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) f j (x 1 ,...,x n ,y 1 ,...y m )=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 y k in terms of n variables x j , i.e. we could write y k = g k (x 1 ,...,x n ), 1 ≤ k ≤ m, where g k are functions of n variables. In order to develop a meaningful discussion we must assume that (1.1) is satisfied at least at one point, that is, there exist x 0 1 ,...,x 0 n ,y 0 1 ,...,y 0 m such that (1.2) f j (x 0 1 ,...,x 0 n ,y 0 1 ,...y 0 m )=0, 1 ≤ j ≤ m. In this case we require that the functions y k = g k (x 1 ,...,x n ), 1 ≤ k ≤ m, satisfy g k (x 0 1 ,...,x 0 n )= y 0 k , 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 x 2 + y 2 + z 2 =1 x 3 + 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 equa- tions. a) The equation x + y =5, 1
Transcript
THE IMPLICIT FUNCTION THEOREM
ALEXANDRU ALEMAN
1. Motivation and statement
We want to understand a general situation which occurs in almostany area which uses mathematics. Suppose we are given number ofequations which involve a higher number of unknown variables, say
Common sense says that we might not be able to solve these equationsuniquely, but we might be able to express the variables yk in terms ofn 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)is satisfied at least at one point, that is, there exist x01, . . . , x
0n, y
01, . . . , y
0m
such that
(1.2) fj(x01, . . . , x
0n, y
01, . . . y
0m) = 0, 1 ≤ j ≤ m.
In this case we require that the functions yk = gk(x1, . . . , xn), 1 ≤ k ≤m, satisfy
gk(x01, . . . , x
0n) = y0k, 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 decidewhether the system
x2 + y2 + z2 = 1
x3 + y3 + z3 = 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 equa-tions.a) The equation
x+ y = 5,1
2 ALEXANDRU ALEMAN
has no unique solution, but each unknown x, y can be expressed in termsof 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 interms of z. On the other hand, we can say that y, z can be expressedin 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 nlinear equations with n+m unknowns then there exists a number k ≤ nsuch that all unknowns can be expressed in terms of k of them. As wehave seen above these k unknowns are not arbitrary.c) In the linear case it is quite easy to decide which variables can beused to express the others. Here is an example which illustrates this.Consider the linear system
a1x+ b1y + c1z + d1u = 1
a2x+ b2y + c2z + d2u = 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
(1.3)
(c1 d1c2 d2
)is non-zero. Then writing the system as
c1z + d1u = 1− a1x− b1yc2z + d2u = 2− a2x− b2y,
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 unknownsare expressed as functions of the given ones. For nonlinear expressionsthis might fail.
2) The equation of a circle. Consider the equation
x2 + y2 = 1, x, y ∈ [−1, 1]
Its general solution is given by
x = ±√
1− y2, 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 3
if x0 =√
34, y0 = 1
2, then for x is close to x0, the function
y = +√
1− x2,satisfies the equation as well as the condition y(x0) = y0. However, ify0 = 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 arelooking for. For example, we could consider them as definedonly in a neighborhood of the point (x01, . . . , x
0n) 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 illustratedin Example 1c).
The second type of condition may appear mysterious, but it is just thenonlinear version of the condition in Example 1c), and it can be easilyexplained using derivatives.
Before we do that let us give a precise formulation of Problem (1.1) forcontinuously differentiable functions.
Assume that the functions f1, . . . , fm have continuous partial deriva-tives in a neighborhood V of p0 = (x01, . . . , x
0n, y
01, . . . , y
0m) ∈ Rn+m and
satisfy:fj(x
01, . . . , x
0n, y
01, . . . y
0m) = 0, 1 ≤ j ≤ m.
Does there exists a neighborhood U of x0 = (x01, . . . , x0n) ∈ Rn and
uniquely determined functions, g1, . . . , gm : U → R with continuouspartial derivatives in U , such that
Let us look at the simplest case when n = m = 1. Suppose thatf : 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, y0) = 0,
for some (x0, y0) ∈ R2, and there exists an interval I containing x0
and a function g : I → R with a continuous derivative on I such thatg(x0) = y0 and
f(x, g(x)) = 0, x ∈ I.
4 ALEXANDRU ALEMAN
By differentiating this equality w.r.t. x and using the chain rule weobtain
∂f
∂x(x, g(x)) +
∂f
∂y(x, g(x))g′(x) = 0, x ∈ I.
Then obviously ∂f∂x
(x, g(x)) = 0 whenever ∂f∂y
(x, g(x)) = 0
This is a very restrictive condition which we definitely want to avoid!In fact, Example 2) shows that when x0 = 1, y0 = 0, there are twofunctions g as above and none of them is differentiable at x0. Thus theassumption we shall use here is that at least near (x0, y0) we have
∂f
∂y(x, y) 6= 0.
Note that in this case we have
g′(x) =∂f
∂x(x, g(x))/
∂f
∂y(x, g(x)),
in particular, g′ is continuous at those points At the cost of workingwith smaller sets we may assume as well that
∂f
∂y(x0, y0) 6= 0,
since the continuity of the partial derivate implies that this holds in aneighborhood of (x0, y0).
and
g′(x) =∂f
∂x(x, g(x))/
∂f
∂y(x, g(x)),
∂f
∂y(x, g(x)) 6= 0,
A similar situation occurs in higher dimensions as well. Suppose thatf1, f2 : R3 → R has continuous partial derivatives, and that Problem(1.1) can be solved, i.e.
fj(x0, y0, z0) = 0, j = 1, 2,
for some (x0, y0, z0) ∈ R3, and there exists an interval I containing x0
and two functions g1, g2 : I → R with continuous derivatives on I, suchthat g1(x
0) = y0, g2(x0) = z0, 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 chainrule we obtain
∂fj∂x
(x, g1(x), g2(x))+∂fj∂y
(x, g1(x), g2(x))g′1(x)+∂fj∂z
(x, g1(x), g2(x))g′2(x) = 0,
THE IMPLICIT FUNCTION THEOREM 5
when x ∈ I, j = 1, 2. This can be rewritten as(∂f1∂x∂f2∂x
) ∣∣∣∣(x,g1(x),g2(x))
+
(∂f1∂y
∂f1∂z
∂f2∂y
∂f2∂z
)∣∣∣∣(x,g1(x),g2(x))
(g′1(x)g′2(x)
)= 0.
If the 2 × 2 matrix above is not invertible, this creates similar com-plications as in the scalar case. For this reason, we shall assume it isinvertible near the point (x0, y0, z0). Again, by continuity it sufficesto assume that this matrix computed at x0 is invertible, in which caseit 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 (seeCC:128, p. 732). Our condition is equivalent to
∂(f1, f2)
∂(y, z)|(x0,y0,z0) 6= 0.
Also note that in this case we can apply Cramer’s rule to find animplicit formula for the derivatives g′1, g
′2. Recall that the Cramer rule
says that if ∣∣∣∣a bc d
∣∣∣∣ 6= 0,
then the unique solution of the linear system
ax+ by = u
cx+ dy = v
is given by
x =
∣∣∣∣u cv d
∣∣∣∣∣∣∣∣a bc d
∣∣∣∣ , y =
∣∣∣∣a ub v
∣∣∣∣∣∣∣∣a bc d
∣∣∣∣In our case this gives
g′1(x) = −∂(f1,f2)∂(x,y)
∂(f1,f2)∂(y,z)
∣∣∣∣(x,g1(x),g2(x))
, g′2(x) = −∂(f1,f2)∂(z,x)
∂(f1,f2)∂(y,z)
∣∣∣∣(x,g1(x),g2(x))
.
We are now ready to state the Implicit Function Theorem.
Theorem 1.1. Assume that the functions f1, . . . , fm have continuouspartial derivatives in a neighborhood V of p0 = (x01, . . . , x
0n, y
01, . . . , y
0m) ∈
Rn+m and satisfy:
(1) fj(x01, . . . , x
0n, y
01, . . . y
0m) = 0, 1 ≤ j ≤ m,
6 ALEXANDRU ALEMAN
(2)∂(f1, . . . , fm)
∂(y1, . . . , ym)
∣∣∣∣p0
6= 0.
Then there exists a neighborhood U of x0 = (x01, . . . , x0n) ∈ Rn and
uniquely determined functions, g1, . . . , gm : U → R with continuouspartial derivatives in U , such that
(3) gk(x01, . . . , x
0n) = y0k, 1 ≤ k ≤ m,
(4) (x1, . . . , xn, g1(x), . . . , gm(x)) ∈ Vand
(5) fj(x, g1(x), . . . , gm(x)) = 0, x = (x1, . . . xn) ∈ U, 1 ≤ j ≤ m.
Moreover, for 1 ≤ j ≤, 1 ≤ k ≤ m
(6)∂gk∂xj
(x) = −∂(f1,...,fm)
∂(y1,...,yk−1,xj ,yk,...,ym)
∂(f1,...,fm)∂(y1,...,ym)
∣∣∣∣(x,g1(x),...,gm(x))
.
2. Exercises
Exercise 2.1. Show that Problem (1.1) is solvable for the system
x2 + y2 + z2 = 1
x3 + y3 + z3 = 2,
and any point (x0, y0, z0) where the equations are satisfied, and y0, z0 6=0, y0 6= z0.
Solution We only need to verify hypothesis (3) in Theorem 1.1 for thefunctions f1(x, y, z) = x2 + y2 + z2, f2(x, y, z) = x3 + y3 + z3, that is,
∂(f1, f2)
∂(y, z)
∣∣∣∣(x0,y0,z0)
6= 0.
Since∂f1∂y
= 2y,∂f1∂z
= 2z,∂f2∂y
= 3y2,∂f2∂z
= 3z2,
the determinant on the left is∣∣∣∣ 2y 2z3y2 3z2
∣∣∣∣(x0,y0,z0)
= 6y0(z0)2 − 6(y0)2z0 = 6y0z0(z0 − y0).
Under the conditions on y0, z0, this determinant does not vanish andthe 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 + y2 + z2 = 1
x3 + y3 + z3 = 2,
Show thatS ∩ {(x, y, z) : 0 < y < z},
is a smooth curve, in th sense that there exists an interval I ⊂ R anda 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 func-tion g : R2 → R2 whose inverse is not differentiable on R2.b) Let V ⊂ R3 be open, and let f : V → R3 have continuous partialderivatives in V . Show that if the Jacobian matrix Df(p0) of f atp0 ∈ V is non-singular then there exists a neighborhood U of f(p0) anda function h : U → V with continuous partial derivatives in U , suchthat 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
h1(x, y) =
(x3
y
),
with inverse
h−11 (x, y) =
(3√xy
),
which is not differentiable at the origin because it does not have apartial derivative in x at that point.b) The equality
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
∂(x1, x2, x3)
∣∣∣∣(p0,f(p0)
6= 0,
we can apply Theorem 1.1 which gives a unique function h defined nearf(p0) with continuous partial derivatives, such that
g(h(y), y) = 0⇐⇒ f(h(y)) = y.
8 ALEXANDRU ALEMAN
�
Question 2.1. Does there exist a neighborhood V1 ⊂ V of p0 such thatthe 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, . . . , x
0n) ∈
Rn and uniquely determined functions, g1, . . . , gm : U → R whose par-tial 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 precedingthe statement of the theorem. Suppose we have found differentiablefunctions g1, . . . , gm : U → R as above. From (5) and the chain rule weobtain for fixed 1 ≤ j ≤ n
∂fk∂y1
((x, g1(x), . . . , gm(x))∂g1∂xj
+ . . .+∂fk∂ym
((x, g1(x), . . . , gm(x))∂gm∂xj
= −∂fk∂xj
((x, g1(x), . . . , gm(x)).
This can be seen as a linear system with matrix (∂fk∂yl
((x, g1(x), . . . , gm(x)))1≤k≤mand right hand side
(−∂f1∂xj
(((x, g1(x), . . . , gm(x), . . . ,−∂fm∂xj
((x, g1(x), . . . , gm(x).
By Cramer’s rule its solution is just (6). Now if (6) holds and each gk iscontinuous, 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 deriva-tives exist in U , such that (3),(4),(5) hold.
We shall prove the claim by induction. We start with the case whenm = 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
∂y(x01, . . . , x
0n, y
0) 6= 0.
THE IMPLICIT FUNCTION THEOREM 9
Also, we can assume without loss that
∂f
∂y(x0, y0) > 0,
otherwise we can change from f to −f . Since V is open and thefunction y → ∂f
∂y(x0, y) is assumed to be continuous, there exists a
cube
Q = [x01 − a, x01 + a]× . . .× [x0n − a, x0n + a]× [y0 − a, y0 + a] ⊂ V,
such that∂f
∂y(x, y) > 0, (x, y) ∈ Q.
In particular, since f(x0, y) vanishes at y0 and is strictly increasing on[y0−, y0 + a], it satisfies
f(x0, y0 − a) < 0, f(x0, y0 + a) > 0.
At this point we use uniform continuity. Recall that a continuous real-valued 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| < δthen |h(u)− h(v)| < ε.This means that for every ε > 0 there exists δ > 0 which can be chosenin (0, a], such that whenever |x− x0| < δ,
|f(x, y)− f(x0, y)| < ε, y ∈ [y0 − a, y0 + a].
If ε is sufficiently small we conclude that for |x− x0| < δ we have
f(x, y0 − a) < 0, f(x, y0 + a) > 0.
Since for each x as above, y → f(x, y) is strictly increasing on [y0 −a, y0 + a] there exists a unique y = g(x) ∈ [y0 − a, y0 + a] withf(x, g(x)) = 0. Clearly, g(x0) = 0 which gives (3), (4) and (5)witha function g defined on
U = {x : |x− x=| < ε},
with values in [y0 − a, y0 + a]. It remains to show that g has partialderivatives in U .
This is the hard part of the proof! Fortunately it isn’t verylong. If you did not already note it, the only informationavailable about g is that f(x, g(x)) = 0, so that all furtherinformation about this function must emerge from it.
10 ALEXANDRU ALEMAN
a) g is continuous on U . Indeed, if we assume the contrary, there existsx ∈ U and a sequence (xm) in U with
limm→∞
xm = x, g(x) 6= limm→∞
g(xm) = y ∈ [y0 − a, y0 + a].
But then f(x, ·) is strictly increasing on [y0 − a, y0 − a] and satisfiesf(x, y) = f(x, g(x)) = 0. this is a contradiction which shows the conti-nuity of g.b) g has partial derivatives in U . If {e1, . . . , en} denotes the canonicalbasis in Rn, we want to show that for each x ∈ U , the limits
limh→0
g(x+ hej)− g(x)
h, 1 ≤ j ≤ n
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 ,
0 =f(x+ hej, g(x+ hej))− f(x, g(x))
h
(3.4)
=f(x+ hej, g(x+ hej))− f(x, g(x+ hej))
h+f(x, g(x+ hej))− f(x, g(x))
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))
h=u(h)− u(0)
h= u′(s)
=∂f
∂x(x+ sej, g(x+ hej),
where s ∈ [0, 1]. For the second term on the right we consider thefunction
The function is well defined and by the mean-value theorem it satisfies
v(1)− v(0) = v′(r), r ∈ [0, 1],
which, by the chain rule implies
f(x, g(x+hej))−f(x, g(x)) =∂f
∂y(x, rg(x+hej)+(1−r)g(x))(g(x+hej)−g(x)),
for some r ∈ [0, 1], and thus
f(x, g(x+ hej))− f(x, g(x))
h=∂f
∂y(x, rg(x+hej)+(1−r)g(x))
g(x+ hej)− g(x)
h.
THE IMPLICIT FUNCTION THEOREM 11
If we replace these in (3.4) we obtain
0 =f(x+ hej, g(x+ hej))− f(x, g(x))
h(3.5)
=∂f
∂x(x+ sej, g(x+ hej) +
∂f
∂y(x, rg(x+ hej)
+ (1− r)g(x))g(x+ hej)− g(x)
h,(3.6)
for some s ∈ [0, h], r ∈ [0, 1], which may depend of h (and x). Nowuse the continuity of g, ∂f
∂x, ∂f∂y
to conclude that
limh→0
∂f
∂x(x+ sej, g(x+ hej)) =
∂f
∂x(x, g(x)),
uniformly on [0, h], and
limh→0
∂f
∂y(x, rg(x+hej)+(1−r)g(x))
g(x+ hej)− g(x)
h=∂f
∂y(x, g(x)) 6= 0, x ∈ U,
uniformly on [0, 1]. Thus
limh→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 thebeginning, we prove the general case by induction. Assume (*) holdsfor some positive integer m. We want to prove that the claim holds form+ 1 as well.Assume that f1, . . . , fm+1 have continuous partial derivatives in a neigh-borhood V of p0 = (x01, . . . , x
0n, y
01, . . . , y
0m+1) ∈ Rn+m+1 and satisfy:
fj(x01, . . . , x
0n, y
01, . . . y
0m+1) = 0, 1 ≤ j ≤ m+ 1,
∂(f1, . . . , fm+1)
∂(y1, . . . , ym+1)
∣∣∣∣p0
6= 0.
Note that∂(f1, . . . , fl−1, fl+1, . . . fm+1)
∂(y1, . . . , yk−1, yk+1, . . . , ym+1)
is obtained from ∂(f1,...,fm+1)∂(y1,...,ym+1)
by deleting the l-th row and the k-th col-
umn, i.e. it is a minor of order m of this determinant. Then at least
12 ALEXANDRU ALEMAN
one of these minors must be non-zero at p0, otherwise
∂(f1, . . . , fm+1)
∂(y1, . . . , ym+1)
∣∣∣∣p0
= 0.
Since the functions and variables can be permuted, there is no harm ifwe assume that
∂(f2, . . . , fm+1)
∂(y2, . . . , ym+1)
∣∣∣∣p0
6= 0.
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 continuouspartial derivatives in a neighborhood V of p0 = (x01, . . . , x
0n, y
01, . . . , y
0m+1) ∈
Rn+1+m and satisfy:
fj(x01, . . . , x
0n, y
01, . . . y
0m+1) = 0, 2 ≤ j ≤ m+ 1,
∂(f2, . . . , fm+1)
∂(y2, . . . , ym+1)
∣∣∣∣p0
6= 0.
But then the induction hypothesis implies that there exists a neigh-borhood U ′ of x0 = (x01, . . . , x
0n, y
01) ∈ Rn+1 and uniquely determined
functions, g2, . . . , gm+1 : U ′ → R whose partial derivatives exist in U ′,such that
Recall from the beginning of the proof that in this case the partialderivatives of g2, . . . , gm+1 are continuous in U ′ and replace this in thefirst equation to obtain
that is, we arrived at the case when m = 1. Since this was solvedabove, it follows that there exists a neighborhood U of x0 and a functiong1 : U → R whose partial derivatives exist in U , such that g(x0) = y01,(x0, g1(x)) ∈ U ′, and
By the chain rule, the partial derivatives of , gk(x, g1(x)) exist in U andwe are done. �
THE IMPLICIT FUNCTION THEOREM 13
4. An Application
Let us return to the situation considered in Exercise 2.3. We have seenthat the inverse of a bijective differentiable function is not necessar-ily differentiable. It is of great importance to find (simple) additionalconditions which prevent this. With help of the Implicit Function The-orem this becomes an easy task. However, the result below is actuallyequivalent to Theorem 1.1.
Theorem 4.1. Let U ⊂ Rn be open and let f : U → Rn be injectiveand have continuous partial derivatives on U . on U . If f = (f1, . . . , fn)satisfies
∂(f1, . . . , fn)
∂(x1, . . . , xn)(x) 6= 0, x ∈ U,
then f(U) is open and f−1 : f(U) → U has continuous partial deriva-tives on f(U).
Proof. Let y0 ∈ f(U), say y0 = f(x0), and consider again the functiong : 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), theequality g(x, y) = 0 is equivalent to
By Theorem 1.1 it follows that there exists a neighborhood W of y0
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 y0 =∈ f(U) was arbitrary, it follows that f(U) isopen (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 contin-uous 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) = u3x− 3x2y − 3zy − z4.
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 onthe set
{(x, y, z) : x2 + 4y2 + ez = 1, −1 ≤ z ≤ 0}.
Exercise 3. Show that the curve C given by r(t) = 2 cos t sin ti +2 cos2 tj + 2 sin tk, 0 ≤ t ≤ π lies on a sphere centered at the origin.Find
∫Czds.
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 thefield lines and the equipotential surfaces of F .
Exercise 5. Let r = xi+yj+zk, and r = |r|2. Show that if f : R→ Ris twice differentiable on R then
∇2f(r) = 4rf ′′(r) + 6f ′(r).
THE IMPLICIT FUNCTION THEOREM 15
Recall that ∇2 is the Laplacian, ∇2u = ∂2u∂x2
+ ∂2u∂y2
+ ∂2u∂z2
.
Exercise 6. Use a line integral to compute the area enclosed by thecurve 4x2/3 + 9y2/3 = 16.
Exercise 7. Use a surface integral to compute the volume of theellipsoid {(x, y, z) : 4x2 + 9y2 + z2 ≤ 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-planeis a disc. Use Stokes’s theorem to evaluate∮
C
F · dr,
where F (x, y, z) = yexi + exj + z2k.
6. Hints
1. Note that on the given set we have ∂f∂u6= 0. Apply the implicit
function theorem.
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 forcesz = 0. Thus the extrema can only occur on the boundary when z = 0or z = −1. If z = 0, x = y = 0 and 2xy + ez = 1. If z = −1 ,x2 + y2 = 1 − e−1 and we can consider the maximum or minimum of
16 ALEXANDRU ALEMAN
2xy+e−1 when x2 +y2 = 1−e−1. Another application of the Lagrangemethod 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∫C
zds = 4
∫ π
0
sin tdt = 8.
4. If φ(x, y, z) = xyz then ∇φ = F . The equipotential surfaces arethen 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
yz=dy
xz=dz
xy,
so that
xdx = ydy = zdz, x2 = y2 + C1, y2 = z2 + C2.
5. We have f(r) = f(x2 + y2 + z2) = u(x, y, z), hence by the chain andproduct rule
∂u
∂x= f ′(r)2x,
∂2u
∂x2= f ′′(r)4x2 + 2f ′(r).
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 becomputed with Green’s theorem and it is given by
∫Cxdy. To compute
this integral we can use the parametrization r = 2 cos3 ti+2 sin3 tj, 0 ≤t ≤ 2π and obtain ∫
C
xdy = 12
∫ 2π
0
cos4 t sin2 tdt.
THE IMPLICIT FUNCTION THEOREM 17
7. By the divergence theorem and symmetry, the volume is given by∫ ∫ ∫div(xi + yj + zk)dV = 2
∫ ∫S(xi + yj + zk) · N̂dS,
where S = {(x, y, z) : 4x2 + 9y2 + z2 = 1, , z ≥ 0}. The surface has aprojection on the xy-plane.
8. It is easy to see that the curve lies in the given plane. Note thatt→ (1+sin t, 1+cos t) parametrizes the the circle of radius one centeredat (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 thedisc D of radius one centered at (1, 1) in the xy-plane. Stokes’s theoremgives ∮
C
F · dr =
∫S
curlF · N̂dS,
where S is described by z = 5−x−y, x, y ∈ D. The outer unit normalis the same as the one of the plane x+ y + z = 5, that is N̂ = (1, 1, 1),and
curlF = exk, 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 + x2y + y + y3.
Compute
g(x, y) = −∂f∂x∂f∂y
.
Show that there exists a solution u of the differential equation
u′(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+y2+z2 when x+y+z = 1.
Exercise 3. If the curve C is given by r(t) = t2 cos ti+t2 sin tj+tk, 0 ≤t ≤ π, evaluate
∫C
(x2 + y2)ds.
Exercise 4. Show that the field
F (x, y, z) =2x
zi +
2y
zj +
(3− x2 + y2
z2
)k,
is conservative and find a potential (find φ with ∇φ = F ). Find thefield lines and the equipotential surfaces of F .
Exercise 5. Let r = xi + yj + zk, and r = |r|2. Let f, g : R → Rbe differentiable on R. Show that the field F (x, y, z) = f(r)i + g(r)jcannot be conservative unless f, g are constant on [0,∞).
Exercise 6. Use a line integral to compute the area enclosed by thecurve 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 + 4y2 = z, 0 ≤ z ≤ 1}.
Exercise 8. For F (x, y, z) = −yi + x2j + zk, use Stokes’s theorem tocompute the circulation of this field around the oriented boundary of{(x, y, z) : z = 16 − (x − 1)2 − y2, z ≥ 0} where the normal pointsoutwards.
THE IMPLICIT FUNCTION THEOREM 19
8. Hints
1. We have
g(x, y) = −∂f∂x∂f∂y
= − ex + 2xy
1 + x2 + 3y2,
in particular, we see that ∂f∂y> 0 on R2. Since f(0, 0) = 1, we can apply
the implicit function theorem to find a continuously differentiable func-tion u defined on a neighborhood of 0 such that f(x, u(x)) = 1, u(0) =0 and
u′(x) = −∂f∂x∂f∂y
(x, u(x)).
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 = 2
3, and
x2 + y2 + z2 = 13. Since for x = 1, y = z = 0, we have x2 +2 +z2 = 1
this must be the minimum.
3. We have
ds =√
(2t cos t− t2 sin t)2 + (2t sin t+ t2 cos t)2 + t2 =√
5t2 + t4,
and on C
x2 + y2 = t4.
Therefore∫C
(x2 + y2)ds =
∫ π
0
t4√
5t2 + t2dt =1
2
∫ π2
0
x2√
5 + xdx.
Integrate by parts.
4. The potential is given by
φ(x, y, z) = 3z +x2 + y2
z,
and therefore the equipotential surfaces are given by
{(x, y, z) : 3z2 + x2 + y2 = Cz}, C ∈ R.
20 ALEXANDRU ALEMAN
The field lines are given by the differential equations
zdx
2x=zdy
2y=
z2dz
3z − x2 − y2.
From the first equation we get x = Cy, and the second becomes
zdz
3z − (C2 + 1)y2=dy
y,
i.e.
z
3+C2 + 1
3ln |3z − (C2 + 1)y2| = ln |y|.
5. We have F (x, y, z) = f(x2 + y2 + z2)i + g(x2 + y2 + z2)j + 0k =F1i + F2j + F3k. If F is conservative we must have
i.e. f ′(x2 +y2 +z2) = g′(x2 +y2 +z2) = 0, whenever z 6= 0. Since everyt ∈ (0,∞) can be written as t = x2 + y2 + z2 for some x, y, z ∈ R with
z 6= 0 (for example z =√
t2, x = 0, y =
√t2,) it follows that f, g are
constant on (0,∞) and the desired result follows from the continuityof f, g.
6. By Green’s theorem, the area is given by
∫C
xdy =
∫ 2π
0
(ecos t + sin t)(− sin t)dt = 0 +
∫ 2π
0
1− cos(2t)
2dt = π.
THE IMPLICIT FUNCTION THEOREM 21
7. I’ll take this out of the collection since the calculations are tootideous. Sorry!
8. This is the surface of a cone and its boundary is the circle C ={(x− 1)2 + y2 = 16} in the xy-plane. Stokes’s theorem gives∮
C
F · dr =
∫S
curlF · N̂dS,
for any surface S with this boundary. The most convenient one is thedisc D = {(x − 1)2 + y2 ≤ 16} in the xy-plane. Therefore taking alsointo account the orientation,∮
