Existence Theorems for Ordinary Differential Equations

CHAPTER 1

The Fundamental Existence Theorems

1. The basic existence theorem

1.1 Let y be an unknown function of x. Frequently our information on y is its initial value y0 for x = x0 and its rate of change dy/dx. If our information on dy/dx is that dy/dx is a continuous function f(x), then we know that there is a unique y, which may be obtained by integration. If, on the other hand, as often happens, our information specifies dy/dx in terms of y as well as x, that is, dy/dx = f(y, x)¹ then even if f depends continuously on y and x the situation is not clear. For this equation on dy/dx does not yield prima facie evidence of the existence of y, since to evaluate f, y must be known, that is, y must exist.

Clearly then the existence of y must be established by some building-up process. Since there is an interval (x0, x0 + h) such that y is well approximated by y0 + f(y0, x0). (x — x0). If x1 = x0 + h and y1 = y0 + f(y0, x0) h we can take a second interval (x1, x1 + h) in which y is approximated by y1 + f(y1, x1) (x — x1). Continuation of this process will yield a polygonal line with slope in the form f(yi, xi) except at the vertices where in general there is not a unique slope. Intuitively, we would expect that if h is taken smaller and smaller the corresponding polygonal lines would converge to a solution y of the differential equation dy/dx = f(y, x).

However, the difficulties of the situation appear as soon as we try to make this procedure precise. Question one is, how do we know these lines will converge ? Secondly, if they do, will the resulting function have a derivative ? (The approximants in general will fail to have derivatives at more and more points as we take h smaller and smaller.) Thirdly, if the limiting function exists and has a derivative, will it satisfy the given differential equation ?

Relative to the first question one can show that if f is continuous one can choose a sequence of h’s approaching zero such that the polygonal lines will converge for some x interval with lower end point x0. However, in general, the size of this interval may be very small and has to be investigated in each individual case.

For questions two and three one has available certain technical devices. One transforms the problem from one in differential equations to an equivalent integral equation

Then if our approximants converge uniformly we obtain the desired results.

Precisely formulated,² the fundamental existence theorem reads as follows:

Theorem 1. Let f(y, x) be a real valued function of the two real variables y, x defined and continuous² on an open region of two dimensional euclidean space. Then for every point (y0, x0) of we can find a b > 0 and a function φ(x) which has a continuous first derivative in the neighborhood , | x — x0 | b such that

and φ(x0) = y0 in this neighborhood .

We shall prove Theorem 1 in Section 1.8. Note that the theorem has been stated for the case of one unknown function y. This will be generalized in later sections to n functions, implicit forms and higher derivatives. However, in order to appreciate clearly the ideas underlying the theory, we shall carry through the initial discussion for the one unknown function y.

It is important to note that Theorem 1 is an existence theorem and not a uniqueness theorem. Under the hypothesis of the theorem, cases exist in which there exist two or more distinct functions satisfying the conclusions of the theorem. Examples will be given later (cf. Section 1.1 of Chapter 3) to illustrate this phenomenon.

1.2 We shall unfold the theory as a logical entity. As we attempt to prove it, it will appear that certain auxiliary lemmas are necessary. These we shall state and prove only after we have seen the necessity for such a digression.

We are concerned with the differential equation

where f(y, x) satisfies the conditions of Theorem 1. Suppose b > 0 is assigned and let be the set of x’s with | x — x0 | b. Suppose further that there exists a function y(x) defined for x ε which satisfies Equation (1) and at x = x0 assumes the value y0. Of necessity, dy/dx must exist and hence y(x) is continuous. Since f is continuous, f(y(x), x) is a continuous function of x in the neighborhood of x = x0. Hence, since dy/dx = f(y(x), x), dy/dx is continuous, and consequently

for x ε . Equation (1) therefore implies

On the other hand, a function y that satisfies Equation (2) must be a continuous function of x. For, if y satisfies Equation (2) it implies that f(y (t), t) is Riemann integrable in and hence y0 + f(y(t), t) dt is continuous. But this expression is precisely y. Now since y is continuous, and f is continuous by hypothesis, f(y(t), t) is continuous for t ε . Thus the derivative of the integral on the right hand side of Equation (2) exists and equals the integrand. Consequently, differentiating Equation (2) will yield Equation (1).

We have thus proved the following lemmas:

Lemma 1. If there exists a function φ(x) defined for x ε which satisfies Equation (1), then dφ/dx exists and is continuous.

Lemma 2. The solutions of Equation (1) which at x0 equal y0 are identical with the solutions of Equation (2).

The study of the differential equation, Equation (1), has thus been reduced to the study of the integral equation, Equation (2). As stated above, this will permit us eventually to answer questions two and three.

1.3 In the statement of Theorem 1, a neighborhood of x0, namely | x — x0 b appeared in the conclusion. The construction of this neighborhood as well as a discussion of its significance will now be given.

Since is an open region, there exists a square of side 2a with center at (x0, y0) such that the entire square including its boundaries is in the region . This square, consisting of the points (y, x) such that | x — x0 | a, |y — y0 | a will be called . (Cf. Figure 1.) Since is a closed point set interior to and f(y, x) is continuous on , f is bounded on . That is, there exists a constant M such that

FIGURE 1

Now let

and consider the set of points | x — x0 | b, | y — y0 | a. We shall call this closed rectangular subregion of , ′. Clearly,

 (i) f(y, x) is continuous in ’ and | f(y, x) | M in ′.

(ii) If | x — x0 | b and | y — y0 | M | x — x0 |, then (y, x) ε ′. To prove (i) we merely note that . To demonstrate (ii), we note that

As will be seen later, the reason for introducing b is to make (ii) a true statement.

Our next task is to construct a polygonal line function which is intended to approximate a solution to Equation (1). Let h > 0 be given. We shall suppose that h is small relative to b. Furthermore, it is convenient to assume that b is an integral multiple of h, that is,

where p is a positive integer.

We define inductively a sequence of pairs of points y1, x1; … ; yp, xp = x0 + b by means of the relations

In order to make this definition valid we must prove that the f(yj, xj) exist. By hypothesis it exists for j = 0, and furthermore | f(y0, x0) | M. Now suppose f(y0, x0), …, f(yj-1, xj-1) exist and are less than M in absolute value. Then since

we have

But by (ii) above this means that (yj, xj) is in ′ and thus f(yj, xj) exists, and by (i), | f(yj, xj) | M.

We now define a function y(x, h) by the equation

for

and

The quantity y(x0, h) is defined as

Note that the x0 and y0 are the initial point of Theorem 1.

The function y(x, h) has thus been defined for all x in the interval x0 x x0 + b. It is a continuous function of x consisting of a finite number of straight line segments. A similar definition of y(x, h) can be given for the interval x0 — b x < x0 by letting y(x, h) equal yj+1 + f(yj+1, xj+1) (x — xj+1) for — h x — xj+1 < 0 where yj = y(xj, h), xj = x0 + jh and j = — 1, — 2, …, — p.

Now in the region x0 x x0 + b define the function F(x, h) by the equation

for 0 < x — xj-1 h, j = 1, 2, …, p and in the region x0 — b x < x0 let F(x, h) be defined by

for — h x — xj+1 < 0, j = — 1, — 2, …, — p.

From Equation (3) we conclude that

Also

while

From these equations and Equations (4) and (5) we see that

As mentioned above, y(x, h) is a continuous function of x. Also, we have shown above that (yj, xj) ε ′ and consequently

But from this result and Equation (6),

which by virtue of (ii) implies

for all x, | x — x0 | b.

Since F(t, h) is a step function with a finite number of jumps, Equation (6) defines a continuous function of x and we may write

where, by definition,

1.4 We recall that since is compact and f(y, x) is continuous on , f is uniformly continuous. Hence given an ε > 0 there exists a δ > 0 such that | f(y′, x′) — f(y″, x″) | < ε for all pairs of points (x′, y′), (x″, y″) in with | x′ — x″ | < δ and |y′ — y″ | < δ. We shall refer to this δ = δ(ε) as the "uniform δ for ε" — referring of course to the function f(y, x) in .

Lemma 3. Let ε > 0 be assigned. Let δ > 0 be the uniform δ for ε/b and let h0 = min (δ, δ/M). Then if h < h0 and | x