146
Chapter 4. Local Properties of Differentiable Mappings
Exercise 4.1.10. Find the form of the Laplace operator
∆u 
∂ 2u ∂ 2 u
+ 2
∂x2
∂y
in the polar coordinates in the set
G = {(x, y) ∈ R2 : x2 + y 2 > 0}
for
u ∈ C 2 (G).
Hint. If v(r, ϕ) = u(r cos ϕ, r sin ϕ), then
(∆u) ◦ Φ =
∂ 2v
1 ∂ 2v
1 ∂v
+
+
∂r2
r2 ∂ϕ2
r ∂r
where Φ(r, ϕ) = (r cos ϕ, r sin ϕ)
is the transformation. Note that we have


∂u ∂u
∂v ∂v
,
,
=
∂x ∂y
∂r ∂ϕ
◦ (Φ )−1 .
It is more comfortable to use this formula once again to compute
∂2 u ∂2u
∂x2 , ∂y 2 .
Exercise 4.1.11. Show that the estimate
%[f (x)]−1 %L(Y,X) ≤ c + d%x%X
is sufficient in Theorem 4.1.7.
Hint. Use the Gronwall Lemma (Exercise 5.1.16) to estimate %ω (t)%.
4.2 Implicit Function Theorem
Let us start with a simple example of f : R2 → R, e.g.,
f (x, y)  x2 + y 2 − 1.
Denote
M = {(x, y) ∈ R2 : f (x, y) = 0},
i.e., M is the unit circle in R2 . We would like to solve the equation
f (x, y) = 0
for the unknown variable y or to express M as the graph of a certain function
y = ϕ(x).
√
We immediately see that for any x ∈ (−1, 1) there is a pair of y’s (y1,2 = ± 1 − x2 )
such that (x, y) ∈ M. In particular, M is not a graph of any function y = ϕ(x).
We can only obtain that M is locally a graph, i.e., for (a, b) ∈ M, a ∈ (−1, 1),
4.2. Implicit Function Theorem
147
there is a neighborhood U of (a, b) such that M ∩ U is the graph of a function
y = ϕ(x). On the other hand, for x = ±1 there is a unique y (y = 0) for which
(x, y) ∈ M. But there is no neighborhood U of (1, 0) such that M ∩ U is the graph
of a function y = ϕ(x). What is the difference between these two cases?
In the former case the tangent line to M ∩ U exists at the point (a, b) with
the slope ϕ (a). Since
f (x, ϕ(x)) = 0
for x ∈ (a − δ, a + δ),
we have (formally by the Chain Rule)
∂f
∂f
(a, b) +
(a, b)ϕ (a) = 0,
∂x
∂y
i.e., ϕ (a) = − ab , since
∂f
∂y (a, b)
(4.2.1)
= 2b .= 0.
In the latter case, where (a, b) = (1, 0), we have ∂f
∂y (1, 0) = 0, and ϕ (1)
cannot be determined from (4.2.1). The tangent line to M at the point (0, 1) is
parallel to the y-axis, which indicates some problems with determining a solution,
i.e., the (implicit) function ϕ. The reader is invited to sketch a figure.
This discussion shows the importance of the assumption
∂f
(a, b) .= 0.
∂y
How can this assumption be generalized to f : RM+N → RN ? A brief inspection
of the linear case leads to the observation that we can compute the unknowns
yM+1 , . . . , yM+N from the equations
fi (y1 , . . . , yM+N ) =
M+N
!
aij yj = 0,
i = 1, . . . , N,
j=1
uniquely as functions of y1 , . . . , yM if and only if
det (aij )
i=1,...,N
j=M+1,...,M +N
.= 0.
∂fi
Nevertheless, aij = ∂y
, and the condition on the regularity of the matrix
j
means that the partial (Fréchet) derivative of f (see Defini(aij )
i=1,...,N
j=M+1,...,M +N
tion 3.2.17) with respect to the last N variables is an isomorphism of RN .
Theorem 4.2.1 (Implicit Function Theorem). Let X, Y , Z be Banach spaces,
f : X × Y → Z. Let (a, b) ∈ X × Y be such a point that
f (a, b) = o.
Let G be an open set in X × Y containing the point (a, b). Let f ∈ C 1 (G) and let
the partial Fréchet derivative f2 (a, b) be an isomorphism of Y onto Z.
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Chapter 4. Local Properties of Differentiable Mappings
Then there are neighborhoods U of a and V of b such that for any x ∈ U
there exists a unique y ∈ V for which
f (x, y) = o.
Denote this y by ϕ(x). Then ϕ ∈ C 1 (U). Moreover, if f ∈ C k (G), k ∈ N, then
ϕ ∈ C k (U).
Proof. We denote A  [f2 (a, b)]−1 and define
F (x, y) = (x, Af (x, y)),
(x, y) ∈ G.
Then F : X × Y → X × Y , F ∈ C 1 (G) and
F (a, b)(h, k) = (h, Af (a, b)(h, k)).
One can verify that F (a, b) is an isomorphism of X × Y onto itself. Hence we can
apply Theorem 4.1.1 to get neighborhoods U × V of (a, b) and Ũ × Ṽ of (a, o) such
that for any ξ ∈ Ũ and η = o ∈ Ṽ there exists a unique (x, y) ∈ U × V such that
F (x, y) = (x, Af (x, y)) = (ξ, o),
i.e.,
x = ξ,
U = Ũ,
and, denoting y = ϕ(x),
f (x, ϕ(x)) = o.
This means that
F −1 (x, o) = (x, ϕ(x)).
Since the inverse F −1 is differentiable, by Theorem 4.1.1 we conclude that ϕ ∈
C 1 (U).

Remark 4.2.2. We can also deduce a formula for ϕ (x): Indeed, since
f (x, ϕ(x)) = o
for every x ∈ U
and both the functions f and ϕ are differentiable, we get from the Chain Rule
f1 (x, ϕ(x)) + f2 (x, ϕ(x)) ◦ ϕ (x) = o,
and therefore
ϕ (x) = −[f2 (x, ϕ(x))]−1 ◦ f1 (x, ϕ(x))
for
x ∈ U1
(4.2.2)
where U1 ⊂ U may be smaller if necessary in order to guarantee the existence of
the inverse
[f2 (x, ϕ(x))]−1
(see Exercise 2.1.33).
Remark 4.2.3. The statement of Theorem 4.2.1 is by no means the best one. If we
have used the Contraction Principle directly we would obtain the existence of a
solution y = ϕ(x) under weaker assumptions. Namely, f (x, y) = o is equivalent to
y = y − Af (x, y)
4.2. Implicit Function Theorem
149
and since x is a parameter here we do not need to assume the differentiability
with respect to x if we content ourselves just with the existence of ϕ (and give up
its differentiability). We recommend that the reader uses directly the Contraction
Principle to obtain the following statement:
Let X be a normed linear space, Y , Z be Banach spaces and let
f : X × Y → Z be continuous at the point (a, b) where
f (a, b) = o.
Assume that the partial Fréchet derivative f2 (a, b) is an isomorphism
of Y onto Z and f2 : X × Y → L(Y, Z) is continuous at (a, b).
Then there are neighborhoods U of a and V of b such that for any
x ∈ U there is a unique y = ϕ(x) ∈ V for which
f (x, ϕ(x)) = o.
Moreover, ϕ is continuous at a.
It is also possible to avoid partly the requirement of invertibility of f2 (a, b)
(see Remark 4.1.6 and references given there).
There are many examples in Calculus where the Implicit Function Theorem
is used. We give one in Exercise 4.2.9, see also exercises in Dieudonné [35]. Our
attention is turned mainly towards more theoretical applications.
Example 4.2.4. Let
P (z) = z n + an−1 z n−1 + · · · + a0
be a polynomial with real or complex coefficients a0 , . . . , an−1 . The famous Fundamental Theorem of Algebra says that if n ≥ 1, then the equation P (z) = 0 has
at least one solution z̃ ∈ C and actually n solutions if all of them are counted with
their multiplicity. This means that P can be factorized as follows:
P (z) = (z − z1 )k1 · · · (z − zl )kl ,
k1 + · · · + kl = n,
where z1 , . . . , zl are different. A natural question arises: How do these solutions
z1 , . . . , zl depend on the coefficients a0 , . . . , an−1 of P ? Let
F (z, y0 , . . . , yn−1 ) = z n + yn−1 z n−1 + · · · + y0 : C × Cn → C.
Then
F (z1 , a0 , . . . , an−1 ) = P (z1 ) = 0
and
If z1 is a simple root, i.e., k1 = 1, then
∂F
(z1 , a0 , . . . , an−1 ) .= 0,
∂z
F ∈ C ∞ (C × Cn ).
150
Chapter 4. Local Properties of Differentiable Mappings
and the Implicit Function Theorem says that z1 depends continuously on
a0 , . . . , an−1 (also in the real case). But what happens if k1 > 1? Notice that
the cases of real and complex roots are different. In the former case, the real root
can disappear (x2 + ε = 0 for ε > 0), and in the latter case, the uniqueness can be
lost. Since the solution z1 ramifies or bifurcates at a0 , . . . , an−1 , this phenomenon
is called a bifurcation. We postpone a basic discussion of this very important

nonlinear phenomenon till the end of the next section.
Example 4.2.5 (dependence of solutions on initial conditions). Suppose that
f : R × RN → RN is continuous in an open set G ⊂ R × RN and has continuous
partial derivatives with respect to the last N variables in G. Denote by ϕ(·; τ, ξ) a
(unique) solution of the initial value problem
"
ẋ = f (t, x),
x(τ ) = ξ
(see Theorem 2.3.4). We are now interested in the properties of ϕ with respect to
the variables (τ, ξ) ∈ G, cf. Remark 2.3.5. Let us define
# t
f (s, ϕ(s)) ds − ϕ(t).
(4.2.3)
Φ(τ, ξ, ϕ)(t)  ξ +
τ
For a fixed (t0 , x0 ) ∈ G the solution ϕ(·; t0 , x0 ) of Φ(t0 , x0 , ϕ) = o is defined
on an open interval J . Choose a compact interval I ⊂ J such that t0 ∈ int I.
Then the mapping Φ given by (4.2.3) is defined on a certain open subset H ⊂
R × RN × C(I, RN ) and takes its values in C(I, RN ). Further,
Φ(t0 , x0 , ϕ(·; t0 , x0 )) = o
and
[Φ1 (τ, ξ, ϕ)](t) = −f (τ, ϕ(τ )),
[Φ2 (τ, ξ, ϕ)η](t) = η,
# t
[Φ3 (τ, ξ, ϕ)ψ](t) =
f2 (s, ϕ(s))ψ(s) ds − ψ(t),
τ
t ∈ I,
η ∈ RN ,
ψ ∈ C(I, RN ) whenever (τ, ξ, ϕ) ∈ H.
Since these partial Fréchet derivatives are continuous, Φ ∈ C 1 (H) (see Proposition 3.2.18). The crucial assumption of the Implicit Function Theorem is the continuous invertibility of Φ3 (t0 , x0 , ϕ(·; t0 , x0 )) in the space C(I, RN ). Put
# t
Bψ(t) =
f2 (s, ϕ(s; t0 , x0 ))ψ(s) ds,
ψ ∈ C(I, RN ).
t0
We have proved in Example 2.3.7 that σ(B) = {0}. In particular,
B − I = Φ3 (t0 , x0 , ϕ(·; t0 , x0 ))
4.2. Implicit Function Theorem
151
is continuously invertible. By Theorem 4.2.1, there exist neighborhoods U of
(t0 , x0 ) and V of ϕ(·; t0 , x0 ) such that for any (τ, ξ) ∈ U there is a unique ϕ ∈ V
such that
Φ(τ, ξ, ϕ) = o.
Moreover, this ϕ is continuously differentiable with respect to τ and ξ, and for the
continuous mappings
Θ(·) 
∂ϕ
(·; t0 , x0 )
∂τ
and
Ξ(·) 
∂ϕ
(·; t0 , x0 )
∂ξ
we have, by Remark 4.2.2,
# t
−f (t0 , x0 ) +
f2 (s, ϕ(s; t0 , x0 ))Θ(s) ds − Θ(t) = o,
t0
η+
#
t
t0
f2 (s, ϕ(s; t0 , x0 ))Ξ(s)η ds − Ξ(t)η = o,
η ∈ RN .
This means that Θ and Ξ solve the so-called equation in variations
ẏ(t) = f2 (t, ϕ(t; t0 , x0 ))y(t)
(4.2.4)
(this is a system of N linear equations for Θ and a system of N × N equations for
Ξ) and fulfil the initial conditions
Θ(t0 ) = −f (t0 , x0 ),
Ξ(t0 ) = I.
In particular, Ξ(·) is a fundamental matrix of (4.2.4).
(4.2.5)

As an application of differentiability with respect to initial conditions we
briefly sketch the approach to orbital stability of periodic solutions.
Example 4.2.6. Assume that we know a non-constant T -periodic solution ϕ0 of an
autonomous system
ẋ = f (x),
and that we are interested in the behavior of other solutions which start at time
t = 0 near ϕ0 (0) = x0 . We assume that f ∈ C 1 (G), G is an open set in RN , and
denote by ϕ(·, ξ) the solution satisfying ϕ(0, ξ) = ξ. Let
M = {x ∈ RN : (x − x0 , f (x0 ))RN = 0}.
In order to show that a solution ϕ(·, ξ) exists on such an interval [0, t(ξ)] that it
meets M ∩ U (U is a neighborhood of x0 ) for the first positive time t(ξ) near T
(T is the period of ϕ0 ), see Figure 4.2.1, we can solve the equation
Φ(t, ξ)  (ϕ(t, ξ) − x0 , f (x0 )) = 0
in the vicinity of the point (T, x0 ). We have
Φ1 (T, x0 ) = (f (x0 ), f (x0 )) > 0