1 Taylor Series: functions of a single variable

Recall that smooth functions f (x) of one variable have convergent Taylor series. The Taylor
series of such a function about x = a is
1 00 1
f (x) = f (a) + f 0 (a)(x − a) + f (a)(x − a)2 + · · · f (n) (a)(x − a)n + Rn (1)
2! n!
where the remainder term Rn → 0 as n → ∞ for all x in the interval of convergence. Such
expansions can be used to tell how a function behaves for x near a. When a = 0 the Taylor
series is also called the MacLaurin series of f (x). Some common series are:

∞
x
X xn x2 x3
e = =1+x+ + + ··· all x
n=0
n! 2! 3!

∞
X (−1)n x2n+1 x3 x5
sin x = =x− + + ··· all x
n=0
(2n + 1)! 3! 5!

∞
X (−1)n x2n x2 x4
cos x = =1− + + ··· all x
n=0
(2n)! 2! 4!

∞
1 X
= xn = 1 + x + x2 + x3 + · · · |x| < 1
1−x n=0

∞
X (−1)n−1 xn x2 x 3 x4
ln(1 + x) = =x− + − + ··· |x| < 1 , x = 1
n=1
n 2 3 4

∞
X (−1)n x2n+1 x3 x5 x7
tan−1 x = =x− + − + ··· |x| ≤ 1
n=0
2n + 1 3 5 7

It is important to emphasize that the Taylor series is ”about” a point. For example, the
Taylor series of f (x) = ln(1 + x) about x = 0 is
x2 x3 x4
ln(1 + x) = x − + − + ···
2 3 4
If you truncate the series it is a good approximation of ln(1 + x) near x = 0. Using (1), the
Taylor series of f (x) = ln(1 + x) about x = 1 is
1 1 1
ln(1 + x) = ln(2) + (x − 1) − (x − 1)2 + (x − 1)3 + · · ·
2 8 24
If you truncate this series it is a good approximation of ln(1 + x) near x = 1.
2 Taylor series: functions of two variables
If a function f : IR2 → IR is sufficiently smooth near some point (x̄, ȳ) then it has an m-th
order Taylor series expansion which converges to the function as m → ∞. Expressions for
m-th order expansions are complicated to write down. For our purposes we will only need
second order expansions so we state a related Theorem here:
Theorem 2.1 Let f : IR → IR and assume that f and all its derivatives up to third-order
are continous on some neighbourhood Nr (a), a = (a1 , a2 ). For each x = (x1 , x2 ) ∈ Nr (a)
f (x) = f (a) + fx1 (a)(x1 − a1 ) + fx2 (a)(x2 − a2 )
1
fx1 x1 (a)(x1 − a1 )2 + 2fx1 x2 (a)(x1 − a1 )(x2 − a2 ) + fx2 x2 (a)(x2 − a2 )2 + R3


+
2!
where the remainder term R3 → 0 as x → a.
Notationally there are many ways to write out Taylor series. For f = f (x1 , x2 ) one can
define the gradient of f as  
∂f ∂f
∇f (x) = ,
∂x1 ∂x2
The Hessian Hf (x) of f is defined as
∂2f ∂2f
" #
∂x21 ∂x1 x2
Hf (x) = ∂2f ∂2f
∂x2 x1 ∂x22

Note that Hf is a symmetric matrix. With these definitions, the expansion in the Theorem
above can be written
1
f (x) = f (a) + ∇f (a) · (x − a) + (ζ − a)T Hf (a)(ζ − a) + R3
2!
where · is the dot product. Here, the linear approximation 1 of f is
f (x) ' f (a) + ∇f (a) · (x − a)
1
and the remainder term is R2 (ζ) = 2!
(ζ − a)T Hf (a)(ζ − a). The second-order Taylor series
of f about a is
1
(x − a)T Hf (a)(x − a) + R3
f (x) = f (a) + ∇f (a) · (x − a) +
2!
where the exact form of the remainder term R3 is a complicated expression. Here we will
simply write:
R3 = O(k x − a k3 ) = O(3).
Example 1 Let f (x) = x21 + ex1 − x1 x2 + 3x2 and a = (0, 0). Then,
∇f (x) = (2x1 + ex1 − x2 , −x1 + 3)
so that
∇f (x) = (1, 3)
The Hessian is  
2 + ex1 −1
Hf (x) =
−1 0
1 or first-order Taylor series approximation
so that  
3 −1
Hf (a) =
−1 0
Since f (a) = 1,
 
1 3 −1
f (x) = 1 + (1, 3) · x + xT x + O(k x k3 )
2! −1 0
Longhand,
3
f (x) = 1 + x1 + 3x2 + x21 − x1 x2 + O(k x k3 )
2
Example 2 Do several in class
3 Implicit Function Theorem (IFT): Single variable
Theorem 3.1 Let f : IR2 → IR, f = f (x, µ) and that f , fx and fµ are continuous on a
neighbourhood (x, µ) = (x∗ , µ∗ ). If
i) f (x∗ , µ∗ ) = 0
ii) fµ (x∗ , µ∗ ) 6= 0
then there is a unique function µ̄(x) such that
f (x, µ̄(x)) = 0 (2)
for all x in a neighbourhood of x = x∗ .

Basically the theorem guarantees f (x, µ) = 0 has a unique solution µ̄(x) for all x near x∗ .
For example, consider
f (x, µ) = eµx − x + µ − 1
Then
fµ (x, µ) = xeµx + 1
from which
f (0, 0) = 0 fµ (0, 0) = 1 6= 0
By the IFT (with (x∗ = 0, µ∗ = 0)
ẋ = f (x, µ)
has a unique fixed point for all x near x∗ = 0. Note that this precludes the possibility of any
transcritical bifurcations for such x values (Why?).
Note that if f is smooth enough one can differentiate (2) in x to get
dµ̄
fx (x, µ̄) + fµ (x, µ̄) =0
dx
Were fµ = 0 then (x, µ) would have to solve f = 0 and fx = 0. That would mean (x, µ)
would a ”point” and µ could not vary in x!! This explains the importance of the (IFT)
hyptothesis.
Lastly, when (IFT) hypotheses are satisfied we know
dµ̄ fx (x, µ̄)
µ̄0 (x) = =−
dx fµ (x, µ̄)
For saddle-nodes with quadratic tangency we need µ̄0 (x∗ ) = 0 hence fx = 0 (nonhyperbolic-
ity) is required at saddle node bifurcations.
4 Hyperbolicity and hyperbolic fixed points

∂f
Definition 4.1 A fixed point x̄ of ẋ = f (x, µ) is hyperbolic if ∂x
(x̄, µ) 6= 0.

Hyperbolicity of fixed points is at the very heart of bifurcation theory. All the basic bifurca-
tion saddle node, transcritical, pitchfork bifurcations occur (respectively) at nonhyperbolic
fixed points (x∗ , µ∗ ) = (0, 0)
ẋ = µ − x2
ẋ = µx − x2
ẋ = x(µ − x2 )
5 Local Bifurcation Theory - some simple results
Here we summarize some theorems for “generic” saddle-node, transcritical and pitchfork
bifurcations of
ẋ = f (x, µ) , x, µ ∈ IR.
Proofs use the implicit function theorem, near identity transformations and various other
transformations. In all of the the theorems below we assume that f has continuous (mixed)
derivatives up to third order (fourth for pitchforks) for all (µ, x) near the bifurcation point
(µ∗ , x∗ ).
Theorem 5.1 If there is a pair (µ∗ , x∗ ) for which

f (x∗ , µ∗ ) = 0 (3)
∗ ∗
fx (x , µ ) = 0 (4)
∗ ∗
fµ (x , µ ) 6= 0 (5)
fxx (x∗ , µ∗ ) 6= 0 (6)

then ẋ = f (x, µ) has a saddle-node bifurcation with quadratic tangency at (µ∗ , x∗ ).

Transcritical (2-branch)

Theorem 5.2 If there is a pair (µ∗ , x∗ ) for which

f (x∗ , µ∗ ) = 0 (7)
∗ ∗
fx (x , µ ) = 0 (8)
∗ ∗
fµ (x , µ ) = 0 (9)
fxµ (x∗ , µ∗ ) 6= 0 (10)
fxx (x∗ , µ∗ ) 6= 0 (11)

then ẋ = f (x, µ) has a ( 2 branch) transcritical bifurcation at (µ∗ , x∗ ).

Pitchfork (Quadratic Tangency)

Theorem 5.3 If there is a pair (µ∗ , x∗ ) for which

f (x∗ , µ∗ ) = 0 (12)
∗ ∗
fx (x , µ ) = 0 (13)
fµ (x∗ , µ∗ ) = 0 (14)
∗ ∗
fxx (x , µ ) = 0 (15)
∗ ∗
fxµ (x , µ ) 6= 0 (16)
fxxx (x∗ , µ∗ ) 6= 0 (17)

then ẋ = f (x, µ) has a pitchfork bifurcation with quadratic tangency at (µ∗ , x∗ ).

5.1 Saddle-Node Local Theory
Suppose that on the locus of equilibria µ = µ(x), i.e. µ is a function of x in a neighbourhood of the suspect
bifurcation point. If the bifurcation occurs at (µ∗ , x∗ ) then by letting

X = x − x∗ , η = µ − µ∗

the differential equation for X is

Ẋ = F (X, η) = f (X + x∗ , η + µ∗ )

which has a bifurcation at (X, η) = (0, 0). Thus, without any loss of generality we will assume that the
bifurcation occurs at (µ, x) = (0, 0).
In this setting f (0, 0) = 0, µ(0) = 0 and

f (x, µ(x)) = 0

for all x near the bifurcation point. Differentiating this expression in x yields
dµ
fx (x, µ(x)) + fµ (x, µ(x)) =0 (18)
dx
dµ
which when solved for µ0 (x) = dx gives

fx (x, µ(x))
µ0 (x) = − .
fµ (x, µ(x))

One condition which must be satisfied at a quadratic tangency is that µ0 (0) = 0. Given the formula above,
this can only happen if

fx (0, 0) = 0 (19)
fµ (0, 0) 6= 0. (20)

Together, the conditions (19)-(20) imply that x = 0 is a nonhyperbolic fixed point. For there to be a
quadratic tangency, however, we also need to have µ00 (0) 6= 0. By differentiating (18) in x, evaluating the
resulting expression at x = 0 and using (19) it can be shown that

fxx (0, 0) + fµ (0, 0)µ00 (0) = 0.

Because fµ (0, 0) 6= 0, µ00 (0) 6= 0 only if

fxx (0, 0) 6= 0. (21)
In conclusion, a saddle-node bifurcation with quadratic tangency will exist if the three conditions (19)-(21)
are satisfied. We can also deduce that the Taylor series expansion of f about such a bifurcation point will
have the form
f (x, µ) = a0 µ + a1 x2 + a2 xµ + a3 µ2 + O(3)
for some constants a0 6= 0,a1 6= 0, a2 and a3 . Here, O(3) is notation to indicate higher order terms in the
Taylor series, i.e., x3 , x2 µ, . . .. In the next section we will discuss how one can simplify this expression to
create what is called the “normal form” for the bifurcation.

Theorem 5.4 Let ẋ = f (x, µ) and assume that for all (µ, x) near some point (µ∗ , x∗ ) f has continuous
(mixed) derivatives up to and including third order, i.e., fx , fµ , fxx , . . . , fxµµ , fµµµ . If

f (x∗ , µ∗ ) = 0 (22)
∗ ∗
fx (x , µ ) = 0 (23)
fµ (x∗ , µ∗ ) 6= 0 (24)
fxx (x∗ , µ∗ ) 6= 0 (25)

then ẋ = f (x, µ) has a saddle-node bifurcation with quadratic tangency at (µ∗ , x∗ ).

Example: Consider
ẋ = f (x, µ) = µ − x − e−x .
A nonhyperbolic equilibria exists at any pair of (µ, x) such that

f = 0 ⇔ µ − x − e−x = 0
fx = 0 ⇔ −1 + e−x = 0

Thus (µ∗ , x∗ ) = (1, 0) is a bifurcation point. Since fxx (x, µ) = −e−x , fxx (0, 1) = −1 6= 0. Lastly, fµ = 1 6= 0
so that a saddle-node bifurcation of quadratic tangency occurs at (µ∗ , x∗ ) = (1, 0), i.e.,

f =0 , fx = 0 , fµ 6= 0 , fxx 6= 0.

The (2-variable) Taylor series of f (x, µ) about (µ∗ , x∗ ) = (1, 0) is

1 1
f (x, µ) = (µ − 1) − x2 + x3 + O(x4 )
2 6
so that
1
ẋ = η − x2 + O(x3 )
2
where η = µ − 1 is a new parameter. For this example, the equation above is in “normal form”.
Note that some saddle-node bifurcations do not result from “quadratic” tangencies. For example,

ẋ = µ + x4

has a saddle-node bifurcation at (µ, x) = (0, 0) even though fxx (0, 0) = 0. This is an example of a “quartic”
tangency. Clearly other more complicated variants can occur as well. For example, just consider what
happens at (µ, x) = (0, 0) if
ẋ = (x2 − µ)(x2 − 4µ).
Also, some bifurcations that are quadratic are not a result of a saddle-node bifurcation. For example, consider
√
ẋ = f (x, µ) = µ−x , µ ≥ 0,
√
which has a sole branch of fixed points x = µ and a quadratic tangency at x = 0. At (µ∗ , x∗ ) = (0, 0) two
branches of fixed points do not coalescece. This example does not violate Theorem 5.4 since fµ (0, 0) is not
defined, let alone continuous for all (µ, x) near (µ∗ , x∗ ) = (0, 0).

5.2 Transcritical Local Theory

Suppose that two disjoint branches of fixed points intersect at a transcritical bifurcation point. From Figure 1
it is easy to see that near (TC), f (x, µ) = 0 does not imply that there is a single function which describes
both branches of fixed points.
Considering the implicit function theorem (IFT) fµ must vanish at a (TC) bifurcation point. Alternately
by swapping x and µ in the theorem, fx must also vanish. The latter implies that fixed points must (like
at saddle node bifurcations) be nonhyperbolic at transcritical bifurcations. Like saddle node bifurcations,
there are theorems which place sufficiency conditions on f assuring that transcritical bifurcations exist.

Theorem 5.5 Let f : IR2 → IR, f = f (x, µ) have continuous derivatives of all orders up to and including
degree three, i.e., f, fµ , . . . fxxµ near (µ∗ , x∗ ). If

f (x∗ , µ∗ ) = 0 (26)
fx (x∗ , µ∗ ) = 0 (27)
∗ ∗
fµ (x , µ ) = 0 (28)
∗ ∗
fxµ (x , µ ) 6= 0 (29)
∗ ∗
fxx (x , µ ) 6= 0 (30)

then ẋ = f (x, µ) has a ( 2 branch) transcritical bifurcation at (µ∗ , x∗ ).

Figure 1: Generic transcritical bifurcation showing that f (x, µ) = 0 does not imply µ = µ(x) or x = x(µ).

Example: Let
ẋ = f (x, µ) = µln(x) + x − 1
The first three conditions of the Theorem are:
f = µln(x) + x − 1 = 0 (31)
µ
fx = +1=0 (32)
x
fµ = ln(x) = 0 (33)
∗ ∗
Solving (32)-(33) one finds (µ , x ) = (−1, 1) as a candidate for a (TC). It is easily verified that (31) is also
satisfied and that
1
fxµ = 6= 0
x
µ
fxx = − 2 6= 0
x
at (µ∗ , x∗ ) = (−1, 1). By the Theorem one concludes there is a (2-branch) transcritical bifurcation point at
(µ∗ , x∗ ) = (−1, 1).
Much like saddle node bifurcations, such Theorems lead to normal forms for (TC) bifurcations. For
example, if (µ∗ , x∗ ) is a pair which satisfies the hypotheses of the Theorem then
1 1
ẋ =fµµ (µ∗ , x∗ )(µ − µ∗ )2 + fxµ (µ∗ , x∗ )(x − x∗ )(µ − µ∗ ) + fxx (µ∗ , x∗ )(x − x∗ )2 + O(3)
2 2
Making the definitions
y = x − x∗ , η = µ − µ∗
one has the normal form
ẏ = aη 2 + bηy + cy 2 + O(3)
where a, b, c are constants and bc 6= 0.
Example: For the previous example
ẋ = f (x, µ) = µln(x) + x − 1
where (µ∗ , x∗ ) = (−1, 1) is a (TC) bifurcation point,
1
ẋ = (x − 1)(µ + 1) + (x − 1)2 + O(3)
2
leading to the normal form  
1
ẏ = y η + y + O(3)
2