Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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12.2 Calculus of Vector-Valued Functions 393

12.2.7. PROPOSITION. Fix f : [a, b] → R

with f(x) = (f

(x), . . . , f

(x))

for all x ∈ [a, b]. Then f is Riemann integrable if and only if each coordinate

function f

is Riemann integrable. In this case,

f(x) dx =

(x) dx,

(x) dx, . . . ,

(x) dx

PROOF. Suppose that f is Riemann integrable. To show that f

is Riemann inte-

grable, for some j between 1 and n, we use condition (d) of Theorem 6.3.8. That

is, there is a real number, call it M, so that for every ε > 0, there is δ > 0 so that

for every partition Q with mesh(Q) < δ and every choice of points X subordinate

to Q, we have |I(f

, Q, X) − M| < ε.

Since f is Riemann integrable, we have (u

, . . . , u

) ∈ R

with

f(x) dx = (u

, . . . , u

Moreover, there is δ > 0 so that for every partition Q with mesh(Q) < δ and every

set of points X subordinate to Q,

I(f, Q, X) −

f(x) dx

< ε.

For any vector x = (x

, . . . , x

), recall that |x

| ≤ kxk for 1 ≤ i ≤ n.

Applying this fact to the vector I(f, Q, X) −

f(x) dx and using Lemma 12.2.5,

we have |I(f

, Q, X) − u

| < ε. Thus, if we choose M = u

, it follows that f

Riemann integrable and, moreover,

(x) dx = u

Conversely, suppose that each f

is Riemann integrable. Let ε > 0 and let

u ∈ R

be given by u = (

(x) dx, . . . ,

(x) dx). Find δ

> 0 satisfying

condition (d) of Theorem 6.3.8 so that |

(x) dx − I(f

, Q, X)| < ε/

√

n. Let

δ be the minimum of δ

, . . . , δ

. Then for any partition Q with mesh(Q) < δ and

any set of points X subordinate to Q, we have

ku − I(f, Q, X)k

≤

j=1

(x) dx − I(f

, Q, X)

≤

j=1

/n = ε

Taking square roots, we are done. ¥

Using this proposition, we can carry over many properties of integration for

real-valued functions to vector-valued functions. For example, linearity of integra-

tion and the Fundamental Theorem of Calculus carry over in this way.

12.2.8. FUNDAMENTAL THEOREM OF CALCULUS II.

Let f : [a, b] → R

be a bounded Riemann integrable function and deﬁne

F (x) =

f(x) dx for a ≤ x ≤ b.

Then F : [a, b] → R

is a continuous function. If f is continuous at a point x

then F is differentiable at x

and F

) = f (x

394 Differential Equations

Finally, we have the following estimate, which says that the norm of an integral

is less than the integral of the norm. This is the higher-dimensional analogue of

taking the absolute value inside the integral. While the proof is similar in spirit to

the scalar case, the argument requires inner products and the Schwarz inequality.

12.2.9. LEMMA. Let F : [a, b] → R

be continuous. Then

F (x) dx

≤

kF (x)kdx.

PROOF. Let y =

F (x) dx. If kyk = 0, then the inequality is trivial. Otherwise,

using the standard inner product on R

kyk

F (x) dx, y

hF (x), yidx

≤

|hF (x), yi|dx ≤

kF (x)k kykdx

= kyk

kF (x)kdx.

The second line follows by the Schwarz inequality. Dividing through by kyk gives

the result. ¥

Exercises for Section 12.2

A. Prove Lemma 12.2.5.

B. Show that if α, β are real numbers and f, g are Riemann integrable functions from

[a, b] to R

, then αf + βg is Riemann integrable. Moreover, show that

f(x) dx + β

g(x) dx =

(αf(x) + βg(x)) dx.

C. Prove Theorem 12.2.8.

D. Every continuous function f : [a, b] → R

is Riemann integrable.

(a) Show this using Proposition 12.2.7.

(b) Show this directly from the deﬁnition, without using Proposition 12.2.7.

HINT: Adapt Theorem 6.3.10 to use Riemann sums.

E. Deﬁne a regular curve to be a differentiable function f : [a, b] → R

so that f

(x) is

never the zero vector.

(a) Given two regular curves f : [a, b] → R

and g : [c, d] → R

, we say that g is

a reparametrization of f if there is a differentiable function h : [c, d] → [a, b] so

that h

(t) 6= 0 for all t and g = f ◦h. Show that reparametrization is an equivalence

relation and that g

(t) = f

(h(t))h

(t).

(b) Deﬁne the length of a regular curve f to be L(f) =

(x)kdx. Show that

the length is not changed by reparametrization. HINT: Consider reparametrizations

where h(t) is always positive or always negative.

12.3 Differential Equations and Fixed Points 395

and a partition P = {x

< ··· < x

} of [a, b],

deﬁne `(f, P ) =

i=1

kf(x

) − f(x

i−1

)k. Show that for each ε > 0, there is some

δ > 0 such that for all partitions P with mesh(P ) < δ, |`(f, P ) − L(f)| < ε.

(d) Given a regular curve f, show there is a reparametrization g with kg

(t)k = 1 for

all t. Such a curve is said to have unit speed.

HINT: Show that x 7→

kf(t)kdt has an inverse function.

(e) Suppose that f : [a, b] → R

and g : [c, d] → R

are regular curves and both have

unit speed. If g is a reparametrization of f, show that either g(x) = f(x + (a −c))

for all x ∈ [c, d] or g(x) = f (−x + (c − b)) for all x ∈ [c, d]. In either case, show

that d − c = b − a.

12.3. Differential Equations and Fixed Points

The goal of this section is to start with a DE of order n, and convert it to the

problem of ﬁnding a ﬁxed point of an associated integral operator.

The ﬁrst step is to take a fairly general form of a higher-order differential equa-

tion and turn it into a ﬁrst-order DE at the expense of making the function vector

valued. We deﬁne an initial value problem, for functions on [a, b] and a point

c ∈ [a, b], as follows:

(n)

(x) = ϕ(x, f (x), f

(x), . . . , f

(n−1)

(x))(12.3.1)

f(c) = γ

(n−1)

n−1

where ϕ is a real-valued continuous function on [a, b] × R

. This is not quite the

most general situation, but it includes many important examples. The ﬁrst equation

is referred to as a differential equation of nth order, and the second set is called

the initial conditions.

A standard trick reduces this to a ﬁrst order differential equation with values in

. As we shall see, this has computational advantages. Replace the function f by

the vector valued function F : [a, b] → R

given by

F (x) = (f (x), f

(x), . . . , f

(n−1)

(x)) .

Then the differential equation becomes

(x) =

(x), . . . , f

(n−1)

(x), ϕ

x, f(x), . . . , f

(n−1)

(x)

and the initial data become

F (c) = (γ

, γ

, . . . , γ

n−1

396 Differential Equations

We further simplify the notation by introducing a function Φ from [a, b] × R

Φ(x, y

, . . . , y

n−1

) =

, y

, . . . , y

n−1

, ϕ(x, y

, . . . , y

n−1

)

and the vector

Γ = (γ

, γ

, . . . , γ

n−1

Note that Φ is continuous since ϕ is continuous. Then (12.3.1) becomes the ﬁrst-

order initial value problem with vector values

(x) = Φ(x, F (x))(12.3.2)

F (c) = Γ.

It is easy to see that a solution of (12.3.1) gives a solution of (12.3.2). To go

the other way, suppose (12.3.2) has a solution

F (x) = (f

(x), f

(x), . . . , f

n−1

(x)).

Then (12.3.2) means

(x) = (f

(x), f

(x), . . . , f

n−2

(x), f

n−1

(x))

= Φ(x, f

(x), f

(x), . . . , f

n−1

(x))

(x), . . . , f

n−1

(x), ϕ

x, f(x), . . . , f

(n−1)

(x)

By identifying each coordinate, we obtain

(x) = f

(x)

(x) = f

(x)

n−2

(x) = f

n−1

(x)

n−1

(x) = ϕ(x, f

(x), f

(x), . . . , f

n−1

(x)) .

Thus f

= f

, f

(2)

= f

, ..., f

(n−1)

= f

n−2

= f

n−1

, and

(n)

(x) = f

n−1

(x) = ϕ

x, f

(x), f

(x), . . . , f

(n−1)

(x)

The initial data become Γ = F (c) = (f

(c), . . . , f

n−1

(c)) so

, f

, . . . , f

(n−1)

n−1

Thus (12.3.1) is satisﬁed by f

(x). Consequently, (12.3.2) is an equivalent formu-

lation of the original problem.

12.3 Differential Equations and Fixed Points 397

12.3.3. EXAMPLE. We will expressthe unknown function as y, instead of f(x).

Consider the differential equation

1 + (y

)

(3)

= y

− xy

y + sinx for − 1 ≤ x ≤ 1

y(0) = 1

(0) = 0

(0) = 2.

It is ﬁrst necessary to reformulate this DE to express the highest-order derivative,

(3)

, as a function of lower-order terms. This yields

(3)

− xy

y + sinx

1 + (y

)

Then the vector function Φ deﬁned from [−1, 1] × R

into R

given by

x, y

, y

− xy

+ sin x

1 + y

The initial vector is Γ =

1, 0, 2

. The DE is now reformulated as a ﬁrst order

vector valued DE looking for a function F (x) =

(x), f

(x)

deﬁned on

[−1, 1] with values in R

such that

(x) =

(x), f

(x),

(x) − xf

(x)f

(x) + sin x

1 + f

(x)

for − 1 ≤ x ≤ 1

F (0) =

1, 0, 2

Now we can integrate this example as before. Deﬁne a mapping T from the

space C([−1, 1], R

) of functions on [−1, 1] with vector values in R

into itself by

sending the vector valued function F (x) =

(x), f

(x)

T F (x) = Γ +

Φ(t, F (t)) dt

1 +

(t) dt ,

(t) dt , 2 +

(t) − tf

(t)f

(t) + sin t

1 + f

(t)

This converts the differential equation into the integral equation T F = F . An argu-

ment similar to the preceding one shows that this ﬁxed-point problem is equivalent

to the differential equation.

Returning to the general case, we need to ﬁnd a suitable framework (i.e., a

complete normed vector space) in which to solve the problem in general. Since

the coordinate functions f

, . . . , f

n−1

of F are all continuous functions on [a, b],

we can think of F as an element of C([a, b], R

), the vector space of continuous

functions from [a, b] into R

. This space has the norm

kF k

∞

= max

a≤x≤b

kF (x)k = max

a≤x≤b

n−1

i=0

(x)|

1/2

398 Differential Equations

A sequence F

= (f

, . . . , f

n−1

) converges to F

∗

= (f

∗

, . . . , f

∗

n−1

) in the

max norm if and only if each of the coordinate functions f

converges uniformly

to f

∗

for 0 ≤ i ≤ n − 1. To see this, notice that for each coordinate i,

− f

∗

k = max

a≤x≤b

(x) − f

∗

(x)|

≤

n−1

j=0

(x) − f

∗

(x)|

1/2

= kF

− F

∗

∞

Therefore, the convergence of F

to F

∗

implies that f

converges to f

∗

for each

0 ≤ i ≤ n − 1. Conversely,

− F

∗

∞

= max

a≤x≤b

n−1

j=0

(x) − f

∗

(x)|

1/2

≤

n−1

j=0

max

a≤x≤b

(x) − f

∗

(x)|

1/2

n−1

j=0

− f

∗

∞

1/2

So if kf

− f

∗

∞

tends to zero for each i, then kF

− F

∗

∞

also converges to

zero.

We need to know that this space of functions is complete. This is a straightfor-

ward consequence of the scalar version Theorem 8.2.2.

12.3.4. THEOREM. C([a, b], R

) is complete.

PROOF. It is clear that C([a, b], R

) is a normed vector space. To prove complete-

ness, we must show that every Cauchy sequence F

converges to a function F

∗

C([a, b], R

). We will write these functions in coordinate form as

(x) = (f

(x), f

(x), . . . , f

n−1

(x)).

If F

is Cauchy, the inequalities just established show that each sequence of

coordinate functions (f

)

∞

k=1

is Cauchy in C[a, b] for each 0 ≤ i ≤ n −1. By The-

orem 8.2.2, C[a, b] is complete. Therefore, the sequence (f

) converges uniformly

to a function f

∗

in C[a, b]. By the preceding remarks, F

converges uniformly to

the function F

∗

= (f

∗

, . . . , f

∗

n−1

). ¥

We have found a setting for our problem (12.3.2) in which the Contraction

Principle is valid, and so we can formulate the problem in terms of a ﬁxed point.

To do this, integrate (12.3.2) from a to x to obtain

F (x) = F (c) +

(t) dt = Γ +

Φ(t, F ) dt.

This suggests deﬁning a map on C([a, b], R

) by

T F (x) = Γ +

Φ(t, F (t)) dt.

12.4 Solutions of Differential Equations 399

A solution of (12.3.2) is clearly a ﬁxed point of T . Conversely, by the Fundamental

Theorem of Calculus, a ﬁxed point of T is a solution of (12.3.2). So the prob-

lem (12.3.1) can be solved by ﬁnding the ﬁxed point(s) of the mapping T from

C([a, b], R

) into itself.

Exercises for Section 12.3

For each of the following three differential equations, convert the DE into a ﬁrst-order

vector-valued DE and then into a ﬁxed-point problem.

A. y

(3)

+ y

− x(y

)

= e

, y(0) = 1, y

(0) = −1 and y

(0) = 0

B. f

(x) =

f(x)

+ x

1/2

− f

(x)

, f(1) = 0 and f

(1) = 1

+ 2x

v = 0, v(1) = 1 and v

(1) = 0

D. Let V be a complete normed vector space. (Think of a few examples!)

(a) Let (f

) be a Cauchy sequence in C([a, b], V ). Show that for each x ∈ [a, b] that

(x)) is Cauchy, and so deﬁne the pointwise limit f(x) = lim

n→∞

(x).

(b) Prove that (f

) converges uniformly. HINT: Use the Cauchy criterion to obtain

an estimate for kf

(x) − f(x)k that is independent of the point x.

E. (a) Solve the DE y

= xy and y(0) = 1. Deduce that the solution is valid on the whole

line. HINT: Integrate y

/y = x.

(b) Deﬁne T f(x) = 1 +

tf(t) dt. Start with f

(x) = 1 and compute f

= T f

n−1

for n ≥ 1. Prove that this converges to the same solution. Where is this conver-

gence uniform?

F. Let ϕ be a continuous positive function on R. Consider the DE f

(x) = ϕ(f (x)) and

f(c) = γ. Let F(y) = c+

ϕ(t)

. Show that f(x) = F

−1

(x) is the unique solution.

HINT: Integrate f

(t)/ϕ(f(t)) from c to x.

G. Consider the DE: xyy

= y

− 1 and y(1) = 1/

√

(a) Solve the DE. HINT: Integrate

−1

(b) The solution only exists for a ﬁnite interval containing 1. Explain why the solution

cannot extend further.

12.4. Solutions of Differential Equations

In this section, we use a modiﬁcation of the Contraction Principle to demon-

strate the existence and uniqueness of solutions to a large class of differential equa-

tions. The basic idea of contraction mappings is that, starting with any function,

iteration of the map T leads inevitably to the solution. As we have seen for New-

ton’s method in Section 11.2, it may well be the case that there is a contraction

400 Differential Equations

provided that we start near enough to the solution. Even when T is not a contrac-

tion, it may still have an attractive ﬁxed point. In this case, if we start at a reasonable

initial approximation, the structure of the integral mapping will force convergence,

which is eventually contractive.

To analyze the map T , we need a computational result.

12.4.1. DEFINITION. A function Φ(x, y) is Lipschitz in the y variable if

there is a constant L so that

kΦ(x, y) − Φ(x, z)k ≤ Lky − zk

for all (x, y) and (x, z) in the domain of Φ.

Note that both the y-variable and the range may be vectors rather than elements

of R. While this estimate does not concern variation in the x-variable, it does

require that the constant L be independent of x.

12.4.2. LEMMA. Let Φbe a continuous function from [a, b] ×R

into R

which

is Lipschitz in y with Lipschitz constant L. Let

T F (x) = Γ +

Φ(t, F (t)) dt.

If F, G ∈ C([a, b], R

) satisfy kF (x) − G(x)k ≤

M(x − a)

, then

kT F (x) − T G(x)k ≤

LM(x − a)

k+1

(k + 1)!

In particular, T is uniformly continuous.

PROOF. Compute

kT F (x) − T G(x)k =

Φ(t, F (t)) dt − y

−

Φ(t, G(t)) dt

Φ(t, F (t)) − Φ(t, G(t)) dt

≤

Φ(t, F (t)) − Φ(t, G(t))

≤

F (t) − G(t)

≤

(t − a)

dt =

(k + 1)!

(x − a)

k+1

In particular, kF − Gk

∞

= kF − Gk

∞

(x−a)

. It follows that

kT F − T Gk

∞

≤ kF − Gk

∞

Lkx − ak

∞

= kF − Gk

∞

L(b − a).

So T has Lipschitz constant L(b − a) and therefore is uniformly continuous. ¥

12.4 Solutions of Differential Equations 401

12.4.3. GLOBAL PICARD THEOREM.

Suppose that Φ is a continuous function from [a, b] ×R

into R

which is Lipschitz

in y. Then the differential equation

(x) = Φ(x, F (x)), F (a) = Γ

has a unique solution.

PROOF. Let T map C([a, b], R

) into itself by

T F (x) = Γ +

Φ(t, F (t)) dt.

As discussed at the beginning of this section, a function F in C([a, b], R

) is a ﬁxed

point for T if and only it is a solution of (12.3.2). Deﬁne a sequence of functions

(x) = Γ and F

k+1

= T F

for k ≥ 0.

Let L be the Lipschitz constant of Φ, and set M = max

a≤x≤b

kΦ(x, Γ)k

. We

have the inequality

(x) − F

(x)k =

Φ(t, Γ) dt

≤

M(x − a)

Therefore, by Lemma 12.4.2,

(x) − F

(x)k ≤

(x − a)

(x) − F

(x)k ≤

(x − a)

and, by induction, we get

k+1

(x) − F

(x)k ≤

k+1

(x − a)

k+1

k + 1!

As in the proof of the Banach Contraction Principle (11.1.6),

n+m

(x) − F

(x)k ≤

m+n−1

k=n

k+1

(x) − F

(x)k

≤

m+n−1

k=n

k+1

(x − a)

k+1

k + 1!

≤ M

∞

k=n+1

L(b − a)

But the series M

∞

k=0

L(b − a)

/k! converges to Me

L(b−a)

. So given any ε > 0,

there is an integer N so that the tail of the series satisﬁes

∞

k=N

L(b − a)

< ε.

402 Differential Equations

Thus if n, n + m ≥ N, it follows that

n+m

− F

∞

< ε.

This means that the sequence (F

) is Cauchy in C([a, b], R

), and hence converges

to a limit function, call it F

∗

Since T is continuous,

T F

∗

= lim

k→∞

T F

= lim

k→∞

k+1

= F

∗

If G is another solution, then G satisﬁes T G = G. An argument similar to the

previous paragraph shows that

∗

− Gk

∞

= kT

∗

− T

∞

≤ kF

∗

− Gk

∞

L(b − a)

The left-hand side is constant and nonnegative, while the right-hand side converges

to 0 as k tends to inﬁnity. Therefore, kF

∗

− Gk

∞

= 0. In other words, G = F

∗

and the solution is unique. ¥

12.4.4. EXAMPLE. Consider the initial value problem

+ y +

+ (y

)

= 0 y(0) = γ

and y

(0) = γ

We set this up by letting Y = (y

, y

) and

Φ(x, y

, y

) =

, −y

−

+ y

, −y

− kY k

Then the DE becomes

(x) = Φ(x, Y ) and Y (0) = Γ := (γ

, γ

Let us verify that Φ is Lipschitz. Let Z = (z

, z

). Recall that the triangle

inequality implies that

kZk − kY k

≤ kZ − Y k.

kΦ(x, Y ) − Φ(x, Z)k = k

− z

, z

− y

+ kZk − kY k

≤ k

− z

, z

− y

k +

kZk − kY k

≤ 2kZ − Y k

Therefore, Picard’s Theorem applies, and this equation has a unique solution that

is valid on the whole real line.

It may be surprising that this DE actually can be solved explicitly. We may

rearrange this equation to look like

y + y

+ (y

)

= −1.

There is no obvious way to integrate the left-hand side. The key observation is that

+ (y

)

= 2yy

+ 2y

= 2y

(y + y