Feng et al. Journal of Inequalities and Applications (2016) 2016:271 DOI 10.1186/s13660-016-1214-x
RESEARCH
Open Access
A note on Cauchy-Lipschitz-Picard theorem Zonghong Feng1 , Fengying Li2* , Ying Lv3 and Shiqing Zhang1 *
Correspondence:
[email protected] School of Economic Mathematics, Southwestern University of Finance and Economics, Chengdu, 61130, China Full list of author information is available at the end of the article 2
Abstract In this note, we try to generalize the classical Cauchy-Lipschitz-Picard theorem on the global existence and uniqueness for the Cauchy initial value problem of the ordinary differential equation with global Lipschitz condition, and we try to weaken the global Lipschitz condition. We can also get the global existence and uniqueness. MSC: 34A34; 34C25; 34C37 Keywords: initial value problems; generalized Lipschitz condition; global existence and uniqueness
1 Introduction In his famous book [], Brezis gave a very sketchy and interesting proof on the classical Cauchy-Lipschitz-Picard theorem. Theorem . Let E be a Banach space and let F : E → E be a Lipschitz map, i.e., there is a constant L such that F(u) – F(v) ≤ Lu – v,
∀u, v ∈ E.
(.)
Then, for any given u ∈ E, there exists a unique solution u ∈ C ([, +∞), E) of the problem:
du(t) dt
= F(u(t)), u() = u .
∀t ∈ (, +∞),
(.)
It is well known that if we only assume the local Lipschitz condition, we can only get the local existence and uniqueness for Cauchy initial value problems. In this paper, we try to weaken the global Lipschitz condition, but we also want to get the global existence and uniqueness; we have the following theorem. Theorem . Let E be a Banach space (with norm · ) and let F : [, +∞) × E → E be a map satisfying F(t, u) – F(t, v) ≤ L(t) · p u – v ,
∀t ∈ [, +∞), u, v ∈ E,
© Feng et al. 2016. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Feng et al. Journal of Inequalities and Applications (2016) 2016:271
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where L : [, +∞) → [, +∞), p : (, +∞) → [, +∞) are continuous and there are ≤ a < +∞ such that L = L(t) ≤ a, p = p(s) ≤ s, and p(s) is an increasing function. Then, for any given u ∈ E, there exists a unique solution u ∈ C ([, +∞); E) for the problem
∀t ∈ (, +∞),
du(t) dt
= F(t, u(t)), u() = u .
(.)
Corollary . In Theorem ., if we take Ł(s) = a > , p(s) = s or p(s) = ln s, then the conditions and the results of Theorem . hold.
2 The proof of Theorem 1.2 Lemma . ([], Banach contraction mapping principle) Let X be a nonempty complete metric space and let T : X → X be a strict contraction, i.e., there is < k < such that d(T(x), T(y)) ≤ kd(x, y), ∀x, y ∈ X, then S has a unique fixed point u = T(u). Lemma . (Gronwall []) Let x ∈ C [a, b]. If R[a, b] denotes the set of Riemann integrable functional on [a, b]; β ∈ R[a, b], and x (t) ≤ β(t) · x(t),
∀t ∈ [a, b],
then t
x(t) ≤ x(a) · e
a β(s) ds
,
∀t ∈ [a, b].
The following lemma can be regarded as a natural generalization of the Gronwall inequality. Lemma . Let x ∈ C [a, b], α, β ∈ R[a, b]. If x (t) ≤ α(t) + β(t)x(t),
∀t ∈ [a, b],
then t s t e– β(τ ) dτ α(s) ds · e a β(s) ds , x(t) ≤ x(a) + a t
Proof Let v = e w (t) =
a β(s) ds
, w = xv , then v (t) = β(t)v(t),
x v – xv (α + βx)v – βxv αv α ≤ = = . v v v v
Then w(t) ≤
t
e– a
s
β(τ ) dτ
α(s) ds + w(a),
∀t ∈ [a, b].
Feng et al. Journal of Inequalities and Applications (2016) 2016:271
x(t) ≤ x(a) + v(t)
t
x(t) ≤ e
a β(s) ds
t
e–
s
β(τ ) dτ
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α(s) ds,
a
t s · x(a) + e– β(τ ) dτ α(s) ds . a
Now to prove Theorem ., we use some similar arguments to Brezis []. Let k > , which is to be determined, and assume
X = u ∈ C [, +∞); E sup e–kt · u(t) < +∞ . t≥
Then it is easy to see that X is a Banach space for the norm uX = sup e–kt · u(t). t≥
For ∀u ∈ X, we define
F s, u(s) ds.
t
(u)(t) = u +
Then u is a solution of (.) if and only if (u) = u, that is, u is a fixed point of . () We now show that, for every u ∈ X, (u) also belongs to X. In fact, uX = sup e–kt (u)(t) t≥
≤ sup e
–kt
· u + sup e
t≥
–kt
t≥
t · F s, u(s) ds .
We only need to prove –kt
sup e
·
t≥
t
F s, u(s) ds < +∞.
Notice that F s, u(s) – F(, u ) ≤ L(s) · p u(s) – u , F s, u(s) ≤ L(s) · p u(s) – u + F(, u ). Since sup e–kt · t≥
t
F(, u ) ds = F(, u ) sup e–kt · t < +∞. t≥
Hence we only need to prove sup e–kt · t≥
t
L(s)p u(s) – u ds < +∞.
Feng et al. Journal of Inequalities and Applications (2016) 2016:271
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Let
t
–kt
ϕ(t) = e
L(s)p u(s) – u ds,
ϕ() = ,
then
t
ϕ(t) ≤
L(s)eks e–ks u(s) + u dse–kt
t –ks ks –kt u(s) + sup e u sup L(s)e dse
–ks
≤ sup e t≥
t≥
t≥
≤ C C = C < +∞. Hence we have proved is a self-mapping from X to X. () We prove the contraction property of . We have
t
F s, u(s) – F s, v(s) ds ≤
u – v ≤
t
u – vX ≤ sup e–kt ·
L(s)p u(s) – v(s) ds,
L(s)p u(s) – v(s) ds
t
≤ sup t≥
t
t≥
L(s)eks e–ks u(s) – v(s) dse–kt
≤ sup e–ks u(s) – v(s) sup e–kt t≥
t≥
t
L(s)eks ds
t
≤ sup e
–kt
t≥
L(s)eks dsu – vX .
Hence we have u – vX ≤
L u – vX , k
∀u, v ∈ X.
We can choose k > L, then we use Banach contraction mapping principle to find that (.) has at least one solution on [, +∞). Furthermore, by Gronwall’s inequality, we can get the uniqueness. In fact, let u , u be two solutions of (.), then u (t) – u (t) ≤
t
F s, u (s) – F s, u (s)
≤
t
L(s)p u (s) – u (s) ds
t
L(s) u (s) – u (s) ds.
≤
By Gronwall’s inequality, we have u (t) – u (t) ≤ . Hence for any t ∈ [, +∞), we have u (t) = u (t).
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3 Examples Example .
du(t) dt
= F(t, u(t)) =
, +t+|u|
∀t ∈ (, +∞),
u() = u .
(.)
Then F t, u(t) – F t, v(t) =
v| – |u ≤ v| – |u. ( + t + |u|)( + t + |v|)
By the triangle inequality, we have F t, u(t) – F t, v(t) ≤ |u – v|. So F(t, u(t)) =
+t+|u|
is Lipschitz with a = and p(s) = s.
Example . Let p (t) : [, +∞) → [, +∞) continuous and there is ≤ a < +∞ such that p (t) ≤ a. Let p (t) : [, +∞) → R continuous. We consider du(t) = F(t, u(t)) = p (t)|u – p (t)|, dt u() = u .
∀t ∈ (, +∞),
(.)
Then F t, u(t) – F t, v(t) = p (t) u – p (t) – v – p (t) . By the triangle inequality, we have F t, u(t) – F t, v(t) ≤ p (t)|u – v|. So F(t, u(t)) = p (t)|u – p (t)| is Lipschitz with L(t) = p (t) and p(s) = s. Competing interests The authors declare that they have no competing interests. Authors’ contributions The research and writing of this manuscript was a collaborative effort from all the authors. All authors read and approved the final manuscript. Author details 1 Department of Mathematics, Sichuan University, Chengdu, 610064, China. 2 School of Economic Mathematics, Southwestern University of Finance and Economics, Chengdu, 61130, China. 3 Department of Mathematics and Statistics, Southwest University, Chongqing, 400715, China. Acknowledgements The authors sincerely thank referee for his/her valuable comments. This paper was partially supported by NSFC (No. 11671278 and No. 11426181) and the National Research Foundation for the Doctoral Program of Higher Education of China under Grant No. 20120181110060. Received: 6 July 2016 Accepted: 18 October 2016
Feng et al. Journal of Inequalities and Applications (2016) 2016:271
References 1. Brezis, H: Functional Analysis, Sobolev Spaces and PDE, pp. 184-185. Springer, New York (2011) 2. Thomas, HG: Note on the derivatives with respect to a parameter of the solutions of a system of differential equations. Ann. Math. 20, 292-296 (1919)
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