Download LECTURE 14: LINEAR SYSTEMS AND EQUILIBRIUM SOLUTIONS

LECTURE 14: LINEAR SYSTEMS AND EQUILIBRIUM SOLUTIONS
0.1. Linear Systems. A linear system of differential equations is of the form:
dx
= ax + by
dt
dy
= cx + dy
dt
~ = [x, y], is given by:
The matrix form of this equation, if Y
~
dY
ax + by
=
cx + dy
dt
Recall that the following equation from matrix algebra:
x
ax + by
a b
=
y
cx + dy
c d
| {z } | {z }
A
~
Y
The linear system may therefore be written in the form:
~
dY
~
= A·Y
dt
Example: Write the following system using the matrix notation described above.
dx
= −2x + 1y
dt
√
dy
2x + 6y
=
dt
~ = 0.
0.2. Equilibrium Solutions: To find equilibrium solutions we need only solve the system A· Y
Or, we need to solve the system of equations:
(1)
ax + by = 0
cx + dy = 0
Certainly, ~0 = [0, 0] is a solution. What are the other equilibrium solutions? We just solve the
system. Assume a 6= 0. Multiply the top equation by −c and the bottom equation by a.
−acx − bcy
acx + ady
=
0
=
0
Adding gives us (ad − bc)y = 0. Thus, if (ad − bc) 6= 0, we have that x = 0, y = 0 is the only solution.
If ad − bc = 0, We can let y be anything we like. We get infinitely many solutions. The case a = 0
is a HW problem.
a b
Definition: The determinant of A =
is given by det(A) = ad − bc.
c d
Theorem 1. Let A be a 2 × 2 matrix as above. Then det(A) 6= 0 if and only if the system of
equations (1) has a unique solution. This is also true for n × n matrices with the corresponding
~ /dt = A · Y~ has only the origin as its
system of equations. Moreover, det(A) 6= 0 if and only if dY
equilibrium solution.
1
2
LECTURE 14: LINEAR SYSTEMS AND EQUILIBRIUM SOLUTIONS
Example: How many equilibrium solutions do the following equations have:
~
~
~
dY
dY
dY
2 2
1 2
0 0
x
x
x
=
=
=
,
,
,
1 1
3 4
0 0
y
y
y
dt
dt
dt
0.3. Linearity Principle. The linear combination of two solutions is also a solution.
~1 (t) and Y~2 (t) are solutions to dY
~ /dt = AY
~ , then αY
~1 (t)+β Y~2 (t) is also a solution.
Theorem 2. If Y
Proof. It is always true that A(B + C) = AB + AC, for any 2 × 2 matrices A, B, C. It is also true
that A(kB) = kAB for any matrices A, B, C. Also, the derivative is linear. Hence:
d~
d~
d ~
Y2 (t)
(αY1 (t) + β Y~2 (t)) = α Y
1 (t) + β
dt
dt
dt
~1 (t) + βAY
~2 (t)
= αAY
~1 (t) + β Y~2 (t))
= A(αY
Example: Find infinitely many solutions to the partially decoupled system given below:
~
dY
1 4
x
=
,
0 3
y
dt