Uniqueness and Existence

7/30/2019 Uniqueness and Existence

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Lecture Notes for Math 524

Dr. Michael Y. Li

September 18, 2009

These notes are based on the lecture notes of Professor James S. Muldowney, the books of Hale,Copple, Coddington and Levinson, and Perko. They are for the use of students in my graduateODE class.

Chapter 1. Existence and Uniqueness Theorems

1.1. Initial value problems

Let D be an open set in Rn+1, and f C(D Rn). Consider an ODE in Rn

x = f(t, x). (1)

A solution to (1) in an interval I R is a differentiable function : I Rn such that

(t) = f(t, (t)).

Example 1. The equation x

= 1 has solutions defined in R of form

x(t) = t + c,

where c is an arbitrary constant. In general, an ODE has infinitely many solutions.

We would like to solve equation (1) subject to conditions

x(t0) = x0 (2)

for (t0, x0) D. Equations (1) and (2) are called an Initial Value Problem, an IVP for short.

In this chapter, we investigate the following questions:

1. Existence of solutions to the IVP (1)-(2).

2. Uniqueness of solutions to the IVP (1)-(2).

3. Maximal interval of existence of a solution.

4. Continuous dependence of solutions on initial conditions and system parameters.

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We first examine some examples.

Example 2. The IVP x = x, x(t0) = x0 has solution

x(t) = x0 e(tt0)

which exists for all t R and is unique.

Example 3. The ODE x = x2 has a solution

x(t) = 1

t + c,

where c can be uniquely determined by the initial condition x(t0) = x0, for x0 = 0. This solutiononly exists near t0 and is unique as long as it exists. When x0 = 0 the IVP has a unique solutionx(t) = 0.

Example 4. Consider the IVP x = |x|1/2, x(t0) = x0. Since x(t) 0, x(t) is increasing.

For x(t) > 0, we have x

x1/2= 1, 2x1/2 = t + c,

where c = 2x1/20 t0. A solution is given by

x(t) =

14(t + c)

2, t c,

14(t + c)2, t < c.

This solution exists in R. When x0 = 0, we have x(t) = 0 for all t. Solutions to the IVP are not

unique over R . For instance, if x0 > 0 and c = 2x1/20 t0, you may check that the function

x(t) =

14(t + c)2, t c,0, t < c,

is another solution with the same initial condition x(t0) = x0. However a solution is unique in anhood of (t0, x0) if x0 = 0.

1.2. An equivalent problem

The IVP (1)-(2) is equivalent to the following integral equation

x(t) = x0 + t

t0

f(s, x(s)) ds. (3)

Namely, the IVP and the integral equation has the same set of solutions.

The following useful result is a simple version of Gronwalls lemma.

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Lemma 1.1 Let g(t) be a continuous real-valued function and g(t) 0 for t [t0, b]. If, forK > 0, C 0,

g(t) C+ K

tt0

g(s) ds

for t [t0, b], then

g(t) CeK(tt0) (4)

for t [t0, b].

Proof. Let G(t) = C+ Ktt0

g(s)ds. Then G(t) g(t) and G(t) 0 for all t [t0, b]. Moreover,

G(t) = Kg(t) KG(t).

ThusG(t) KG(t) 0

andG(t)eKt G(t)KeKt 0,

namely[G(t)eKt ] 0,

and thus G(t)eKt G(t0)eKt0 = CeKt0, it follows g(t) G(t) CeK(tt0).

Corollary 1.2 If g(t) 0, K > 0 satisfy

g(t) K

tt0

g(s)ds, t t0.

Then g(t) = 0 for t t0.

1.3. Uniqueness under Lipschitz conditions

A function f C(D Rn) is said to satisfy Lipschitz condition with respect to x if there exists(Lipschitz) constant K > 0, such that

|f(t, x1) f(t, x2)| K|x1 x2| (5)

for all (t, x1), (t, x2) D. Here, | | denote any norm in Rn.

Example 5. f(x) = sin x is Lipschitz continuous on (, +).

Example 6. f(x) = x2 is not globally Lipschitz continuous over R. It is, however, Lipschitz

continuous over any compact subset of R. In this case, we say it is locally Lipschitz continuousover R.

Example 7. f(x) = 1x2 is globally Lipschitz continuous on [, ) for any > 0, but onlylocally Lipschitz continuous on (0, ).

Example 8. If f(x) is C1 in an open subset U if Rn, then f is locally Lipschitz in U.

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Theorem 1.3 (Uniqueness) Suppose

(1) f C(D Rn).

(2) f satisfies Lipschitz condition in D with respect to x.

Then the IVP has at most one solution in D.

More specifically, if there are two solutions 1(t) and 2(t) to the IVP, both defined fort [a, b], such that (t, i(t)) D for t [a, b] and i = 1, 2, then 1(t) = 2(t) for all t [a, b].

Proof. (Proof # 1) Suppose 1(t) and 2(t) are both solutions to the IVP. Then

i(t) = x0 +

tt0

f(s, i(s)) ds, i = 1, 2.

and thus, if t t0,

|1(t) 2(t)| = |tt0

[f(s, 1(s)) f(s, 2(s))] ds|

tt0

|f(s, 1(s)) f(s, 2(s))| ds (t t0)

tt0

K|1(s) 2(s)| ds (Lipschitz condition)

(6)

Set l(t) = |1(t) 2(t)| 0. Then, if t t0,

l(t)

tt0

Kl(s) ds,

which implies l(t) = 0 for t t0 by Corollary 1.2. A similar argument can show l(t) = 0 for allt t0.

Proof. (Proof # 2) Set = maxt[t0,b]|1(t) 2(t)|. From (6), we have

|1(t) 2(t)|

tt0

Kds = K(t t0).

Using (6) again, we get

|1(t) 2(t)|

tt0

K2(s t0) ds = K2(t t0)

2/2.

Repeat the process, we get, for any integer n,

|1(t) 2(t)| Kn(t t0)

n/n! 0 as n .

This implies 1(t) = 2(t) for t [t0, b]. (Proof for t [a, t0] is similar.)

Questions:

(1) Can we weaken the global Lipschitz assumption in the theorem to local Lipschitz?

(2) Can you explain the observations in Example 4 in the context of the uniqueness theorem?Does Example 7 contradict the theorem?

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1.4. Local existence under Lipschitz conditions: Picards Method of Successive

Approximation

We prove the following result.

Theorem 1.4 (Picard Local Existence Theorem) Suppose

(1) f C(D Rn).


Then there exists > 0 such that the IVP has a solution for t [t0 , t0 + ].

Proof. Let (t0, x0) D. Then there exists a1, a2 > 0 such that the closed rectangle R = {(t, x) :|t t0| a1, |x x0| a2} D ( D is open). Let

M = max{|f(t, x)| : (t, x) R}.

We will show that = min{a1,a2M} satisfies the conclusion in the theorem. The following method

is called Picard Method of Successive Approximation.Consider the equivalent integral equation

x(t) = x0 +

tt0

f(s, x(s)) ds. (7)

Step 1: Set up the successive approximation

x0(t) = x0xn+1(t) = x0 +

tt0

f(s, xn(s)) ds, n = 0, 1, 2, .(8)

Step 2: Verify (use induction) that the sequence {xn(t)} are well-defined for |t t0| , andthat (t, xn(t)) R, |t t0| . In fact,

|x1(t) x0|

tt0

|f(s, x0)|ds M|t t0| a2,

|xn(t) x0|

tt0

|f(s, xn1(s))| ds M|t t0| a2, n = 2, 3, ,

if |t t0| a2/M.

Step 3: Show that {xn(t)} is a uniform Cauchy sequence for |t t0| .

|xn+1(t) xn(t)| tt0 |f(s, xn(s)) f(s, xn1(s))| ds

(Lipschitz) K

tt0 |xn(s) xn1(s)| ds, n = 1, 2,...

and

|x1(t) x0(t)| = |

tt0

f(s, x0) ds| M|t t0|, |t t0| .

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Thus|x2(t) x1(t)| M K|t t0|

2/2!, |t t0| ,

and|x3(t) x2(t)| MK

2|t t0|3/3!, |t t0| ,

and, for n = 1, 2,...

|xn+1(t) xn(t)| M Kn|t t0|

n+1

(n + 1)!

M

K

(K)n+1

(n + 1)!.

Thus, for 0 m n, we have

|xn(t) xm(t)| M

K

n1k=m

(K)k+1

(k + 1)!, |t t0| . (9)

Step 4: The solution to the IVP is obtained as the uniform limit of {xn(t)}.Cauchys criterion and (9) imply that

limt

xn(t) = x(t)

uniformly for t [t0, t0 + ], and x(t) is continuous. In the equation (8), let t , we verifythat x(t) is a solution to the integral equation (7). Thus x(t) is differentiable and satisfies theIVP.

Summarize the results in Sections 1.3 and 1.4, we have the following existence and uniquenesstheorem under Lipschitz conditions.

Theorem 1.5 Suppose

(1) f C(D Rn).


Then there exists > 0 such that the IVP has a unique solution for t [t0 , t0 + ].

1.5. The existence and uniqueness under Lipschitz conditions: a proof using fixed

point theorem

The following is an abstract version of the proof of Picards existence and uniqueness theorem usinga fixed point theorem.

Let (B, | |) be a Banach space. Let

V = { B : || b}

be a ball in B with radius b.

Lemma 1.6 Suppose that T : V B is a transformation that satisfies:

(a) |T()| b for V, i.e., T(V) V.

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(b) 0 < K1 < 1, such that |T() T()| K1 | |, for , V.

Then, there exists a unique V such that T() = , namely, T has a unique fixed point inV.

Proof. Exercise. (Hint: use Picard successive approximation.)

A mapping that satisfies assumption (b) is called a contraction, and the fixed point theoremin the lemma is also called the contraction mapping theorem. In the following, we prove Picardstheorem using this fixed point result.

Let B = C[t0 a1, t0 + a1] and V = {x B : |x x0| a2}. Define T using the integralequation (3) as follows:

T()(t) = x0 +

tt0

f(s, (s)) ds.

Choose a1 small enough so that a1K < 1, where K is the Lipschitz constant of f in D, then

we can show that T satisfies

(a) B implies T() B.

(b) |T() T()| a1K| |, for ||, || B.

Then by the previous lemma, T has a unique fixed point B and x = (t) is a solution to theintegral equation (3), and hence a solution to the IVP.

1.6. Peano existence theorem

In Theorem 1.5, if the assumption (2) on the Lipschitz condition is removed, the conclusion stillholds, except for the uniqueness.

Theorem 1.7 Suppose that f C(D Rn). Then the IVP has a solution which exists on amaximal interval (, +).

Note: and + depend on t0, x0 in general.

To establish this result, we need some preliminary propositions.

Let F be a family of real-valued (or Rn valued) functions on a closed interval I R.

(a) F is uniformly bounded if there exists M > 0 such that

|f(t)| M, f F, t I.

(b) F is equi-continuous if for any > 0, there exists > 0 such that t1, t2 I, |t1 t2| 0. Choose > 0 such that t1, t2 I, |t1 t2| < , f F imply

|f(t1) f(t2)| < /3. (11)

Choose r1, r2, , rN such that {(ri , ri + ) : i = 1, , N} covers I (I compact). Thereexists H such that for m, n H,

|fmm(ri) fnn(ri)| /3, i = 1, 2, , N. (12)

If t I, choose ri such that |ri t| < . Then m, n H implies

|fmm(t) fnn(t)|

|fmm(t) fmm(ri)| + |fmm(ri) fnn(ri)| + |fnn(ri) fnn(t)|

< /3 + /3 + /3 = by (11), (12).

Therefore, {fkk} is a subsequence of {fk} and is uniformly convergent on I (Cauchy criterion).

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Proposition 1.9 (Peano Local Existence Theorem) Suppose that f C(D Rn). Thenthe IVP has a solution which exists in a neighborhood of t0.

Proof. (Proof # 1, Tonelli). We show the existence of a solution on a right half-neighborhood oft0. (left half-neighborhood is similar)

For > 0, letx(t) = x0, t0 t t0,

x(t) = x0 +

tt0

f(s, x(s)) ds, t0 < t(13)

Let a1, a2, M , be as in the Picard Local Existence Theorem (Theorem1.4). Then x(t) is welldefined by (13) for t0 t t0 + , since

|x(t) x0| = |

tt0

f(s, x(s)) ds| M|t t0| M a1 a2.

This also shows that {x : 0 < } is uniformly bounded on [t0, t0 + ]. The set of functions is

also equi-continuous, since

|x(t1) x(t2)| = |

t1t2

f(s, x(s)) ds| M|t1 t2| , if |t1 t2| < /M.

By Arzela-Ascoli Theorem, there exists sequence k 0+, such that {xk} is uniformly convergent

on [t0, t0 + ]. Therefore (13) implies that the limit function x(t) satisfies

x(t) = x0 +

tt0

f(s, x(s)) ds

and x(t) is a solution of the IVP on [t0, t0 + ].

Proof. (Proof #2, Peano). Construct an -approximate solution (Euler polygon) as follows:Let ti = t0 + i, i = 0, 1, 2, , n. ( n /. )If ti t ti+1, letx(t) = xi + (t ti)f(ti, xi),where xi = x(ti), i = 1, 2, , n.Thus x(t) satisfies an integral equations

x(t) = x0 +

tt0

F(s) ds,

where F(s) = f(ti, xi), ti s ti+1, i = 1, 2, , n, is a step function.Note

|x1 x0| = |f(t0, x0)|(t1 t0) M(t1 t0),|xi+1 x0| = |x(ti+1) x0| |x(ti+1) x(ti)| + + |x(t1) x0|

M(ti+1 t0) a2, (since t0 ti t0 + )

Thus (use induction) (ti, xi) R for i = 1, 2, , n.

Show

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(i) |x(t) x0| M(t t0), t0 t t0 + ;

(ii) |x(t) x(s)| M|t s|, t0 t, s t0 + ;

(iii) There is a sequence k 0+ such that {xk(t)} is uniformly convergent on [t0, t0 + ] to a

function x(t);

(iv) Fk(t) converges uniformly on [t0, t0 + ] to f(t, x(t));

(v) x(t) = f(t, x(t)), x(t0) = x0.

The rest of the section concerns the maximal interval of existence.

Proposition 1.10 Suppose

(i) f C(D Rn);

(ii) K D compact;

(iii) (t, x(t)) K, t0 t < < for some solution x(t) of the IVP.

Then x1 = limt x(t) exists and (, x1) K. Furthermore, [t0, ) is not a right-maximalinterval of existence for x(t).

Proof. To show x1 = limt x(t) exists, we show that x(t) is uniformly continuous on [t0, ).From the integral equation we have

|x(t1) x(t2)| = |

t2t1

f(t, x(t))dt| M|t1 t2|

for all t1, t2 [t0, ), where M = max {|f(t, x)| : (t, x) K}. The Peano Existence Theorem(Theorem1.9) implies that x(t) can be continued to the right of .

Exercise. Suppose that x1(t) is a solution of the IVP

x(t) = f(t, x), x(t0) = x0,

in [t0, a) and that limta x1(t) = x1 exists and that (a, x1) D. Let x2(t) be a solution to theIVP

x = f(t, x), x(a) = x1

in [a, b). Define x(t) = x1(t) for t [t0, a) and x(t) = x2(t) for t [a, b) Show that x(t) is a

solution of the IVPx = f(t, x), x(t0) = x0

in [t0, b).

Proposition 1.11 Suppose

(i) f C(D Rn);

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(ii) x(t) is a solution of the IVP defined for [t0, ) ;

(iii) t0 < tk < , limk tk = and limk x(tk) = x1;

(iv) (, x1) D.

Then x1 = limt

x(t) and [t0, ) is not a right-maximal interval of existence for x(t).

Proof. Let > 0 be sufficiently small that J D, where J = {(t, x) : |t| , |xx1| }.Let M = max{|f(t, x)| : (t, x) J}. Then N 0 < tN < /2M and |x(tN) x1| < /2.We claim that

|x(t) x(tN)| < M( tN) < /2, tN t < . (14)

Then, tN t < implies

|x(t) x1| |x(t) x(tN)| + |x(tN) x1| < /2 + /2 = .

and thus limt x(t) = x1. Moreover, (, x1) D implies that [t0, ) is not the right-maximal

interval of existence.To show claim (14): suppose it is false. Then there exists a least T > tN such that

|x(T) x(tN)| = M( tN) < /2. (15)

Therefore,|x(t) x1| |x(t) x(tN)| + |x(tN) x1| < /2 + /2 = ,

for tN t < T. Thus (t, x(t)) J, tN t < T. Therefore

|x(T) x(tN)| = |

TtN

f(s, x(s))ds|

TtN

|f(s, x(s))|ds (16)

M(T tN) < M( tN) (17)

which contradicts (15), proving the claim. (Question: why cant we use (refBBB) to prove (14)directly?)

Corollary 1.12 Assume that D = (a, b) Rn, t0 (a, b), and (, +) is a maximal intervalof existence for a solution of x = f(t, x). Then

= a or limt+

|x(t)| = ,

+ = b or limt

+

|x(t)| = .

Proof. Suppose that + < b. Then the uniform continuity of x(t) in [t0, +), now a finite interval,implies that limt

+|x(t)| exists (though may not be finite). Suppose that limt

+|x(t)| 0 andtn + such that limn d((t, x(tn)), D) 0. First of all, this relation implies that x(tn) isbounded, since otherwise, D must be unbounded and {} D, and there exists a subsequencetk such that limk |x(tk)| = . This implies that limk d((t, x(tk)), D) = 0 < 0. Now|x(tn)|, being bounded, must have a convergent subsequence, x(tm) such that tm + andx(tm) x1, and thus d((+, x1), D) 0. This implies that (+, x1) D, and the solutioncan then be further continued. This contradiction shows that (18) must hold on a maximal interval.

However, for a solution x(t) on its maximal interval of existence (, +), the limit limt x(t)may not exist, as the following example shows.

Example. Consider

x(t) =1

t2cos

1

t.

The function f(t, x) = 1t2 cos1t is continuous in D = (0, ) R. A solution x(t) = c sin

1t has

a left maximal interval (0, t0). We see that limt0+ d((t, x(t)), D) 0, but limt0+ x(t) doesnot exist.

1.7. Continuous dependence on initial conditions and parametersConsider the IVP

x = f(t,x,) x(t0) = x0, (19)

where t R, x Rn, Rm. We make the following assumptions:

(i) f C(G Rn), G open in Rn+m+1.

(ii) Solutions to the IVP is unique for each (t0, x0, ) G.

Denote the solution to the IVP as x = x(t; t0, x0, ).

We will show that x(t; t0, x0, ) is continuous in its domain.

Theorem 1.13 Assume that x(t; t0, x0, 0) exists for t [a, b]. Then, > 0, > 0 suchthat

|t0 t0| < , |x0 x0| < , |

0 0| <

imply

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(1) x(t; t0, x0,

0) exists for t [a, b].

(2) supatb |x(t; t0, x

0,

0) x(t; t0, x0, 0)| < .

W.o.l.g., we may drop the dependence on . The IVPx = f(t,x,)

x(t0) = x0,x = x(t; t0, x0, )

is equivalent to the IVP

x = f(t,x,y)y = 0

x(t0) = x0y(t0) = ,

x = x(t; t0, x0, )y = y0 (= ).

Proof of Theorem 1.13. Equivalently, we prove the following. Let

(tk, xk) (t0, x0), as k .

Then for k > K,

(1) x(t; tk, xk) exists for t [a, b].

(2) limk supatb |x(t; tk, xk) x(t; t0, x0)| = 0, or equivalently, x(t; , tk, xk) x(t, t0, x0) ask uniformly for t [a, b].

Step 1. Choose D0 D0 D1 D1 D such that (t, x(t)) D0 for t [a, b]. Set

M > sup(t,x)D1 |f(t, x)|.

Choose such that

R(t0, x0) = {(t, x) : |t t0| , |x x0| M } D1.

Then x(t; t0, x0) is defined for [t0 , t0 + ], and K > 0, such that k > K implies (tk, xk) R(t0, x0) and

R(tk, xk) D1,

and thus xk(t) = x(t; tk, xk) is defined for t [t0 0, t0 + 0] for some 0 < 0 . From theintegral equation

xk(t) = xk +ttk

f(s, xk(s)) ds

we conclude that {xk(t)} is uniformly bounded and equi-continuous in [t0 0, t0 + 0]. Arzela-Ascoli Theorem implies that any subsequence of {xk(t)} has a subsequence uniformly convergentto the same limit x(t, t0, x0) (uniqueness of solutions). This implies that

x(t; tk, xk) x(t; t0, x0) as k

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uniformly in [t0 0, t0 + 0].

Step 2. Suppose [a, b] [t0 0, t0 + 0]. Let t10 = t0 + 0, x

10 = x(t

10). Choose 1 such that

R1(t10, x

10) D1.

Repeating the preceding argument, we conclude: K1 > 0 such that k > K1 implies x(t; tk, xk)is defined for t [t0, t0 + 0 + 1], and

x(t; tk, xk) x(t; t0, x0) as k

uniformly in [t0, t0 + 0 + 1]. Proceed like this, after finite number of steps, we obtain [a1, b1] [a, b] and K > 0 such that k > K implies

(1) x(t; tk, xk) exists for t [a1, b1].

(2) limk supa1tb1 |x(t; tk, xk) x(t; t0, x0)| = 0.

Remark. The result also includes the continuous dependence on the vector field f(t, x). Supposexk(t) solves the IVP

x = fk(t, x), x(tk) = xk,

and, as k , (tk, xk) (t0, x0), fk(t, x) f(t, x) on any compact subset of D Rn+1.

Then, for k > K > 0, x(t; tk, xk, fk) exists in [a, b] and

limk

supatb

|x(t; tk, xk, fk) x(t; t0, x0, f)| = 0.

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Uniqueness and Existence