lambda {ϵ}^2/2$, givingThe bead initially accelerates outward and reaches the end. Letting $ϵ\to 0^{+}$ recovers exactly the stated exit speed. The conserved $J$ is a Jacobi integral associated with the imposed rotation; ordinary mechanical energy need not be conserved because the apparatus driving the wire can do work.

Work on an interval with $x\ne 0$. The homogeneous first-order differential equation becomes a separable differential equation under $v=y/x$: substituting $y=xv$ gives $v+x{v}^{\prime }=v^2+v$, hence $x{v}^{\prime }=v^2$.

For $v\ne 0$, integration gives $-1/v=\ln |x|+C$. Renaming the arbitrary constant producesFor the nonzero family the denominator must remain nonzero, so take any connected interval avoiding both $x=0$ and $|x|=e^C$. The zero solution must be restored after division by $v^2$; by the Picard-Lindelöf theorem away from $x=0$, a solution cannot switch between zero and a nonzero branch there.

First retain the equilibrium solution $\boxed{y\equiv 0}$. Any nonzero real solution has constant sign on its interval of existence, since the polynomial right-hand side is locally Lipschitz and Picard-Lindelöf theorem prevents crossing the zero solution.

For a nonzero solution, the Bernoulli differential equation substitution $u=y^{-4}$ givesThis linear ordinary differential equation has solution $u+2=C{e}^{-8x}$. Therefore every nonzero real solution isEither sign may be chosen and remains fixed. The restriction ensures $y^{-4}>0$; nonpositive $C$ gives no nonzero real solution. These are maximal real intervals, and their restrictions are also solutions. As the right endpoint is approached, $|y|$ diverges in finite-time blowup. The substitution would have lost $y=0$ if it had not been checked separately.

The zero solution satisfies both the ordinary differential equation and the initial condition. The right-hand side is continuously differentiable in $y$ near $(x,y)=(2,0)$, so the Picard-Lindelöf theorem gives local uniqueness; equivalently the nonzero family in part (a) can never take the value zero at $x=2$. ThusThe restriction $x>0$ is the maximal interval containing $2$ on which the original ordinary differential equation is defined.

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

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set, $d<\infty$, and $f:U\to {\mathbb{R}}^d$ a continuous function. Through every $(t_0,y_0)\in U$ there is a local solution of a differential equation: a continuously differentiable function $y:I\to {\mathbb{R}}^d$ on an open interval containing $t_0$, satisfying the displayed initial value problem pointwise. No Lipschitz condition is assumed.^[ref]

Continuity of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and initial condition actually be followed? Prediction also needs uniqueness. The delayed solutions of a square-root differential equation below show how the same rule and initial condition can leave a departure time undetermined. The Picard-Lindelöf theorem supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.

For the initial value problem $y^{\prime }=2\sqrt{|y|}$, $y(0)=0$, a forward solution of a differential equation can stay zero until any $c\ge 0$, then follow $y=(t-c{)}^2$. The delayed solutions of a square-root differential equation verify this freedom exactly.

Local existence gives neither a global solution of a differential equation nor a bound on all future values. The continuous function $f(t,y)=y^2$ already permits finite-time blowup; the finite-dimensional continuation criterion explains the additional issue of extending a trajectory within its domain.

Proof outline: choose a compact set $K=[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$. By the extreme value theorem, $M={\max }_K\Vert f\Vert <\infty$. Take $0<h\le a$ with $Mh\le r$. Forward Euler method polygons remain in the closed ball and satisfy $\Vert y_n(t)-y_n(s)\Vert \le M|t-s|$. The Arzelà-Ascoli theorem yields a convergent subsequence with uniform convergence. Uniform continuity of $f$ on $K$ makes their integral equation defect tend uniformly to zero, so the limit satisfiesThe fundamental theorem of calculus gives a continuously differentiable function. Repeat backward and join at $t_0$, where both derivatives equal $f(t_0,y_0)$.^[ref]

The compact set here uses a finite-dimensional vector space; bounded closed balls in an arbitrary Banach space need not be compact. When $M=0$, the constant solution of a differential equation $y=y_0$ works on the entire time interval in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.

Consider the original analytic initial value problemHere $x$ and $y$ are dimensionless variables in a stipulated mathematical model. The trajectories below are exact solutions of a differential equation, with no measured data or empirical parameter fit. For $a\ne 0$, the Bernoulli power substitution $z=1/y$ givesThe integrating factor is $e^{-x}$. Direct differentiation gives $y^{\prime }=-y+e^x{y}^2$, independently checking the transformation. The linear $z$ exists for every real $x$, while the nonlinear solution of a differential equation uses only the interval containing $0$ where $z\ne 0$:For $a<0$, $y^{\prime }>0$, so its absolute value decreases toward zero. The lost zero solution of a differential equation is recovered by substitution into the original equation. Since $f(x,y)=-y+e^x{y}^2$ is locally Lipschitz in $y$ everywhere, the Picard-Lindelöf theorem ensures uniqueness through every finite point, including $y=0$. Thus a nonzero trajectory never meets or crosses zero at a finite $x$.

For $a=1/2$, the exact solution of a differential equation is $y=e^{-x}/(2-x)$ on $(-\infty ,2)$. It first decreases, reaches a local minimum $y(1)=e^{-1}$, then increases without bound. At the endpoint,The finite-time blowup comes from a simple zero of $z$ at $2$, where $z^{\prime }(2)=-e^2$, rather than from a singular coefficient in the first-order linear differential equation. The same algebraic expression for $y$ is negative on $(2,\infty )$, but that is a separate solution of a differential equation on a disjoint interval; infinity supplies no finite joining value. For $a=-1$, $y=-e^{-x}/(1+x)$ on $(-1,\infty )$, giving a distinct decaying branch whose finite blowup is instead in the past.

The direction field also explains the positive turning point. Its nullclines are $y=0$ and $y=e^{-x}$, becauseAbove the positive nullcline, $y^{\prime }>0$; between it and zero, $y^{\prime }<0$; below zero, $y^{\prime }>0$. The moving nullcline $y=e^{-x}$ is not a solution of a differential equation: its derivative is $-e^{-x}$, whereas the prescribed slope there is zero. The $a=1/2$ trajectory crosses it at $x=1$, exactly its local minimum.

A streamline at a specified instant $T$ is a curve whose tangent line follows the velocity field at that same frozen instant everywhere along the curve. One convenient integral curve parameterization isHere $s$ is an auxiliary curve parameter, with units of seconds in this convention; advancing $s$ does not advance physical time. Any regular reparameterization preserves the tangent geometry. Where $u_x\ne 0$, a planar streamline satisfies $dy/dx=u_y/u_x$ evaluated at time $T$. At a stagnation point the velocity field has no direction; under the local Lipschitz condition of the Picard-Lindelöf theorem, regular streamlines at the same time cannot cross. A streamline answers which directions fit together now, making it useful for describing local flow geometry. ^[ref]

In the analytical velocity field $(U,at)$ introduced under pathline, the streamline through the origin at time $T$ isAt $T=2\mathrm{s}$ this is $y=2x$ when both coordinates are expressed in metres. At $x=1\mathrm{m}$ its height is $2\mathrm{m}$, whereas the earlier pathline of the origin's first released fluid element has height $0.5\mathrm{m}$. Its instantaneous streamline at the final position $(2,2)\mathrm{m}$ is instead $y-2\mathrm{m}=2(x-2\mathrm{m})$: the pathline is tangent to that streamline at its current endpoint, although its recorded history does not lie along it.

The green streamline uses the velocity field at $T=2\mathrm{s}$ throughout; the orange pathline uses times $0\le t\le 2\mathrm{s}$; the purple streakline uses different release times at the single observation time $T=2\mathrm{s}$. The grey arrows all show the instantaneous velocity direction at $T$. These are exact analytical curves with chosen units, not measured dye shapes.

In a steady flow in a fixed reference frame, a pathline and a streamline solve the same autonomous ordinary differential equation. Uniqueness makes each particle released from a fixed point follow the same integral curve, so its pathline, the streakline from that point and the streamline share one geometrical curve. Finite recording or release intervals usually show different segments of that curve; the comparison assumes particles stay in the domain and treats stationary fluid elements separately. Neither two-dimensionality nor incompressibility is needed for this coincidence. Steadiness is sufficient, but is not necessary: $\mathbf{u}(\mathbf{x},t)=g(t)\mathbf{v}(\mathbf{x})$ with smooth $g(t)>0$ changes speed while preserving the same integral curves. In a general time-dependent velocity field, however, the three curves differ. Kline explains why continuous dye release in steady flow can therefore display streamlines. ^[ref]

For two-dimensional incompressible flow, the convention $\mathbf{u}=({\partial }_y\psi ,-{\partial }_x\psi )$ gives $\mathbf{u}\cdot \nabla \psi =0$, so regular streamlines are level sets of the stream function at each frozen time. In time-dependent incompressible flow, that does not force a pathline to stay on one numerical value of $\psi$: its material derivative is $D\psi /Dt={\partial }_t\psi$. The distinction is the difference between a geometric snapshot and a parcel's history.

Compare the displayed ordinary differential equation with $u^{\prime }=1+u^2$, $u(0)=1$, whose solution is $u=\tan (x+\pi /4)$. Their difference satisfies $w^{\prime }=(y+u)w+x^2$, $w(0)=0$. The integrating factor formula makes $w\ge 0$ on the common forward existence interval. Since $u$ diverges at $\pi /4$, the increasing solution $y$ has a maximal forward endpoint $0<x_{\ast }\le \pi /4$. If $y$ remained bounded at this finite endpoint, local existence with the smooth right side would continue it, a contradiction. Hence $y\to +\infty$. The Picard-Lindelöf theorem guarantees local existence, not existence through such a finite-time blow-up. On $|x|\le 1$, $|y|\le 10$, a bound $|1+x^2+y^2|\le 102$ and distance nine to the nearer horizontal box edge give the explicit guaranteed interval $|x|\le 3/34$.

For the initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, continuity of $f$ and local Lipschitz continuity in $y$ ensure a unique local solution. The Lipschitz condition compares different state values at a common time: $\Vert f(t,u)-f(t,v)\Vert \le L\Vert u-v\Vert$ on a suitable neighborhood. Subtracting two integral equations and applying Gronwall's inequality proves uniqueness.

Continuity alone does not suffice. The scalar equation $y^{\prime }=2\sqrt{|y|}$ with $y(0)=0$ has the zero solution and, for every $a\ge 0$, the solution $y(t)=0$ up to $a$ and $y(t)=(t-a{)}^2$ thereafter. The non-Lipschitz behavior at zero permits an arbitrary waiting time. This illustrates why the hypotheses of the Picard-Lindelöf theorem carry real information.

For the autonomous differential equation $y^{\prime }=-|y-y_{\ast }{|}^p$, the displayed family stays at $y_{\ast }$ until time $c$ and then decreases below it. Here $[z{]}_{+}=\max (z,0)$. Direct differentiation verifies the equation on each side; since $1/(1-p)>1$, the derivative tends to zero at departure, making the joined function continuously differentiable. Distinct departure times give different solutions from the same equilibrium initial condition. The ratio $|f(y_{\ast }+\epsilon )-f(y_{\ast })|/\epsilon ={\epsilon }^{p-1}$ is unbounded, so the local Lipschitz continuity hypothesis of the Picard-Lindelöf theorem fails.

A solution starting above $y_{\ast }$ reaches it in finite time. It can then use any of these continuations. For $p=1/2$, $y_{\ast }=-1$, and initial value $y(0)=0$, arrival occurs at time two; departure at $c\ge 2$ gives $y=-1-(t-c{)}^2/4$ afterwards. Waiting-time solutions of a sublinear Bernoulli equation give a related mechanism for departures in the opposite direction.

Let q be continuous and strictly positive, with a constant a. A nonnegative solution of a differential equation may stay zero until any prescribed point $x_0$ and become positive afterwards. For $x>x_0$, defineand set $y=0$ before $x_0$. The Bernoulli power substitution verifies the equation after $x_0$. Continuity of q gives $y=O((x-x_0{)}^{1/(1-p)})$, so the joining derivative is zero and the equation also holds at the join. These different waiting times give nonunique solutions from zero. The right side lacks Lipschitz continuity as a function of y near zero, so the uniqueness conclusion of the Picard-Lindelöf theorem does not apply there.