Hirsch M., Smale S., Devaney R., Differential Equations, Dynamical Systems, and an Introduction to Chaos

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396 Chapter 17 Existence and Uniqueness Revisited

This theorem says that if a solution Y (t ) cannot be extended to a larger time

interval, then this solution leaves any compact set in

O. This implies that, as

t → β, either Y (t ) accumulates on the boundary of

O or else a subsequence

|Y (t

)| tends to ∞ (or both).

Proof: Suppose Y (t) ⊂

C for all t ∈ (α, β). Since F is continuous and C is

compact, there exists M>0 such that |F(X )|≤M for all X ∈

Let γ ∈ (α, β). We claim that Y extends to a continuous function

Y :[γ , β]→

C. To see this, it sufﬁces to prove that Y is uniformly continuous

on J . For t

∈ J we have

|Y (t

) − Y (t

)|=







(s) ds



≤



|F(Y (s))|ds

≤ (t

− t

)M.

This proves uniform continuity on J . Hence we may deﬁne

Y (β) = lim

t→β

Y (t ).

We next claim that the extended curve Y :[γ , β]→

is differentiable at

β and is a solution of the differential equation. We have

Y (β) = Y (γ ) + lim

t→β





(s) ds

= Y (γ ) + lim

t→β



F(Y (s)) ds

= Y (γ ) +



F(Y (s)) ds.

where we have used uniform continuity of F(Y (s)). Therefore

Y (t ) = Y (γ ) +



F(Y (s)) ds

for all t between γ and β. Hence Y is differentiable at β, and, in fact, Y



(β) =

F(Y (β)). Therefore Y is a solution on [γ , β]. Since there must then be a

solution on an interval [β, δ) for some δ > β, we can extend Y to the interval

(α, δ). Hence (α, β) could not have been a maximal domain of a solution. This

completes the proof of the theorem.



17.4 Extending Solutions 397

The following important fact follows immediately from this theorem.

Corollary. Let

C be a compact subset of the open set O ⊂ R

and let

F :

O → R

be C

. Let Y

∈ C and suppose that every solution curve of the

form Y :[0, β]→

O with Y (0) = Y

lies entirely in C. Then there is a solution

Y :[0, ∞) →

O satisfying Y (0) = Y

, and Y (t ) ∈ C for all t ≥ 0, so this

solution is deﬁned for all (forward) time.



Given these results, we can now give a slightly stronger theorem on the

continuity of solutions in terms of initial conditions than the result discussed

in Section 17.3. In that section we assumed that both solutions were deﬁned on

the same interval. In the next theorem we drop this requirement. The theorem

shows that solutions starting at nearby points are deﬁned on the same closed

interval and also remain close to each other on this interval.

Theorem. Let F :

O → R

be C

. Let Y (t ) be a solution of X



= F(X) that is

deﬁned on the closed interval [t

, t

], with Y (t

) = Y

. There is a neighborhood

U ⊂

of Y

and a constant K such that, if Z

∈ U , then there is a unique

solution Z (t ) also deﬁned on [t

, t

] with Z (t

) = Z

. Moreover Z satisﬁes

|Y (t ) − Z(t)|≤K |Y

− Z

|exp(K (t − t

))

for all t ∈[t

, t

]. 

For the proof we will need the following lemma.

Lemma. If F :

O → R

is locally Lipschitz and C ⊂ O is a compact set, then

C is Lipschitz.

Proof: Suppose not. Then for every k>0, no matter how large, we can ﬁnd

X and Y in

C with

|F(X ) − F(Y )| >k|X − Y |.

In particular, we can ﬁnd X

, Y

such that

|F(X

) − F(Y

)|≥n|X

− Y

| for n = 1, 2, ....

Since

C is compact, we can choose convergent subsequences of the X

and

. Relabeling, we may assume X

→ X

∗

and Y

→ Y

∗

with X

∗

and Y

∗

in C.

Note that we must have X

∗

= Y

∗

, since, for all n,

∗

− Y

∗

|= lim

n→∞

− Y

|≤n

−1

|F(X

) − F(Y

)|≤n

−1

2M,

398 Chapter 17 Existence and Uniqueness Revisited

where M is the maximum value of |F(X)| on C. There is a neighborhood O

of X

∗

on which F |O

has Lipschitz constant K . Also there is an n

such that

∈ O

if n ≥ n

. Therefore, for n ≥ n

|F(X

) − F(Y

)|≤K |X

− Y

which contradicts the assertion above for n>n

. This proves the

lemma.



Proof: The proof of the theorem now goes as follows: By compactness of

, t

], there exists  > 0 such that X ∈ O if |X − Y (t )|≤ for some

t ∈[t

, t

]. The set of all such points is a compact subset C of O. The C

map

F is locally Lipschitz, as we saw in Section 17.2. By the lemma, it follows that

C has a Lipschitz constant K .

Let δ > 0 be so small that δ ≤  and δ exp(K |t

− t

|) ≤ . We claim that

if |Z

− Y

| < δ, then there is a unique solution through Z

deﬁned on all of

, t

]. First of all, Z

∈ O since |Z

− Y (t

)| < , so there is a solution Z (t )

through Z

on a maximal interval [t

, β). We claim that β >t

. For suppose

β ≤ t

. Then, by Gronwall’s inequality, for all t ∈[t

, β), we have

|Z(t) − Y (t )|≤|Z

− Y

|exp(K |t − t

≤ δ exp(K |t − t

≤ .

Thus Z (t ) lies in the compact set

C. By the results above, [t

, β) could

not be a maximal solution domain. Therefore Z(t) is deﬁned on [t

, t

The uniqueness of Z (t ) then follows immediately. This completes the

proof.



17.5 Nonautonomous Systems

We turn our attention brieﬂy in this section to nonautonomous differ-

ential equations. Even though our main emphasis in this book has been

on autonomous equations, the theory of nonautonomous (linear) equa-

tions is needed as a technical device for establishing the differentiability of

autonomous ﬂows.

Let

O ⊂ R × R

be an open set, and let F : O → R

be a function that

is C

in X but perhaps only continuous in t . Let (t

, X

) ∈ O. Consider the

nonautonomous differential equation



(t) = F(t , X), X(t

) = X

17.5 Nonautonomous Systems 399

As usual, a solution of this system is a differentiable curve X (t)inR

deﬁned

for t in some interval J having the following properties:

1. t

∈ J and X(t

) = X

2. (t , X (t )) ∈

O and X



(t) = F(t , X(t)) for all t ∈ J .

The fundamental local theorem for nonautonomous equations is as follows:

Theorem. Let

O ⊂ R × R

be open and F : O → R

a function that

is C

in X and continuous in t . If (t

, X

) ∈ O, there is an open interval J

containing t and a unique solution of X



= F(t , X) deﬁned on J and satisfying

X(t

) = X

. 

The proof is the same as that of the fundamental theorem for autonomous

equations (Section 17.2), the extra variable t being inserted where appropriate.

An important corollary of this result follows:

Corollary. Let A(t ) be a continuous family of n × n matrices. Let (t

, X

) ∈

J ×

. Then the initial value problem



= A(t )X , X(t

) = X

has a unique solution on all of J . 

For the proof, see Exercise 14 at the end of the chapter.

We call the function F(t , X ) Lipschitz in X if there is a constant K ≥ 0 such

that

(

t, X

)

− F

(

t, X

)

|≤K |X

− X

for all (t , X

) and (t , X

)inO. Locally Lipschitz in X is deﬁned analogously.

As in the autonomous case, solutions of nonautonomous equations are

continuous with respect to initial conditions if F(t, X) is locally Lipschitz in

X. We leave the precise formulation and proof of this fact to the reader.

A different kind of continuity is continuity of solutions as functions of the

data F(t , X). That is, if F :

O → R

and G : O → R

are both C

in X ,

and |F − G| is uniformly small, we expect solutions to X



= F(t , X) and



= G(t , Y ), having the same initial values, to be close. This is true; in fact,

we have the following more precise result:

Theorem. Let

O ⊂ R × R

be an open set containing (0, X

) and suppose

that F, G :

O → R

are C

in X and continuous in t . Suppose also that for all

(t, X) ∈

|F(t , X) − G(t, X)| < .

400 Chapter 17 Existence and Uniqueness Revisited

Let K be a Lipschitz constant in X for F(t , X).IfX(t) and Y (t ) are solutions of

the equations X



= F(t , X ) and Y



= G(t, Y ), respectively, on some interval J,

and X (0) = X

= Y (0), then

|X(t ) − Y (t)|≤





exp(K |t |) − 1



for all t ∈ J.

Proof: For t ∈ J we have

X(t ) − Y (t) =





(s) − Y



(s)) ds



(F(s, X(s)) − G(s, Y (s))) ds.

Hence

|X(t ) − Y (t)|≤



|F(s, X(s)) − F(s, Y (s))|ds



|F(s, Y (s)) − G(s, Y (s))|ds

≤



K |X (s) − Y (s)|ds +



 ds.

Let u(t ) =|X(t) − Y (t )|. Then

u(t ) ≤ K





u(s) +





ds,

so that

u(t ) +



≤



+ K





u(s) +





ds.

It follows from Gronwall’s inequality that

u(t ) +



≤



exp

(

K |t |

)

which yields the theorem.



17.6 Differentiability of the Flow

Now we return to the case of an autonomous differential equation X



= F(X)

where F is assumed to be C

. Our aim is to show that the ﬂow φ(t , X ) = φ

(X)

17.6 Differentiability of the Flow 401

determined by this equation is a C

function of the two variables, and to

identify ∂φ/∂X. We know, of course, that φ is continuously differentiable in

the variable t , so it sufﬁces to prove differentiability in X .

Toward that end let X(t) be a particular solution of the system deﬁned for

t in a closed interval J about 0. Suppose X(0) = X

. For each t ∈ J let

A(t ) = DF

X(t)

That is, A(t ) denotes the Jacobian matrix of F at the point X(t ). Since F is C

A(t ) is continuous. We deﬁne the nonautonomous linear equation



= A(t )U .

This equation is known as the variational equation along the solution X(t ).

From the previous section we know that the variational equation has a solution

on all of J for every initial condition U (0) = U

. Also, as in the autonomous

case, solutions of this system satisfy the linearity principle.

The signiﬁcance of this equation is that, if U

is small, then the function

t → X(t) + U (t )

is a good approximation to the solution X (t ) of the original autonomous

equation with initial value X (0) = X

+ U

To make this precise, suppose that U (t , ξ) is the solution of the variational

equation that satisﬁes U (0, ξ) = ξ where ξ ∈

.Ifξ and X

+ ξ belong to

O, let Y (t , ξ ) be the solution of the autonomous equation X



= F (X) that

satisﬁes Y (0) = X

+ ξ .

Proposition. Let J be the closed interval containing 0 on which X(t) is

deﬁned. Then

lim

ξ→0

|Y (t , ξ ) − X(t) − U (t , ξ)|

|ξ|

converges to 0 uniformly for t ∈ J.



This means that for every  > 0, there exists δ > 0 such that if |ξ |≤δ, then

|Y (t , ξ ) − (X(t) + U (t , ξ))|≤|ξ |

for all t ∈ J . Thus as ξ → 0, the curve t → X(t ) + U (t , ξ ) is a better and

better approximation to Y (t , ξ ). In many applications X(t) + U (t , ξ ) is used

in place of Y (t , ξ); this is convenient because U (t , ξ ) is linear in ξ .

402 Chapter 17 Existence and Uniqueness Revisited

We will prove the proposition momentarily, but ﬁrst we use this result to

prove the following theorem:

Theorem. (Smoothness of Flows) The ﬂow φ(t , X ) of the autonomous

system X



= F (X) isaC

function; that is, ∂φ/∂t and ∂φ/∂X exist and are

continuous in t and X.

Proof: Of course, ∂φ(t , X)/∂t is just F(φ

(X)), which is continuous. To

compute ∂φ/∂X we have, for small ξ,

φ(t , X

+ ξ ) − φ(t, X

) = Y (t , ξ) − X (t ).

The proposition now implies that ∂φ(t, X

)/∂X is the linear map ξ → U (t , ξ ).

The continuity of ∂φ/∂X is then a consequence of the continuity in initial

conditions and data of solutions for the variational equation.



Denoting the ﬂow again by φ

(X), we note that for each t the derivative

Dφ

(X) of the map φ

at X ∈ O is the same as ∂φ(t , X )/∂X. We call this the

space derivative of the ﬂow, as opposed to the time derivative ∂φ(t, X )/∂t .

The proof of the preceding theorem actually shows that Dφ

(X) is the solu-

tion of an initial value problem in the space of linear maps on

: for each

∈ O the space derivative of the ﬂow satisﬁes the differential equation

(

Dφ

)

= DF

)

Dφ

with the initial condition Dφ

) = I . Here we may regard X

as a parameter.

An important special case is that of an equilibrium solution

X so that

(

X) ≡

X. Putting DF

= A, we get the differential equation

(Dφ

(

X)) = ADφ

(

X),

with Dφ

(

X) = I . The solution of this equation is

Dφ

(

X) = exp tA.

This means that, in a neighborhood of an equilibrium point, the ﬂow is

approximately linear.

We now prove the proposition. The integral equations satisﬁed by X(t),

Y (t , ξ ), and U (t , ξ ) are

X(t ) = X



F(X (s)) ds,

17.6 Differentiability of the Flow 403

Y (t , ξ ) = X

+ ξ +



F(Y (s, ξ )) ds,

U (t , ξ) = ξ +



X(s)

(U (s, ξ )) ds.

From these we get, for t ≥ 0,

|Y (t ,ξ )−X (t )−U (t ,ξ )|≤



|F(Y (s,ξ))−F(X(s))−DF

X(s)

(U (s,ξ))|ds.

The Taylor approximation of F at a point Z says

F(Y ) =F(Z )+DF

(Y −Z)+R(Z ,Y −Z ),

where

lim

Y →Z

R(Z ,Y −Z)

|Y −Z|

uniformly in Z for Z in a given compact set. We apply this to Y =Y (s,ξ ), Z =

X(s). From the linearity of DF

X(s)

we get

|Y (t ,ξ )−X (t )−U (t ,ξ )|≤



|DF

X(s)

(Y (s,ξ )−X(s)−U (s,ξ ))|ds



|R(X (s),Y (s,ξ )−X (s))|ds.

Denote the left side of this expression by g (t ) and set

N =max{|DF

X(s)

||s ∈J }.

Then we have

g (t) ≤N



g (s)ds +



|R(X (s),Y (s,ξ )−X (s))|ds.

Fix  > 0 and pick δ

> 0 so small that

|R(X (s),Y (s,ξ )−X (s))|≤|Y (s,ξ )−X(s)|

if |Y (s,ξ )−X (s)|≤δ

and s ∈J .

From Section 17.3 there are constants K ≥0 and δ

> 0 such that

|Y (s,ξ )−X(s)|≤|ξ |e

≤δ

if |ξ |≤δ

and s ∈J .

404 Chapter 17 Existence and Uniqueness Revisited

Assume now that |ξ |≤δ

. From the previous equation, we ﬁnd, for t ∈J ,

g (t) ≤N



g (s)ds +



|ξ|e

ds,

so that

g (t) ≤N



g (s)ds +C|ξ |

for some constant C depending only on K and the length of J . Applying

Gronwall’s inequality we obtain

g (t) ≤Ce

|ξ|

if t ∈J and |ξ|≤δ

. (Recall that δ

depends on .) Since  is any positive

number, this shows that g (t )/|ξ |→0 uniformly in t ∈J , which proves the

proposition.

EXERCISES

1. Write out the ﬁrst few terms of the Picard iteration scheme for each of

the following initial value problems. Where possible, use any method to

ﬁnd explicit solutions. Discuss the domain of the solution.

(a) x



=x −2;x(0) =1

(b) x



4/3

;x(0) =0



4/3

;x(0) =1

(d) x



=cosx;x(0) =0

(e) x



=1/2x;x(1)=1

2. Let A be an n×n matrix. Show that the Picard method for solving X



AX,X(0) =X

gives the solution exp(tA)X

3. Derive the Taylor series for cost by applying the Picard method to

the ﬁrst-order system corresponding to the second-order initial value

problem



=−x; x(0) =1, x



(0) =0.

4. For each of the following functions, ﬁnd a Lipschitz constant on the

region indicated, or prove there is none:

(a) f (x) =|x|,−∞<x<∞

(b) f (x) =x

1/3

,−1 ≤x ≤1

Exercises 405

(d) f (x,y) =(x +2y,−y),(x,y) ∈

(e) f (x,y) =

1+x

≤4

5. Consider the differential equation



1/3

How many different solutions satisfy x(0) =0?

6. What can be said about solutions of the differential equation x



=x/t?

7. Deﬁne f :

R →R by f (x) =1ifx ≤1; f (x) =2ifx>1. What can be said

about solutions of x



=f (x) satisfying x(0) =1, where the right-hand side

of the differential equation is discontinuous? What happens if you have

instead f (x) =0ifx>1?

8. Let A(t) be a continuous family of n ×n matrices and let P(t )bethe

matrix solution to the initial value problem P



=A(t )P,P(0) =P

. Show

that

det P(t)=(det P

)exp





TrA(s)ds



9. Suppose F is a gradient vector ﬁeld. Show that |DF

|is the magnitude of

the largest eigenvalue of DF

.(Hint: DF

is a symmetric matrix.)

10. Show that there is no solution to the second-order two-point boundary

value problem



=−x, x(0) =0, x(π) =1.

11. What happens if you replace the differential equation in the previous

exercise by x



=−kx with k>0?

12. Prove the following general fact (see also Section 17.3): If C ≥0 and

u,v :[0,β]→

R are continuous and nonnegative, and

u(t ) ≤C +



u(s)v(s)ds

for all t ∈[0,β], then u(t ) ≤Ce

V (t)

, where

V (t) =



v(s)ds.

13. Suppose

C ⊂R

is compact and f :C →R is continuous. Prove that

f is bounded on

C and that f attains its maximum value at some

point in