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Near $(\bar{u},\bar{v})$ in a graph, this property requires neighborhoods $U,V$ and a finite $\kappa$ such that $S(u)\cap V\subset S(u^{\prime })+\kappa \Vert u-u^{\prime }\Vert \mathbb{B}$ for all $u,u^{\prime }\in U$. Nearby solutions can be matched under nearby perturbations with linear displacement control. Single-valued maps reduce to local Lipschitz continuity.

For arbitrary real $n\ne 0,1$, work first on an interval where $y>0$. Define exponentiation by $y^r=\exp (r\log y)$, using the real logarithm. The change of variable is differentiable and invertible there, and the chain rule givesMultiplying the Bernoulli differential equation by $(1-n)y^{-n}$ yields the first-order linear differential equationThis is an exact transformation of nonzero solutions of a differential equation on the chosen branch; it is valid far from any reference state and does not discard higher powers as a local linearization would.

Given the initial condition $y(x_0)=y_0>0$, the integrating factor and transformed solution of a differential equation areThe product rule verifies $(\mu z{)}^{\prime }=(1-n)\mu Q$. Recover $y=z^{1/(1-n)}$ only on the connected component of $\{x:z(x)>0\}$ containing $x_0$, inside the original coefficient interval. The first-order linear differential equation can have a longer existence interval than this inverse transformation allows.^[ref]

Negative values need a separately justified real branch. For integer $n$, the required powers are real and differentiable for $y<0$ too, but the inverse must retain the sign selected by the initial condition. For $n=3$, $z=y^{-2}>0$ represents either $y=+1/\sqrt{z}$ or $y=-1/\sqrt{z}$; both square roots are taken of positive $z$. For $n=2$, $z=1/y$ retains the sign directly. For a general noninteger exponent, use the positive branch unless a specific differentiable real negative branch has been established. Never evaluate a real fractional power of a negative value merely because the linear formula has become negative.

Division by $y^n$ excludes $y=0$. When $n>0$ and the original real power is defined at zero, $y\equiv 0$ is an additional solution of a differential equation and must be checked in the original equation. When $n<0$, the equation as written is undefined at zero. On a positive branch the right-hand side is locally Lipschitz in $y$, so the Picard-Lindelöf theorem gives local uniqueness. At zero, the same hypothesis needs a fresh check: it holds for the polynomial $n=2$ example, whereas for $0<n<1$ a nonzero power coefficient prevents local Lipschitz continuity there. If $z\downarrow 0$, the inverse sends $y\to \infty$ for $n>1$, but sends $y\downarrow 0$ for $0<n<1$. The latter endpoint can lead to finite-time extinction in a sublinear Bernoulli equation, where the original equation determines how to join to zero.

A causal curve has an everywhere nonspacelike tangent of consistent time orientation; its smooth segments are timelike or null. More generally, curves with local Lipschitz continuity and future causal tangent almost everywhere are allowed. The causal future and causal past are defined by reachability along these curves.

For a one-step approximation $y_{n+1}=y_n+h\varphi (t_n,y_n,h)$ to an ordinary differential equation, insert the exact solution and define the unscaled local truncation errorConsistency of a numerical method requires ${\tau }_{n+1}/h\to 0$ uniformly on the finite interval being approximated, under the stated smoothness assumptions. Merely asking ${\tau }_{n+1}\to 0$ is too weak: the useless update $y_{n+1}=y_n$ has defect $h{y}^{\prime }(t_n)+O(h^2)$, which tends to zero but generally fails the scaled test. Here $O$ denotes Big O notation. MIT uses this division-by-step convention.^[ref]

For the Euler method, Taylor's theorem gives ${\tau }_{n+1}=h^2{y}^{\prime \prime }({\xi }_n)/2$. A bounded second derivative therefore gives local truncation error $O(h^2)$ and a scaled defect $O(h)$. Under a uniform Lipschitz continuity assumption on $f$ in $y$, within a region containing the exact and numerical solutions, numerical stability controls accumulation over about $T/h$ steps. This yields global error $O(h)$ with exact starting data. Consistency of a numerical method checks each step; numerical stability controls their combination.

Every characteristic function is uniformly continuous and satisfies $\phi (0)=1$ and $|\phi (t)|\le 1$. Continuity follows from the dominated convergence theorem applied to $e^{itX}$.

More explicitly, if $t_n\to t$, then $e^{i{t}_{n}X}\to e^{itX}$ pointwise and $|e^{i{t}_{n}X}|=1$. Apply the dominated convergence theorem to the real part and imaginary part separately. There is no finite moment hypothesis. For uniform continuity, the stronger estimatefollows by factoring out $e^{itX}$, whose absolute value is one. The integrand tends pointwise to zero and is bounded by two, so the dominated convergence theorem applies again. The bound is independent of $t$, which supplies uniform continuity on the entire real line.^[ref]

If the first absolute moment is finite, $|e^{iu}-1|\le |u|$ additionally gives $|{\phi }_X(t+h)-{\phi }_X(t)|\le |h|\mathbb{E}[|X|]$, a Lipschitz continuity estimate. The earlier argument matters because continuity survives even when this expected value is infinite. For convergence in distribution, the separate Lévy continuity theorem concerns limits of whole sequences of characteristic functions; the argument here establishes continuity of one characteristic function.

Solutions depend continuously on their forcing when convergence of the forcing in a specified norm implies convergence of solutions in a specified solution norm, with boundary conditions fixed. A uniform estimate $\Vert u-v{\Vert }_X\le C\Vert f-g{\Vert }_Y$ is a stronger quantitative form, namely Lipschitz continuity of the solution map. The monotonicity energy estimate for a semilinear Dirichlet problem is an example.

A real convex function locally bounded on an open convex set has a continuous supporting linear functional at every interior point. Apply the dominated Hahn-Banach theorem to its sublinear directional derivative $p_x$ to obtain $l\le p_x$. Increasing secant slopes imply $p_x(z)\le f(x+z)-f(x)$ whenever the endpoint is in the domain. A local Lipschitz continuity bound gives both $l(y)\le L\Vert y\Vert$ and $-l(y)=l(-y)\le L\Vert y\Vert$, proving continuity.

A contraction mapping $T:X\to X$ on a nonempty complete metric space has exactly one fixed point $x_{\ast }$, and the sequence $x_{n+1}=T(x_n)$ converges to $x_{\ast }$ from every $x_0\in X$. The same constant $0\le L<1$ must satisfy $d(Tx,Ty)\le Ld(x,y)$ for all $x,y\in X$. The statement applies to metric spaces, including closed subsets of a Banach space; neither linearity nor a vector space structure is required.^[ref]

The force of the result is that existence, uniqueness, and a practical approximation procedure arrive together. Repeated application produces corrections bounded by a geometric series, hence a Cauchy sequence; completeness supplies its limit. The Lipschitz continuity of $T$ makes that limit a fixed point. Two different fixed points would have their positive distance multiplied by at most $L<1$ while remaining unchanged, which is impossible.

The displayed estimate, for all $x$ and real $t$, defines a coordinate Lipschitz constant $L_i$ for the partial derivative in coordinate $i$. It implies the coordinate descent lemmaby the fundamental theorem of calculus and integration of the Lipschitz continuity estimate. Subtracting an affine function leaves these constants unchanged.

The apparently weaker estimate along the singleton negative gradient direction is equivalent when every real step size is allowed: if the initial partial derivative is nonzero, any displacement can be written $-\alpha {\partial }_{i}f(x)e_i$; if it is zero but the final partial derivative is nonzero, reverse the endpoints; if both vanish the estimate is trivial.

Target: gradient-descent.bigb, under Lipschitz gradient.

This initial value problem uses a square root and absolute value to define a continuous function on all of ${\mathbb{R}}^2$. For every finite waiting time $c\ge 0$, a forward solution of a differential equation on $t\ge 0$ isThe constant solution of a differential equation $y_{\infty }(t)=0$ is another choice. For example, at $t=3$, waits $c=0,1,2,3$ give values $9,4,1,0$. These are dimensionless mathematical trajectories; no measured physical waiting time is being supplied.

Verification uses the original ordinary differential equation, including the join. Before departure, both $y_c^{\prime }$ and $2\sqrt{|y_c|}$ vanish. For $t>c$, their common value is $2(t-c)$. For $c>0$, the difference quotient at the join isThus the derivative exists at $c$, equals the rate law there, and is continuous across the join. For $c=0$ the same computation gives the right derivative at the starting endpoint. For $c>0$, the left and right second derivatives at departure are $0$ and $2$: these forward solutions of a differential equation are continuously differentiable functions, without being twice differentiable at departure. The local two-sided statement of the Peano existence theorem is separate from this explicitly forward family.

The rate law tends to zero as $y\to 0$, but its secant slopes from zero are unbounded:No finite local Lipschitz constant exists at zero. At $\epsilon =10^{-2},10^{-4},10^{-6}$ the quotient is $20,200,2000$, even though the rates are shrinking to zero. Continuity controls the rate itself; Lipschitz continuity would control its change relative to the state displacement.

There is also a numerical lesson. The forward Euler method starting exactly at zero gives $Y_{k+1}=Y_k+2\Delta t\sqrt{|Y_k|}=0$ at every step. It reproduces the constant solution of a differential equation for every positive step size, while the moving trajectories remain equally valid. A computation producing one trajectory cannot establish uniqueness. Likewise, the separable differential equation calculation that divides by $\sqrt{|y|}$ only describes the moving pieces: checking the original equation recovers the zero pieces and their joins.

If $E\subseteq [0,1]$ is a Lebesgue measurable set and $0\le b\le \lambda (E)$, some measurable subset $B\subseteq E$ has Lebesgue measure $b$. The function $F(t)=\lambda (E\cap [0,t])$ has Lipschitz continuity with constant $1$, starts at $0$ and ends at $\lambda (E)$; the intermediate value theorem proves the assertion. Applying the assertion successively to remaining portions of $E$ gives pairwise disjoint measurable pieces with any finite list of nonnegative measures whose sum is at most $\lambda (E)$. This is a concrete consequence of non-atomic measure structure, which fails for general measures containing an atom of a measure.

The Euler method approximates an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, by evaluating the derivative at the current point and moving along that tangent line for a time $h$. With $t_n=t_0+nh$, its displayed update needs one evaluation of $f$ per step. This simple rule turns local slope information into a whole approximate trajectory. MIT's course notes derive it by dropping the second-order Taylor remainder.^[ref]

Take $y^{\prime }=-2y$, $y(0)=1$, whose exact solution is the exponential function $y(t)=e^{-2t}$. The Euler method givesAt $h=0.2$, the first values are $1,0.6,0.36,0.216$. At $t=1$, five steps give $0.6^5=0.07776$, compared with $e^{-2}\approx 0.135335$. The result decays but is too small: absolute stability alone does not ensure accuracy. At $h=1.2$, the first values are $1,-1.4,1.96,-2.744$, and at $t=6$ five steps give $-5.37824$, while the exact solution is about $0.00000614421$.

For the Dahlquist test equation $y^{\prime }=\lambda y$, the Euler method has stability function $R(z)=1+z$ with $z=h\lambda$. Its absolute stability region is $|1+z|\le 1$, the closed disk centred at $-1$ with radius one. For the present real decay problem and $h>0$, bounded numerical values require $0<h\le 1$; strict decay requires $0<h<1$. At $h=1$ the values alternate $1,-1,1,\dots$ without decay. These conditions follow directly from the amplification factor $1-2h$, matching MIT's criterion for $y^{\prime }=-ay$.^[ref]

The Euler method is first order: with bounded second derivative and suitable Lipschitz continuity, its global error is $O(h)$ on a fixed interval. In this example the logarithm gives $\log (1-2h)=-2h-2h^2+O(h^3)$ for small positive $h$. At $t=nh$, this means $y_n=e^{-2t}(1-2th+O(h^2))$. It explains both the downward bias and why halving the step roughly halves the error once $h$ is sufficiently small.

This explicit numerical method is a two-stage Runge-Kutta method and a one-step method. For a sufficiently smooth ordinary differential equation, the chain rule gives $y^{\prime \prime }=f_x+f_{y}f$, and Taylor's theorem shows that its local truncation error is $O(h^3)$. Uniform Lipschitz continuity of $f$ in its solution argument gives an increment-function Lipschitz constant at most $L(1+hL/2)$. The Lipschitz one-step error bound then proves second-order numerical convergence on a fixed time interval. This method uses an intermediate prediction from $y_n$; the explicit midpoint linear multistep method instead requires a separate previous value $y_{n-1}$.

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.

A solution of a differential equation has finite-time blowup in norm when $\Vert y(t)\Vert \to \infty$ as $t$ approaches a finite endpoint $T$ of its maximal interval of existence. This is stronger than merely approaching an excluded boundary of an open state domain while staying bounded. The finite-dimensional continuation criterion distinguishes these obstructions; for a locally Lipschitz continuous ordinary differential equation on $I\times {\mathbb{R}}^d$, a finite maximal endpoint inside $I$ necessarily entails finite-time blowup.

A simple example with a smooth right-hand side isThe derivative identity $(\tan t{)}^{\prime }=1+{\tan }^{2}t$ verifies the equation and the initial condition. Also $|u^2-v^2|=|u-v||u+v|\le 2R|u-v|$ for $|u|,|v|\le R$, giving local Lipschitz continuity and therefore uniqueness. Yet the positive blowup time is $T=\pi /2\approx 1.57079633$. Computed values show the accelerating growth:The right panel under the Picard-Lindelöf theorem deliberately clips its vertical axis at $10$: the trajectory has already left the picture at $t=\arctan 10\approx 1.47112767$, before reaching $T$. A plot ceiling is a display limit, not a finite value at blowup.

The tangent function also quantifies the last part of the trajectory. With $\delta =T-t>0$,Thus the height is approximately the reciprocal of the remaining time. At remaining times $\delta =0.1,0.01,0.001$, the heights are approximately $9.96664442,99.99666664,999.99966667$. These follow from the sine and cosine limits at zero and explain the steep rise without relying on the appearance of a clipped function graph.

The distinction is useful when interpreting a solution of a differential equation: uniqueness certifies a trajectory while it exists, while growth estimates decide how far it can be followed. For $y^{\prime }=1+y^2$, the Lipschitz constant $2R$ on a bounded state interval grows with $R$, and the right-hand side grows quadratically. There is no global Lipschitz constant or linear-growth bound of the kind used by the finite-dimensional continuation criterion. Smoothness at every finite state is entirely compatible with finite-time blowup.

Let $X$ be a nonempty compact metric space and let $f:X\to X$ satisfy $d(f(x),f(y))<d(x,y)$ whenever $x\ne y$. Then $f$ has exactly one fixed point. The distance condition implies continuity, and $x\mapsto d(f(x),x)$ attains a minimum by compactness. If a minimizer $p$ were not a fixed point, then $d(f^2(p),f(p))<d(f(p),p)$ would contradict minimality. Two distinct fixed points would contradict the same strict distance inequality directly. A uniform Lipschitz constant below $1$ is unnecessary; compactness replaces the quantitative hypothesis of the contraction mapping theorem.

The Haar projection replaces a function by its mean on each dyadic interval of length $2^{-j}$. For a function satisfying Lipschitz continuity with constant $L$, every value in a cell differs from its cell mean by at most $L{2}^{-j}$. This bound holds for all points with a consistent endpoint convention, and requires no global L2 norm assumption when the projection is defined by local averages.

Here Heun's method means the explicit endpoint-averaging Runge-Kutta method: take an Euler predictor $Y_2=y_n+hf(t_n,y_n)$, sample $k_2=f(t_n+h,Y_2)$ there, and average it with $k_1=f(t_n,y_n)$ for the output. The second Runge-Kutta stage is evaluated at the predictor, not at the corrected endpoint. In particular it requires no implicit solve as the ODE trapezoidal rule does. Naming conventions vary between books; the formulas specify the method unambiguously.^[ref]

For smooth $f$, a Taylor expansion about $(t_n,y_n)$ gives $k_2=f+h(f_t+f_{y}f)+O(h^2)$, where $f_y$ is the state Jacobian matrix. Consequently the output is $y_n+hf+h^2(f_t+f_{y}f)/2+O(h^3)$, matching the exact solution through degree two by the chain rule. This gives local truncation error $O(h^3)$ and, with the usual bounded smoothness and Lipschitz continuity hypotheses on a fixed finite interval, global error $O(h^2)$. The Euler-Heun embedded pair compares this output with its Euler predictor.

For $y^{\prime }=y$, $y_n=1$ and $h=0.2$, the sampled slopes are $1$ and $1.2$. Transporting the second slope to the starting point would end at $1.24$; the average slope $1.1$ instead ends at $1.22$. The exact exponential function reaches $e^{0.2}\approx 1.22140275816$. The magnified endpoint comparison separates the pair difference $0.02$ from the actual Heun local defect $0.00140275816$.

The Hölder exponent in Hölder continuity measures the power in the bound $|f(x)-f(y)|\le C|x-y{|}^{\alpha }$. For $0<\alpha \le 1$, a larger exponent imposes stronger local regularity on bounded distance ranges. Exponent one is Lipschitz continuity.

Let $v\le 0$ be $C^1$ on a connected domain and $C^2$ away from its zero set, with $\Delta v+v=0$ wherever $v<0$. If negative and zero points coexist, an interior Euclidean ball in $\{v<0\}$ touches its zero set. On that Euclidean ball $\Delta v=-v>0$, so the Hopf boundary point lemma forces a nonzero derivative at an interior maximum, contrary to $Dv=0$. Thus $v$ is identically zero or strictly negative. Mere Lipschitz continuity does not suffice: $-\max \{\sin x_1,0\}$ on $(-1,1{)}^n$ has both negative and zero regions.