MATH2111 Higher Several Variable Calculus
Inverse and Implicit Function Theorems
Dr. Jonathan Kress
School of Mathematics and Statistics
University of New South Wales
Semester 1, 2016 [updated: February 29, 2016]
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MATH2111 Analysis
Inverse function theorem for
a
a
b
f 0 (c ) = 0, f not invertible on (a, b)
a
b
f 0 (c ) = 0, f invertible on (a, b)
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:R→R
b
f invertible on (a, b)
a
f
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MATH2111 Analysis
b
f not invertible on (a, b)
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2 / 24
Inverse function theorem for
f
:R→R
From rst year. . .
Theorem (Inverse function theorem)
If f : R → R is dierentiable on an interval I ⊂ R and f 0 (x ) 6= 0 for all x ∈ I ,
then f is invertible on I and the inverse f −1 is dierentiable with
(f −1 )0 (x ) =
1
f 0 (f −1 (x ))
.
That is, if y = f (x ) then f −1 exists and is dierentiable with x = f −1 (y ) and
dx
1
.
=
dy
dy
dx
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Inverse function theorem
Consider an ane function T : R → R given by
T (x ) = mx + b.
T is dierentiable on R with T 0 (x ) = m. When m 6= 0, T is invertible and
T (x ) = mx + b ⇒ x = m−1 T (x ) − m−1 b
so
T −1 (x ) = m−1 x − m−1 b.
T −1 is dierentiable and
(T −1 )0 (x ) = m−1 .
If T is a good ane approximation to f near c then it seems plausible that on a
small enough interval around c, the existence of T −1 guarantees the existence of
f −1 with good ane approximation T −1 .
Note that
(f −1 )0 (f (x )) = m−1 and f 0 (x ) = m.
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Inverse function theorem
Consider an ane function T : Rn → Rn given by
T(x) = Lx + b
T is dierentiable on Rn with D T = L. When detL 6= 0, T is invertible and
T(x) = Lx + b ⇒ x = L−1 T(x) − L−1 b
so
T−1 (x) = L−1 x − L−1 b.
If T is a good ane approximation to f near c then it seems plausible that on a
small enough ball around c, the existence of T−1 guarantees the existence of f −1
with good ane approximation T−1 .
We would expect
Dc f = L and Df (c) f
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−1
= L−1 .
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Inverse function theorem
Theorem
Let Ω ⊂ Rn be open, f : Ω → Rn be C 1 and suppose a ∈ Ω.
If D f (a) is invertible (as a matrix) then f is invertible on an open set U containing
a. That is,
f −1 : f (U ) → U
exists.
Furthermore, f −1 is C 1 and for x ∈ U,
D f (x ) f
−1
= Dx f
−1
.
Note that this says f −1 has a good ane approximation at f (a) given by
f
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−1
(x) ' a + Da f
−1 MATH2111 Analysis
x − f (a) .
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Inverse function theorem
Example: Can the map x = r cos θ, y = r sin θ be inverted?
Dene f :
→
by
x
r
r cos θ
=f
=
y
θ
r sin θ
R2
Eg, at a =
So
R2
∂x
Df = 
 ∂∂yr
∂r

∂x

∂θ  = cos θ
∂y
sin θ
∂θ
−r sin θ
r cos θ
4
, f (a) = (1, 1).
2
D f
and det(D f ) = r cos2 θ + r sin2 θ = r 6= 0.
So f is locally invertible away from r = 0.
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2,
π
 1
√
√ π  2
D f 2,
= 1
4
√
Away from (x , y ) = (0, 0) (ie r = 0) f is
dierentiable with D f = J f so

√
−1
√
(1, 1) = D f

1
√

= 2
1
−
MATH2111 Analysis

−1


1
π −1
2,
4
1
√
2
1 .
2 2
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Inverse function theorem
We can check that this matches what we get from directly inverting f . In the rst
quadrant away from 0,
p
p
2 + y2
x
x
r = x 2 + y 2 , θ = tan−1 (y /x ) ⇒ f −1
=
y
tan−1 (y /x )



x
y
1
1 
p
p
√
√
2 + y2 
 x2 + y2
 2
x
−1
−1 1
2
⇒ Df = 
=
 ⇒ Df
−y
x
1 1 .
1
−
x2 + y2
x2 + y2
2 2
An ane approximation to f −1 near
f −1
x
1
1 x −1
' f −1
+ D ( f −1 )
y
1
1 y −1

1
1 
√ !
√
√
2
x
−
1


2
= π + 2
1 1  y −1 .
−
4
2 2
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1
is
1
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inverse-polar.canvas
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Inverse function theorem
Suppose f : R2 → R2 is dened by
x
x 3 e y + y − 2x
f
=
.
y
2xy + 2x
1
−1
Note f
=
.
0
2
Show that f hasa dierentiable
inverse near (1, 0) and hence nd an approximate
−1.2
solution to f −1
, that is, an approximate solution to
2.1
x 3 e y + y − 2x = −1.2,
2xy + 2x = 2.1.
The partial derivatives of the components of f exist and are continuous
everywhere. Hence f is dierentiable on R2 and
3x 2 e y − 2 x 3 e y + 1
Df = Jf =
2y + 2
2x
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1 2
D f (1, 0) =
.
2 2
⇒
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Inverse function theorem
1
= −2 6= 0, the Inverse Function Theorem says that f has a
0
C 1 local inverse near (1, 0) with derivative
D f −1
Since det D f
−1
2
= Df
1 −1
0
1 2 −2
−1 1
=
.
=−
1 − 12
2 −2 1
Now, the best ane approximation to f −1 is
f −1
u
−1
−1
' f −1
+ D f −1
v
2
2
−1 1
u − (−1)
1
=
+
v −2
0
1 − 12
u+1
v −2
So now the approximate solution is
−1 1
x
−1.2
1
= f −1
'
+
y
2.1
0
1 − 12
−0.2
1 1 −0.6
1.3
=
=
−
0.1
0 2 0.5
−0.25
1.3
−1.14
−1.02
1.03
1.03
−1.019
f
'
, f −1
'
, f
'
.
−0.25
1.95
2.01
−0.025
−0.025
2.009
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Implicit function theorem
Consider a C 1 function g : R2 → R, its 0 contour,
S = {(x , y ) ∈ R2 : g (x , y ) = 0}
and a point (x0 , y0 ) ∈ S. When does S dene y as a function of x near the point
(x0 , y0 )?
At A and B but not C.
y
A
B
C
x
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Implicit function theorem
Given
g : R2 → R is C 1 ,
g (x0 , y0 ) = 0 and
∂g
(x0 , y0 ) 6= 0,
∂y
we want to show that there
is a δ such that for
y0 + δ
g (x , y ) = 0
y
y0
x0
x0 − δ
x ∈ (x0 − δ, x0 + δ)
x
x0 + δ
there is a unique
y ∈ (y0 − δ, y0 + δ)
satisfying
y0 − δ
g (x , y ) = 0.
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Implicit function theorem
y0 + a
Assume that
∂g
(x0 , y0 ) > 0.
∂y
As g is C 1 , there are a > 0 and
b > 0 such that for
g (x , y ) = 0
x ∈ (x0 − a, x0 + a)
y0
y ∈ (y0 − a, y0 + a),
x0
x0 − a
x0 + a
for some M > 0.
y0 − a
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∂g
(x , y ) > b
∂y
∂g
(x , y ) < M
∂x
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Implicit function theorem
y0 + a
y0 + δ
Choose positive a0and δ sothat
ba
a0 < a, δ < min a0 , 0 .
M
(x , y0 + a0 )
g (x , y0 + a0 ) > 0 − M δ + ba0
> −ba0 + ba0 = 0.
g (x , y0 − a0 ) < 0 + M δ − ba0
< ba0 − ba0 = 0.
g (x , y ) = 0
y
y0
x0
x0 − a
x0 − δ
y0 − δ
y0 − a
x
x0 + a
x0 + δ
IVT ⇒ ∃y ∈ (y0 − a0 , y0 + a0 )
such that g (x , y ) = 0.
∂g
> 0 ⇒ y is unique.
∂y
(x , y0 − a0 )
MVT ⇒ g (x , y0 ± a0 ) = g (x0 , y0 ) +
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∂g
∂g ±
(c , y0 )(x − x0 ) +
(x , d ± )(a0 − y0 ).
∂x
∂y
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Implicit function theorem
We have shown that given
g : R2 → R is C 1 ,
g (x0 , y0 ) = 0 and
∂g
(x0 , y0 ) 6= 0,
∂y
there is a δ such that for
So, there is f : (x0 − δ, x0 + δ) → R such that
g (x , f (x )) = 0.
It can also be shown that f is C 1 . Assuming
f is dierentiable, we can nd f 0 by implicit
dierentiation and nd
d
g (x , f (x )) = 0
dx
∂ g dx
∂ g dy
⇒
+
=0
∂ x dx
∂ y dx
∂g
∂g 0
⇒
+
f (x ) = 0
∂x
∂y
x ∈ (x0 − δ, x0 + δ)
there is a unique
y ∈ (y0 − δ, y0 + δ)
satisfying
g (x , y ) = 0.
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⇒ f (x0 ) = −
0
MATH2111 Analysis
∂g
(x0 , y0 )
∂y
−1
∂g
(x0 , y0 ).
∂x
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Implicit Function Theorem
For the Implicit Function Theorem in higher dimensions, consider the following.
Near which points does
2
2
2
x + y + z = 1 ⇔ z = ± 1 − x2 − y2
p
dene z as a function of x and y? That is, when does there exist f such that
z = f (x , y )?
Given
y +z =6−x
x +y +z =6
1 1
⇔
A=
−1 2
−y + 2z = 8 − 2x
2x − y + 2z = 8,
you can nd y and z given
just the value of x. So there is a function
y
f : R → R2 such that
= f (x ).
z
Typically, if there are n equations and r variables, we expect to be able to solve
for n of variables in terms of the remaining n − r variables near most points.
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Implicit Function Theorem
Let x ∈ Rm denote our known variables and let u ∈ Rn denote our unknown
variables. To solve for u in terms of x we expect to need n equations:
g1 (x1 , . . . , xm , u1 , . . . , un ) = 0
g2 (x1 , . . . , xm , u1 , . . . , un ) = 0
..
..
.
.
gn (x1 , . . . , xm , u1 , . . . , un ) = 0.
We can write this more succinctly as
where g : Rm+n → Rn is
g(x, u) = 0
g(x, u) = (g1 (x, u), . . . , gn (x, u)).
Solving this system of equations means nding a way of specifying what u is if we
know x. That is, we need to nd a continuous function f : Rm → Rn satisfying
g(x, f (x)) = 0.
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Implicit Function Theorem
Dene the n × m matrix A and n × n matrix B in terms of D g.
∂ g1
 ∂ x1
 .
Dg = 
 ..
 ∂g

n
∂ x1
∂ g1
∂ x2
..
.
∂ gn
∂ x2
∂ g1
∂ xm
···
..
.
∂ gn
···
∂ xm
..
.
∂ g1
∂ u1
|
···
..
.
∂ gn
|
∂ u1
..
|
.
···

∂ g1
∂ un 
.. 
. 
 = [A|B ]
∂g 
n
∂ un
Theorem (Implicit Function Theorem)
Suppose that (x0 , u0 ) is on the surface g(x, u) = 0. If B (x0 , u0 ) is an invertible
matrix, then there is an open set V around x0 on which u is dened implicitly as a
function of x. That is, there exists a continuously dierentiable function
f : Rm → Rn such that for all x ∈ V
g(x, f (x)) = 0.
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Implicit Function Theorem
To nd D f in terms of D g use the chain rule. Let h : Rm → Rm+n be dened by

x1
..
.
xm









x

h(x) =
=

f
(
x
,
.
.
.
,
x
)
f (x)
m
1 1



.
..


f n ( x1 , . . . , xm )
I
Dx h = m
Dx f
⇒
where Im is the m × m identity matrix. Dierentiating the equation
g(x, f (x)) = (g ◦ h)(x) = 0
gives
0 = Dh(x) g Dx h =
(A(h(x)) | B (h(x)))
Im
= A(h(x)) + B (h(x))Dx f .
Dx f
Rearranging this gives Dx f = −B (h(x))−1 A(h(x)).
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Implicit Function Theorem
Show that there are open sets U ⊂
R2
containing
1
and V ⊂ R2 containing
2
1
so that the equations
5
x 2 + xy + yu + u 2 − xv − 1 = 0,
y 2 + xy − u 2 − v = 0,
dene a dierentiable
function
f : U → V for which (x , y , u , v ) satises the
u
x
equations when
=f
.
v
y
1
and hence nd an approximate
2
u
x
1.2
solution
to these equations when
=
.
v
y
1.9
Find the ane approximation to f near
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Implicit Function Theorem
x
u
0
The equations can be written in the form g
,
=
. Then
y
v
0
2x + y − v x + u y + 2u −x
Dg =
.
y
x + 2y −2u −1
1
1
At the known point x0 =
, u0 =
, this gives
2
5
D g(x0 , u0 ) =
−1 2
4 −1
= [A | B ].
−2 −1
|
5 |
2
B is invertible and so there is a C 1 function f
x
x
x0 so that g
,f
= 0, and
y
x
y
dened on an open set around
y
1 −1 1
D f (x0 ) = −B −1 A = −
−6 2 4
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−1
MATH2111 Analysis
2
1 3 3
2
=
.
5
6 6 24
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Implicit Function Theorem
x
Thus the ane approximation to f
y
near
1
is
2
1
1 3 3
f (x) ≈ f (x0 ) + D f (x0 )(x − x0 ) =
+
5
6 6 24
x −1
.
y −2
In particular
1.2
1
1 3 3
f
≈
+
1.9
5
6 6 24
0.2
1.05
=
.
−0.1
4.8
You can check whether this is any good by calculating g
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1.2
1.05
,
1.9
4.8
MATH2111 Analysis
.
Semester 1, 2016
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Implicit Function Theorem
What we have done is to replace the original equations
g(x , y , u , v ) = 0
with the equations
x −1

−1 2 4 −1 
y
−
2

= 0
T(x , y , u , v ) = g(1, 2, 1, 5) +
2 5 −2 −1 u − 1
0
v −5


where T is the best ane approximation to g near (1, 2, 1, 5). That is, since
g(1, 2, 1, 5) = 0, the given equations are approximately
−(x − 1) + 2(y − 2) + 4(u − 1) − (v − 5)
0
2(x − 1) + 5(y − 2) − 2(u − 1) − (v − 5) = 0
=
which simpies to the pair of linear equations
−x + 2y + 4u − v
=
−2
2x + 5y − 2u − v = −5.
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