Find the Equilibrium Solution for Equilibrium Xe1
11/12/2016 1
EEE 285 Applied Differential Equations
Chapter 3: Equilibrium Solution and Stability of First Order Differential Equations Prof. Dr. AhmetUçar
© Dr. Ahmet UçarEEE 285 Chapter 31
Equilibrium Solution and Stability of First Order Differential Equations
© Dr. Ahmet UçarEEE 285 Chapter 32
In previous chapter the methods are given to find all solutions of the differential equations in the following forms;
) , (
y t f dt dy
(1) to answer specific numerical questions such as the general solution (time response)
y
c
(
t
)
. When a given differential equation is difficult or impossible to solve explicitly then it is important to extract
qualitative
information about general properties of its solutions.
) (
t f dt dy
) (
y f dt dy
(2) (3)
11/12/2016 2
Equilibrium Solution and Stability of First Order Differential Equations
© Dr. Ahmet UçarEEE 285 Chapter 33
Consider a given differential equations in the following form
0
) 0 ( ); , (
y y y t f dt dy
The time response of the particular solution
y
p
(
t
)
obtained from general solution
y
c
(
t
)
may grow without bound as
t
, or approaches to a finite limit, or is a periodic function of
t
. (1)
y
p
(
t
) 0
t
(
time
)
y
(
0
)
a
)
The time response
y
p
(
t
)
grows without bound as
t
.
b
)
The time response
y
p
(
t
)
approaches a finite limit as
t
.
y
p
(
t
) 0
t
(
time
)
y
(
0
)
c
)
The time response
y
p
(
t
)
is a periodic function of
t.
y
p
(
t
) 0
t
(
time
)
y
(
0
)
In this chapter we introduce some of the more important
qualitative
questions that can sometimes be answered for equations that are difficult or impossible to solve.
Equilibrium Solution and Stability of First Order Differential Equations
© Dr. Ahmet UçarEEE 285 Chapter 34
Example 3.1:
Let start with the simple first order differential equations given in Eq.(1) that can be solved explicitly which represents the temperature of a body. (1) where
y
(
t
)
denote the temperature of a body with initial temperature
y
(0)
=
y
0
and
A
is constant and representsthe temperature of the body when it is in
thermal equilibrium
with the surrounding medium.The general solution is
0
) 0 ( ), 0 ( ), (
y y k A y k dt dy
0 , ) ( , ) ( ) ln( ) ( ) (
1 1
1
t Ce A t y e C Ce A y e A y C kt A y kdt A y dy kdt A y dy
kt c C kt C kt
The constant
C
for particular solution of above initial value problem is
A y C Ce A y y
k
0 ) 0 ( 0
) 0 (
0 , ) ( ) (
0
t e A y A t y
kt p
Substituting
C
in the general solution leads to following particular solution
y
p
(
t
)
;
11/12/2016 3
Equilibrium Solution and Stability of First Order Differential Equations
© Dr. Ahmet UçarEEE 285 Chapter 35
0 , ) ( ) (
0
t e A y A t y
kt p
Example 3.1:
From the particular solution;
y
eq
=A y
01
y
02
y
p
(
t
)
y
03
y
04
=0
t
it follows immediately that
A t y
p t
) ( lim
so the temperature of the body approaches that of the surrounding medium. Thus the constant function
y
p
(
t
)
A
is a solution of Eq. (1) and corresponds to the temperature of the body when it is in
thermal equilibrium
with the surrounding medium.
Remark:
For different body initial temperature
y
(0)
=
y
0
the temperature of the body approaches
y
p
(
t
)
reach to a certain limit which is
y
p
(
t
)
A
and is called
equilibrium point
. If the initial temperature start at
y
(0)
=
A
the temperature of body stay there for ever. This particular solution is called
the equilibrium solution
,
y
p
(
t
)
A
.
Equilibrium Solution and Stability of First Order Differential Equations
© Dr. Ahmet UçarEEE 285 Chapter 36
Example 3.1:
The particular solutions of
linear
first order differential equations in Eq.(1) are shown in Figure 1 for different initial conditions
.
However plotting the vector field
dy
/
dt
as a function of
y
itselfcontains many dynamic properties of the given first order differential equation and is called
the phase diagram
(vector field). (1)
0
) 0 ( ), 0 ( ), ( ) (
y y k A y k y f dt dy
y
eq
=A y
01
y
02
y
p
(
t
)
y
03
y
04
=0
t
Figure 1:
Time response
f
(
y
) = -
k
(
y
-
A
)
y
eq
=
A y y
2
(0)
y
3
(0)
Figure 2:
Phase diagram
0
The differential in Eq. (1) has the
phase diagram
shown in Figure 2 where arrows on the
y
axis indicates the corresponding velocity vector for each
y
(0)
. The
velocity arrows
point to the
right
when
velocity is positive
f
(
y
)
=
y
'>0 and to the
left
when
velocity is negative
f
(
y
)
=
y
'<0. At the
equilibrium point
f
(
y
)
=
y
'=
A
there is no flow and has the constant solution
y
(
t
)
A
.The particular solution for
y
(
0
) =
A
is the
equilibrium solution
and is
y
p
(
t
)
A
.
Find the Equilibrium Solution for Equilibrium Xe1
Source: https://www.scribd.com/document/352827874/Chapter-3-Equilibrium-Solution-and-Stability-of-FO-DE-pdf