Download THE IMPLICIT FUNCTION THEOREM 1. Motivation and statement

THE IMPLICIT FUNCTION THEOREM
ALEXANDRU ALEMAN
1. Motivation and statement
We want to understand a general situation which occurs in almost
any area which uses mathematics. Suppose we are given number of
equations which involve a higher number of unknown variables, say
(1.1)
fj (x1 , . . . , xn , y1 , . . . ym ) = 0,
1 ≤ j ≤ m.
Common sense says that we might not be able to solve these equations
uniquely, but we might be able to express the variables yk in terms of
n variables xj , i.e. we could write
yk = gk (x1 , . . . , xn ), 1 ≤ k ≤ m,
where gk are functions of n variables.
In order to develop a meaningful discussion we must assume that (1.1)
0
is satisfied at least at one point, that is, there exist x01 , . . . , x0n , y10 , . . . , ym
such that
(1.2)
0
fj (x01 , . . . , x0n , y10 , . . . ym
) = 0,
1 ≤ j ≤ m.
In this case we require that the functions yk = gk (x1 , . . . , xn ), 1 ≤ k ≤
m, satisfy
gk (x01 , . . . , x0n ) = yk0 , 1 ≤ k ≤ m.
Let us refer to this question as Problem (1.1).
As is to expect, this is a tricky problem; Just imagine we want to decide
whether the system
x2 + y 2 + z 2 = 1
x3 + y 3 + z 3 = 2,
has solutions of the form x, y(x), z(x).
Some more tractable examples are listed below.
Examples 1.1. Linear equations. 1) Let us start with linear equations.
a) The equation
x + y = 5,
1
2
ALEXANDRU ALEMAN
has no unique solution, but each unknown x, y can be expressed in terms
of the other
x = 5 − y, y = 5 − x.
b) This is not always the case as the following linear system shows:
x+y+z =2
x + y + 2z = 3
which has solutions x = 1 − y, z = 1, i.e. x, y cannot be expressed in
terms of z. On the other hand, we can say that y, z can be expressed
in terms of x , or x, z can be expressed in terms of y.
There is a general result in Linear Algebra which says that given n
linear equations with n+m unknowns then there exists a number k ≤ n
such that all unknowns can be expressed in terms of k of them. As we
have seen above these k unknowns are not arbitrary.
c) In the linear case it is quite easy to decide which variables can be
used to express the others. Here is an example which illustrates this.
Consider the linear system
a1 x + b 1 y + c 1 z + d 1 u = 1
a2 x + b2 y + c2 z + d2 u = 2,
with unknowns x, y, z, u, and constant coefficients a1 , a2 , b1 , b2 , c1 , c2 , d1 , d2 .
Suppose that the determinant of the matrix
c1 d 1
(1.3)
c2 d 2
is non-zero. Then writing the system as
c1 z + d1 u = 1 − a1 x − b1 y
c2 z + d2 u = 2 − a2 x − b2 y,
it follows by Cramer’s rule that z, u can be expressed in terms of x, y.
It is important to note that in all linear cases, the remaining unknowns
are expressed as functions of the given ones. For nonlinear expressions
this might fail.
2) The equation of a circle. Consider the equation
x2 + y 2 = 1,
Its general solution is given by
p
x = ± 1 − y2,
x, y ∈ [−1, 1]
√
or y = ± 1 − x2 .
Thus we need two functions to express x in terms of y , or vice versa.
This problem can be removed using the assumption (1.3). For example,
THE IMPLICIT FUNCTION THEOREM
if x0 =
q
3
,
4
3
y0 = 12 , then for x is close to x0 , the function
√
y = + 1 − x2 ,
satisfies the equation as well as the condition y(x0 ) = y 0 . However, if
y 0 = 1 then there are always two solutions to Problem (1.1).
These examples reveal that a solution of Problem (1.1) might require:
• To restrict the domains of definition of the functions gk we are
looking for. For example, we could consider them as defined
only in a neighborhood of the point (x01 , . . . , x0n ) considered in
(1.2) (see Example 2).
• Some additional conditions regarding the independence of the
equations (1.1) with respect to the ”y-variables” as illustrated
in Example 1c).
The second type of condition may appear mysterious, but it is just the
nonlinear version of the condition in Example 1c), and it can be easily
explained using derivatives.
Before we do that let us give a precise formulation of Problem (1.1) for
continuously differentiable functions.
Assume that the functions f1 , . . . , fm have continuous partial deriva0
) ∈ Rn+m and
tives in a neighborhood V of p0 = (x01 , . . . , x0n , y10 , . . . , ym
satisfy:
0
fj (x01 , . . . , x0n , y10 , . . . ym
) = 0, 1 ≤ j ≤ m.
Does there exists a neighborhood U of x0 = (x01 , . . . , x0n ) ∈ Rn and
uniquely determined functions, g1 , . . . , gm : U → R with continuous
partial derivatives in U , such that
gk (x01 , . . . , x0n ) = yk0 ,
1 ≤ k ≤ m,
and
fj (x, g1 (x), . . . , gm (x)) = 0,
x = (x1 , . . . xn ) ∈ U, 1 ≤ j ≤ m?
Let us look at the simplest case when n = m = 1. Suppose that
f : R2 → R has continuous partial derivatives, and that Problem (1.1)
can be solved in the terms described above. In other words, f satisfies
f (x0 , y 0 ) = 0,
for some (x0 , y 0 ) ∈ R2 , and there exists an interval I containing x0
and a function g : I → R with a continuous derivative on I such that
g(x0 ) = y 0 and
f (x, g(x)) = 0, x ∈ I.
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ALEXANDRU ALEMAN
By differentiating this equality w.r.t. x and using the chain rule we
obtain
∂f
∂f
(x, g(x)) +
(x, g(x))g 0 (x) = 0, x ∈ I.
∂x
∂y
Then obviously
∂f
(x, g(x))
∂x
= 0 whenever
∂f
(x, g(x))
∂y
=0
This is a very restrictive condition which we definitely want to avoid!
In fact, Example 2) shows that when x0 = 1, y 0 = 0, there are two
functions g as above and none of them is differentiable at x0 . Thus the
assumption we shall use here is that at least near (x0 , y 0 ) we have
∂f
(x, y) 6= 0.
∂y
Note that in this case we have
∂f
∂f
g 0 (x) =
(x, g(x))/ (x, g(x)),
∂x
∂y
in particular, g 0 is continuous at those points At the cost of working
with smaller sets we may assume as well that
∂f 0 0
(x , y ) 6= 0,
∂y
since the continuity of the partial derivate implies that this holds in a
neighborhood of (x0 , y 0 ).
and
g 0 (x) =
∂f
∂f
(x, g(x))/ (x, g(x)),
∂x
∂y
∂f
(x, g(x)) 6= 0,
∂y
A similar situation occurs in higher dimensions as well. Suppose that
f1 , f2 : R3 → R has continuous partial derivatives, and that Problem
(1.1) can be solved, i.e.
fj (x0 , y 0 , z 0 ) = 0,
j = 1, 2,
for some (x0 , y 0 , z0 ) ∈ R3 , and there exists an interval I containing x0
and two functions g1 , g2 : I → R with continuous derivatives on I, such
that g1 (x0 ) = y 0 , g2 (x0 ) = z 0 , and
fj (x, g1 (x), g2 (x)) = 0,
x ∈ I , j = 1, 2.
By differentiating these equalities w.r.t. x and using again the chain
rule we obtain
∂fj
∂fj
∂fj
(x, g1 (x), g2 (x))+
(x, g1 (x), g2 (x))g10 (x)+
(x, g1 (x), g2 (x))g20 (x) = 0,
∂x
∂y
∂z
THE IMPLICIT FUNCTION THEOREM
when x ∈ I, j = 1, 2. This can be rewritten as
!
∂f1 ∂f1
∂f1
∂y
∂z
∂x
+
∂f
∂f2
∂f
2
2
(x,g1 (x),g2 (x))
∂x
∂y
∂z
(x,g1 (x),g2 (x))
5
g10 (x)
= 0.
g20 (x)
If the 2 × 2 matrix above is not invertible, this creates similar complications as in the scalar case. For this reason, we shall assume it is
invertible near the point (x0 , y 0 , z 0 ). Again, by continuity it suffices
to assume that this matrix computed at x0 is invertible, in which case
it will be invertible in a (possibly smaller) neighborhood of this point.
Since a matrix is invertible if and only if it has a nonzero determinant,
we can reformulate this in terms of the Jacobian determinants (see
CC:128, p. 732). Our condition is equivalent to
∂(f1 , f2 )
|(x0 ,y0 ,z0 ) 6= 0.
∂(y, z)
Also note that in this case we can apply Cramer’s rule to find an
implicit formula for the derivatives g10 , g20 . Recall that the Cramer rule
says that if
a b c d 6= 0,
then the unique solution of the linear system
ax + by = u
cx + dy = v
is given by
u
v
x = a
c
c d
,
b d
a
b
y = a
c
u
v
b d
In our case this gives
g10 (x)
=
∂(f1 ,f2 ) ∂(x,y) − ∂(f1 ,f2 ) ,
(x,g
(x),g
(x))
1
2
∂(y,z)
g20 (x)
=
∂(f1 ,f2 ) ∂(z,x) − ∂(f1 ,f2 ) .
(x,g
(x),g
(x))
1
2
∂(y,z)
We are now ready to state the Implicit Function Theorem.
Theorem 1.1. Assume that the functions f1 , . . . , fm have continuous
0
partial derivatives in a neighborhood V of p0 = (x01 , . . . , x0n , y10 , . . . , ym
)∈
n+m
R
and satisfy:
(1)
0
fj (x01 , . . . , x0n , y10 , . . . ym
) = 0,
1 ≤ j ≤ m,
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ALEXANDRU ALEMAN
(2)
∂(f1 , . . . , fm ) 6= 0.
∂(y1 , . . . , ym ) p0
Then there exists a neighborhood U of x0 = (x01 , . . . , x0n ) ∈ Rn and
uniquely determined functions, g1 , . . . , gm : U → R with continuous
partial derivatives in U , such that
(3)
gk (x01 , . . . , x0n ) = yk0 ,
(4)
(x1 , . . . , xn , g1 (x), . . . , gm (x)) ∈ V
1 ≤ k ≤ m,
and
(5)
x = (x1 , . . . xn ) ∈ U, 1 ≤ j ≤ m.
fj (x, g1 (x), . . . , gm (x)) = 0,
Moreover, for 1 ≤ j ≤, 1 ≤ k ≤ m
(6)
∂(f1 ,...,fm )
∂gk
∂(y1 ,...,yk−1 ,xj ,yk ,...,ym ) (x) = −
.
∂(f1 ,...,fm )
∂xj
(x,g1 (x),...,gm (x))
∂(y1 ,...,ym )
2. Exercises
Exercise 2.1. Show that Problem (1.1) is solvable for the system
x2 + y 2 + z 2 = 1
x3 + y 3 + z 3 = 2,
and any point (x0 , y 0 , z 0 ) where the equations are satisfied, and y 0 , z 0 6=
0, y 0 6= z 0 .
Solution We only need to verify hypothesis (3) in Theorem 1.1 for the
functions f1 (x, y, z) = x2 + y 2 + z 2 , f2 (x, y, z) = x3 + y 3 + z 3 , that is,
∂(f1 , f2 ) 6= 0.
∂(y, z) 0 0 0
(x ,y ,z )
Since
∂f1
∂f1
∂f2
∂f2
= 2y,
= 2z,
= 3y 2 ,
= 3z 2 ,
∂y
∂z
∂y
∂z
the determinant on the left is
2y 2z 0 0 2
0 2 0
0 0 0
0
2
3y 3z 2 0 0 0 = 6y (z ) − 6(y ) z = 6y z (z − y ).
(x ,y ,z )
Under the conditions on y 0 , z 0 , this determinant does not vanish and
the solution is complete.
A more challenging related problem is the following:
THE IMPLICIT FUNCTION THEOREM
7
Exercise 2.2. Let S be the set of solutions of the system
x2 + y 2 + z 2 = 1
x3 + y 3 + z 3 = 2,
Show that
S ∩ {(x, y, z) : 0 < y < z},
is a smooth curve, in th sense that there exists an interval I ⊂ R and
a continuously differentiable function u : I → R3 with
u(I) = S ∩ {(x, y, z) : 0 < y < z}.
Exercise 2.3. a) Construct a continuously differentiable bijective function g : R2 → R2 whose inverse is not differentiable on R2 .
b) Let V ⊂ R3 be open, and let f : V → R3 have continuous partial
derivatives in V . Show that if the Jacobian matrix Df (p0 ) of f at
p0 ∈ V is non-singular then there exists a neighborhood U of f (p0 ) and
a function h : U → V with continuous partial derivatives in U , such
that f (h(y)) = y.
Solution a) The typical function on R with this property is h(x) = x3 .
Indeed, h is strictly increasing
√ on R with h(R) = R, i.e. h is bijective.
On the other hand, h−1 = 3 x which is not differentiable at the origin.
On R2 we can consider the function
3
x
h1 (x, y) =
,
y
with inverse
√
3
x
=
,
y
which is not differentiable at the origin because it does not have a
partial derivative in x at that point.
b) The equality
h−1
1 (x, y)
f (y) = x ⇔ g(x, y) = f (x) − y = 0
leads to system of the form (1.1) with 3 equations and 6 unknowns.
Since g(p0 , f (p0 )) = 0, and
∂g
6 0,
=
∂(x1 , x2 , x3 ) (p0 ,f (p0 )
we can apply Theorem 1.1 which gives a unique function h defined near
f (p0 ) with continuous partial derivatives, such that
g(h(y), y) = 0 ⇐⇒ f (h(y)) = y.
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ALEXANDRU ALEMAN
Question 2.1. Does there exist a neighborhood V1 ⊂ V of p0 such that
the function f from b) is injective on V1 and f (V1 ) is open?
3. Proof of the Implicit Function Theorem
Let us note from the start the following:
If we can show that there exists a neighborhood U of x0 = (x01 , . . . , x0n ) ∈
Rn and uniquely determined functions, g1 , . . . , gm : U → R whose partial derivatives exist in U , such that (3),(4),(5) hold, then gk , 1 ≤ k ≤
m have continuous partial derivatives and (6) holds.
This s just a repetition of the argument in the last example preceding
the statement of the theorem. Suppose we have found differentiable
functions g1 , . . . , gm : U → R as above. From (5) and the chain rule we
obtain for fixed 1 ≤ j ≤ n
∂fk
∂g1
∂fk
∂gm
((x, g1 (x), . . . , gm (x))
+ ... +
((x, g1 (x), . . . , gm (x))
∂y1
∂xj
∂ym
∂xj
∂fk
=−
((x, g1 (x), . . . , gm (x)).
∂xj
k
((x, g1 (x), . . . , gm (x)))1≤k≤m
This can be seen as a linear system with matrix ( ∂f
∂yl
and right hand side
(−
∂f1
∂fm
(((x, g1 (x), . . . , gm (x), . . . , −
((x, g1 (x), . . . , gm (x).
∂xj
∂xj
By Cramer’s rule its solution is just (6). Now if (6) holds and each gk is
continuous, then the partial derivatives of gk are obviously continuous.
Thus we need to prove the following claim:
(*) There exists a neighborhood U of x0 = (x01 , . . . , x0n ) ∈ Rn and
uniquely determined functions, g1 , . . . , gm : U → R whose partial derivatives exist in U , such that (3),(4),(5) hold.
We shall prove the claim by induction. We start with the case when
m = 1. Thus we want to solve Problem (1.1) for one equation
f (x1 , . . . , xn , y) = 0
assuming that f (x01 , . . . , x0n , y 0 ) = 0, and
∂f 0
(x , . . . , x0n , y 0 ) 6= 0.
∂y 1
THE IMPLICIT FUNCTION THEOREM
9
Also, we can assume without loss that
∂f 0 0
(x , y ) > 0,
∂y
otherwise we can change from f to −f . Since V is open and the
function y → ∂f
(x0 , y) is assumed to be continuous, there exists a
∂y
cube
Q = [x01 − a, x01 + a] × . . . × [x0n − a, x0n + a] × [y 0 − a, y 0 + a] ⊂ V,
such that
∂f
(x, y) > 0, (x, y) ∈ Q.
∂y
In particular, since f (x0 , y) vanishes at y 0 and is strictly increasing on
[y 0 −, y 0 + a], it satisfies
f (x0 , y 0 − a) < 0, f (x0 , y 0 + a) > 0.
At this point we use uniform continuity. Recall that a continuous realvalued function h on a compact subset K of Rn is uniformly continuous,
i.e. given ε > 0 there exists δ > 0 such that if u, v ∈ K and |u − v| < δ
then |h(u) − h(v)| < ε.
This means that for every ε > 0 there exists δ > 0 which can be chosen
in (0, a], such that whenever |x − x0 | < δ,
|f (x, y) − f (x0 , y)| < ε,
y ∈ [y 0 − a, y 0 + a].
If ε is sufficiently small we conclude that for |x − x0 | < δ we have
f (x, y 0 − a) < 0, f (x, y 0 + a) > 0.
Since for each x as above, y → f (x, y) is strictly increasing on [y 0 −
a, y 0 + a] there exists a unique y = g(x) ∈ [y 0 − a, y 0 + a] with
f (x, g(x)) = 0. Clearly, g(x0 ) = 0 which gives (3), (4) and (5)with
a function g defined on
U = {x : |x − x= | < ε},
with values in [y 0 − a, y 0 + a]. It remains to show that g has partial
derivatives in U .
This is the hard part of the proof ! Fortunately it isn’t very
long. If you did not already note it, the only information
available about g is that f (x, g(x)) = 0, so that all further
information about this function must emerge from it.
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ALEXANDRU ALEMAN
a) g is continuous on U . Indeed, if we assume the contrary, there exists
x ∈ U and a sequence (xm ) in U with
lim xm = x,
m→∞
g(x) 6= lim g(xm ) = y ∈ [y 0 − a, y 0 + a].
m→∞
But then f (x, ·) is strictly increasing on [y 0 − a, y 0 − a] and satisfies
f (x, y) = f (x, g(x)) = 0. this is a contradiction which shows the continuity of g.
b) g has partial derivatives in U . If {e1 , . . . , en } denotes the canonical
basis in Rn , we want to show that for each x ∈ U , the limits
g(x + hej ) − g(x)
, 1≤j≤n
h→0
h
exist. Recall that f (x, g(x)) = 0, and write for fixed x and h 6= 0 with
|h| sufficiently small (we need that x − tej , x + tej , t ∈ [0, h] ∈ U ,
lim
(3.4)
f (x + hej , g(x + hej )) − f (x, g(x))
0=
h
f (x + hej , g(x + hej )) − f (x, g(x + hej )) f (x, g(x + hej )) − f (x, g(x))
+
=
h
h
Apply the mean value theorem to the function
u(t) = f (x + tej , g(x + hej )) t ∈ [0, h],
to conclude that
f (x + hej , g(x + hej )) − f (x, g(x + hej )) u(h) − u(0)
=
= u0 (s)
h
h
∂f
(x + sej , g(x + hej ),
=
∂x
where s ∈ [0, 1]. For the second term on the right we consider the
function
v(t) = f (x + hej , tg(x + hej ) + (1 − t)g(x)) t ∈ [0, 1].
The function is well defined and by the mean-value theorem it satisfies
v(1) − v(0) = v 0 (r),
r ∈ [0, 1],
which, by the chain rule implies
∂f
f (x, g(x+hej ))−f (x, g(x)) =
(x, rg(x+hej )+(1−r)g(x))(g(x+hej )−g(x)),
∂y
for some r ∈ [0, 1], and thus
f (x, g(x + hej )) − f (x, g(x))
∂f
g(x + hej ) − g(x)
=
(x, rg(x+hej )+(1−r)g(x))
.
h
∂y
h
THE IMPLICIT FUNCTION THEOREM
11
If we replace these in (3.4) we obtain
f (x + hej , g(x + hej )) − f (x, g(x))
h
∂f
∂f
=
(x + sej , g(x + hej ) +
(x, rg(x + hej )
∂x
∂y
g(x + hej ) − g(x)
+ (1 − r)g(x))
(3.6)
,
h
for some s ∈ [0, h], r ∈ [0, 1], which may depend of h (and x). Now
, ∂f to conclude that
use the continuity of g, ∂f
∂x ∂y
(3.5)
0=
∂f
∂f
(x + sej , g(x + hej )) =
(x, g(x)),
h→0 ∂x
∂x
lim
uniformly on [0, h], and
∂f
g(x + hej ) − g(x)
∂f
(x, rg(x+hej )+(1−r)g(x))
=
(x, g(x)) 6= 0,
h→0 ∂y
h
∂y
lim
uniformly on [0, 1]. Thus
lim
h→0
g(x + hej ) − g(x)
,
h
must exist for all x ∈ U .
Thus we have verified the claim (*) for m = 1. As pointed out at the
beginning, we prove the general case by induction. Assume (*) holds
for some positive integer m. We want to prove that the claim holds for
m + 1 as well.
Assume that f1 , . . . , fm+1 have continuous partial derivatives in a neigh0
) ∈ Rn+m+1 and satisfy:
borhood V of p0 = (x01 , . . . , x0n , y10 , . . . , ym+1
0
fj (x01 , . . . , x0n , y10 , . . . ym+1
) = 0, 1 ≤ j ≤ m + 1,
∂(f1 , . . . , fm+1 ) 6= 0.
∂(y1 , . . . , ym+1 ) p0
Note that
∂(f1 , . . . , fl−1 , fl+1 , . . . fm+1 )
∂(y1 , . . . , yk−1 , yk+1 , . . . , ym+1 )
∂(f1 ,...,fm+1 )
is obtained from ∂(y
by deleting the l-th row and the k-th col1 ,...,ym+1 )
umn, i.e. it is a minor of order m of this determinant. Then at least
x ∈ U,
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ALEXANDRU ALEMAN
one of these minors must be non-zero at p0 , otherwise
∂(f1 , . . . , fm+1 ) = 0.
∂(y1 , . . . , ym+1 ) p0
Since the functions and variables can be permuted, there is no harm if
we assume that
∂(f2 , . . . , fm+1 ) 6= 0.
∂(y2 , . . . , ym+1 ) p0
Then we can treat y1 as an x-variable, and look at the last m equations,
in order to arrive at the conclusion that f2 , . . . , fm+1 have continuous
0
)∈
partial derivatives in a neighborhood V of p0 = (x01 , . . . , x0n , y10 , . . . , ym+1
n+1+m
R
and satisfy:
0
fj (x01 , . . . , x0n , y10 , . . . ym+1
) = 0, 2 ≤ j ≤ m + 1,
∂(f2 , . . . , fm+1 ) 6= 0.
∂(y2 , . . . , ym+1 ) p0
But then the induction hypothesis implies that there exists a neighborhood U 0 of x0 = (x01 , . . . , x0n , y10 ) ∈ Rn+1 and uniquely determined
functions, g2 , . . . , gm+1 : U 0 → R whose partial derivatives exist in U 0 ,
such that
gk (x01 , . . . , x0n , y1 ) = yk0 , 2 ≤ k ≤ m + 1,
(x1 , . . . , xn , y1 , g2 (x, y1 ), . . . , gm+1 (x, y1 )) ∈ V, (x, y1 ) = (x1 , . . . xn , y1 ) ∈ U 0
and for 2 ≤ j ≤ m + 1,
fj (x1 , . . . , xn , y1 , g2 (x, y1 ), . . . , gm+1 (x, y1 )) = 0, (x, y1 ) = (x1 , . . . xn , y1 ) ∈ U 0 .
Recall from the beginning of the proof that in this case the partial
derivatives of g2 , . . . , gm+1 are continuous in U 0 and replace this in the
first equation to obtain
f2 (x1 , . . . , xn , y1 , g2 (x, y1 ), . . . , gm+1 (x, y1 )) = 0, (x, y1 ) = (x1 , . . . xn , y1 ) ∈ U 0 ,
that is, we arrived at the case when m = 1. Since this was solved
above, it follows that there exists a neighborhood U of x0 and a function
g1 : U → R whose partial derivatives exist in U , such that g(x0 ) = y10 ,
(x0 , g1 (x)) ∈ U 0 , and
fk (x1 , . . . , xn , g1 (x)1 , g2 (x, g1 (x)), . . . , gm+1 (x, g1 )(x)) = 0, x = (x1 , . . . xn ) ∈ U.
By the chain rule, the partial derivatives of , gk (x, g1 (x)) exist in U and
we are done. THE IMPLICIT FUNCTION THEOREM
13
4. An Application
Let us return to the situation considered in Exercise 2.3. We have seen
that the inverse of a bijective differentiable function is not necessarily differentiable. It is of great importance to find (simple) additional
conditions which prevent this. With help of the Implicit Function Theorem this becomes an easy task. However, the result below is actually
equivalent to Theorem 1.1.
Theorem 4.1. Let U ⊂ Rn be open and let f : U → Rn be injective
and have continuous partial derivatives on U . on U . If f = (f1 , . . . , fn )
satisfies
∂(f1 , . . . , fn )
(x) 6= 0, x ∈ U,
∂(x1 , . . . , xn )
then f (U ) is open and f −1 : f (U ) → U has continuous partial derivatives on f (U ).
Proof. Let y 0 ∈ f (U ), say y 0 = f (x0 ), and consider again the function
g : Rn × U → Rn defined by
g(x, y) = f (x) − y,
as in the solution of Exercise 2.3. If we write g = (g1 , . . . , gn ), the
equality g(x, y) = 0 is equivalent to
gj (x1 , . . . , xn , y1 , . . . , ym ) = 0,
We also have that g(x0 , y 0 ) = 0, and
∂(g1 , . . . , gn ) ∂(x1 , . . . , xn ) 1 ≤ j ≤ n.
6= 0.
(x0 ,y 0 )
By Theorem 1.1 it follows that there exists a neighborhood W of y 0
and a function h : W → U with continuous partial derivatives on W ,
such that
g(y, h(y)) = 0, y ∈ W.
Equivalently,
f (h(y)) = y, y ∈ W.
In particular, since y 0 =∈ f (U ) was arbitrary, it follows that f (U ) is
open (any point in f (U ) has a neighborhood W contained in f (U )).
We obviously have also
f (f −1 (y)) = y, y ∈ W, ⇒ f −1 (y) = h(y),
for all y ∈ W , because f is injective. But then f −1 must have continuous partial derivatives in W , hence in all of f (U ).
14
ALEXANDRU ALEMAN
5. Preparation for the written exam- set 1
Exercise 1. Consider the function f : R4 → R defined by
f (x, y, z, u) = u3 x − 3x2 y − 3zy − z 4 .
Show that if (x0 .y0 , z0 , u0 ) ∈ {(x, y, z, u) : f (x, y, z, u) = 0, xu 6= 0},
there exists a neighborhood U ⊂ R3 of (x0 , y0 , z0 ) and a function g :
U → R with continuous partial derivatives such that
f (x, y, z, g(x, y, z)) = 0,
(x, y, z) ∈ U.
Exercise 2. Find the maximum and minimum values of 2xy + ez on
the set
{(x, y, z) : x2 + 4y 2 + ez = 1, −1 ≤ z ≤ 0}.
Exercise 3. Show that the curve C given by r(t) = 2 cos t sin ti +
2 cos2Rtj + 2 sin tk, 0 ≤ t ≤ π lies on a sphere centered at the origin.
Find C zds.
Exercise 4. Show that the field
F (x, y, z) = yzi + xzj + xyk
is conservative and find a potential (find φ with ∇φ = F ). Find the
field lines and the equipotential surfaces of F .
Exercise 5. Let r = xi+yj+zk, and r = |r|2 . Show that if f : R → R
is twice differentiable on R then
∇2 f (r) = 4rf 00 (r) + 6f 0 (r).
THE IMPLICIT FUNCTION THEOREM
Recall that ∇2 is the Laplacian, ∇2 u =
∂2u
∂x2
+
∂2u
∂y 2
+
15
∂2u
.
∂z 2
Exercise 6. Use a line integral to compute the area enclosed by the
curve 4x2/3 + 9y 2/3 = 16.
Exercise 7. Use a surface integral to compute the volume of the
ellipsoid {(x, y, z) : 4x2 + 9y 2 + z 2 ≤ 1, }.
Exercise 8. Consider the curve C given by r(t) = (1 + sin t)i + (1 +
cos t)j+(3−cos t−sin t)k in R3 . Show it lies in the plane {(x, y, z) : x+
y + z = 5}, and it encloses a surface whose projection on the xy-plane
is a disc. Use Stokes’s theorem to evaluate
I
F · dr,
C
where F (x, y, z) = yex i + ex j + z 2 k.
6. Hints
1. Note that on the given set we have
function theorem.
∂f
∂u
6= 0. Apply the implicit
2. Note that the gradient of the Lagrange function has the form
(2y, 2x, ez ) − λ(2x, 2y, ez )
so the critical point occurs when λ = 1, and x = y = 0, which forces
z = 0. Thus the extrema can only occur on the boundary when z = 0
or z = −1. If z = 0, x = y = 0 and 2xy + ez = 1. If z = −1 ,
x2 + y 2 = 1 − e−1 and we can consider the maximum or minimum of
16
ALEXANDRU ALEMAN
2xy + e−1 when x2 + y 2 = 1 − e−1
p. Another application of the Lagrange
method gives that these are ± (1 − e−1 )/2
3. We have
|r|2 = 4 cos2 t sin2 t + +4 cos4 t + 4 sin2 t = 4.
Therefore ds = 2dt and
Z
Z
π
sin tdt = 8.
zds = 4
0
C
4. If φ(x, y, z) = xyz then ∇φ = F . The equipotential surfaces are
then given by
{(x, y, z) : φ(x, y, z) = c} = {(x, y, z) : xyz = c}.
The field lines are given by the differential equations
dx
dy
dz
=
=
,
yz
xz
xy
so that
xdx = ydy = zdz,
x2 = y 2 + C1 , y 2 = z 2 + C2 .
5. We have f (r) = f (x2 + y 2 + z 2 ) = u(x, y, z), hence by the chain and
product rule
∂ 2u
∂u
= f 0 (r)2x,
= f 00 (r)4x2 + 2f 0 (r).
2
∂x
∂x
Since these relations are symmetric in x, y, z the result follows.
6. If C denotes the given curve then the area enclosed
by it can be
R
computed with Green’s theorem and it is given by C xdy. To compute
this integral we can use the parametrization r = 2 cos3 ti+2 sin3 tj, 0 ≤
t ≤ 2π and obtain
Z
Z 2π
xdy = 12
cos4 t sin2 tdt.
C
0
THE IMPLICIT FUNCTION THEOREM
17
7. By the divergence theorem and symmetry, the volume is given by
Z Z Z
Z Z
(xi + yj + zk) · N̂ dS,
div(xi + yj + zk)dV = 2
S
where S = {(x, y, z) : 4x2 + 9y 2 + z 2 = 1, , z ≥ 0}. The surface has a
projection on the xy-plane.
8. It is easy to see that the curve lies in the given plane. Note that
t → (1+sin t, 1+cos t) parametrizes the the circle of radius one centered
at (1, 1) in the xy-plane, so that the function G(x, y, z) = (x, y, 5−x−y)
maps this circle onto C. The surface enclosed is the image by G of the
disc D of radius one centered at (1, 1) in the xy-plane. Stokes’s theorem
gives
I
Z
F · dr =
curlF · N̂ dS,
S
C
where S is described by z = 5 − x − y, x, y ∈ D. The outer unit normal
is the same as the one of the plane x + y + z = 5, that is N̂ = (1, 1, 1),
and
curlF = ex k, curlF · N̂ = ex .
√
Finally, dS = 3.
7. Preparation for the written exam- set 2
Exercise 1. Consider the function f : R2 → R defined by
f (x, y) = ex + x2 y + y + y 3 .
Compute
∂f
∂x
g(x, y) = − ∂f
.
∂y
Show that there exists a solution u of the differential equation
u0 (x) = −
ex + 2xu(x)
1 + x2 + 3u2 (x)
18
ALEXANDRU ALEMAN
defined on an interval (−a, a), a > 0 with u(0) = 0.
Exercise 2. Find the minimum value of x2 +y 2 +z 2 when x+y +z = 1.
Exercise 3. If the
C is given by r(t) = t2 cos ti+t2 sin tj+tk, 0 ≤
R curve
t ≤ π, evaluate C (x2 + y 2 )ds.
Exercise 4. Show that the field
2x
2y
x2 + y 2
F (x, y, z) =
i+ j+ 3−
k,
z
z
z2
is conservative and find a potential (find φ with ∇φ = F ). Find the
field lines and the equipotential surfaces of F .
Exercise 5. Let r = xi + yj + zk, and r = |r|2 . Let f, g : R → R
be differentiable on R. Show that the field F (x, y, z) = f (r)i + g(r)j
cannot be conservative unless f, g are constant on [0, ∞).
Exercise 6. Use a line integral to compute the area enclosed by the
curve r = (ecos t + sin t)i + cos tj, 0 ≤ t ≤ 2π.
Exercise 7. Use a surface integral to compute the volume of the cone
{(x, y, z) : x2 + 4y 2 = z, 0 ≤ z ≤ 1}.
Exercise 8. For F (x, y, z) = −yi + x2 j + zk, use Stokes’s theorem to
compute the circulation of this field around the oriented boundary of
{(x, y, z) : z = 16 − (x − 1)2 − y 2 , z ≥ 0} where the normal points
outwards.
THE IMPLICIT FUNCTION THEOREM
19
8. Hints
1. We have
g(x, y) = −
∂f
∂x
∂f
∂y
=−
ex + 2xy
,
1 + x2 + 3y 2
> 0 on R2 . Since f (0, 0) = 1, we can apply
in particular, we see that ∂f
∂y
the implicit function theorem to find a continuously differentiable function u defined on a neighborhood of 0 such that f (x, u(x)) = 1, u(0) =
0 and
∂f
∂x
u0 (x) = − ∂f
(x, u(x)).
∂y
2. The gradient of the Lagrange function is (2x, 2y, 2z, 0)−λ(1, 1, 1, x+
y + z − 1) and it vanishes when 2x = 2y = 2z = λ. Then λ = 23 , and
x2 + y 2 + z 2 = 31 . Since for x = 1, y = z = 0, we have x2 +2 +z 2 = 1
this must be the minimum.
3. We have
p
√
ds = (2t cos t − t2 sin t)2 + (2t sin t + t2 cos t)2 + t2 = 5t2 + t4 ,
and on C
x2 + y 2 = t4 .
Therefore
Z
2
2
Z
4
t
(x + y )ds =
C
π
0
√
1
5t2 + t2 dt =
2
Z
π2
√
x2 5 + xdx.
0
Integrate by parts.
4. The potential is given by
x2 + y 2
,
z
and therefore the equipotential surfaces are given by
φ(x, y, z) = 3z +
{(x, y, z) : 3z 2 + x2 + y 2 = Cz},
C ∈ R.
20
ALEXANDRU ALEMAN
The field lines are given by the differential equations
zdx
zdy
z 2 dz
=
=
.
2x
2y
3z − x2 − y 2
From the first equation we get x = Cy, and the second becomes
zdz
dy
,
=
3z − (C 2 + 1)y 2
y
i.e.
z C2 + 1
+
ln |3z − (C 2 + 1)y 2 | = ln |y|.
3
3
5. We have F (x, y, z) = f (x2 + y 2 + z 2 )i + g(x2 + y 2 + z 2 )j + 0k =
F1 i + F2 j + F3 k. If F is conservative we must have
∂F1
= f rac∂F2 ∂x, f rac∂F1 ∂z = f rac∂F3 ∂x, f rac∂F2 ∂z = f rac∂F3 ∂y.
∂y
This gives
2yf 0 (x2 +y 2 +z 2 ) = 2xg 0 (x2 +y 2 +z 2 ), 2zf 0 (x2 +y 2 +z 2 ) = 0, 2zg 0 (x2 +y 2 +z 2 ) = 0,
i.e. f 0 (x2 + y 2 + z 2 ) = g 0 (x2 + y 2 + z 2 ) = 0, whenever z 6= 0. Since every
t ∈ (0, ∞) can be writtenq
as t = x2 + y 2 +q
z 2 for some x, y, z ∈ R with
z 6= 0 (for example z = 2t , x = 0, y = 2t ,) it follows that f, g are
constant on (0, ∞) and the desired result follows from the continuity
of f, g.
6. By Green’s theorem, the area is given by
Z
Z
xdy =
C
2π
cos t
(e
0
Z
+ sin t)(− sin t)dt = 0 +
0
2π
1 − cos(2t)
dt = π.
2
THE IMPLICIT FUNCTION THEOREM
21
7. I’ll take this out of the collection since the calculations are too
tideous. Sorry!
8. This is the surface of a cone and its boundary is the circle C =
{(x − 1)2 + y 2 = 16} in the xy-plane. Stokes’s theorem gives
I
Z
F · dr =
curlF · N̂ dS,
S
C
for any surface S with this boundary. The most convenient one is the
disc D = {(x − 1)2 + y 2 ≤ 16} in the xy-plane. Therefore taking also
into account the orientation,
I
Z
F · dr = (2x + 1)k · kdxdy = π.
C
D