The Picard–Lindelöf theorem

Theorem

The Picard–Lindelöf theorem

T-063Home MU-205Threads change

Statement

Let \( D \subseteq \mathbb{R} \times \mathbb{R}^n \) be open, let \( (t_0,y_0) \in D \), and let \( f : D \to \mathbb{R}^n \) be continuous on \( D \) and Lipschitz in its second argument: there exists \( L \gt 0 \) and a closed box \( R = \{(t,y) : |t-t_0| \le a,\ \|y-y_0\| \le b\} \subseteq D \) such that \( \|f(t,y_1)-f(t,y_2)\| \le L\|y_1-y_2\| \) for all \( (t,y_1),(t,y_2) \in R \). Let \( M = \sup_{(t,y)\in R} \|f(t,y)\| \) and \( h = \min\{a, b/M\} \) (interpreting \( b/M = \infty \) if \( M=0 \)). Then the initial value problem \[ y' = f(t,y), \qquad y(t_0) = y_0 \] has a solution \( y \in C^1([t_0-h,t_0+h],\mathbb{R}^n) \) with graph contained in \( R \), and this solution is the unique such solution on \( [t_0-h,t_0+h] \): any other \( C^1 \) solution of the IVP on that interval with graph in \( R \) coincides with \( y \) identically.

Why it matters

Continuity of \( f \) alone (Peano's theorem) guarantees a local solution but not uniqueness — trajectories can bifurcate from a single initial point, which is useless for prediction. The Picard–Lindelöf theorem identifies the extra hypothesis, a Lipschitz bound on \( f \) in \( y \), that rules this out: it is the theorem that lets you treat an ODE as a genuine deterministic law, and it underwrites every existence-uniqueness statement used downstream in dynamical systems, control theory, and numerical analysis (it is exactly what justifies calling the flow map \( \phi_t \) a well-defined function).

It is also the prototype application of the Banach fixed-point theorem to a function space rather than \( \mathbb{R}^n \): the proof method (Picard iteration as a contraction on a complete metric space of curves) is a template reused throughout analysis, from the inverse function theorem to existence theorems for integral and delay equations.

Hypotheses

\( f \) continuous on \( R \).Without continuity, \( f \) need not even have an antiderivative candidate; e.g. \( f(t,y) = \mathrm{sgn}(t) \) is discontinuous at \( t=0\) and the "solution" \( y(t) = |t| \) fails to be \( C^1\) there — existence in the classical (differentiable) sense can fail outright. \( f \) Lipschitz in \( y \), uniformly for \( t \) in the box.Drop this and existence survives (Peano) but uniqueness breaks: \( f(t,y) = \sqrt{|y|} \) is continuous but not Lipschitz near \( y=0\), and \( y(t)=0 \) and \( y(t) = t^2/4 \) (for \( t \ge 0 \)) both solve \( y'=\sqrt{y},\ y(0)=0\). The box \( R \) is compact (closed and bounded) with \( f \) bounded on it, giving finite \( M \).If \( R \) is not bounded, or \( f \) is unbounded on it, \( M=\infty \) and the shrinking radius \( h=\min\{a,b/M\}\) collapses to \(0\); the argument bounding the iterates' escape from \(R\) (Step 3) has no room to run, so even the existence half of the proof stalls before uniqueness is ever tested. \( D \) is open and \( (t_0,y_0) \) an interior point.Openness is what lets a box \( R\) of positive side-lengths \(a,b\) be chosen around \((t_0,y_0)\) inside \(D\) at all; without it (e.g. \((t_0,y_0)\) on the boundary of \(D\)) the construction of \(R\) in Step 1 has no room on one side and the theorem's hypotheses cannot even be instantiated.

Proof

\text{Fix } R = \{(t,y): |t-t_0|\le a,\ \|y-y_0\|\le b\} \subseteq D,\quad L,\ M,\ h = \min\{a,\,b/M\} \text{ as in the statement.}

Such \( R \) exists because \( D \) is open and \((t_0,y_0)\in D\); \(L\) and finite \(M\) exist by the Lipschitz and continuity hypotheses together with compactness of \(R\) (continuous functions attain a finite supremum on compact sets, extreme value theorem). A

y \text{ solves the IVP on } I=[t_0-h,t_0+h] \iff y \in C(I,\mathbb{R}^n) \text{ and } y(t) = y_0 + \int_{t_0}^{t} f(s,y(s))\,ds \ \ \forall t\in I.

(\(\Rightarrow\)) If \(y\) is \(C^1\) and solves \(y'=f(t,y)\), integrate both sides from \(t_0\) to \(t\) using the fundamental theorem of calculus. (\(\Leftarrow\)) If \(y\) is continuous and satisfies the integral identity, the integrand \(s\mapsto f(s,y(s))\) is continuous (composition of continuous maps), so by the fundamental theorem of calculus the right-hand side is \(C^1\) in \(t\) with derivative \(f(t,y(t))\); evaluating at \(t=t_0\) gives \(y(t_0)=y_0\). This equivalence converts a differential problem into a fixed-point problem for an integral operator. B

X = \{ y \in C(I,\mathbb{R}^n) : \|y(t)-y_0\| \le b \ \ \forall t \in I \}, \qquad d(y,z) = \sup_{t\in I}\|y(t)-z(t)\|.

\((X,d)\) is a closed subset of the Banach space \((C(I,\mathbb{R}^n), \|\cdot\|_\infty)\) — closed because it is defined by non-strict pointwise inequalities preserved under uniform limits — hence \((X,d)\) is itself a complete metric space (a closed subset of a complete space is complete). A

(Ty)(t) := y_0 + \int_{t_0}^t f(s,y(s))\,ds \quad \text{defines a map } T : X \to X.

For \(y\in X\), \((t,y(t))\in R\) for all \(t\in I\) since \(\|y(t)-y_0\|\le b\), so \(f(s,y(s))\) is defined and \(\|f(s,y(s))\|\le M\); \(Ty\) is continuous (integral of a continuous function) and \(\|(Ty)(t)-y_0\| = \left\|\int_{t_0}^t f(s,y(s))\,ds\right\| \le M|t-t_0| \le Mh \le b\) by the choice \(h\le b/M\), using the triangle inequality for integrals. So \(Ty \in X\). B

\|(Ty)(t)-(Tz)(t)\| \le L|t-t_0|\, d(y,z) \quad \text{for } y,z\in X,\ t\in I.

\(\|(Ty)(t)-(Tz)(t)\| = \left\|\int_{t_0}^t \big(f(s,y(s))-f(s,z(s))\big)\,ds\right\| \le \left|\int_{t_0}^t \|f(s,y(s))-f(s,z(s))\|\,ds\right| \le \left|\int_{t_0}^t L\|y(s)-z(s)\|\,ds\right| \le L|t-t_0|\,d(y,z)\), using the Lipschitz hypothesis on \(f\) pointwise inside the integral and then bounding \(\|y(s)-z(s)\|\) by its supremum. B

\|(T^k y)(t) - (T^k z)(t)\| \le \frac{L^k|t-t_0|^k}{k!}\, d(y,z) \quad \forall k\ge 0,\ t \in I.

Induction on \(k\). Base case \(k=0\) is trivial. Inductive step: apply the same integral estimate as Step 5 to \(T^{k+1}y - T^{k+1}z = T(T^ky) - T(T^kz)\), using the inductive hypothesis to bound \(\|(T^ky)(s)-(T^kz)(s)\| \le \frac{L^k|s-t_0|^k}{k!}d(y,z)\) inside the integrand and integrating \(|s-t_0|^k\) from \(t_0\) to \(t\). C t3

d(T^ky, T^kz) \le \frac{(Lh)^k}{k!}\, d(y,z), \qquad \text{and } \frac{(Lh)^k}{k!} \to 0 \text{ as } k\to\infty, \text{ so some } T^N \text{ is a contraction.}

Take the supremum over \(t\in I\) in Step 6 (\(|t-t_0|\le h\)). Since \(\frac{(Lh)^k}{k!}\) is the general term of the (convergent) exponential series \(e^{Lh}=\sum \frac{(Lh)^k}{k!}\), its terms tend to \(0\); choose \(N\) with \(\frac{(Lh)^N}{N!} \lt 1\), so \(T^N:X\to X\) is a contraction with constant \(\frac{(Lh)^N}{N!} \lt 1\). B

T \text{ has a unique fixed point } y^\ast \in X, \text{ and } T^N \text{ has none other.}

Banach fixed-point theorem applied to \(T^N\) on the complete metric space \((X,d)\) (Step 3, Step 7): \(T^N\) has a unique fixed point \(y^\ast\). If \(Ty^\ast \ne y^\ast\), then applying \(T^N\) to \(Ty^\ast\) gives \(T^N(Ty^\ast) = T(T^Ny^\ast) = Ty^\ast\), so \(Ty^\ast\) is also a fixed point of \(T^N\); by its uniqueness \(Ty^\ast = y^\ast\), a contradiction avoided only because in fact \(Ty^\ast=y^\ast\) — i.e. \(y^\ast\) is a fixed point of \(T\) itself, and any fixed point of \(T\) is automatically a fixed point of \(T^N\), so \(T\) and \(T^N\) share exactly the same (unique) fixed point. C t3

y^\ast \text{ is a } C^1 \text{ solution of the IVP on } I \text{ with graph in } R, \text{ and it is the only one.}

By Step 2, \(Ty^\ast=y^\ast\) is exactly the statement that \(y^\ast\) solves the integral equation, hence (Step 2, \(\Leftarrow\) direction) \(y^\ast\) is a \(C^1\) solution of the IVP; its graph lies in \(R\) because \(y^\ast\in X\). Conversely any \(C^1\) solution \(y\) of the IVP on \(I\) with graph in \(R\) lies in \(X\) and, by Step 2, satisfies \(Ty=y\), i.e. is a fixed point of \(T\) in \(X\); by Step 8 the fixed point of \(T\) in \(X\) is unique, so \(y=y^\ast\). A

Result

f \text{ continuous and Lipschitz in } y \text{ on } R \implies \exists! \ y\in C^1(I,\mathbb{R}^n): y'=f(t,y),\ y(t_0)=y_0,\ \text{graph}(y)\subseteq R,\ I=[t_0-h,t_0+h]

Reading. If the "velocity field" \( f \) is continuous and does not vary too wildly (Lipschitz) with respect to the state \( y \), then starting from any point \( y_0 \) at time \( t_0 \) there is exactly one trajectory obeying the law \( y'=f(t,y) \), at least for a short enough time \( h \) determined explicitly by the size of the box and the bounds \( L \), \( M \).

Scope. The conclusion is local: \( h \) can be much smaller than \( a \), and the theorem makes no claim about behaviour outside \( I \) or outside \( R \) (the solution may leave \( R \) and cease to be governed by this particular Lipschitz estimate, or blow up in finite time). It applies verbatim to systems (\( n\gt1\)) and, via the equivalence of a Lipschitz condition with a bounded partial derivative \( \partial f/\partial y \) on a convex domain (mean value inequality), covers essentially every \( C^1\) right-hand side encountered in practice.

Corollaries & converses

Global uniqueness on overlaps. If two solutions of \(y'=f(t,y)\) (with \(f\) locally Lipschitz on all of \(D\)) agree at one point of a common interval of definition, they agree on the whole common interval — proved by covering the interval with finitely many local boxes and chaining Step 9.
Maximal interval of existence. Local solutions patch together (by the uniqueness corollary above) into a unique maximal solution on an open interval \((\alpha,\beta)\), which either equals \(\mathbb{R}\) or has the solution leaving every compact subset of \(D\) as \(t\to\alpha^+\) or \(t\to\beta^-\) (no "silent" non-uniqueness ever reappears inside \(D\)).
\(C^1\) right-hand side suffices. If \(f\) is \(C^1\) in \(y\) (with continuous \(t\)-dependence) on \(D\), it is automatically locally Lipschitz in \(y\) there, by the mean value inequality applied on a convex neighbourhood with \(L=\sup\|\partial f/\partial y\|\); so the theorem applies to any \(C^1\) vector field without checking Lipschitz continuity by hand.
The converse fails. Existence and uniqueness of a solution do not imply \(f\) is Lipschitz: uniqueness can hold under weaker one-sided conditions (e.g. Osgood's or Nagumo's uniqueness criteria) that are strictly more general than a Lipschitz bound, so "unique solution exists" does not logically force the Lipschitz hypothesis to hold.

Fails without

No Lipschitz bound (only continuity): \(y'=\sqrt{|y|}\), \(y(0)=0\) has infinitely many solutions on any interval around \(0\) — for every \(c\ge0\), \(y_c(t)=0\) for \(t\le c\) and \(y_c(t)=\tfrac14(t-c)^2\) for \(t\gt c\) all solve the IVP; \(f(t,y)=\sqrt{|y|}\) is continuous but its "derivative" \(\tfrac{1}{2\sqrt{|y|}}\) is unbounded near \(y=0\), so no Lipschitz constant exists there.
No continuity of \(f\): take \(f(t,y)=2t\) for \(t\ne0\) redefined as \(f(0,y)=1\) (a jump discontinuity in \(t\) alone, independent of \(y\), hence trivially "Lipschitz in \(y\)" with \(L=0\)). The candidate antiderivative \(y(t)=t^2\) satisfies \(y'(t)=2t=f(t,y(t))\) for \(t\ne0\) but \(y'(0)=0\ne1=f(0,y(0))\), so no \(C^1\) function solves the IVP \(y(0)=0\) at all — existence in the classical sense breaks, illustrating why continuity of \(f\) cannot be dropped even though it plays no role in the uniqueness estimate itself.
Unbounded \(f\) on the box (\(M=\infty\)): if \(f\) is Lipschitz in \(y\) but not bounded on any box around \((t_0,y_0)\) (impossible for genuinely continuous \(f\) on a compact box, but the point is structural), the escape estimate of Step 4 gives no positive \(h\), and the theorem's existence conclusion becomes vacuous even before uniqueness is invoked.

Common errors

Quoting the theorem with the weaker hypothesis "\(f\) continuous" and expecting uniqueness — that gives Peano's theorem (existence only); confusing the two theorems is the single most common error.
Treating "Lipschitz in \(y\)" as if it were "Lipschitz in \((t,y)\) jointly" — the theorem only needs the Lipschitz estimate in the \(y\)-variable, with continuity (not a Lipschitz bound) in \(t\); imposing joint Lipschitz continuity is stronger than needed and can mislead students into thinking \(f\) must be jointly smooth.
Forgetting that \(h\) is generally strictly smaller than the half-width \(a\) of the box in \(t\) — students often assume the local solution automatically survives on the entire interval where \(f\) was checked to be Lipschitz.
Applying the global uniqueness corollary without first checking \(f\) is (at least locally) Lipschitz on the whole domain in play — patching only works because each local piece individually satisfies the hypothesis.
Mistaking "unique local solution" for "solution exists for all time" — blow-up in finite time (e.g. \(y'=y^2\), \(y(0)=1\), which is Lipschitz on every bounded box yet only exists on \([0,1)\)) is entirely compatible with the theorem.

Discussion

The theorem is named for Émile Picard, whose method of successive approximations (the iterates \(y_0, Ty_0, T^2y_0,\dots\) converging to the solution) predates and inspired the general Banach fixed-point theorem, and Ernst Lindelöf, who sharpened the hypotheses and the domain of validity around the turn of the twentieth century. The proof given here is the modern, streamlined version: rather than estimating the Picard iterates termwise as Picard originally did, it recognises the whole scheme as an instance of Banach's abstract contraction principle, with Step 6–7 supplying exactly the "eventually contractive" estimate the abstract theorem needs when \(T\) itself need not be a contraction on the first pass.

Conceptually the shift from Step 2 onward is the heart of the argument: differentiation is a badly-behaved operation to iterate (it is unbounded, discontinuous, hard to control uniformly), whereas integration is smoothing and contractive on short intervals. Converting the differential equation into the equivalent integral equation trades a hard analytic problem for a fixed-point problem in a Banach space of continuous functions, and this "integrate the ODE first" trick reappears throughout numerical ODE theory (it is precisely why explicit Runge–Kutta and multistep methods are analysed via their agreement with this integral form) and in the abstract Cauchy problem for evolution equations in infinite dimensions.

A genuine subtlety is the gap between local and global existence, and between existence-with-uniqueness and existence-for-all-time — three logically independent questions that this theorem answers only for the first pair, only locally. The interval length \(h=\min\{a,b/M\}\) is not a technical artefact to be dismissed: it is the quantitative statement that the solution cannot outrun the box faster than the bound \(M\) on its own speed allows, and sharper (a priori) bounds on solutions are exactly what is needed to extend \(h\) to a maximal interval of existence, which is a separate piece of theory built on top of, not inside, this proof.

Common misconception: that Lipschitz continuity is necessary for uniqueness. It is sufficient but not necessary — Osgood's criterion replaces the linear bound \(\|f(t,y_1)-f(t,y_2)\|\le L\|y_1-y_2\|\) with the weaker \(\|f(t,y_1)-f(t,y_2)\|\le \omega(\|y_1-y_2\|)\) for a modulus \(\omega\) with \(\int_{0^+}\!\frac{dr}{\omega(r)}=\infty\) (e.g. \(\omega(r)=r\log(1/r)\) near \(0\), which is not Lipschitz), and uniqueness still follows from a Gronwall-type comparison argument rather than from the contraction mapping theorem used here; the Picard–Lindelöf proof is the easiest sufficient route to uniqueness, not the most general one.

Worked examples

y' = y,\quad y(0) = 1.\ \text{Take } R = \{|t|\le a,\ |y-1|\le b\} \text{ for any } a,b\gt0.

\(f(t,y)=y\) is continuous everywhere and Lipschitz in \(y\) with \(L=1\) on any box (in fact on all of \(\mathbb{R}\times\mathbb{R}\)), since \(|f(t,y_1)-f(t,y_2)|=|y_1-y_2|\). A

M = \sup_{|y-1|\le b}|y| = 1+b, \qquad h = \min\{a,\ b/(1+b)\}.

Direct evaluation of \(M\) on the box, then substitution into the formula for \(h\) from the Statement; note \(h\lt1\) for every choice of \(a,b\gt0\), so the theorem only promises the solution on a bounded interval even though \(f\) is Lipschitz on all of \(\mathbb{R}^2\). A

y_0(t)\equiv1,\quad y_1(t) = 1+\int_0^t 1\,ds = 1+t,\quad y_2(t) = 1+\int_0^t(1+s)\,ds = 1+t+\tfrac{t^2}{2},\ \dots

Direct computation of the Picard iterates \(y_{k+1}=Ty_k\) from Step 4 of the proof; by induction \(y_k(t)=\sum_{j=0}^k t^j/j!\), the degree-\(k\) Taylor polynomial of \(e^t\). B

y_k(t) \to e^t \text{ uniformly on compacts, and } y(t)=e^t \text{ is the unique solution guaranteed by the theorem on } [-h,h].

Reading. The abstract contraction-mapping argument, carried out concretely, reproduces exactly the familiar exponential — and confirms that no other function agrees with \(y_0=1\) and \(y'=y\) near \(t=0\).

Scope. The theorem itself only certifies uniqueness on the short interval \([-h,h]\); that \(e^t\) actually solves the IVP for all \(t\in\mathbb{R}\) requires the separate patching corollary (global uniqueness on overlaps), since \(f(t,y)=y\) is Lipschitz with the same constant on every box.

y' = t^2+y^2,\quad y(0)=0.\ \text{Take } R=\{|t|\le1,\ |y|\le1\}.

\(f(t,y)=t^2+y^2\) is continuous (polynomial) and, on the bounded box \(R\), Lipschitz in \(y\) because \(\partial f/\partial y = 2y\) is bounded there by \(2\) (mean value inequality on the convex box, giving \(L=2\) — see the Corollaries). This is the typical situation: \(f\) is not globally Lipschitz (e.g. as \(|y|\to\infty\), \(L\) would need to grow), but the theorem only needs a Lipschitz bound on the fixed box \(R\). A

M=\sup_R|t^2+y^2| = 2, \qquad h=\min\{1,\ 1/2\}=\tfrac12.

\(M\) is attained at the corner \((\pm1,\pm1)\) of the compact box \(R\) (extreme value theorem); substitute \(a=b=1\) into \(h=\min\{a,b/M\}\). A

\text{No closed form for } y \text{ is available, but existence and uniqueness on } [-\tfrac12,\tfrac12] \text{ are certified without solving the equation.}

This is precisely the situation the theorem is built for: it guarantees a well-defined, unique local solution to a nonlinear equation for which no elementary closed-form solution exists, purely from the estimates \(L=2\), \(M=2\), \(a=b=1\) computed above, without ever exhibiting \(y\) explicitly. B

\exists!\ y \in C^1\big([-\tfrac12,\tfrac12]\big):\ y' = t^2+y^2,\ y(0)=0

Reading. Even though \(y'=t^2+y^2\) cannot be solved by elementary functions, the theorem certifies that exactly one solution curve passes through the origin, valid at least on \([-\tfrac12,\tfrac12]\); numerical methods (e.g. Runge–Kutta) can then be trusted to approximate this specific, unique curve rather than one of many possible branches.

Scope. The bound \(h=\tfrac12\) came from the particular box \(R\) chosen; a larger box (larger \(a,b\)) generally changes \(M\) and can shrink or (rarely) grow \(h\) — the interval of guaranteed existence is not intrinsic to the equation alone but depends on the box used in the estimate, though the true maximal interval of existence (via the patching corollary) is intrinsic.

Problems

Show that \(f(t,y)=y^{1/3}\) is continuous but not Lipschitz on any box containing \(y=0\), and exhibit two distinct solutions of \(y'=y^{1/3}\), \(y(0)=0\).
Solution
Continuity: \(y\mapsto y^{1/3}\) is continuous on all of \(\mathbb{R}\) (composition/inverse of continuous odd power). Non-Lipschitz near \(0\): for \(y_1=h\to0^+\), \(y_2=0\), \(\dfrac{|f(y_1)-f(y_2)|}{|y_1-y_2|} = \dfrac{h^{1/3}}{h} = h^{-2/3}\to\infty\) as \(h\to0^+\), so no finite \(L\) bounds the difference quotient on any box containing \(0\). Two solutions: \(y_1(t)\equiv0\) clearly solves \(y'=y^{1/3}\), \(y(0)=0\). Separating variables for \(t\ge0\), \(y^{-1/3}dy = dt \Rightarrow \tfrac32y^{2/3}=t+C\); with \(y(0)=0\) taken as a limiting boundary one checks \(y_2(t) = \left(\tfrac23t\right)^{3/2}\) for \(t\ge0\), \(y_2(t)=0\) for \(t\lt0\) satisfies \(y_2'(t) = \left(\tfrac23t\right)^{1/2}\) and \(y_2(t)^{1/3} = \left(\tfrac23t\right)^{1/2}\) for \(t\ge0\), matching, and both sides are \(0\) for \(t\lt0\); at \(t=0\) both one-sided derivatives are \(0\). So \(y_1\ne y_2\) both solve the IVP — uniqueness fails exactly because the Lipschitz hypothesis fails.
For \(y'=1+y^2\), \(y(0)=0\), using the box \(R=\{|t|\le a,\ |y|\le b\}\), compute \(L\), \(M\), and \(h\) in terms of \(a,b\), then choose \(a=b=2\) numerically. Compare \(h\) to the true maximal interval of existence, given that the exact solution is \(y=\tan t\).
Solution
\(f(t,y)=1+y^2\), \(\partial f/\partial y = 2y\), so on \(R\), \(L=\sup_{|y|\le b}|2y| = 2b\) (mean value inequality). \(M=\sup_R(1+y^2) = 1+b^2\). \(h=\min\{a,\ b/(1+b^2)\}\). With \(a=b=2\): \(h=\min\{2,\ 2/5\} = 2/5\). The theorem thus guarantees a unique solution on \([-2/5,2/5]\). The true solution \(y=\tan t\) is defined and unique on the much larger interval \((-\pi/2,\pi/2)\) and blows up at \(t=\pm\pi/2\approx\pm1.571\); the theorem's \(h=0.4\) is a valid but conservative lower bound coming from the specific (non-optimal) box chosen, illustrating that the guaranteed interval is generally far smaller than the true maximal interval of existence.
Prove the global-uniqueness corollary stated in the Corollaries section: if \(f\) is continuous on a strip \(D=\mathbb{R}\times\mathbb{R}^n\) and Lipschitz in \(y\) with a constant \(L\) independent of the box (globally Lipschitz), and \(y_1,y_2\) both solve \(y'=f(t,y)\) on an interval \(J\) with \(y_1(t^\ast)=y_2(t^\ast)\) for some \(t^\ast\in J\), then \(y_1\equiv y_2\) on \(J\).
Solution
Let \(S=\{t\in J: y_1(t)=y_2(t)\}\); \(t^\ast\in S\) so \(S\ne\varnothing\). \(S\) is closed in \(J\) since \(y_1-y_2\) is continuous and \(S=(y_1-y_2)^{-1}(0)\). \(S\) is open in \(J\): for any \(t_1\in S\), apply the local theorem at \((t_1,y_1(t_1))=(t_1,y_2(t_1))\) with any box \(R\) around it — both \(y_1\) and \(y_2\) restricted to \([t_1-h,t_1+h]\cap J\) are \(C^1\) solutions of the same IVP with graph eventually inside \(R\) (by continuity, shrinking \(h\) if necessary so the graphs stay in \(R\)), so Step 9 of the proof forces them to agree on a neighbourhood of \(t_1\) in \(J\); hence a neighbourhood of \(t_1\) lies in \(S\). Since \(J\) is connected (an interval) and \(S\) is nonempty, open, and closed in \(J\), \(S=J\).
Let \(f(t,y)=A(t)y\) where \(A(t)\) is an \(n\times n\) matrix with entries continuous on \([\alpha,\beta]\). Show that the hypotheses of Picard–Lindelöf hold on any box \(R=[\alpha,\beta]\times\{\|y-y_0\|\le b\}\) with \(L\) independent of \(b\), and explain why this shows every linear IVP \(y'=A(t)y\), \(y(t_0)=y_0\) has a solution on the whole interval \([\alpha,\beta]\), not merely locally.
Solution
Continuity: \((t,y)\mapsto A(t)y\) is continuous since \(A\) is continuous and matrix-vector multiplication is continuous (bilinear, finite-dimensional). Lipschitz: \(\|A(t)y_1-A(t)y_2\| = \|A(t)(y_1-y_2)\| \le \|A(t)\|_{op}\|y_1-y_2\|\), and \(\|A(t)\|_{op}\) is continuous in \(t\) (entries continuous \(\Rightarrow\) operator norm continuous) hence bounded on the compact interval \([\alpha,\beta]\) by some \(L=\sup_{t\in[\alpha,\beta]}\|A(t)\|_{op}\), which does not depend on \(y\) or on \(b\) at all — this is the hallmark of a linear (or affine) right-hand side. Since \(L\) is independent of \(b\), enlarging \(b\) does not change \(L\); the only obstruction to \(h=\min\{a,b/M\}\) reaching the full interval is \(M=\sup_R\|A(t)y\|\), but since \(L\) is fixed and \(b\) can be taken arbitrarily large without changing \(L\), a standard Gronwall estimate on the integral equation of Step 2 (bounding \(\|y(t)\|\) a priori, independent of any box) shows the solution cannot escape to infinity in finite time, so the local solution extends step by step (via the global uniqueness corollary) across the whole compact interval \([\alpha,\beta]\).
Construct an explicit example showing that the local existence interval \(h\) from the theorem can be made arbitrarily short by an unfortunate choice of box, even for an \(f\) that is in fact globally Lipschitz — and explain why this does not contradict the existence of a much longer true solution.
Solution
Take \(f(t,y)=y\) again (globally Lipschitz, \(L=1\), true solution \(y=y_0e^{t-t_0}\) exists for all \(t\)). Choose the box \(R=\{|t-t_0|\le a,\ |y-y_0|\le b\}\) with \(b\) tiny, say \(b=\varepsilon\), and \(y_0=1\): then \(M=\sup_{|y-1|\le\varepsilon}|y| = 1+\varepsilon\), so \(h=\min\{a,\ \varepsilon/(1+\varepsilon)\}\); letting \(\varepsilon\to0^+\) forces \(h\to0\) regardless of how large \(a\) is chosen. This does not contradict the true (much longer) existence: the theorem's \(h\) is only a lower bound on the existence interval coming from a specific, possibly wasteful, choice of box — it certifies that the solution cannot leave a box of half-width \(b\) faster than its speed bound \(M\) allows, which is a genuinely weaker statement than the true behaviour of the specific solution once it is known. The freedom to choose \(R\) (and hence optimize \(h\)) is part of applying the theorem well; the theorem itself makes no claim that the \(h\) it produces from one particular box is sharp.