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Chapter 2 Theory of First Order Differential Equations. Shurong Sun University of Jinan Semester 1, 2011-2012. 1 Existence-uniqueness theorem ( Picard’s Method of Successive Approximations ). Consider the initial-value problem. (1). - PowerPoint PPT Presentation
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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
y ydt
2( ) ,( , 0.)
0,
t c t cy c
t 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 ( , ( )) .
t
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( ) ( , ( )) .
t
ty t y f s y s ds
Then we iterate: define
02 0 1( ) ( , ( )) .
t
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( ) ( , ( )) .
t
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( ) ( , ( ))
t
n nty t y f s y s ds
But then 0 0( ) .y t yand( , )dy
f t ydt
00lim ( ) lim ( , ( ))
t
n ntn ny t y f s y s ds
00( ) ( , ( ))
t
ty t y f s y s ds
Remark: Was that a foolproof proof?
00lim ( ) lim ( , ( ))
t
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 ( , ( ))
t
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
Let
where
( , )min( , ) with max | ( , ) | .
t y A
bh a M f t y
M
• satisfies a Lipschitz condition with respect to y.
0 0( , )t y
0t a0t a
0y b
0y b
min{ , }b
h a aM
MM
0 0( , )t y0t a0t a
0y b
0y b
min{ , }b b
h aM M
MM
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 ( , ( )) .
t
ty y f s y s ds
(1)
(2)
(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
is equivalent to the integral equation
00( ) ( , ( )) .
t
ty t y f s y s ds (2)
0 0
( )( , ( )) .
t t
t t
dy sds f s y s ds
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).
then
00( ) ( , ( )) .
t
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( ) ( , ( )) .
t
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( ) ( , ( )) .
t
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
0
0 0 0
0 1 0
( ) ,| | ,
( ) ( , ( )) ,| | , 1.t
n nt
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
0
| ( , ( )) |t
jtf s y s ds
for 0 0 .t t t h
0 0
(| ( ) | | ( ) | )t t
t tf s ds f s ds
0 0
( ( ) ( ) ( ) | ( ) )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
n nty t y f s y s ds
0
| ( , ( )) |t
jtf s y s ds
b
.
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 .
if,
it suffices
11
| ( ) ( ) | .n nn
y t y t
Observe that
01 1 2| ( ) ( ) | | [ ( , ( )) ( , ( ))] |
t
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| ( , ( )) ( , ( )) |
t
n ntf s y s f s y s ds
01 2| ( ) ( ) | ,
t
n ntL y s y s ds
to show that
01 0 0 0| ( ) ( ) | | ( , ) | ( )
t
ty t y t f s y ds M t t
02 1 1 0| ( ) ( ) | | ( ) ( ) |
t
ty t y t L y s y s ds
20( )
,2
LM t t
00( )
t
tLM s t ds
This, in turn, implies that
03 2 2 1| ( ) ( ) | | ( ) ( ) |
t
ty t y t L y s y s ds
0
22 0( )
2
t
t
s tML ds
2 30( )
.3!
ML t t
Setting n=1 in (5) gives
Proceeding inductively, we see that1
01
( )| ( ) ( ) | ,
!
n n
n n
ML t ty t y t
n
0 0for .t t t h
Therefore, for 0 0 ,t t t h
1
1| ( ) ( ) | .!
n n
n n
ML hy t y t
n
Consequently, converges uniformly ( )ny t
on the interval 0 0 ,t t t h since1
.!
n nML h
n
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 ( , ( ))
t
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( ) ( , ( )) .
t
n nty t y f s y s ds
Taking limits of both sides of (6) gives
(6)
00( ) lim ( , ( )) .
t
ntny t y f s y s ds
equation
( , ( ))nf s y s converges uniformly
since
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.
Thus
0 0 0
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
10( )
| ( ) ( ) | .( 1)!
n n
n
ML t tz t y t
n
We have
00( ) ( ) ( , ( ))
t
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
(7)
(v) Show that the solution is unique.
If (7) holds for n − 1 we have
01| ( ) ( ) | | [ ( , ( )) ( , ( ))] |
t
n ntz t y t f s z s f s y s ds
and this proves1
0( )| ( ) ( ) | .
( 1)!
n n
n
ML t tz t y t
n
0
10( )
!
n nt
t
ML s tL ds
n
10( )
,( 1)!
n nML t t
n
01| ( ) ( ) |
t
ntL z s y s ds
for all .n N
Consequently,1
| ( ) ( ) | .( 1)!
n n
n
ML hz t y t
n
This implies that
1
lim 0,( 1)!
n n
n
ML h
n
since
1
0
.( 1)!
n n
n
ML h
n
( )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
k
We have
0
( ) ( ) [ ( , ( )) ( , ( ))]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
(7)
If (7) for k − 1 holds we have
0
| ( ) ( ) | | [ ( , ( )) ( , ( ))] |t
ty t z t f s y s f s z s ds
and this proves1
02 | || ( ) ( ) | .
( 1)!
k kML t ty t z t
k
0
102 ( )
!
k kt
t
ML s tL ds
k
102 ( )
,( 1)!
k kML t t
k
0
| ( ) ( ) |t
tL y s z s ds
for all .k N
Consequently,12
| ( ) ( ) | .( 1)!
k kML hy t z t
k
This implies that
12lim 0,
( 1)!
k k
k
ML h
k
since
1
0
2.
( 1)!
k k
k
ML h
k
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
0y b
0y b
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
0y b
0y b
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
y
is continuous in the rectangle A ,
then f satisfies a Lipschitz condition,
1 2 1 2 1 2
1 2
| ( , ) ( , ) | | ( ) | | |
for all ( , ), ( , ) ,
ff t y f t y y y L y y
y
t y t y A
where
( , )max .t y A
fL
y
Note that f satisfies a Lipschitz condition
f
y
is continuous.
For example: f(t,y)=|y|.
since
Corollary : Let f and f
y
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
M
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
dt
1 1,
2 2t | ( ) | 1.y t
Solution: Let 1
{( , ) :| | ,| | 1}.2
A t y t y
Computing 22
( , )
5max ,
4y
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( , ) ,
52 24
h
interval,
Remark (4) The initial value problem (IVP)
where
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
Proof
The unique solution is to the IVP
is
0 0( ) ( ), ( )dy
p t y q t y t ydt
0 0
0
( ) ( )
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
(i) Show that the initial value problem
0 0( ) ( ), ( )dy
p t y q t y t ydt
is equivalent to the integral equation
00( ) [ ( ) ( ) ( )] .
t
ty t y p s y s q s ds
(1)
(2)
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].
(ii) Construction of the approximating sequence
0
0 0
0 1
( ) ,
( ) [ ( ) ( ) ( )] , 1.t
n nt
y t y
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
Then
are well defined on [a,b] for all ( )ny t 0;n
and only if, the infinite series
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 10
1
( )| ( ) ( ) | , [ , ], 1.
! !
n n n n
n n
ML t t ML hy t y t t a b n
n n
Then
[ , ]max | ( ) | .x a b
L p x
(iv) Show that the limit of the sequence { ( )}ny t
is a continuous solution to the integral equation (2) on [a,b].
(v) Show that this solution is unique.
[ , ]max | ( ) ( ) ( ) | .x a b
M p x z s q x
Take1 1
0( ) ( )| ( ) ( ) | , 0.
( 1)! ( 1)!
n n n n
n
ML t t ML b az t y t n
n n
Then
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
Then the Cauchy problem
0 0( , ), ( )dy
f t y y t ydt (1)
has a solution on 0 0[ , ].t h t h
Let
( , )max | ( , ) | .t y A
M f t y
Set min( , ) .b
h aM
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( ) ( , ( )) , ( ) .
t
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
| ( ) ( ) | .( 1)!
n n
n
ML hy t y t
n
( ),ny t then the error
is called the nth approximate solution to the( )ny t
IVP.
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
dx
(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, } .
2 2
bh a
M
Thus, the interval of solution is 1 1
[ , ].2 2
2L
1| ( ) ( ) |( 1)!
nn
n
MLx x h
n
11 1
( ) 0.05( 1)! ( 1)!
nMLh
L n n
we can take
| | | 2 | 2,( , ) ,yf y x y R Since
Take then
1
( 1)!n
1 1 10.05
4! 24 20 .
3,n
The successive approximates are:
0 ( ) 0x 3
2 21 00( ) [ ( )]
3
x xx x x dx 3 7
2 22 10( ) [ ( )]
3 63
x x xx x x dx
3( )x is the approximate solution desired,
6 10 142 2 2
3 20
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
3
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
2 10( ) 1 ( ) 1 ,
2
x xy x y t dt x
100
( 1), ( ) 1 ( ) ,
!
i inx
n ni
xy x y t dt
i
Recalling Taylor’s series expansion of
100
( 1), ( ) 1 ( ) ,
!
i inx
n ni
xy x y t dt
i
,xe
The function
we see that
lim ( ) .xnn
y x e
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
(1)
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
0t
0y
O t
y
0 1t h 0 1t h
M
N
S0 1( )y t h
U
0 1 2t h 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
(1)
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
Then
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
(b)
max 0{ is an open interval, t , there exists a solution
of IVP (1) on }.
I I I
I
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 ( ).
t I I
y t
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
U
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.
2 1
2
dy y
dt
The general solution to the given DE is
Thus the solution passing through (0,0) is
So the maximal interval is
1
1
t
t
cey
ce
1
1
t
t
ey
e
.t
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
1
1
t
t
ey
e
0 .t
tO
y
1
1
(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
t ydt
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
y a
y a
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 (- , ).
U R R
f U R
y f t y
Remark:
0 0( , )t y
0 0( )y y M t t
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.
2dyy
dt
2. Prove that for any
The solution to the IVP
0 02 2
( 1), ( )
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
K U L
f t y f t x L x y
t x
( , ) .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
(1)
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
Mx y
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
and
Proof: Then
and
Therefore
or, equivalently,
0
( ) ( ) .t
tG t a b g s ds
0( ) ( ), [ , ]g t G t t t T
0( ) ( ), [ , ].G t bg t t t T
Set
00 0( ) ( ) 0, [ , ].btbte G t e G t t t T
0
( ) ( )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
Hence
or
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
0 0
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
0 0
0 0
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
0ta b
VK
1t
1y P
0y
1 1( , )t y
t
( )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
0
We choose a closed interval [ , ] ( , ) such
that [ , ].
a b I t y
t a b
0 0
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 t
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| - ( , , ) | | ( , , ) - ( , , ( , , )) |
t
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| - ( , , ) | | ( , , ) - ( , , ) |
t
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
y
Corollary
0 0(ii) ( , , ) is a continuous function on .y t t y G
0 0
0 0 0 0
0 0 0 0
For each ( , ) , let the maximal interval of the
solution ( , , ) to the IVP ( , ), ( )
be ( ( , ), ( , )).
t y U
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
a
b
0 0( , )t y
0 0( , )t y
U
t
0t
0yO
G
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
(i)
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 2
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
U R R
0 0
0 0
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
Theor
on
em:
( , , , ) to
f G R U R R
y t y G
y t t y
0 0 0 0 0 0
0 0
0 0 0 0 0 0
the
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
0 0
We turn next to the question whether supplementary
conditions on ( , ) such as differentiability do imply
that the solution ( , , ) is differentiable.
f t y
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
0
( , , )y t t y
y
we write the Cauchy problem as
0and differentiate formally with respect to .y
0 Exchanging the derivatives with respect to and
we find
t y
0
0 0 0
0
0 0 0
( , , )( , ( , , )), ( , , )
y t t yf t y t t y y t t y y
t
0 0 0
0 0
0
0
0 0 0
0 0
( , , ) ( , ( , , )) ( , , )
( , , )1
y t t y f t y t t y y t t y
t y y y
y t t y
y
This formal calculation shows that the function0 0
0
( , , )y t t y
y
is a solution of the linear equation
0 0
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.
00
0
( , ( , , )), ( ) 1
f t y t t yz z z t
y
0 0
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
y t t y y
0 0
0 0
( , ), ( )
is continuously differentiable with respect to , , .
f t y y t y
t t y
Theorem 3.2
Proof: Let
0 0 0 0
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
00
0 0 0 0( , , ) ( , ( , , )) ,t
ty t t y y f s y s t y ds
00
0 0 0 0( , , ) ( , ( , , )) .t
ty t t y y f s y s t y ds
0
( , ( ))( )t
t
f s dsy
Using the theorem of the mean, we have
where 0 1. Note that the continuity of , and ,
f
y
we have
1
( , ( )) ( , ),
f s f sr
y y
1 1where 0 as 0, and 0 0.r r as Thus, for 0, we have
0
( )1 ( , ( ))
t
t
f s dsy
This shows that is the solution of the IVP
1 0 0
( , )[ ] , ( ) 1 .
dz f tr z z t z
dt y
This shows that
is the solution
1 0 0
( , )[ ] , ( ) 1 .
dz f tr z z t z
dt y
From the theorem of continuous dependence, we get
is a continuous function of 0 0, , , ,t t z so the limit
0 0
00
( , , )lim
y t t y
y
exists, 0 0
0
( , , )y t t y
y
is the solution of the IVP
0
( , )[ ] , ( ) 1.
dz f tz z t
dt y
and
of the IVP
Clearly, it is a continuous function of
Hence
0
0 00 0
0
( , , )exp{ ( , ( , , ) }.
t
t
y t t yf s y s t y ds
y y
0 0, , .t t y
In the same manner, we can prove that 0 0
0
( , , )y t t y
t
Clearly, it is continuous function of
exists and is the solution of the IVP
0 0 0
( , )[ ] , ( ) ( , ).
dz f tz z t f t y
dt y
0 0, , .t t y
Hence
0
0 00 0 0 0
0
( , , )- ( , )exp{ ( , ( , , ) }.
t
t
y t t yf t y f s y s t y ds
y y
0 0
0 0
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
t
Exercise 3.3
1. Proof:
0 0 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
f
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
x
y
O0x 0x
0y0y
(1) (2)
1 0 1
0 0 0
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
0 0
0 0
Since the solution to y = f(x, y),y(x )
is continuous dependence on x ,y ,
y
1 2 20
0 0 0 10 0
00
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
2 20
00 0
0, 0, | |
| ( , , ) | .
such that y implies that
y x x y for all x x
0 0
0 0
Since the solution to y = f(x, y),y(x )
is continuous dependence on x ,y ,
y
1 0 2
0 0 0 0 0 1
00
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
1 0 1
0 0 0
0, 0, | |
| ( , , ) | ,
such that y implies that
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
x
y
O0x 0x
0y0y
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
3 23
, we have | | ( ) .2
dyy y x c
dx
So all solutions passing through (0,0) are
3
2
0,| | , ( 0).
( ) ,
x cy c
x c x c
5.
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 [ ( ) ( )] .
x
xy y P s y Q s ds
(1)
(2)
(ii) Set up a sequence of Picard iterates
(iii)
0
0 0
0 1
( ) ,
( ) [ ( ) ( ) ( )] , 1,2, .x
n nx
y x y
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
Then1
1| ( ) ( ) | ( ) .( 1)!
nn
n n
MLy x y x
n
(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
Then
and
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
or
or
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
0
0
| ( ) ( ) | | [ ( , ( )) ( , ( ))] |
| ( ) ( ) | .
t
t
t
t
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, .
t
tLds
y t z t y t z t e t t
0 ,If t t then
0
0 0
| ( ) ( ) | | [ ( , ( )) ( , ( ))] |
| | ( ) ( ) | | | ( ) ( ) | .
t
t
t t
t t
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
so
00 0 0| ( ) ( ) | | ( ) ( ) | 0, .
t
tL ds
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.
(1)
But this follows immediately, for
21| | ,k
k kx x N M
then1
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)
1 12
, ( )n nn
Hence x x x
This implies that { } .nx converges
(5) Let * *( )x f x and ( ),x f x
then* * *| | | ( ) ( ) | | |x x f x f x N x x
implies * ,x x since N<1.
* *1
* *
| ( ) ( ) | | | ( ),
( ).
n n n nFrom f x f x N x x and x f x
we have x f x
(4)
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
a b
M f x L K x
Then 1 0 0,b
ax x K x d , ( )
b
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
if 1
,L b a
then
lim ( ) ( ), ( ) [ , ].nset x x then x is continuous on a b
( , ) ( )nK x x converges uniformly
since
| ( , ) ( ) ( , ) ( ) | | ( ) ( ) |n nK x x K x x L x x
and ( )n x
[ , ],a bon
[ , ].a b
(iii) Note that
to
converges uniformly to
Then one can exchange integral and the limit.
( , ) ( )K x x
( )x on
Thus
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
then
0| ( ) ( ) | | ( , ) ( ) |b
ax x K x d
| | ( ).L N b a
where [ , ]
max .x a b
N x
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,
Then the Cauchy problem
0 0( , ), ( )dy
f t y y t ydt (1)
has a unique solution on0 0[ , ],t h t h
Let
where
( , )min( , ) with max | ( , ) | .
t y A
bh a M f t y
M
• satisfies a Lipschitz condition with respect to y.
Proof of the TheoremWe 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( ) ( , ( )) .
t
ty t y f s y s ds
(1)
(2)
Specifically, y(t) is a solution of (1) if, and only if, it is a continuous solution of (2).
(ii) Construction of the approximating sequence
0
0 0
0 1 0
( ) ,
( ) ( , ( )) ,| | , 1.t
n nt
y t y
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
and only if, the infinite series
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).
(v) Show that this solution is unique.
