Ch 3. Fundamental Theory of ODEs

§3.1. Existence-Uniqueness Theorem

(Braun: Differential equations and their applications 1.10; Agarwal: Es-

sentials of ODE:Lecture 6-9; Simmons: sec 68-69; Williamson: Appendix

C)

We consider the initial value problem

x = f(t, x), x(t0) = x0, 3.1.1

where x = dxdt . We shall use the notation

Bδ(t0, x0) = (t, x) :√|t− t0|2 + |x− x0|2 < δ,

Bδ(t0, x0) = (t, x) :√|t− t0|2 + |x− x0|2 ≤ δ.

Definition Let G be a subset in R2. f(t, x) : G → R is said to satisfy

the Lipschitz condition with respect to x in G if there exists a constant

L > 0 such that, for any (t, x), (t, y) ∈ G,

|f(t, x)− f(t, y)| ≤ L|x− y|.

f is said to be locally Lipschitz in G if for each point (t0, x0) ∈ G there exist

δ > 0, L(t0, x0) > 0, such that Bδ(t0, x0) ⊂ G, and for all x, y ∈ Bδ(t0, x0)

we have

|f(t, x)− f(t, y)| ≤ L(t0, x0)|x− y|.1

2

Example (1) f(t, x) = |xt| on [−2, 2]× [−3, 3] ;

(2) g is continuously differentiable on R2;

(3) h : R2 → R such that h and ∂h∂x(t, x) are continuous on [a, b]× [c, d].

Exercises: Show that all the following functions satisfy the Lipschitz

condition with respect to x in (t, x) ∈ [−1, 3]× [−2, 5]

(1) x3t2, (2) ex cos t, (3) sin xt.

3

Remark 3.1 Assume G is an open subset of R2 and f(t, x) is continu-

ously differentiable in G. Then f is locally Lipschitz in G.

The following is usually known as local existence and uniqueness theorem.

Theorem 3.1 (Picard) Assume G is an open subset of R2 containing

(t0, x0) and f(t, x) is continuously differentiable in G. Then there exists

a > 0 such that (3.1.1) has a unique solution on the interval [t0−a, t0 +a].

4

Instead of proving the above directly, we will indeed prove a stronger

version of that first.

Theorem 3.2 (Picard, local existence) Suppose:

(1) f(t, x) is continuous on the rectangle S = [t0 − a, t0 + a] × [x0 −b, x0 + b] and hence there exists M > 0 such that |f(t, x)| ≤ M on

the rectangle S.

(2) f(t, x) satisfies a uniform Lipschitz condition with respect to x on

the rectangle S, that is, there exists L > 0 such that

|f(t, x1)− f(t, x2)| ≤ L|x1 − x2| for all

x1, x2 ∈ [x0 − b, x0 + b] and t ∈ [t0 − a, t0 + a].

Then, the initial value problem (3.1.1) has a unique solution on the interval

[t0 − h, t0 + h] where h = mina, b/M.

5

Proof of Theorem 3.2: A special case

First, let us assume h = mina, b/M < 1/L.

Step 1. Consider the following integral equation

x(t) = x0 +∫ t

t0f(s, x(s))ds. 3.1.2

x(t) is a continuously differentiable solution of (3.1.1), if and only if x(t)

is a continuous solution of (3.1.2).

We shall use successive iteration method (Picard’s iteration) to show that

(3.1.2) has a solution. Let

x0(t) = x0,

xn+1(t) = x0 +∫ t

t0f(s, xn(s))ds.

Check that by induction, we have |xn+1(t)− x0| ≤ Mh ≤ b for all n.

6

Step 2. We will show by induction that, for all n ≥ 1, xn(t) is continuous

on [t0 − h, t0 + h], and

|xn+1(t)− xn(t)| ≤ (Lh)nb.

First, clearly,

|x2(t)− x1(t)| = |∫ t

t0f(s, x1(s)− f(s, x0)ds|

≤ |∫ t

t0L|x1(s)− x0|ds| ≤ Lhb.

Next, note that by induction hypothesis,

|xn+2(t)− xn+1(t)| = |∫ t

t0f(s, xn+1(s)− f(s, xn(s))ds|

≤ |∫ t

t0L|xn+1(s)− xn(s)|ds| ≤ (Lh)n+1b.

Hence for m > n and |t− t0| ≤ h we have

|xm(t)− xn(t)| ≤m∑

k=n

|xk+1(t)− xk(t)|

≤m∑

k=n

(Lh)kb <(Lh)n

1− Lhb → 0 as m,n →∞.

By Cauchy criterion, there exists a function x(t) such that xn(t) con-

verges to x(t) uniformly on [t0−h, t0+h] as n →∞, and x(t) is continuous.

7

Hencex(t) = lim

n→∞xn+1(t) = limn→∞

[x0 +

∫ t

t0f(s, xn(s))ds

]

= x0 +∫ t

t0f(s, x(s))ds.

So x(t) is a continuous solution of (3.1.2) on [t0 − h, t0 + h].

Step 3. We show (3.1.2) has only one solution. Suppose x(t) and y(t)

are solutions of (3.1.2) on [t0 − h, t0 + h]. Then, for t0 < t ≤ t0 + h,

|x(t)− y(t)| = |∫ t

t0

[f(s, x(s))− f(s, y(s))

]ds|

≤∫ t

t0|f(s, x(s))− f(s, y(s))|ds

≤L∫ t

t0|x(s)− y(s)|ds ≤ Lh max

t0≤s≤t0+h|x(s)− y(s)|.

Hence

maxt0≤t≤t0+h

|x(t)− y(t)| ≤ Lh maxt0≤s≤t0+h

|x(s)− y(s)|.Since Lh < 1 we find x(t)− y(t) = 0 for all t0 ≤ t ≤ t0 + h.

Similarly we show x(t)− y(t) = 0 for all t0 − h ≤ t ≤ t0. ¤

8

Example 1.dx

dt= x, x(0) = 1

Example 2.dx

dt= sin4(tx), x(0) = 0

Example 3.dx

dt= x + t, x(0) = 0.

9

Proof of Theorem 3.2: (actual proof)

Again, let

x0(t) = x0,

xn+1(t) = x0 +∫ t

t0f(s, xn(s))ds.

Again by induction, we have |xn+1(t) − x0| ≤ Mh ≤ b for all n. For

simplicity let t0 = 0. Then, (assuming 0 ≤ t ≤ h)

|x2(t)− x1(t)|

≤∫ t

0|f(s, x1(s))− f(s, x0)|ds

≤∫ t

0L|x1(s)− x0|ds ≤

∫ t

0Lb = Lbt

since max|x1(s)− x0| : s ∈ [0, h] ≤ b. Hence,

|x3(t)− x2(t)| ≤∫ t

0|f(s, x2(s))− f(s, x1(s))|ds

≤∫ t

0L|x2(s)− x1(s)|ds ≤

∫ t

0L(Lbs)ds = bL2t2/2.

Continue the process, we have

|x4(t)− x3(t)| ≤ bL3t3/6.

By induction, we have

|xn+1(t)− xn(t)| ≤ bLntn/n!.

It is then easy to see that xn(t) is uniformly Cauchy on [0, h]. We can

then show the existence just as before.

10

Finally, suppose x(t) and y(t) are solutions of (3.1.2) on [−h, h]. Then,

for 0 < t ≤ h,

|x(t)− y(t)| = |∫ t

0

[f(s, x(s))− f(s, y(s))

]ds|

≤∫ t

0|f(s, x(s))− f(s, y(s))|ds

≤L∫ t

0|x(s)− y(s)|ds ≤ Ltε

where ε = sup|x(s)− y(s)| : s ∈ [0, h]. Repeat the argument,

|x(t)− y(t)| ≤∫ t

0L2sεds ≤ L2εt2/2 ≤ · · · ≤ εLntn/n!.

It then follows that ε ≤ ε(Lh)n/n! for all n and hence ε = 0.

11

A useful inequality (Gronwall’s Inequality)

Let u(x), p(x) and q(x) be nonnegative continuous functions in [x0 −a, x0 + a] and

u(x) ≤ p(x) + |∫ x

x0

q(t)u(t)dt| for all x ∈ [x0 − a, x0 + a]. (∗)

Then for all x ∈ [x0 − a, x0 + a],

u(x) ≤ p(x) + |∫ x

x0

p(t)q(t) exp(|∫ x

tq(s)ds|

)dt|.

Proof.

We will only prove the case x ∈ [x0, x0 + a]. Define

r(x) =∫ x

x0

q(t)u(t)dt.

Then r(x0) = 0 and r′(x) = q(x)u(x) by the fundamental theorem of

Calculus. By (*), we have

r′(x) ≤ p(x)q(x) + q(x)r(x).

Hence, r′(x)− q(x)r(x) ≤ p(x)q(x), and

d

dx

(e−

∫ x

x0q(s)dsr(x)

)≤ e−

∫ x

x0q(s)dsp(x)q(x).

The Gronwall inequality then follows by integrating both side on [x0, x].

12

Proof of Theorem 3.1

Theorem 3.3 Under the conditions of Theorem 3.2, suppose b = ∞,

then the initial value problem (3.1.1) has a unique solution x(t) on [t0 −a, t0 + a].

13

Example (1) dxdt = sin3(xt), (2) dx

dt = e−t2 cos xt,

(3) dxdt = f(t)g(x) + h(t), f and h are continuous on [a, b] and |g(x1) −

g(x2)| ≤ L|x1 − x2| for all x1, x2 ∈ R.

14

Theorem 3.4 (Peano) Assume G is an open subset of R2 containing

(t0, x0) and f(t, x) is continuous in G. Then there exists a > 0 such that

(3.1.1) has at least one solution on the interval [t0 − a, t0 + a].

15

Example. y = y2/3, y(0) = 0.

y = 0 and y = 127t

3 are solutions. Moreover, for any a < 0 and b ≥ 0,

y =

127(t− a)3 if t < a,

0 if t ≥ a,

y =

0 if t < b,

127(t− b)3 if t ≥ b,

and

y =

127(t− a)2 if t < a,

0 if a ≤ t < b,

127(t− b)3 if t ≥ b,

are also solutions.

Example. y =√

1− y2, y(0) = 0.

The solution is

y(t) =

−1 if −∞ < t < −π2 ,

sin t if − π2 ≤ t ≤ π

2 ,

1 if π2 < t < ∞.

16

§3.2. Dependence on Initial Conditions and Parameters (Lecture

10-11, Essentials of ODE)

Theorem 3.5 (Dependence on initial conditions) Let G be an open

subset of R2 containing (t00, x00) and f(t, x) be continuously differentiable

in G. Then there exist positive constants a and δ such that, if |t0− t00| < δ

and |x0 − x00| < δ, the initial value problem

x = f(t, x), x(t0) = x0 3.2.1

has a unique solution x(t; t0, x0) defined for t ∈ [t00−a, t00+a]. As a function

of (t, t0, x0), x(t; t0, x0) is continuously differentiable in (t00−a, t00+a)×(t00−δ, t00+δ)×(x0

0−δ, x00+δ), and, for fixed (t0, x0) ∈ (t00−δ, t00+δ)×(x0

0−δ, x00+δ),

is twice continuously differentiable in t on (t00 − a, t00 + a).

17

Example

18

Theorem 3.6 (Dependence on parameters) Let U be an open

subset of R3 containing (t00, x00, µ0), and let f(t, x, µ) be continuously dif-

ferentiable in U . Then there exist positive constants a and δ (a > 2δ) such

that, if |t0−t00| < δ, |x0−x00| < δ and |µ−µ0| < δ, the initial value problem

x = f(t, x, µ), x(t0) = x0 3.2.2

has a unique solution x(t; t0, x0, µ) defined for t ∈ [t00−a, t00+a]. x(t; t0, x0, µ)

is continuously differentiable in (t00 − a, t00 + a) × (t00 − δ, t00 + δ) × (x00 −

δ, x00 + δ)× (µ0 − δ, µ0 + δ).

19

Example

20

§3.3. Differential Systems and Higher Order Equations

Consider a system of differential equations

x1 = f1(t, x1, · · · , xn),

x2 = f2(t, x1, · · · , xn),

· · · · · · · · · · · · · · ·xn = fn(t, x1, · · · , xn),

where xj = dxj

dt . Let us introduce notations

x =

x1

· · ·xn

, x =

x1

· · ·xn

, f(t,x) =

f1(t,x)

· · ·fn(t,x)

.

Then the system can be written in a vector form :

x = f(t,x). 3.3.1

Example

21

Differential equations of higher order can be reduced to equivalent sys-

tems. Let us consider

dny

dtn= F (t, y, y, · · · ,

dn−1y

dtn−1 ). 3.3.2

Let

x1 = y, x2 =dy

dt, · · · , xn =

dn−1y

dtn−1 .

Then (3.3.2) is equivalent to the following system

x1 = x2,

x2 = x3,

· · · · · · · · ·xn = −F (t, x1, x2, · · · , xn).

It can be written in the form of (3.3.1) if we let

x =

x1

· · ·xn

, f(t,x) =

x2,

x3,

· · ·F (t, x1, · · · , xn)

.

22

Recall the norms of vectors and matrices:

x =

x1

· · ·xn

, ‖x‖ =

√√√√√n∑

j=1|xj|2,

A =

a11 · · · a1n

· · · · · · · · ·an1 · · · ann

, ‖A‖ =

√√√√√n∑

i,j=1|aij|2.

We have ‖x + y‖ ≤ ‖x‖+ ‖y‖ and

‖Ax‖ ≤ ‖A‖‖x‖.

To prove, let y = Ax, i.e. yi =∑n

j=1 aijxj, then

‖Ax‖ = ‖y‖ =

√√√√n∑

i=1|yi|2 =

√√√√√n∑

i=1|

n∑

j=1aijxj|2

≤√√√√√

n∑

i=1(

n∑

j=1|aij|2)(

n∑

k=1|xk|2) = ‖A‖‖x‖.

LetBδ(t0,x0) ≡ (t,x) ∈ R1+n :

√(t− t0)2 + ‖x− x0‖2 < δ,

Bδ(t0,x0) ≡ (t,x) ∈ R1+n :√

(t− t0)2 + ‖x− x0‖2 ≤ δ.

23

A vector-valued function f : G ⊂ R1+n → Rn is said to be differentiable

in x if every fj is differentiable in x1, · · · , xn. We write

Dxf(t,x) =

∂f1

∂x1· · · ∂f1

∂xn

. . . . . . . . .∂fn

∂x1· · · ∂fn

∂xn

.

Write

‖f(t,x)‖ =( n∑

j=1|fj(t,x)|2

)1/2,

‖Dxf(t,x)‖ =( n∑

i,j=1| ∂fi

∂xj(t,x)|2

)1/2.

24

Definition Let G be a subset in R1+n. f(t,x) : G → Rn is said to satisfy

the Lipschitz condition with respect to x in G if there exists a constant

L > 0 such that, for all (t,x), (t,y) ∈ G,

‖f(t,x)− f(t,y)‖ ≤ L‖x− y‖.

f is said to be locally Lipschitz with respect to x in G if for each point

(t0,x0) ∈ G there exist δ = δ(t0,x0) > 0 and L(t0,x0) > 0, such that

Bδ(t0,x0) ⊂ G, and for all (t,x), (t,y) ∈ Bδ(x0),

‖f(t,x)− f(t,y)‖ ≤ L(t0,x0)‖x− y‖.

Lemma Assume G is an open subset of R1+n and f : G → Rn. If f(t,x)

is continuous in G and differentiable with respect to x, and Dxf(t,x) is

continuous in G, then f is locally Lipschitz in G.

25

Consider the initial value problem

x = f(t,x), x(t0) = x0, 3.3.3

where x0 ∈ Rn. We can generalized the conclusions in §3.1, 3.2 to the

case of systems. Here we only state the Existence-Uniqueness Theorem for

systems.

Theorem 3.7 (local existence) Suppose:

(1) f(t,x) is continuous on the Ω = (t,x) : |t − t0| ≤ a, ‖x − x0‖ ≤ b

and hence there exists M > 0 such that ‖f(t,x)‖ ≤ M on Ω.

(2) f(t,x) satisfies a uniform Lipschitz condition with respect to x on

Ω, that is, there exists L > 0 such that

‖f(t,x)− f(t,y)| ≤ L‖x− y‖ for all

(t,x), (t,y) ∈ Ω.

Then, the initial value problem (3.3.3) has a unique solution on the interval

[t0 − h, t0 + h] where h = mina, b/M.

26

For higher order equations, we have similar conclusions. Consider the

following initial value problem

dny

dtn= F (t, y,

dy

dt, · · · ,

dn−1y

dtn−1 ),

y(t0) = b1,dy

dt(t0) = b2, · · · ,

dn−1y

dtn−1 (t0) = bn.

3.3.4

27

Theorem 3.8 Under the conditions of Theorem 3.7, suppose b = ∞,

then the initial value problem (3.3.3) has a unique solution x(t) on [t0 −a, t0 + a].

28

Example