2. Equilibrium Solutions and Stability

Example 2.1. Newton’s law of cooling

dx

dt= �k(x� A), (k > 0 constant)

x(0) = x0

By separation of variables, we have the explicit solution

x(t) = A+ (x0 � A)e�kt

It follows that

limt!1

x(t) = A

(a) Typical solution curve.

(b) Phase diagram.

1

,

,

•

→environmental temperature ( coast)

*¥÷÷¥:* - fof anobject

¥b= At

+→ o ,-ht → -o, e-

"→ o

Definition 2.1. Any first-order ODE which can be written:

(2.1)dx

dt= f(x),

where the independent variable t does not appear explicitly, is called autonomous.

Any autonomous ODE is automatically separable, with the (implicitly defined)

general solution:

(2.2)

Z1

f(x)dx = t+ c.

Definition 2.2. The solutions of the equation f(x) = 0 are called critical pointsof the autonomous di↵erential equation (2.1).

Definition 2.3. If x = c is a critical point of Eq. (2.1), then the di↵erential

equation has the constant solution x(t) = c. A constant solution of a di↵erential

equation is sometimes called an equilibrium solution.

Example 2.2.

(2.3)dx

dt= kx (M � x)

Find the critical points.

2

=( separable but not

autonomous) dxIt

=✗'

→ Separable equation

f- is not dependingont

.

independent

± ☐pits of the ODE -

= ( Po - Ax - Vol , p

- =f

use the definition to find thethe

critical points

are actually theconstant

critical pts .solutions of the ODE .

some fi check,

b. ✗ ( in-14--0 ✗12=1"

⇒ {✗1=0

→" ""

uts -_d¥= 0→ii.Hiiiii pints

✗2 = M of the ODERHS = b. ii. in,

"

= 0

⇒ × ,= inis asolati

.

of the ODE.

Definition 2.4. The critical point c is stable if, for each ✏ > 0, there exists � > 0

such that

|x0 � c|< � implies that |x(t)� c|< ✏

for all t > 0. The critical point x = c is unstable if it is not stable.

Example 2.3. Determine the stability of critical points in Example 2.2.

3

if there is an errorin the initial condition,

but the solution will go to thescene value

¥.-0 -

•Cic initial

y,wud:t:n of

Hut"" ")glue line :

solution✗

✓ of +↳ efuat""

12

c.is,€¥É- > > > > as time ct, → 0

g.adit""

1in 1 , =- Iim 12=/im ↳

÷

T it:me)

Harvesting a Logistic Population:

The autonomous di↵erential equation

(2.4)dx

dt= ax� bx2 � h

(with a, b, and h all positive) may be considered to describe a logistic population

with harvesting. It can be re-written in the form

(2.5)dx

dt= kx (M � x)� h

Example 2.4. Solve the di↵erential equation (2.5). Find the critical points of

(2.5).

4

~

= fix) →auto .

= 1i±_Fykcritical points .

N= MÉ4h/kkxlm

-x) -hiopts

= %fµg-}'cat:c,

=

- hx' -1km

-X- h :O

=

_kr±kFÉ¥_1° <

H.ee am

HIN fix = KHS of 4)É

-2k dx = him -4.1×-1-1 )

✗solution It

✗CH Y %. stable :

01¥÷•÷És>↳ > s > > •

no matterwhat 7-'C' is

the solut!- will finally

↳*

convey tothe witted

to

point .

whenF-°

A

⇒µµµ,⇒*N is stable

.

②✗e NI

µ-✗ so,

X-H >°

¥ = klN-×t)

⇒pits

> 0⇒

so

when f- 0 ,

⇒ ✗is}

① ✗ > Nj"*kHs, w-✗co, X-H"

e-

⇒ pasco ,⇒ ¥+10 -7×4

Harvesting a Logistic Population:

The autonomous di↵erential equation

(2.4)dx

dt= ax� bx2 � h

(with a, b, and h all positive) may be considered to describe a logistic population

with harvesting. It can be re-written in the form

(2.5)dx

dt= kx (M � x)� h

Example 2.4. Solve the di↵erential equation (2.5). Find the critical points of

(2.5).

4

1-1--4×1① critical points .

solutions of f-1×1=0

② critical points → equilibrium

solution of the ODE .

③ state

coast k>0

✗ o- H > 0

DX

It=kTÉH -1×-1--1)

µ-X o 20

µ ,H are

2Cbittcul points

⇒ RHS > 0

DX

µ ,-N >

H ⇒j+

we start from a pointstudy ✗

= H

✗o > ✗=H, however

the

①xd = Xo

< H

functionisincreasing

✗ o- H

< °

⇒ ×=Hisuotstaµ-to

>° It co

⇒ pitsco⇒ It

÷÷¥ > >''¥

⇒✗ I

⇒ *µ is

not state' ×÷ci >so If

÷

④ >✗a)=×

Example 2.5. The di↵erential equation

(2.6)dx

dt= x (4� x)� h

(with x in hundreds) models the harvesting of a logistic population with k = 1 and

limiting population M = 4 (hundred). For di↵erent harvesting level h, what arethe equilibrium solutions and critical points?

5

Éh=o ① h > 4 .

4 -h - o ② when

=) no real solution . ✗(of =XoC2

✗'-4×-14--0

② h < 4 , N>4 ⇒ base

-1×-25

Hiv = ÉÉ <0

③ h= 4⇒ ¥+0

=z t Fh ⇒N=H=z-

⇒ ✗ toµ : 2 -1 4Th ① ✗c) = Xo > 2 .

g. ⇒ µ, , ,,,, , ⇒ ×,

,

notstable

.

¥: mi⇒ ¥+0 ⇒ ✗ o

v

+"' "

. > > . > >✗ =L

•

xd N,

÷