Mathematics 3F03 Advanced Differential Equations, Fall 2013
INSTRUCTOR’S SOLUTIONS TO ASSIGNMENT 1
September 22, 2013 @ 23:11
1
Stability on the Phase Line
For four functions f (x), we are asked to find all equilibria of x0 = f (x), determine their
stability, and sketch the phase line. Equilibria are obtained by finding all x∗ ∈ R such that
f (x∗ ) = 0. If f (x) is differentiable at x∗ and f 0 (x∗ ) 6= 0 then we can use the derivative test
to determine stability at x∗ , i.e., x∗ is stable (a sink) if f 0 (x∗ ) < 0 and x∗ is unstable (a
source) if f 0 (x∗ ) > 0. If f 0 (x∗ ) = 0 or f if not differentiable at x∗ then the derivative test is
not applicable.
If the derivative test does not apply then we must examine the sign of f (x) in a neighbourhood of x∗ to determine stability. For the five functions f (x) in question, there is always
a sufficiently small neighbourhood of x∗ such that f (x) changes sign only at x∗ . In such a
neighbourhood, the sign of f (x) on either side of x∗ determines its stability:
• x∗ is stable if f (x) > 0 for x < x∗ and f (x) < 0 for x > x∗ ;
• x∗ is unstable if f (x) < 0 for x < x∗ and f (x) > 0 for x > x∗ ;
• x∗ is semi-stable if f (x∗ ) = 0 but f (x) is non-zero and has the same sign on either side
of x∗ . Thus, in such cases, x∗ is stable on one side and unstable on the other.
Note that at a semi-stable equilibrium point f (x) must either be non-differentiable or have
a zero derivative. In either case, the derivative test does not apply, so it can never be used
to detect semi-stable equilibria.
Note that while there is no requirement that the function f (x) be differentiable (or even
continuous) at all points x, we do need f (x) to be defined for all x in the domain of interest.
Thus, the case of f (x) = tan x was not well-posed since tan x is not defined for all x ∈ R.
Part (d) should have stated
(
tan x, x 6= k + 12 π, k ∈ Z,
dx
=
dt
0,
otherwise.
These considerations lead us to the following results:
f (x)
(a) x6 − x4
(b) x(x − 1)(x − 2)(x − 3)
(c) arctan x
(d) tan x
Equilibria x∗
−1
0
1
0
1
2
3
0
kπ, k ∈ Z
k + 12 π, k ∈ Z
1
Derivative Test
f 0 (x∗ ) < 0 =⇒ stable
f 0 (x∗ ) = 0
f 0 (x∗ ) > 0 =⇒ unstable
f 0 (x∗ ) < 0 =⇒ stable
f 0 (x∗ ) > 0 =⇒ unstable
f 0 (x∗ ) < 0 =⇒ stable
f 0 (x∗ ) > 0 =⇒ unstable
f 0 (x∗ ) > 0 =⇒ unstable
f 0 (x∗ ) > 0 =⇒ unstable
f 0 (x∗ ) undefined
In the graphs below, the phase line (x-axis) is indicated in red. The function f (x) is plotted
(in blue) so direction of motion on the phase line is clear from the sign of f (x) (which settles
all the cases above where the derivative test told us nothing). Equilibria are identified with
circles. Unstable equilibria are unfilled, stable equilibria are filled, and semi-stable equilibria
are half-filled on the stable side.
(a) f (x) = x6 − x4
(b) f (x) = x(x − 1)(x − 2)(x − 3)
2
2
1
1
#
0
-1
#
H
#
0
-1
-2
#
#
#
0
1
2
3
-2
-1.0
0.0 0.5 1.0
x
x
(c) f (x) = arctan(x)
(d) f (x) = tan(x)
2
2
1
1
## # # # #
# # # # # #
0
0
-1
-1
-2
-2
-4
-2
0
2
4
-4
x
2
#
-2
0
2
4
x
Bifurcation Diagrams
Each of the following families of differential equations depends on a parameter a ∈ R. Sketch
the corresponding bifurcation diagrams. In each case, we refer to the RHS of the equation
as fa (x).
(a)
dx
dt
= x2 − ax4
√
3
Solution: fa (x) = 0 =⇒ x = 0 or (if a > 0) x = ±1/ √a. f 0 (x) =
√ 2x − 4ax =
2
0
0
2x(1 − 2ax ). Therefore, f (0) = 0 and (if a > 0) f (±1/ a) = ∓2/ a. Thus, the
2
derivative test tells us nothing
about√ the equilibrium at x = 0, but for a > 0 implies
√
that the equilibria at −1/ a and 1/ a unstable and stable, respectively. If a ≤ 0 then
the equilibrium at x = 0 is semi-stable (stable on the left), since fa (x) ≥ 0 for all x ∈ R.
On the other hand, noting that fa00 (x) = 2 − 12ax2 , we that fa00 (0) > 0 for all a ∈ R, so
there is a local minimum at x = 0 for all a ∈ R, and hence x = 0 is always stable on the
left and unstable on the right. Using solid, dashed and dotted lines for stable, unstable
and semi-stable equilibria, respectively, these considerations yield:
2
x
1
0
-1
-2
-1
0
1
2
3
a
We see that there is exactly one bifurcation (at a = 0).
dx
dt
= arctan (x − a)
Solution: fa (x) = arctan (x − a) = 0 =⇒ x − a = 0 =⇒ x = a. f 0 (a) = 1 > 0, so
there is always a unique and unstable equilibrium at x = a. Thus, we obtain:
x
(b)
3
2
1
0
-1
-2
-3
-3
-2
-1
0
a
3
1
2
3
We see that this family of ODEs does not display any bifurcations.
(c)
dx
dt
= tan (ax)
Solution: fa (x) = tan (ax) = 0 =⇒ ax = kπ, k ∈ Z. If a = 0 then fa (x) = f0 (x) ≡ 0
and every point x ∈ R is a neutrally stable equilibrium. If a 6= 0 then there are equilibria
at x∗,k = kπ/a for each k ∈ Z. For any a, fa0 (x) = a sec2 (ax), so for a 6= 0 the sign
of fa0 (x∗,k ) = fa0 (kπ/a) is simply equal to the sign of a. If we define tan x to be zero at
points where tan is not normally defined (following the assumption
in problem 1(a)),
π
1
then we have another infinite set of equilibria at x∗∗,k = a k + 2 . The sign of tan (ax) on
either side of these equilibria implies that they are stable if a > 0 and unstable if a < 0.
Putting all this information together, we obtain the following bifurcation diagram.
4
x
2
0
-2
-4
-4
-2
0
2
4
a
We see that this family of ODEs has an extraordinarily impressive bifurcation at a = 0.
As a → 0, all non-zero equilibria move off to infinity and disappear at a = 0, where the
equilibrium at x = 0 explodes into a continuum of neutrally stable equilibria (indicated
by a dotted line). The stability of all corresponding equilibria on either side of the
bifurcation at a = 0 is opposite.
3
Exercise 11, page 18
First-order differential equations need not have solutions that are defined for all time.
(a) Find the general solution of the equation dx/dt = x2 .
Solution: Consider the associated initial value problem
dx
= x2 ,
dt
x(0) = x0 .
4
(∗)
For x0 6= 0, we can solve by separation:
t=
Z
t
dt =
0
Z
x(t)
x(0)
Solving for x(t), we have
x(t) =
x(t)
dx
1
1
1 +
.
=− =−
2
x
x x0
x(t) x0
1
x0
1
,
−t
t 6=
1
.
x0
(∗∗)
Differentiating this, we find
x0 (t) =
2
2 = x(t) ,
−t
1
1
x0
which confirms that we have truly found the solution to (*) for x0 6= 0. For the special
case x0 = 0, the unique solution is simply x(t) ≡ 0, as is easily verified. It is worth
noting that although the derivation of the solution for x0 6= 0 is not valid for x0 = 0, if
we formally multiply the solution by x0 /x0 we obtain
x(t) =
x0
,
1 − x0 t
t 6=
1
,
x0
(∗ ∗ ∗)
which is now valid for all initial conditions x0 ∈ R.
(b) Discuss the domains over which each solution is defined.
Solution: If x0 = 0 then the solution (an equilibrium at the origin) is valid for all
times t ∈ R. If x0 6= 0 then the solution (**) is valid only on the semi-infinite time
intervals −∞ < t < 1/x0 and 1/x0 < t < ∞.
(c) Give an example of a differential equation for which the solution satisfying x(0) = 0 is
defined only for −1 < t < 1.
Solution: Consider the simplest possible generalization of the example above, i.e.,
dx
= x2 + a ,
dt
x(0) = x0 ,
(♥)
where a ∈ R. Repeating the analysis above, for the situation where a > 0, we have
x(t)
Z t
Z x(t)
dx
1
x 1
x(t)
x0
t=
dt =
= √ arctan √ =√
arctan √ − arctan √
.
2
a
a x0
a
a
a
0
x(0) x + a
Restricting attention to the situation of interest (i.e., x0 = 0), this simplifies to
1
x(t)
t = √ arctan √ ,
a
a
i.e.,
x(t) =
√
√
a tan ( at) .
5
The largest interval containing t = 0 on which this solution is valid is
π √
π
− < at < ,
2
2
i.e.,
π
π
− √ <t< √ .
2 a
2 a
Therefore, choosing a such that
π
√ = 1,
2 a
i.e.,
π2
,
a=
4
we have an example, namely
dx
π2
= x2 +
,
x(0) = x0 ,
(♠)
dt
4
for which the solution including x(0) = 0 is valid only on the time interval −1 < t < 1,
as required.
4
Extension of Exercise 13, page 18
Let dx/dt = f (x) be an autonomous, first-order, one-dimensional differential equation with
an equilibrium point at x0 .
(a) Suppose f 0 (x0 ) = 0. What can you say about the behaviour of solutions near x0 ? Give
examples.
Solution: Nothing can be said in general. If f (x) ≡ 0 then every point near x0 is
fixed (every x ∈ R is an equilibrium point). If f (x) = (x − x0 )2 then x0 is a semi-stable
equilibrium (stable on the left), whereas if f (x) = −(x − x0 )2 then x0 is semi-stable and
stable on the right. If f (x) = (x − x0 )3 then x0 is unstable, whereas if f (x) = −(x − x0 )3
then x0 is stable. So, anything can happen!
(b) Suppose f 0 (x0 ) = 0 and f 00 (x0 ) 6= 0. What can you say now?
Solution: These are sufficient conditions for the point x0 to be a maximum or mininum point of the function f (x). If f 00 (x0 ) > 0 then f 0 (x) is increasing at x0 so x0 is
a minimum point; hence, since f (x0 ) = 0, f (x) is positive in a neighbourhood of x0
(excluding x0 ), implying that x0 is a semi-stable equilibrium that is stable on the left. If
f 00 (x0 ) < 0 then we have the opposite situation, a semi-stable equilibrium that is stable
on the right.
(c) Suppose f 0 (x0 ) = f 00 (x0 ) = 0 but f 000 (x0 ) 6= 0. What can you say now?
Solution: x0 is a point of inflection of f (x). If f 000 (x0 ) > 0 thenf (x) is strictly increasing at x0 (the behaviour of f (x) near x0 is similar to the function (x − x0 )3 ); hence,
x0 is an unstable equilibrium point of the differential equation. Similarly, if f 000 (x0 ) < 0
then x0 is a stable equilibrium point of the differential equation.
6
(d) (Bonus.) Suppose f (k) (x0 ) = 0 ∀k ∈ N, i.e., f is infinitely differentiable at x0 and all
derivatives of f vanish at x0 . What can you say now?
Solution: Nothing can be said in this case. Consider, for example, the functions
(
2
e−1/x , x 6= 0
f1 (x) =
0,
x=0

−1/x2

,
x>0
e
f2 (x) = 0,
x=0

 −1/x2
−e
, x<0
Both these functions have vanishing derivatives of all orders at the origin (a full solution
would include a proof of this fact). However, f1 (x) [or −f1 (x)] has a semi-stable equilibrium at 0 that is stable on the left [or right] and f2 (x) [or −f2 (x)] has an unstable
[or stable] equilibrium at 0. In addition, the zero function f (x) ≡ 0 also has vanishing
derivatives of all orders at 0 (and everywhere else) and yields neutrally stable equilibria
everywhere.
5
Action of matrices in the plane
For each of the following 2 × 2 matrices A, find the eigenvalues and eigenvectors of A. Also
interpret A geometrically, i.e., what does the action of A do to an arbitrary vector in the
plane?
1 2
(a)
2 1
Solution: det (A − λI) = (λ − 3)(λ + 1) =⇒ eigenvalues are 3 and −1. Solving
(A − λI)V = 0 for each λ we find eigenvectors (1, 1) and (−1, 1), respectively. Thus, A
stretches the component of vectors along (1, 1) by a factor of 3 and reverses the direction
of the component along (−1, 1). Graphically, we find:
7
1 2
After applying A =
2 1
Original Grid
λ1 = 3,
λ2 = −1
1 1
(b)
1 1
Solution: det (A − λI) = λ(λ − 2) =⇒ eigenvalues are 2 and 0. Solving (A − λI)V = 0
for each λ we find eigenvectors (1, 1) and (−1, 1), respectively. Thus, A stretches the
component of vectors along (1, 1) by a factor of 2 and anihilates the component along
(−1, 1). Graphically, we find:
1 1
After applying A =
1 1
Original Grid
λ1 = 2,
8
λ2 = 0
1 1
(c)
0 1
Solution: det (A − λI) = (λ − 1)2 =⇒ λ = 1 is the unique eigenvalue (and has
algebraic multiplicity alg(1) = 2). Solving (A − I)V = 0 we find a unique linearly independent eigenvector, namely (1, 0) (hence geom(1) = 1). Since we do not have a basis
of eigenvectors, to determine the action of A on the plane we must consider the action
of A on a vector that is linearly independent of the eigenvector. The simplest choice is
(0, 1). Thus, we consider
1 1
0
1
=
.
0 1
1
1
In general, A maps the vector (x, y) to (x + y, y), skewing the space:
1 1
After applying A =
0 1
Original Grid
λ1 = 1,
9
λ2 = 1