MULTIVARIABLE AND VECTOR ANALYSIS
WWL CHEN
c W W L Chen,1997,2008.
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Chapter 3
IMPLICIT AND
INVERSE FUNCTION THEOREMS
3.1.Implicit Function Theorem
The Implicit function theorem,in its generality,is rather complicated to state and dicult to prove.
Here,we shall rst motivate the result by studying some simple examples.
Example 3.1.1.Consider the unit circle x
2
+z
2
= 1.Write F(x;z) = x
2
+z
2
1.Then the circle can
be described by the equation F(x;z) = 0.Consider now the point (x
0
;z
0
) = (1;0).
Concentrate on a small neighbourhood near the point x
0
= 1 and a small neighbourhood near the point
z
0
= 0.If we are given some value x near x
0
= 1 such that x < 1,can we nd a value z near z
0
= 0 such
that F(x;z) = 0?The answer in this case is not only yes,but we can nd two solutions z =
p
1 x
2
.
Chapter 3:Implicit and Inverse Function Theorems
page 1 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
In view of the lack of uniqueness,the answer for z cannot be described as a function of x.Next,observe
that the partial derivative
@F
@z
= 2z;so that
@F
@z
(1;0) = 0:
Consider next a point (x
0
;z
0
) where z
0
> 0.
Again concentrate on a small neighbourhood near the point x
0
and a small neighbourhood near the
point z
0
.If we are given some value x near x
0
,can we nd a value z near z
0
such that F(x;z) = 0?The
answer in this case is not only yes,but we can precisely one solution z =
p
1 x
2
,given by the positive
square root.The point z =
p
1 x
2
,which also satised the equation,is nowhere near the value z
0
.
It follows that near (x
0
;z
0
),the function z = g(x),where g(x) =
p
1 x
2
,gives the correct value of z
which satises F(x;z) = 0.Consider nally a point (x
0
;z
0
) where z
0
< 0.
Again concentrate on a small neighbourhood near the point x
0
and a small neighbourhood near the
point z
0
.If we are given some value x near x
0
,can we nd a value z near z
0
such that F(x;z) = 0?
The answer in this case is not only yes,but we can precisely one solution z =
p
1 x
2
,given by the
negative square root.The point z =
p
1 x
2
,which also satised the equation,is nowhere near the
value z
0
.It follows that near (x
0
;z
0
),the function z = h(x),where h(x) =
p
1 x
2
,gives the correct
value of z which satises F(x;z) = 0.Observe that in both of these last two cases,we have z
0
6= 0,and
so
@F
@z
(x
0
;z
0
) 6= 0:(1)
Indeed,it appears from observing the pictures that this last condition is sucient to guarantee that in
suitable neighbourhoods of x
0
and z
0
,the value of z can be given as a function of x.
Chapter 3:Implicit and Inverse Function Theorems
page 2 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
Example 3.1.2.Consider the parabola x +z
2
= 0.Write F(x;z) = x +z
2
.Then the parabola can be
described by the equation F(x;z) = 0.Consider now the point (x
0
;z
0
) = (0;0).
Concentrate on a small neighbourhood near the point x
0
= 0 and a small neighbourhood near the point
z
0
= 0.The reader should have little diculty in concluding that z cannot be described as a function
of x,and that the partial derivative
@F
@z
= 2z;so that
@F
@z
(0;0) = 0:
Example 3.1.3.Consider the curve xz
3
= 0.Write F(x;z) = xz
3
.Then the curve can be described
by the equation F(x;z) = 0.Consider now the point (x
0
;z
0
) = (0;0).
Concentrate on a small neighbourhood near the point x
0
= 0 and a small neighbourhood near the point
z
0
= 0.The reader should have little diculty in concluding that z = g(x),where g(x) = x
1=3
,gives the
correct value of z which satises F(x;z) = 0.Note here,though,that the partial derivative
@F
@z
= 3z
2
;so that
@F
@z
(0;0) = 0:
Let us analyze the situation more carefully.From Example 3.1.1,it appears that a condition such as
(1) is sucient to guarantee that in suitable neighbourhoods of x
0
and z
0
,the value of z can be described
as a function of x.However,from Examples 3.1.2 and 3.1.3,a condition such as
@F
@z
(x
0
;z
0
) = 0 (2)
does not give us any rm conclusion as to whether in suitable neighbourhoods of x
0
and z
0
,the value of
z can be described as a function of x.Indeed,we shall concentrate on situations analagous to (1) and
not those analogous to (2).
Chapter 3:Implicit and Inverse Function Theorems
page 3 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
We shall state our result in full,but will then continue to examine only special cases of it.
Throughout this section,(x;z) 2 R
n
R
m
denotes x = (x
1
;:::;x
n
) 2 R
n
and z = (z
1
;:::;z
m
) 2 R
m
.
For functions F
i
:R
n
R
m
!R,where i = 1;:::;m,we write
@(F
1
;:::;F
m
)
@(z
1
;:::;z
m
)
= det
0
B
B
B
B
B
B
B
B
B
@
@F
1
@z
1
:::
@F
1
@z
m
.
.
.
.
.
.
@F
m
@z
1
:::
@F
m
@z
m
1
C
C
C
C
C
C
C
C
C
A
whenever the partial derivatives in the matrix exist.
THEOREM 3A.(IMPLICIT FUNCTION THEOREM) Suppose that for every i = 1;:::;m,the
function F
i
:R
n
R
m
!R has continuous partial derivatives.Suppose further that there exists a point
(x
0
;z
0
) 2 R
n
R
m
such that
F
i
(x
0
;z
0
) = 0 for every i = 1;:::;m;
and
@(F
1
;:::;F
m
)
@(z
1
;:::;z
m
)
(x
0
;z
0
) 6= 0:
Then the following three conclusions hold:
(a) There exist neighbourhoods U R
n
of x
0
and V R
m
of z
0
such that there are unique functions
z
i
= g
i
(x) = g
i
(x
1
;:::;x
n
);where i = 1;:::;m;(3)
dened for every x 2 U and z 2 V and satisfying
F
i
(x;g
1
(x);:::;g
m
(x)) = 0 for every i = 1;:::;m:(4)
(b) If x 2 U and z 2 V satisfy F
i
(x;z) = 0 for every i = 1;:::;m,then (3) must hold.
(c) The functions (3) are continuously dierentiable,and for every i = 1;:::;m and j = 1;:::;n,we
have
@g
i
@x
j
=
@(F
1
;:::;F
m
)
@(z
1
;:::;z
i1
;x
j
;z
i+1
;:::;z
m
)
,
@(F
1
;:::;F
m
)
@(z
1
;:::;z
m
)
:(5)
Remark.Note that
@(F
1
;:::;F
m
)
@(z
1
;:::;z
m
)
(x
0
;z
0
) = det
0
B
B
B
B
B
B
B
B
B
@
@F
1
@z
1
(x
0
;z
0
):::
@F
1
@z
m
(x
0
;z
0
)
.
.
.
.
.
.
@F
m
@z
1
(x
0
;z
0
):::
@F
m
@z
m
(x
0
;z
0
)
1
C
C
C
C
C
C
C
C
C
A
:
Chapter 3:Implicit and Inverse Function Theorems
page 4 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
Note also that
@(F
1
;:::;F
m
)
@(z
1
;:::;z
i1
;x
j
;z
i+1
;:::;z
m
)
= det
0
B
B
B
B
B
B
B
B
B
@
@F
1
@z
1
:::
@F
1
@z
i1
@F
1
@x
j
@F
1
@z
i+1
:::
@F
1
@z
m
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
@F
m
@z
1
:::
@F
m
@z
i1
@F
m
@x
j
@F
m
@z
i+1
:::
@F
m
@z
m
1
C
C
C
C
C
C
C
C
C
A
:
We are interested in the special case m= 1.Here,(x;z) 2 R
n
R denotes x = (x
1
;:::;x
n
) 2 R
n
and
z 2 R.
THEOREM 3A'.Suppose that the function F:R
n
R!R has continuous partial derivatives.
Suppose further that there exists a point (x
0
;z
0
) 2 R
n
R such that
F(x
0
;z
0
) = 0 and
@F
@z
(x
0
;z
0
) 6= 0:
Then the following three conclusions hold:
(a) There exist neighbourhoods U R
n
of x
0
and V R of z
0
such that there is a unique function
z = g(x) = g(x
1
;:::;x
n
) (6)
dened for x 2 U and z 2 V and satisfying F(x;g(x)) = 0.
(b) On the other hand,if x 2 U and z 2 V satisfy F(x;z) = 0,then (6) must hold.
(c) Furthermore,the function (6) is continuously dierentiable,and for every j = 1;:::;n,we have
@g
@x
j
=
@F
@x
j
,
@F
@z
:(7)
Example 3.1.4.Consider the function
F:R
2
R!R:(x;y;z) 7!x
2
+y
2
+z
2
1:
Here n = 2 and m= 1.It is easy to see that F(x
0
;y
0
;z
0
) = 0 on the surface of a sphere centred at the
origin (0;0;0) and with radius 1.On the other hand,
@F
@z
(x
0
;y
0
;z
0
) = 2z
0
6= 0
whenever z
0
6= 0;in other words,this partial derivative does not vanish on the surface of the sphere
except on the\equator"z
0
= 0 and x
2
0
+y
2
0
= 1.Clearly the function
z = g(x;y) =
p
1 x
2
y
2
satises the requirements in a suciently small neighbourhood of (x
0
;y
0
;z
0
) if z
0
> 0,and the function
z = g(x;y) =
p
1 x
2
y
2
satises the requirements in a suciently small neighbourhood of (x
0
;y
0
;z
0
) if z
0
< 0.On the other
hand,in a neighbourhood of (x
0
;y
0
;z
0
) on the\equator",it is not clear whether we should take the
Chapter 3:Implicit and Inverse Function Theorems
page 5 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
positive or negative root.But then the partial derivative above vanishes here,so the theorem does not
apply in this case.Let us investigate the case z
0
> 0 further.We have
@g
@x
=
x
p
1 x
2
y
2
and
@F
@x
,
@F
@z
=
x
z
=
x
p
1 x
2
y
2
:
We also have
@g
@y
=
y
p
1 x
2
y
2
and
@F
@y
,
@F
@z
=
y
z
=
y
p
1 x
2
y
2
:
Example 3.1.5.Consider the functions
F
1
:R
2
R
2
!R:(x;y;z;w) 7!x
2
+y
2
+z
2
+w
2
2;
F
2
:R
2
R
2
!R:(x;y;z;w) 7!x
2
y
2
+z
2
w
2
:
Here n = 2 and m= 2.Note that
@(F
1
;F
2
)
@(z;w)
(z
0
;w
0
) = det
0
B
B
B
B
@
@F
1
@z
(z
0
;w
0
)
@F
1
@w
(z
0
;w
0
)
@F
2
@z
(z
0
;w
0
)
@F
2
@w
(z
0
;w
0
)
1
C
C
C
C
A
= det
2z
0
2w
0
2z
0
2w
0
= 8z
0
w
0
6= 0
if z
0
6= 0 and w
0
6= 0.Suppose now that F
1
(x
0
;y
0
;z
0
;w
0
) = F
2
(x
0
;y
0
;z
0
;w
0
) = 0 with z
0
> 0 and
w
0
> 0.Then it is easy to check that the functions
z = g
1
(x;y) =
p
1 x
2
and w = g
2
(x;y) =
p
1 y
2
satisfy F
1
(x;y;g
1
(x;y);g
2
(x;y)) = 0 and F
2
(x;y;g
1
(x;y);g
2
(x;y)) = 0.Furthermore,it is easily checked
(the reader is advised to ll in the details) that
@g
1
@x
=
@(F
1
;F
2
)
@(x;w)
,
@(F
1
;F
2
)
@(z;w)
=
x
p
1 x
2
;
@g
1
@y
=
@(F
1
;F
2
)
@(y;w)
,
@(F
1
;F
2
)
@(z;w)
= 0;
@g
2
@x
=
@(F
1
;F
2
)
@(z;x)
,
@(F
1
;F
2
)
@(z;w)
= 0;
@g
2
@y
=
@(F
1
;F
2
)
@(z;y)
,
@(F
1
;F
2
)
@(z;w)
=
y
p
1 y
2
:
y Sketch of Proof of Theorem 3A'.We shall show here how one may prove the special case n = 2
and m= 1.For convenience,we shall use the notation x = (x;y) and x
0
= (x
0
;y
0
).Since
@F
@z
(x
0
;z
0
) 6= 0;
we shall assume that it is positive (otherwise we simply consider F instead of F).By the continuity
of the partial derivative @F=@z,there exist a > 0 and b > 0 such that
@F
@z
(x;z) > b whenever kx x
0
k < a and jz z
0
j < a:
Chapter 3:Implicit and Inverse Function Theorems
page 6 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
In view of continuity,we may also assume that there exists M > 0 such that
@F
@x
(x;z)
< M and
@F
@y
(x;z)
< M
in the same region.Note next that since F(x
0
;z
0
) = 0,it follows that
F(x;z) = (F(x;z) F(x
0
;z)) +(F(x
0
;z) F(x
0
;z
0
)):
To study the term F(x;z) F(x
0
;z),we note that the line segment in R
3
from (x
0
;z) to (x;z) can be
described by the function
L:[0;1]!R
3
:t 7!(tx +(1 t)x
0
;z) = (tx +(1 t)x
0
;ty +(1 t)y
0
;z):
If we let h = F L:[0;1]!R,then
F(x;z) F(x
0
;z) = h(1) h(0) = h
0
()
for some 2 (0;1),in view of the Mean value theorem.On the other hand,it follows from the Chain
rule that
h
0
() = (DF)(L())(DL)() =
@F
@x
(L())
@F
@y
(L())
@F
@z
(L())
0
@
x x
0
y y
0
0
1
A
=
@F
@x
(x +(1 )x
0
;z)
(x x
0
) +
@F
@y
(x +(1 )x
0
;z)
(y y
0
):
To study the term F(x
0
;z) F(x
0
;z
0
),we note that the line segment in R
3
from (x
0
;z
0
) to (x
0
;z) can
be described by the function
L:[0;1]!R
3
:t 7!(x
0
;tz +(1 t)z
0
) = (x
0
;y
0
;tz +(1 t)z
0
):
If we let h = F L:[0;1]!R,then
F(x
0
;z) F(x
0
;z
0
) = h(1) h(0) = h
0
()
for some 2 (0;1),in view of the Mean value theorem.On the other hand,it follows from the Chain
rule that
h
0
() = (DF)(L())(DL)() =
@F
@x
(L())
@F
@y
(L())
@F
@z
(L())
0
@
0
0
z z
0
1
A
=
@F
@z
(x
0
;z +(1 )z
0
)
(z z
0
):
Hence
F(x;z) =
@F
@x
(x +(1 )x
0
;z)
(x x
0
) +
@F
@y
(x +(1 )x
0
;z)
(y y
0
)
+
@F
@z
(x
0
;z +(1 )z
0
)
(z z
0
)
for some ; 2 (0;1).We now choose
a
0
2 (0;a) and < min
a
0
;
ba
0
2M
:
Chapter 3:Implicit and Inverse Function Theorems
page 7 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
If kx x
0
k < ,then it is easily seen that
@F
@x
(x +(1 )x
0
;z)
(x x
0
) +
@F
@y
(x +(1 )x
0
;z)
(y y
0
)
< ba
0
;
so that
F(x;z
0
+a
0
) > 0 and F(x;z
0
a
0
) < 0:
It now follows from the Intermediate value theorem that there exists z 2 (z
0
a
0
;z
0
+ a
0
) such that
F(x;z) = 0.Furthermore,this value of z is unique,since a function with positive derivative cannot have
more than one zero.In other words,if we take U = D(x
0
;) and V = (z
0
a
0
;z
0
+a
0
),then for every
x 2 U,there exists a unique z 2 V such that F(x;z) = 0.We can write z = g(x;y) and this completes
the proof of the rst two parts.To prove the last part,note that since F(x;z) = 0,we have
g(x) g(x
0
) = z z
0
=
@F
@x
(x +(1 )x
0
;z)
(x x
0
) +
@F
@y
(x +(1 )x
0
;z)
(y y
0
)
@F
@z
(x
0
;z +(1 )z
0
)
:
In particular,
g(x
0
+h;y
0
) g(x
0
;y
0
)
h
=
@F
@x
(x
0
+h;y
0
;z)
,
@F
@z
(x
0
;y
0
;z +(1 )z
0
):
Letting h!0,we have x!x
0
and z!z
0
,and so
@g
@x
(x
0
;y
0
) = lim
h!0
g(x
0
+h;y
0
) g(x
0
;y
0
)
h
=
@F
@x
(x
0
;y
0
;z
0
)
,
@F
@z
(x
0
;y
0
;z
0
):
Similarly
@g
@y
(x
0
;y
0
) =
@F
@y
(x
0
;y
0
;z
0
)
,
@F
@z
(x
0
;y
0
;z
0
):
The argument can be repeated for every (x;y;z) 2 U V.This completes the proof.
Remarks.(1) Suppose that it has been established that the functions (3) are continuously dierentiable.
We shall show here how we may deduce (5) by the use of the Chain rule.Consider the functions
G:U!R
n
R
m
:x 7!(x;g
1
(x);:::;g
m
(x))
and
F:R
n
R
m
!R
m
:(x;z) 7!(F
1
(x;z);:::;F
m
(x;z)):
In view of (4),it is clear that the composite function H = F G:U!R
m
is identically zero,so that
(DH)(x) is the zero mn matrix.On the other hand,we have,by the Chain rule,that
(DH)(x) = (DF)(x;z)(DG)(x);
Chapter 3:Implicit and Inverse Function Theorems
page 8 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
where z = (z
1
;:::;z
m
) = (g
1
(x);:::;g
m
(x).It follows that
(DH)(x) =
0
B
B
B
B
B
B
B
B
B
@
@F
1
@x
1
:::
@F
1
@x
n
@F
1
@z
1
:::
@F
1
@z
m
.
.
.
.
.
.
.
.
.
.
.
.
@F
m
@x
1
:::
@F
m
@x
n
@F
m
@z
1
:::
@F
m
@z
m
1
C
C
C
C
C
C
C
C
C
A
0
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
@
@x
1
@x
1
:::
@x
1
@x
n
.
.
.
.
.
.
@x
n
@x
1
:::
@x
n
@x
n
@g
1
@x
1
:::
@g
1
@x
n
.
.
.
.
.
.
@g
m
@x
1
:::
@g
m
@x
n
1
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
A
=
0
B
B
B
B
B
B
B
B
B
@
@F
1
@x
1
:::
@F
1
@x
n
@F
1
@z
1
:::
@F
1
@z
m
.
.
.
.
.
.
.
.
.
.
.
.
@F
m
@x
1
:::
@F
m
@x
n
@F
m
@z
1
:::
@F
m
@z
m
1
C
C
C
C
C
C
C
C
C
A
0
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
@
1 0 0:::0
0 1 0:::0
.
.
.
.
.
.
.
.
.
0:::0 1 0
0:::0 0 1
@g
1
@x
1
:::
@g
1
@x
n
.
.
.
.
.
.
@g
m
@x
1
:::
@g
m
@x
n
1
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
A
:
Hence for every j = 1;:::;n,the jth (zero) column of (DH)(x) is equal to
0
B
B
B
B
B
B
B
B
B
@
@F
1
@x
j
.
.
.
@F
m
@x
j
1
C
C
C
C
C
C
C
C
C
A
+
0
B
B
B
B
B
B
B
B
B
@
@F
1
@z
1
:::
@F
1
@z
m
.
.
.
.
.
.
@F
m
@z
1
:::
@F
m
@z
m
1
C
C
C
C
C
C
C
C
C
A
0
B
B
B
B
B
B
B
B
B
@
@g
1
@x
j
.
.
.
@g
m
@x
j
1
C
C
C
C
C
C
C
C
C
A
=
0
B
B
B
B
B
B
B
B
@
0
.
.
.
0
1
C
C
C
C
C
C
C
C
A
;
so that
0
B
B
B
B
B
B
B
B
B
@
@F
1
@z
1
:::
@F
1
@z
m
.
.
.
.
.
.
@F
m
@z
1
:::
@F
m
@z
m
1
C
C
C
C
C
C
C
C
C
A
0
B
B
B
B
B
B
B
B
B
@
@g
1
@x
j
.
.
.
@g
m
@x
j
1
C
C
C
C
C
C
C
C
C
A
=
0
B
B
B
B
B
B
B
B
B
@
@F
1
@x
j
.
.
.
@F
m
@x
j
1
C
C
C
C
C
C
C
C
C
A
:(8)
We can now deduce (5) from (8) by applying Cramer's rule.
(2) Note that in the special case when m= 1,the condition (5) reduces to (7).
Chapter 3:Implicit and Inverse Function Theorems
page 9 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
3.2.Inverse Function Theorem
In the theory of real valued functions of one real variable,we appreciate the importance of inversion.For
example,we know that the exponential function has as its inverse the logarithmic function.In this last
section,we shall study this problemmore closely.More precisely,we shall deduce the following result from
the Implicit function theorem.Throughout this section,(x;z) 2 R
n
R
n
denotes x = (x
1
;:::;x
n
) 2 R
n
and z = (z
1
;:::;z
n
) 2 R
n
.For functions f
i
:R
n
!R,where i = 1;:::;n,we write
@(f
1
;:::;f
n
)
@(z
1
;:::;z
n
)
= det
0
B
B
B
B
B
B
B
B
B
@
@f
1
@z
1
:::
@f
1
@z
n
.
.
.
.
.
.
@f
n
@z
1
:::
@f
n
@z
n
1
C
C
C
C
C
C
C
C
C
A
(9)
whenever the partial derivatives in the matrix exist.We also write
@(f
1
;:::;f
n
)
@(z
1
;:::;z
i1
;e
j
;z
1+1
;:::;z
n
)
for the determinant when the ith column of the matrix in (9) is replaced by the vector
e
j
= (0;:::;0

{z
}
j1
;1;0;:::;0

{z
}
nj
);
written as a column.
THEOREM 3B.(INVERSE FUNCTION THEOREM) Suppose that for every i = 1;:::;n,the func
tion f
i
:R
n
!R has continuous partial derivatives.Suppose further that the system of n equations
f
1
(z
1
;:::;z
n
) = x
1
.
.
.(10)
f
n
(z
1
;:::;z
n
) = x
n
has a solution (x
0
;z
0
) 2 R
n
R
n
,where
@(f
1
;:::;f
n
)
@(z
1
;:::;z
n
)
(z
0
) 6= 0:
Then the following two conclusions hold:
(a) There exist neighbourhoods U R
n
of x
0
and V R
n
of z
0
such that the system (10) can be solved
uniquely as z = g(x) for x 2 U and z 2 V.In other words,there exist unique functions
z
i
= g
i
(x) = g
i
(x
1
;:::;x
n
);where i = 1;:::;n;(11)
such that the equations (10) hold.
(b) Furthermore,the functions (11) are continuously dierentiable,and for every i = 1;:::;n and
j = 1;:::;n,we have
@g
i
@x
j
=
@(f
1
;:::;f
n
)
@(z
1
;:::;z
i1
;e
j
;z
1+1
;:::;z
n
)
,
@(f
1
;:::;f
n
)
@(z
1
;:::;z
n
)
:(12)
Chapter 3:Implicit and Inverse Function Theorems
page 10 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
Remark.Note that
@(f
1
;:::;f
n
)
@(z
1
;:::;z
n
)
(z
0
) = det
0
B
B
B
B
B
B
B
B
B
@
@f
1
@z
1
(z
0
):::
@f
1
@z
n
(z
0
)
.
.
.
.
.
.
@f
n
@z
1
(z
0
):::
@f
n
@z
n
(z
0
)
1
C
C
C
C
C
C
C
C
C
A
:
Note also that
@(f
1
;:::;f
n
)
@(z
1
;:::;z
i1
;e
j
;z
1+1
;:::;z
n
)
= det
0
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
@
@f
1
@z
1
:::
@f
1
@z
i1
0
@f
1
@z
i+1
:::
@f
1
@z
n
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
@f
j1
@z
1
:::
@f
j1
@z
i1
0
@f
j1
@z
i+1
:::
@f
j1
@z
n
@f
j
@z
1
:::
@f
j
@z
i1
1
@f
j
@z
i+1
:::
@f
j
@z
n
@f
j+1
@z
1
:::
@f
j+1
@z
i1
0
@f
j+1
@z
i+1
:::
@f
j+1
@z
n
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
@f
n
@z
1
:::
@f
n
@z
i1
0
@f
n
@z
i+1
:::
@f
n
@z
n
1
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
A
:
Sketch of Proof of Theorem 3B.We use Theorem 3A with m= n and
F
i
:R
n
R
n
!R:(x;z) 7!f
i
(z) x
i
;where i = 1;:::;n:
Note now that the conditions for F
i
in Theorem 3A are satised.Note also that for i;j = 1;:::;n,we
have
@F
i
@z
j
=
@f
i
@z
j
and
@F
i
@x
j
=
1 if i = j,
0 if i 6= j.
It is easy to see that (12) now follows from (5).
Definition.The determinant given by (9) is called the Jacobian determinant of f = (f
1
;:::;f
n
).
Example 3.2.1.Consider the equations
z
2
+w
2
z
= x and e
z
sinw = y:
Here the functions
f
1
(z;w) =
z
2
+w
2
z
and f
2
(z;w) = e
z
sinw
Chapter 3:Implicit and Inverse Function Theorems
page 11 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
have continuous partial derivatives except when z = 0.Now
@(f
1
;f
2
)
@(z;w)
(z
0
;w
0
) = det
0
B
B
B
B
@
@f
1
@z
(z
0
;w
0
)
@f
1
@w
(z
0
;w
0
)
@f
2
@z
(z
0
;w
0
)
@f
2
@w
(z
0
;w
0
)
1
C
C
C
C
A
= det
0
B
B
B
B
@
1
w
2
0
z
2
0
2w
0
z
0
e
z
0
sinw
0
e
z
0
cos w
0
1
C
C
C
C
A
= e
z
0
1
w
2
0
z
2
0
cos w
0
2w
0
z
0
sinw
0
:
It follows that we can solve for z and w in terms of x and y near any point (z
0
;w
0
) for which
1
w
2
0
z
2
0
cos w
0
6=
2w
0
z
0
sinw
0
:
Example 3.2.2.Consider the equations
r cos = x and r sin = y:
Here the functions
f
1
(r;) = r cos and f
2
(r;) = r sin
have continuous partial derivatives everywhere.Now
@(f
1
;f
2
)
@(r;)
(r
0
;
0
) = det
0
B
B
B
B
@
@f
1
@r
(r
0
;
0
)
@f
1
@
(r
0
;
0
)
@f
2
@r
(r
0
;
0
)
@f
2
@
(r
0
;
0
)
1
C
C
C
C
A
= det
cos
0
r
0
sin
0
sin
0
r
0
cos
0
= r
0
:
It follows that we can solve for r and in terms of x and y near any point (r
0
;
0
) for which r
0
6= 0.
Furthermore,we have
@r
@x
=
det
0
B
B
B
B
@
1
@f
1
@
0
@f
2
@
1
C
C
C
C
A
@(f
1
;f
2
)
@(r;)
=:::= cos and
@r
@y
=
det
0
B
B
B
B
@
0
@f
1
@
1
@f
2
@
1
C
C
C
C
A
@(f
1
;f
2
)
@(r;)
=:::= sin;
as well as
@
@x
=
det
0
B
B
B
B
@
@f
1
@r
1
@f
2
@r
0
1
C
C
C
C
A
@(f
1
;f
2
)
@(r;)
=:::=
sin
r
and
@
@y
=
det
0
B
B
B
B
@
@f
1
@r
0
@f
2
@r
1
1
C
C
C
C
A
@(f
1
;f
2
)
@(r;)
=:::=
cos
r
:
These can be checked by using the formulae
r =
p
x
2
+y
2
and = tan
1
y
x
:
Chapter 3:Implicit and Inverse Function Theorems
page 12 of 13
Multivariable and Vector Analysis
c W W L Chen,1997,2008
Problems for Chapter 3
1.Consider the function
F:R
2
R!R:(x;y;z) 7!x
3
z
2
z
3
yx:
a) Explain why there is no neighbourhoods U of (0;0) and V of 0 such that there exists a function
z = g(x;y) dened for (x;y) 2 U and z 2 V and satisfying F(x;y;g(x;y)) = 0.
b) Explain why the equation is soluble for z as a function of (x;y) near the point (1;1;1).Compute
the partial derivatives @z=@x and @z=@y at this point by using the partial derivatives of F.
2.Consider the functions
F
1
:R
3
R
3
!R:(x
1
;x
2
;x
3
;z
1
;z
2
;z
3
) 7!3x
1
+2x
2
+x
2
3
+z
1
+z
2
2
;
F
2
:R
3
R
3
!R:(x
1
;x
2
;x
3
;z
1
;z
2
;z
3
) 7!4x
1
+3x
2
+x
3
+z
2
1
+z
2
+z
3
+2;
F
3
:R
3
R
3
!R:(x
1
;x
2
;x
3
;z
1
;z
2
;z
3
) 7!x
1
+x
3
+z
2
1
+z
3
+2:
Discuss the solubility of the system of equations
F
i
(x
1
;x
2
;x
3
;z
1
;z
2
;z
3
) = 0;where i = 1;2;3;
for z
1
;z
2
;z
3
in terms of x
1
;x
2
;x
3
in a neighbourhood of the point (0;0;0;0;0;2).
3.In R
2
,rectangular coordinates (x;y) and polar coordinates (r;) are related by x = r cos and
y = r sin.Discuss when we can solve for (r;) in terms of (x;y).
4.Spherical coordinates r;; and rectangular coordinates x;y;z in R
3
are related by
x(r;;) = r sincos ;
y(r;;) = r sinsin;
z(r;;) = r cos :
a) Show that
@(x;y;z)
@(r;;)
= r
2
sin:
b) When can we solve for r;; in terms of x;y;z?
Chapter 3:Implicit and Inverse Function Theorems
page 13 of 13
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