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2.3 (Complete)

This section discusses the existence and uniqueness of solutions for non-linear differential equations, particularly focusing on initial value problems (IVPs). It outlines the Existence and Uniqueness Theorems, stating that if the function is continuous in a certain region, at least one solution exists, and if both the function and its derivative are continuous, the solution is unique. Several examples illustrate these concepts, demonstrating when solutions exist and their uniqueness based on the continuity of the functions involved.

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8 views3 pages

2.3 (Complete)

This section discusses the existence and uniqueness of solutions for non-linear differential equations, particularly focusing on initial value problems (IVPs). It outlines the Existence and Uniqueness Theorems, stating that if the function is continuous in a certain region, at least one solution exists, and if both the function and its derivative are continuous, the solution is unique. Several examples illustrate these concepts, demonstrating when solutions exist and their uniqueness based on the continuity of the functions involved.

Uploaded by

lagerscasey
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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2.

3: Existence and Uniqueness of Solutions

In this section, we look at non-linear di!erential equations.

Consider the following initial value problem:


↑
y → = y, y(0) = 0.

It is easy to see that y(x) = 0 (a constant function) is a solution. However,


if we did not notice this and proceeded to solve this separable di!erential
equation we see that:

y is also a solution

This equation is separable

Two distinct solutions?

In some cases, a non-linear first order DE may not have a solution.

Existence and Uniqueness Theorems:


Suppose we have a first order DE y → = f (x, y) with initial value y(x0 ) = y0 .
existence
(a) If f is continuous on an open rectangle R : {(x, y) : a < x < b, c < y < d}
containing (x0 , y0 ), then the initial value problem has at least one solution
on some subinterval of (a, b) containing x0 .
Uniqueness(b) If both f and f are continuous on R, then the initial value problem has
y
a unique solution on on some subinterval of (a, b) containing x0 .

Part a is about the existence of solutions, and part b is about the uniqueness.

Ex. y → =
fixin
↑
y, y(0) = 0

a Existence 9ᵗʰ
Fy 4101 0 not ok
f is continuousfor all
Not continuous when 7 0
and
720 yco 0 is Ok 1
Uniqueness not guaranteed
At least one solution
Ex. Determine the interval on which the following IVP (initial value problem)
has a unique solution:

ty → + 2y = 4t2 , y(1) = 2

Y 41 EY
a f continuous
b fy continuous forall 40
when to 1 Yo 2
o
At least
→
one solution
1/3
x 0 o
Ex. y = y , y(0) = 0
to 1
a f continuous for all y
y 1
b Fy Not continuous at yo 0
Uniqueness not guaranteed
Y t

Ex. Find the interval on which solution of the IVP y → = y 3 , y(0) = 1 exists
and is unique. m

continuous for all x Atleast one


f x y y y
solution
continuous solution
Unique
Fy 342

Solvethe separable DE Y
FI 1 2 70

Also see Examples 1, 2, 3, 4, and 5.


Tdd
mm
2
Ex. Solve the initial value problem t2 y → = 1 + y, y(1) = ↓1.

y 2

a
Exists
b fy Unique

1 Y 1
Guessing y
Doesthis satisfy the DE Yest
2.0 0 1 411 0
4 0

Also see Examples 7 and 8.

Ex. Solve the initial value problem xy → = ↓y ↓ y 2 , y(5) = 0.

y separablets tiny
Continuous at 5,0
Y
y fy
nnous at 15.0

Fy Y
Partial Fractions Y f
t.ge checksthat
and think about one easily
Stop here This works
an easier way 3

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