MAS 201 - Applied DE
D. K. etc
ii
Contents
11 System of Nonlinear Diff. Equation
11.1 Autonomous System . . . . . . . . . . . .
11.2 Stability of Linear System . . . . . . . . .
11.3 Nonlinear system-Linearization . . . . . .
11.4 Autonomous system as Math. Model . . .
11.4.1 Nonlinear Oscillation: Sliding bead
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157
157
161
169
174
176
iv
CONTENTS
156
CONTENTS
Chapter 11
System of Nonlinear Diff.
Equation
11.1
Autonomous System
Autonomous System
The DE. of the form
dx1
dt
dx2
dt
= g1 (x1 , x2 , · · · , xn )
= g2 (x1 , x2 , · · · , xn )
..
.
dxn
dt
= gn (x1 , x2 , · · · , xn )
..
.
(11.1)
is called autonomous. Notice that the equation does not have t explicitly.
Second Order DE as a System
A second order autonomous DE can be written as a system of first order autonomous DE. For example, given
x′′ = F (x, x′ ),
we let
dx
dt
= y. Then y ′ = x′′ and hence
dx
dt
dy
dt
= y
= F (x, y).
This is a system of first order autonomous system in x, y.
157
(11.2)
158
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
θ
ℓ
mℓθ′′
mg sin θ
mg
Figure 11.1: Undamped pendulum
Example 11.1.1. [Undamped pendulum]
Consider the movement of a free(no external force except gravity) undamped(ignoring friction) pendulum in section 6.
mℓθ ′′ + mg sin θ = 0.
(11.3)
Dividing by mℓ we get
θ ′′ + k sin θ = 0,
Sol.
k=
g
.
ℓ
(11.4)
This is an autonomous system. Let x = θ, y = x′ . Then
x′ = y
y ′ = −k sin x.
In vector form, we have X′ (t) = g(X) where




x1 (t)
g1 (x1 , · · · , xn )




..
X′ (t) =  ...  , g(X) = 
.
.
xn (t)
gn (x1 , · · · , xn )
When the variable t is interpreted as time, the DE is called a dynamical system
and the solution X(t) is called the state of the system or response of the
system. The solution is also called path or trajectory.
159
11.1. AUTONOMOUS SYSTEM
Vector Field interpretation
Consider an autonomous system of DE when n = 2.
dx
dt
dy
dt
= P (x, y)
= Q(x, y).
This is called a plane autonomous system. The vector V(x, y) = (P (x, y), Q(x, y))
defines a vector fields in the (x, y)-plane. If X(t) = (x(t), y(t)) denotes the position of the particle at time t, X′ (t) = (x′ (t), y ′ (t)) is the velocity vector.
Type of Solutions
Consider solution of the plane autonomous system satisfying x(0) = x0 . (The
solution exists uniquely).
(1) The constant solution x(t) = x0 = (x, y) for all t. Then we have
P (x, y) = 0
Q(x, y) = 0.
The point x0 is also called critical point, stationary point or equilibrium solution. The solution cannot self intersect(by uniqueness).
(2) A solution (x(t), y(t)) is called an arc if it does not cross itself.
(3) A periodic solution (x(t), y(t)) is called a cycle. If p is a periodic, then
(x(t + p), y(t + p)) = (x(t), y(t)).
Example 11.1.2 (Critical points).
(2)
(1)
x′ = −x + y
y′ = x − y
x′ = x2 + y 2 − 6
y ′ = x2 − y
x′ = 0.01x(100 − x − y)
y ′ = 0.05y(60 − y − 0.2x)
√
Sol. (1) (0, 0). (2) (± 2, 2). (3) (100, 0) and (50, 50).
(3)
Consider
Example 11.1.3 (Periodic solutions-Check the fig. in the book).
x′ = 2x + 8y
y ′ = −x − 2y
(1)
160
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
(2)
x′ = x + 2y
y ′ = − 12 x + y
In each case sketch graph when x(0) = (2, 0).
Sol. (1). In section 10, we have seen the solution as
x = c1 (2 cos 2t − 2 sin 2t) + c2 (2 cos 2t + 2 sin 2t)
y = c1 (− cos 2t − 2) − c2 sin 2t.
with IC, we get
x = 2 cos 2t + 2 sin 2t, y = − sin 2t.
These are clearly periodic. For (2) we see by eigenvector method,
x = c1 (2et cos t) + c2 (2et sin t), y = c1 (−et sin t) + c2 (et cos t).
There is no periodic (cycle) solutions, but the solution satisfying IC is
x = 2et cos t, y = −et sin t.
See figure in the book.
Changing to Polar coordinates
dr
1
=
dt
r
dx
dy
x
+y
dt
dt
,
dθ
1
= 2
dt
r
dx
dy
−y
+x
dt
dt
(11.5)
Example 11.1.4. Solve
p
x x2 + y 2
x′ = −y −p
y ′ = x − y x2 + y 2
(11.6)
with x(0) = (3, 3).
Sol. Change to polar corrdinate using (11.5) we obtain
dr
dt
dθ
dt
=
=
1
2
r [x(−y − xr) + y(x − yr)] = −r
1
[−y(−y − xr) + x(x − yr)] = 1.
r2
(11.7)
This is an uncoupled system which√can be solved by separation of variables. The
IC. by the polar coordi. is r(0) = 2, θ(0) = π/4. Hence
r=
1
,
t + c1
θ = t + c2 .
11.2. STABILITY OF LINEAR SYSTEM
c1 =
√
2/6,
r=
1
√
.
θ + 2/6 − π/4
161
θ = π/4.
Eliminating we get spiral
Example 11.1.5.
dr
dt
dθ
dt
= 0.5(3 − r)
= 1.
(11.8)
r = 3 + c1 e−0.5t , θ = t + c2 .
With IC. x(0) = (0, 1), we get c1 = −2, c2 = π/2. The solution is the spiral
r = 3 − 2e−0.5(θ−π/2) . As θ → ∞ the path approaches a circle.
Exercise 11.1.6. Write the following nonlinear second order DE into a system
autonomous equations. Find all critical points.
(1) y ′′ + 4y = 0
(4) yy ′′ + 4(y ′ )2 = 0
(2) y ′′ = 1 + y 2
(5) y ′′ + 3y = 0
(3) y ′′ − y ′ = 0
(6) y ′′ = (y ′ )2
11.2
Stability of Linear System
We again consider a plane autonomous DE.
dx
dt
dy
dt
= P (x, y)
= Q(x, y).
(11.9)
Assuming the initial position X0 is close to a critical point, we are interested in
the behavior of the solutions.(That is, the stability of the critical points of these
system.) We may interpret the solution X(t) as the path of a particle starting at
X0 .
Fundamental questions-See Figure
Suppose X1 is a critical point of a plane autonomous system and X(t) is a solution
of the system satisfying X(0) = X0 .(close to X1 )
(1) Will the particle return to the critical point? i.e., limt→∞ X(t) = X0 ?
stable.
(2) Or the particle remain close to the critical point or moves away ? stable,
unstable.
162
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
Stability Analysis
For the stability of the generally nonlinear system (11.9), we first study the
stability of the linear system.
dx
= ax + by
dt
,
(11.10)
x′ = Ax or
dy
= cx + dy
dt
where ∆ := det(A) 6= 0 so that the only critical point is 0. With ∆ = ad − bc and
trace τ = a + d, its charact. equation is
λ2 − τ λ + ∆ = 0.
The behavior depends on the eigenvalues.
Example 11.2.1. Consider
dx
dt
dy
dt
= −x + y
(11.11)
= cx − y
−1 − λ
1 = 0.
c
−1 − λ
p
√
−2 ± 4 − 4(1 − c)
λ=
= −1 ± c.
2
The behavior of solution depend on the value of c. (See the phase portrait in
phase plane)
(1) 0 ≤ c < 1: both λ are negative, so the solutions approach the origin.
(2) c = 0: multiple solution, both solutions approach the origin.
(3) 1 < c: λ1 > 0, λ2 < 0, one solution approach the origin the other diverges.
(4) c < 0: both λ are complex, solution will approach the origin like a spiral.
Case I: Real and distinct eigenvalues τ 2 − 4∆ > 0.
Consider
dx
dt
dy
dt
= ax + by
= cx + dy.
(11.12)
The solution of (11.13) is given by the following form:
x(t) = c1 ξeλ1 t + c2 ηeλ2 t .
This is again classified as follows:
(11.13)
163
11.2. STABILITY OF LINEAR SYSTEM
y
Y
x
X
Figure 11.2: Example 11.2.2, Saddle
• Stable node if both λ1 , λ2 are negative (τ 2 − 4∆ > 0, τ < 0, and ∆ > 0)
• Unstable node if both λ1 , λ2 are positive (τ 2 − 4∆ > 0, τ > 0, and ∆ > 0)
• Saddle if λ1 λ2 < 0 (τ 2 − 4∆ > 0, and ∆ < 0) Saddle is unstable.
Example 11.2.2 (Real and distinct : different sign). Investigate the behavior
of critical point (0, 0).
′
x = x + 2y
(11.14)
y ′ = 3x + 2y
Sol.
1 − λ
2 = λ2 − 3λ − 4 = 0 ⇒ λ = 4, −1.
3
2 − λ
For λ = 4, ξ = (2, 3)T , and for λ = −1, η = (1, −1)T . Thus the solution
is
x
2 4t
1
= c1
e + c2
e−t
y
3
−1
(1) If c1 = 0 we see y = −x.
(2) If c2 = 0 we see y = 32 x.
(3) If c1 6= 0, c2 6= 0 (1, −1) direction is X-axis, (2, 3) direction is
Y -axis. Then we have Y = Xc4 , a hyperbola.
It is a saddle.
164
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
Example 11.2.3 (Real and distinct : same sign). Investigate the stability of
the critical point of DE
′
x = −2x + y
(11.15)
y ′ = x − 2y
y
X
Y
x
Figure 11.3: Example 11.2.3, Node
Sol. The critical point is (0, 0). Charac. eq. is
−2 − λ
1 = 0 ⇒ λ2 + 4λ + 3 = 0, λ = −1, −3.
1
−2 − λ
and eigenvectors are
ξ1 =
1
,
1
ξ2 =
1
.
−1
The general solution is
1 −t
1
x
e + c2
e−3t .
= c1
1
−1
y
Depending on the values of c1 , c2 we have
(1) c1 = 0 : Then y = x.
11.2. STABILITY OF LINEAR SYSTEM
165
(2) c2 = 0 : Then y = −x. Both case the solution graph in phase plane is
straight lines.
(3) c1 6= 0, c2 6= 0. Rotate the axis, and assume (1, 1) direction is X-axis,
(1, −1) direction is Y -axis then it holds that Y = cX 3 . So it is a (Stable)
node.
Repeated real eigenvalues (τ 2 − 4∆ = 0)
Degenerate nodes:
(1) Two linearly independent eigenvectors
X(t) = c1 ξeλ1 t + c2 ηeλ1 t
If λ1 < 0 then stable, otherwise unstable.
(2) Single linearly independent eigenvector
X(t) = c1 ξeλ1 t + c2 (ξteλ1 t + ηeλ1 t )
where (A − λ1 I)η = ξ. If λ1 < 0 then stable, otherwise unstable.
Example 11.2.4.
x′ = x − y
y ′ = x + 3y
(11.16)
λ = 2 and ξ = (1, −1) is the only eigenvector. With η = (−1, 0), the solution is
X(t) = c1
1
1
−1 2t
2t
2t
e + c2
te +
e .
−1
−1
0
This is unstable node.
Complex Eigenvalues (τ 2 − 4∆ < 0)
Example 11.2.5.
x′ = αx + βy
y ′ = −βx + αy (α, β real, β > 0)
(11.17)
166
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
y
x
Figure 11.4: spiral
Sol.
Eigenvalues are α ± iβ. Use substitution r 2 = x2 + y 2 , tan θ = xy . So
y ′ x − yx′
.
x2
(11.18)
rr ′ = x(αx + βy) + y(−βx + αy) = αr 2 .
(11.19)
rr ′ = xx′ + yy ′ , (sec2 θ)θ ′ =
Substitute into (11.17) we obtain
So,
r = Ceαt .
Also from the second equation of (11.18) we obtain
Here sec2 θ =
r2
x2
(11.20)
r = Ceαt
θ = −βt + θ0 .
(11.21)
, θ ′ = −β.
Hence this is a spiral.
If α = 0 we have
It is a center.
βr 2
.
x2
(sec2 θ)θ ′ = −
r = C
θ = −βt + θ0 .
167
11.2. STABILITY OF LINEAR SYSTEM
y
µ>0
x
Figure 11.5: center
More generally, when the eigenvalues are α ± iβ with corresponding eigenvectors a1 , a2 , then
x1 (t) = (a1 cos βt − a2 sin βt)eαt ,
x2 (t) = (a2 cos βt + a1 sin βt)eαt .
x(t) = (c11 cos βt + c12 sin βt)eαt ,
y(t) = (c21 cos βt + c22 sin βt)eαt .
So
(1) α = 0. Pure imaginary, center
(2) α < 0. Stable, spiral
(3) α > 0. Unstable, spiral
Stability: Linear case
168
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
Roots of Char. eq.
r1 > r2 > 0
r1 < r2 < 0
r1 · r2 < 0
r1 = r2 < 0
r1 = r2 > 0
α ± iβ, α > 0
α ± iβ, α < 0
±iβ
Critical point(Linear)
node
node
saddle
node
node
spiral
spiral
center
Stability (Linear)
unstable
stable, attr.
unstable
stable, attr.
unstable
unstable
stable, attr.
stable
∆
spiral
spiral
b
τ 2 = 4∆
b
b
center
node
node
τ
saddle
Figure 11.6: Classification-linear case
Classification of Critical Points- Linear Case
x′ = ax + by
y ′ = cx + dy.
(11.22)
Its charact. equation is
λ2 − (a + d)λ + ad − bc = 0.
Let τ = a + d, and the determinant ∆ be ad − bc. We classify the critical points
according to τ 2 − 4∆:
11.3. NONLINEAR SYSTEM-LINEARIZATION
169
(1) Real roots of same sign ∆ > 0, τ 2 − 4∆ ≥ 0 : node
(2) Real roots of opposite sign ∆ < 0 : saddle
(3) Complex roots τ 6= 0, τ 2 − 4∆ < 0 : spiral
(4) Pure imaginary roots τ = 0,
11.3
∆ > 0 : center
Nonlinear system-Linearization
First consider 2nd order nonlinear autonomous differential equation.
y ′′ = F (y, y ′ ).
We can write it as a system of first order equations:
dy
dt = v
dv
dt = F (y, v).
Generally, autonomous nonlinear system is given as
dx
= F (x, y)
dt
dy
dt = G(x, y).
(11.23)
(11.24)
(11.25)
The zeros of F (x, y) = G(x, y) = 0 are called critical points.
Definition 11.3.1. Let x1 be a critical point of a nonlinear system. It is called
a stable critical point if for any radius ρ > 0 there exists a radius r > 0 such
that if the initial position satisfies |x0 − x1 | < r then the corresponding solution
x(t) satisfies |x(t) − x1 | < ρ for all t > 0. If, in addition, the solution satisfies
limt→∞ x(t) = x1 whenever |x0 − x1 | < r then xl is called an asymptotically
stable critical point.
Otherwise, a critical point x1 is called an unstable critical point.
Example 11.3.2. Find the critical point of the following and investigate the
stability.
p
x′ = −y −p
x x2 + y 2
(11.26)
y ′ = x − y x2 + y 2
Sol. We can easily see (0, 0) is the only critical point. We have seen in early
example, that
dr
dt
dθ
dt
=
=
1
2
r [x(−y − xr) + y(x − yr)] = −r
1
[−y(−y − xr) + x(x − yr)] = 1.
r2
(11.27)
170
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
The solution is
r=
1
,
t + c1
θ = t + c2 .
Stable spiral.
Example 11.3.3.
dr
dt
dθ
dt
= 0.05(3 − r)
= −1.
(11.28)
Show that (0, 0) is an unstable critical point. Solving the system directly in
terms of r and θ,
1
r=
e−0.15t .
3 + c0
With IC, r(0) = r0 , we get c0 = (3 − r0 )/r0 . As t → ∞ we see r(t) → 0.
r = 3 − 2e−0.5(θ−π/2) . As θ → ∞ the path approaches a circle of radius 3. Hence
the circle is stable.(limit cycle)
Exercise 11.3.4.
′
x =
(a)
y′ =
′
x =
(b)
y′ =
′
x =
(c)
y′ =
′
x =
(d)
y′ =
(1) Classify the critical point.
x+y
4x + y
x − 3y
x+y
x+y
−2x − y
x − 3y
2x − 3y
Linearization
Consider the following nonlinear system of DE.
′
x = P (x, y)
, or x′ = g(x)
y ′ = Q(x, y)
(11.29)
where P (x, y), Q(x, y) are C 2 -functions. Assume xl = (x0 , y0 ) is a critical point.
We linearize this using the Taylor expansion at (x0 , y0 ).
P (x, y) = P (x0 , y0 ) + Px (x0 , y0 )(x − x0 ) + Py (x0 , y0 )(y − y0 ) + PR (x, y)
Q(x, y) = Q(x0 , y0 ) + Qx (x0 , y0 )(x − x0 ) + Qy (x0 , y0 )(y − y0 ) + QR (x, y)
171
11.3. NONLINEAR SYSTEM-LINEARIZATION
We use up to linear terms. Since P, Q are C 2 , it is known that
p
p
|PR |/ (x − x0 )2 + (y − y0 )2 , |QR |/ (x − x0 )2 + (y − y0 )2 → 0.
For simplicity assume xl = (x0 , y0 ). Then the vector form of the system of
equation is
x′ = g(x) = x1 + A(x − x1 ) + o(kxk),
where A is the Jacobian matrix
A=
Px Py
Qx Qy
(x0 ,y0 )
The system x′ = Ax is called the linearization of (11.29).
Theorem 11.3.5. Assume x1 is a critical point of the plane autonomous system
x′ = g(x).
(1) If the eigenvalues of A = g′ (x1 ) has negative real part, then x1 is an asymptotically stable critical point.
(2) If A = g′ (x1 ) has an eigenvalue with positive real part, then x1 is an unstable critical point.
Theorem 11.3.1. Stability: comparison between the linearized system and
nonlinear system
char. value
r1 > r2 > 0
r1 < r2 < 0
r1 · r2 < 0
r1 = r2 < 0
r1 = r2 > 0
α ± iβ, α > 0
α ± iβ, α < 0
±iβ
point (linear)
node
node
saddle
node
node
spiral
spiral
center
Stab.(linear)
unstable
stable, attr.
unstable
stable, attr.
unstable
unstable
stable, attr.
stable
point(nonlin)
node
node
saddle
node, spiral
node, spiral
spiral
spiral
center, spiral
Example 11.3.6. Classify the critical points of
′
x = x2 + y 2 − 6
y ′ = x2 − y
√
√
The critical points are ( 2, 2) and (− 2, 2).
2x 2y
′
g (x) =
2x −1
Stab.(nonlin)
unstable
stable, attr.
unstable
stable, attr.
unstable
unstable
stable, attr.
indeterm.
172
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
and
√
√
2√2 4
−2√2 4
A1 =
, A2 =
2 2 −1
−2 2 −1
√
√
√
A1 . τ 2 − 4∆ = (2 2 − 1)2 − 4(−2 2 − 8 2) > 0, τ > 0. So A1 has positive real
eigenvalue, so unstable.√
√
√
A2 . τ 2 − 4∆ = (−2 2 − 1)2 − 4(2 2 − 8 2) > 0, τ < 0. So both eigenvalues
have negative real, so stable.
Example 11.3.7. Linearize
x′ = F (x, y) = x + x2 + 2y 2
y ′ = G(x, y) = 2x + y + xy
at a critical point (0, 0)
Sol.
(0, 0) is a critical point. Differentiate F, G, we get x′ = Ax where
1 0
1 + 2x
4y
A = DF (0, 0) =
=
2 1
2 + y 1 + x (0,0)
Example 11.3.8 (Soft Spring). mx′′ + kx + k1 x3 , k > 0, k1 < 0. m = 1. The
system can be written as
′
x = y
y ′ = x3 − x.
Find and classify critical points.
Sol.
x(x2 −1) = 0. so the criticals are (0, 0), (1, 0) and (−1, 0). Differentiating
g, we get x′ = Ax where
0 1
0 1
A1 = Dg(0, 0) =
, A2 = Dg(1, 0) = Dg(−1, 0) =
−1 0
2 0
The eigenvalues of A1 are ±i.
√ So we are not sure about the stability.
The eigenvalues of A2 are ± 2. So saddle.
11.4. AUTONOMOUS SYSTEM AS MATH. MODEL
173
The phase plane method
When the initial position is NOT close to a critical point, we use different method:
From
dy
Q(x, y)
=
dx
P (x, y)
we try to understand the relation between x and y.
Example 11.3.9.
x′ = y 2
y ′ = x2
The critical point is (0, 0). Since the determinant of linearization
g′ =
0 2y
2x 0
is zero, the nature of the critical point is in doubt. Solving
dy
x2
= 2,
dx
y
√
we get y 3 = x3 + c or y = 3 x3 + c. If X(0) = (0, y0 ) then y 3 = x3 + y03 .
From the figure, we conclude it is unstable.
Example 11.3.10 (Phase plane analysis of Soft Spring). mx′′ + x − x3 . x′ = y,
y ′ = (x3 − x)/m.
dy
x3 − x
=
, m = 1.
(11.30)
dx
my
Separation of var.
y2
x4 x2
=
−
+ c.
2
4
2
y 2 = (x2 − 1)2 /2 + c0 .
If X(0) = (x0 , 0) then
0 = (x20 − 1)2 /2 + c0 ⇒ c0 = −(x20 − 1)2 /2
So
y2 =
(x2 − 1)2 (x20 − 1)2
−
.
2
2
174
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
θ̇
−3π
π
−π
3πθ
Figure 11.7: Phase plane of Pendulum
11.4
Autonomous system as Math. Model
Example 11.4.1. [Pendulum- no friction] Recall the movement of the Pendulum in section 1.
θ ′′ +
g
sin θ = 0.
ℓ
Let x = θ, y = x′ . Then the movement of the Pendulum is described by
x′ = y
y ′ = − gℓ sin x.
(11.31)
11.4. AUTONOMOUS SYSTEM AS MATH. MODEL
Sol.
175
The critical points are (kπ, 0), k = ±1, ±2, · · · . Linearizing at (0, 0) gives
0
1
0 1
A=
.
=
− gℓ cos x 0 (0,0)
− gℓ 0
The char. value is pure imaginary, so it is a center. By theorem 11.3.1
the original point is either center or spiral. Also, it is periodic in θ, the
system has similar behavior at (2nπ, 0), n = 1, 2, 3, · · · . Now consider
the critical points ((2n + 1)π, 0), n = ±1, ±2, · · · .
The linearized equation is
0
1
0 1
A=
= g
− gℓ cos x 0 ((2n+1)π,0)
ℓ 0
Its char. value are distinct real. Hence critical points are saddle. Original nonlinear system is also saddle.
Example 11.4.2. [Pendulum- with friction]
176
Sol.
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
We may assume the friction is proportional to the velocity hence friction
is cℓθ ′ . Hence we have
mℓ2 θ ′′ + cℓθ ′ + mgℓ sin θ = 0.
Dividing by mℓ2
θ ′′ + aθ ′ + b sin θ = 0,
a=
c
,
mℓ
Let x = θ, y = θ ′ . Then we get
′
x = y
y ′ = −b sin x − ay.
b=
g
.
ℓ
(11.32)
Its critical points are (nπ, 0), n = ±1, ±2, · · · . Linearizing at (0, 0) we
get
0
1
x.
x′ = Ax =
−b −a
The char. values are r1 , r2 =
we have
√
−a± a2 −4b
.
2
According to location of (a, b)
(1) a2 − 4b > 0: char. values are distinct negative real: So the critical
points are stable and node. Same for original.
(2) a2 − 4b = 0: char. values are double negative real:So the critical
points are stable and node. The original has node or spiral(stable).
(3) a2 − 4b < 0: char. values are complex with negative real part. So
the critical points are stable. Same for original system.
At (2mπ, 0), m = 1, 2, 3, · · · we can show the same behavior. Now look
0 1
at ((2m − 1)π, 0), m = ±1, ±2, · · · . Linearizing, we get A =
.
b −a
In this case the char. values are
for original system.
11.4.1
√
−a± a2 +4b
.
2
In this case, saddle. Same
Nonlinear Oscillation: Sliding bead
Suppose a bead is sliding along a wire forming a curve described by the function
z = f (x).
11.4. AUTONOMOUS SYSTEM AS MATH. MODEL
177
z = f (x)
mg sin θ
b
P (x, y)
θ
w = mg
Figure 11.8: Forces on sliding bead
Example 11.4.3. [ Sliding bead] The tangential force due to weight W = mg
is mg sin θ cos θ. Note that tan θ = f ′ (x). Hence
Fx = −mg sin θ cos θ = −mg
f ′ (x)
.
1 + [f ′ (x)]2
Assume a damping force −βx′ . Then the movement of the sliding bead is describe
by
f ′ (x)
mx′′ = −mg
− βx′ .
(11.33)
1 + [f ′ (x)]2
Hence we have
(
x′ = y
f ′ (x)
y ′ = −g 1+[f
′ (x)]2 −
β
m y.
The critical points (x1 , y1 ) satisfy y1 = 0, f ′ (x1 ) = 0.
g′ (xℓ ) =
0
1
−gf ′′ (x1 ) −β/m
So τ = −β/m, ∆ = gf ′′ (x1 ), τ 2 − 4∆ = β 2 /m2 − 4gf ′′ (x1 ).
(1) f ′′ (x1 ) < 0 rel. max. as point on the graph of f (x) and unstable.
(2) f ′′ (x1 ) > 0 and β > 0. rel. min. If β 2 /m2 − 4gf ′′ (x1 ) > 0, overdamped and
stable node. If β 2 /m2 − 4gf ′′ (x1 ) < 0, underdamped and stable sprial.
(3) f ′′ (x1 ) > 0 and β = 0. undamped. osc.
178
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
Lotka-Volterra Predator prey Model
One species(predator) feeds on a second species(prey) : Example: Assume snakes
eat frogs and only frogs. If x is the number of predator and y is the number of
prey, then
x′ = −ax + bxy = x(−a + by)
y ′ = −cxy + dy = y(−cx + d),
where a, b, c, d are positive constants. Note that
(1) If there were no predator(x = 0) the prey y ′ = by would grow exponentially.
(2) If there were no prey(y = 0) the predator (x′ = −ax) would extinct.
(3) The term −cxy is the death rate by predation(Posik).
Critical points are (0, 0) and (d/c, a/b). Jacobians are
−a 0
0
bd/c
′
′
A2 = g ((d/c, a/b)) =
A1 = g ((0, 0)) =
0 d
−ac/b
0
The critical point (0, 0) is a saddle.
√
Now the critical point (d/c, a/b). A2 has pure imaginary eigenvalues ± adi.
Need more investigation.
dy
y(−cx + d)
=
.
dx
x(−a + by)
Thus
so
Z
−a + by
dy =
y
Z
−cx + d
dx
x
−a ln y + by = −cx + d ln y + c1 , or (xd e−cx )(y a e−by ) = c0 .
Lotka-Volterra Competition Model
Two (or more) species compete for resources(food, light, etc.)(predator) of ecosystem.: Example: Assume snakes eat frogs and only frogs. If x is the number of
predator and y is the number of prey, then
r1
x′ =
x(K1 − x − α12 y)
K1
r2
y′ =
y(K2 − y − α21 x)
K2
Note that
179
11.4. AUTONOMOUS SYSTEM AS MATH. MODEL
x1
d/c
x2
y1
a/b
y2
Figure 11.9: Graph of F (x) = xd e−cx and G(x) = xa e−bx
(1) If there were no second species (y = 0), then x′ = r1 /K1 (K1 − x) so the
first species grow logistically and approach the steady state(section 2)
(2) If there were no first species (x = 0), then y ′ = r2 /K2 (K2 − y) so the second
species show similar behavior.
Example 11.4.4.
x′ = 0.004x(50 − x − 0.75y)
y ′ = 0.001y(100 − y − 3.0x)
Limit Cycle
Van der Pol’s equation.
y ′′ − µ(1 − y 2 )y ′ + y = 0.
(11.34)
Here µ(1 − y 2 )y ′ is a damping term. If |y| > 1, we have large damping and if
|y| < 1, we have negative damping. With the substitution v = y ′ , y ′′ = v dv
dy we
get
dv
v
− µ(1 − y 2 )v + y = 0.
dy
If µ is small we approximate it by v dv
dy + y = 0. So the limit cycle is closer to a
circle. But if µ is large, situation changes.
Exercise 11.4.5.
(1) Study the stability of the system at (0, 0):
180
CHAPTER 11. SYSTEM OF NONLINEAR DIFF. EQUATION
v
2
-2
2
y
-2
Figure 11.10: µ = 0.1
(a)
(b)
(c)
(d)
(e)
x′ = x − 2y + sin x
y ′ = 3x − xy + y 2
x′ = y + x2 x + y 2
y ′ = −x + y + xy + y 2
x′ = 2y
2
y ′ = 1 − x − ex+y
x′ = sin(x − 2y)
y ′ = 1 + x − e−x+y
x′ = y + x(x2 + y 2 )
y ′ = −x + y(x2 + y 2 )
(f)
(g)
(h)
(i)
(j)
x′ = −y + x(x + y)
y ′ = x + y − xy + y 2
x′ = x − x2 y − xy
y ′ = 0.2x − 0.5xy + .001y 2
x′ = ex−y − 1
y ′ = sin(x − 2y)
x′ = x + xy + x2 x + y 2
y ′ = y − xy + y 2
x′ = y sin(1 + x)
y ′ = 1 − cos y − x
181
11.4. AUTONOMOUS SYSTEM AS MATH. MODEL
v
2
-2
2
-2
Figure 11.11: µ = 1.2
y