Section 3.4 Implicit FUnction Theorems

III )

Section 3.4 Section 3.4 Implicit FUnction Theorems

Then

Xl

₌

Xo - f'(XO)-ly

=

Xo

-(I + B)y ⁼

Xo - Y -

By. Hence

3 [ ¹

²

] ^t [ ¹

²

]

Xl

(8) = 2 2

-

"18 - 8 ^-

10

^{(38t 2 -}"It - t dt b ^�

.0063968616)

=

1

+ 8^{2 +}

(.0071279315)

8

Problems 3.3

135

•

1. For the one-dimensional version of Newton's method, prove that if _r is a root of multi­

plicity m, then quadratic convergence in the algorithm can be preserved by defining Xn+! = Xn -mI(xn)/I'(xn)

2. Prove the corollaries, giving in each case the precise assumptions that must be made concerning the starting points.

3. Let eO, el . ..

.

be a sequence of positive numbers satisfying en+l � ce�. Find necessary and sufficient conditions for the convergence limn en = O.

4. Let I be a function from IR to IR that satisfies the inequalities f' > 0 and I" > O. Prove that if I has a zero, then the zero is unique, and Newton's iteration, started at any point, converges to the zero.

5. How must the analysis in Theorem 1 be modified to accommodate functions from C to C? (Remember that the Mean Value Theorem in its real-variable form is not valid. ) 6 . If r i s a zero o f a function I , then the corresponding "basin o f attraction ⁿ is the set

of all x such that the Newton sequence starting at x converges to r. For the function I(z) = z2 + 1, ^{z E} C, and the zero r = i, prove that the basin of attraction contains the disk of radius � ^about r.

3.4 Implicit Function Theorems

In this section we give several versions of the Implicit Function Theorem and prove its corollary, the Inverse Function Theorem. Theorems in this broad cate­

gory are often used to establish the existence of solutions to nonlinear equations of the form

f(x)

= y. The conclusions are typically

local

in nature, and describe how the solution

X

depends on y in a neighborhood of a given solution

(XO , yo).

Usually, there will be a hypothesis involving invertibility of the derivative

f'(xo).

The intuition gained from examining some simple cases proves to be com­

pletely reliable in attacking very general cases. Consider, then, a function

F

^:

IR.2

^{-+ JR.} We ask whether the equation

F(x,

y) =

0

defines y to be a unique function of

x.

For example, we can ask this question for the equation

X

+ y2 -1 =

0 (X,

y E

JR)

This can be "solved" to yield y = vT=X. The graph of this is shown in the accompanying Figure

3.2.

It is clear that we cannot let

X

be the point

A

^{in the}

figure, because there is no corresponding y for which

F(x,

y) ⁼⁼

X

+ y2

- 1

₌ o.

One must start with a point

(xo, Yo)

like B in the figure, where we already have

F(xo, Yo)

₌

O.

Finally, observe that at the point

C

there will be a difficulty, for there are values of

X

near

C

to which no y's correspond. This is a point

13

6

Chapter 3 Calculus in Banach Spaces

where

dy/dx

= 00. Recall that if

y

=

y(x)

and if

F(x, y(x))

= ^{0, then}

y'

can be obtained from the equation

{In this equation,

Di

is partial differentiation with respect to the ith argu­

ment.) Thus

y'

⁼

-DJ F/ D2F,

and the condition

y'(xo)

= 00 corresponds to

D2F(xo, Yo)

= o. In this example, notice t.hat another function arises from Equation ( 1 ), namely

y = -

^vT=X

In a neighborhood of ( 1 , 0), both functions solve Equation ( 1 ), and there is a failure of uniqueness.

- 3

Figure 3.2

C A

In the classical implicit function theorem we have a function

F

of two real variables in class C 1 • That means simply that

EJF/EJx

^and

EJF/EJy

exist and are continuous. It is convenient to denote these partial derivatives by

Fl

^and

F2.

Theorem 1 . Classical Implicit Function Theorem. Let

F

^be

a C 1 -function on the square

{(x, y) : Ix - xol

� 6 ,

Iy - Yol

� 6

} C ]R2

If

F(xo, Yo)

= ^{0 and}

F2(xo, Yo)

I- 0, then there is a continuously differentiable function

J

defined in a neighborhood of

Xo

such that

Yo

=

J(xo)

^and

F(x, J(x))

= 0 in that neighborhood. Furthermore,

f'(x)

=

-Fl (x, y)/ F2(x, y),

^where

y

⁼

J(x).

Proof. Assume that

F2(xo, Yo)

^{> o.} Then by continuity,

F2(X, y)

^{> Q >} 0 in a neighborhood of

(xo, Yo),

which we assume to be the original 6-neighborhood.

The function

y

^H

F(xo, y)

is strictly increasing for

Yo -

6 �

y

�

Yo

+ 6. Hence

F(xo, Yo

^{-6) <}

F(xo, Yo)

= ^{0 <}

F(xo, Yo

^{+ 6)}

By continuity there is an ^c in (0, 6) such that

F(x, Yo -

6) < 0 <

F(x, Yo +

6) if

Ix - xol

^{< c.} By continuity, there corresponds to each such

x

a value of y such that

F(x, y)

= 0 and

Yo -

6 < y <

Yo +

6. If there were two such y's, then by Rolle's theorem,

F2(x,y)

= 0 at some point, contrary to hypothesis. Hence

y

is unique, and we may put

y

=

J(x).

Then we have

F(x, J(x)) =

^{0 and}

^Yo

⁼

^J(xo).

Section 3.4 Implicit Function The

or

ems

137 Now fix

XI

in the c-neighborhood of

Xo.

^Put

YI

⁼

f(xI).

^Let

�x

be a small number and let

YI ⁺ �Y ⁼ f(xI ⁺ �x).

^Then

0 = F(xI + �X' YI + �y)

=

FI (XI + ()�x, YI + ()�y)�x ⁺ F2(xI + ()�x, YI + ()�y)�y

for an appropriate () satisfying

0 � () �

1. (This is the Mean Value Theorem for a function from

JR2

^{to lR.} See Problem 3.2.8, page 1 24.) This equation gives us

�Y FI (xI + ()�x, YI + ()�y)

�x F2(xI + ()�x, YI ⁺ ()�y)

As

�x

^{--+ 0} the right side remains bounded. Hence, so does the left side. This proves that ^as

�x

converges to

0, �Y

also converges to o. Hence

f

is continuous at

XI.

After all,

�Y ⁼ f(xI ⁺ �x) - f(xI ).

Furthermore,

1'(xd ⁼

^lim

�Y =

^_

FI(xI , yd

�x F2(xI , yJ)

Therefore,

f

is differentiable at

XI.

The formula can be written

1'(x) ⁼ - FI (x, f(x)) I F2 (x, f(x))

and this shows that l' is continuous at

x,

provided that

X

is in the open interval

(xo - c,Xo +

^{c). •}

Theorem 2. Implicit Function Theorem for Many Variables.

Let F : JRn

^x

JR

^--+

JR, and suppose that F(xo, Yo) = 0 for some Xo

^E

lRn and Yo

^{E lR.}

If all

n

+

1

partial derivatives DiF exist and are continuous in a neighborhood of (xo, Yo) ^{and if} Dn+1 F(xo, Yo) i= 0, then there is a continuously differentiable function f defined on a neighborhood of Xo such that F(x, f(x))

⁼

^0, f(xo) = Yo, ^and

D;J(x)

⁼

-DjF(x, f(x))IDn+IF(x, f(x))

^{( 1}

�

i

� _n)

Proof. This is left ^as a problem (Problem 3.4.4). • Example 1.

F(x, y)

=

x2 +y2 +

^{l or}

x2 +y2

^or

x2 +y2 -

1 . (Three phenomena

are illustrated. ) •

If we expect to generalize the preceding theorems to normed linear spaces, there will be several difficulties. Of course, division by

F2

will become multi­

plication by

F2-1,

and the invertibility of the F'rechet derivative will have to be hypothesized. A more serious problem occurs in defining the value of

y

^corre­

sponding to

x.

The order properties of the real line were used in the preceding proofs; in the more general theorems, an appeal to a fixed point theorem will be substituted.

Definition. Let X, Y, Z be Banach spaces. Let

F :

^{X x Y --+ Z} be a mapping.

The Cartesian product X x Y is also a Banach space if we give it the norm

lI(x, y)1I ⁼ IIxli + IIYII·

If they exist, the partial derivatives of

F

^at

(xo,Yo)

^are

bounded linear operators

DI F(xo, Yo)

^and

D2F(xo, Yo)

^{such that}

lim

IIF(xo + h,yo) - F(xo,yo) - DIF(xo, yo)hll/llhll ^{= 0} (h

^{E X,}

h

^{--+ 0)}

and

lim

IIF(xo, Yo + k) - F(xo, Yo) - D2F(xo, yo)kll/llkil ^{= 0} (k

^{E Y,}

k

^{--+ 0)}

Thus

DIF(xo, Yo)

^{E C(X, Z) and}

D2F(xo, Yo)

^{E C(Y, Z).} We often use the notation

Fi

in place of

DiF.

138

Chapter 3 Calculus in Banach Spaces

Theorem 3. General Implicit Function Theorem.

Let

X, Y,

and Z be normed linear spaces,

Y

being assumed complete. Let fI be an open set in

X ^x Y.

Let F : ^fI

^{-+ Z.}

^Let (xo,Yo)

^{E fl.}

Assume that

F is continuous at (xo, Yo), ^that F(xo, Yo)

⁼⁼

^{0, that} D2F exists in fI, that D2F is continuous at (xo, Yo), ^{and that} D2F(xo, Yo) is invertible.

Then there is a function f defined on a neighborhood of Xo such that

F(x,f(x)) ^{== 0,} f(xo)

⁼⁼

Yo, f is continuous at xo, ^and f ^{is unique}

in the sense that any other such function must agree with f ^{on some}

neighborhood of Xo.

Proof. We can assume that

(xo, Yo)

^{== (0, 0)}

^.

^Select

a ^{> °}

^{so that}

{(x, y) : IIxll ^�

^0,

IIYII ^� a}

^{c fI}

Put

A

⁼⁼

D2F(O, ⁰

⁾

^.

^Then

A

^{E .c(y, Z) and}

A

^{- I E £(Z,} Y). For each

x

satisfying

II xii ^� a

we define

G.,(y)

⁼⁼

y - A-I F(x, y).

^Here

lIyli ^{� o.}

Observe that if

G.,

has a fixed point

y.,

^then

y.

⁼⁼

G.,(y·)

⁼⁼

y' - A-I F(x, y')

from which we conclude that

F(x,y·)

^{== O.} Let us therefore set about proving that

G.,

has a fixed point. We shall employ the Contraction Mapping Theorem.

(Chapter 4, Section 2, page

177).

We have

By the continuity of

D2F

^at

^{(0, 0)}

we can reduce

a

if necessary such that

Now

G.,(O)

⁼⁼

-A-1F(x,O)

⁼⁼

_A-l{F(x,O) - F(O,O)}.

^Let

^°

^<

^e

^{< O. By}

the continuity of

F

^{at (}

^{0, 0}

⁾ we can find

o

€ E

( ⁰ ,

⁰

)

^{so that}

If

/lxll ^�

^{o€ and}

lIyli ^{� e,}

then by the Mean Value Theorem III of Section 2, page 123,

IIG.,(y)11 ^� IIG.,(O)/I ⁺ IIG.,(y) - G.,(O) II

� �e +

0';;>'';; 1 _sup

IIG�('\Y)II . IIYil

e e

� - + - == e

2 2

Define

U

=

{y

^{E Y :}

IIYII ^{� e}.}

We have shown that, for each

x

satisfying

IIxll ^�

O€ , the function

G.,

^maps

^U

^into

^{U .}

We also know that

IIG�(Y)II ^{� � .}

^By

I'>roblem 1 ,

G.,

has a unique fixed point

y

ⁱⁿ

^U.

Since this fixed point depends on

x,

^{we write}

y

⁼

f(x),

thus defining

f.

From the observations above we infer that

F(x,f(x))

⁼⁼

⁰

Section 3.4 Implicit Function Theorems

139 Since

F(O,O)

⁼

^0,

it follows that

Go(O)

⁼

O.

By the uniqueness of the fixed point,

0

=

f(O).

^{Since c} was arbitrary in

(0, 8)

we have this conclusion: For each

c in

(0, 8)

there is a

8E

such that

Our analysis then showed that

y

⁼

f(x) E ^U,

^or

Ilf(x)11 !!S;

^c. As a consequence,

II xII !!S; ^8€

^=?

IIf(x)lI !!S;

^c, showing continuity at

O.

For the uniqueness, suppose that

7

is another function defined on a neighborhood of

0

^{such that}

7

is continuous at

0, 7(0)

⁼

^0,

^and

F(x,f(x))

⁼

O.

^If

⁰

^{< c <}

^8,

^{find () >}

⁰

^such

that () <

8E

and

IIxll !!S;

^{() =?}

117(x)lI !!S;

^c

Then

7(x) E ^{U .}

So we have apparently two fixed points,

f(x)

^and

7(x),

^{for the}

function

Gx•

Since this is not possible,

f(x)

⁼

7(x)

^whenever

IIxll !!S;

⁽⁾

^.

^•

Theorem 4. Second Version of the Implicit Function Theorem.

In the preceding theorem, assume further that F is continuously dif­

ferentiable in n ^{and that} D2F(xo, Yo) is invertible. Then the function

f will be continuously differentiable and

Furthermore, there will exist a neighborhood of Xo ^{in which} f is unique.

This theorem can be found in [Dieu] , page 265.

Theorem 5. Inverse Function Theorem I.

Let f be a continu­

ously differentiable map from an open set n in a Banach space into a normed linear space. Ifxo E n ^{and if} f'(xo) is invertible, then there is a continuously differentiable function

9

defined on a neighborhood N

of f(xo) ^{such that} f(g(y))

⁼

y ^{for all} Y E N.

Proof.

For

x

^{in n and}

y

in the second space, define

F(x,y)

⁼

f(x) - y.

^Put

Yo

⁼

f(xo)

^{so that}

F(xo, Yo)

⁼

O.

Note that

DJ F(x, y)

⁼

f'(x),

^{and thus}

DJ F(xo, Yo)

is invertible. By Theorem 4, there is a neighborhood

N

^of

Yo

^and

a continuously differentiable function 9 defined on

N

such that

F(g(y),y)

⁼

^0,

or

f(g(y)) - y

⁼

⁰

^{for all}

Y E N.

^•

Theorem 6. Surjective Mapping Theorem I.

Let

X

and

Y

be Banach spaces, n an open set in

X .

Let f : n

^{-+ Y}

be a continuously differentiable map. Let Xo E n ^and Yo

⁼

f(xo).

^1£

f'(xo) is invertible, as an element of .c

(

x,

V), then

f(n) is a neighborhood of Yo.

Proof.

Define

F : n»

^{Y -+} Y by putting

F(x,y)

⁼

f(x)-y.

^Then

F(xo, Yo)

⁼

o

^and

DJF(xo,yo)

⁼

f'(xo). (DJ

is a partial derivative, ^as defined previously.) By hypothesis,

Dl F(xo, Yo)

is invertible. By the Implicit Function Theorem (with the roles of

x

^and

y

reversed!), there exist a neighborhood

N

^of

Yo

^and

a. function

g : N

^-+

n

such that

g(yo)

⁼

Xo

^a.nd

F(g(y),y)

⁼

⁰

^{for all}

y E N.

From the definition of

F

^{we have}

f (g(y)) - y

⁼

⁰

^{for all}

Y E N.

In other words, each element

y

^of

N

is the image under

f

of some point in

n,

^namely,

g(y).

^•

140 Chapter 3 Calculus in Banach Spaces

Theorem 1. A Fixed Point Theorem. Let n be an open set in a Banach space X. and let

G

be a differentiable map from n to X.

Suppose that there is a closed ball B ==

B(xo.

r) ⁱⁿ n such that

(

ⁱ

)

^{k == sup}

IIG'(x)1I

^{< 1}

(

ⁱⁱ

) IIG(xo) - xoll

xEB ^{< r( 1 - k)}

Then

G

has a unique fixed point in

B.

Proof. First, we show that

GIB

is a contraction. If

XI

and

X2

are in B, _then by t.he Mean Value Theorem

(

^{Theorem 4} in Section 3.2, page 123

)

IIG(xd - G(x2)11

� 0";..\"; 1 sup

IIG'(xI

+

A(X2 - xdll "XI - x211

Second, we show that

G

maps

B

into

B.

If

X

E B , then

IIG(x) - xoll

^�

IIG(x) - G(xo)II

+

IIG(xo) - xoll

�

kllx - xoll

+ r( 1 -. k)

� kr + ( I - k)r = r

Since X is complete,

B

is a complete metric space. By the Contractive Mapping Theorem

(

^page

^177). G

has a unique fixed point in

B.

•

Theorem 8. Inverse Function Theorem II. Let n be an open set in a Banach space _X. Let

f

be a differentiable map from n to a normed space Y. Assume that n contains a closed ball

B

⁼⁼

B(xo,

^r

)

such that

(

ⁱ

)

The linear transformation

A

⁼⁼

f'(xo)

is invertible.

(

ⁱⁱ

)

k ⁼⁼ SUPxE B

III - A-I f'(x)II

^{< 1}

Then for each

y

ⁱⁿ Y satisfying

IIY - f(xo)1I

^<

^{( 1}

^-k

)

r

IlA-I r

^{l the}

equation

f(x)

=

y

has a unique solution in

B.

Proof. Let

y

be as hypothesized, and define

G(x)

⁼

X - A-I [f(x)

-

y].

It is clear that

f(x)

=

y

if and only if

X

is a fixed point of

G.

^{The map}

G

^is

differentiable in n, and

G'(x)

⁼

I - A-I f'(x).

To verify the hypothesis (i

)

ⁱⁿ

the preceding theorem, write

IIG'(x)1I

=

III - A-I f'(x)1I

� k

(X

E

B)

By the assumptions made about

y,

we can verify hypothesis

(

ⁱⁱ

)

of the preceding theorem by writing

IIG(xo) - xoll = IIxo - A-I [J(xo)

^-

y] - xoll

=

IIA-I II IIf(xo) - YII

� IIA-I II(l - k)rIIA-I II-I

⁼ (1 - k)r

By the preceding theorem,

G

has a unique fixed point in

B,

which is the unique

solution of

f(x)

= y in

B.

•

Section

3.4

Implicit Function Theorems Example 2. Consider a nonlinear Volterra integral equation

x(t} - 2x(0}

+ � lt

cos(st)[x(sW

ds = y(t}

⁽⁰

^�

^t

^{� I)}

141

in which

y

^{E qO,}

^I ]

. Notice that when

y =

0 the integra) equation has the solution x

= o.

We ask: Does the equation have solutions when

IIYII

i s small?

Here, we use the usual sup-norm on CIO,

1],

as this makes the space complete.

(Weighted sup-norms would have this property, too.) Write the integral equation as

f(x} = y,

^where

f

has the obvious interpretation. Then

f'(x}

is given by

[J'(x}h](t} = h(t} - 2h(0}

⁺

lt cos(st}x(s}h(s}ds

Let

A = I'(O},

^{so that}

Ah = h -

2h(0}. One verifies easily that

A2h = h,

^from

which it follows that

A-I = A.

In order to use the preceding theorem, with

Xo =

0, we must verify its hypotheses. We have just seen that

A

is invertible.

Let

IIxll ^� r,

^where

r

is to be chosen later so that

III - A-I I'(x} II ^{� k <} 1.

From an equation above,

It follows that

I (J'(x}h](t) - (Ah}(t) 1 = 11t

^cos

^{( s}

^t

^}x(

^s

^}h(

^s

^{} ds l}

� IIhll llxll

and that

1IJ'(x}h - Ahll ^� IIhll llxll 1IJ'(x} - All ^� Ilxll ^� r

Since

IIAII = IIA-III = 3,

^{we have}

III - A-IJ'(x}1I = IIA-I(A - J'(x}}11 ^� IIA-Illr = 3r

The hypothesis of the preceding theorem requires that

3r ^{� k <} 1,

^where

^k

^is

to be chosen later. By the preceding theorem, the equation

f(x} = y

^{will have}

a unique solution if

In order for this bound to be as generous as possible, we let

k = ^!,

arriving at

the restriction

IIYII ^{< -k.}

^•

Lemma. Let X and Y be Banach spaces. Let n be an open set in X, and let

f

^{: n � Y} be a continuously differentiable mapping. If

Xo

^{E n and E > 0,} then there is a 6 > 0 such that

Proof. The map x ^>--+

f'(x}

is continuous from n to .c₍X, Y}. Therefore, in correspondence with the given ^E, there is a 6 > 0 such that

IIx - xoll ^<

^{6 ==>}

1IJ'(x} - !'(xo}11 ^{< E}

142 Chapter 3 Calculus in Banach Spaces

(We may assume also that

B(xo,6)

^{e n.) If}

II XI - xoll ^<

^{6 and}

IIx2 - xoll ^<

^6,

then the line segment

S

^joining

XI

^to

X2

^satisfies

^S

^C

^B (xo , ^S)

^c n. By Problem

2,

^{page 145, we have}

IIf(xd - f(X2) - !'(XO)(XI - x2)11 � IIxl - x211 .

^sup _xES

II!'(x) - !'(xo) II

� ellXI - X211

^•

Theorem 9. Surjective Mapping Theorem II.

Let

X

and Y be Banach spaces. Let

n

be an open set in

X.

Let f

^{: n -+}

^{Y be}

continuously differentiable.

If

Xo

^{E n}

^and f'(xo) has a right inverse in .c

(

Y,

X),

then f(

ⁿ

) is a neighborhood of f(xo)·

Proof. Put

A

⁼

f' (xo)

^{and let}

L

be a member of

.c

(

Y,

X) such that

AL

^{= I,}

where I denotes the identity map on

Y.

Let

c

=

IILII.

By the preceding lemma, there exists 6

>

0 such that

B(xo,6)

e n and such that

lIu - xoll �

^6,

IIv - xoll �

⁶

^=? IIf(u) - f(v) - A(u - v)11 � to lIu - vII

Let

Yo

⁼

f(xo)

^and

y

^E

B(yo, 6/2c).

We wiII find

X

E n such that

f(x)

⁼

y.

The point

X

is constructed as the limit of a sequence

{xn}

defined inductively as follows. We start with the given

Xo.

^Put

XI

⁼

Xo

⁺

L(y - Yo).

From then on we define

Xn+1

⁼

Xn - L[J(xn) - f(xn-d - A(xn - Xn-d

By induction we establish that

IIXn - xn-dl � 6/2n

^and

IIxn - xoll � 6.

^Here

are the details of the induction:

IIxl - xoll

⁼

IIL(Y - Yo II � cllY - Yo II � c6/(2c)

⁼

6/2 . IIXn+1 - xnll � cllf(Xn) - f(xn-d - A(xn - xn-dll

� c(i/2C)IIXn - xn_III � 6/2n+1

IIxn+1 - xoll � IIxn+1 - xnll

⁺

IIXn - Xn-III

^{+ . . . +}

IIXI - xoll

6 6 6

�

--

+ - + · · · + - � 6

"" 2n+1 2n 2 '"

Next we observe that the sequence

(xn!

has the Cauchy property, since (for

m > n)

IIXn -xmll � IIxn -Xn+1 II

+ . . . +

IIXm-1 -Xmll � 6 ( ^{2n � 1}

⁺

_{2n � 2}

^{+ . . .}

) � 6/2n

Since X is complete, we can define

X

^{= Hm}

xn

. All that remains is to prove

x

^{E n and}

f(x)

⁼

y.

^Since

IIXn - xoll �

6, we have

IIx - xoll �

^{6 and}

X

^{E n.}

From the equation defining

Xn+1

^{we have}

A(Xn+1 - Xn)

⁼

-AL{J(xn) - f(Xn-l) - A(xn - xn-d}

=

A(xn - xn-d - {J(Xn) - f(xn-d}

By using this equation recursively we reach finally

A(Xn+1 - Xn)

⁼

A(xi - XO) - {J(Xn) - f(Xn-l)}

- {J(xn-d - f(Xn-2)} - ... - {J(xd - f(xo)}

=

AL(y - Yo) - f(xn)

⁺

f(xo)

=

y - Yo - f(xn)

⁺

Yo

⁼

Y - f(xn)

Let

n -+

⁰⁰ in this equation to get 0 =

y - f(x).

^•