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Continuous and Lipschitz dependence on the initial condition on a Picard-Lindelof cylinder
Statement
Let , let be open and let be continuous and locally Lipschitz in the state variable (Local Lipschitz continuity in the state variable, locally uniform in time and parameters). Let , let and satisfy (Open ball, closed ball and sphere in a metric space), put
so that for all (The Euclidean inner product on ), and let satisfy
Then for every the initial value problem
has exactly one solution on whose graph lies in , and for all and all :
- ;
- ;
- ; in particular the map is Lipschitz, hence continuous, on the compact set .
Facts & Assumptions
Given: The open set , the continuous and locally Lipschitz field , the point , the numbers with , the constants and the radius with and .
Picard-Lindelöf: if , if and has state-Lipschitz constant on that cylinder, and if and , then exactly one solution of , exists on with graph in the cylinder; any two solutions through the same initial data agree on every common subinterval containing the initial time (Picard-Lindelöf local existence and uniqueness for first-order systems, First-order systems, initial value problems, and solutions on intervals).
is locally Lipschitz in the state variable: every compact time-state cylinder carries a finite state-Lipschitz constant (Local Lipschitz continuity in the state variable, locally uniform in time and parameters); in particular as chosen in the statement is finite and the inequality holds on .
For the notation means the singleton , as follows from the inequality defining a closed ball. is closed and bounded, hence compact in (Open ball, closed ball and sphere in a metric space, Heine-Borel in : with the Euclidean metric a subset of is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), so the continuous function attains its maximum there and (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).
A curve solves the initial value problem on an interval exactly when it is continuous there, has graph in , and satisfies the Volterra equation ; the oriented integral is additive over subintervals (A first-order initial value problem is equivalent to its Volterra integral equation, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, For : is integrable on if and only if it is integrable on and on , and then ; with the oriented form for arbitrary , Intervals of : the nine order-convex forms, nondegeneracy, and length).
The Euclidean norm satisfies the triangle inequality (Cauchy-Schwarz with its equality case, the triangle inequality for , the parallelogram law and polarisation). For an integrable -valued curve on an interval and reals one has , and if a real integrable satisfies on a nondegenerate interval then (For and integrable when , ; for , is integrable, If on and both are integrable then ; and ).
Gronwall: if continuous and constants , satisfy for , then there; the reflected form for an integral oriented as gives for (Gronwall's integral inequality with variable and constant coefficients).
Proof
Fix and put , so that . Then : if then . Moreover and , and the bounds and hold on the larger cylinder by [F2] and [F3]. By [F1] with , , , exactly one solution on exists whose graph lies in .
For put . By [F4], for , The integrand is continuous because and the two solutions are continuous, so [F5] gives : the Lipschitz estimate applies on since both graphs lie there by step 1.1.
For , [F4] gives , and the graph of lies in by step 1.1, where ; hence by [F5]. This is clause 2.
Applying [F6] with , gives for . For apply the same argument to the reflected curves , on : their difference obeys the same integral inequality, so . This is clause 1.
Combining clauses 1 and 2 with the triangle inequality, for all ; this is clause 3. The domain of the map is a closed bounded subset of , hence compact (Heine-Borel in : with the Euclidean metric a subset of is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), and the displayed inequality exhibits it as Lipschitz.
Depends on
First-order systems, initial value problems, and solutions on intervals
Local Lipschitz continuity in the state variable, locally uniform in time and parameters
Picard-Lindelöf local existence and uniqueness for first-order systems
A first-order initial value problem is equivalent to its Volterra integral equation
Gronwall's integral inequality with variable and constant coefficients
The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral
For and integrable when , ; for , is integrable
If on and both are integrable then ; and
For : is integrable on if and only if it is integrable on and on , and then ; with the oriented form for arbitrary
The Euclidean inner product on
Open ball, closed ball and sphere in a metric space
Intervals of : the nine order-convex forms, nondegeneracy, and length
Heine-Borel in : with the Euclidean metric a subset of is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line
A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value
Cauchy-Schwarz with its equality case, the triangle inequality for , the parallelogram law and polarisation
Dependency tree · 2 levels
Sources
- G. Teschl, Ordinary Differential Equations and Dynamical Systems (AMS Graduate Studies in Mathematics 140, online edition), Lemma 2.7 (generalized Gronwall inequality) and Theorem 2.8 with estimate (2.43), printed pages 42-44
- J. Arino, MATH 8430 Fundamental Theory of Ordinary Differential Equations (University of Manitoba), Section 1.4 (continuous dependence on the initial data), printed pages 32-36, and Appendix A.7 (Gronwall)