EXISTENCE AND UNIQUENESS THEOREMS FOR
FIRST ORDER DIFFERENTIAL EQUATIONS
1. Existence Theorem, page 64 of Blanchard/Devaney/Hall
I have written a statement equivalent to the statement of the Existence Theorem (pp 64)
in our textbook below. Key components of the theorem statement are highlighted in various
colors. Below the theorem statement, I have provided observations about each highlighted
component of the theorem statement.
Theorem Suppose the two-variable function f (t, y) is continuous on the open rectangle
(a, b) × (c, d) . If (t0 , y0 ) is a point in (a, b) × (c, d), then there exists an
the initial value problem given by
dy
= f (t, y),
dt
> 0 such that
y(t0 ) = y0
has a solution y = y(t) defined on (t0 − , t0 + ) .
Observations:
Yellow highlighted part of Existence Theorem A two variable function is continuous on the
open rectangle (a, b) × (c, d) if and only if it is continuous at each point (t0 , y0 ) in the open
rectangle. In turn, the function f is continuous at the point (t0 , y0 ) if and only if the following
holds:
lim
f (t, y) = f (t0 , y0 )
(t,y)→(t0 ,y0 )
Our training in multivariable calculus allows us to verify that a two-variable function is
continuous for a wide range of possible functions. It should be noted that the function f
in the theorem is an auxiliary function that defines the differential equation. The function
f plays a very different role than the function y(t) that solves the differential equation.
Specification of f defines a differential equation. Once f is specified, we can talk about
solutions y of the differential equation.
It is remarkable that f need only be a continuous function; for instance, it does not have
to be differentiable or smooth. A continuous function can still have very bizarre behavior; for
example, mathematicians have identified continuous functions that are differentiable nowhere
or ‘everywhere jagged.’ Even when the right hand side is defined in terms of a highly irregular
function like that, a solution to the differential equation still exists!
Purple highlighted part of Existence Theorem First, we assume only that f is continuous
on the rectangle (a, b) × (c, d), so we can only hope to define a solution to this differential
equation on (t0 − , t0 + ) for values of small enough to ensure that (t0 − , t0 + ) is
contained in (a, b).
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EXISTENCE AND UNIQUENESS THEOREMS FOR FIRST ORDER DIFFERENTIAL EQUATIONS
Also, solutions of the differential equation may also cease to exist for values of t near
t0 ; the theorem only guarantees that the solution y(t) exists in some (perhaps small) open
interval that contains t0 . On pages 64-65 of the textbook, it is shown that the initial value
problem given by dy
= 1 + y 2 , y(0) = 0, has a solution defined only on (− π2 , π2 ) even though
dt
the function f (t, y) = 1 + y 2 is continuous on the entire ty−plane.
Green highlighted part of Existence Theorem Implicitly, any function y(t) that solves the
differential equation with initial value y(t0 ) = y0 must be differentiable on (t0 − , t0 + );
otherwise, we would not be able to check if the differential equation is satisfied.
Also, y(t) solves the differential equation on (t0 − , t0 + ) only if the left hand side and
right hand side of the differential equation are equivalent for all values of t in (t0 − , t0 + ).
Exercise 3 on Homework 1 gives an example of a function that solves a differential equation
for one value of the independent variable, x = 1, but does not satisfy the differential equation
in any open interval of x = 1 so it is not a solution of the differential equation given in that
exercise.
2. Uniqueness Theorem, page 66 of Blanchard/Devaney/Hall
Below, I have written a statement equivalent to the statement of the Uniqueness Theorem
(pp 66) in our textbook. Below the theorem statement, I have provided observations about
each highlighted component of the theorem statement.
Theorem Suppose the two-variable function f (t, y) is continuous in the rectangle (a, b) ×
df
is a continuous function of y and t on (a, b) × (c, d) . If (t0 , y0 ) is a point in
(c, d) and
dy
(a, b) × (c, d) and there exists an > 0 such that the initial value problem given by
dy
= f (t, y),
y(t0 ) = y0
dt
has two solutions y1 = y1 (t) and y2 = y2 (t) defined on (t0 − , t0 + ). Then, necessarily,
y1 (t) = y2 (t) for all t in (t0 − , t0 + ) .
Observations:
Yellow highlighted part of Uniqueness Theorem This part of the theorem is the only additional assumption about the differential equation itself made in the Uniqueness Theorem
that has not already been used in the Existence Theorem above.
df
We now also assume that f (t, y) is differentiable with respect to y and that dy
(t, y) is
continuous at each point in (a, b) × (c, d); that is, for all (t0 , y0 ) ∈ (a, b) × (c, d),
df
df
lim
(t, y) = (t0 , y0 )
(t,y)→(t0 ,y0 ) dy
dy
df
Using multivariable calculus, we can differentiate f with respect to y and verify dy
is continuous for a wide range of functions f . To verify the necessity of this additional assumption,
df
we would like an example of a continuous function f for which dy
does not exist or is not
continuous that exhibits non-unique solutions to the corresponding differential equation. On
EXISTENCE AND UNIQUENESS THEOREMS FOR
FIRST ORDER DIFFERENTIAL EQUATIONS
3
page 67 of our textbook, an example of such a differential equation is given. It is shown that
the differential equation dy
= 3y 2/3 does not satisfy the differentiability assumption of the
dt
uniqueness theorem and exhibits non-unique solutions through (0, 0).
Green highlighted part of Uniqueness Theorem This part of the theorem could be written
“if y = y(t) is a solution of the differential equation on (t0 − , t0 + ), then y(t) is unique”
because the highlighted statement, by definition, is what we mean when we write that a
solution is unique.
This uniqueness condition is often helpful when qualitatively analyzing the general solution
to a differential equation. The “Uniqueness and Qualitative Analysis” subsection on pages
69-70 of our textbook provides an example of using the uniqueness result for the qualitative
analysis of solutions of the differential equation given by dy
= (y − 2)(y − 1).
dt