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

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386 Chapter 17 Existence and Uniqueness Revisited

As earlier, the function F is said to be continuously differentiable, or C

,if

all of the partial derivatives of the f

exist and are continuous. We will assume

for the remainder of this chapter that F is C

. For each X ∈ R

, we deﬁne the

norm |DF

| of the Jacobian matrix DF

|DF

|= sup

|U |=1

|DF

(U )|

where U ∈

. Note that |DF

|is not necessarily the magnitude of the largest

eigenvalue of the Jacobian matrix at X.

Example. Suppose





Then, indeed, |DF

|=2, and 2 is the largest eigenvalue of DF

. However, if





then

|DF

|= sup

0≤θ≤2π







cos θ

sin θ





= sup

0≤θ≤2π



(cos θ + sin θ )

+ sin

= sup

0≤θ≤2π



1 + 2 cos θ sin θ +sin

> 1

whereas 1 is the largest eigenvalue.



We do, however, have

|DF

(V )|≤|DF

|| V |

for any vector V ∈

. Indeed, if we write V = (V /|V |) |V |, then we have

|DF

(V )|=





V /|V |





|V |≤|DF

|| V |

since V /|V | has magnitude 1. Moreover, the fact that F :

→ R

is C

implies that the function R

→ L(R

), which sends X → DF

,isa

continuous function.

17.2 Proof of Existence and Uniqueness 387

Let O ⊂ R

be an open set. A function F : O → R

is said to be Lipschitz

O if there exists a constant K such that

|F(Y ) − F (X)|≤K |Y − X |

for all X , Y ∈

O. We call K a Lipschitz constant for F . More generally, we say

that F is locally Lipschitz if each point in

O has a neighborhood O



in O such

that the restriction F to



is Lipschitz. The Lipschitz constant of F |O



may

vary with the neighborhoods



Another important notion is that of compactness. We say that a set

C ⊂ R

is compact if C is closed and bounded. An important fact is that, if f : C → R

is continuous and C is compact, then ﬁrst of all f is bounded on C and,

secondly, f actually attains its maximum on

C. See Exercise 13 at the end of this

chapter.

Lemma. Suppose that the function F :

O → R

is C

. Then F is locally

Lipschitz.

Proof: Suppose that F :

O → R

is C

and let X

∈ O. Let  > 0 be so small

that the closed ball



of radius  about X

is contained in O. Let K be an upper

bound for |DF

| on O



; this bound exists because DF

is continuous and O



is compact. The set O



is convex; that is, if Y , Z ∈ O



, then the straight-line

segment connecting Y to Z is contained in



. This straight line is given by

Y + sU ∈



, where U = Z − Y and 0 ≤ s ≤ 1. Let ψ(s) = F(Y + sU ).

Using the chain rule we ﬁnd



(s) = DF

Y +sU

(U ).

Therefore

F(Z ) − F(Y ) = ψ(1) − ψ(0)





(s) ds



Y +sU

(U ) ds.

Thus we have

|F(Z ) − F(Y )|≤



K |U |ds = K |Z − Y |.



The following remark is implicit in the proof of the lemma: If O is convex,

and if |DF

|≤K for all X ∈ O, then K is a Lipschitz constant for F |O.

388 Chapter 17 Existence and Uniqueness Revisited

Suppose that J is an open interval containing zero and X : J → O satisﬁes



(t) = F(X (t ))

with X (0) = X

. Integrating, we have

X(t ) = X



F(X (s)) ds.

This is the integral form of the differential equation X



= F(X ). Conversely,

if X : J →

O satisﬁes this integral equation, then X (0) = X

and X satisﬁes



= F(X ), as is seen by differentiation. Thus the integral and differential

forms of this equation are equivalent as equations for X : J →

O. To prove

the existence of solutions, we will use the integral form of the differential

equation.

We now proceed with the proof of existence. Here are our assumptions:

is the closed ball of radius ρ > 0 centered at X

2. There is a Lipschitz constant K for F on

3. |F(X)|≤M on

4. Choose a<min{ρ/M ,1/K } and let J =[−a, a].

We will ﬁrst deﬁne a sequence of functions U

, U

, ...from J to O

. Then we

will prove that these functions converge uniformly to a function satisfying the

differential equation. Later we will show that there are no other such solutions.

The lemma that is used to obtain the convergence of the U

is the following:

Lemma from Analysis. Suppose U

: J → R

, k = 0, 1, 2, . . . is a sequence

of continuous functions deﬁned on a closed interval J that satisfy: Given  > 0,

there is some N > 0 such that for every p, q>N

max

t∈J

(t) − U

(t)| < .

Then there is a continuous function U : J →

such that

max

t∈J

(t) − U (t )|→0ask →∞.

Moreover, for any t with |t |≤a,

lim

k→∞



(s) ds =



U (s) ds.



17.2 Proof of Existence and Uniqueness 389

This type of convergence is called uniform convergence of the functions U

This lemma is proved in elementary analysis books and will not be proved

here. See [38].

The sequence of functions U

is deﬁned recursively using an iteration

scheme known as Picard iteration. We gave several illustrative examples of

this iterative scheme back in Chapter 7. Let

(t) ≡ X

For t ∈ J deﬁne

(t) = X



F(U

(s)) ds = X

+ tF(X

Since |t |≤a and |F(X

)|≤M, it follows that

(t) − X

|=|t ||F(X

)|≤aM ≤ ρ

so that U

(t) ∈ O

for all t ∈ J. By induction, assume that U

(t) has been

deﬁned and that |U

(t) − X

|≤ρ for all t ∈ J . Then let

k+1

(t) = X



F(U

(s)) ds.

This makes sense since U

(s) ∈ O

so the integrand is deﬁned. We show that

k+1

(t) − X

|≤ρ so that U

k+1

(t) ∈ O

for t ∈ J ; this will imply that the

sequence can be continued to U

k+2

, U

k+3

, and so on. This is shown as follows:

k+1

(t) − X

|≤



F(U

(s))

≤



Mds

≤ Ma < ρ.

Next, we prove that there is a constant L ≥ 0 such that, for all k ≥ 0,



k+1

(t) − U

(t)



≤ (aK )

Let L be the maximum of |U

(t) − U

(t)| over −a ≤ t ≤ a. By the above,

L ≤ aM. We have

(t) − U

(t)|=





F(U

(s)) − F(U

(s)) ds



390 Chapter 17 Existence and Uniqueness Revisited

≤



K |U

(s) − U

(s)|ds

≤ aKL.

Assuming by induction that, for some k ≥ 2, we have already proved

(t) − U

k−1

(t)|≤(aK )

k−1

for |t |≤a, we then have

k+1

(t) − U

(t)|≤



|F(U

(s)) − F(U

k−1

(s))|ds

≤ K



(s) − U

k−1

(s)|ds

≤ (aK )(aK )

k−1

= (aK )

Let α = aK , so that α < 1 by assumption. Given any  > 0, we may choose N

large enough so that, for any r>s>Nwe have

(t) − U

(t)|≤

∞



k=N



k+1

(t) − U

(t)



≤

∞



k=N

≤ 

since the tail of the geometric series may be made as small as we please.

By the lemma from analysis, this shows that the sequence of functions

, U

, ...converges uniformly to a continuous function X : J → R

. From

the identity

k+1

(t) = X



F(U

(s)) ds,

we ﬁnd by taking limits of both sides that

X(t ) = X

+ lim

k→∞



F(U

(s)) ds

17.2 Proof of Existence and Uniqueness 391

= X





lim

k→∞

F(U

(s))



= X



F(X (s)) ds.

The second equality also follows from the Lemma from analysis. Therefore

X : J →

satisﬁes the integral form of the differential equation and hence

is a solution of the equation itself. In particular, it follows that X : J →

is C

This takes care of the existence part of the theorem. Now we turn to the

uniqueness part.

Suppose that X, Y : J →

O are two solutions of the differential equation

satisfying X (0) = Y (0) = X

, where, as above, J is the closed interval [−a, a].

We will show that X (t) = Y (t ) for all t ∈ J . Let

Q = max

t∈J

|X(t ) − Y (t)|.

This maximum is attained at some point t

∈ J . Then

Q =|X (t

) − Y (t

)|=







(s) − Y



(s)) ds



≤



|F(X (s)) − F(Y (s))|ds

≤



K |X (s) − Y (s)|ds

≤ aKQ.

Since aK < 1, this is impossible unless Q = 0. Therefore

X(t ) ≡ Y (t ).

This completes the proof of the theorem.



To summarize this result, we have shown: Given any ball O

⊂ O of radius

ρ about X

on which

1. |F(X)|≤M ;

2. F has Lipschitz constant K ; and

3. 0 <a<min{ρ/M ,1/K };

there is a unique solution X :[−a, a]→

O of the differential equation such

that X (0) = X

. In particular, this result holds if F is C

on O.

392 Chapter 17 Existence and Uniqueness Revisited

Some remarks are in order. First note that two solution curves of X



= F(X)

cannot cross if F satisﬁes the hypotheses of the theorem. This is an imme-

diate consequence of uniqueness but is worth emphasizing geometrically.

Suppose X : J →

O and Y : J

→ O are two solutions of X



= F(X ) for

which X (t

) = Y (t

). If t

= t

we are done immediately by the theorem.

If t

= t

, then let Y

(t) = Y (t

− t

+ t ). Then Y

is also a solution of

the system. Since Y

) = Y (t

) = X(t

), it follows that Y

and X agree

near t

by the uniqueness statement of the theorem, and hence so do X(t )

and Y (t).

We emphasize the point that if Y (t ) is a solution, then so too is Y

(t) =

Y (t + t

) for any constant t

. In particular, if a solution curve X : J → O

of X



= F (X) satisﬁes X (t

) = X(t

+ w) for some t

and w>0, then that

solution curve must in fact be a periodic solution in the sense that X (t +w) =

X(t ) for all t .

17.3 Continuous Dependence on Initial

Conditions

For the existence and uniqueness theorem to be at all interesting in any physical

or even mathematical sense, the result needs to be complemented by the

property that the solution X (t) depends continuously on the initial condition

X(0). The next theorem gives a precise statement of this property.

Theorem. Let

O ⊂ R

be open and suppose F : O → R

has Lipschitz

constant K . Let Y (t ) and Z (t ) be solutions of X



= F(X) which remain in O

and are deﬁned on the interval [t

, t

]. Then, for all t ∈[t

, t

], we have

|Y (t ) − Z(t)|≤|Y (t

) − Z(t

)|exp(K (t − t

)).



Note that this result says that, if the solutions Y (t ) and Z (t ) start out close

together, then they remain close together for t near t

. While these solutions

may separate from each other, they do so no faster than exponentially. In

particular, we have this corollary:

Corollary. (Continuous Dependence on Initial Conditions) Let

φ(t , X) be the ﬂow of the system X



= F (X) where F is C

. Then φ is a

continuous function of X .



The proof depends on a famous inequality that we prove ﬁrst.

17.3 Continuous Dependence on Initial Conditions 393

Gronwall’s Inequality. Let u :[0, α]→R be continuous and nonnegative.

Suppose C ≥ 0 and K ≥ 0 are such that

u(t ) ≤ C +



Ku(s) ds

for all t ∈[0, α]. Then, for all t in this interval,

u(t ) ≤ Ce

Proof: Suppose ﬁrst that C>0. Let

U (t ) = C +



Ku(s) ds > 0.

Then u(t ) ≤ U (t). Differentiating U ,weﬁnd



(t) = Ku(t ).

Therefore,



(t)

U (t )

Ku(t )

U (t )

≤ K .

Hence

(log U (t)) ≤ K

so that

log U (t) ≤ log U (0) + Kt

by integration. Since U (0) = C, we have by exponentiation

U (t ) ≤ Ce

and so

u(t ) ≤ Ce

If C = 0, we may apply the above argument to a sequence of positive c

that

tends to 0 as i →∞. This proves Gronwall’s inequality.



394 Chapter 17 Existence and Uniqueness Revisited

Proof: We turn now to the proof of the theorem. Deﬁne

v(t ) =|Y (t) − Z (t )|.

Since

Y (t ) − Z(t) = Y (t

) − Z(t

) +





F(Y (s)) − F(Z (s))



ds,

we have

v(t ) ≤ v(t

) +



Kv(s) ds.

Now apply Gronwall’s inequality to the function u(t ) = v(t + t

)toget

u(t ) = v(t + t

) ≤ v(t

) +



t+t

Kv(s) ds

= v(t

) +



Ku(τ ) dτ

so v(t + t

) ≤ v(t

) exp(Kt )orv(t ) ≤ v(t

) exp(K (t − t

)), which is just the

conclusion of the theorem.



As we have seen, differential equations that arise in applications often

depend on parameters. For example, the harmonic oscillator equations depend

on the parameters b (the damping constant) and k (the spring constant); cir-

cuit equations depend on the resistance, capacitance, and inductance; and so

forth. The natural question is how do solutions of these equations depend on

these parameters? As in the previous case, solutions depend continuously on

these parameters provided that the system depends on the parameters in a

continuously differentiable fashion. We can see this easily by using a special

little trick. Suppose the system



= F

(X)

depends on the parameter a in a C

fashion. Let’s consider an “artiﬁcially”

augmented system of differential equations given by



= f

, ..., x

, a)



= f

, ..., x

, a)



= 0.

17.4 Extending Solutions 395

This is now an autonomous system of n +1 differential equations. While this

expansion of the system may seem trivial, we may now invoke the previous

result about continuous dependence of solutions on initial conditions to verify

that solutions of the original system depend continuously on a as well.

Theorem. (Continuous Dependence on Parameters) Let X



= F

(X)

be a system of differential equations for which F

is continuously differentiable

in both X and a. Then the ﬂow of this system depends continuously on a as

well as X .



17.4 Extending Solutions

Suppose we have two solutions Y (t ), Z (t ) of the differential equation X



F(X ) where F is C

. Suppose also that Y (t) and Z (t ) satisfy Y (t

) = Z (t

) and

that both solutions are deﬁned on an interval J about t

. Now the existence

and uniqueness theorem guarantees that Y (t ) = Z (t ) for all t in an interval

about t

which may a priori be smaller than J . However, this is not the case. To

see this, suppose that J

∗

is the largest interval on which Y (t ) = Z(t). If J

∗

= J ,

there is an endpoint t

of J

∗

and t

∈ J . By continuity, we have Y (t

) = Z (t

Now the uniqueness part of the theorem guarantees that, in fact, Y (t ) and

Z(t) agree on an interval containing t

. This contradicts the assertion that J

∗

is the largest interval on which the two solutions agree.

Thus we can always assume that we have a unique solution deﬁned on a

maximal time domain. There is, however, no guarantee that a solution X(t )

can be deﬁned for all time. For example, the differential equation



= 1 + x

has as solutions the functions x(t ) = tan(t − c) for any constant c. Such a

function cannot be extended over an interval larger than

c −

<t<c+

since x(t ) →±∞as t → c ± π/2.

Next, we investigate what happens to a solution as the limits of its domain

are approached. We state the result only for the right-hand limit; the other

case is similar.

Theorem. Let

O ⊂ R

be open, and let F : O → R

be C

. Let Y (t ) be a

solution of X



= F(X) deﬁned on a maximal open interval J = (α, β) ⊂ R with

β < ∞. Then, given any compact set

C ⊂ O, there is some t

∈ (α, β) with

Y (t

) ∈ C.