The aim of existence theory is to specify conditions under which one can be sure that there is a solution to a differential equation such as
(5.8.1)
There is no point in wasting analytical and computational effort on trying to find a solution when there is not one. Basically, there are two ways of demonstrating existence, non-constructive in which no attempt is made to show how one might arrive at a solution, and constructive in which a method for building up the solution is described. We shall consider only a constructive approach, one that lays the foundation for a numerical attack when that is desired.
The initial value problem for (5.8.1) seeks a solution such that y=y0 at t=t0 and this is the problem that we shall discuss in some detail when y0 is a prescribed constant.
The purpose of the analysis is to show that, under specified conditions, when t does not stray too far from t0 there is a solution y which does not differ from y0 by more than a certain amount. So we consider what happens
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FIGURE 5.8.1: The domain of existence.
as t ranges from t0 to t0+h where h is positive (similar considerations apply when h is negative). In this range we are prepared to consider deviations of y from y0 of magnitude k, i.e., we expect y to lie between y0−k and y0+k. The points that originate values of f(t, y) are then in the domain D of Figure 5.8.1. Suppose that f is bounded in D, say |f|<M; then we shall impose the restriction h<k/M. This can always be arranged by reducing h if necessary. The constraint is an expression of our expectation that the more t departs from t0 the more y will deviate from y0. Further conditions are placed on f in the following theorem.
EXISTENCE THEOREM I
Let f(t, y) be a single-valued continuous function of t and y in D such that
K being a finite constant, for any pair of points (t, y) and (t, y′) in D. Then, for h<k/M, the differential equation (5.8.1) possesses one and only one continuous solution y(t) in t0≤t≤t0+h such that y(t0)=y0.
PROOF The proof starts by observing that the problem is equivalent to showing that
(5.8.2)
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has a solution. For, since f is bounded, the integral exists and tends to zero as t→t0 to with the consequence that y(t0)=y0. Also, a derivative with respect to t of (5.8.2) returns, because of the assumed continuity of f, to (5.8.1). So it is sufficient to discuss (5.8.2).
Now solve (5.8.2) by iteration by making a series of approximations. First put y(u)=y0 in the integral to generate y1(t) given by
(5.8.3) Now produce the sequence yn(t) defined by
(5.8.4)
In each approximation the right-hand side can be calculated and a practical mechanism of solution has been erected, provided that the iteration converges to a solution.
Note firstly that (5.8.3) implies that y1(t) is a continuous function of t and, since (u, y0) is in D for t<t0+h, thus (t, y1(t)) is in D for t<t0+h. The reasoning can now be repeated to show that y2(t) is a continuous function of t such that (t, y2(t)) is in D for t<t0+h. It is then clear from (5.8.4) that yn(t) is a continuous function of t such that (t, yn(t)) is in D for every n while t<t0+h.
Suppose now that
(5.8.5) for t0≤t≤t0+h, a result already proved for n=1. Then, from (5.8.4),
by the Lipschitz condition. Invoking our hypothesis, we have
which is the same as (5.8.5) except that n is replaced by n+1. Since it is true for n=1 it follows by induction that (5.8.5) holds for every n.
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which reveals that the series on the left is absolutely and uniformly convergent (by the Weierstrass M-test) in t0≤t≤ t0+h. But the sum of a uniformly convergent series of continuous functions is itself continuous and, since
it follows that limn→∞yn(t) exists and is a continuous function in t0≤ t≤t0+h. It remains to identify as a solution of (5.8.2). Now
By the Lipschitz condition, the magnitude of the last integral does not exceed which tends to zero as n→∞. Equation (5.8.2) has been recovered.
The existence of a continuous solution has now been verified and to complete the theorem it is necessary to show there is no other. Suppose, in fact, there were another solution Y(t) such that Y(t0)=y0, which is continuous in t0≤t≤t0+H with H≤h and |Y(t)−y0|<k. Then
Hence, if
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and so the result for n=1 holds. Letting n→∞ in the inequality we have and uniqueness has been established.
(b) System of first-order equations
The theory of the preceding section generalises to the system
(5.8.6)
under the initial conditions ym=ym0 at t=t0. The region D is not so simple to depict since it is a rectangular parallelepiped in space of n+1 dimensions, because each ym may change by a different amount from its initial value as t moves from t0. So D is defined by t0≤t≤t0+h, |ym−ym0|≤km (m= 1,…,n). Analogous to the conditions for a single equation the restrictions |fm|<M and h<km/M for m=1, 2,… are imposed.
EXISTENCE THEOREM II
Let fm(t, y1,…, yn) be single-valued continuous functions of t, y1,…, ynin D such that for m=1,…, n (a) |fm(t, y1,…, yn)|<M in D,
(b) (Lipschitz condition)
K1,…, Knbeing finite constants, for any (t, y1,…, yn) and in D. Then, for h<km/M (m=1,…, n) the system (5.8.6) possesses one and only one set of continuous solutions y1(t)…, yn(t) in t0≤t≤t0+h such that ym(t0) =ym0(m=1,…, n).
PROOF The method of proof runs parallel to that for a single equation. It begins with
and an iteration is performed according to
In view of the similarity to the single equation. details will be omitted.
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(c) Differential equation of order n
The above theory can be applied to the n differential equation
(5.8.7) with the initial conditions
at t=t0. Make the substitutions The system
is obtained. When this is compared with (5.8.6) we see that
The fm for m=1,…,n−1 obviously satisfy the conditions stated in Existence Theorem II. Therefore, if we make fn comply with these conditions, that theorem is available for (5.8.7). Accordingly, we have the next theorem. EXISTENCE THEOREM III
If f(t, y1,…,yn) is continuous and
the differential equation (5.8.7) has one and only one continuous solution y(t) such that dy/dt,…,dn−1y/dtn −1are continuous for t0≤t≤t0+h and such that y, dy/dt,…,dn−1y/dtn−1take given values at t=t0.
It may be remarked that, if (5.8.7) is linear,
of t where one or more of g, a0,…,an−1 are not continuous. Hence, a linear differential equation has a unique continuous
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solution to the initial value problem, provided that t0 is not a point where there is lack of continuity on the part of g, a0,…,an−1.