An Existence and Uniqueness Theorem

Chapter VI: Boundary Value and Eigenvalue Problems

W. Integrating Factors (or Euler Multipliers). The differential equa- tion

II. Parametric Representation with y' as the Parameter. In the following sections we will discuss some examples of implicit differential equations

6. An Existence and Uniqueness Theorem

All of the functions in this section are assumed to be real valued. We consider the following initial value problem

y'=f(x,y)

for (1)

The main assumptions in the following theorem are that f is continuous in the strip S =Jx IR with J = a Lipschitz condition with respect to y in S:

If(x,y)—f(x,yi)l ⁽²⁾

No restrictions are placed on the value of the Lipschitz constant L 0.

I.

Existence and Uniqueness Theorem. Let I E

C(S) satisfy the Lipschitz condition (2). Then the initial value problem (1) has exactly one so- lution y(x). The solution exists in the interval J:

x + a.

The proof is essentially an application of the fixed point theorem SIX. As a preliminary step, the initial value problem is transformed into an equivalent fixed point equation y =Ty. Let y(x) be differentiable on the interval J and satisfy the initial value problem (1). Because of the continuity of u(x) f(x, y(x)) is continuous in J, so y(x) is actuaily continuously differentiable. Therefore, by the fundamental theorem of calculus

y(x) =

+

JX

f(t,

y(t)) dt. (3)

§ 6. An Existence and Uniqueness Theorem 63

x

Conversely, if y(x) is a solution of (3) that is merely continuous in J, then the right-hand side of (3) is continuously differentiable; hence so is y(x), and

= f(x,y) holds. Furthermore, y satisfies the initial condition

Therefore, the initial value problem (1) is equivalent to the integral equation (3), which can be written in the form of an operator equation

y—Ty with

(Ty)(x) (3')

The integral operator T maps each function y from the Banach space C(J) of continuous functions (cf. Example 5.111. (c)) to a function Ty in the same space.

It follows that the solutions to the initial value problem (1) are precisely the fixed points of the operator T, considered as a mapping B B with B =C(J).

To complete the proof of Theorem I, we show that the operator T satisfies a Lipschitz condition (5.3) with a Lipschitz constant q < ^1, and we then apply the fixed point theorem 5.IX.

If the space C(J) is normed with the maximum norm ilyllo = xE J}, then (2) implies that for x,y E C(J),

(Ty)(x) - (Tz)(x)l=

y(t)) - f(t,

z(t))}

dt

-

^ZIIO(x

-

e

andhence, because x — a,

llTy— TzlIo ^— zilo.

Hence T satisfies a Lipschitz condition with Lipschitz constant La. Note, how- ever, that this Lipschitz constant is less than 1 only if the interval is small, since La < ¹ implies that a < Oneway to handle the case where a is to find an n such that b = ^andthen use the above procedure to determine the

n ^L

solution successively on the mtervals

y

To do this one needs the result in VI. (b).

A more elegant way is to work with a weighted maximum norm:

111111 = ^{: x E J}} (a > 0). (5)

The last integral in (4) is nowestimated as follows:

U y(t) — dt

Lily

— zil f ^dt

Lily

^— _a

Then we conclude from (4) that

I(Ty)(x) — —

andhence

IlTy —TzII ^{— zIl.}

Thus if one chooses a =2L, for example, then T satisfies a Lipschitz condition with Lipschitz constant Thisvariant of the proof gives existence for the whole

interval (of arbitrary length) in a single step.

I

II. Comments.

(a) The theorem shows that starting with an arbitrary function yo(x) E C(J) and calculating the sequence of "successive approxima- tions" given by

yk+1(x) = + f

(k =0,1,2,...), (6)

one obtains a sequence that converges in the norm, and hence uniformly in J, to the y(x) of the initial value problem. This iteration procedure can also be used to determine a numerical approximation to the solution. In numerical approximations, it is a good idea to start with a function yo(x) that is as close as possible to the solution. However, if nothing is known about the solution, then yo(x) = ^r,iis not a bad choice.

(b) The following is a sufficient condition for the Lipschitz condition (2) to hold: f is differentiable with respect to y, and _y)l L (the proof uses the mean value theorem).

(c) Existence and Uniqueness Theorem to the Left of the Initial Value. Let

J =[e—a,e] (a>0). fff is continuous in the strip S_ :=

J_

x Rand the

Lipschitz condition (2) holds in S_, then the initial value problem

y'=f(x,y)

^{for (1_)}

has exactly one solution in J_.

This result can be proved by

(d) Reflection about the Line x = e. Weintroduce the functions := y(2e—

x),

f(x,y) :=

—f(2e—x,y) to transform the problem (1_) in J_ into the initial value problem in J

for (1k)

§ 6. An Existence and Uniqueness Theorem 65

Clearly, f satisfies the hypotheses of Theorem I. In addition, one can see at once that ^i—p — x) defines a bijective mapping of C(J_) onto C(J) that maps solutions of (1_) into solutions of (1+) (and conversely). The conclusion then follows from Theorem I.

(e) We note that an alternative approach is to carry the original proof directly over to the present case. Existence to the left and to the right can both be proved using the norm

max

(equation (3) holds in both cases).

Frequently, f is not defined in the whole strip, but only in a neighborhood of the point Thefollowing result deals with this situation.

III.

Theorem. Let R be the rectangle x e+a, b (a, b> 0) and let f E C(R) satisfy a Lipschitz condition (2) in R. Then there exists exactly one solution to the initial value problem (1). The solution exists (at least) in an interval

x

^{+ a, where}

( b\

with A=maxlfl.

A corresponding statement holds for ^— a x with the rectangle R lying to the left of the point

For the proof, we extend f continuously to the strip x + a,

<y

<oo; for example, by setting

f(x,'q—b) for

y<ri—b, J(x,y)= ^f(x,y)

^{in R,} -

for

y>'q+b.

The function J is clearly continuous in the strip and satisfies the same Lipschitz condition (2) as f with the same Lipschitz constant. By Theorem I, there exists exactly one solution y(x) of the initial value problem with right-hand side f.

As long as this solution remains in R, it is also a solution of the original initial value problem. Since If I A, we have Iy'I ( A; i.e., the solution remains in the angular region formed by the two lines through the point with slopes ±A (see the figure). Hence the solution does not leave R as long as x

+ a,

where a is the smaller of the two numbers a and b/A.

I

Remark. We sketch another proof that does not require a continuous exten-

sion of f outside of R. Let J'

+ a]. One considers the Banach space

B =

C(J') and the operator T as defined in (3') on the subset D of all B with _— b. In order to apply the fixed point theorem 5.IX, one has to show that D is closed, T(D) C D, and T is a contraction (proof as in Theorem I). The details of this proof are recommended as an exercise.

-

^U

-

IV.

Local Lipschitz Condition.

Definition. The function f(x,y) is said to satisfy a local Lipschitz condition with respect to y in D C if for every (x0,y0) D there exists a neighborhood U = U(xo,yo) and an L = L(xo,yo) such that in U fl D the function f satisfies the Lipschitz condition

(2)

Criterion. If D is open and if f e C(D) has a continuous derivative

f

in this set.

— Suppose, namely, that U is a circular neighborhood of (x0, Yo) E D with U C D; then is bounded in U, say _çL, and for (x,y), (x,y) E U, the relation

f(x,y)

^— f(x,y) ₌(y— y_)fy(x,y*) with y*

E

follows from the mean value theorem. Hence, (2) holds.

I

A local Lipschitz condition is a weak requirement compared to the global Lipschitz condition in Theorem I. For instance, f(x,y) =y2 satisfies

f(x,y)

^—f(x,y)I

I

satisfies a local Lipschitz condition in R2 (or in the strip J x IR) but does not satisfy a Lipschitz condition in this set.

Theorem on Local Solvability. If D is open and f E C(D) satisfies a lo- cal Lipschitz condition in D, then the initial value problem (1) is locally uniquely solvable for E D; i.e., there is a neighborhood I such that exactly one solution exists in I.

This follows iinmedliately from Theorem III. A rectangle like the one that appears in iii is constructed to the right of the point If the rectangle is chosen small enough, then a Lipschitz condition holds in this rectangle and Theorem III applies. A corresponding argument holds to the left. ^U

Our next objective is to derive some global statements on the unique exten- sion of these local solutions. We begin with a lemma that looks awkward at first sight. It wifi later be used in different situations in connection with global existence.

§ 6. An Existence and Uniqueness Theorem 67 V. Lemma. Let f be defined in D and let = (A be a set of functions, where is a solution to the initial value problem (1) in the interval Ja containing Assume that the following property (U) holds:

for

(a,/3EA).

(U)

Then there exists exactly one solution j defined in the interval J = U _Ja with

czEA

the property cb for every a E A.

This solution can be constructed as follows: For each x e J determine an a E A such that x E and then define = If /3 is another index with x E Jp, then by hypothesis = i.e., q5(x) is well-defined.

If a is an arbitrary point from J, then there exists an a E A such that

a E Ja. Thus a] C Jo, and qS(x) q5o,(x) holds in a]. It follows that is a solution of (1) in J.

If this lemma is applied to the set of all solutions to the initial value problem, then (U) implies the uniqueness of the solution. In summary, one obtains the following

Corollary. If the initial value problem (1) has at least one solution and if the uniqueness statement (U) holds for every pair of solutions, then there exists a solution of (1) which cannot be extended. All other solutions are restrictions

of this solution.