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1.6 Existence and Uniqueness Questions 41

The differential equation has general solution y = (x + c) , but there is no real number c such
2
that y(0) =−1.
An initial value problem also may have more than one solution. In particular, the initial value
problem
√

y = 2 y; y(1) = 0
has the zero solution y = ϕ(x) = 0 for all x. But it also has the solution

0 for x ≤ 1
y = ψ(x) =
(x − 1) 2 for x ≥ 1.
Because existence and/or uniqueness can fail for even apparently simple initial value prob-
lems, we look for conditions that are sufficient to guarantee both existence and uniqueness of a
solution. Here is one such result.

THEOREM 1.2 Existence and Uniqueness

Let f (x, y) and ∂ f/∂y be continuous for all (x, y) in a rectangle R centered at (x 0 , y 0 ). Then
there is a positive number h such that the initial value problem

y = f (x, y); y(x 0 ) = y 0
has a unique solution defined at least for x 0 − h < x < x 0 + h.

A proof of Theorem 1.2 is outlined in the remarks preceding Problem 6.
The theorem gives no control over h, hence it may guarantee a unique solution only on a
small interval about x 0 .

EXAMPLE 1.19

The problem
2

y = e x y − cos(x − y); y(1) = 7
2
has a unique solution on some interval (1 − h,1 + h), because f (x, y) = e x y − cos(x − y) and
∂ f/∂y are continuous for all (x, y), hence, on any rectangle centered at (1,7). Despite this, the
theorem does not give us any control over the size of h.

EXAMPLE 1.20
The initial value problem
2

y = y ; y(0) = n
in which n is a positive integer has the solution
1
y(x) =− 1 .
x −
n
This solution is defined only for −1/n < x < 1/n, hence, on smaller intervals about x 0 = 0as n
is chosen larger.

For this reason, Theorem 1.2 is called a local result, giving a conclusion about a solution
only on a perhaps very small interval about the given point x 0 .

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October 14, 2010 14:9 THM/NEIL Page-41 27410_01_ch01_p01-42