Download MAT 274 HW 12 Hints

MAT 274 HW 12 Hints
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MAT 274 HW 12 Hints
1. Problem 15 on Page 395.
Solution. To find critical point (equilibrium) solution, set dx/dt = dy/dt = 0 and get
an algebraic system
x−y =0
5x − 3y − 2 = 0
First equation gives x = y. Plug it into the second equation 5y − 3y − 2 = 0. So
y = 1 and thus x = 1. We found the only critical point
(x0 , y0 ) = (1, 1).
To classify (x0 , y0 ) as to type and stability, we calculate the Jacobian matrix. Let
f (x, y) = x − y and g(x, y) = 5x − 3y − 2 so that the original system is of the standard
form
dx
= f (x, y)
dt
dy
= g(x, y)
dt
Then, the Jacobian matrix, by definition, is
!
fx fy
J(x, y) =
gx gy
Plug in the definitions of f and g to get
J(x, y) =
!
1 −1
.
5 −3
(Remark: J is a constant matrix, which is expected for linear systems. The Jacobian
matrix for nonlinear system is expected to depend on x, y.)
Now, evaluate J(x0 , y0 ) at the critical point (1,1) and get the same constant matrix
!
1 −1
J(1, 1) =
.
5 −3
Then, calculate the eigenvalues of the Jacobian matrix at (1, 1) to determine the
type and stability of (1, 1).
!
1−λ
−1
det
= λ2 + 2λ + 2 = 0
5
−3 − λ
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MAT 274 HW 12 Hints
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=⇒ λ1,2 = −1 ± i
The eigenvalues are complex with negative real part. Therefore, it is a spiral point
(by complex) and stable (by negative real part), i.e. a spiral sink.
2. Problem 17 on Page 395.
Solution. Follow the same steps as above. Set dx/dt = dy/dt = 0 and get an algebraic
system. Solve it to get the critical point (2.5,-0.5). Calculate the Jacobian matrix
and evaluate it at this critical point
!
1 −5
J(2.5, −0.5) =
.
1 −1
The eigenvalues are ±2i. It is purely imaginary, i.e. complex with zero real part. Since
the original system is linear, purely imaginary eigenvalues DO lead to a conclusion;
that is, (2.5, −0.5) is a center, and it is stable.
3. Problem 12 on Page 410.
Solution. A framework for solving this type of problem, i.e., to determine the type
and stability of equilibria/critical points for
dx
= f (x, y)
dt
dy
= g(x, y)
dt
is to start with finding all equilibria/critical points by setting dx/dt = dy/dt = 0 and
solving the resulting algebraic system.
Then, deal with each critical point one by one. It is useful to calculate the Jacobian
matrix first
!
fx fy
.
J(x, y) =
gx gy
Note that for nonlinear system, expect J to depend on x, y, not a constant matrix
any more!
For a given critical point (x0 , y0 ), evaluate the Jacobian matrix J(x0 , y0 ) so that the
original nonlinear system has a linear approximation near (x0 , y0 )
dx
= fx (x − x0 ) + fy (y − y0 )
dt
dy
= gx (x − x0 ) + gy (y − y0 )
dt
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MAT 274 HW 12 Hints
Upon substitution u = x − x0 , v = y − y0 , it becomes
du
= fx u + fy v
dt
i.e.
dv
= gx u + gy v
dt
!
!
d u
u
= J(x0 , y0 )
.
dt v
v
Now, calculate the eigenvalues λ1 , λ2 of J(x0 , y0 ). Then, always relate positive real
part to unstable, and negative real part to stable. Saddle point is unstable because
one positive real part is enough to destroy the stability.
• Real eigenvalues. 3 cases.
(1) both postive: nodal souce, unstable.
(2) both negative: nodal sink, stable.
(3) opposite signs: saddle point, unstable.
• Complex eigenvalues. 3 cases.
(1) postive real part: spiral souce, unstable.
(2) negative real part: spiral sink, stable.
(3) zero real part: inconclusive for a genuinely nonlinear system; center and
stable for a linear system.
Now, let’s look at Problem 12 on Page 410. The equilibria are all given in the
problems. For Problem 12, we look at (x0 , y0 ) = (5, 0). The Jacobian matrix is
!
!
∂
∂
2 − xy),
2 − xy)
(5x
−
x
(5x
−
x
fx fy
∂y
J(x, y) =
= ∂x ∂
.
∂
gx gy
(−2y
+
xy),
∂x
∂y (−2y + xy)
that is
J(x, y) =
5 − 2x − y,
−x
y,
−2 + x
!
Plug in the critical point (5,0) for (x, y)
J(5, 0) =
!
−5, −5
.
0,
3
In other words, the linear approximation of the original system near (5, 0) should be
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MAT 274 HW 12 Hints
du
= −5u − 5v
dt
dv
= 0u + 3v.
dt
The characteristic equation
"
det(λI − J(5, 0)) = det
λ 0
0 λ
!
−
−5, −5
0,
3
!#
= (λ + 5)(λ − 3) = 0
Thus, eigenvalues λ1 = −5, λ2 = 3. Real and opposite signs. So,
conclusion: the critical point at (5,0) is a saddle point, and it is unstable.
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