Cole’s Notes for Autonomous Scalar ODEs
for Mech 2 Math 255
Cole Zmurchok
⇤
November 30, 2015
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Introduction
The goal of this document is to provide notes and exercises for the discussion of twodimensional nonlinear systems of ODEs for Math 255 for Mech 2.
We begin with a review of autonomous scalar equations with a focus on qualitative
analysis. Part II of this document will focus on systems.
References
[1] L. Edelstein-Keshet. Mathematical Models in Biology. SIAM, 2005.
[2] E. Kreyszig. Advanced Engineering Mathematics. John Wiley and Sons, Inc., 2006.
[3] S. H. Strogatz. Nonlinear Dynamics and Chaos: With Application to Physics, Biology,
Chemistry, and Engineering. Westview Press, 2014.
Strogatz (3) is easy to read and very instructive (available as an eBook from UBC Library), Kreyszig (2) is an engineering textbook (so you’ll like it, sadly not an eBook), and
Edelstein-Keshet (1) does this analysis with a Math Biology context (available as an eBook
from the UBC Library), which I include because I’m biased.
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Autonomous scalar ODEs
Consider the first-order ODE x0 = f (t, x). If the right-hand side, f (t, x) = f (x) does not
depend on time t, the ODE is called autonomous.
Example 1. The ODE x0 = sin x is autonomous, but the ODE x0 = x + t is not.
⇤
Department of Mathematics, University of British Columbia, Vancouver, Canada V6T 1Z2:
zmurchok@math.ubc.ca
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Although some of these autonomous one-dimensional ODE can be solved using separation
of variables (or the solution can be at least written in terms of an integral), it is often hard
to understand the qualitative behaviour of these solutions.
Exercise 1. Use separation of variables to show that the solution to x0 = sin x, x(0) = x0
can be written as
csc x0 + cot x0
t = ln
.
csc x + cot x
Exercise 1 highlights this problem. Although the solution to x0 = sin x, x(0) = x0 can be
found, the solution is impossible to understand. What does the graph of x(t) look like for
an arbitrary initial condition? What occurs as t ! 1?
To answer these questions we can use qualitative methods.
2.1
Steady-states and phase-line analysis
To qualitatively analyze the behaviour, we plot the phase-portrait, that is, x0 versus x.
x gives the position of a particle moving along the x-axis, with x0 the velocity of the
particle. The ODE x0 = sin x, thus describes a vector field on the line. Right or left facing
arrows indicate the direction of the particles motion–the velocity. Right arrows correspond to
positive velocity: x0 > 0 equivalently f (x) > 0. Left arrows correspond to negative velocity:
x0 < 0 equivalently f (x) < 0. Points where x0 = 0 are called steady-states.
Definition 1. x⇤ is called a steady-state if
dx
dt x=x⇤
= f (x⇤ ) = 0.
On the diagram, solid black dots represent stable steady-states as the particle moves
toward them over time. Open circles represent unstable steady-states as the particle moves
away from them over time. A caveat: if x(t) = x⇤ for some time t, then x0 (x⇤ ) = 0 and
the solution will forever remain at the steady-state even if the steady-state is unstable. The
words unstable and stable refer to eventual behaviour of trajectories near the steady-state.
Example 2. Find the steady-states and determine their stability for x0 = x(1 x), x(0) = x0 .
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The steady-states satisfy f (x⇤ ) = 0. x⇤ is a steady-state if and only if x⇤ (1 x⇤ ) = 0.
We find that x⇤ = 0, 1 are the steady-states.
Thus, x⇤ = 0 is unstable, and x⇤ = 1 is stable. Moreover, by looking at the shape of f (x)
in the phase-portrait, we can infer the qualitative shape of solutions and sketch them in the
xt-plane.
2.1.1
Exercises
For each of the following ODEs, find the steady-states, sketch the phase-portrait, qualitatively determine the stability of the steady-states, and sketch sample trajectories (if possible).
1. (Terminal Velocity)
2. (Circuits)
dQ
dt
=
dv
dt
V0
R
3. x0 = x2
9.
4. x0 = (x
1)(x + 1)(x
=g
Q
,
RC
kv, g, k 2 R.
R, V0 , C 2 R. Hint: is this the same as question 1?
2).
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5. x0 = x
cos x. Hint: plot x and cos x separately since x
cos x is hard to sketch.
6. x0 = x2 . Hint: does the idea of a semi-stable steady-state seem applicable here?
7. Suppose that some ODE x0 = f (x) has the following phase-portrait:
Find f (x) that is consistent with this phase-portrait.
8. Find an ODE x0 = f (x) whose solutions are consistent with those trajectories sketched
below.
2.2
Linear stability analysis
In the previous section, we learned to qualitatively determine the behaviour of solutions
to an autonomous first-order ODE. It is sometimes possible to find the stability of fixed
points mathematically.
For intuition on this topic, consider the following exercise.
Exercise 2. For each of the phase-portraits sketched in the above exercises, can you associate
the slope of f (x) at each steady-state x⇤ with the stability of that steady-state?
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To formalize this idea, we consider the linearization about the steady-state.
Let x⇤ be a steady-state of x0 = f (x), and let ⌘(t) = x(t) x⇤ be a small perturbation
away from x⇤ . That is, we assume that x(t) a small distance ⌘(t) away from x⇤ . If this
distance ⌘(t) shrinks, then x(t) will tend toward x⇤ . If this distance ⌘(t) grows, then x(t)
will tend away from x⇤ .
To determine if ⌘(t) is increasing or decreasing, we derive a di↵erential equation for ⌘(t):
⌘0 =
d
(x(t)
dt
x⇤ ) = x0 .
Since x0 = f (x),
⌘ 0 = f (x).
However, we can write x = ⌘ + x⇤ , so that
⌘ 0 = f (x⇤ + ⌘).
Using a Taylor expansion we thus obtain
⌘ 0 = f (x⇤ ) + f 0 (x⇤ )⌘ + higher order terms involving ⌘ 2 etc.
If we neglect the higher order terms and use the fact that f (x⇤ ) = 0, we find
⌘ 0 = f 0 (x⇤ )⌘.
Provided f 0 (x⇤ ) 6= 0, this equation describes how the perturbation ⌘(t) grows or shrinks. If
f 0 (x⇤ ) = 0, then neglecting the higher-order terms was a bad idea, and we need to do more
work.
Definition 2. ⌘ 0 = f 0 (x⇤ )⌘ is called the linearization about the steady-state x⇤ .
Returning to the earlier intuition about the slope of the tangent line to f (x) at x⇤ , we
see that the linearized system gives the same result:
• If f 0 (x⇤ ) > 0 (positive slope at x⇤ ) then ⌘(t) grows exponentially, and x⇤ is unstable.
• If f 0 (x⇤ ) < 0 (negative slope at x⇤ ) then ⌘(t) decays exponentially, and x⇤ is stable.
Example 3. For x0 = f (x) = x(1 x), we found the steady-states to be x⇤ = 0, 1. Di↵erentiating f (x) gives f 0 (x) = 1 2x. Thus:
• f 0 (0) = 1 > 0. This confirms that x⇤ = 0 is unstable.
• f 0 (1) =
1 < 0. This confirms that x⇤ = 1 is stable.
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2.2.1
Exercises
Use linear stability analysis to determine the stability of steady-states to the following
ODEs. If f 0 (x⇤ ) = 0 for a steady-state x⇤ , try to find the stability using a qualitative method.
1. x0 = sin x.
2. x0 = x2
9.
3. x0 = log x, with x > 0 only.
4. x0 = tan x, with x 2 (
5. x0 = x2 (6
6. x0 = ax
3
⇡ ⇡
, ).
2 2
x).
x3 , where a < 0, a = 0, or a > 0. Check all three cases.
Review of two-dimensional linear systems
It would also be a good idea to look back at your notes for linear systems.
Keywords: steady-state, nullcline, saddle point, centre, node, spiral, romeo and juliet.
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