0387986995Basic TheoryC

328X. THE SECOND-ORDER DIFFERENTIAL EQUATION

(iii)the function Y(x) is a solution of the differential equation

satisfying the conditions y(O) = Y(0) and ly'(0) - Y'(0)l l< p. Then,

ly(x) - Y(x)I < {h, + a(L + k)}exp[ Lx]

on the interval 0 < x < if c and A are sufficiently small.

Proof.

We prove this theorem in four steps.

Step 1. Setting

Do = {(x, u, v) : 0 < x < t, Jul <- a(x), lv) < pe-`= + b(x)}.

Also,

Step 2. Suppose that two functions wl (x, .A) and w2(x, A) satisfy the following conditions:

(X.7.3)

10 < wl(x,.A) < a(x), 0 < w2(x,A) < pe-' + b(x),

wi (x, A) > W&, A), Aw2(x, A) > -Lw2(x, A) + Kwl (x, A) + (E + AM)

on the interval 0 < x < e. Then, as long as (x, u, v) is in the region

Dl = {(x,u,v): 0 <_ x < e, IuI < wi(x,A), IvI < w2(x,A)},

it holds that

IvI < wi (x, A),

.F(x, u, w2(x, A), A) < Aw2(x, )1),

F(x. u, -w2(x, A), .A) > -aw2(x, I\).

Look at the right-hand side of (X.7.1) on the boundary of V. Then, it can be easily seen that if a solution of (X.7.1) starts from Dl, it will stay in Dl on the interval 0 < x < e.

Step 3. Show that two functions

and a positive constant 62 are sufficiently small. Observe that two roots of AX2 +

LX -K=0 are -L -(o and (o = L +O(A).

Step 4. Note that

wl (x, A) = (L / (1 - e-Lx/A) + bK I (eKx/L - 1) + 82(x + 1).

To complete the proof, look at Jul < wl (x, A) as 62 -4 0. 0

Note that

X-8. A singular perturbation problem

In this section, we look at behavior of solutions of the van der Pol equation

(X.4.1) as a --+ +oo more closely. Setting t = Er and A = c2, let us change (X.4.1) to

Solving (X.8.2) with an initial value x(O) = xo < -1, we obtain x = 0(r), where

2 )2 - In I0(r) I = -r + 2 - In(-xo).

Observe that

2

Note also that setting ro = 2 - In(-xo) -- 2 > 0, we obtain ,(ro) = -1 and

45(7-)2 - 1 > 0 for 0:5 r < ro. The graph of 0(r) is given in Figure 22.

FIGURE 22.

(i) Behavior for 0 < r < 7-o -bo, where do > 0: Let us denote by x(r, A) the solution to the initial-value problem

where the prime is d7and q is a fixed constant. Using Theorem X-7-1 (Na6J, we

derive

Ix(r, A) - O(T)I 5 AK,

I Ix'(r, A) - 4'(r)l 5 I7-7 - 0'(0)le-P'1x + AK,

for 0 < T < To - 60, where K and a are suitable positive constants.

where A0 is a sufficiently small positive number. It is important to notice that if we make 60 small and if we make A0 also small accordingly, we can make p1 small.

Set

This is a contradiction if al and A are small. Therefore, if Ao > 0 is sufficiently small, there exists an x(A) such that

It can be shown that Urn x(A) = 2.

ao*

Setting r(x(A)) = r(A), it can be shown that liM +r(A) = ro. Now again, apply

Theorem X-7-1 (Na6) to the initial-value problem

Figure 23 shows the general behavior of x(r, A) for small positive A.

TO

FIGURE 23.

has a solution for any real numbers A and B.

Hint. A counterexample for (2) is

and 3 are sufficiently large positive constants.

X-2. Find the global phase portrait of each of the following two differential equa- tions:

X-3. Suppose that

(i)f (x, y, t) is continuously differentiable everywhere in the (x, y, t)-space (i.e., in p3),

(ii)g(x) is continuously differentiable in -oo < x < +oo,

(iii)e(t) is continuous on 0 < t < +oo,

(iv)f (x, y, t) 0 for x2 + y2 > r2 and t > 0, where r is a positive number r,

(vi) 1 je(r)Idr is bounded for 0 < t < +oo.

0 `

Show that every solution (x(t), y(t)) of the system

is well defined and bounded for 0 < t < +oo.

Hint. See [Sat].

The main ideas are as follows:

(1)If there are more than one nontrivial periodic orbits, then at least one of them should not be orbitally asymptotically stable.

(2)For a nontrivial periodic orbit (x(t), x'(t)) of period T > 0, the inequality fT

f (x(t))dt > 0 implies that this orbit is orbitally asymptotically stable

o

(cf. Exercise IX-5).

(3) Set

Now, investigate the quantity (x'(t))2 as a function of A along the periodic orbit

(x(t), x'(t)). For a fixed value of A, compare (x'(t))2 for different values of x(t), using the assumptions on f and g.

Step 1. First the following remarks are very important.

(a) If we set y = x', the given differential equation is reduced to the system

A careful observation shows that the index of the critical point (0, 0) is 1, while the index of the critical point (-2,0) is -1. There are only two critical points of (S).

(c)The periodic orbit and any line {x = a constant) intersect each other at most twice.

(d)The critical point (-2,0) should not be contained in the domain bounded by the periodic orbit. To show this, use the fact that the index of any periodic orbit is 1.

These facts imply that the periodic orbit should be confined in the half-plane x >

-2.

totally confined in the strip jxj < 1. In particular, (1, 0) and (-1, 0) are on the same

jx(t)I < 1, the periodic orbit cannot intersect the level curve V(x,y) = / ,

0

Therefore, the periodic orbit must intersect the lines x = 1 (as well as the line x = -1) twice. Denote these four points by (1,s1(A)), (1,-rt(B)), (-1,-q(C)),

and (-1, 77(D)), where r)(A), n(B), q(C), and q(D) are positive numbers.

Z(B) = { A : AB < A < min(AA, )'C) },

Z(C) = { A : Ac > A > max(AB, AD) },

Z(D) = { A: AD < A < min(AA, Ac) }.

Note that the interval lo = JA: min(AA, Ac) > A > max(AB, AD)} is contained in

Z(A) n Z(B) n Z(C) n I(D).

Now,

(a)on the arc between (1, rl(A)) and (1, -rl(B)) of the periodic orbit, regard y as a function of A and denote it by yi (A), where AB < A 5 AA,

(Q) on the arc between (1, -rl(B)) and (-1, -r1(C)) of the periodic orbit, regard y as a function of A and denote it by y2 (A), where AB < A < AC,

(ry) on the are between (-1, -rl(C)) and (-1, rl(D)) of the periodic orbit, regard y as a function of A and denote it by yi (A), where AD < A < AC,

(6) on the arc between (-1, rl(D)) and (1, rl(A)) of the periodic orbit, regard y as a function of A and denote it by yz (A), where AD :5 A < AA.

Then, it can be shown that

Step 4. Now, fixing a AO E Z(A) n Z(B) n 1(C) n T(D), evaluate the integral (I) as follows:

yi (A)2 = - JaD yi (A)2 - ao yi

(A)2.