LECTURE 8: AUTONOMOUS EQUATIONS
MINGFENG ZHAO
September 19, 2014
To solve differential equation M (x, y) + N (x, y) y 0 = 0:
M y − Nx
N
2) Compute the integrating factor µ(x):
1) Compute My (x, y), Nx (x, y) and
µ(x) = e
My −Nx
N
R
dx
.
3) Understand the meaning of the integrating factor, that is, [µM ] + [µN ] y 0 = 0 is exact, then there exists some
φ(x, y) such that
φx (x, y) = µ(x)M (x, y),
and φy (x, y) = µ(x)N (x, y).
4) Since φx (x, y) = µ(x)M (x, y), then
Z
φ(x, y) =
µ(x)M (x, y) dx + f (y).
5) Since φy (x, y) = µ(x)N (x, y), take the partial derivative with respect to y on the both sides of φ(x, y) =
Z
µ(x)M (x, y) dx + f (y), then
Z
µ(x)N (x, y) = φy (x, y) =
µ(x)My (x, y) dx + f 0 (y).
6) Compute f (y), that is,
Z f (y) =
Z
µ(x)N (x, y) −
µ(x)My (x, y) dx dy.
7) Write down the solution:
Z
Z Z
φ(x, y) = My (x, y) dx +
µ(x)N (x, y) − µ(x)My (x, y) dx dy = C.
Example 1. Solve ydx + (x2 y − x)dy = 0, x > 0.
Rewrite the equation:
y + (x2 y − x)y 0 = 0.
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MINGFENG ZHAO
Let M (x, y) = y and N (x, y) = x2 y − x, then
and Nx = 2xy − 1.
My = 1,
So
2 − 2xy
2
M y − Nx
= 2
=− .
N
x y−x
x
So the integrating factor is:
µ(x) = e
That is, the equation
My −Nx
N
dx
= e−
R
2
x
dx
= e−2 ln x =
1
.
x2
y
x2 y − x 0
+
y = 0 is exact. So there exist some φ(x, y) such that
x2
x2
φx =
Since φx =
R
y
,
x2
and φy =
x2 y − x
1
=y− .
2
x
x
y
, then
x2
Z
y
y
dx + f (y) = − + f (y).
x2
x
φ(x, y) =
Take the partial derivative with respect to y, since φy = y −
y−
So f 0 (y) = y. We can take f (y) =
1
, then
x
1
1
= φy = − + f 0 (y).
x
x
y2
. So the solution is:
2
−
y
y2
+
=C .
x
2
Use Euler’s Method to approximate solutions of y 0 = f (x, y): Given the initial condition y(x0 ) = y0 and the step size h,
compute the point (xk+1 , yk+1 ) from the previous point (xk , yk ) as follows:
1) Use the differential equation to compute the slope f (xk , yk ).
2) Calculate the next point (xk+1 , yk+1 ) using the formula
xk+1
= xk + h
yk+1
= yk + hf (xk , yk )
LECTURE 8: AUTONOMOUS EQUATIONS
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Let xk = x0 + kh, and
x1 = x0 + h
y1 = y0 + hf (x0 , y0 )
x2 = x1 + h
y2 = y1 + hf (x1 , y1 )
x3 = x2 + h
..
.
y3 = y2 + hf (x2 , y2 )
..
.
xk+1 = xk + h
..
.
yk+1 = yk + hf (xk , yk )
..
.
Theorem 1 (Picard’s Theorem on Existence and Uniqueness). If f (x, y) is continuous with respect to x and y, and
∂f
(x, y) exists and is continuous near some (x0 , y0 ), then a solution to
∂y
y 0 = f (x, y),
y(x0 ) = y0 ,
exists for some small interval containing x0 , and is unique.
AUTONOMOUS EQUATIONS
An autonomous equation is of the form:
dy
= f (y).
dx
Then
1) y(x) ≡ a for some constant a such that f (a) = 0. In this case, y(x) ≡ a is called an equilibrium solution, and a
is called a critical point of f .
2) If f (y) 6= 0, then
Z
x=
1
dy.
f (y)
For the slope field of an autonomous equation, the slopes are the same as long as they have same y-value. That is to
say, if we know the slope field along a single vertical line(such single vertical line is called phase line/diagram/portrait),
then we know the slope field in the entire xy-plane. On this vertical line, we mark all critical points, and then
we draw arrows between critical points: If f (y) > 0, we draw an up arrow “↑”; If f (y) < 0, we draw a
down arrow “↓”.
Lemma 1. Let f be differentiable, f 0 is continuous, and y be a solution to y 0 = f (y). If y 0 (x0 ) = 0 for some x0 , then
y(x) ≡ y(x0 ),
for all x.
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MINGFENG ZHAO
Proof. Since y 0 (x0 ) = 0 and y 0 (x) = f (y(x)), then 0 = f (y(x0 )), which implies that y(x) ≡ y(x0 ) is a solution to
y 0 = f (y). Since f is differentiable and f 0 is continuous, by Picard’s Theorem on Existence and Uniqueness, Theorem
1, we know that
y(x) ≡ y(x0 ),
for all x.
Theorem 2. Let f be differentiable, f 0 is continuous, and y be a solution to y 0 = f (y), then either y 0 (x) ≡ 0 for all x,
or, y 0 (x) > 0 for all x, or y 0 (x) < 0 for all x.
Proof. If y 0 (x) ≡ 0 for all x, we are done. If y 0 (x) 6≡ 0, by Lemma 1, we must have that either y 0 (x) > 0 for all x, or
y 0 (x) < 0 for all x.
Remark 1. Theorem 2 tells us that any solution to an autonomous equation must be either constant function, or
strictly increasing, or strictly decreasing.
Example 2. Find equilibrium points of y 0 = y(1 − y), draw the phase diagram, and describe the solutions y1 , y2 , y3 ,
y4 satisfy the following initial conditions:
1) y1 (0) = 0,
2) y(0) = −1,
3) y(0) = 2,
4) y(0) =
1
.
2
Let f (y) = y(1 − y). Solve f (y) = y(1 − y) = 0, then y = 0 or y = 1, that is, all equilibrium points of y 0 = y(1 − y)
are:
0
and
1.
For the initial value problems:
1) Since f (0) = 0(1 − 0) = 0, then y1 is a constant function, that is,
y1 (x) ≡ 0.
2) Since f (−1) = (−1)[1 − (−1)] = −2 < 0, then
y2 (x) is strictly decreasing .
3) Since f (2) = 2(1 − 2) = −2 < 0, then
y3 (x) is strictly decreasing.
LECTURE 8: AUTONOMOUS EQUATIONS
1
1
1
1
4) Since f ( ) = (1 − ) = > 0, then
2
2
2
4
y4 (x) is strictly increasing.
Figure 1. Phase diagram of y 0 = y(1 − y)
Figure 2. Slop field of y 0 = y(1 − y)
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6
MINGFENG ZHAO
Figure 3. Slop field of y 0 = y(1 − y)
Remark 2. In general, for any initial day y(x0 ) = y0 if x0 is between two critical points, then the solution must be the
whole real line, like the case y(0) = 1/2 in Example 2.
If x0 is not between two critical points, for example, consider the problem: y 0 = y 2 , y(0) = 1. It’s easy to see that
1
. So the solution interval is (−∞, 1). However, we say that y(x) → ∞ as x → 1− .
the solution is: y(x) =
1−x
Definition 1. Consider the autonomous equation:
dy
= f (y).
dx
Let a be a critical point of f , that is, f (a). 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 4. Stable or sink
LECTURE 8: AUTONOMOUS EQUATIONS
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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 5. Source
III. We say a is a node if the equilibrium point a is neither a source nor a sink.
Figure 6. Node
If a is either a source or a node, we say a is unstable.
Example 3. Consider Logistic population mode:
dP
= kP
dt
P
1−
N
,
where k, N > 0 are constants.
1) Equilibrium points:
P = 0 and P = 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) .
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MINGFENG ZHAO
4) Initial value problems:
– If the initial value is P (0) =
N
, then P (t) is strictly increasing, P (t) → N as t → ∞, and P (t) → 0 as
3
t → −∞.
– If the initial value is P (0) = N , then P (t) ≡ N for all t, i.e., the population will stay as N .
– If the initial value is P (0) = N + 5, then P (t) is strictly decreasing, P (t) → N as t → ∞, and P (t) → ∞
as t → −∞.
Theorem 3 (Linearization Theorem). Suppose a is an equilibrium point of the differential equation:
dy
= f (y),
dx
where f is differentiable and f 0 is continuous. Then
1) If f 0 (a) < 0, then a is a sink (stable).
2) If f 0 (a) > 0, then a is a source (unstable).
3) If f 0 (a) = 0, then undecidable. (Need more information to decide the type of a)
Example 4. y 0 = (2 − y) sin(y).
Let f (y) = (2 − y) sin(y), then
1) Equilibrium points:
y = 2, y = 0, ±π, ±2π, · · · .
2) Types of equilibrium points: Since f 0 (y) = − sin(y) + (2 − y) cos(y), then
f 0 (2) = − sin(2) < 0,
f 0 (0) = 2 > 0
For any positive integer k, we have
f 0 (2kπ)
f 0 (−2kπ)
f 0 ((2k − 1)π)
=
(2 − 2kπ) cos(2kπ)
=
2 − 2kπ < 0
=
(2 + 2kπ cos(−2kπ)
=
2 + 2k > 0
=
[2 − (2k − 1)π] cos((2k − 1)π)
= −[2 − (2k − 1)π]
LECTURE 8: AUTONOMOUS EQUATIONS
f 0 (−(2k − 1)π)
=
(2k − 1)π − 2 > 0
=
[2 + (2k − 1)π] cos(−(2k − 1)π)
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= −[2 + (2k − 1)π]
= −2 − (2k − 1)π < 0.
Then

 P = 2, 2kπ, −(2k − 1)π are sink (stable),
 P = 0, −2kπ, (2k − 1)π are source (unstable),
k is any positive integer. .
3) Initial value problems:
– If y(0) = 0.4 ∈ (0, 2), then y(x) is strictly increasing, y(x) → 2 as x → ∞, and y(x) → 0 as x → −∞.
– If y(0) = 3 ∈ (2, π), then y(x) is strictly decreasing, y(x) → 2 as x → ∞, and y(x) → π as x → −∞.
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