Jim Lambers
MAT 285
Spring Semester 2012-13
Lecture 14 Notes
These notes correspond to Section 2.8 in the text.
The Existence-Uniqueness Theorem
So far, we have learned how to solve certain specific types of first-order ODE. We now prove that
under certain assumptions, a general first-order ODE has a unique solution.
Theorem (Existence-Uniqueness) Let the function f and ∂f /∂y be continuous in some rectangle α ≤ t ≤ β, γ ≤ y ≤ δ containing the point (t0 , y0 ). Then, in some interval t0 − h ≤ t ≤ t0 + h
contained in (α, β), the initial value problem
y 0 = f (t, y),
y(t0 ) = y0
has a unique solution y(t).
Proof We first note that if we integrate both sides of the equation y 0 (s) = f (s, y(s)) with respect
to s, from t0 to t, then we obtain
Z t
y(t) = y0 +
f (s, y(s)) ds.
t0
Then, we define a sequence of functions, known as the Picard iterates, as follows:
Z t
y0 (t) = y0 , yn+1 (t) = y0 +
f (s, yn (s)) ds, n = 0, 1, 2, . . . .
t0
Our goal is to prove that for some interval |t − t0 | ≤ h contained within [α, β], the Picard iterates
converge to a function y(t) that is a solution of the initial value problem. To that end, we first note
that from the continuity of f on R, we have
Z t
|y1 (t) − y0 | = f (s, y0 ) ds ≤ M |t − t0 |,
t0
where |f (t, y0 )| ≤ M on [t0 − h, t0 + h].
Then, we let n ≥ 1 and examine the difference
Z t
Z t
|f (s, yn (s)) − f (s, yn−1 (s))| ds.
|yn+1 (t) − yn (t)| = f (s, yn (s)) − f (s, yn−1 (s)) ds ≤
t0
t0
Because of the continuity of ∂f /∂y in the rectangle R = [α, β] × [γ, δ], we have, for a, b ∈ [γ, δ] and
|s − t0 | ≤ h, by the Mean Value Theorem,
∂f
|f (s, b) − f (s, a)| = (s, ψ)(b − a) ≤ K|b − a|,
∂y
where ψ lies between a and b and |∂f /∂y| ≤ K on R. It follows that
Z t
|yn+1 (t) − yn (t)| ≤ K |yn (s) − yn−1 (s)| ds .
t0
1
Setting n = 1 yields
Z t
Z t
KM |t − t0 |2
|y2 (t) − y1 (t)| ≤ K |y1 (s) − y0 (s)| ds ≤ KM |s − t0 | ds ≤
.
2
t0
t0
Continuing by induction, we obtain
|yn+1 (t) − yn (t)| ≤
M K n |t − t0 |n+1
M K n hn+1
≤
.
(n + 1)!
(n + 1)!
It follows that as n → ∞, |yn+1 (t) − yn (t)| → 0, thus establishing convergence of the Picard iterates
to a solution
y(t) = lim yn (t).
n→∞
Now that we have proved that a solution exists, we show that it is also unique. We assume that
there are two distinct solutions, y(t) and ỹ(t). Then we have
Z t
Z t
|y(t) − ỹ(t)| = f (s, y(s)) − f (s, ỹ(s)) ds ≤ K |y(s) − ỹ(s)| ds .
t0
t0
Let z(t) = |y(t) − ỹ(t)|. Then, z(t) ≥ 0, z(t0 ) = 0, and for t ≥ t0 , we have, by the Fundamental
Theorem of Calculus,
z 0 (t) ≤ K|y(t) − ỹ(t)| = Kz(t).
Rearranging and using the integrating factor µ(t) = e−Kt , we obtain
e−Kt z 0 (t) − Ke−Kt z(t) = [e−Kt z(t)]0 ≤ 0.
Integration from t0 to t yields e−Kt z(t) ≤ 0, and therefore z(t) ≤ 0. However, we also have z(t) ≥ 0.
We conclude that z(t) = 0, which implies that y(t) = ỹ(t), contradicting our assumption that these
solutions are distinct. We conclude that the solution is unique. 2
We now make two noteworthy observations about the preceding proof.
• The proof relies on the fundamental result that a continuous function on a compact set, such
as a closed interval [a, b] or a closed rectangle [a, b] × [c, d], has a maximum and minimum.
• The use of Picard iterates is an example of fixed-point iteration, in which an equation of the
form g(x) = x is solved by choosing an initial guess x0 and then computing xn+1 = g(xn ) for
n = 0, 1, 2, . . .. This iteration provides both a practical and theoretical foundation for various
methods of solving nonlinear equations or systems of equations.
2