Chapter 2 Theory of First Order Differential Equations Shurong Sun University of Jinan Semester 1,...

Chapter 2 Theory of First Order Differential

Equations

Shurong Sun

University of Jinan

Semester 1, 2011-2012

Consider the initial-value problem

0 0( , ), ( )dy

f t y y t ydt (1)

Q1. How to know that the IVP (1) actually has a solution if we can’t exhibit it?

1 Existence-uniqueness theorem (Picard’s Method of Successive Approximations)

2 2 , (0) 0;dy

t y ydt

Q2. How do we know that there is only one solution y(t) of (1)?

2 , (0) 0.dy

2( ) ,( , 0.)

t c t cy c

The short answer to the question, how do you know the equation has a solution, is ‘Picard iteration’.

Iteration means to do something over and over.

Picard’s Existence and Uniqueness Theorem is one of the most important theorems in ODE!

Why is Picard’s Theorem so important?

( ) 0f x ( )x g x 0take x 1 0( )x g x

1( )n nx g x If *, nx x n * ( *).x g x

00 ( , ( )) .

ty y f s y s ds

(ii) Pick a function 0( )y t

Pick 0 0( ) .y t y

Define 1( ) byy t

01 0 0( ) ( , ( )) .

ty t y f s y s ds

Then we iterate: define

02 0 1( ) ( , ( )) .

ty t y f s y s ds

0 0( ( , ), ( ) )dy

f t y y t ydt

satisfying the initial condition.

(i) Rewrite the differential equation as an integral equation:

and having defined 1( ), , ( )ny t y t define

01 0( ) ( , ( )) .

n nty t y f s y s ds

This gives a sequence of functions

1 2( ), ( ),y t y t

The solution is ( ) lim ( ),nny t y t

when Picard

Intuitively, is the solution since( )y t

iteration works.

Picard iteration

01 0( ) ( , ( ))

But then 0 0( ) .y t yand( , )dy

f t ydt

00lim ( ) lim ( , ( ))

n ntn ny t y f s y s ds

00( ) ( , ( ))

ty t y f s y s ds

Remark: Was that a foolproof proof?

00lim ( ) lim ( , ( ))

n ntn ny t y f s y s ds

Almost, but the formula

is called ‘taking the limit under the integral sign’ and it is not always correct.

We need to add some further conditions on f to make it work. But basically this is the idea.

00 lim ( , ( ))

nt ny f s y s ds

Theorem 1.1 (Existence and uniqueness for the Cauchy problem)

and suppose that

0 0{( , ) :| | ,| | }A t y t t a y y b

:f A R• is continuous,

Then the Cauchy problem

0 0( , ), ( )dy

f t y y t ydt (1)

has a unique solution on0 0[ , ],t h t h

( , )min( , ) with max | ( , ) | .

bh a M f t y

• satisfies a Lipschitz condition with respect to y.

0 0( , )t y

0t a0t a

min{ , }b

h a aM

0 0( , )t y0t a0t a

min{ , }b b

h aM M

Proof of the Theorem

We will prove the theorem in five steps:

(i) Show that the initial value problem

0 0( , ), ( )dy

f t y y t ydt

is equivalent to the integral equation

00 ( , ( )) .

ty y f s y s ds

(ii) Set up a sequence of Picard iteration and show that they are well defined.

(iii) Show that our sequence of Picard iteration converges uniformly.

(iv) Show that the function to which our sequence converges is a solution to the IVP.

(v) Show that this solution is unique.

(i) The initial-value problem

0 0( , ), ( )dy

f t y y t ydt (1)

Specifically, if y(t) satisfies (1), then

00( ) ( , ( )) .

ty t y f s y s ds (2)

( )( , ( )) .

dy sds f s y s ds

Hence 0

0( ) ( , ( )) .t

ty t y f s y s ds (2)

Conversely, if y(t) is continuous and satisfies (2),

( , ).dy

f t ydt

Moreover, 0 0( ) .y t y

Therefore, y(t) is a solution of (1) if, and only if, it is a continuous solution of (2).

00( ) ( , ( )) .

ty t y f s y s ds

(ii) Construction of the approximating sequence

( ).ny tLet us start by guessing a solution 0( )y t of (2).

The simplest possible guess is 0 0( ) .y t y

To check whether 0( )y t is a solution,

01 0 0( ) ( , ( )) .

ty t y f s y s ds

we compute

If 1 0( ) ( ),y t y t then 0( ) ( )y t y t is indeed a solution

of (2). 1( )y t as our next guess.If not, then we try

To check whether 1( )y t is a solution, we compute

02 0 1( ) ( , ( )) .

ty t y f s y s ds

and so on.

In this manner, we define a sequence of functions

1 2( ), ( ), ,y t y t

where 0

1 0( ) ( , ( )) .t

n nty t y f s y s ds (3)

These functions ( )ny t

or Picard iteration,

are called successive approximation,

who first discovered them.

after the French mathematician Picard

Define the sequence of Picard iteration as

( ) ,| | ,

( ) ( , ( )) ,| | , 1.t

y t y t t h

y t y f s y s ds t t h n

To ensure that Picard iteration are well defined it is sufficient to ensure that each ( )ny t that is

0 0| ( ) | ,| | .ny t y b t t h (4)

stays within [-b,b],

0;t t0.t t

It’s convenient to assume a similar proof will hold for

We prove (4) by induction on n.

Observe first that (4) is obviously true for n=0, since

0 0( ) .y t yNext, we must show that (4) is true for n=j+1 if it is true for n=j.

But this follows immediately, for

0| ( ) | ,jy t y b then

1 0| ( ) |jy t y

| ( , ( )) |t

jtf s y s ds

for 0 0 .t t t h

(| ( ) | | ( ) | )t t

t tf s ds f s ds

( ( ) ( ) ( ) | ( ) )t t

t tf t g t f s ds g s ds

0( )M t t Mh

0 0 0| ( ) | , .ny t y b t t t h

01 0( ) ( , ( )) .

| ( , ( )) |t

jtf s y s ds

Clearly, the sequence

0 0[ , ]t t h

(iii) We now show that our sequence of Picard iteration converges uniformly for t in the interval

This is accomplished by writing ( )ny t in the form

0 1 0 1( ) ( ) [ ( ) ( )] [ ( ) ( )]n n ny t y t y t y t y t y t ( )ny t converges

and only if, the infinite series

1 0 1[ ( ) ( )] [ ( ) ( )]n ny t y t y t y t

converges .

it suffices

| ( ) ( ) | .n nn

y t y t

Observe that

01 1 2| ( ) ( ) | | [ ( , ( )) ( , ( ))] |

n n n nty t y t f s y s f s y s ds

0 0 .t t t h (5)

To prove the infinite series converges,

01 2| ( , ( )) ( , ( )) |

n ntf s y s f s y s ds

01 2| ( ) ( ) | ,

n ntL y s y s ds

to show that

01 0 0 0| ( ) ( ) | | ( , ) | ( )

ty t y t f s y ds M t t

02 1 1 0| ( ) ( ) | | ( ) ( ) |

ty t y t L y s y s ds

LM t t

tLM s t ds

This, in turn, implies that

03 2 2 1| ( ) ( ) | | ( ) ( ) |

ty t y t L y s y s ds

22 0( )

s tML ds

2 30( )

ML t t

Setting n=1 in (5) gives

Proceeding inductively, we see that1

( )| ( ) ( ) | ,

ML t ty t y t

0 0for .t t t h

Therefore, for 0 0 ,t t t h

1| ( ) ( ) | .!

ML hy t y t

Consequently, converges uniformly ( )ny t

on the interval 0 0 ,t t t h since1

n nML h

We denote the limit of the sequence ( )ny t

(iv) Prove that y(t) satisfies the initial-value problem (1)

We will show that y(t) satisfies the integral equation

0 0[ , ].t t h

00 ( , ( ))

ty y f s y s ds

which is continuous on

by y(t),

To this end, recall that the Picard iteration ( )ny t

are defined recursively through the

01 0( ) ( , ( )) .

Taking limits of both sides of (6) gives

00( ) lim ( , ( )) .

ntny t y f s y s ds

equation

( , ( ))nf s y s converges uniformly

0 0| ( , ( )) ( , ( )) | | ( ) ( ) |, .n nf s y s f s y s L y s y s t s t h and ( )ny t

0 0[ , ],t t hon

0 0[ , ].t t h

Note that to f(s,y(s))

converges uniformly to y(t) on

Then one can exchange integral and the limit.

lim ( , ( )) lim ( , ( )) ( , ( )) ,t t t

n nt t tn nf s y s ds f s y s ds f s y s ds

and y(t) satisfies0

0 ( , ( )) .t

ty y f s y s ds

(i) Let z(t) be also a solution of (2). By induction we show that

| ( ) ( ) | .( 1)!

ML t tz t y t

We have

00( ) ( ) ( , ( ))

tz t y t f s z s ds

and therefore

which (7) holds for n = 0.

0 0| ( ) ( ) | ( )z t y t M t t

(v) Show that the solution is unique.

If (7) holds for n − 1 we have

01| ( ) ( ) | | [ ( , ( )) ( , ( ))] |

n ntz t y t f s z s f s y s ds

and this proves1

0( )| ( ) ( ) | .

ML t tz t y t

ML s tL ds

,( 1)!

n nML t t

01| ( ) ( ) |

ntL z s y s ds

for all .n N

Consequently,1

| ( ) ( ) | .( 1)!

ML hz t y t

This implies that

lim 0,( 1)!

.( 1)!

( )ny t converges uniformly

to z(t) on 0 0 .t t t h

And so 0 0( ) ( ), .y t z t t t t h

(v) (ii) It remains to prove uniqueness of the solution. Let y(t) and z(t) be two solutionsof (2). By recurrence we show that

102 ( )

| ( ) ( ) | .( 1)!

k kML t ty t z t

We have

( ) ( ) [ ( , ( )) ( , ( ))]t

ty t z t f s y s f s z s ds

and therefore

which (7) for k = 0.

0| ( ) ( ) | 2 ( )y t z t M t t

If (7) for k − 1 holds we have

| ( ) ( ) | | [ ( , ( )) ( , ( ))] |t

ty t z t f s y s f s z s ds

and this proves1

02 | || ( ) ( ) | .

k kML t ty t z t

102 ( )

ML s tL ds

102 ( )

,( 1)!

k kML t t

| ( ) ( ) |t

tL y s z s ds

for all .k N

Consequently,12

| ( ) ( ) | .( 1)!

k kML hy t z t

This implies that

12lim 0,

0 0( ) ( ), .y t z t t t t h

Remark: (1) Geometrically, this means the graph

of ( )y t

region shown below.

is constrained to lie in the shaded

0 0( , )t y 0t a0t a

min{ , }b

h a aM

0 0( )y y M t t 0 0( )y y M t t

0 0( , )t y0t a0t a

min{ , }b b

h aM M

0 0( )y y M t t 0 0( )y y M t t

Remark: (2) Geometrically, this means the graph

is constrained to lie in the shaded( )ny tof

region shown below.

(3) If f

is continuous in the rectangle A ,

then f satisfies a Lipschitz condition,

1 2 1 2 1 2

| ( , ) ( , ) | | ( ) | | |

for all ( , ), ( , ) ,

ff t y f t y y y L y y

t y t y A

( , )max .t y A

Note that f satisfies a Lipschitz condition

is continuous.

For example: f(t,y)=|y|.

Corollary : Let f and f

be continuous in the

rectangle0 0{( , ) :| | ,| | }.A t y t t a y y b

Compute ( , )max | ( , ) | and set min( , ) .t y A

bM f t y h a

Then the initial problem

0 0( , ), ( )dy

f t y y t ydt

has a unique solution y(t) on the interval 0 0[ , ].t h t h

Show that the solution y(t) of the initial-value problem

exists for and in this interval,

22 , (0) 0ydyt e y

2 2t | ( ) | 1.y t

Solution: Let 1

{( , ) :| | ,| | 1}.2

A t y t y

Computing 22

5max ,

t y AM t e

we see that y(t) exists for 1| | ,

2t h and in this

| ( ) | 1.y t

Example 1.1

1 1 1min( , ) ,

interval,

Remark (4) The initial value problem (IVP)

has a unique solution valid on [a, b].

0 0, [ , ], [ , ]p q C a b t a b and y R

0 0( ) ( ), ( ) ,dy

p t y q t y t ydt

The unique solution is to the IVP

0 0( ) ( ), ( )dy

p t y q t y t ydt

( ) ( )

0( ( ) ),t u

x xP s ds P s dst

xy e q u e du y

[ , ].t a b

(Method I)

Method II Successive approximation

0 0( ) ( ), ( )dy

p t y q t y t ydt

00( ) [ ( ) ( ) ( )] .

ty t y p s y s q s ds

Specifically, y(t) is a solution of (1) on [a,b] if, and only if, it is a continuous solution of (2) on [a,b].

( ) [ ( ) ( ) ( )] , 1.t

y t y p s y s q s ds n

(iii) Show that the sequence of Picard iteration

converges uniformly on the interval

{ ( )}ny t

[ , ];a b

are well defined on [a,b] for all ( )ny t 0;n

1 0 1[ ( ) ( )] [ ( ) ( )]n ny t y t y t y t

{ ( )}ny t converges if,

converges .

Take 0[ , ]

max | ( ) ( ) |,x a b

M p x y q x

( )| ( ) ( ) | , [ , ], 1.

n n n n

ML t t ML hy t y t t a b n

[ , ]max | ( ) | .x a b

(iv) Show that the limit of the sequence { ( )}ny t

is a continuous solution to the integral equation (2) on [a,b].

[ , ]max | ( ) ( ) ( ) | .x a b

M p x z s q x

Take1 1

0( ) ( )| ( ) ( ) | , 0.

( 1)! ( 1)!

n n n n

ML t t ML b az t y t n

Theorem 1.2 (Peano 1890)

and suppose that

0 0{( , ) :| | ,| | }A t y t t a y y b

:f A R is continuous with

0 0( , ), ( )dy

f t y y t ydt (1)

has a solution on 0 0[ , ].t h t h

( , )max | ( , ) | .t y A

M f t y

Set min( , ) .b

The method of Picard iteration not only give the solution but also found functions which approximate it well, i.e. Let y(t) be the exact solution of the IVP

Estimates of error and approximate calculation

00 1 0 0( ) ( , ( )) , ( ) .

n nty t y f s y s ds y t y

0 0( , ), ( )dy

f t y y t ydt

and the Picard scheme for solving this is

If we make in approximating the solution y(t) by

| ( ) ( ) | .( 1)!

ML hy t y t

( ),ny t then the error

is called the nth approximate solution to the( )ny t

Example1.2 Let

(i) Determine the interval of existence that the Existence Theorem predicts for the solution y(x) of the initial-value problem

2 2dyx y

(ii) Determine an approximate solution of the IVP above such that the error is less than 0.05.

(0) 0y

2 2dyx y

dx , ( , ) :[ 1,1] [ 1,1].x y R

Solution: Since , ( ),yf f C R the existence and

From the Existence Theorem, 2 2dyx y

dx (0) 0y

has a unique solution on [-h,h].

Here ( , )max | ( , ) | 2,x y R

M f x y

uniqueness theorem holds.

1 1min{ , } min{1, } .

Thus, the interval of solution is 1 1

[ , ].2 2

1| ( ) ( ) |( 1)!

MLx x h

( ) 0.05( 1)! ( 1)!

we can take

| | | 2 | 2,( , ) ,yf y x y R Since

Take then

( 1)!n

1 1 10.05

4! 24 20 .

The successive approximates are:

0 ( ) 0x 3

2 21 00( ) [ ( )]

x xx x x dx 3 7

2 22 10( ) [ ( )]

x x xx x x dx

3( )x is the approximate solution desired,

6 10 142 2 2

2( ) [ ( )] ( )

9 189 3969

x x x xx x x dx x dx

3 7 11 152

3 63 2079 59535

x x x x

1 1| ( ) ( ) | 0.05, [ , ].

2 2x x x

Example 1.3 The initial value problem

, (0) 1y y y is equivalent to solving the integral equation

01 ( ) .

xy y t dt

Let 0( ) 1.y x We obtain

1 00( ) 1 ( ) 1 ,

xy x y t dt x

2 10( ) 1 ( ) 1 ,

x xy x y t dt x

( 1), ( ) 1 ( ) ,

xy x y t dt

Recalling Taylor’s series expansion of

( 1), ( ) 1 ( ) ,

xy x y t dt

The function

we see that

lim ( ) .xnn

xy e the given initial value problem in R.

is indeed the solution of

2. Continuation of solutions

So far we only considered local solutions,

0 0( , ).t y

To extend the solution we solve the Cauchy problem locally, say from

0 0tot t hcontinue the solution by solving the Cauchy problem

( , )dy

f t ydt

with new initial condition 0( ).y t h

and then we can try to

which are defined in a neighborhood of

i.e. solutions

0 0( , ( ))t h y t h

0 0( , )t y .

0t h 0t h

In order to do this we should be able to solve it locally everywhere and we will assume that f satisfy a local Lipschitz condition.

Definition A function f(t,y) (where U is an open set of R×R) satisfies a local Lipschitz condition

0 0( , )t y U V U

Remark: If the function f is of class 1C

if for anythere exist a neighborhood

f satisfies a Lipschitz condition on V.

such that

satisfies a local Lipschitz condition.

in U, then it

Theorem 2.1 (Theorem on local solvability)

If U is open and ( )f C U

Lipschitz condition in U,

satisfies a local

0 0( , ) ;t y U i.e., there is a neighborhood I of

0t such that exactly one solution exists in I.

problem (1) is locally uniquely solvable for

then the initial value

0 0( , )P t yQ

0 1t h 0 1t h

S0 1( )y t h

0 1 2t h h

0 1 0 1MQ : ( ),y y t t h t t h

0 1 2 0 1 2NS : ( ),z z t t h h t t h h

0 1 0 1

0 1 0 1 2

( ),MS : ( )

y t t h t t ht

z t t h t t h h

0 1 0 1

0 1 2 0 1

( ) ( ),

y t h z t h

z t y t t h h t t h

Consider the differential equation

( , )dy

f t ydt (1)

If y(t) is a solution of (1) defined on an interval I, we say that z(t) is a continuation or extension of y(t) if z(t) is itself a solution of (1) defined on an interval J which properly contains I and z restricted to I equals y .

A solution is non-continuable or saturated if no such extension exists; i.e., I is the maximal interval on which a solution to (1) exists.

Lemma 2.1Let U R R :f U R

0 0( , )t y U

max 0( , ), withI t

such that (i) the Cauchy problem

0 0( , ), ( )dy

f t y y t ydt

Lipschitz condition. be an open set and assume that

is continuous and satisfies a localThen for any there exists an open interval

has a unique solution y on max .I

(2) If :z I R is a solution of (1), then max ,I I

and | .Iz y

Proof: (a) Let : , :y I R z J R

of the Cauchy problem with 0 , .t I Jy(t) = z(t) .t I J

be two solutions

Suppose it is not true, 1 such thatt

1 1( ) ( ).y t z tthere is point

Consider the first point where the solutions separate.

The local existence theorem 2.1 shows that it is impossible.

0 0( , )t y

max 0{ is an open interval, t , there exists a solution

of IVP (1) on }.

This interval is open and we can define the solution onmax as follows:I

maxif , then there exists where the Cauchy

problem has a solution and we can define ( ).

maxThe part (a) shows that ( ) is uniquely defined on .y t I

Define

Theorem 2.2

:f U R

( , )dy

f t ydt can be extended to

Let U R R be an open set and assume that is continuous and satisfies a local Lipschitz condition.Then every solution of

the left and to the right up to the boundary of U.

right in an interval 0t t b ( is allowed),band one of the following cases applies:(i) ;b the solution exists for all

0 .t t(ii) and lim sup | ( ) | ;

t bb t

the solution “becomes infinite”.(iii) and , (( , ( )), ) 0, .b t t U t b

Remark: solution can be extended to the right

up to the boundary of U means that

exists to the

where (( , ), )t y U denotes the distance from the point (t,y) to the boundary of U.

If U is a bounded region, then (iii) is the only case.

exists to the

0a t t ( is allowed),a and one of the following cases applies:

(i) ;a the solution exists for all 0 .t t

(ii) lim sup | ( ) | ;t a

a and t

the solution “becomes infinite”.(iii) lim (( , ( ), ) 0.

t aa and t t U

can be extended to the left up to

the boundary of means that

solution

left in an interval

If U is a bounded region, then (iii) is the only case.

Determine the maximal intervals on which the solutions to the differential equation

passing through (0,0) and (ln2,-3) exist.

Example 1

Solution: The function2 1

( , )2

yf t y

is defined on

the whole ty-plane and satisfies the conditions of existence Theorem and continuation theorem.

The general solution to the given DE is

Thus the solution passing through (0,0) is

So the maximal interval is

The solution passing through (ln2,-3) is

Hence, the maximal interval of this solution is

The domain of this solution is ( ,0) (0, ).

Note that 0 ln 2 and , as 0 .y t

(ln 2, 3)

Determine the maximal intervals on which the solutions to the IVP exists.

Example 2

Solution: The function1 ln t is defined on

the right-half-plane t>0 and the existence Theorem and continuation theorem hold.

1 ln , (1) 0dy

The solution to the IVP is

The solution is well-defined, continuous on t>0, and

So the maximal interval is

ln 0 0.y t t as t (0, ).

ln .y t t

Example 3： Prove that for any 0 0and | |t R y a

The solution to the IVP2 2

0 0( ) ( , ), ( )dy

y a f t y y t ydt

exists on 2( , ), where , ( ).yf f C R

Proof: The function

the whole ty-plane and satisfies the conditions of existence Theorem and continuation theorem.

2 2( ) ( , )y a f t y is defined on

It is easy to see that y aDE.

are solutions to the given

From the continuation theorem, the solution to 2 2

0 0( ) ( , ), ( )dy

y a f t y y t ydt

the solution can not pass through .y aSo it can only be extended to the left and to the right

0 0( , )t yt

infinitely.

moves away from the origin, but from existence theorem,

Let and let us assume that

: is continuous, bounded, and

satisfies a local Lipschitz condition.

Then every solution of ( , )

can be extended to (- , ).

y f t y

Remark:

0 0( , )t y

0 0( )y y M t t

Homework:

1. Determine the maximal intervals on which the solutions to the differential equation

passing through (1,1) and (3,-1) exist.

2. Prove that for any

The solution to the IVP

0 02 2

( 1), ( )

dy y yy t y

dt t y

exists on ( , ).

0 00 1t R and y

3 Dependence on initial conditions and parameters

The first-order equation y y has the solution0

0 0 0( ) through the point ( , ).t tt y e t y

(1) Continuous dependence on initial conditions

We first investigate the continuous dependence of the solution 0 0 0 0( , , ) on ( , ).y t t y t y

Consider the IVP

0 0( , ), ( ) .dy

f t y y t ydt

Let : ( an open set of ) be

continuous and satisfy a local Lipschitz

condition. Then for any bounded closed

region there exists 0 such that

| ( , ) - ( , ) | | - |, for all

( , ),

f U R U R R

f t y f t x L x y

( , ) .t y K

Lemma 3.1

Proof:

( , ) and ( , ) such thatn n n nt x t y

| ( , ) ( , ) | | | .n n n n n nf t x f t y n x y

By Bolzano-Weierstrass, the sequence ( , )n nt y

accumulation point (t, y). By assumption the function

Since f is bounded on K with( , )max | ( , ) |t y K

M f t y

has an

Let us assume the contrary. Then there

exist sequences

f(t, y) satisfies a Lipschitz condition in a neighborhood V of (t, y).

it follows from (1) that 2| | .n n

exists infinitely many indices n such that( , ) , ( , )n n n nt x t y V

Then (1) contradicts the Lipschitz condition on V .

Therefore there

Lemma 3.2 (Gronwall inequality) Suppose that g(t) is a continuous function with ( ) 0g t

Then we have0

0( ) ( ) , [ , ].t

tg t a b g s ds t t T

0( )0( ) , [ , ].b t tg t ae t t T

that there exits constants 0, 0a b such that

Proof: Then

Therefore

or, equivalently,

( ) ( ) .t

tG t a b g s ds

0( ) ( ), [ , ]g t G t t t T

0( ) ( ), [ , ].G t bg t t t T

00 0( ) ( ) 0, [ , ].btbte G t e G t t t T

( ) ( )t

tg t a b g s ds

0( ) ( ), [ , ]G t bG t t t T

0( ) ( ), [ , ]bt bte G t be G t t t T

that is0[ ( )] 0, [ , ].bte G t t t T

which implies that

0 0( ) ( )0 0( ) ( ) , [ , ]b t t b t tG t G t e ae t t T

0( )0( ) , [ , ].b t tg t ae t t T

Assume that : ( an open set of ) is

continuous and satisfy a local Lipschitz condition.

f U R U R R Theorem 3.1

00 0 0 0 0

0, a neighborhood ( , , ) of ( , ),s.t.

ˆ ˆ ˆ| ( , , ) ( , , ) | , [ , ], ( , ) .

V a b t y

y t t y y t t y t a b t y V

Let the maximal interval of the solution ( , , )

to the Cauchy problem ( , ), ( )

be ( , ),

y t t y

y f t y y t y

then for any , : , a b a b

0 0( , )t y

1 1 0 0( , ( , , ))P t y t t y0 0 1 1 0 0( , , ) ( , , ( , , ))y t t y y t t y t t y

0 0( , )t y

1 1( , )t y

( )1 0 0 0 0 1 0 0[| - | | ( , , ) - ( , , ) |]L b ae y y y t t y y t t y

1| - |1 1 0 0 1 1 0 0| ( , , ) - ( , , ) | | - ( , , ) |L t ty t t y y t t y e y y t t y

0 0 0 0( , , )y y t t y

Proof:

max 0 0

We choose a closed interval [ , ] ( , ) such

that [ , ].

a b I t y

We choose small enough such that the stripregion

of ( , , )K y t t y

0 0 {( , ) : [ , ], | ( , , ) | },K t y t a b y y t t y

is contained in the open set U.

By Lemma 3.1, f(t, y) satisfies a Lipschitz condition on K with a Lipschitz constant L.

implies- ( - )

1 1 1 0 1 0Set {( , ) : | | , | | / 2} .L b aV t y t t y y e

0 0Then ( , ) .t y V K U

1 1If ( , ) we havet y V1 1 0 0 1 1 1 1 0 0| ( , , ) - ( , , ) | | ( , , ) - ( , , ( , , )) |y t t y y t t y y t t y y t t y t t y

1 11 1 1 1 0 0 1 1 0 0| ( , ( , , )) -[ ( , , ) ( , ( , , ( , , )) ] |

t ty f s y s t y ds y t t y f s y s t y t t y ds

11 1 0 0 1 1 1 1 0 0| - ( , , ) | | ( , , ) - ( , , ( , , )) |

ty y t t y L y s t y y s t y t t y ds

From Gronwall inequality we conclude that1| - |

1 1 0 0 1 1 0 0| ( , , ) - ( , , ) | | - ( , , ) |L t ty t t y y t t y e y y t t y

and this concludes the proof.

11 1 0 0 1 1 0 0| - ( , , ) | | ( , , ) - ( , , ) |

ty y t t y L y s t y y s t y ds

Since 0 0( , , )y t t y is continuous at 0 ,t t 00,s.t. | - |t t ( )

0 0 0 0 0| ( , , ) - ( , , ) | / 2. L b ay t t y y t t y e

( )1 0 0 0 0 1 0 0 [ | | | ( , , ) ( , , ) |]L b ae y y y t t y y t t y

Let : ( an open set of ) be continuous

and satisfy a local Lipschitz condition with respect to .

f U R U R R

Corollary

0 0(ii) ( , , ) is a continuous function on .y t t y G

0 0 0 0

For each ( , ) , let the maximal interval of the

solution ( , , ) to the IVP ( , ), ( )

be ( ( , ), ( , )).

y t t y y f t y y t y

t y t y

0 0 0 0 0 0Then (i) G: ( , ) ( , ), ( , ) is a region;t y t t y t y U

0 0( , )P t y

0 0( , )t y

0 0( , , )Q t t y

Proof:

0 0 0 0 0 0 0 0( , , ) , i.e., ( , ) ( , ), ( , )t t y G t y t t y t y U

Choose a,b, such that

0 0 0 0( , ) ( , )t y a t b t y

From Theorem 3.1, 1 1( , , ) is defined on ,y t t y a t b

It follows that 0 0( , , )t t y is an interior point of G.This means that G is an open set.

1 1 0 0whenever ( , ) sufficiently close to ( , ).t y t y

From the connectivity of U, it is easy to see G is a connected set, and so G is a region.

0 0 0 0 2

Again ( , , ) is continuous at , so 0, . .

| ( , , ) ( , , ) | ,whenever | | .2

y t t y t s t

y t t y y t t y t t

(ii) From Theorem 3.1, we have that

1 1 1 0 00, 0, . . | ( , , ) ( , , ) | .2

s t y t t y y t t y

1 1 0 0

1 1 0 0 0 0 0 0

1 0 1 0

| ( , , ) ( , , ) |

| ( , , ) ( , , ) | | ( , , ) ( , , ) |

, whenever | | ,| | ,| | .

y t t y y t t y

y t t y y t t y y t t y y t t y

t t t t y y

Setting 1 2min{ , }, we have

The proof is complete.

2. Continuous Dependence On Parameters

Consider the IVP

0 0( , , ), ( ) ,dy

f t y y t ydt

where is parameter, ( , , ) ( , ),

an open set of .

t y G U

Let : ( an open set of )

be continuous and satisfy a local Lipschitz condition

with respect to . For each ( , , ) , let the

maximal interval of the soluti

( , , , ) to

f G R U R R

y t y G

y t t y

0 0 0 0 0 0

IVP ( , , ), ( ) is ( ( , , ), ( , , )).

Then ( , , , ) is a continuous function on

: ( , , ) ( , , ), ( , , ) .

y f t y y t y t y t y

y t t y

t y t t y t y G

0 0(2) Smooth dependence with respect to , and .t t y

We turn next to the question whether supplementary

conditions on ( , ) such as differentiability do imply

that the solution ( , , ) is differentiable.

y t t y

0We considerfirst the differentiability with respect to .x

In order to obtain an idea of the form of the derivative0 0

( , , )y t t y

we write the Cauchy problem as

0and differentiate formally with respect to .y

0 Exchanging the derivatives with respect to and

we find

( , , )( , ( , , )), ( , , )

y t t yf t y t t y y t t y y

( , , ) ( , ( , , )) ( , , )

( , , )1

y t t y f t y t t y y t t y

t y y y

y t t y

This formal calculation shows that the function0 0

( , , )y t t y

is a solution of the linear equation

This equation is called the variational equation for

the Cauchy problem ( , ), ( ) . y f t y y t y

All the following proof does is to justify this procedure.

( , ( , , )), ( ) 1

f t y t t yz z z t

Let : ( an open set of ) be continuous

and satisfy a local Lipschitz condition. Assume that

( , ) exists and is continuous on . Then the solution

( , , ) of the Cauchy problem

f U R R U R R

f t yU

y t t y y

( , ), ( )

is continuously differentiable with respect to , , .

f t y y t y

Theorem 3.2

Proof: Let

0 0 0 0

( , , ) and ( , , ) (| | ,

is small enough) be the solutions to the Cauchy problem

( , ), ( ) and ( , ), ( ) ,

respectively.

y t t y y t t y

y f t y y t y y f t y y t y

0 0 0 0The integral equations for ( , , ), ( , , ) arey t t y y t t y

0 0 0 0( , , ) ( , ( , , )) ,t

ty t t y y f s y s t y ds

0 0 0 0( , , ) ( , ( , , )) .t

ty t t y y f s y s t y ds

( , ( ))( )t

f s dsy

Using the theorem of the mean, we have

where 0 1. Note that the continuity of , and ,

we have

( , ( )) ( , ),

f s f sr

1 1where 0 as 0, and 0 0.r r as Thus, for 0, we have

( )1 ( , ( ))

f s dsy

This shows that is the solution of the IVP

( , )[ ] , ( ) 1 .

dz f tr z z t z

This shows that

is the solution

( , )[ ] , ( ) 1 .

dz f tr z z t z

From the theorem of continuous dependence, we get

is a continuous function of 0 0, , , ,t t z so the limit

( , , )lim

y t t y

exists, 0 0

( , , )y t t y

is the solution of the IVP

( , )[ ] , ( ) 1.

dz f tz z t

of the IVP

Clearly, it is a continuous function of

0 00 0

( , , )exp{ ( , ( , , ) }.

y t t yf s y s t y ds

0 0, , .t t y

In the same manner, we can prove that 0 0

( , , )y t t y

Clearly, it is continuous function of

exists and is the solution of the IVP

( , )[ ] , ( ) ( , ).

dz f tz z t f t y

0 0, , .t t y

0 00 0 0 0

( , , )- ( , )exp{ ( , ( , , ) }.

y t t yf t y f s y s t y ds

Noting that ( , , ) is the solutionof the Cauchy problem

( , ), ( ) ,

y t t x

y f t y y t y

is continuous function of 0 0, , .t t y

0 0( , , )it follows that

y t t y

Exercise 3.3

1. Proof:

0 0 0 0

Since is continuous and satisfy a local Lipschitz condition

on , so ( , ) , the solution ( , , ) to the

Cauchy problem ( , ), ( )

exists, and is unique and continuous depend

G x y G y x x y

y f x y y x y

0 0ence on , .x y

Note that ( ,0) 0, so 0 is the solution to ( , ).f x y y f t y

O0x 0x

(1) (2)

0, 0, | |

| ( , , ) | ,

such that y implies that

y x x y for all x x

Assume that

0 00 0 000 , ( , , ).and x x set y y x x y

Since the solution to y = f(x, y),y(x )

is continuous dependence on x ,y ,

1 2 20

0 0 0 10 0

0, 0, | |

| ( , , ) ( , ,0) | | ( , , ) |

so for the such that y implies that

y x x y y x x y x x y

for all x x x

00 0 10, | ( , , ) | | | .In particularly y x x y y

Hence, 0 0 0| ( , , ) | ,y x x y whenever x x

Thus 00 0 0| ( , , ) | .y x x y whenever x x x

(2) (1) Assume that

0, 0, | |

| ( , , ) | .

y x x y for all x x

Since the solution to y = f(x, y),y(x )

is continuous dependence on x ,y ,

0 0 0 0 0 1

0, 0, | |

| ( , , ) ( , ,0) | | ( , , ) |

so such that y implies that

y x x y y x x y x x y

for all x x x

0, 0, | |

| ( , , ) | ,

y x x y for all x x

Note that the solution to 0 0 ( , ), ( ) y f t y y t y

is unique, so

O0x 0x

Exercise 3.1

4. 1, are continuous in 0,

2yf f y

So conditions of the existence theorem are satisfied.

It is easy to see that 0y is a solution to the

equation passing through (0,0).

Solving for 1 3

, we have | | ( ) .2

dyy y x c

So all solutions passing through (0,0) are

0,| | , ( 0).

x cy c

x c x c

Proposition 1: y=y(x) is the solution to the initial value problem

0 0( ) ( ), ( )dy

P x y Q x y x ydx

is equivalent to y=y(x) is a continuous solution to the integral equation

00 [ ( ) ( )] .

xy y P s y Q s ds

(ii) Set up a sequence of Picard iterates

( ) [ ( ) ( ) ( )] , 1,2, .x

y x y P s y s Q s ds n

( ) [ , ] .ny x is well defined and continuous on

( )ny x ( ), [ , ].y x x

0[ , ] [ , ]

max | ( ) ( ) |, max | ( ) | .x x

Set M P x y Q x L P x

1| ( ) ( ) | ( ) .( 1)!

MLy x y x

(iv) y(x) is a continuous solution to the integral equation.

(v) uniqueness.

[ , ]max | ( ) ( ) ( ) | .x

M P x z x Q x

6. Proof: Set

Therefore

or, equivalently,

( ) ( ) ( ) .t

G t K f s g s ds

( ) ( ), [ , ]f t G t t

( ) ( ) ( ).G t f t g t

( ) ( ) ( ) ( ) ( )( ), [ , ]

( ) ( ) ( )

G t f t g t g t G tg t t

G t G t G t

which implies that

ln ( ) ( ), [ , ]d

G t g t tdt

ln ( ) ln ( ) ( ) , [ , ]t

G t G g s ds t

( ) ( )( ) ( ) , [ , ]

t tg s ds g s ds

G t G e Ke t

( )( ) , [ , ].

tg s ds

g t Ke t

Let y(t) and z(t) are both solutions to the given IVP, then

| ( ) ( ) | | [ ( , ( )) ( , ( ))] |

| ( ) ( ) | .

y t z t f s y s f s z s ds

L y s z s ds

From Gronwall inequality, we get

00 0 0| ( ) ( ) | | ( ) ( ) | 0, .

y t z t y t z t e t t

0 ,If t t then

| ( ) ( ) | | [ ( , ( )) ( , ( ))] |

| | ( ) ( ) | | | ( ) ( ) | .

y t z t f s y s f s z s ds

L y s z s ds L y s z s ds

00 0 0| ( ) ( ) | | ( ) ( ) | 0, .

y t z t y t z t e t t

7. Assume that the contrary, then there exist two different solutions

1 2 1 2 0 1

1 1 2 1

( ), ( ) ( ) ( ),

( ) ( ).

y x y x with x x x x x

and x x

1 2 0 1( ) ( ) ( ), ( ) 0, ( ) 0,set y x x x then y x y x

This is contrary to

1 2 0 1( , ( )) ( , ( )) 0, .dy

f x x f x x x x xdx

9. (1) For any fixed 0 1, ( ), 0,1,2,n nx set x f x n

(2)1 1

1 1 0| | | | 1.n nn nx x N x x N M for all n

We show inductively that

2 1 1 0 1 0| | | ( ) ( ) | | | .x x f x f x N x x NM

Observe first that (1) is obviously true for n=1, since

Next, we must show that (1) is true for n=k+1 if it is true for n=k.

But this follows immediately, for

21| | ,k

k kx x N M

1 1 1| | | ( ) ( ) | | | .kk k k k k kx x f x f x N x x N M

10 1, nSince N so N M (3)

, ( )n nn

Hence x x x

This implies that { } .nx converges

(5) Let * *( )x f x and ( ),x f x

implies * ,x x since N<1.

| ( ) ( ) | | | ( ),

n n n nFrom f x f x N x x and x f x

we have x f x

10. (i) Define the successive approximations

0( ) ( ),x f x and

1( ) ( ) ( , ) ( ) , 1.b

n nax f x K x d n

By induction on n, we can show that ( )n x

is well defined and continuous on [a,b], since

( ) [ , ] ( , ) ([ , ] [ , ]).f x C a b and K x C a b a b

Clearly, the sequence

[ , ].a b

(ii) We now show that our successive approximations converges uniformly for t on the interval

This is accomplished by writing ( )n xin the form

0 1 0 1( ) ( ) [ ( ) ( )] [ ( ) ( )]n n nx x x x x x

( )n x converges

if, and only if, the infinite series

1 0 1[ ( ) ( )] [ ( ) ( )]n nx x x x

converges .

Set [ , ] [ , ]

max , max ,x a b x a b

M f x L K x

Then 1 0 0,b

ax x K x d , ( )

aK x f d ML b a

2 1 1 0,b

ax x K x d 2 2 2

1 0, ( )b

aK x d ML b a

Thus, in turn, implies that

Proceeding inductively, we have that

n 1 ( ) [ , ].n n n

nx x ML b a for x a b

Consequently,

converges uniformly

( )n x

on the interval [ , ],a b

,L b a

lim ( ) ( ), ( ) [ , ].nset x x then x is continuous on a b

( , ) ( )nK x x converges uniformly

| ( , ) ( ) ( , ) ( ) | | ( ) ( ) |n nK x x K x x L x x

and ( )n x

[ , ],a bon

[ , ].a b

(iii) Note that

converges uniformly to

Then one can exchange integral and the limit.

( , ) ( )K x x

( )x on

lim ( , ) ( ) ( , ) ( ) ,b b

na anK x d K x d

and ( ) ( ) ( , ) ( ) .b

ax f x K x d

(iV) If ( ) ( ) ( , ) ( ) ([ , ]).b

ax f x K x d C a b

0| ( ) ( ) | | ( , ) ( ) |b

ax x K x d

| | ( ).L N b a

where [ , ]

max .x a b

Hence, ( ) ( ), [ , ].x x x a b

1 0| ( ) ( ) | | ( , )( ( ) ( )) |b

ax x K x d 2 2 2| | ( ) .L N b a

Proceeding inductively, we have that1 1 1| ( ) ( ) | | | ( )n n n

nx x L N b a

which implies that converges uniformly1

[ , ], | | .( )

a b whenL b a

( )n xon the interval

Theorem 1.1 (Existence and uniqueness for the Cauchy problem)

and suppose that

0 0{( , ) :| | ,| | }A t y t t a y y b

:f A R• is continuous,

0 0( , ), ( )dy

f t y y t ydt (1)

has a unique solution on0 0[ , ],t h t h

( , )min( , ) with max | ( , ) | .

bh a M f t y

• satisfies a Lipschitz condition with respect to y.

Proof of the TheoremWe will prove the theorem in five steps:

0 0( , ), ( )dy

f t y y t ydt

00( ) ( , ( )) .

ty t y f s y s ds

Specifically, y(t) is a solution of (1) if, and only if, it is a continuous solution of (2).

( ) ( , ( )) ,| | , 1.t

y t y f s y s ds t t h n

(iii) Show that the sequence of Picard iteration

converges uniformly on the interval

{ ( )}ny t

0 0[ , ];t h t h

and show that are well defined for all ( )ny t 0;n

1 0 1[ ( ) ( )] [ ( ) ( )]n ny t y t y t y t

{ ( )}ny t converges if,

converges .

(iv) Show that the limit of the sequence { ( )}ny t

is a continuous solution to the integral equation (2).

Chapter 2 Theory of First Order Differential Equations Shurong Sun University of Jinan Semester 1,...

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