Transcript The existence-uniqueness theorem

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
dy
 f ( t , y ), y( t0 )  y0
(1)
dt
2

(
t

c
)
,t  c
dy
dy
, c  0.)
 t 2  y 2 , y(0)  0;
 2 y , y(0)  0. ( y  
dt
dt
 0, t  c
Q1. How to know that the IVP (1) actually has a solution if
we can’t exhibit it?
Q2. How do we know that there is only one solution y(t) of
(1)?
The short answer to the question, how do you know
the equation has a solution, is ‘Picard iteration’.
Picard’s Existence and Uniqueness Theorem is one of the
most important theorems in ODE!
Why is Picard’s Theorem so important?
Iteration means to do something over and over.
f ( x )  0  x  g( x ) take x0 x1  g( x0 )
xn  g( xn1 ) If xn  x*, n    x*  g( x*).
(i) Rewrite the differential equation as an integral
equation:
t
dy
y  y0  f ( s, y( s))ds. (  f ( t , y ), y( t0 )  y0 )
t0
dt
(ii) Pick a function y0 ( t ) satisfying the initial condition.

Pick y0 (t )  y0 .
Define y1 (t ) by
y1 (t )  y0   f ( s, y0 ( s))ds.
t
t0
Then we iterate: define
y2 (t )  y0   f ( s, y1 ( s))ds.
t
t0
and having defined y1 (t ),
, yn (t ) define
yn1 (t )  y0   f ( s, yn ( s))ds.
t
t0
This gives a sequence of functions
y1 (t ), y2 (t ),
Picard iteration
The solution is y( t )  lim yn ( t ), when Picard
n 
iteration works.
Intuitively, y( t ) is the solution since
yn1 (t )  y0   f ( s, yn ( s))ds
t
t0
 lim yn (t )  y0  lim  f ( s, yn ( s))ds
n
t
n t0
 y(t )  y0   f ( s, y( s))ds
t
t0
dy
But then
 f ( t , y ) and y(t0 )  y0 .
dt
Remark: Was that a foolproof proof?
Almost, but the formula
lim yn (t )  y0  lim  f ( s, yn ( s))ds
t
n
n t0
t
 y0   lim f ( s, yn ( s))ds
t0 n
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.
Theorem 1.1 (Existence and uniqueness for the
Cauchy problem)
Let A  {(t , y) :| t  t0 | a,| y  y0 | b}
and suppose that f : A  R
• is continuous,
• satisfies a Lipschitz condition with respect to y.
Then the Cauchy problem
dy
 f ( t , y ), y( t0 )  y0
(1)
dt
has a unique solution on [t0  h, t0  h], where
b
h  min(a , ) with M  max | f ( t , y ) | .
( t , y ) A
M
y0  b
M
t0  a
( t 0 , y0 )
t0  a
y0  b
b
h  min{a , }  a
M
M
y0  b
M
t0  a
( t 0 , y0 )
y0  b
b
b
h  min{a , } 
M
M
M
t0  a
Proof of the Theorem
We will prove the theorem in five steps:
(i) Show that the initial value problem
dy
 f ( t , y ), y( t0 )  y0
dt
(1)
is equivalent to the integral equation
y  y0   f ( s, y( s))ds.
t
t0
(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
dy
 f ( t , y ), y( t0 )  y0
(1)
dt
is equivalent to the integral equation
y(t )  y0   f ( s, y( s))ds.
t
(2)
t0
Specifically, if y(t) satisfies (1), then
t
dy( s )
t0 ds ds  t0 f ( s, y( s ))ds.
t
Hence y(t )  y0 

t
t0
f ( s, y( s))ds.
(2)
Conversely, if y(t) is continuous and satisfies (2),
dy
 f ( t , y ).
then
dt
Moreover, y(t0 )  y0 .
Therefore, y(t) is a solution of (1) if, and only
if, it is a continuous solution of (2).
y(t )  y0   f ( s, y( s))ds.
t
t0
(ii) Construction of the approximating sequence
yn ( t ).
Let us start by guessing a solution y0 ( t ) of (2).
The simplest possible guess is y0 (t )  y0 .
To check whether y0 ( t ) is a solution, we compute
y1 (t )  y0   f ( s, y0 ( s))ds.
t
t0
If y1 (t )  y0 (t ), then y(t )  y0 (t ) is indeed a solution
of (2). If not, then we try y1 ( t ) as our next guess.
To check whether y1 ( t ) is a solution, we compute
y2 (t )  y0   f ( s, y1 ( s))ds.
t
and so on.
t0
In this manner, we define a sequence of functions
y1 (t ), y2 (t ),
yn1 ( t )  y0   f ( s, yn ( s ))ds.
t
where
,
(3)
t0
These functions yn ( t ) are called successive approximation,
or Picard iteration, after the French mathematician Picard
who first discovered them.
Define the sequence of Picard iteration as
y0 (t )  y0 ,| t  t0 | h,
yn (t )  y0   f ( s, yn1 ( s ))ds,| t  t0 | h, n  1.
t
t0
To ensure that Picard iteration are well defined it is sufficient
to ensure that each yn ( t ) stays within [-b,b], that is
| yn (t )  y0 | b,| t  t0 | h.
(4)
It’s convenient to assume t  t0 ; a similar proof will hold for
t  t0 .
We prove (4) by induction on n.
Observe first that (4) is obviously true for n=0, since
t
y0 (t )  y0 .
yn1 ( t )  y0   f ( s, yn ( s ))ds.
t0
Next, we must show that (4) is true for n=j+1 if it is
true for n=j.
| yn (t )  y0 | b, t0  t  t0  h.
t
t
But this follows immediately, for
(|  f ( s )ds |  | f ( s ) | ds )
t
t
| y j (t )  y0 | b, then
.
0
0
| y j 1 (t )  y0 | |  f ( s, y j ( s ))ds |
t
t0
  | f ( s, y j ( s)) | ds  M (t  t0 )  Mh  b
t0
t
for t0  t  t0  h.
( f (t )  g(t )  f ( s)ds |
t

t0

t
t0
g( s)ds)
(iii) We now show that our sequence of
Picard iteration converges uniformly for t
in the interval [t0 , t0  h]
This is accomplished by writing yn ( t ) in the form
yn (t )  y0 (t )  [ y1 (t )  y0 (t )]   [ yn (t )  yn1 (t )]
Clearly, the sequence yn ( t ) converges if,
and only if, the infinite series
[ y1 (t )  y0 (t )] 
converges .
 [ yn (t )  yn1 (t )] 
To prove the infinite series converges, it suffices

to show that
 | y (t )  y
n 1
n
n 1
(t ) | .
Observe that
| y1 (t )  y0 (t ) |  | f ( s, y0 ) | ds  M (t  t0 )
t
t0
| yn (t )  yn1 (t ) ||  [ f ( s, yn1 ( s))  f ( s, yn 2 ( s))]ds |
t
t0
  | f ( s, yn1 ( s))  f ( s, yn 2 ( s)) | ds
t
t0
 L | yn1 ( s)  yn 2 ( s) | ds, t0  t  t0  h. (5)
t
t0
Setting n=1 in (5) gives
| y2 (t )  y1 (t ) | L | y1 ( s)  y0 ( s) | ds
t0
t
LM ( t  t0 )2
 LM  ( s  t0 )ds 
,
t0
2
t
This, in turn, implies that
| y3 (t )  y2 (t ) | L | y2 ( s)  y1 ( s) | ds
t
t0
2
2
3
(
s

t
)
ML
(
t

t
)
0
0
 ML2 
ds 
.
t0
2
3!
t
Proceeding inductively, we see that
MLn1 ( t  t0 )n
| yn ( t )  yn1 ( t ) |
, for t0  t  t0  h.
n!
Therefore, for t0  t  t0  h,
MLn1hn
| yn ( t )  yn1 ( t ) |
.
n!
Consequently, yn ( t ) converges uniformly
on the interval t0  t  t0  h,
MLn1hn
since  n !  .
We denote the limit of the sequence yn ( t ) by y(t),
which is continuous on [t0 , t0  h].
(iv) Prove that y(t) satisfies the initial-value problem (1)
We will show that y(t) satisfies the integral equation
y  y0   f ( s, y( s))ds
t0
To this end, recall that the Picard iteration yn ( t )
are defined recursively through the equation
t
yn1 (t )  y0   f ( s, yn ( s))ds.
t0
Taking limits of both sides of (6) gives
t
y(t )  y0  lim  f ( s, yn ( s))ds.
t
n t0
(6)
Note that f ( s, yn ( s)) converges uniformly to f(s,y(s))
on [t0 , t0  h], since
| f ( s, yn ( s))  f ( s, y( s)) | L | yn ( s)  y( s) |, t0  s  t0  h.
and yn ( t ) converges uniformly to y(t) on [t0 , t0  h].
Then one can exchange integral and the limit.
Thus
t
t
t
lim  f ( s, yn ( s))ds   lim f ( s, yn ( s))ds   f ( s, y( s))ds,
n t0
t0 n
t0
and y(t) satisfies y  y0   f ( s, y( s ))ds.
t
t0
(v) Show that the solution is unique.
(i) Let z(t) be also a solution of (2). By induction
we show that
MLn (t  t0 )n1
(7)
| z(t )  yn (t ) |
.
(n  1)!
We have
t
z(t )  y0 (t )   f ( s, z( s))ds
t0
and therefore | z(t )  y0 (t ) | M (t  t0 )
which (7) holds for n = 0.
If (7) holds for n − 1 we have
| z(t )  yn (t ) ||  [ f ( s, z( s))  f ( s, yn1 ( s))]ds |
t
t0
 L | z( s)  yn1 ( s) | ds  Lt
t
t
t0
0
MLn1 ( s  t0 )n
ds
n!
MLn ( t  t0 )n1

,
( n  1)!
MLn (t  t0 )n1
and this proves | z(t )  yn (t ) |
.
(n  1)!
for all n  N .
n
n 1
ML h
Consequently, | z( t )  yn ( t ) |
.
(n  1)!
n n1
n n 1

ML
h
ML h
 .
lim
 0, since 
n ( n  1)!
n  0 ( n  1)!
This implies that yn ( t ) converges uniformly
to z(t) on t0  t  t0  h.
And so y(t )  z(t ), t0  t  t0  h.
(v) (ii) It remains to prove uniqueness of the
solution. Let y(t) and z(t) be two solutions
of (2). By recurrence we show that
2 MLk (t  t0 )k 1
| y( t )  z( t ) |
.
( k  1)!
(7)
We have
y(t )  z(t )   [ f ( s, y( s))  f ( s, z( s))]ds
t
t0
and therefore | y(t )  z(t ) | 2 M (t  t0 )
which (7) for k = 0.
If (7) for k − 1 holds we have
| y(t )  z(t ) ||  [ f ( s, y( s))  f ( s, z( s))]ds |
t0
k 1
k
t 2 ML
t
( s  t0 )
ds
 L | y( s)  z( s) | ds  Lt
t0
0
k!
k
k 1
2 ML ( t  t0 )

,
( k  1)!
k
k 1
t
2 ML | t  t0 |
and this proves | y(t )  z(t ) |
( k  1)!
for all k  N .
.
2 MLk hk 1
Consequently, | y( t )  z( t ) |
.
( k  1)!
k k 1

k k 1
2
ML
h
2 ML h
 .
lim
 0, since 
k  ( k  1)!
k  0 ( k  1)!
This implies that y(t )  z(t ), t0  t  t0  h.
Remark: (1) Geometrically, this means the graph
of y( t ) is constrained to lie in the shaded
region shown below.
y  y0  M (t  t0 )
t0  a
y0  b
( t 0 , y0 )
y0  b
b
h  min{a , }  a
M
y  y0  M ( t  t0 )
t0  a
y  y0  M ( t  t0 )
t0  a
y0  b
( t 0 , y0 )
y0  b
b
b
h  min{a , } 
M
M
y  y0  M ( t  t0 )
t0  a
Remark: (2) Geometrically, this means the graph
of yn ( t ) is constrained to lie in the shaded
region shown below.
(3) If
f
y
is continuous in the rectangle A ,
then f satisfies a Lipschitz condition, since
f
| f ( t , y1 )  f ( t , y2 ) || ( y1  y2 ) | L | y1  y2 |
y
f
.
for all ( t , y1 ),( t , y2 )  A, where L  (max
t , y ) A
y
Note that f satisfies a Lipschitz condition

f
 y is continuous.
For example: f(t,y)=|y|.
f
Corollary : Let f and
be continuous in the
y
rectangle A  {(t , y) :| t  t0 | a,| y  y0 | b}.
b
Compute M  max | f ( t , y ) | and set h  min(a , ) .
( t , y ) A
M
Then the initial problem
dy
 f ( t , y ), y( t0 )  y0
dt
has a unique solution y(t) on the interval [t0  h, t0  h].
Example 1.1
Show that the solution y(t) of the initial-value problem
dy
2
 y2
 t  e , y(0)  0
dt
1
1
exists for   t  , and in this interval, | y(t ) | 1.
2
2
1
Solution: Let A  {(t , y ) :| t | ,| y | 1}.
2
5 h  min( 1 , 1 )  1 ,
2
y
Computing M  max t  e  ,
2 5
2
( t , y ) A
4
4
2
we see that y(t) exists for
interval, | y(t ) | 1.
| t | h 
1
,
2
and in this
Remark (4) The initial value problem (IVP)
dy
 p( t ) y  q(t ), y(t0 )  y0 ,
dt
where p, q  C[a, b], t0  [a, b] and y0  R
has a unique solution valid on [a, b].
Proof
(Method I)
The unique solution is to the IVP
dy
 p(t ) y  q(t ), y(t0 )  y0
dt
is
ye

t
 x0 P ( s )ds
t  [a, b].
u
(  q( u)e
t
x0
 x0 P ( s )ds
du  y0 ),
Method II Successive approximation
(i) Show that the initial value problem
dy
 p( t ) y  q( t ), y( t 0 )  y0
dt
(1)
is equivalent to the integral equation
y(t )  y0   [ p( s) y( s)  q( s)]ds. (2)
t
t0
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
y0 (t )  y0 ,
yn (t )  y0   [ p( s ) yn1 ( s )  q( s )]ds, n  1.
t
t0
Then yn ( t ) are well defined on [a,b] for all n  0;
(iii) Show that the sequence of Picard iteration { yn (t )}
converges uniformly on the interval [a , b];
{ yn ( t )} converges if, and only if, the infinite series
[ y1 (t )  y0 (t )] 
converges .
 [ yn (t )  yn1 (t )] 
| p( x ) y0  q( x ) |, L  max | p( x ) | .
Take M  max
x[ a ,b ]
x[ a , b ]
MLn1 ( t  t0 )n MLn1hn

, t  [a , b], n  1.
Then | yn (t )  yn1 (t ) |
n!
n!
(iv) Show that the limit of the sequence { yn (t )}
is a continuous solution to the integral equation (2)
on [a,b].
(v) Show that this solution is unique.
| p( x ) z( s )  q( x ) | .
Take M  max
x[ a ,b ]
Then
MLn (t  t0 )n1 MLn (b  a )n1
| z(t )  yn (t ) |

, n  0.
(n  1)!
( n  1)!
Theorem 1.2 (Peano 1890) Let
A  {(t , y) :| t  t0 | a,| y  y0 | b}
and suppose that f : A  R
| f (t , y ) | .
is continuous with M  (max
t , y ) A
b
h  min(a , ) .
M
Set
Then the Cauchy problem
dy
 f ( t , y ), y( t0 )  y0
dt
has a solution on [t0  h, t0  h].
(1)
Estimates of error and approximate calculation
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
dy
 f ( t , y ), y( t0 )  y0
dt
and the Picard scheme for solving this is
yn (t )  y0   f ( s, yn1 ( s))ds, y0 (t )  y0 .
t
t0
If we make in approximating the solution y(t) by
yn (t ), then the error
n
n 1
ML h
| yn ( t )  y( t ) |
.
( n  1)!
yn ( t ) is called the nth approximate solution to the
IVP.
Example1.2 Let
dy
 x 2  y 2 ,( x, y )  R :[1,1]  [1,1].
dx
(i) Determine the interval of existence that the Existence
Theorem predicts for the solution y(x) of the initial-value
problem
dy
 x 2  y 2 y(0)  0
dx
(ii) Determine an approximate solution of the IVP above
such that the error is less than 0.05.
Solution: Since f , f y  C( R), the existence and
uniqueness theorem holds.
From the Existence Theorem,
dy
 x 2  y 2 y(0)  0
dx
has a unique solution on [-h,h].
Here M  max | f ( x, y ) | 2,
( x , y )R
b
1
1
h  min{a , }  min{1, }  .
M
2
2
1 1
Thus, the interval of solution is [ , ].
2 2
Since | f y || 2 y | 2,( x, y)  R,
we can take L  2
MLn n1
|  n ( x)   ( x) |
h
(n  1)!
M 1
1
n 1

( Lh) 
 0.05
L (n  1)!
(n  1)!
Take n  3, then
1
1
1
1



 0.05 .
(n  1)!
4! 24 20
The successive approximates are:
0 ( x)  0
3
x
1 ( x)   [ x 2   02 ( x)]dx 
0
3
3
7
x
x
x
 2 ( x)   [ x 2  12 ( x)]dx  
0
3 63
x
x
3 ( x)   [ x 2   22 ( x)]dx  
0
6
10
14
x
2
x
x
( x2 


)dx
9 189 3969
x3 x 7
2 x11
x15




3 63 2079 59535
3 ( x) is the approximate solution desired,
1 1
|  3 ( x )   ( x ) | 0.05, x  [  , ].
2 2
Example 1.3 The initial value problem
y   y, y(0)  1
is equivalent to solving the integral equation
y  1   y(t )dt .
x
0
Let
y0 ( x )  1.
We obtain
y1 ( x )  1   y0 (t )dt  1  x,
x
0
x
2
x
y2 ( x )  1   y1 (t )dt  1  x  ,
0
2
i i
n
x
( 1) x
, yn ( x )  1   yn1 ( t )dt  
,
0
i!
i 0
( 1)i x i
, yn ( x )  1   yn1 ( t )dt  
,
0
i!
i 0
x
Recalling Taylor’s series expansion of e , we see that
n
x
lim yn ( x )  e  x .
n
The function
ye
x
is indeed the solution of
the given initial value problem in R.
2. Continuation of solutions
So far we only considered local solutions, i.e. solutions
which are defined in a neighborhood of (t0 , y0 ).
To extend the solution we solve the Cauchy problem
locally, say from t0 to t0  h and then we can try to
continue the solution by solving the Cauchy problem
dy
 f (t , y )
dt
with new initial condition y(t0  h).
(t0  h, y(t0  h))
(1)
( t0 , y0 )
.
t0  h
t0  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 if for any
(t0 , y0 )  U there exist a neighborhood V  U such that
f satisfies a Lipschitz condition on V.
Remark: If the function f is of class C 1 in U, then it
satisfies a local Lipschitz condition.
Theorem 2.1 (Theorem on local solvability)
If U is open and f  C (U ) satisfies a local
Lipschitz condition in U, then the initial value
problem (1) is locally uniquely solvable for
(t0 , y0 )  U ; i.e., there is a neighborhood I of
t 0 such that exactly one solution exists in I.
y
MQ : y  y(t ), t0  h1  t  t0  h1
U
NS : z  z(t ), t0  h1  h2  t  t0  h1  h2
y(t0  h1 )
Q
P (t0 , y0 )
N
y0
S
y(t0  h1 )  z(t0  h1 ),
z(t )  y(t ), t0  h1  h2  t  t0  h1
 y( t ), t0  h1  t  t0  h1
MS :  ( t )  
 z( t ), t0  h1  t  t0  h1  h2
M
t0  h1  h2
O
t0  h1
t 0 t0  h1
t0  h1  h2
t
Consider the differential equation
dy
 f (t , y )
dt
(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 2.1
Let U  R  R be an open set and assume that f : U  R
is continuous and satisfies a local Lipschitz condition.
Then for any (t0 , y0 )  U there exists an open interval
Imax  ( ,  ), with      t0    
such that (i) the Cauchy problem
dy
 f ( t , y ), y( t0 )  y0
dt
(1)
has a unique solution y on Imax .
(2) If z : I  R is a solution of (1), then I  Imax ,
and z  y |I .
Proof: (a) Let y : I  R, z : J  R be two solutions
of the Cauchy problem with t0  I , J . Then
y(t) = z(t) t  I J .
Suppose it is not true, there is point t1 such that
y(t1 )  z(t1 ).
Consider the first point where the solutions separate.
The local existence theorem 2.1 shows that it is
impossible.
( t0 , y0 )
(b) Define
Imax  { I is an open interval, t 0  I , there exists a solution
of IVP (1) on I }.
This interval is open and we can define the solution on
Imax as follows:
if t  Imax , then there exists I where the Cauchy
problem has a solution and we can define y(t ).
The part (a) shows that y(t ) is uniquely defined on Imax .
Theorem 2.2
Let U  R  R be an open set and assume that f : U  R
is continuous and satisfies a local Lipschitz condition.
dy
Then every solution of  f ( t , y ) can be extended to
dt
the left and to the right up to the boundary of U.
Remark: solution  can be extended to the right
up to the boundary of U means that  exists to the
right in an interval t0  t  b (b   is allowed),
and one of the following cases applies:
(i) b  ; the solution exists for all t  t0 .
(ii) b   and lim sup |  ( t ) | ;
t b
the solution “becomes infinite”.
(iii) b   and ,  ((t ,  (t )), U )  0, t  b .
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.
solution  can be extended to the left up to
the boundary of U means that  exists to the
left in an interval a  t  t0 (a   is allowed),
and one of the following cases applies:
(i) a  ; the solution exists for all t  t0 .
(ii) a   and lim sup |  ( t ) | ;
t a
the solution “becomes infinite”.
(iii) a   and lim  (( t ,  ( t ), U )  0.
t a
If U is a bounded region, then (iii) is the only case.
Example 1
Determine the maximal intervals on which the
solutions to the differential equation
dy y 2  1

dt
2
passing through (0,0) and (ln2,-3) exist.
y2  1
Solution: The function f ( t , y ) 
is defined on
2
the whole ty-plane and satisfies the conditions
of existence Theorem and continuation theorem.
The general solution to the given DE is
1  ce t
y
1  ce t
Thus the solution passing through (0,0) is
1 e
y
t
1 e
t
So the maximal interval is
  t   .
The solution passing through (ln2,-3) is
1 e
y
t
1 e
t
The domain of this solution is (, 0)  (0, ).
Note that

0  ln 2 and y  , as t  0 .
Hence, the maximal interval of this solution is
0  t  .
y
1
t
O
1
(ln 2, 3)
Example 2
Determine the maximal intervals on which the solutions
dy
to the IVP  1  ln t , y(1)  0 exists.
dt
Solution: The function 1  ln t is defined on
the right-half-plane t>0 and the existence Theorem
and continuation theorem hold.
The solution to the IVP is y  t ln t .
The solution is well-defined, continuous on t>0, and
y  t ln t  0 as t  0.
So the maximal interval is (0, ).
Example 3： Prove that for any t  R and | y | a
0
0
The solution to the IVP
dy
 ( y 2  a 2 ) f ( t , y ), y( t0 )  y0
dt
exists on (, ), where f , f y  C( R2 ).
Proof: The function ( y2  a2 ) f (t , y) is defined on
the whole ty-plane and satisfies the conditions of
existence Theorem and continuation theorem.
It is easy to see that y   a are solutions to the given
DE.
From the continuation theorem, the solution to
dy
 ( y 2  a 2 ) f ( t , y ), y( t0 )  y0
dt
moves away from the origin, but from existence theorem,
the solution can not pass through y   a .
So it can only be extended to the left and to the right
infinitely.
ya
t
( t0 , y0 )
y  a
Remark:
Let U  R  R and let us assume that
f : U  R is continuous, bounded, and
satisfies a local Lipschitz condition.
Then every solution of y  f ( t , y )
can be extended to (-,  ).
y  y0  M (t  t0 )
( t0 , y0 )
y  y0  M (t  t0 )
Homework:
1. Determine the maximal intervals on which
the solutions to the differential equation
dy
2
y
dt
passing through (1,1) and (3,-1) exist.
2. Prove that for any t0  R and 0  y0  1
The solution to the IVP
dy
y( y  1)

, y(t0 )  y0
2
2
dt 1  t  y
exists on
( ,  ).
3 Dependence on initial conditions and parameters
The first-order equation y  y has the solution
t  t0
 (t )  y0e through the point (t0 , y0 ).
(1) Continuous dependence on initial conditions
Consider the IVP
dy
 f ( t , y ), y( t 0 )  y0 .
dt
We first investigate the continuous dependence
of the solution y(t , t0 , y0 ) on (t0 , y0 ).
Lemma 3.1
Let f : U  R (U an open set of R  R ) be
continuous and satisfy a local Lipschitz
condition. Then for any bounded closed
region K  U there exists L  0 such that
| f ( t , y ) - f ( t , x ) | L | x - y |, for all
( t , x ), ( t , y )  K .
Proof: Let us assume the contrary. Then there
exist sequences (tn , xn ) and (tn , yn ) such that
| f (tn , xn )  f (tn , yn ) | n | xn  yn | .
(1)
By Bolzano-Weierstrass, the sequence (tn , yn ) has an
accumulation point (t, y). By assumption the function
f(t, y) satisfies a Lipschitz condition in a neighborhood
| f (t , y) |
V of (t, y). Since f is bounded on K with M  (max
t , y )K
2M
it follows from (1) that | xn  yn |
. Therefore there
n
exists infinitely many indices n such that
(tn , xn ) , (tn , yn ) V
Then (1) contradicts the Lipschitz condition on V .
Lemma 3.2 (Gronwall inequality) Suppose that
g(t) is a continuous function with g( t )  0 and
that there exits constants a  0, b  0 such that
g(t )  a  b g( s)ds, t  [t0 , T ].
t
t0
Then we have
g(t )  ae
b ( t  t0 )
, t [t0 , T ].
Proof: Set G(t )  a  bt g( s)ds. Then
g(t )  G(t ), t [t0 , T ]
t
and
g(t )  a  b g( s )ds
t
0
G(t )  bg(t ), t  [t0 , T ].
Therefore G(t )  bG(t ), t [t0 , T ]
t0
or, equivalently, e bt G(t )  be bt G(t ), t [t0 , T ]
that is [e bt G(t )]  0, t [t0 , T ].
Hence e bt G(t )  e bt G(t0 )  0, t [t0 , T ].
or G(t )  G(t0 )eb( t t )  aeb( t t ) , t [t0 , T ]
0
0
0
which implies that
b ( t  t0 )
g(t )  ae
, t [t0 , T ].
Theorem 3.1
Assume that f : U  R (U an open set of R  R) is
continuous and satisfy a local Lipschitz condition.
Let the maximal interval of the solution y( t , t0 , y0 )
to the Cauchy problem y  f ( t , y ), y( t0 )  y0
be (  ,   ), then for any a, b:  a  b   ,
  0, a neighborhood V ( , a , b) of ( t 0 , y0 ),s.t.
| y( t , tˆ0 , yˆ 0 )  y( t , t 0 , y0 ) |  , t  [a , b], ( t 0 , yˆ 0 )  V .
(t0 , y0 )


P(t1 , y(t1 , t0 , y0 )) y(t , t0 , y0 )  y(t , t1 , y(t1 , t0 , y0 ))
y0  y(t0 , t0 , y0 )
y1
y0
K
V
( t1 , y1 )
( t0 , y0 )
a

P
t 0 t1
t
b
|y(t , t1 , y1 ) - y(t , t0 , y0 ) | e L|t -t1| | y1 - y( t1 , t0 , y0 ) |
 e L( ba ) [| y1 - y0 |  | y(t0 , t0 , y0 ) - y(t1 , t0 , y0 ) |]
Proof:
We choose a closed interval [a, b]  Imax ( t0 , y0 ) such
that t0  [a, b].
We choose  small enough such that the stripregion
K of y(t , t0 , y0 )
K  {(t , y) : t [a, b], | y  y(t , t0 , y0 ) |  },
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.
Since y(t , t0 , y0 ) is continuous at t  t0 ,   0,s.t. | t - t0 | 
implies |y(t , t0 , y0 ) - y(t0 , t0 , y0 ) | e  L( ba ) / 2.
Set V  {(t1 , y1 ) : |t1  t0 |  , | y1  y0 |  e- L( b-a ) / 2} .
Then (t0 , y0 )  V  K  U .
If (t1 , y1 ) V we have
|y(t , t1 , y1 ) - y(t , t0 , y0 ) |  | y(t , t1 , y1 ) - y( t , t1 , y( t1 , t0 , y0 )) |
| y1   f ( s, y( s, t1 , y1 ))ds -[ y(t1 , t0 , y0 )   f ( s, y( s, t1 , y( t1 , t0 , y0 ))ds]|
t
t1
t
t1
| y1 - y(t1 , t0 , y0 ) |  L | y( s, t1 , y1 ) - y( s, t1 , y( t1 , t0 , y0 )) | ds
t
t1
t
| y1 - y(t1 , t0 , y0 ) |  L | y( s, t1 , y1 ) - y( s, t0 , y0 ) | ds
t1
From Gronwall inequality we conclude that
|y(t , t1 , y1 ) - y(t , t0 , y0 ) | e L|t -t1| | y1 - y( t1 , t0 , y0 ) |
 e L( ba ) [ | y1  y0 |  | y(t0 , t0 , y0 )  y(t1 , t0 , y0 ) |] 
and this concludes the proof.
Corollary
Let f : U  R (U an open set of R  R) be continuous
and satisfy a local Lipschitz condition with respect to y.
For each ( t0 , y0 )  U , let the maximal interval of the
solution y( t , t0 , y0 ) to the IVP y  f ( t , y ), y( t 0 )  y0
be (  ( t0 , y0 ),   ( t0 , y0 )).
Then (i) G: (t0 , y0 )  t   (t0 , y0 ),(t0 , y0 )  U is a region;
(ii) y(t , t0 , y0 ) is a continuous function on G.
G
 (t0 , y0 )
t
b
Q(t , t0 , y0 )
U
O
P (t0 , y0 )
y0
t0
a
 (t0 , y0 )
Proof: (i)
(t , t0 , y0 )  G,i.e.,  (t0 , y0 )  t   (t0 , y0 ),(t0 , y0 ) U
Choose a,b, such that
 (t0 , y0 )  a  t  b   (t0 , y0 )
From Theorem 3.1, y(t , t1 , y1 ) is defined on a  t  b,
whenever (t1 , y1 ) sufficiently close to (t0 , y0 ).
It follows that (t , t0 , y0 ) is an interior point of G.
This means that G is an open set.
From the connectivity of U, it is easy to see G is a
connected set, and so G is a region.
(ii) From Theorem 3.1, we have that
  0,  1  0, s.t . | y( t , t1 , y1 )  y( t , t 0 , y0 ) |

.
2
Again y( t , t0 , y0 ) is continuous at t , so  2  0, s.t .
| y( t , t0 , y0 )  y( t , t0 , y0 ) |

, whenever | t  t |  2 .
2
Setting   min{ 1 ,  2 }, we have
| y( t , t1 , y1 )  y( t , t0 , y0 ) |
| y( t , t1 , y1 )  y( t , t0 , y0 ) |  | y( t , t0 , y0 )  y( t , t 0 , y0 ) |
  , whenever | t  t |  ,| t1  t0 |  ,| y1  y0 |  .
The proof is complete.
2. Continuous Dependence On Parameters
Consider the IVP
dy
 f ( t , y,  ), y( t0 )  y0 ,
dt
where  is parameter, (t , y,  )  G  U  ( ,  ),
U an open set of R  R.
Theorem: Let f : G  R (U an open set of R  R)
be continuous and satisfy a local Lipschitz condition
with respect to y . For each ( t0 , y0 ,  )  G , let the
maximal interval of the solution y( t , t 0 , y0 ,  ) to the
IVP y  f ( t , y,  ), y( t 0 )  y0 is (  ( t 0 , y0 ,  ),   ( t 0 , y0 ,  )).
Then y( t , t0 , y0 ,  ) is a continuous function on
 :   ( t0 , y0 ,  )  t    ( t0 , y0 ,  ),( t0 , y0 ,  )  G .
(2) Smooth dependence with respect to t , t0 and y0 .
We turn next to the question whether supplementary
conditions on f ( t , y ) such as differentiability do imply
that the solution y(t , t0 , y0 ) is differentiable.
We considerfirst the differentiability with respect to x0 .
In order to obtain an idea of the form of the derivative
y( t , t0 , y0 )
y0
we write the Cauchy problem as
y( t , t0 , y 0 )
t
 f ( t , y( t , t0 , y 0 )), y(t 0 , t 0 , y 0 )  y 0
and differentiate formally with respect to y0 .
Exchanging the derivatives with respect to t and y0
we find
 y( t , t0 , y 0 ) f ( t , y( t , t 0 , y 0 )) y( t , t 0 , y 0 )

t
y 0
y
y 0
y ( t 0 , t 0 , y 0 )
y 0
1
This formal calculation shows that the function
y(t , t0 , y0 )
y0
is a solution of the linear equation
z 
f (t , y(t , t0 , y 0 ))
y
z , z( t 0 )  1
This equation is called the variational equation for
the Cauchy problem y  f (t , y ), y( t0 )  y0 .
All the following proof does is to justify this
procedure.
Theorem 3.2
Let f : U  R  R (U an open set of R  R ) be continuous
and satisfy a local Lipschitz condition. Assume that
f ( t , y )
exists and is continuous on U . Then the solution
y
y( t , t 0 , y0 ) of the Cauchy problem y   f ( t , y ), y( t 0 )  y0
is continuously differentiable with respect to t , t 0 , y0 .
Proof: Let
y( t , t 0 , y0 )   and y( t , t 0 , y0   )   (| |  , 
is small enough) be the solutions to the Cauchy problem
y  f ( t , y ), y( t 0 )  y0 and y   f ( t , y ), y( t 0 )  y0   ,
respectively.
The integral equations for y(t , t0 , y0   ), y(t , t0 , y0 ) are
y(t , t0 , y0   )  y0     f ( s, y( s, t0 , y 0   ))ds,
t
t0
y(t , t0 , y0 )  y0   f ( s, y( s, t0 , y 0 ))ds.
t
t0
Using the theorem of the mean, we have
t

       f ( s ,    (   ))(   )ds
y
t0
where 0    1.
f
,  and  , we have
Note that the continuity of
y
f ( s,    (   )) f ( s,  )

 r1 ,
y
y
where r1  0 as   0, and r1  0 as   0.
Thus, for   0, we have
t
 

(   )
 1   f ( s,    (   ))
ds

y

t
0
This shows that    is the solution of the IVP

dz f (t ,  )
[
 r1 ]z , z(t0 )  1  z0 .
dt
y
This
 
shows that
is the solution of

dz f (t ,  )
[
 r1 ]z , z(t0 )  1  z0 .
dt
y
the IVP
From the theorem of continuous dependence, we get
 
is a continuous function of t , t0 , z0 ,  , so the limit

   y(t , t0 , y0 ) exists, and y(t , t0 , y0 ) is the solution
lim

y0
 0

y0
dz f ( t ,  )
Hence
of the IVP
[
]z , z( t0 )  1.
dt
y
t
y ( t , t 0 , y 0 )

 exp{  f ( s, y( s, t 0 , y0 )ds }.
 y0
y
t0
Clearly, it is a continuous function of t , t0 , y0 .
In the same manner, we can prove that
exists and is the solution of the IVP
y( t , t0 , y0 )
t0
dz f ( t ,  )
[
]z , z( t0 )   f ( t0 , y0 ).
dt
y
Hence
t
y ( t , t 0 , y 0 )

 - f ( t0 , y0 )exp{  f ( s, y( s, t 0 , y0 )ds }.
 y0
y
t0
Clearly, it is continuous function of t , t0 , y0 .
Noting that y( t , t0 , x0 ) is the solutionof the Cauchy problem
y( t , t0 , y0 )
y  f (t , y ), y( t0 )  y0 , it follows that
t
is continuous function of t , t0 , y0 .
Exercise 3.3
1. Proof:
Since f is continuous and satisfy a local Lipschitz condition
on G , so ( x0 , y0 )  G , the solution y( x, x0 , y0 ) to the
Cauchy problem y  f ( x, y ), y( x0 )  y0
exists, and is unique and continuous dependence on x0 , y0 .
Note that f ( x , 0)  0, so y  0 is the solution to y  f ( t , y ).

y
y0
y0
O
x0
x0
x
(1)  (2)
Assume that
  0, 1  0, such that | y0 | 1 implies that
| y( x, x0 , y0 ) |  for all x  x0 ,
  0 and x0  x0 , set y0  y( x0 , x0 , y0 ).
Since the solution to y = f(x, y),y(x0 )  y0
is continuous dependence on x0 ,y 0 ,
so for the 1  0,  2  0, such that | y 0 |  2 implies that
| y( x, x 0 , y 0 )  y( x, x 0 ,0) || y( x, x 0 , y 0 ) | 1
for all x0  x  x0 .
In particularly, | y( x0 , x0 , y0 ) || y0 | 1.
Hence,
Thus
| y( x, x0 , y0 ) |  whenever x  x0 ,
| y( x, x0 , y0 ) |  whenever x  x0  x0.
(2)  (1)
Assume that
  0,  2  0, such that | y 0 |  2 implies that
| y ( x, x0 , y0 ) |  for all x  x 0 .
Since the solution to y = f(x, y),y(x0 )  y0
is continuous dependence on x0 ,y 0 ,
so   0, 1  0, such that | y0 |  2 implies that
| y ( x, x0 , y0 )  y ( x, x0 , 0) || y ( x, x0 , y0 ) | 1
for all x0  x  x 0 .
Note that the solution to y  f (t , y), y(t0 )  y0
is unique, so
  0, 1  0, such that | y0 | 1 implies that
| y( x, x0 , y0 ) |  for all x  x0 ,
y

y0
y0
O
x0
x0
x
Exercise 3.1
4.
f , fy 
1
are continuous in y  0,
2
So conditions of the existence theorem are satisfied.
It is easy to see that
y0
is a solution to the
equation passing through (0,0).
1
3
dy
3
Solving for  y 3 , we have | y | ( x  c ) 2 .
dx 2
So all solutions passing through (0,0) are
0, x  c


| y | 
,(c  0).
3
2

(
x

c
)
,xc

5.
Proposition 1: y=y(x) is the solution to the
initial value problem
dy
 P ( x ) y  Q( x ), y( x0 )  y0
dx
(1)
is equivalent to y=y(x) is a continuous solution
to the integral equation
y  y0   [ P( s) y  Q( s )]ds.
x
x0
(2)
(ii) Set up a sequence of Picard iterates
y0 ( x )  y0 ,
yn ( x )  y0   [ P ( s ) yn1 ( s )  Q( s )]ds, n  1, 2,
x
x0
yn ( x) is well defined and continuous on [ ,  ] .
(iii)
yn ( x )
y( x ), x  [ ,  ].
Set M  max | P ( x ) y0  Q( x ) |, L  max | P ( x ) | .
x[ ,  ]
n
x[ ,  ]
ML
n 1
|
y
(
x
)

y
(
x
)
|

(



)
.
Then n1
n
(n  1)!
.
(iv) y(x) is a continuous solution to the integral
equation.
(v) uniqueness.
M  max | P ( x ) z( x )  Q( x ) | .
x[ ,  ]
6. Proof: Set G( t ) 
Then
and
K   f ( s ) g( s )ds.
G(t )  f (t ) g(t ).
t

f (t )  G(t ), t  [ ,  ]
Therefore
G( t ) f ( t ) g( t ) g(t )G(t )


 g( t ), t  [ ,  ]
G( t )
G( t )
G( t )
or, equivalently,
d
ln G( t )  g( t ), t  [ ,  ]
dt
or
ln G(t )  ln G( )   g( s )ds, t  [ ,  ]
t

or
t
t
g ( s ) ds
g ( s ) ds




G(t )  G( )e
 Ke
, t  [ ,  ]
which implies that
t
g ( 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 |
t
t0
 L | y( s ) z( s ) | ds.
t
t0
From Gronwall inequality, we get
t
| y(t )  z(t ) || y(t0 )  z(t0 ) | e
t0 Lds
 0, t  t0 .
If t  t0 , then
| y( t )  z( t ) ||  [ f ( s, y( s ))  f ( s, z( s ))]ds |
t
t0
 L |  | y( s ) z( s ) | ds |  L | y( s ) z( s ) | ds.
t
t
t0
t0
so
| y(t )  z(t ) || y(t0 )  z(t0 ) | e
L
t
t0 ds
 0, t  t0 .
7. Assume that the contrary, then there exist
two different solutions
y  1 ( x ), y   2 ( x ) with 1 ( x )   2 ( x ), x0  x  x1
and 1 ( x1 )   2 ( x1 ).
set y( x)  1 ( x)  2 ( x), then y( x0 )  0, y( x1 )  0,
This is contrary to
dy
 f ( x , 1 ( x ))  f ( x ,  2 ( x ))  0, x0  x  x1 .
dx
9.
(2)
(1) For any fixed x0 , set xn1  f ( xn ), n  0,1, 2,
We show inductively that
| xn  xn1 | N n1 | x1  x0 | N n1 M for all n  1.
Observe first that (1) is obviously true
for n=1, since
| x2  x1 || f ( x1 )  f ( x0 ) | N | x1  x0 | NM .
Next, we must show that (1) is true for
n=k+1 if it is true for n=k.
But this follows immediately, for
(1)
| xk  xk 1 | N k 2 M ,
then
| xk 1  xk || f ( xk )  f ( xk 1 ) | N | xk  xk 1 | N k 1 M .
(3)
Since 0  N  1, so  N n1 M 

Hence, x1   ( xn  xn1 )  
n 2
This implies that { xn } converges.
*
*
From
|
f
(
x
)

f
(
x
)
|

N
|
x

x
| and xn1  f ( xn ),
(4)
n
n
we have x*  f ( x* ).
(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.
10. (i) Define the successive approximations
0 ( x)  f ( x),
and
 n1 ( x )  f ( x )    K ( x,  ) n ( )d , n  1.
b
a
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]).
(ii) We now show that our successive
approximations converges uniformly for t
on the interval [a , b].
This is accomplished by writing n ( x )
in the form
n ( x)  0 ( x)  [1 ( x)  0 ( x)] 
 [n ( x)  n1 ( x)]
Clearly, the sequence n ( x ) converges
if, and only if, the infinite series
[1 ( x)  0 ( x)]   [n ( x)  n1 ( x)] 
converges .
M  max f  x  , L  max K  x,  
Set
x[ a ,b ]
x[ a ,b ]
 [ a ,b ]
b
Then 1  x   0  x    a K  x,   0   d


b
a
K  x,   f   d   ML(b  a)
Thus, in turn, implies that
 2  x   1  x     K  x,   1    0    d
b
a


b
a
K  x,   1    0   d   ML2 (b  a)2
2
Proceeding inductively, we have that
n  x   n 1  x    MLn (b  a)n for x  [a, b].
n
1
Consequently, if   L b  a , then n ( x )


converges uniformly on the interval [a, b],
set limn ( x)   ( x), then  ( x) is continuous on[a, b].
(iii) Note that
to
K ( x,  )n ( x )
converges uniformly
K ( x,  ) ( x ) on [a, b],
since
| K ( x,  )n ( x)  K ( x,  ) ( x) | L | n ( x)   ( x) |
and n ( x ) converges uniformly to
 ( x ) on
[a, b].
Then one can exchange integral and the
limit.
Thus
lim  K ( x,  ) n ( )d   K ( x,  ) ( )d ,
b
b
n a
a
and  ( x )  f ( x )  

b
a
K ( x,  ) ( )d .
(iV) If  ( x )  f ( x )   a K ( x,  ) ( )d  C ([a, b]).
then
b
| ( x )   0 ( x ) ||   K ( x ,  ) ( )d |
b
a
 L |  | N (b  a ).
  x .
where N  xmax
[ a ,b ]
| ( x )  1 ( x ) ||   K ( x ,  )( ( )   0 ( ))d |
b
a
 L2 |  |2 N (b  a)2 .
Proceeding inductively, we have that
| ( x)  n ( x) | Ln1 |  |n1 N (b  a)n1
which implies that n ( x ) converges uniformly
1
on the interval [a , b], when |  |
.
Hence,  ( x )   ( x ), x  [a, b].
L(b  a )
Theorem 1.1 (Existence and uniqueness for the
Cauchy problem)
Let A  {(t , y) :| t  t0 | a,| y  y0 | b}
and suppose that f : A  R
• is continuous,
• satisfies a Lipschitz condition with respect to y.
Then the Cauchy problem
dy
 f ( t , y ), y( t0 )  y0
(1)
dt
has a unique solution on [t0  h, t0  h], where
b
h  min(a , ) with M  max | f ( t , y ) | .
( t , y ) A
M
Proof of the Theorem
We will prove the theorem in five steps:
(i) Show that the initial value problem
dy
 f ( t , y ), y( t0 )  y0
dt
(1)
is equivalent to the integral equation
y(t )  y0   f ( s, y( s))ds.
t
(2)
t0
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
y0 (t )  y0 ,
yn (t )  y0   f ( s, yn1 ( s ))ds,| t  t0 | h, n  1.
t
t0
and show that yn ( t ) are well defined for all n  0;
(iii) Show that the sequence of Picard iteration { yn (t )}
converges uniformly on the interval [t0  h, t0  h];
{ yn ( t )} converges if, and only if, the infinite series
[ y1 (t )  y0 (t )] 
converges .
 [ yn (t )  yn1 (t )] 
(iv) Show that the limit of the sequence { yn (t )}
is a continuous solution to the integral equation (2).
(v) Show that this solution is unique.