Chapter 3
The Implicit Function Theorem
and Its Applications
We present the Inverse Mapping Theorem first (Theorem 3.18 in the text)
and then the Implicit Function Theorem (Theorem 3.9 in the text)
Theorem 3.1
(The inverse mapping theorem). Let U and V be open sets
in R
n
and a ∈ U. Let f : U → V be a mapping of class C
1
and b =
f (a). Suppose Df (a) is invertible, that is, det Df (a) 6= 0. Then there exist
neighborhoods M ⊂ U of a and N ⊂ V of b, so that f is a one-to-one map
from M onto N, and the inverse map f
−
1
from N to M is also of class C
1
.
Moreover, if x ∈ M and y = f (x) ∈ N, then D(f
−
1
)(y) = [Df (x)]
−
1
.
Read the example on p. 138 of the textbook.
Let F : R
n
× R
k
→ R
k
. For x ∈ R
n
, y ∈ R
k
, F (x, y) = (F
1
, F
2
, . . . , F
k
) ∈
R
k
. We use the following notation
D
x
F =
∂
x
j
F
i
1≤i≤k
1≤j≤n
=
∂
x
1
F
1
∂
x
2
F
1
. . . ∂
x
n
F
1
∂
x
1
F
2
∂
x
2
F
2
. . . ∂
x
n
F
2
...
...
...
...
∂
x
1
F
k
∂
x
2
F
k
. . . ∂
x
n
F
k
,
35
36
CHAPTER 3.
THE IMPLICIT FUNCTION THEOREM
D
y
F =
∂
y
j
F
i
1≤i,j≤k
=
∂
y
1
F
1
∂
y
2
F
1
. . . ∂
y
k
F
1
∂
y
1
F
2
∂
y
2
F
2
. . . ∂
y
k
F
2
...
...
...
...
∂
y
1
F
k
∂
y
2
F
k
. . . ∂
y
k
F
k
.
Note that D
x
F is a k ×n matrix, D
y
F is a k ×k matrix and the derivative
of F is
DF = ( D
x
F
D
y
F ),
a k × (n + k) matrix.
Theorem 3.2.
Let U ⊂ R
n
× R
k
be open and F : U → R
k
is of class C
1
.
Let a ∈ R
n
, b ∈ R
k
such that (a, b) ∈ U. Suppose F (a, b) = 0 and the matrix
B = D
y
F (a, b) is invertible, that is, det B 6= 0. Then there are positive
numbers r
0
and r
1
such that
(i) For all x ∈ B(r
0
, a), there exists unique y ∈ B(r
1
, b) such that (x, y) ∈
U and F (x, y) = 0.
We define the function f : B(r
0
, a) → B(r
1
, b) as follows: for each x ∈
B(r
0
, a), f (x) is that unique y ∈ B(r
1
, b).
(ii) The function f above is of class C
1
and F (x, f (x)) = 0 for all x ∈
B(r
0
, a). Consequently, for x ∈ B(r
0
, a) and y = f (x), we have
Df (x) = −[D
y
F (x, y)]
−
1
D
x
F (x, y),
whenever D
y
F (x, y) is invertible.
Example 3.3.
Consider the problem of solving
x − yu
2
= 0,
xy + uv = 0
(3.1)
for u and v as functions of x and y.
Let n = k = 2. Set F = (F
1
, F
2
) = (x − yu
2
, xy + uv). We have
A = D
(x,y)
F =
1 −u
2
y
x
!
,
37
B = D
(u,v)
F =
−
2yu 0
v
u
!
.
We have det D
(u,v)
F (x, y, u, v) = −2yu
2
. Therefore, for any solution (x
0
, y
0
, u
0
, v
0
)
of (3.1) such that y
0
u
0
6
= 0, we can solve (3.1) for (u, v) = f (x, y) =
(u(x, y), v(x, y)) nearby the given point (x
0
, y
0
, u
0
, v
0
).
For example, let (x
0
, y
0
, u
0
, v
0
) = (1, 1, 1, −1) be a solution of (3.1). We
want to find also Df (1, 1). We have
A = D
(x,y)
F (1, 1, 1, −1) =
1 −1
1
1
!
,
B = D
(u,v)
F (1, 1, 1, −1) = D
(u,v)
F =
−
2 0
−
1 1
!
.
It is known that if ad − bc 6= 0, the inverse matrix of
a b
c d
!
is
a b
c d
!
−
1
=
1
ad − bc
d
−
b
−
c
a
!
.
We have det B = −2 6= 0 and hence B
−
1
=
−
1/2 0
−
1/2 1
!
. Thus
Df (1, 1) = −B
−
1
A = −
−
1/2 0
−
1/2 1
!
1 −1
1
1
!
= −
−
1/2 1/2
1/2
3/2
!
=
1/2
−
1/2
−
1/2 −3/2
!
.
This implies
∂
x
u(1, 1) = 1/2, ∂
y
u(1, 1) = −1/2, ∂
x
v(1, 1) = −1/2, ∂
y
v(1, 1) = −3/2.