Functions of Several Variables

14 Functions of Several Variables "And thick and fast they came at last, and more, and more, and more. " Lewis Carroll 14.1 Continuity There are evidently many mathematical quantities which depend on more than one variable so we must augment our analysis to include the consideration of functions of several variables. Much of what we have established for functions of one variable remains true in the new situation, but some features change. To oversimplify, the ε - δ techniques remain much the same, the extra variables adding a certain longwindedness, but there is an underlying change arising from the fact that the geometry of η-dimensional space is rather different from that of one dimension. The first issue is to decide what we mean by limits and continuity. If A and Β are sets we denote by A B the set of all ordered pairs (JC, y) where JC e A and y e Β , that is, AyB = {(JC, y): x e A , y e β ) . A^BJZ is the set of triples (JC, y, z) with JC e A , y e Β , z e C , and so on. We denote by IR" (- R R ... R with η occurrences of IR) the set of all η-tuples (JCI, JC„) of real numbers. Where helpful, we shall abbreviate (JCI, JC„) as x. To define a limit as χ tends to a we need an idea of distance, which we generalise from that used in two- and three-dimensional geometry. The definitions of limit and continuity are then obtained from the idea that / ( x ) —> L as χ —> a if |f(x) - L\ can be made as small as desired by choosing χ 'near enough' to a. X

X

Definitions Let χ = (jc quantity ||x|| =

=

X

JC„) e IR". The norm of x, denoted by ||x||, is the

h

Χ" ||^;

X

|

2

• The distance apart of two points

be | | x - y | | , where χ - y = (x, - y , , . . . , x„ - y„).

χ

and y is said to

Sec. 14.1]

Continuity

Let a = ( a

a)

h

n

203

and suppose that / : A - » R where A is some subset of

containing | x e R" : 0 < ||x — a|| < /?} χ —» a if and only if

for some

R>0.

R"

We say / ( x ) - > L

as

Ve > 0 3 δ > 0 such that Vx e R" 0 < ||x - a|| < δ => |/(x) - L\ < ε . Suppose that A c R" and that / : A - » R . Then if a e A we say that / continuous a t a if and only if

is

ν ε > 0 3 δ > 0 such that Vx e A ||x - a|| < δ => |/(x) - / ( a ) | < ε . / is said to be continuous if it is continuous at each point of A. If a is 'interior to' A in the sense that there is a positive h for which {x: | | x - a | | < A ] cA, then continuity at a is exactly the same as the property that / ( x ) — » / ( a ) as x—»a. If, however, a e A but a is not interior to A then continuity at a does not imply that / ( x ) - ¥ / ( a ) as χ —> a , since the definition of continuity only pays heed to those points χ near a which happen to lie in A. This is akin to the definition of a continuous function on the interval [0, 1] where we make a slightly less stringent demand for continuity at the endpoints. Since the geometry of the 'edge' of a set in R" is more complicated than that in one dimension, we cannot just formulate what we wish in terms of one-sided continuity. It is easy to see that if A is an interval in R , the new definition of a continuous function from A to R coincides with our old one, though the new one also allows us to consider less regular sets than intervals. • 1

2

Example 14.1 Let / : R -> R be given by f(x,y)=xy. (We shall label the variables χ and y in place of x and x to accord with tradition.) Then / ( * , y ) - > 1 as ( * , y ) - > ( L 1). t

Solution:

2

Let ε > 0 . Let δ = min(l, ε/3). Then ||(jc,y)-(l,l)||<ô

=>

2

2

((*-1) +(ν-1) )<δ

= > | J C - 1 | < Ô and | y - l | < δ => I χ I < 2 and | y |< 2 (since δ< 1). Therefore, |(jt,y)-(l l)||<8=>|/(jc,>)-l | = | ^ - 1 1 ^ | f

(JT-%| + | 3^-1 | < 2 δ + δ ^ ε .

Since ε > 0 was arbitrary we have shown the result. From this it follows that / is continuous at ( 1 , 1 ) and a similar argument shows / is continuous everywhere. • Example 14.2 More generally,

let g, h: A —» R

be continuous, where A

is an

interval in R. Define / : A A -* R by / ( J C , y) = g(x)h(y); f is continuous. X

Solution:

To show this, let

(a,b)e

AA X

and let

ε > 0. Since g

and

continuous,

Also,

3δ| > 0 such that Vy e A |y - b\ < δι => \g(y) - g(b)\ < ε and 3δζ > 0 such that V y e A | y - o | < Ô 2 = > \h(y) - h(b)\ < ε . 3Ô3 > 0 such that V y e A [y - fo| < 83 = > \h(y) - h(b)\ < 1 .

h are

[Ch.14

Functions of Several Variables

204 Then

(x,y)-(a,

b)

< δ = ηιίη(δ,, δ , δ ) => | χ - α \ < (x,y)-(a,b) 2

<δ

3

and

I y - b I < (x, y) - (α, b) < δ , so II (jc,y)-(a,b) Il < δ

I f{x,y)-

f{a,b)

\ = \ g(x)h(y)-

g(a)h(b)

\


Jh(y)I +1


\g(a)\)

+


h(y)-h(b)\g(a)\

+ \g(a)\)

Since the factor multiplying ε is a constant and the above is true for all ε > 0, we could have substituted ε/(1 + \h(b)\

+ \g(a)\)

for ε throughout; this shows that /

is continuous at (a, b) and, since (a, b) was typical, / is continuous.

•

L e m m a 14.1 Let / g: A -> R be continuous, where A c R" and let λ e R. Then fg> V

/+£>

and

A\(x:/(x) = 0).

|/|

are all continuous, and

1//

is continuous on the set

•

We leave the proof of the Lemma as an exercise. From this and the example preceding it, we see that functions which can be defined by adding, multiplying and dividing functions of one variable are continuous except perhaps where their denominator is zero. Thus, for example, (x\X + JC )/(JC| + x% + x^) is a continuous function of (X],x , xi) except possibly at ( 0 , 0 , 0 ) . 2

2

3

2

Suppose that / : R" —> R, so that f(x , ...,x ) is a function of the η independent variables X\,...,x . By fixing some of these variables we can obtain a related function dependent on fewer variables, which raises the question of how we relate the continuity of the various functions so obtained. Suppose a , a e IR and g\-. R —> IR is defined by g\{x) =f(x, a ,a„). Then if / is continuous at (a\,a ), g\ is continuous at a\. To see this, let ε > 0 ; t

n

n

2

n

2

n

3 δ > 0 such that ||x - a|| < δ => |/(x) -f(a)\

<ε )

whence | χ - α, | < δ => \\(x, α ,..., α„) - (α,,..., α„)|| < δ => | g,(x) - g (a ) 2

x

x

\ < ε.

Similarly g given by g {x)=f(a ,...,a - ,x,a ,...,a ), is continuous at a,. This result we can paraphrase as saying that if / is continuous in the η variables JCI, ....,x„ jointly, then / is continuous in each variable separately (i.e. fixing all but one of the variables). The converse result is false. h

l

l

2

i l

M

n

2

2

Example 14.3 D e f i n e / : IR - » IR by f(x, y) = 2xy/(x + y ) (for (JC, y) * (0, 0)) and / ( 0 , 0) = 0. / is continuous at all points other than (0, 0), since at all such points, JC + y Φ 0. Now for χ Φ 0 2

2

/ ( J C , χ) - / ( 0 , 0) = 1 while f(x, -JC) - / ( 0 , 0) = - 1 , so, whatever δ > 0 we choose, there are points (JC, y) with ||(JC, y)|| < δ for which |/(jc, y ) - / ( 0 , 0) I = 1 so / is not continuous at (0,0), nor could a different choice of / ( 0 , 0) produce a continuous function. The result here is more easily visualised by changing to polar co-ordinates and letting jc = r c o s 0 , y = r s i n 0 so /(JC, y) —/(0, 0) = 2 cosOsinO. This does not tend to 0 as r = ||(JC, y)|| —> 0 (unless cosO sinO = 0).

Sec. 14.1]

Continuity

205

However, f(x, 0) = 0 and / ( 0 , y) = 0 so the functions obtained from / by fixing one of the variables are continuous. We conclude that 'joint' continuity with respect to several variables is stronger than continuity with respect to each variable separately. This difference can be significant and occasionally awkward. The reader may take heart from the knowledge that, in the development of analysis, even such an eminent mathematician as Cauchy initially failed to notice this! • We shall prove an analogue of Theorem 10.5 which shows that if / : [a, b] -> R is continuous, it is bounded. For this we need to find a suitable type of set to form the domain of our function. m

Definitions Let ( x j be a sequence of points of R . We say that x„ —» χ as η -> °° if and only if V ε > 0 3 Ν such that Vn > /V ||x„ - x | | < ε . Let A c R . We say A is closed if, whenever (x„) is a sequence of elements of A and x„ —» χ as η —> °° , then χ e A. • m

m

1

Example 14.4 R itself is closed. Also [a, b] is a closed subset of R (since if V n , x„e [a, b] and x —> JC as η —» °°, then χ e [a, b]). The set (0, 1 ) is not closed since Vn e Ν 1/n e (0, 1) and 1/n - » 0 but 0 e (0, 1). Ο n

Lemma 14.2 Let A

be a subset of

m

R

and / : A —>R

be continuous.

If

Vn e Ν x„ e A and x„ - » a e Λ as η —» °° then / ( x „ ) - > / ( a ) as η —» <». • This result, whose proof is a direct adaptation of that for sequences and functions defined in R, leads us to a useful technique for spotting closed sets. m

Let f.R ^>R be continuous and A = { x : / ( x ) e [a, A]} = / " ' ( [ a . fc]); A is closed. Notice that / has the whole of R as its domain. The proof is easy: let x e I " and suppose that V n e Ν x„ e A and x„ —» χ . Then / ( x „ ) —»/(x) as η —> °° so, since α < / ( x „ ) for all n, a
l

The same argument shows that f~ ([a, °°)) a n d / ~ ' ( ( - o o , a]) are closed, and a slight change shows that the intersection of two closed sets, or of any collection of closed sets, is closed. Thus we can see easily that, for example, the 'disc' {xe R :||x||
2

2

m

m

L e m m a 14.3 Let / : R -» R be continuous. The sets / " ' ( [ a , b]) ,f'\[a, °°)) and / " ' ( ( - o o , a]) are closed, and the intersection of a collection of closed sets is closed. • A useful intuitive view of a closed set is one which includes the points 'at its edges'; {x e 0Γ:||χ|| < 1} is not closed but ( x e R :||x|| < 1} is closed. m

To proceed further, we need a short discussion on sequences.

[Ch. 14

Functions of Several Variables

206

Definition Let (a„) be a sequence of real numbers. A subsequence of (a„) is a sequence consisting of some of the terms of (a„) in their original order, that is, a sequence of the form (a , a , a ,...) 2

4

{a Yk=\

where

nk

V h Ν n >n M

and n e N. Thus

k

k

is a subsequence of (a„). Notice that n, > 1, and n > ri\ so n > 2

b

2

2

and, in general, n* > k. • Lemma 14.4 Every sequence of real numbers has either an increasing subsequence or a decreasing one. Proof: Let (a„) be a sequence of real numbers. We first try to find an increasing subsequence. satisfying

If we just proceed naively, choosing

a„ > a 2

H{

an,

and seeking

n > ri\ 2

, we shall become unable to proceed if any of our suffices rij

has the property that

Vn > η j

a < a ,. n

This suggests that we look at the set of

n

such points, so let S = [n e M: Vm > n, a < a„), the set of suffices such that the corresponding term is greater than all subsequent terms. m

If S is finite (including the case S = 0 ) then choose n, to be greater than all the elements of 5. Thus «ι « S. By definition, then, 3n >n such that a „ > a . 2

Since n > n \ , n £ S and 3 n > n 2

2

3

2

[

2

0 |

such that a ^ > α„ . Proceeding in this way n

Ί

we obtain an increasing subsequence ( a „ ) of (a ) . t

n

If 5 is infinite let n e 5. For n e S let n \ be chosen to belong to S and satisfy n > n . This is possible for all k since S is infinite. This gives a subsequence (a„ ) with the property that V i e Ν a,i < « n , (since n e 5), a t

M

k

k+

k

k

k+l

k

strictly decreasing sequence. • Lemma 14.5 A subsequence of a convergent sequence converges to the same limit as the whole sequence. Proof: Let x —>x as η —> °°, where (x„) is a sequence of real numbers. Let (xn ) be a subsequence of (x„), so V& e Ν η* > n

k

Let ε > 0. 3N such that Vn>N \ x„ - χ \ < ε, whence, for this N, k> Ν => n^ > f c > / V = ^ | j c - χ | < ε . njt

Thus jr,i —> x as /: —> °° . t

m

This result is equally true of a sequence in R . • Definition A sequence (x„) in W" is said to be bounded if (||x„||) is a bounded sequence in RL A subset A c l * is said to be bounded if [||x||: x e A] is a bounded subset of R.

Sec. 14.1]

Continuity

207

Theorem 14.6 Bolzano-Weierstrass Theorem. has a convergent subsequence.

m

Every bounded sequence in

R

Proof: We first take the simpler case of a sequence of real numbers. Let (a ) be a bounded sequence of real numbers. Then there is a subsequence (an ) which is n

k

either increasing or decreasing. Since (a„) is bounded, so is {an ) , hence (an ) , k

k

being bounded and either increasing or decreasing, is convergent. m

Now let (x„) be a bounded sequence in R ; suppose that Vn e Ν ||χ || < K. For π

each n, let x„ = ( j | " ' , . . . , ^ ) . The sequence (x\"^)

(as η varies) is a sequence

Vn, | J C [ " ' |< || x„ || ^ Κ

it is a bounded sequence.

of real numbers, and since

(ΛΓ"'^')

Choose a subsequence ]

paragraph. Now | x" ^

which is convergent, which is possible by the last

\< || x

II ^ Κ whence ( x " ' ^ ' )

is a bounded sequence

2

2

of real numbers, so it has a convergent subsequence (x^ ^).

Since (JC" **') is a

subsequence of the convergent sequence ( J C " ^ ) , it also converges. Proceeding in 1

this way we obtain after m steps a sequence j=\,...,m

each

m

sequence

(x" ^)

(n (£))£L]

of integers such that for

m

converges, to

Xj

say.

Since

Il x " ™ ' * ' - χ II —» 0 as k —» °°, where χ = (x\,..., x„), the result is proved. • Theorem 14.7 Uniform Continuity. Let A be a closed, bounded subset of

R

m

and / : A —» R be continuous. Then / is uniformly continuous, that is, Ve > 0 3 δ > 0 such that Vx, y e A ||x - y|| < δ => | / ( x ) - / ( y ) | < ε. Proof

Suppose the result were false, so that 3ε > 0 such that Vô > 0 3x, y e A

suchthat | | x - y | | < 6 and |f(x) - / ( y ) | > ε. Let ε > 0 have the stated property. For each

ne Ν

\f(*n) ~fyn)\ let

\n

xn

=(x"

l/n>0,

3 x „ , y„e A

such that

as

k — S i n c e a

, . · · , Xm ) rtjk

"d

-y«

A

yn = ( k

II < —

A y

y

is closed, %)

|| < 1 In

- » x as &—>«>.

However, /

x e A.

and

Also, letting

we have, for each

/' and

i

Ii - 4 o o

is continuous at

x,

so 3 δ > 0 (

such that

H y - x H < δ, => II/·(y) - / ( x ) | | < ε/2. Choose k so large that || x „ , - χ || < δ , Il y "* - χ | | < δ ι , I/(*«.

then

)-f(yn )\<\f(xn )~ k

k

k.

so that y"* —» x = lim jcf* and it follows that

ft h

k

n

n

I <|| x yn

|| x „ - y

^ ε. Since A is bounded, (x„) has a convergent subsequence, ( x , ) ;

—>x

k

k

k

since

/ ( χ ) I +1 / ( x ) - / ( y

This is a contradiction, since by definition | f(xn

k

) I < ε / 2 + ε / 2.

) - / ( y n ) | ^ ε. • k

and

208

Functions of Several Variables

[Ch. 14

In common with the corresponding result for functions of one variable, Theorem 13.6, the significant result here is that the δ above can be chosen to be independent of χ and y. Let A = [a, b] χ [c, d] and / : A —> IR be continuous. A is closed and bounded; to see the closure, notice that DS. —> IR. given by Pi(x\, x ) - Xi is continuous and A = p ~~\[a, fc])n p{~\[c, d]) is closed by Lemma 14.3. Therefore the Theorem applies and / is uniformly continuous. Let ε > 0 . Then we can choose δ > 0 such that Vx, ye A ||x - y|| < δ => |f(x) - / ( y ) | < ε. In particular, if \x - JC | < δ/2 and |y, - ν | < δ/2 so that ||(JC,, yi) - (x , y )|| < δ, then |/"(xi, y\) —f{xz, yi}\ < ε, so that we can use the same value of δ for all y e [c, d] to ensure that \f(x\, y) -f(xi, y)| < ε. One use of this is: 2

2

t

x

2

2

2

2

T h e o r e m 14.8 Let / : [a, b] [c, d] —> IR be continuous and define F: fa, b] —» IR by x

F ( x ) = j ^ / ( x , y ) d y . Then F is continuous. Proof: Notice first that, for each x e [a, b], y t-¥ f(x, y) integrable, so the expression for F is meaningful.

is continuous, hence

Let ε > 0. By the uniform continuity of / on fa, b] [c, d], 3 δ > 0 such that Vx, x'e [a, b] [c, d] ||x - x'|| < δ => | / ( x ) - / ( χ ' ) | < d(d - c). Therefore, let x,x'e [a,b] with p : - x j < ô . Then Vye[c,d] \f (x, y) -f (x', y)\ < tl(d - c), so x

x

that

\F(x)-F(x')\
Since ε > 0

was arbitrary we

have shown that νε > 0 3 δ > 0 such that Vx, x'e [a, b] I χ - x'l < δ => I F(x) - F(x') I < ε so F is (uniformly) continuous. • Obviously, the analogous result, obtaining a function of y by integrating with respect to x, is true. More generally, if / is a continuous function of the m variables x x in | a , b]] ... [a , b ], the function obtained by integrating with respect to one of these is continuous with respect to the remaining variables. u

m

t

x

x

m

m

That a continuous function on a closed, bounded set is bounded is now easy: 1

Theorem 14.9 Let Λ be a closed, bounded subset of I R " and / : A - » IR be continuous. Then / is a bounded function and / attains its supremum and infimum at points of A . Proof By Theorem 14.7, / is uniformly continuous, so 3 δ > 0 Vx, ye A , ||x - y|| < δ => | / ( x ) - / ( y ) | < 1 => | / ( x ) | < | / ( y ) | + 1 .

such that

Since A is bounded, there is an M for which χ e A => ||x|| < M so χ = ( x i , x ) e A => I x,;| < M i = 1 , 2 , m . Thus A is a subset of the 'cube' [- M, M] [- M, M] ... [- M, M). Let a = - Μ < α , < ...
X

X

t

X

0

S

—

m

(See Fig. 14.1.)

r

Sec. 14.1]

Continuity

209

M

M

—

'

-Ma\a ...

a _\M

2

n

Fig 14.1 Choose

z, = ( z [ ' \ . . . , 2 ^ ' ) e Λ,

I zf

|< δ / m

-xj

VxeA,

for each

/.

If

χ = (jc ...,jc )e A 1(

m

t

then

for each j , ||x - z || < δ and so |/(x)| < |/(z, )| + 1 . Thus ;

|/(x)|
since χ must belong to some A,. It follows

that / is bounded. To show the supremum is attained, let a = sup(/(x): χ e A). Then if a is not attained and we set g(x) = l/(a - / ( x ) ) g is continuous on A and thus bounded. This gives a contradiction in exactly the same way as in the case of a function of one variable. • 14.2 Differentiation As with continuity, we may consider the differentiation of a function of several variables either by fixing all but one of them and differentiating the resulting function of one variable or by considering a differentiation process where all the variables are treated simultaneously. As with continuity again, these two processes differ. m

Definitions Suppose that A is a subset of R , /: A -> R and that a e A. If the function g, defined by g,{x) =f(a\,..., α,-ι, JC, a , a ) is differentiable at a M

then we say the partial derivative

m

h

exists at a = (a\,a ) m

and that its value

is g'/(a,). / is said to be differentiable at a if for some h > 0 A contains the set ( x e R : | | x - a | | < h} and, for some constants βι,..., β„ , n

m

/ ( χ ) = / ( 3 ) + Σ β · ( χ , · - α , ) + θ(χ) 1

ι=1

where θ(χ)/ x - a

—»0as

x->a.

Partial derivatives are the type of derivative we have met before, of functions of one variable, and we can apply our existing theorems to them. Notice that there is no need for all partial derivatives to exist: if / ( J C x ) = X\ + \x \ then df/dx\ exists at (0, 0) but 3//3JC does not. 1;

2

2

2

[Ch. 14

Functions of Several Variables

210

Differentiability of / as a whole is about the approximation of f(x) - / ( a ) by a linear function of χ - a , the error θ being 'small' in the sense stated. This is often expressed in terms of 'differentials' as d / = Σ ) " | β/(<1ϊ,) where, as we shall see, β, turns out to equal df/dx

at a. The more precise statement is the one we have made

about a p p r o x i m a t i n g / ( x ) - / ( a ) by the linear function

β

i(x,aï).

Example 14.5 Let f (x\, x ) = X \ x . It is easy to see that at the point df/dx]=a and df/dx = a], so / possesses partial derivatives. Also 2

2

2

2

A

fixi, Xi) ~A \> 2) = X\Xi - a\a = (x, - α,)α + (xj - a )a\ + d(x a

2

where

(a\,a ),

2

-(x\ -a )(x -a ).

Q(X\,x )

x

2

2

2

|2«v| < u + v

so

2

2

2

2|(;q - a )(x

that

2

x)

u

2

At this point we recall that if t

2

|0Ui, x )\l\\x - a|| < 1 ((JC, - a , ) + (JC - a f) 2

2

- a )\ < (x\ - a\) + (x - a )

2

2

2

u, v e Κ

2

2

0(JC,, JC )/||X - a|| - * 0

whence

2

and

2

2

χ —> a. / is differentiable at (a\, a ) and the constants βι, β equal (dfldx^){a and (3//dx )(ci], a ) respectively. 2

2

2

u

as a) 2

2

In practice, writing X\ - a = r cosO, x - a - r sinO is illuminating in the above calculation. • t

2

2

Lemma 14.10 Let A be a subset of W and (x: ||x - a|| < h) c A, where h>0. If / : A —> \Ά is differentiable at a then all the partial derivatives at a exist and (a ,a ) t

= β, (/'= 1,2,..., m)

m

where

β,

is the number occurring in the

definition of differentiability. Conversely, if all the partial derivatives of / continuous at a, then / is differentiable at a.

exist at all points of A, and are

Proof: Suppose that / is differentiable at a and that / ( Χ ) = / ( 8 ) + Σ * β , · ( Λ · - ΰ · ) + θ(Χ) 1

Ι

1

where 0(x)/ x - a - > 0 as x - » a . Then let χ = (a ,..., α,·_ι, t, a g M =f(a , α,_ι, t, a a ). Since ||x - a|| = ||f - α,·|| we have t

i+u

M

, a

m

)

and

m

*,-(0 = */(α ·) + β,-(/-α,·) + θ ·(0 ι

ι

where θ,(/) = θ(χ) and θ,(ί)/|ί - a\ - » 0 as t -> a,. From this we see that g, is differentiable at a,, and that β, = g\(aï) =

(α,,..., a ). m

Now suppose that all the partial derivatives exist, and that they are continuous at a. To deduce information about / ( x ) - / ( a ) wc need to re-express this in a form where in each part only one variable changes, so if ||x - a|| < h m

f(x ,...,x )-f(a ,...,a ) i

m

l

m

=

Y,(f(x\,...,x ,a ,...,a )-f(x ,...,x _ a ...,a j) i

i+l

m

]

i

u

h

m

i=l

= Σ™ 1 0 * 1 ' - ·

a

' ξ/' i+\ - - ' «m)(*i -

(Ο

Sec. 14.2]

Differentiation

211

obtained by applying the one-variable Mean Value Theorem to the function n - » / ( j c J C , _ I , r, a

i +

x ...,Xi).

this

l t

In

u

/(x)=/(a)+ θ

m

t

notation,

χ

< *l ' - '

ε > 0.

have,

for

||x-a||<«,

A

' M)- £

α

α

( 1. - .

Then, by continuity of

1

a m ))(*/

at

a

1

i

χ

I ( ) I ^ Σ JJT(*1 · ·•·'

dXj i + 1

-

)

a, 3 δ > 0

| | L (y) - | L ( ) | < — . Choose

dï, with ξ, as in (1), | | ( x , . . . , x , _ ^ , a

that

where

' ;+1 ' -

'

that H y —a || < δ => for all i,

θ

we

e(x)

ZM^-(»K*i-fli) +

( ) =Σ

Let

i , a ) ; ξ, will be between a, and x (and will depend on

such

|| x - a [| < δ , so,

m

, . . . , a ) - ( a , . . . , a ) | | < | | x - a | | < 5 and m

· ξι ·

1

m

U i )- 0 / - - - m ) - J[- («1. — «M a

I

i=l

M <—ΣΙ·*ί ϊ Ι^ ΙΙ - ΙΙ _ί,

ε

χ

2

(

s i n c e

U i - a - , I <|| x - a I).

m

i=i

So 0 < ||x - a|| < δ => |θ(χ)|/||χ - a|| < ε , and, since θ(χ)/||χ - a|| -> 0 as χ -> a. •

ε>0

was arbitrary,

The condition for differentiability may be put in a version which is co-ordinate-free χ

Σφ=\^ί( ί

by observing that

a

~ i)

is the form of a typical linear function from R

m

to IR evaluated at x - a ; we shall not pursue this here. Notice that differentiability is a stronger condition than the existence of the partial derivatives.

It is

differentiability that we require in order to establish the chain rule for functions of several variables, but we have a few items to tidy up first. Example 14.6 Let f(x, y) = xyN(x + y ) ((JC, y) Φ (0, 0)) and / ( 0 , 0) = 0. Then (3//3JC)(0, 0) = lim (/(JC, 0) - / ( 0 , 0))/JC = 0 and (df/dy)(0, 0) = 0. / is not, 2

2

however, differentiable at (0,0), since /(x,y)-/(0,0)-P,Jc-P y = xy/(jc +y ) 2

2

2

where

β, = (3//3JC)(0, 0) and

β = (3/73ν)(0, 0) 2

and

2

Jcy/V(jc + y ) -A> 0 2

as

•

(jc,y)->(0,0).

The existence of partial derivatives of / exists at a point

a,

does not imply that /

this means that, as a function of

is continuous. If X\ only, f

is

differentiable and so / is continuous with respect to the variable X\. Therefore, the existence of all the partial derivatives guarantees only that / is continuous with respect to each variable separately. Consideration of the function defined by / ( J C , y) = jçy/(jc + y ) for (JC, y) Φ ( 0 , 0 ) , / ( 0 , 0) = 0, gives a function which is not continuous at (0, 0) but (of /dx) and (of/dy) exist at all points. 2

2

212 If /

[Ch. 14

Functions of Several Variables is differentiable at

a,

then

differentiability, / ( x ) = / ( a ) +

/

is continuous at

a.

By definition of

β, (JC, - α,) + θ(χ) where θ(χ)/||χ - a|| -> 0 as

χ —» a. Since for each /', \x, - a , \ < ||x - a||,

β, (χ, - α,) —> 0 as χ —> a, and

also θ(χ) = (θ(χ)/||χ - a||)||x — a|| —> 0 so / is continuous at a. Theorem 14.11 The Chain Rule. Suppose that b e f i c T and that f-.B-^M. is differentiable at b. Let g \ , g : A —> R be m functions, differentiable at a e / l c K " and with g,(a) = ft, (/' = 1, 2 , m ) . Then the function h: A -> R where h(x\,...,x )=f(g\(x\,...,x ),...,g (x\,...,x )) is differentiable at a and m

n

n

m

n

(where g(a) = ( g i ( a ) , g ( a ) ) ) . Here / is regarded as a function of ( j , , y ) . m

m

Proof: By the assumption of differentiability, f(y)-f and

g (x) = g (a) + Y aij(Xj-aj)+^i(x) i

i

(b) + £™ p,-(_y - ft,) + 9(y) (

f

where G(y)/||y - b|| -> 0 as y - ) b

i

and φ,(χ)/||χ - a|| —> 0 as χ —» a ((' = I,.... m). Then A(x) = / ( g ( x ) ) = / ( g ( a ) ) + Σ™, β / (*/ (x) - * / (a)) + 6(g(x)) = / ( 8 ( a ) ) + ΣΓ=ι β«

(Σ"=,

«(/

= /(8(3)) +Σ"= Υ;(^-";)

" « y )) +

ι

where

γ, =

+ Σ™ ι β. Φ/ ( > + χ

θ

χ

<8( )

χ

Ψ( ) - 0)

β,α^· and ψ(χ) = ]Γ™,β,φ, (χ) + 0(g(x)).

We have to show

ψ(χ)/||χ - a|| —> 0 as χ —> a. From what we already know of φ,·, we need only prove 0(g(x))/||x - a||-> 0. Let M = (1 + max Yf. . \α-Λ ) /m~. 3δ, > 0 such that A

i

'

0 < ||y - b|| < δ, => |0(>O|/||y - b|| < ε/Μ. since e(y)/||y - b|| -> 0 , so jjy-bjj<δ, = > | 0 ( y ) | < | | y - b | | / / W . £

(2)

Since φ,(χ)/ x - a - » 0 as x—» a, 3 δ > 0 such that 0 < ||x-a|| < δ => | φ,(χ) | /1| x - a || < I (/' = 1,2 m) 2

2

=> | g ( x ) - ( a ) | < Σ"=,Ι «// Ill xj - α |+ |φ,·(χ)| /

g /

}

< 0 + t l « , y 1)11 x - a | | =>||g(x)-g(a)|| ||g(x) - g(a)| < δ, by (3)

(3)

2

=> |θ( (χ))| < ε ||g(x) - g(a)|| / Μ < ε |x - a | by (2) and (.3). 8

Therefore, 6(g(x))/||x - a||-> 0 as χ - » a so h is differentiable, by (1).

Sec. 14.2]

Differentiation

213

Finally, we see that y-,, which is J^-(a) , equals Σ ^ β , α , · , , that is,

*_

(

a

)

dxj

=

£i

(

i=\

8

(

a

)

A

tyi

(

a

)

.

D

dx j

The Chain Rule for functions of several variables is easiest remembered in the form where / is a function of the variables y \ , y and each of y i , y is a function of JCI, x giving, in the standard form, m

m

n

dxj

^ /

the rate of change of / with Xj being the sum of the rate of change of / with each co-ordinate times the rate of change of that co-ordinate with JC,. We have not chosen to state the theorem in this way because the symbol / is used here with two distinct meanings, and the same is true of y,-. This is clearer if we take an example. Let fiy y ) = yiy and let g,(x,, x ) = JC, + J C , g (x , x ) = x - x . Then let A ( J C , , x ) =figi(x\, J C ) , g2(x\y X2)) - X\ - xi • The First thing to notice is that A is not the same function as / (e.g. A ( l , 2) 2)). The confusion usually arises in applications where the value of / represents some physical quantity and A u

2

2

2

2

2

x

2

x

2

2

2

2

represents the same quantity as a function of different variables, so that by ' may really mean

' we

dh .

The second issue with the traditional notations is that the y

x

in

J^-

is spurious;

-—- is a function of two variables, being that function obtained from / by fixing the second variable and differentiating with respect to the first. Since these variables are 'dummy' variables the fact that we have to attach a name to them is unfortunate. above is the function whose value at (a, b) is b. The Newtonian notation of for

f

y

is no better in this respect, while the various attempts to introduce a more

rational notation have not penetrated far into practice so we shall just have to make the best of the situation. The principal analytical advice is to be quite clear which functions are distinct and, where possible, to use different symbols to distinguish them. Example 14.7

Let

/

be a function of the two variables

F(x, y) = / ( y , JC). Show that

(a, b)=^

( A , a).

(x, y)

and define

(This is, of course, deliberately

awkward, although we do need to be able to deal with functions which are symmetric in JC and y . ) Solution:

Let g,(jc, y) = y and g (x, y)=x 2

so that F(JC, y) =fyg\(x, y), g (jc, y)) 2

[Ch. 14

Functions of Several Variables

214

and ^ ( ^ ) = f ( g ( « ^ ) ) ^ ( ^ )

^ ( g ( ^ ) ) ^ ( ^ )

+

|-(£,α)0 + | ^ , ) 1 = | - ( Α α ) . α

dx

ay

(If it helps, you can write

ay

in place of | £ and

in place of | £ .) •

Armed with the Chain Rule, it is now easy to apply earlier results. Theorem 14.12. Mean Value Theorem. Let / : A -» R be continuous, a, b e A c R and suppose that / is differentiable at all points on the straight line joining a to b. Then, for some θ e (0, 1 ) m

/(b) - /(a) =

J- (a + 0(b - a))(fti -a,)

.

Proof: Define F(t) = / ( a + f(b - a)) for te [0, 1]. F is continuous on [0,1] and differentiable on (0, 1 ) and Rule. By the ordinary F(l) - F(0) = F ' ( 0 ) . • m

Definition Let A <= R partial derivatives

differentiable. We define

F\t) = Mean

| £ - (a + f (b -a))(A - - α,· )

Value

Theorem

and f: A —>R. If / is a function from ζ

lobe

djc jdXj

by the Chain

(

that

is differentiable then each of the

A

to

-^-(4^·), oXj

3Θ e (0, 1) such

R

and it may in turn be

the partial derivative of

oXj

with respect to the y'th variable, when this exists. differentiable if / and each of its partial derivatives

-1^oXj

F

is said to be twice is differentiable. The

analogous definitions apply to higher order derivatives. 3 f d f The 'mixed' partial derivatives -, and -, i are not necessarily equal, since 2

2

Λ

ox\OX2

J

ax ox 2

n

l

the two limiting processes involved are taken in the opposite order in the two cases. Under very mild additional assumptions, however, they are equal, as we shall show shortly. 14.3 Results Involving Interchange of Limits We shall devote a little effort to several results, all of which are essentially about reversing the order of limiting processes. Theorem 14.13 Differentiation f. [a, b] [c, d] —> R , and that x

under

the

Integral

Sign.

Suppose

that

exists and is continuous (with respect to both

Sec. 14.3]

Results Involving Interchange of Limits

215

variables jointly).

Then, with

and

F(x) = |f f(x,t)dt,

F

is differentiable

d

F'(x) =

i ^(x,t)d . c

t

Proof: We have to show that the derivative of the integral and the integral of the derivative are equal. Fix x e (a, b)\ if JC is an endpoint, the usual minor changes are needed. Then, applying the Mean Value Theorem to x h > / ( J C , t) we see that, for each r, there is some ξ, between χ and x satisfying Q

0

0

f(x,t)-f(x ,t)

= (x-x )-£-(t„,t)

Q

and therefore

0

F(x) - F(x

0

)= Γ

(JC - x ) £ ( ξ , , 0 dt 0

Let ε > 0. Then, by the uniform continuity of that

3/

on [a, b]x[c, d] , 3δ > 0 such

||(jc,y)-(x',/)ll<6^>|f (Jc,y)-|(x',y')|
Let

0 < | J C - J C | < δ . Then 0

^)-^o>-f^(jc ,,)d,

| ( ξ , . 0 - | ( , . 0 di < ε ,

0

J C °

X-XQ

t, ξ, is between JC and x,

since for each whence

0

Corollary With )

=

< χ - JC

0

< δ,

•

c

*

0

1

a

(

(ξ,, ί) - (JC , i)

so

V( e [c, rf] | -dx ^ - ( ξ , , r) - | ^ ( * o , 01 < ε/(α* - c) . This establishes that

F'(x) = \ ^(x,t)dt.

F

0

X

C?

/

U

i

/ )

d

as above, if

n

h

e

φ, ψ: [a, b] —» [c, d]

are differentiable and

n

*"(*) =

^-U,i)dr + /(x, (x)) '(x)-/(x^(xWW. V

V

^ Φ ( Λ ) OJC

Prao/: Let ^

G(JC, y, ζ) =

= -/(x, y), and ^

/(jc,/)df

. Then

exists by the Theorem, while

=/(x, z) by the Fundamental Theorem of Calculus. AH three

partial derivatives are continuous, F(JC) = G(JC, φ(χ), ψ(χ)). •

so

we

may

apply

the

Theorem 14.14 Let / : [a, b] [c, d] —> R be continuous. Then x

^/(x,y)dy)dx = j;(i>,y)dx)dy.

Chain

Rule

to

[Ch. 14

Functions of Several Variables

216

z

Proof: Define g(x, z) = | / ( x , y ) d y . Then ^--f, G(z)

-

^g(x,z)dx

,

G'{z) = f f{x,z)a\x.

so,

by

which is continuous. Now let

differentiation

under

the

integral

sign,

Since G ' is continuous (Theorem 14.8),

u

tfe/(*.z)dx}k

= tG'=

G(d)-G(c)

=G(d) = £ | f f(x,y)dy\ix.

•

The result about inverting the order of integration remains true under more general circumstances, but it may fail for improper integrals. To see this, consider inverting the order of integration in the two improper integrals Jf (Jf (x-y)(x+ or i (i (x-y)(x 0

3

+

0

3

y)~ dx)ày

yy dx)dy.

Theorem 14.15 Suppose that / is defined in the neighbourhood of the point (a, b) in R , that is, on the set {(JC, y)e R : ||(JC, y) - (a, b)\\ < h) for some h > 0, that 2

2

^f- and M- are continuous on this set and that one of ^-4- and -|^r- exists and ox ay dray dydx ^2

^2

ι·

r

is (jointly) continuous on this set. Then both -g^-- and

exist and are equal

on this set. 2

3 f Proof

Suppose that

is continuous in the neighbourhood of (a, b). Regarding

y as fixed, we see, by differentiating under the integral sign, that

2

d f a n a

n e

(This uses the continuity of g^jjy " ' Fundamental Theorem of Calculus.) Since this is true of all y sufficiently close to b, and since, for fixed JC, the lefthand side is differentiable with respect to y, the right-hand side is differentiable with respect to y and 3* f d *d ' f — J-(x,t)dt=-^-(x,y)-0. • U y)=dxdy ' ' dy J dxdy ^' ' dydx 2

r y

a/

v

2

2

a

J

b

There are several possible minor variations on Theorem 14.15, which may easily be made to apply to functions of more than two variables, and to higher order partial derivatives. For most purposes it is sufficient to know that if the two mixed derivatives are (jointly) continuous they are equal. m

Theorem 14.16 Taylor's Theorem. Suppose that a e R and that f.A—>R where A = {x e R : ||x - a|| < δ} for some δ > 0 . Then if / possesses all partial derivatives of orders up to and including η and if all these partial derivatives are continuous, we have, for \h\ < δ , m

Sec. 14.3]

Results Involving Interchange of Limits

217

3

/ < . + » . / < . Η | * , | ^ ( . Η - + Σ & , ^ - ^ A, "^ '

'Ί

«+«.
where, for some θ e (0, 1 ), A. κ

"=ς:,

-A,

n

f

„•

;

}

.

x

i

•

:

Proof. Because all the various partial derivatives are continuous, all up to order η - 1 are differentiable as functions of m variables, so the Chain Rule, applied η times, shows that if F ( f ) = / ( a + fh) ( 0 < î < l ) t h e n F is η-times differentiable. The Chain Rule shows that F\t) = Y The rest is messy but routine.

m

m

,h:^-,

F"(t) = Y .

A-A,

, etc.

• 2

Example 14.8 Suppose that / : R —» R is differentiable. If / has a local maximum at

(a, A) then the functions

x-a

and y-b

x\->f(x,

b) and y\-*f{a,

respectively so

and

y) have local maxima at

are both zero at (a, b). This, of

course, is true if / has a local minimum, so a necessary condition for a local maximum or minimum is that both (or all if there are more than two variables) the first order partial derivatives should be zero at that point. Now suppose

| ^ ( a > b) = ^(a,

b) = 0 .

To find sufficient conditions for a

maximum, minimum etc., we presume that / has partial derivatives of order two and that these are continuous. Then, for ||h|| small enough, /(

(

a

2

+ A,,A + A ) = / ( a , A ) + i A, 2

^U-

(χ) + 2A, A

2

|if

(x) + A

2

where

2

2

2

9JC bay for some point χ = (a + ΘΛ], b + ΘΛ ) with 0 < θ < 1. sufficiently small, the term on the right will be close to fa,

2

(x) dy ) If we choose

2

h

2

b)+ 1(ΓχΑ, + 2βΛ Α + γΑ ) 1

2

2

α = - ^ - , β = ^ ^ - a n d γ = -^-£-, all evaluated at (a,b), 2

so we investigate

2

expressions of the form ccA) + 2βΑ|Α + γΑ . Such expressions are called quadratic forms and their algebraic theory is elegant and complete; we shall merely poach some fragments from it. Notice that ocAi + 2βΑ|Α + γΑ = α(Α, + (β/α)Α ) + ( ( α γ - β )/α)Α so that (i) if α γ - β > 0 and α > 0 then ah + 2βΑ,Α + γΛ > 0 unless Α| = Α = 0 , whence fa, b) + (ah + 2βΛ|Α + γΑ )/2 has a minimum at A = A = 0 . (ii) if α γ - β > 0 and α < 0 then αΑ, + 2βΑιΑ + γΛ < 0 unless h\ = Λ = 0 so fa, b) + (ah + 2βΛ]Λ + γΑ )/2 has a maximum at h\ — A = 0. 2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

(

2

2

2

2

2

2

2

2

2

2

2

2

2

2

[Ch. 14

Functions of Several Variables

218 2

(iii)

2

2

if α γ - β < 0 then f(a, b) + (α/ι, + 2 β Μ + γ/ι )/2 has neither a maximum nor a minimum. If α * 0 this may be seen by considering the cases /i ) = (λ, 0) and ( - λ β / α , λ), while re-expressing by taking a y(h + (β/γ)«ι) term shows the same situation if γ * 0 . The case α = γ = 0, β * 0 is easy. 2

2

2

2

2

2

Since α γ - β > 0 implies α * 0 , we have considered all cases where α γ - β * 0 . In case (iii), / is said to have a saddle point. 2

These are the cases we can deal with, since the expression for f(a + h\,b + h ) 2

2

2

d /'

involves

d f

—{- etc. evaluated not at (a, b) but at (a + ΘΛι, b + θ/ι->). If —i- is 2

dx

continuous

V

and α = — y (a,

b) * 0 ,

then

I

for

^

||h||

^

small

2

enough,

2

dx

^-4r-(a + Qh b + Qh ) u

2

will have the same sign as α ; similarly, if α γ - β * 0

2

dx

(•^r)(-^-4-)-(4-T-) 2

dx

2

v

dy

2

will have constant sign near (a, b).

Since our arguments

« ' 2

2

about the sign of ah + 2$h\h

2

+ yh\ depend only on the sign of α and α γ - β ,

the conclusions about the sign of f(a + h\,b + h ) -f (a, b) are valid for small ||h||. 2

For

the remaining

ah\ + 2 β « | Λ +Y«2 2

values of h\, h

2

cases,

αγ-β

2

= 0,

and we see, for example,

that

s

' always non-negative if α > 0 , but that there are non-zero

for which the expression is zero. For these values of h\,h , 2

if we

substitute the appropriate derivatives evaluated at (a + θ/ι,, a + θ/ι ) we cannot be 2

sure of the sign of f (a + h b + h ) -f (a, b). [y

2

The results for a function of three or more variables are similar but a little more complicated. • At this point the techniques we have used are becoming somewhat strained, largely because in handling functions of several variables there is a degree of technicality and complication which can obscure matters. This is particularly true if one tries to consider whether a function whose domain is a subset of R ' and whose image is in IR has an inverse function, where we need to treat simultaneously the two realvalued functions corresponding to the first and second co-ordinates of the image. This sort of issue is best tackled using a rather more abstract and sophisticated viewpoint which eliminates some of the technicalities, so we shall not pursue it here. A good source is Apostol (1974). 2

14.4 Solving Differential Equations Ordinary differential equations (that is, differential equations involving only one independent variable) are important in many aspects of mathematical modelling, from biology to economics and physics. They are also quite difficult to solve! An elementary treatment of the subject introduces various ingenious techniques for

Sec. 14.4]

Solving Differential Equations

219

finding explicit solutions of particular classes of differential equations - for example, for solving

(*)

where a and b are given constants and / is a given function. The technique of these calculus-based methods is to construct explicit solutions, often by interesting and ingenious means. These methods do not, however, deal with all cases and if we modified equation (*) so that the constants a and b were replaced by functions we would have an example where the calculus-based methods would be useful only in certain special cases. What can we do in these awkward cases? If we expect that, perhaps because the functions involved are awkward, we will not find an explicit solution, is there any progress to be made? We shall use our analysis to show that in fairly general cases the differential equation does have a solution. Now the practical person may not be too impressed by knowing that there is a solution if he or she cannot find it, but the techniques used will actually produce a means of approximating this elusive solution. Taking another perspective, we shall also produce conditions under which our differential equation has a unique solution, which will lead us to a much-used technique for constructing all solutions out of a known number of specific solutions. We shall begin with a result which contains all the hard work. Lemma 14.16 Suppose that F: [a, b]%[a, b] —> R variables, and that / : [a, b] —> R

is a continuous function of two

is also continuous. Then the equation x

z(x) = f(x) + \ F(x, t)z(t)at

(a
a

(1)

has a unique continuous solution z: [a, b] —> R . Comment: Notice that this result is so general that we should be able to apply it in many specific cases. It also has the property that, because of the generality, we are unlikely to find a neat formula for the solution. Proof. Define a sequence of functions z„ by z (JC) = /(JC), z 0

(JC) = f(x) + \ F(x,t)z (t)dt X

n + 1

a

(a
n

(2)

Now zo is continuous, and (by Problem 16) if z is continuous so is z„+\ so it follows that for all ne N u { 0 } , z„ is continuous. We need to show that (z„) tends to a limit as η -» °°. n

Let M = sup{ \F(x,t)\: a
, and

d„ = sup{ | z„(t) - z -\it) | : n

(JC) - z (JC)| = \j F(x,t)(z X

\z

n+]

n

a

<£|

(t) - z„_, (0)df by (2)

n

F(x,i)|z„(i)-z„-i(0|di

As this holds for all χ e [a, b], Vn e Ν d

n+]

< M(b - a)d . n

(3)

a
[Ch. 14

Functions of Several Variables

220 Therefore, d
(b-a)d\,

2

d
<(M (b-a)yd

2

and so on.

t

To simplify matters we shall consider the case where M(b - a) < 1 first. Then from l

(3),

d„ <(M(b-a))"' d

and, by comparison, the series

]

observing that for all ζ

χ

χ

Σ( η( )~Ζη-\( ))

|z (x) - z „ - i W | ^d„

xe [a,b]

n

Y4

converges. By

n

we see in turn that

also converges. Now

Σ^=! (*» ( ) χ

ί*)) = Z W - Z N

0

W .

so the convergence of the series implies that the sequence (z„ (x)) tends to a limit as η -> oo. Let z(x) = lim z„ (x) • We need to show that

Choose JC e \a, b] and ε > 0.

ζ is continuous.

0

xe[a,b]

|z(x)-z W|= ΣΓ=Ν+ΐ^"^~ «-ι

so that

ΣΓ=/ν+ΐ

ζ

-ΣΓ=/ν+ΐ^η

( χ ) )

w

<

^

t n e

y

continuity of

z,

, so choose /V

3δ > 0

N

For all

such that

\\x - x \\ < δ => \z (x) - z (xo)\ < ε/3. Then 0

N

N

jjc — JCQ j < δ =>

I z(x) - z(x ) I ^ I z(x) -z (x)\ 0

+ \z (x)-z (x )\

N

N

^2ΣΓ=^ ,«η

N

+ε/3

+

+ \z (x )-

0

N

0

z(x ) I 0

ε

< ·

Since ε > 0 was arbitrary we have shown ζ is continuous at JC , and since Xo was typical, ζ is continuous. 0

x

Because ζ is continuous,

j F(x,t)z(t)dt

this is the limit of j*F(x,t)z

(t)dt

n

is continuous, and we might hope that

as η —> °°. As this conclusion is equivalent to

interchanging the operations of integrating and letting η —» °° , we need to prove its truth. Choose

ΣΓ=ΛΜ-Ι^Λ

Ν so that

V/n > Ν \z(t) - Z (t)\ < Σ ~ m

<

ε

·

s

o

m a t

<

a

s

above Vi e [a, b] and

l I Z« ( 0 - Zn-\ ( 0 I * Σ Γ = « Ι n < d

m +

ε

+

4

()

Then, if χ e [a, ft] and m>N, z (x)-f(x)-\*F(x,t)z(t)dt

x

<

m+i

\ F(x,t)(z (t)-z(t))dt m

x

<\ M\z (t)-z(t)\dt m

x

<\ Mtdt0

is arbitrary here , z \ m+

x

(x) —> f(x) + \ , F(x,t)z(t)dt

as m —» °° , and

so, by the definition of ζ as the limit of ( z ) , z satisfies (1). We have shown that a continuous solution of (1) exists. m

To show that the solution is unique, suppose that ζ and w are continuous solutions o f ( l ) , and let d= sup[|z(r) - w{t)\: t e [a,b]}. Then, for all xe [a,b] \z(x)~

w(x)I< j * \ F{x,0||

z(t)-w(t)\dt
Sec. 14.4]

Solving Differential Equations

221

Because this is true for all xs [a,b] , d
f\ (·*) - / M

+h

x

l"

F( > t)z(t)dt

for the solution ζ just found. By the result above

the equation z(x) = f\(x) +$* F(x,t)z(t)dt

has a unique continuous solution ζ on

+h

[a + h,a + 2h] , which extends the function ζ already found to [a, a + 2h]. Proceeding in this way, extending in steps of less than MM , we extend ζ to the whole of [a, b\. • Lemma 14.17 actually tells us a little more than we have said. We know a unique solution ζ exists, and that z „ , as defined in the proof, tends to ζ as η —» °°. We can therefore ensure that z„ is as close to the true solution as we wish if η is large enough. In this case, we can even say how large is enough. We saw in equation (4) above that \/te[a,b]

d

\ z(t)-z (t)\<

T^= \ n

m

•

m+

Now, considering the simpler case where M(b - a) < 1 , we have for all η > m + 1 , d < M(b -a)d ., < ... < (Μφ - a))"^d , so n

n

m

z:= ^

m b

±Y° ^w(b-a))»->»d = ^ d

m+1

n

m

x M

(on summing the geometric series). We can calculate Zi, z ,

m

in turn and from

2

these find d .

a)

—^—d < ε , we know \ — M(b-a) that the approximation z has the property that for all / e [a, b] , z(t) differs from the true solution z(f) by less than ε. We can relate all this immediately to the general second-order, "linear" differential equation. m

Then once we find an m such that

m

m

Theorem 14.18 Suppose that ρ , q and g: [a, b] -> R are continuous, and that a, β ε R. Then the differential equation problem ^ + de 2

p(x$-

+ q(x)y(x)

= g(x) (a
y(a) = a, / ( β ) = β

dx

has a unique solution y. Proof. If y satisfies the differential equation then it must be twice differentiable, hence both y and y ' are continuous. From this, and the differential equation, we see that y " is continuous. Therefore, letting ζ denote y",z is continuous, %

y(x) = a + p (x-a)

+ l*(x-t)z(t)dt

{a
(1)

(by Corollary 2 to Theorem 13.12) and ζ satisfies X

z(x) = f(x) + j F(x,t)z(t)dt a

{a
(2)

[Ch. 14

Functions of Several Variables

222

where fix) = g(x) - βρ(χ) - (α + β(χ - a))q(x)

and F{x, t) = -(p(x) + {x-

t)q(x)).

Moreover, if ζ is continuous and satisfies (2), then the Fundamental Theorem of Calculus shows us that y given by (1) is twice differentiable and satisfies the original differential equation problem. It follows that the differential equation has a unique solution y if and only if equation (2) has a unique continuous solution z. By Lemma 14.17 this is true, and wc have finished. • Suppose now that ρ and q arc continuous on [a, b| and we let yi and y (unique) solutions of ^-+p(x)^ dx

+ q(x)y{x)

2

be the

= 0(a
(*)

2

with

yj(a)=l,

y[(a)=0,

y (a)=0,

y>)=1.

2

Then if y is another solution of (*), let α = y(a) and β = y'(a). We now notice that the function y , where y (x) = ayi(x) + βy (x) , satisfies (*) (because y, and y do), y (a) = α , and y '(a) = β . Therefore, by the uniqueness part of Theorem 14.18, y and v must be equal, that is, y(x) = ayi(x) + βy (x) . 3

3

3

2

2

3

3

2

In other words each solution of (*) can be expressed in the form of a constant times yi plus a constant times y . This can be extended slightly (see Problem 22) to yield the result that if we can find two independent solutions yi and y of (*) then all other solutions are of the form ayi + β y 2 for constants α and β. 2

2

The results about differential equations can be extended in many directions, but we shall be content with what we have achieved, which we could not have done without some form of analysis. However, our arguments have become somewhat lengthy, and this length tends to obscure the structure of the proofs. This can be remedied by placing the analysis in a more elegant framework, taking account of algebraic and other mathematical structures available. Good examples are in the books by Brown and Page, Porter and Stirling, Sutherland, and Conway. Problems 1. In each case below, define f(x, y) to have the value stated for (x, y) * (0, 0) and to have the value 0 at (0, 0). Determine whether / is continuous at x

(0,0):

2

y

X

2

2 '

x+y 2.

{

2

2 '

x+y

X

+

y

2

x+y

)

2

x

2 '

2

y

A

~ 4 x+y

2

2

Let / , g], g : IR —> Κ be three continuous functions and define F by F(x, y) =f (g\(x, y), g (x, y))- Show that F is continuous. Prove also that if h: R —> IR is continuous so is Η , where H(x, y) - h(f(x, y)). 2

2

3.

Let / and g be defined as follows: [sin Jt-sin y

sin(x+y) fix,y)

=

x +y

(x+y*0), ( x + y = 0),

g{x,y) =

(x* χ-y cosx

y)

(x=y).

Sec. 14.4]

Solving Differential Equations

223

Show that both functions are continuous at (0, 0). m

4.

Let / : R -» R be continuous at a. For b e l " , define F: R -> R by F (t) — F (a + fb) and show that F is continuous at 0. The converse result is 2

false; to see this d e f i n e / : R -> R

by

l

f(x,y)

2

2

=\ x
/ is not continuous at ( 0 , 0 ) but lim / ( f b ) = 0 /->0

5.

((JC,,JC ): JC, > 0 ) , {(x 2

6.

2

Which of the following subsets of 2

2

x ): x

u

R

2

+x

2

are closed? 2

< 1 or(jc, +jc =l

2

{xe R:

χ

< 1) ,

and j c > 0 ) } . t

Suppose that (JC„) is a sequence of real numbers with the property that Ve > 0 3N such that Vw , n>N x -x„ < ε; such a sequence is called a Cauchy sequence. Prove that if (JC„) has a convergent subsequence then (x ) is itself convergent. By showing that (x ) is a bounded sequence, deduce that it converges. (Thus every Cauchy sequence in R converges.) m

n

n

m

7.

Let A = { x e R : χ - a < r) and suppose that f:A->R is continuous. If b, c e A and / ( b ) < y < / ( c ) , prove that there is a point x e A for which / ( χ ) = γ. (Hint: define F: [0, \]^>R such that F(0) = / ( b ) and F(I)=/(c)?)

8.

Calculate the partial derivatives

and

of the following functions at

(0,0) and decide whether / is differentiable at (0,0). In all cases, fiO, 0) = 0 and for other values of (JC, y) , f(x, y) is given by the formula .2..

_..

r—.

v-2,,2

2

**y

,x(x

+

y

2

m

Let f:R ->R and a e l * If b e T and | | b | | = l we say that the directional derivative of / at a in the direction b is F ' ( 0 ) where F(t) -f(a

+ rb), when this exists. (So

the direction ( 1 , 0 , 0 ) . )

-gj- is the directional derivative in

Show that if /

is differentiable and J|b|| = 1, the

directional derivative in the direction b = (b\, b , b 2

2

(X

- ν

XV

10.

Let /(JC, y) = - A ;

γ

2

m

)

a

is

^"( ) ·

)

1

if

y) * (0, 0) and / ( 0 , 0) = 0. Calculate the

JC + y 2

first- and second-order partial derivatives and show that

and

are

g (*i, x ) = X\

and set

unequal at (0, 0). 11.

2

Define

g,,g :!R ->R

by

2

C(JC,, JC ) = gi(gi(jc,, x ), 2

2

gi(JCi, JC ) = J C , 2

2

2

2

g (jci, JC )). Use the Chain Rule (carefully!) to show 2

that ψ- = 1 and | ^ = 0 .

2

Functions of Several Variables

[Ch.14

Let / : R* -> R be differentiable and let y, ζ: R -> R also be differentiable. Calculate F ' ( x ) where F(x) = / ( x , v(x), z(x)). (If this is confusing, let x(t) = ί and find (d/dt)(f(x(t), y(/), z(/))).) The internal energy, u, of a gas can be expressed as a function of the pressure, volume and temperature of unit mass of the gas, as u(p, v, t). The pressure ρ may be expressed as a function of volume and temperature, ρ = p(v, t). Thus the internal energy can be expressed in terms of ν and / as U(v, t) = u(p(v, t), ν, I). Show that

=

+

/ is twice differentiable and Vx, y e R fix, y) =fiy, x) • Show that

*f(a,b) = !f (b,a),

^f

OX

^X

Oy

andiff (ajh) = iff

(a,b) = ^-f(b,a) ^2

fady

y

(b,a).

dyd

X

m

Let A = (x e R : ||x||< 11 and f:A-*R be continuous and be differentiable at all points a satisfying a < 1 . Prove that if / attains a maximum or minimum at an internal point of A (i.e. at a point a with a

= -M- =•··•= f£— — 0

< 1 ). then

there. Hence prove that if / ( x )

is constant on the set { x : | | x | | = l } then there is a point y with which J ^ ( y ) = 0 (1= 1.2,

y

<1

for

m) .

Let / : [a, b] \c, d] -» R be continuous and define F: [a, b] [c, d\ —» R x

x

F(x,y)=j*f(x,t)dt.

by

Prove that F is continuous. (Modify Theorem 14.7

and consider small changes in χ and y separately.) Show that if / ( x , y) is continuous in the two variables separately and exists and is jointly continuous, then / / ( x , y) -f(x,

is jointly continuous.

^

(Hint: use

y„) = Γ | ; ( x , / ) d / . )

Let / : [a, b\ \a, b\ —> R

be continuous, and define

x

G(z)=l

*"(z) =a(a f(x,y}ay)dx,

( ;,/(x,y)dx)dy .

F, G: [a, b] —» R

by

By showing that the

result of the inner integration in both cases is (jointly) continuous, and differentiating under the integral sign, deduce that F'-G' and hence that F = G. Sketch the sets over which integration is taken in the (x, y)-plane. 2

A r

Given that f"' -^—, integral

sign

.,

— = ^rr f n ' ——Γ"

«

2

2

2

-,—--—V

2

that

J

rn/2 dt •ta (o sin .ï +

show by differentiating under the

2

(a~ sin " Λ + b~ cos x)~

and

find

4ab

2

fc cos ^)

Decide whether the functions given by the formulae below have a maximum, minimum or saddle point at (0, 0): xy, x + y , x + xy + y , x + xy + y . 2

2

2

2

4

2

Sec. 14.41 21.*

Solving Differential Equations

225

2

Let A = {xe R : χ R have continuous first-order partial derivatives. By considering F\(t) = / i ( a + /(b - a)) show that if = / i ( b ) where a and b are two distinct points of A, then there is a 2

point X| e A forwhich (b\-a\)-^-(xi) If, in addition,

/ (a) = / (b) 2

2

a

+ ( £ ~ 2>d-^-(x\) 2

= 0.

show that there is a second point x e A for 2

which G ( x , , x ) = 0 where G(x,y) = ^ - ( x ) ^ - ( y ) - ^ - ( x ) ^ - ( y ) 2

.

Deduce that if G(0, 0) Φ 0, then for r sufficiently small, the function χ H > (/I(X),/ (X)) is injective on A. (Notice that the condition G(0, 0 ) * 0 is 2

equivalent to the statement that the matrix

at, 2

of partial derivatives

2

2 2

is non-singular at 0.) 22.*

3JC


Suppose that ρ and q are continuous functions on [a, b] and that y, and y

2

satisfy ^-+p(x)^

+ q(x)y(x)

=0

{a
(*)

Let y (a) = a and ν '(α) = β, (/'= 1.2). j

i

;

(i) Show that if there is a constant λ such that α, = λ α and βι = λ β , then >ι = λν · In this case one of the solutions is a constant multiple of the other (and ο.|β - α β ! = 0). (ii) Show that if ο.|β - α β ι * 0 then y, and y are independent in the sense that neither of them is a constant multiple of the other. (iii) By noticing that if α ι β - α β ι * 0 then, given two real numbers α and β, there are constants λ and μ such that λαι + μ α = α λβ, + μ β = β deduce that if α ι β - α β , * 0 and y is a solution of (*), then there are constants λ and μ such that y = Xy + μ ν . (Use Part (iii) and Theorem 14.18.) 2

2

2

2

2

2

2

2

2

2

2

2

2

2

x

23.

Let

/ : [a, b] -> R

2

be continuous and define F: [a, fc] —» R

F{x) = £ U - t | f(t)dt.

by

Show that F is twice differentiable, F" = 2f and

F'(0) + F'(1) = 0 . 24.*

Let

f.\a,

b] -*

R

]

F(x) = \ (xt-m\n(x,

t))f(t)dt.

Q

other y* = /

words

be continuous and define F: [a, b] - » R

y

=

F

by

Show that F" =f and F(0) = F ( l ) = 0 . In satisfies

the

"boundary-value

problem"

y(0) = y(l) = 0 . Show conversely that if y satisfies this boundary

value problem then y = F .