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Theory of Ordinary Differential Equations
Basic Existence and Uniqueness
John A. Burns
Center for Optimal Design And Control
Interdisciplinary Center for Applied Mathematics
Virginia Polytechnic Institute and State University Blacksburg, Virginia 24061-0531
MATH 5245 - FALL 2012
1st Order Scalar Equations
( ) ( , ( ))x t f t x t
1st order ordinary differential equation
2( ) ( ) 5[ ( )]x t tx t x t
( ) 3 ( )d
p t p tdt
3( ) 5[ ( )] cos( )y t y t t
3( ) sin( ( ))y t y t t
( , ) 3f t p p
( , ) :f t x D R R R
2( , ) 5f t x t x x
3( , ) sin( )f t y y t
3( , ) 5 cos( )f t y y t
1st Order Autonomous DEs
2( ) ( ) 5[ ( )]x t tx t x t
( ) 3 ( )d
p t p tdt
3( ) 5[ ( )] cos( ( ))y t y t y t
3( ) sin( ( ))y t y t t
( , ) 3f t p p
2( , ) 5f t x t x x
3( , ) sin( )f t y y t
3( , ) 5 cos( )f t y y y
( ) ( , ( ))x t f t x t ( ) ( ( ))x t f x t
Auto
Non-auto
Non-auto
Auto
0( ) [1 (1/ ) ( )] ( )p t r k p t p t 0( ) [1 (1/ ) ]f p r k p p
Logistic Equation
Autonomous
Definition of Solution
A solution to the ordinary differential equation (Σ) is a
differentiable function
( ) , ( )x t f t x t(Σ)
( ) : ( , )x t a b R
defined on a connected interval (a,b) such that x(t)
satisfies (Σ) for all t (a,b).
2
( ) ( )d
x t x tdt
1( )
1x t
t
1( ) , 1
1lx t t
t
1( ) , 1
1rx t t
t
TWO SOLUTIONS
Solutions
-3 -2 -1 0 1 2 3 4-20
-15
-10
-5
0
5
10
15
20
1( ) , 1
1lx t t
t
1( ) , 1
1rx t t
t
Initial Condition
-3 -2 -1 0 1 2 3 4-20
-15
-10
-5
0
5
10
15
20
1( )0lx
1( ) , 1
1lx t t
t
1( ) , 1
1rx t t
t
1( )2rx
EXISTENCE AND UNIQUENESS
SCALAR PROBLEMS
Ordinary Differential Equations
1R
t
x0
t0
)(tx
1
0 0( )x t x R (IC)
( ) , ( )x t f t x t(Σ) {(IVP)
( , ) : Df t x R R R
D
D
Solutions to First Order Equations
A solution to the ordinary differential equation (Σ) is a
differentiable function
( ) , ( )x t f t x t(Σ)
1( ) : ( , )x t a b Rdefined on a connected interval (a, b) such that
1. (t, x(t)) D for all t (a, b),
2. x(t) satisfies (Σ) for all t (a, b).
1 ,1
1)(
t
ttxl
tt
txr 1 ,1
1)(
HAS TWO
GENERAL
SOLUTIONS
2
( ) ( )x t x t
( ) 3 ( )x t x t3( ) tx t e k ( , ) ( , )a b k
( , ) :f t x D R R R
Multiple Solutions
-3 -2 -1 0 1 2 3 4-20
-15
-10
-5
0
5
10
15
20
1 ,1
1)(
t
ttxl
tt
txr 1 ,1
1)(
( , ) ( ,1)l la b
( , ) (1, )r ra b
Initial Value Problem
1
0 0( )x t x R (IC)
A solution to the initial value problem (IVP) is a
differentiable function 1( ) : ( , )x t a b R
defined on a connected interval (a, b) such that
1. t0 (a, b),
2. x(t0 ) = x0 ,
3. (t, x(t)) D for all t (a, b),
4. x(t) satisfies (Σ) for all t (a, b).
( ) , ( )x t f t x t(Σ) {(IVP)
( , ) :f t x D R R R
Initial Value Problem
1R
t
x0
t0
)(tx
1
0 0( )x t x R (IC)
( ) , ( )x t f t x t(Σ) {(IVP)
( , ) :f t x D R R R
D
-3 -2 -1 0 1 2 3 4-20
-15
-10
-5
0
5
10
15
20
1 ,1
1)(
t
ttxl
2
( ) ( )x t x t (0) 1x
tt
txr 1 ,1
1)(
Initial Condition
1
0 0( )x t x R (IC)
( ) , ( )x t f t x t(Σ) {(IVP)
( , ) :f t x D R R R
)(tx
1R
t
x0
t0
D
2/3
( ) ( )x t x t 2/3( , )f t x x D R R
f(t, x) is continuous EVERYWHERE!!!
Theorem 1. If f: D ---> R is a continuous function on
a domain D and (t0, x0 )D, then there exists at least
one solution to the initial value problem (IVP).
( )y t
Existence Theorem
2/3
( ) ( )x t x t (0) 0x
1R
t
( ) 0x t
c
3
0,( )
([ ] / 3) ,
t cx t
t c t c
{
c
!! INFINITE NUMBER OF SOLUTIONS !!
Non-Uniqueness
Uniqueness Theorem
1R
t
x0
t0
)(txD
Theorem 2. If there is an open rectangle D
about (t0, x0) such that
are continuous at all points (t, x) , then there a
unique solution to the initial value problem (IVP).
( , )f t x ( , )x
f t x
and
First Order Linear
( ) ( ) ( ) ( )x t a t x t g t
( )
00
( ) ( )t
at a t sx t e x e g s ds
VARIATION OF PARAMETERS FORMULA
( )a t given ( )g tand
homogenous ( ) 0g t non-homogenous ( ) 0g t
constant coefficient ( ) constanta t a
( ) ( ) ( )x t ax t g t 0(0)x x
First Order Linear
( ) ( , ( )) ( ) ( ) ( )x t f t x t a t x t g t
( , ) ( ) ( )f t x a t x g t ( , ) ( )f t x a tx
( ) ( ) ( ) ( )x t a t x t g t
Note: If is continuous on an interval (a, b) and
t0 (a, b), then there a unique solution to the initial
value problem
( )a t
0 0( ) .x t x
( ) ( ) ( ) ( )x t a t x t g t
1( ) ( ) ( ) cos( )
( ) 1
tx t x t t
x
sin( )1( ) ( )t
t tx t
(0, )t
Gronwall – Reid – Bellman Inequality
0
0 ( ) ( ( )) ( )
t
btt a b s ds t ate Gronwall (1919)
Reid (1929) Dissertation - page 296
( ) 0, ( ) 0,
0 ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )exp ( )
k s s a s b
tt t k s s ds
a
t tt t s k s k d ds
a s
Re-discovered by Bellman in 1943
W. T. Reid, “Properties of solutions of an infinite system of ordinary differential equations of the first order4 with auxiliary boundary conditions”, Transactions of the AMS, vol. 32 (1930), pp. 284 – 318.
Bellman’s Version
Bellman in 1943
( )
( ) 0, 0,
0 ( ) ( ) ( )
( ) exp ( ) e
t
a
t k s ds
a
k s a s b
tt k s s ds
a
t k s ds
Peano - Reid Version
( ) 0, ( ) 0,
0 ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )exp ( )
k s s a s b
tt t k s s ds
a
t tt t s k s k d ds
a s
0 ( ) ( ) ( ) ( )
( ) ( )exp ( )
tt t k s s ds
a
tt t k s ds
a
and if is non-decreasing, then ( )s
Homework
PROVE THESE INEQUALITIES
EXISTENCE AND UNIQUENESS
SCALAR PROBLEMS
Proof of Main Theorems
Theorem 1. If f: D ---> R is a continuous function on
a domain D and (t0, x0 )D, then there exists at least
one solution to the initial value problem (IVP).
1
0 0( )x t x R (IC)
( ) , ( )x t f t x t(Σ) {(IVP)
1R
t
x0
t0
D
Peano Existence Theorem
Two step process (Coddington & Levinson: Pages 1-8)
Peano Existence Theorem
Peano Existence Theorem
ˆ :1, 2,...,S t p
Peano Existence Theorem
( ) , ( )x t f t x t
An -approximate solution of () on a connected interval I is a PWS function
such that 1: I
( ) ( , ( )) , ,
( ) ( ) ( , ( )) , .
i t t D t I
ii t f t t t I S
1
0 0( )x t x R
1R
t
D x0
t0
Let R be a closed rectangle inside D as
shown below
R
0t a0t a
0x b
0x b
Peano Existence Theorem 1R
t
D x0
t0
R
0t a0t a
0x b
0x b
0 0 ( , ) : , t x t t a x x b R
max ( , ) : ( , )M f t x t x R
min ,b
aM
Peano Existence Theorem
Lemma 1.1 Assume f: D ---> R1 is a continuous function on a domain D and
(t0, x0 )R D is the closed rectangle above, then there exists an
-approximate solution of (),
where . 0 I :t t t
1: I
Proof Since f: R ---> R1 is continuous, it is uniformly continuous so there
exists a = () >0 such
1 1 2 2 1 1 2 2( , ) ( , ) , if ( , ) , ( , )f t x f t x t x t x R R
1 2 1 2 ( ), ( ).t t x x and
t0 t0+ t2 t1 t3 tn t0+a …
1
( )max min ( ), k kt t
M
Peano Existence Theorem
t0 t0+ t2 t1 t3 tn t0+a …
x0 + b
x0 - b
slope = M
slope = -M
x0
0 0slope ( , )f t x
1 1slope ( , )f t x
x1
x2
2 2slope ( , )f t x
Peano Existence Theorem
0 0( )t x 1 1 1 1 1( ) ( ) ( , )( ), k k k k k kt t f t x t t t t t
1If , then ( ) andk k kt t t t t
0 0ˆ ˆ ˆ( ) ( ) , ,t t M t t t t t t
1 1 1 1( ) ( , ( )) ( , ) ( , ( )) ( , ( )) ( , ( ))k k k kt f t t f t x f t t f t t f t t
1 1 1 1 1
( )ˆ ˆ( ) ( ) ( , )( ) ( )k k k k kt t f t x t t M t t M
M
and hence
Thus, is an -approximate solution on . 1: I 0 0[ , ]t t
The same construction yields an -approximate solution on
and connecting the two pieces produces the piecewise smooth -
approximate solution 1
0 0:[ , ]t t
0 0[ , ]t t
Peano Existence Theorem
A set of functions defined on a bounded interval I is equicontinuous on I if
for any > 0, there is a () > 0 such that if
then
1 2 1 2, , satisfy ( ) and ( )t t I t t
1 2 ( ) ( ) t t
Lemma 1.2 (Ascoli) Assume that for each n=1,2,3,… , n : I ---> R1 is a
sequence of uniformly bounded and equicontinuous functions defined on the
interval I. Then there is a subsequence
and a continuous function : I ---> R1 such that as
uniformly on I.
1:kn I
( ) ( )kn
k
Proof of Main Theorems
Theorem 1. (Peano) If f: D ---> R is a continuous
function on a domain D and (t0, x0 )D, then there
exists at least one solution to the initial value
problem (IVP).
Proof Let and be the -approximate
solution of the IVP computed as above . Clearly, so the
sequence is uniformly bounded and
Thus, is a uniformly bounded equicontinuous
set.
1 2 1 2( ) ( ) 1/ .n n nt t M t t n
1/n n 1
0 0:[ , ]n t t
0( )n t x b
( ) : 1, 2,3,...n n
( ) : 1,2,3,...n n
Pick the subsequence uniformly convergent on
to the continuous function .
( ) ( )kn 0 0[ , ]t t
( )
Proof of Main Theorems
Let ( ) ( ) and ( ) ( )zk nx x
( ) ( , ( )) , kk k nx t f t x t t I S
0
0( ) [ ( , ( )) ( )] where
t
k k k
t
x t x f s x s s ds ( ) ( , ( )) ( ) on k k kx t f t x t t I S
( )kk nt
( ) ( ) uniformly ( , ( )) ( , ( )) uniformlyk kx x f t x t f t x t
0 0 0
0 0 0( ) ( , ( )) [ ( , ( )) ( )] - ( , ( ))
t t t
k k k
t t t
x t x f s x s ds x f s x s s ds x f s x s ds
0
[ ( , ( )) ( , ( )) ( )]
t
k k
t
f s x s f s x s s ds
0
( , ( )) ( , ( ))k
t
k n
t
f s x s f s x s ds
Proof of Main Theorems
0
0 0
0( ) ( , ( )) ( , ( )) ( , ( )) 0k
tt
k k n
t t
x t x f s x s ds f s x s f s x s ds
( ) ( ) uniformly ( , ( )) ( , ( )) uniformlyk kx x f t x t f t x t
0
0( ) ( ) and ( ) ( , ( ))
t
k k
t
x x x x f s x s ds
0
0( ) ( , ( ))
t
t
x x f s x s ds
Since is continuous is differentiable and ( )x 0
0 ( , ( ))
t
t
x f s x s ds
( ) ( , ( ))x t f t x t
Proof of Main Theorems
The function f: D ---> R is said to satisfy a Lipschitz
condition on D if there is a constant k > 0 such that for
every (t, x1 ) and (t, x2 ) in D,
1 2 1 2( , ) ( , ) .f t x f t x k x x
Theorem 2. Assume that the function f: D ---> R
satisfies a Lipschitz condition on D. If
are two solutions to the (IVP) on the interval (a, b),
then
In particular, solutions are unique.
1 2( ) and ( )x x
1 2( )= ( ), for all ( , ).x t x t t a b
Proof of Main Theorems
Proof first note that f: D ---> R1 is continuous, so there exists at least one
solution. Assume that are two solutions defined on (a,b) so
that a < t0 < b and
then
which implies
00( ) ( ) ( , ( )) , 1,2, .
t
i i it
x t x t f s x s ds i a t b
1 2( ) and ( )x x
0 01 2 1 2( ) ( ) ( , ( )) ( , ( ))
t t
t tx t x t f s x s ds f s x s ds
01 2[ ( , ( )) ( , ( ))]
t
tf s x s f s x s ds
01 2( , ( )) ( , ( ))
t
tf s x s f s x s ds
01 2( ) ( )
t
tk x s x s ds
Proof of Main Theorems
01 2 1 2( ) ( ) ( ) ( )
t
tx t x t k x s x s ds
Define so that 1 2( ) ( ) ( )t x t x t
0 0
0 ( ) ( ) 0 ( ) , ( , ).t t
t tt k s ds k s ds t a b
0( ) 0, 0 0, k t k t t b
Apply the Gronwall Inequality with
0 0
0
( ) ( ) ( ) exp ( ) e 0e 0
t
k t t k t t
t
t k s ds
First consider the case 0t t b
Proof of Main Theorems
0 0
0
( ) ( )
0 ( ) exp ( ) e 0e 0,
t
k t t k t t
t
t k s ds t t b
Hence,
1 2 0( ) ( ) ( ) 0, x t x t t t t b
and the solution is unique on . 0t t b
Homework: Complete the proof and show that
1 2 0( ) ( ) ( ) 0, x t x t t a t t
so that 1 2( ) ( ), x t x t a t b
Continuation of Solutions
1R
t
x0
t0
)(tx
1
0 0( )x t x R (IC)
( ) , ( )x t f t x t(Σ) {(IVP)
( , ) : Df t x R R R
DD
a b a1 b1
Can’t extend beyond a certain point
am bM
Continuation of Solutions
Theorem 2. Assume that the function f: D ---> R is
continuous on D and there is a constant such that
If
If then the solution
may be extended to the left of a [or to the right of b].
M
( , ) , for all ( , ) .f t x M t x D R R
( ) is a solution to the (IVP) on the interval ( , ), thenx a b
lim ( )= ( ) and lim ( )= ( ) exist.t a t b
x t x a x t x b
( , ( ) ) [or ( , ( )) ], a x a D b x b D ( )x t
00( ) ( ) ( , ( )) , .
t
tx t x t f s x s ds a t b
00( ) ( ) ( , ( ))
b
tx b x t f s x s ds
00( ) ( ) ( , ( ))
a
tx a x t f s x s ds
Continuation of Solutions
1R
t
x0
t0
)(tx
DD
a b a1 b1
R
R