Differential Equations
for Engineers: the Essentials
Supplement to Class 6:
Proof of Existence and Uniqueness
Theorem
Appendix
The following pages contain a detailed proof of the existence and
Appendix:
Equivalent Integral Equation
Let be continuous in R.
A function is a (differentiable) solution of the ODE
on an interval of the t-axis containing if and only if it is a
(continuous) solution on to the integral equation
)(t
),( ytf
I
0
t
I
Appendix:
Successive Approximations
Consider the sequence of functions defined by
This sequence of successive approximations converges to
the solution of the integral equation (and therefore the ODE).
Note that
)(t
i
Appendix:
Outline of Theorem Proof
We must show that
(1) the integral equation is equivalent to the ODE
(5) the limit of the series satisfies
Appendix: Theorem Proof Part 1: Integral
Equation and ODE are Equivalent
Suppose is a solution to
Then
by the fundamental theorem of calculus, and hence is also a
solution to the ODE.
)(t
)(t
Appendix: Theorem Proof Part 1: Integral
Equation and ODE are Equivalent (2)
Suppose is a solution to the ODE
Hence we can integrate the ODE from to and obtain
)(t
tdfyt
0
To complete the proof we must show that the function
0
is continuous.
t
t
Since is continuous on and is continuous on R,
the function is continuous on .
)(t
I
),( ytf
))(,()( tytftF =
I
Appendix: Theorem Proof Part 1: Integral
Equation and ODE are Equivalent (3)
Recall the definition of continuity:
A function is said to be continuous at if for every
In our case
)(tx
=tt
0
By the mean value theorem of integral calculus , because
is continuous on :
for some in the closed interval .
t
tt ,
))(,()( tytftF =
I
Appendix: Theorem Proof Part 1: Integral
Equation and ODE are Equivalent (4)
Setting where
and recalling that we have stated that is bounded by , then
−
tt
),( ytf
M
Then
Appendix: Theorem Proof Part 2:
are continuous in
Clearly is continuous on . The function
is continuous on because the function is continuous.
We just proved this in proving the equivalence of the ODE and integral
equation solutions. In the same way, defined as
00 )( yt =
I
I
),()( 00 ytftF =
)(t
n
I
)(
2t
I
Appendix:
Core of Proof that is in R if is in
R
by +
0
y
)(
1t
n+
Illustrating the
geometry that bounds
slope
M
))(,( 1tt n+
t
I
df
t
tn))(,(
0
Appendix: Theorem Proof Part 3:
is in R if is in
By definition of R, the point is in R if
Obviously, satisfies Equation 2. Now consider
so satisfies Equation 2 and thus is in R. We
can continue in this way and show by induction that all the in
the sequence are in R.
))(,( 1tt n+
t
I
))(,( 1tt n+
00 )( yt =
M
byt − 01 )(
)(
1t
)(t
n