Ch02 4

Ch 2.4: Differences Between Linear and
Nonlinear Equations
Recall that a first order ODE has the form y' = f (t,y), and is
linear if f is linear in y, and nonlinear if f is nonlinear in y.
Examples: y' = ty - et
, y' = ty2
.
In this section, we will see that first order linear and
nonlinear equations differ in a number of ways, including:
The theory describing existence and uniqueness of solutions, and
corresponding domains, are different.
Solutions to linear equations can be expressed in terms of a general
solution, which is not usually the case for nonlinear equations.
Linear equations have explicitly defined solutions while nonlinear
equations typically do not, and nonlinear equations may or may not
have implicitly defined solutions.
For both types of equations, numerical and graphical
construction of solutions are important.

Theorem 2.4.1
Consider the linear first order initial value problem:
If the functions p and g are continuous on an open interval
(α, β) containing the point t = t0, then there exists a unique
solution y = φ(t) that satisfies the IVP for each t in (α, β).
Proof outline: Use Ch 2.1 discussion and results:
0)0(),()( yytgytp
dt
dy
==+
∫
=
+
=
∫ t
t
dssp
t
t
et
t
ydttgt
y 00
)(0
)(where,
)(
)()(
µ
µ
µ

Theorem 2.4.2
Consider the nonlinear first order initial value problem:
Suppose f and ∂f/∂y are continuous on some open rectangle
(t, y) ∈ (α, β) x (γ, δ) containing the point (t0, y0). Then in
some interval (t0- h, t0+ h) ⊆ (α, β) there exists a unique
solution y = φ(t) that satisfies the IVP.
Proof discussion: Since there is no general formula for the
solution of arbitrary nonlinear first order IVPs, this proof is
difficult, and is beyond the scope of this course.
It turns out that conditions stated in Thm 2.4.2 are sufficient
but not necessary to guarantee existence of a solution, and
continuity of f ensures existence but not uniqueness of φ.
0)0(),,( yyytf
dt
dy
==

Example 1: Linear IVP
Recall the initial value problem from Chapter 2.1 slides:
The solution to this initial value problem is defined for
t > 0, the interval on which p(t) = -2/t is continuous.
If the initial condition is y(-1) = 2, then the solution is given
by same expression as above, but is defined on t < 0.
In either case, Theorem 2.4.1
guarantees that solution is unique
on corresponding interval.
( ) 222
2ln521,52 tttyytyyt +=⇒==−′

Example 2: Nonlinear IVP (1 of 2)
Consider nonlinear initial value problem from Ch 2.2:
The functions f and ∂f/∂y are given by
and are continuous except on line y = 1.
Thus we can draw an open rectangle about (0, -1) on which f
and ∂f/∂y are continuous, as long as it doesn’t cover y = 1.
How wide is rectangle? Recall solution defined for t > -2,
with
( ) ( )
,
12
243
),(,
12
243
),( 2
22
−
++
−=
∂
∂
−
++
=
y
xx
yx
y
f
y
xx
yxf
( )
1)0(,
12
243 2
−=
−
++
= y
y
xx
dx
dy
4221 23
+++−= xxxy

Example 2: Change Initial Condition (2 of 2)
Our nonlinear initial value problem is
with
which are continuous except on line y = 1.
If we change initial condition to y(0) = 1, then Theorem 2.4.2
is not satisfied. Solving this new IVP, we obtain
Thus a solution exists but is not unique.
( ) ( )
,
12
243
),(,
12
243
),( 2
22
−
++
−=
∂
∂
−
++
=
y
xx
yx
y
f
y
xx
yxf
( )
1)0(,
12
243 2
−=
−
++
= y
y
xx
dx
dy
0,221 23
>++±= xxxxy

Example 3: Nonlinear IVP
Consider nonlinear initial value problem
Thus f continuous everywhere, but ∂f/∂y doesn’t exist at y = 0,
and hence Theorem 2.4.2 is not satisfied. Solutions exist but
are not unique. Separating variables and solving, we obtain
If initial condition is not on t-axis, then Theorem 2.4.2 does
guarantee existence and uniqueness.
3/23/1
3
1
),(,),( −
=
∂
∂
= yyt
y
f
yytf
( )00)0(,3/1
≥==′ tyyy
0,
3
2
2
3
2/3
3/23/1
≥





±=⇒+=⇒=−
ttyctydtdyy

Example 4: Nonlinear IVP
Consider nonlinear initial value problem
Thus f and ∂f/∂y are continuous at t = 0, so Thm 2.4.2
guarantees that solutions exist and are unique.
Separating variables and solving, we obtain
The solution y(t) is defined on (-∞, 1). Note that the
singularity at t = 1 is not obvious from original IVP statement.
yyt
y
f
yytf 2),(,),( 2
=
∂
∂
=
1)0(,2
==′ yyy
t
y
ct
yctydtdyy
−
=⇒
+
−
=⇒+=−⇒= −−
1
1112

Interval of Definition: Linear Equations
By Theorem 2.4.1, the solution of a linear initial value
problem
exists throughout any interval about t = t0 on which p and g
are continuous.
Vertical asymptotes or other discontinuities of solution can
only occur at points of discontinuity of p or g.
However, solution may be differentiable at points of
discontinuity of p or g. See Chapter 2.1: Example 3 of text.
Compare these comments with Example 1 and with previous
linear equations in Chapter 1 and Chapter 2.
0)0(),()( yytgytpy ==+′

Interval of Definition: Nonlinear Equations
In the nonlinear case, the interval on which a solution exists
may be difficult to determine.
The solution y = φ(t) exists as long as (t,φ(t)) remains within
rectangular region indicated in Theorem 2.4.2. This is what
determines the value of h in that theorem. Since φ(t) is usually
not known, it may be impossible to determine this region.
In any case, the interval on which a solution exists may have
no simple relationship to the function f in the differential
equation y' = f (t, y), in contrast with linear equations.
Furthermore, any singularities in the solution may depend on
the initial condition as well as the equation.
Compare these comments to the preceding examples.

General Solutions
For a first order linear equation, it is possible to obtain a
solution containing one arbitrary constant, from which all
solutions follow by specifying values for this constant.
For nonlinear equations, such general solutions may not
exist. That is, even though a solution containing an arbitrary
constant may be found, there may be other solutions that
cannot be obtained by specifying values for this constant.
Consider Example 4: The function y = 0 is a solution of the
differential equation, but it cannot be obtained by specifying
a value for c in solution found using separation of variables:
ct
yy
dt
dy
+
−
=⇒=
12

Explicit Solutions: Linear Equations
By Theorem 2.4.1, a solution of a linear initial value
problem
exists throughout any interval about t = t0 on which p and g
are continuous, and this solution is unique.
The solution has an explicit representation,
and can be evaluated at any appropriate value of t, as long
as the necessary integrals can be computed.
,)(where,
)(
)()(
00
)(0 ∫
=
+
=
∫ t
t
dssp
t
t
et
t
ydttgt
y µ
µ
µ
0)0(),()( yytgytpy ==+′

Explicit Solution Approximation
For linear first order equations, an explicit representation
for the solution can be found, as long as necessary
integrals can be solved.
If integrals can’t be solved, then numerical methods are
often used to approximate the integrals.
∑∫
∫
=
∆≈
∫
=
+
=
n
k
kkk
t
t
dssp
t
t
ttgtdttgt
et
t
Cdttgt
y
t
t
1
)(
)()()()(
)(where,
)(
)()(
0
00
µµ
µ
µ
µ

Implicit Solutions: Nonlinear Equations
For nonlinear equations, explicit representations of solutions
may not exist.
As we have seen, it may be possible to obtain an equation
which implicitly defines the solution. If equation is simple
enough, an explicit representation can sometimes be found.
Otherwise, numerical calculations are necessary in order to
determine values of y for given values of t. These values can
then be plotted in a sketch of the integral curve.
Recall the following example from
Ch 2.2 slides:
1sinln1)0(,
31
cos 3
3
+=+⇒=
+
=′ xyyy
y
xy
y

Direction Fields
In addition to using numerical methods to sketch the
integral curve, the nonlinear equation itself can provide
enough information to sketch a direction field.
The direction field can often show the qualitative form of
solutions, and can help identify regions in the ty-plane
where solutions exhibit interesting features that merit more
detailed analytical or numerical investigations.
Chapter 2.7 and Chapter 8 focus on numerical methods.

Ch02 4

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