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A smooth vector field along a map $\gamma :[a,b]\to M$ is a smooth section of the pullback tangent bundle ${\gamma }^{\ast }TM$: it assigns $V(t)\in T_{\gamma (t)}M$, with smooth coefficients in each local frame. It need not be the restriction of one ambient vector field, since a curve may return to the same point with different values of $V$.

An affine connection, also called a Koszul connection, induces a pullback connection on this bundle. In coordinates, write $V=V^i(t){\partial }_i$ and ${\nabla }_{{\partial }_j}{\partial }_k={\Gamma }^i{}_{jk}{\partial }_i$. Its covariant derivative along a curve isThe connection transformation law, or the Leibniz rule applied to a change of frame, makes this expression independent of the frame. In particular it is defined even when $\dot{\gamma }=0$; it differentiates the section's coefficients as well as the frame.

The field is parallel precisely when $D_{t}V=0$. In a frame along a coordinate segment this is the linear ordinary differential equationThe Picard-Lindelöf theorem gives a unique solution with prescribed initial value. Smooth coefficients on compact subintervals are bounded, and the usual norm estimate for a linear ordinary differential equation prevents finite-time blowup there. A finite sequence of coordinate segments covers the image of the compact interval $[a,b]$; solve successively and use uniqueness on overlaps. This gives a unique parallel field on the whole interval for every $V_a\in T_{\gamma (a)}M$.

Define parallel transport by ${\tau }_t(V_a)=V(t)$. Linearity of the equation and uniqueness give ${\tau }_t(c{V}_a+d{W}_a)=c{\tau }_t(V_a)+d{\tau }_t(W_a)$. Solving along the reversed curve supplies its inverse. Therefore each ${\tau }_t:T_{\gamma (a)}M\to T_{\gamma (t)}M$ is a linear isomorphism.

For two affine connections, put $\Delta (X,Y)={\nabla }_{X}Y-{\tilde{\nabla }}_{X}Y$. Their rules givesince the two $X(f)Y$ terms cancel. The tensoriality of this operation proves that the difference of affine connections is a tensor, of type $(1,2)$, with smooth coordinate coefficients ${\Gamma }^i{}_{jk}-{\tilde{\Gamma }}^i{}_{jk}$.

A parametrized geodesic satisfies $D_t\dot{\gamma }=0$, or the geodesic equationWith $v^i={\dot{\gamma }}^i$, this becomes the first-order system ${\dot{\gamma }}^i=v^i$, ${\dot{v}}^i=-{\Gamma }^i{}_{jk}(\gamma )v^j{v}^k$. Its right side is smooth, so the Picard-Lindelöf theorem gives a unique local solution for each initial point and tangent vector. This establishes uniqueness with the parametrization fixed.

The two accelerations differ byIf $\Delta (v,v)=0$ for every tangent vector, either acceleration vanishes exactly when the other does. Conversely, start the $\nabla$-geodesic with arbitrary initial vector $v$ at an arbitrary point. If it is also a $\tilde{\nabla }$-geodesic with the same parameter, evaluating this identity initially gives $\Delta (v,v)=0$. HencePolarization makes the latter condition equivalent to $\Delta (u,v)+\Delta (v,u)=0$: parametrized geodesics determine the symmetric part of an affine connection. In particular torsion-free connections are determined by their parametrized geodesics, since their difference is also symmetric and must then vanish.

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 $[0,T]$.

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.