23. On the convergence of solutions to nabla dynamic equations on time scales

The convergent rate of solutions is estimated when f satisfies the Lipschitz condition in both variables.. By a general view, we derive a new approach to the approximation of dynamic equ

Article in Qualitative Theory of Dynamical Systems · October 2015

DOI: 10.1007/s12346-015-0166-8

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Vietnam National University, Hanoi

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Qualitative Theory of Dynamical

Systems

ISSN 1575-5460

Qual Theory Dyn Syst.

DOI 10.1007/s12346-015-0166-8

Dynamic Equations on Time Scales

Nguyen Thu Ha, Nguyen Huu Du, Le Cong Loi & Do Duc Thuan

1 23

and shall not be self-archived in electronic repositories If you wish to self-archive your article, please use the accepted manuscript version for posting on your own website You may further deposit the accepted manuscript version in any repository, provided it is only made publicly available 12 months after official publication or later and provided acknowledgement is given to the original source of publication and a link is inserted

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DOI 10.1007/s12346-015-0166-8 of Dynamical Systems

On the Convergence of Solutions to Dynamic Equations

on Time Scales

Nguyen Thu Ha 1 · Nguyen Huu Du 2 ·

Le Cong Loi 2 · Do Duc Thuan 3

Received: 23 February 2015 / Accepted: 15 September 2015

© Springer Basel 2015

Abstract In this paper, we study the convergence of solutions to dynamic equations

x  = f (t, x) on time scales {Tn}∞

n=1when this sequence converges to the time scaleT

The convergent rate of solutions is estimated when f satisfies the Lipschitz condition

in both variables By a general view, we derive a new approach to the approximation

of dynamic equations on time scales, especially the Euler method for differential equations Some examples are given to illustrate our results

Keywords Dynamic equations· Time scales · Grownall–Bellman inequality · Convergence of solutions

Mathematics Subject Classification 06B99· 34D99 · 47A10 · 47A99 · 65P99

B Nguyen Huu Du

nhdu@viasm.edu.vn; dunh@vnu.edu.vn

Nguyen Thu Ha

ntha2009@yahoo.com

Le Cong Loi

loilc@vnu.edu.vn

Do Duc Thuan

ducthuank7@gmail.com

1 Department of Basic Science, Electric Power University, 235 Hoang Quoc Viet Str., Hanoi, Vietnam

2 Department of Mathematics, Mechanics and Informatics, Vietnam National University,

334 Nguyen Trai Str., Hanoi, Vietnam

3 School of Applied Mathematics and Informatics, Hanoi University of Science and Technology,

1 Dai Co Viet Str., Hanoi, Vietnam

1 Introduction

Ordinary differential equations (ODEs) occur in many scientific disciplines, for instance in physics, chemistry, biology, and economics Therefore, finding solutions

of ODEs is important both in theory and practice Unfortunately, almost ODEs can not

be solved analytically, which causes in science and engineering, to find a numerical approximation to the solutions The Euler method is well-known because it is simple and useful, see [4,8,9,17] For solving the stiff initial value problem



˙x(t) = f (t, x(t)),

at each step[tm , t m+1], the explicit Euler approximation of (1.1) is

x m+1= xm + h f (tm , x m ), (1.2)

where tm+1 = tm + h and xm is the approximative value of x (t) at t = t m The

quantity en := |x(tm ) − x m| is called the error of this method after n time steps which

characterizes the difference between the approximative solution and the exact solution

The interesting problem is how the error en can be estimated when the mesh step h tends to zero With some added assumptions on f , we have known that entends to

zero as h tends to zero, see [8,9,13]

In recent years, in order to unify the continuous and discrete analysis, the theory

of the analysis on time scales was introduced by Stefan Hilger in his PhD thesis (supervised by Bernd Aulbach) [14] and has received a lot of attention, see [1,2,6,10,

11,15] By using the notation of the analysis on time scales, Eqs (1.1) and (1.2) can

be rewritten by the form 

x  (t) = f (t, x(t)),

with the time t belongs to the time scalesT = R or Th= hZ Thus, in view of analysis

on time scale, using the Euler method means we consider Eq (1.1) on the timeTh, which is “close” toT = R in some sense Then the problem of the error estimation above can be restated as follows: How do the solutions of (1.3) onThconverge to the solution of (1.3) onT = R as the mesh step h tends to zero?

Following this idea in a more general context, in this paper, we will consider the behavior of solutions of Eq (1.3) on time scales{Tn}∞

n=1whenTntends toT by the Hausdorff distance Assume that on time scales{Tn}∞

n=1Eq (1.3) has the solutions

{xn (t)}∞

n=1and on time scaleT Eq (1.3) has the solution x (t) Then, we will prove

that

under the assumption that f (t, x) satisfies the Lipschitz condition in the variable x.

Moreover, if f satisfies the Lipschitz condition in both variables t and x then the

convergent rate of solutions is estimated as the same degree as the Hausdorff distance betweenTnandT, i.e.,

xn (t) − x(t)  C2d H (T, T n ), for all t ∈ T ∩ T n : t0 t  T. (1.5)

Using these results, we obtain the convergence of the Euler method as a con-sequence It can be considered as a new and general approach to the convergence problems of the approximative solutions

This paper is organized as follows Section2summarizes some preliminary results

on time scales In Sect.3, we study the convergence of solutions of dynamic equations

on time scales The main results of the paper are derived here In Sect.4, we give some examples and show the convergence of the Euler method The last section deals with some conclusions and open problems

2 Preliminaries

LetT be a closed nonempty subset of R, endowed with the topology inherited from the standard topology onR Let σ(t) = inf{s ∈ T : s > t}, μ(t) = σ(t) − t and

ρ(t) = sup{s ∈ T : s < t}, ν(t) = t − ρ(t) (supplemented by sup ∅ = inf T, inf ∅ =

supT) A point t ∈ T is said to be right-dense if σ (t) = t, right-scattered if σ(t) > t,

left-dense if ρ(t) = t, left-scattered if ρ(t) < t and isolated if t is simultaneously

right-scattered and left-scattered A function f defined on T is called rd-continuous

provided it is continuous at right-dense points inT and its left-sided limits exist (finite)

at left dense points f is ld-continuous if it is continuous at every left-dense point and

if the right-sided limit exists in every right-dense point It is easy to see that a function

is continuous if and only if it is both r d-continuous and ld-continuous.

Definition 2.1 (Delta derivative) A function f : T → Rd is called delta differentiable

at t if there exists a vector f  (t) such that for all  > 0

 f (σ(t)) − f (s) − f  (t)(σ(t) − s)  |σ (t) − s|

for all s ∈ (t − δ, t + δ) ∩ T and for some δ > 0 The vector f  (t) is called the delta derivative of f at t.

IfT = R then the delta derivative is f(t) from continuous calculus; if T = Z then

the delta derivative is the forward difference,f (t) = f (t + 1) − f (t), from discrete

calculus

For any r d-continuous function p (·) from T to R, the solution of the dynamic

equa-tion x  = p(t)x, with the initial condition x(s) = 1, defines a so-called exponential function We denote this exponential function by e p (T; t, s) For the properties of

the exponential function ep (T; t, s) the interested reader can see [2,6] To simplify

notation, we write ep (T; t, s) for e p (t, s) if there is no confusion.

Let f be a rd-continuous function and a , b ∈ T Then, the Riemann integral

b

a f (s)s exists (see [12]) In case b /∈ T, writing b

a f (s)s meansb

a f (s)s,

where b = max{t < b : t ∈ T}.

Consider the dynamic equation on the time scaleT



x  (t) = f (t, x),

where f : T × Rd → Rd For the existence, uniqueness and extensibility of the solution of Eq (2.1) we refer to [5] In particular, for any positive regressive number

α, the Cauchy problem

x  (t) = αx(t); x(t0) = 1

has a unique solution, namely e α (t, t0) satisfying the estimate

(see [1,2])

It is easy to see that if f (t, x) is a continuous function in (t, x) then x(t) is a solution

to (2.1) if and only if

x(t) = x0+

 t

t0

f (s, x(s)) s.

Lemma 2.2 (Gronwall–Bellman lemma, see [5]) Let x (t) be a continuous function and k > 0, x0∈ R Assume that x(t) satisfies the inequality

x (t)  x0+ k

 t

t0

Then, the relation

x(t)  x0e k (t, t0) for all t ∈ T, t  t0 (2.4)

holds.

Fix t0∈ R Let T = T(t0) be the set of all time scales with bounded graininess such

that t0∈ T for all T ∈ T We endow T with the Hausdorff distance, that is Hausdorff distance between two time scalesT1andT2defined by

d H (T1, T2) := max

 sup

t1 ∈T 1

d(t1, T2), sup

t2 ∈T 2

d(t2, T1)



where

d (t1, T2) = inf

t∈T |t1− t2| and d(t2, T1) = inf

t ∈T |t2− t1|.

For properties of the Hausdorff distance, we refer the interested readers to [3,7,16] Throughout this paper, all considered time scales belong toT

3 Convergence of Solutions

Let{Tn}n∈N⊂ T be a sequence of time scales satisfying:

lim

n→∞Tn= T,

by the Hausdorff distance We define the time scale

Assume that f is continuous on T and satisfies the Lipschitz condition in the variable

x, that is there exists a constant k > 0 such that

 f (t, x) − f (t, y)  kx − y, for all t ∈  T : t0 t  T and x, y ∈ R d

(3.2)

By these assumptions, the initial value problems (IVPs)

x 

n (t) = f (t, x n (t)), t ∈ T n , x n (t0) = x0, n = 1, 2, , (3.3) and

have a unique solution xn (t) defined on T n (respectively solution x (t) defined on T).

It is clear that the solutions of the IVPs (3.3) and (3.4) are given by

x n (t) = x0+

 t

t0

and

x(t) = x0+

 t

t0

respectively, where

f  n s denotes the integral on the time scaleTn.

The following lemma gives the uniformly bounded property of solutions of the IVPs (3.3) and (3.4) on different time scales

Lemma 3.1 Let xS(t) be the solution to the dynamic equation

x  (t) = f (t, x(t)), t ∈ S, x(t0) = x0 Then, for any T > t0one has

sup

S∈T;S⊂ T

sup

Proof LetS ∈ T; S ⊂ T For any t ∈ S, we have

xS(t) =

x0+

 t

t0

f (s, xS(s))s



 x0 +

 t

t0

f (s, 0)s

 + t

t0

f (s, xS(s))s −

 t

t0

f (s, 0)s



 x0 +

 t

t0  f (s, 0)s +

 t

t0  f (s, xS(s)) − f (s, 0)s. (3.8)

By virtue of continuity of f on  T, one has C = sup s∈T;t0sT  f (s, 0)  ∞.

Hence,

 t

t0

 f (s, 0)s  C(t − t0)  C(T − t0).

Moreover, since f satisfies the Lipschitz condition (3.2),

 t

t0

 f (s, xS(s)) − f (s, 0)s  k

 t

t0

xS(s)s.

Therefore,

xS(t)  x0 + C(T − t0) + k

 t

t0

xS(s)s.

By using the Gronwall–Bellman lemma, we get

xS(t) x0 + C(T − t0) e k (S; t, t0),

where ek (S; t, t0) is the exponential function defined on S Thus, by (2.2), we obtain

sup

S∈ T

sup

t∈S:t 0tT xS(t)  (x0 + C(T − t0)) e k (T −t0) < ∞.

Let n ∈ N, we denote by σnthe forward jump operator on the time scaleTn For

any t ∈ T, there exists a unique s ∈ Tn, say s = γ T,T n (t), such that either s = t or

t ∈ (s, σn (s)) It is easy to check that the function γ T,T n (t) is rd-continuous on T.

Also, there exists t∗

n = t∗

n (t) ∈ T nsatisfying

|t − t∗

n | = d(t, Tn ) = inf{|t − s| : s ∈ T n}. (3.9)

We choose t∗

n = γ T,T n (t) if |t − γ T,T n (t)| = d(t, T n ), otherwise t∗

n =

σ n (γ T,T n (t)) Define

f n (t, x) = f (γ T,T n (t), x), t ∈ T; x ∈ R d , (3.10)

x n (t) = x(γ T,T n (t)), t ∈ T. (3.11) Assume thatTn⊂ T Then, by the definition of Riemann integral on time scales,

we have

 t

t0

f (s, x(s)) n s=

 t

t0

f n x n (s))s,

for any t ∈ Tn(see, e.g [12])

Since dH (T, T n ) → 0 as n → ∞, we can assume that t∗

n (t) < T + 1 when t  T

By Lemma3.1, A= supS∈T supt∈S:t 0tT +1 xS(t) < ∞, and hence let

M = sup{ f (t, x) : t0 t  T + 1, x < A}.

Now, we need the following lemmas for proving the convergence of the solution sequence{xn (t)} of the IVPs (3.3) whenTntends toT.

Lemma 3.2 Let x n (t), n = 1, 2, be solutions to the IVPs (3.3) and x (t) be the solution to the IVP (3.4) Assume thatTn⊂ T Then,

x(t) − xn (t)  δ (n) T e k (T n; t, t0), t ∈ T n : t0 t  T, (3.12)

and

x(t) − xn (t∗

n )  δ (n) T+1e k (T n; t∗

n , t0) + Md H (T, T n ), t ∈ T : t0 t  T, (3.13)

where t∗

n is defined by (3.9) and

δ t (n)=

 t

t0

 f (s, x(s)) − fn x n (s))s. (3.14)

Proof For any t ∈ Tn, t  T we have

x(t) − x n (t) =

 t

t0

f (s, x(s))s −

 t

t0

f (s, x n (s)) n s

=

 t

t0

f (s, x(s))s −

 t

t0

f (s, x(s)) n s

+

 t

t0[ f (s, x(s)) − f (s, xn (s))] n s

=

 t

t0

[ f (s, x(s)) − fn x n (s))]s

+

 t

t [ f (s, x(s)) − f (s, xn (s))] n s

By virtue of the Lipschitz condition

 f (s, x(s)) − f (s, xn (s))  kx(s) − x n (s),

it follows that

x(t) − xn (t) 

 t

t0

 f (s, x(s)) − fn x n (s))s + k

 t

t0

x(s) − xn (s) n s

 δ (n) T + k

 t

t0

x(s) − xn (s) n s.

By using Gronwall–Bellman lemma, we obtain (3.12)

If t ∈ T, t  T then

x(t) − xn (t n∗)  x(t) − x(t n∗) + x(t n∗) − x n (t n∗).

Since t∗

n  T + 1 and supt0tT +1 x(t))  A,

x(t n∗) − x n (t n∗)  δ (n) T+1e k (T n; t n∗, t0), t ∈ T : t0 t  T.

Further,

x(t) − x(t n∗) =





 t∗

n

t

f (s, x(s))s



 M|t − t n∗|  Md H (T, T n ).

Summing up, (3.13) holds The proof is complete

Lemma 3.3 Assume thatTn ⊂ T For each  and T ∈ T, there exists θ = θ(, T )

such that if d H (T, T n ) < θ then

δ T (n)  (T − t0) + 2M (T − t0)

where δ T (n) is defined by (3.14).

Proof Since f is continuous, f is uniformly continuous on [t0, T ] × B(0, A) where B(0, A) is the ball with the center 0 and radius A Therefore, for each , there exists

δ = δ() such that if |t1− t2| + x1− x2 < δ then

 f (t1, x1) − f (t2, x2)   on [t0, T ] × B(0, A).

We chooseθ = θ() = δ()

M+ 1 If t − γ T,T n (t) < θ then

x n (t) =

γ t T,Tn (t) f (s, x(s))s

  M(t − γ T,T n (t)) < Mθ,

|t − γ T,T n x n (t) < (M + 1)θ = δ.

This implies that if t − γ T,T n (t) < θ then

 f (t, x(t)) − fn x n (t)) < .

We see that the number of values s∈ Tnsatisfying t0 s  T and

{t ∈ T : s < t < σn (s), t − s  θ} = ∅,

is less than or equal to[T −t0

θ ] Assume that these values are s1 < s2 < · · · < s r with r  [T −t0

θ ] In case dH (T, T n ) < θ, we see that if t ∈ T such that s i < t <

σ n (s i ), t − s i  θ then

σ n (s i ) − t = d(t, T n )  d H (T, T n ).

Let

τ i = min{t ∈ T : si < t < σ n (s i ), t − s i  θ}; i = 1, r.

It is clearσ n (s i ) − τ i  dH (T, T n ) Further,

δ (n) T =

 T

t0  f (s, x(s)) − fn x n (s))s

=

 τ1

t0  f (s, x(s)) − fn x n (s))s

+

r−1

i=1

 T ∧τ i+1

σ n (s i )  f (s, x(s)) − fn x n (s))s

+

r

i=1

 T ∧σ n (s i )

T ∧τ i

 f (s, x(s)) − fn x n (s))s,

where a ∧ b = min{a, b}.

Since f (s, x(s)) − fn (s, x n (s)) <  for all s ∈ [t0, τ1) ∪ [σ n (s i ), τ i+1), 1  i 

r − 1,

 τ1

t0

 f (s, x(s)) − fn x n (s))s  (τ1− t0),

 T ∧τ i+1

σ (s )  f (s, x(s)) − fn x n (s))s  (T ∧ τ i+1− σn (s i ))(T ∧ τ i+1−τi ).

Proof Since f is continuous, f is uniformly continuous on [t0, T ] × B(0, A) where B(0, A) is the ball with the center and radius A Therefore, for each , there exists... T,T n (t) < θ then

 f (t, x(t)) − fn x n (t)) < .

We see that the number of values s∈ Tnsatisfying t0... x2 < δ then

 f (t1, x1) − f (t2, x2)   on [t0,