LECTURE 8: AUTONOMOUS EQUATIONS AND EULER’S METHOD
MINGFENG ZHAO
September 25, 2015
Phase diagram of the autonomous equation
An autonomous equation is of the form y 0 = f (y), if f (a) = 0, y(x) ≡ a is called an equilibrium solution, and a is
called a critical point of y 0 = f (y).
To draw the phase diagram of y 0 = f (y):
1) Find all critical points of y 0 = f (y), that is, find all zeros of f (y) = 0.
2) Mark all critical points on a vertical line.
3) For any two neighboring critical points a and b, choose any point c which is between a and b, compute f (c):
– If f (c) > 0, draw an up arrow “↑” between a and b.
– If f (c) < 0, draw a down arrow “↓” between a and b.
Example 1. x0 = (x − 1)(x − 2)x2 .
• Critical Points: Solve (x − 1)(x − 2)x2 = 0, then x = 0, 1, 2 .
• Phase Diagram:
Figure 1. Phase diagram of x0 = (x − 1)(x − 2)x2
1
2
MINGFENG ZHAO
• For the initial value problem: x(0) = 0.5. Since 0 < x(0) = 0.5 < 1, then x0 (0) = (0.5 − 1)(0.5 − 2)(0.5)2 > 0,
and
lim x(t) = 1 .
t→∞
Classification of critical points of the autonomous equation
Definition 1. Consider the autonomous equation y 0 = f (y), let a be a critical point of y 0 = f (y), that is, f (a) = 0.
Then
I. We say a is stable or a sink if any solution with initial condition close to a is asymptotic to a as x increases.
Figure 2. Stable or sink
II. We say a is a source if all solutions start close to a tend toward y as x decreases, and tend away from a as x
increases.
Figure 3. Source
III. We say a is a node if the equilibrium point a is neither a source nor a sink.
Figure 4. Node
IV. If a is either a source or a node, we say a is unstable.
LECTURE 8: AUTONOMOUS EQUATIONS AND EULER’S METHOD
Example 2. The phase diagram in Example 1 shows that:
Phase Diagram:
Figure 5. Phase diagram of x0 = (x − 1)(x − 2)x2
So we know that 2 is a source(unstable), 1 is a sink(stable), and 0 is a node(unstable).
Example 3. Consider Logistic population mode:
dP
= kP
dt
1−
P
N
,
where k, N > 0 are constants.
1) Critical points: Solve kP
P
1−
, we get P = 0 and P = N .
N
2) Determine the signs:
P > N,
0 < P < N,
P < 0,
dP
= (+)(−) < 0
dt
dP
= (+)(+) > 0
dt
dP
= (−)(+) < 0
dt
↓
↑
↓
3) Types of equilibrium points:
0 is a source (unstable);
N is a sink (stable) .
4) Initial value problems:
– If the initial value is P (0) =
N
, then P (t) is strictly increasing, and
3
lim P (t) = N
t→∞
and
lim
t→−∞
P (t) = 0.
– If the initial value is P (0) = N , then P (t) ≡ N for all t, i.e., the population stays as N at all time t.
3
4
MINGFENG ZHAO
– If the initial value is P (0) = N + 5, then P (t) is strictly decreasing.
Example 4. y 0 = sin(y).
Let f (y) = sin(y), then
1) Critical points: Solve f (y) = sin(y) = 0, then y = kπ for k ∈ Z, that is,
y = 0, ±π, ±2π, · · · .
2) Types of critical points:
..
.
2π
π
0
−π
−2π
..
.
So we know that 0 is unstable, ±π are stable. In general, 2kπ are unstable, (2k + 1)π are stable, k ∈ Z.
3) Initial value problems:
– If y(0) = 0.4 ∈ (0, π), then y(x) is strictly increasing, y(x) → π as x → ∞, and y(x) → 0 as x → −∞.
– If y(0) = 4 ∈ (π, 2π), then y(x) is strictly decreasing, y(x) → π as x → ∞, and y(x) → 2π as x → −∞.
Numerical methods: Euler’s method
Consider the initial value problem:
y 0 = f (x, y),
y(x0 ) = y0 .
Recall Taylor expansion:
y(x + ∆x) = y(x) + y 0 (x)∆x +
y 00 (x)
(∆x)2 + · · · .
2
Let’s take the linear approximation at x, then
(1)
y(x + ∆x) ≈ y(x) + y 0 (x)∆x.
LECTURE 8: AUTONOMOUS EQUATIONS AND EULER’S METHOD
5
Euler’s method:
Given the initial condition y(x0 ) = y0 and the step size ∆x, for any k = 0, 1, 2, · · · , let yk = y(xk ) and
xk = xk−1 + ∆x = x0 + (k − 1)∆x.
Now we want to approximate yk : Replace x by xk in (1), then we get
yk+1 = yk + f (xk , yk )∆x.
Example 5. Use Euler’s method to approximate y(3) with step size ∆x = 1, where y 0 =
y2
, y(0) = 1.
3
Recall the linear approximation, we get
y(x + ∆x) ≈ y(x) + y 0 (x)∆x.
Let yk = y(xk ), put x = xk and ∆x = 1 in the above formula, so we have
yk+1 = yk +
yk2
.
3
Then
k
xk
yk
0
0
1
1
1
y1 = y0 +
2
2
3
3
y02
12
4
=1+
=
3
3 3 2
y12
4 1
4
52
y2 = y1 +
= + ·
=
≈ 1.925
3
3 3
3
27
2
52 1
52
y2
5980
+ ·
≈ 2.734
y3 = y2 + 2 =
=
3
27 3
27
2187
Problems you can do:
Lebl’s Book [2]: All exercises on Page 40, Page 41 and Page 46.
Braun’s Book [1]: All Exercises on Page 100 and Page 67. Read all materials in Section 1.13.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca