Sharp conditions for rapid nonlinear oscillations

Nonlinear Analysis 39 (2000) 519 – 533

www.elsevier.nl/locate/na

Sharp conditions for rapid nonlinear oscillations Yu. A. Klokov, F.Zh. Sadyrbaev ∗ Institute of Mathematics and Computer Science, University of Latvia, LV-1459 Riga, Latvia Received 11 January 1998; accepted 1 February 1998

Keywords: Nonlinear equations; Superlinearity; Multiplicity of solutions; Sharp conditions

1. Introduction We consider the equation x00 + g(x) = f(t; x; x0 );

(1)

where g : R → R and f : [a; b] × R × R → R are C 1 -functions, xg(x) ¿ 0 for | x| large. A nonlinearity g(x) is called usually superlinear if g(x) → +∞ x

as | x| → ∞

(2)

and associated boundary-value problems are referred to as superlinear also. Superlinear BVP were considered, for instance, in [1, 2, 4–7, 10–13, 16]. A survey before the year 1983 could be found in [11]. The semi-superlinear case (which is characterized by nonlinearities g(x) exhibiting superlinear behaviour only for values of x of de nite sign, such as exp(x)) was considered by Ruf and Solimini [14] and Ruf and Srikanth [15]. Superlinear second-order problems with separated boundary conditions are often known to have in nite number of solutions. This fact can be easily explained geometrically. Consider the BVP x00 + x3 = 0;

(3)

x(0) = 0;

(4)

∗ Corresponding

x(T ) = 0: author. E-mail: [email protected]

0362-546X/99/$ - see front matter ? 1999 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 8 ) 0 0 2 1 8 - 1

520

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

Solutions of Eq. (3) with initial values x0 (0) = 

x(0) = 0;

(5)

do not have zeros in the interval (0; T ) for small enough . On the other hand, the number of zeros in the same interval tends to in nity as  → ∞. Since zeros are continuous functions of , for any N there exists a solution of Eqs. (3) and (5) with exactly N zeros in (0; T ). It is shown in [4], for instance, that Eq. (1) with Sturm–Liouville boundary conditions a1 x(0) + a2 x0 (0) = A; b1 x(T ) + b2 x0 (T ) = B;

(6)

a21 + a22 ¿ 0; b21 + b22 ¿ 0 has in nite number of solutions if g(x) is superlinear (in the sense of the condition (2)) and |f(t; x; y)| ¡ c1 |x| + c2 |y| + c3

(7)

for any (t; x; y) considered. On the other hand, Eq. (1) may not have the above-mentioned property for nonlinear f(t; x; y) growing quadratically with respect to y even if g(x) grows exponentially. As an instance, consider the equation x00 + g(x) = x02 ;

(8)

where g(x) is of odd type and g(x) = e(2−) x for x ¿
, where
is a large number. Rewriting Eq. (8) as y2 − g(x) dy = ; dx y

(9)

where y = x0 , one gets for z = y2 that dz = 2z − 2g(x): dx

(10)

The solution of Eq. (10), satisfying the initial condition z(x0 ) = 0, has the form z(x) = 2 e2x ·

Z x0 x

e−2x g(x) d x:

Noting that  z(x(t)) =

dx dt

2 ;

(11)

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

521

one obtains that time T (x0 ) needed for a solution x(t) of Eq. (8) to decrease from x = x0 ; x0 = 0 to zero value satis es the inequality Z x0 dx q : (12) T (x0 ) ¿ T (x0 ;
) = R x0
2e2x · x e−2s · g(s) ds Integrating Eq. (12) one gets r Z x0  dx p · T (x0 ) ¿ T (x0 ;
) ≥ 2x −x 2
e (e − e−x0 ) r Z x0  dx √ · ≥ 2
e(2−)x √
2 e(−2)=2
− e(−2)=2 x0 : = 2−

(13)

The last expression tends to a nite limit as x0 → ∞. Hence, T (x0 ) is separated from zero as x0 → +∞. Remark. One can show, however, that if in Eq. (8) |g(x)| ≥ e2x for large values of x then the number of zeros in a nite interval of a solution with the maximal value x0 tends to in nity as x0 → ∞. Thus, the growth rate of g(x) and the growth rate of f(t; x; x0 ) with respect to x0 at in nity should correlate. The elucidation of this interdependence is the main goal of this paper. 2. S-property We say that a nonlinear equation x00 + g(x) = 0

(14)

with g(x) of class C 1 and of odd type (xg(x) ¿ 0 for |x| large) has the S-property in the interval [a; b] if the number of zeros of a solution x(t), satisfying the initial conditions x(a) = x0 ;

x0 (a) = x00 ;

(15)

increases unboundedly if kx0 k := |x0 | + |x00 | → ∞: Remark. Eq. (14) is then easily shown to have an in nite number of essentially different solutions satisfying the boundary conditions (6). Solutions are distinguished by a number of zeros. We say that the perturbed Eq. (1) possesses the S-property in a ÿnite interval (a; b) if: 1. all solutions extend to [a; b];

522

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

2. for any N there exists
such that the number of zeros in [a; b] of any solution x(t) of the initial-value problem (1) and (15) is greater than N , if kx0 k ¿
. The following proposition explains our interest in the S-property. Proposition. Eq. (1) possessing the S-property has inÿnite number of solutions satisfying Sturm–Liouville boundary conditions (6). The proof is based on geometrical considerations involving the integral funnel of solutions of Eq. (1), satisfying the former of the boundary conditions (6) and the general scheme is described, for instance, in [9]. 3. Preliminary results We start with the de nition of a separatrix for the rst-order equation dz = a(x) (f(z) − b(x)); dx

(16)

where a(x) and b(x) are positive valued continuous functions, f(z) is a continuous monotonically increasing function. A zero isocline i(x) is de ned by the equation f(i(x)) = b(x): Denote by z(x; x0 ) a solution of Eq. (16), satisfying the initial condition z(x0 ) = i(x0 ):

(17)

Now let x be xed and x0 tend to +∞. A solution z(x; x0 ) is increasing for x ¡ x0 and decreasing for x ¿ x0 . Suppose that there exists a nite limit z ∗ (x) = limx0 →+∞ z(x; x0 ):

(18)

We will call z ∗ (x) a separatrix. In the sequel the rst-order equation dz = 2cz =( +1) − 2k 2 x dx

(19)

will play an important role. Lemma 3.1. A separatrix for Eq. (19) does not exist if   2 c c 2 : k ¿ c∗ :=

+ 1 ( + 1)2 Proof. A zero isocline for Eq. (19) is given by  2  =( +1) k x +1 : i(x) = c

(20)

(21)

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

523

Eq. (19) is reduced to x

du = −(2k 2 + ( + 1)u − 2cu =( +1) dx

(22)

by the variable change z = x +1 u(x);

u ≥ 0:

(23)

Suppose that there exists a separatrix s(x). Then all solutions z(x; x0 ) of the initialvalue problem (19) and (17) are bounded from above at the point x, and the ratio z(x; x0 )=i(x) is also bounded along with s(x)=i(x): Consider solutions z(x; x0 ) of Eqs. (19) and (17) or, respectively, solutions u(x; x0 ) of Eq. (22), satisfying the initial condition  2  =( +1) k : (24) u(x0 ; x0 ) = c We have, using Eqs. (21) and (23), that z(x; x0 )  c  =( +1) = u(x; x0 ): i(x) k2 Let x be xed. We will show that u(x; x0 ) is not bounded from above as x0 → +∞. Integrating Eq. (22) we obtain Z u(x;x0 ) x0 ds = ln : (25) 2

=( +1) x u(x0 ;x0 ) 2k + ( + 1)s − 2cs The equation with respect to D 2k 2 + ( + 1)D = 2cD =( +1) :

(26)

does not have real solutions, if k 2 ¿ c∗ : Under the integral sign, the denominator is positive. Let x0 → +∞. Then the right-hand side tends to in nity and hence u(x; x0 ) → +∞ also. This means that solutions z(x; x0 ) are unbounded from above and a separatrix s(x) does not exist. Consider the equation x00 + k 2 |x| sign x = c(x02 )p sign x:

(27)

Since the right-hand side of Eq. (27) is discontinuous with respect to x, orbits on the (x; x0 )-plane may not be smooth at the points of intersection with the x0 -axis. We prove the following assertions with respect to Eq. (27). Lemma 3.2. Eq. (27) does not have oscillatory solutions if p = =( + 1) and k 2 ≤ c∗ ;

(28)

524

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

where c∗ is as in Eq. (20). Eq. (27) possesses the S-property if p = =( + 1) and k 2 ¿ c∗ . All nontrivial solutions are periodic and the period of a solution with the initial values x(a) = x0 ;

x0 (a) = 0

(29)

is given by the formula T (x0 ) = x0(1− )=2 T (1):

(30)

Proof. Rewrite Eq. (27) as x0 = y; y0 = c(y2 ) =( +1) − k 2 x

(31)

and c(y2 ) =( +1) − k 2 x dy = : dx y

(32)

Then the new variable z = y2 satis es the equation dz = 2cz =( +1) − 2k 2 x : dx

(33)

The function z = Dx +1 solves this equation, where D can be found from 2k 2 + ( + 1)D = 2cD =( +1) :

(34)

We have three possibilities for nite Eq. (34): k 2 ¡ c∗ implies that there exist two solutions D2 ¿ D1 ¿ 0; k 2 = c∗ implies the existence of exactly one solution D1 ¿ 0; k 2 ¿ c∗ implies that Eq. (34) does not have real solutions. In the rst case, Eq. (27) does not have oscillatory solutions since the respective solutions of Eq. (33) cannot intersect integral curves z = D1 x +1 and z = D2 x +1 : By the same reason the second case is also excluded. Consider the case of k 2 ¿ c∗ . Then the zero isocline is given by Eq. (21). Consider the solution z(x) of Eq. (27) which satis es the initial condition z(x0 ) = z0 : We will show that z(0) ¿ 0: Suppose that z(0) = 0 (the case z(0) ¡ 0 is excluded, in view of the form of the zero isocline). Eq. (27) is reduced to Eq. (22) by the variable change given by the Eq. (23). One has by integrating Eq. (22) that Z

u(x) u0

(2k 2

x0 ds = ln :

=( +1) + ( + 1)s − 2cs ) x

(35)

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

525

The above equality implies that u(x) → +∞ as x → 0 if z(x) → 0 as x → 0. Rewrite Eq. (35) as Z

u(x)

u0

ds + ( + 1)s

Z

u(x)

u0

x0 (−2k 2 + 2cs =( +1) ) d s = ln : x ( + 1)s(2k 2 + ( + 1)s − 2cs =( +1) )

(36)

The second integral tends to a nite limit L as u(x) tends to +∞: We get from Eq. (36), making use of Eq. (23), that 1=( +1) x0 z(x)  x0  +1 · + L + (x) = ln ; ln z(x0 ) x x 

(37)

where (x) → 0 as x → 0: Passing to the limit in Eq. (37) as x → 0 one deduces that ln(z(0)=z(x0 ))1=( +1) + L = 0 or, nally, z(0) = z(x0 ) e−L( +1) ¿ 0. Taking into account the result of Lemma 3.1 and the symmetry Eq. (27) with respect to x and x0 , one concludes that the (x; x0 )-plane is lled with orbits representing periodic solutions of Eq. (27). To compute periods, let x(t) be a solution of Eq. (27), de ned by the initial data x(a) = 1; x0 (a) = 0 and let T (1) stand for its period. Then easy computations show that the function z(t) = x0 · x(x0( −1)=2 t) is a solution of Eq. (27) with the initial values Eq. (29) and its period is T (x0 ) =

T (1) x0( −1)=2

:

4. Main results Theorem 4.1. Suppose that all solutions of Eq. (1) extend to the interval [a; b] and there exists a continuous function ! : (0; +∞) → (0; +∞) and numbers
¿ 0; k; c ¿ 0 such that: (1) |g(x)| ≥ !(x) + k 2 |x| for |x| ¿
; (2) |f(t; x; y)| ≤ !(x) + c y2 =( +1) for t ∈ [a; b] and x2 + y2 ¿
2 ; (3) k 2 ¿ c∗ ; where c∗ is as in Lemma 3.1. Then Eq. (1) possesses the S-property. First, we prove two lemmas. Lemma 4.1. Let conditions of Theorem 4.1 hold. Then for  ¿ 0 given there exists
such that the solutions of Eq. (1) with the initial values x(t0 ) = x0 ;

x0 (t0 ) = 0;

t0 ∈ [a; b)

(38)

have a zero t1 ∈ (t0 ; t0 + ); if x0 ¿
. A number
can be chosen uniformly with respect to t0 ∈ [a; b − ].

526

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

Proof. Consider a solution x(t) of Eq. (1), satisfying the initial conditions (38). Extendability of the solutions of Eq. (1) and Theorem 15.11 in [9] imply that there exists a function m(x0 ) such that m(x0 ) → ∞ and | x(t)| + |x0 (t)| ¿ m(x0 )

in the interval [a; b]:

(39)

Then one gets, using the conditions (1) and (2) of the Theorem 4.1, that in some interval [t0 ; t
] where x(t) ≥
¿ 0 (
is as in the conditions of Theorem 4.1) x00 (t) + g(x(t)) = x00 (t) + k 2 |x(t)| + !(x(t)) + 1 (t); = !(x(t)) + c x0 (t)2 =( +1) − 2 (t); where 1 (t) ≥ 0 and 2 (t) ≥ 0. Finally, x00 (t) + k 2 |x(t)| = c x0 (t)2 ( +1) − (t);

(40)

where (t) ≥ 0. Since x0 (t0 ) = 0, x(t) is a decreasing function in some right neighbourhood of t = t0 . Let (t0 ; tM ) be the maximal interval in which x(t) decreases and is positive. Then either x(tM ) ¿ 0 and x0 (tM ) = 0, or x(tM ) = 0 and x0 (t) ¡ 0 for any t ∈ (t0 ; tM ). The rst case is ruled out by the following argument. Obviously, tM ≤ t
. Then x(tM ) ≥
and Eq. (40) holds. Let y(t) be a solution of the shortened equation y00 + k 2 |y| = c y02 ( +1)

(41)

subject to initial conditions (38). Since k 2 ¿ c∗ (c∗ is de ned in Eq. (20)), Lemma 3.2 is applicable and it states that Eq. (41) possesses the S-property. Besides, d y (42) d t ¿ c() for a −  ¿ t ≥ t1 ; where t1 is the rst zero of y(t) to the right of t = t0 ,  is small positive number. We will show that dy dx (43) d t (tx ) ≥ d t (ty ) ; where x(tx ) = y(ty ) ∈ [
; x0 ). For this, proceeding like in the proof of Lemma 3.2, one has for new variables z1 = x02 (t) and z2 = y02 (t) the rst-order equations: d z1 = 2cz1 =( +1) − 2k 2 x − E(x); dx

E(x) = (t(x));

d z2 = 2cz2 =( +1) − 2k 2 x ; dx

(44) (45)

where E(x) ≤ 0, t(x) : [xM ; x0 ) → R is the inverse function with respect to x(t): We will show that z1 (x) ≥ z2 (x)

in the interval [
; x0 ):

(46)

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

527

If z1 (x∗ ) ¡ z2 (x∗ ) for some x∗ ∈ [
; x0 ), then z10 (x) ¡ z20 (x) on the whole interval where z1 (x) ¡ z2 (x). This follows from the monotonicity of the right-hand sides of Eqs. (44) and (45) with respect to z. Hence, the dierence z2 (x) − z1 (x) is increasing and z2 (x0 ) ¿ z1 (x0 ). Therefore, Eq. (46) holds along with Eq. (43) and hence, dx d t (
) ¿ 0; which is absurd. Since |x0 (t)| ≥ m(x0 ) −
for values of t, in which 0 ¡ x(t) ¡
, x(t) is monotone and attains zero value if x0 is suciently large. The interval (t0 ; t
) unboundedly decreases as x0 → ∞, as well as the interval (t0 ; T1 ), where T1 is the rst zero of x(t) to the right of t = t0 . Lemma 4.2. Let the conditions of Theorem 4.1 hold. Then given  ¿ 0 there exists
such that any solution of Eq. (1) with the initial values x(t0 ) = x0 ;

x0 (t0 ) = x00 ;

t0 ∈ [a; b)

(47)

satisfying the inequality kx0 k ¿
; has a zero t1 ∈ (t0 ; t0 + ). A number
can be chosen uniformly with respect to t0 ∈ [a; b − ]. Proof. To be speci c, suppose that x0 ≥ 0; x00 ¿ 0: We will show now that x(t) has an extremum (positive maximum) to the right of t0 if kx0 k is large enough. Suppose that x(t) is monotone. Consider Eq. (40) and the comparitively shortened Eq. (41). Choose t2 ∈ (t0 ; b) so that x0 (t2 ) ¿ 0 and a solution y(t) of Eq. (41), satisfying the initial conditions y(t2 ) = ;

y0 (t2 ) = 0;

(48)

vanishes at some t1 ∈ (t0 ; t2 ). This is possible, in view of Lemma 3.2. Graphs of x(t) and y(t) intersect at least twice, at the point t = t2 and at some a1 ∈ (t1 ; t2 ); x0 ¿ 0; y0 ¿ 0; t ∈ [a1 ; t2 ). We will show now that this is impossible. Since the functions x(t) and y(t) are monotone in [a1 ; t2 ), the respective inverse functions t(x) and (x) are well de ned. Their graphs intersect also in the interval [; ], where  = x(a1 ) = y(a1 ) (t() = () = t2 ; t() = () = a1 ; t(x) ¿ (x) for x ∈ (; ); tx0 ; 0x ¿ 0 for x ∈ (; )). One gets the following equations for t(x) and (x): −

00 1 txx + k 2 x = c 02 − (t(x)); 03 tx t

(49)

−

1 00xx + k 2 x = c 02 ; 03  x

(50)

528

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

where = =( + 1). Subtracting Eq. (50) from Eq. (49) we get  0   1 1 1 1 1 − − − (t(x)): = c 02 2 t 02 (x) 02 (x) x t (x) 02 (x)

(51)

Using the intermediate value formula v1 − v2 = ( v −1 )(v1 − v2 );

v1 ≤ v ≤ v2

(52)

and substituting 1=t 02 (x) and 1=02 (x) for v1 and v2 , respectively, we rewrite Eq. (51) as  0   1 1 1 1 1 1− − − − (t(x)); (53) = c z 2 t 02 (x) 02 (x) x t 02 (x) 02 (x) where t 02 (x) ≤ z(x) ≤ 02 (x): Introducing variables u = 1=t 02 (x) − 1=02 (x) and q(x) = z 1− (x), one gets the equation ux0 = 2cq(x)u − 2(t(x)):

(54)

Resolving Eq. (54) and taking into account that u(x∗ ) = 0 for some x∗ ∈ [; ], we obtain Z x e(x)−(s) (t(s)) ds; (55) u(x) = −2 where (x) = u() =

Rx x∗

x∗

2cq(s) ds. It follows from Eq. (55) that u() ≤ 0. On the other hand,

1 1 1 − = 02 = (x0 (t0 ))2 ¿ 0 tx02 () 02 () tx ()

and we arrive at the contradiction. Other cases can be treated similarly. The proof is completed by the application of Lemma 4.1. Proof of the theorem. Let an integer N be given. De ne  = (b − a)=(N + 1). Let
1 be such that a solution of Eq. (1) with kx(a)k ¿
1 vanishes at some point t1 to the right of a and t1 − a ¡ . This is possible by Lemma 4.1. Let
2 be such that a solution of Eq. (1) with |x0 )| ¿
2 changes from an extremal point x0 to zero in time less than a quarter of . This is possible by Lemma 4.2. Finally, let
3 be such that any solution of Eq. (1) with kx(a)k ¿
3 satis es the inequality |x(t)| + |x0 (t)| ¿max{
1 ;
2 } in [a; b]. This is possible in view of the extendability of the solutions of Eq. (1). Then the distance between consecutive zeroes of any solution of Eq. (1) is less than  and the total number of zeros is greater than N .

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

529

Remark. It is well known ([8]) that equations of the type x00 + a(t)x2n+1 = 0 with the coecient a(t) continuous and positive valued, but not of bounded variation, can have solutions, which do not exist on the interval [a; b]. It was shown (Lemma 4 in [4]) that solutions of equations of the type x00 + g(x) = f(t; x; x0 );

(56) 0

where g(x) is superlinear (2) and f(t; x; x ) is sublinear (6), can be extended to the whole interval [a; b]. We give an example in the Appendix showing that Eq. (56), where the rate of growth of f(t; x; x0 ) with respect to x0 is between 1 and 2, generally has solutions of oscillatory type which are not extendable to [a; b]. Corollary 4.1. Suppose that all the solutions of Eq. (1) extend to the interval [a; b]; there exist a continuous function ! : (0; +∞) → (0; +∞) and numbers
¿ 0; k; c ¿ 0 such that: (1) |g(x)| ¿ !(x) + k 2 |x| for |x| ¿
; (2) |f(t; x; y)| ≤ !(x) + c y2p for t ∈ [a; b] and x2 + y2 ¿
2 ; where 0 ¡ p ¡

=( + 1). Then Eq. (1) possesses the S-property. Proof. Rewrite the inequality for f as |f(t; x; y)| ≤ !(x) + c

y2p y =( +1)

· y =( +1) :

(57)

For y large enough |f(t; x; y)| is less than !(x) + c1 y =( +1) , where c1 can be made arbitrarily small and hence   2 c1 c1 2 : k ¿ c1∗ :=

+ 1 ( + 1)2 Then apply Theorem 4.1. Theorem 4.2. Suppose that all the solutions of Eq. (1) extend to the interval [a; b] and the following conditions are fulÿlled (1) xg(x) ¿ 0 for x large; (2) |g(x)| ≤ !(x) + k 2 |x| for large |x|; k – a number; (3) f(t; x; y) ≥ !(x) + c y2 =( +1) for t ∈ [a; b] and large x2 + y2 ; x ¿ 0; or f(t; x; y) ≤ −!(x) − c y2 =( +1) for t ∈ [a; b] and large x2 + y2 ; x ¡ 0; c ¿ 0;   2 c c (4) k 2 ≤ c∗ := + 1 ( + 1)2 . Then the S-property is not fulÿlled for Eq. (1).

530

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

Proof. Consider solutions of the initial-value problem (1) and (38). To be speci c, suppose that x0 is positive. In view of the conditions 1–3 a particular solution x(t; x0 ) of Eqs. (1) and (38) in the interval of positivity satis es the equation x00 (t) + k 2 |x(t)| = c x0 (t)2 =( +1) + (t);

(58)

where (t) ≥ 0. Suppose that Eq. (1) possesses the S-property. Then the distance between two consecutive zeros of x(t; x0 ) tends to zero as x0 tends to in nity. Now, we would like to compare x(t; x0 ) with the respective solution y(t; x0 ) of problem (41) and (38). Changing the roles of Eqs. (58) and (41), one can nd like in the proof of Lemma 3.2 that y(t; x0 ) must vanish in the interval [t0 ; t1 ], where t1 is a zero of x(t; x0 ) to the right of t0 . But this is impossible in view of the condition (4) and Lemma 3.2. Corollary 4.2. Suppose that all the solutions of Eq. (1) extend to the interval [a; b] and: (1) xg(x) ¿ 0 for x large; (2) |g(x)| ≤ !(x) + r |x| for large |x|; r ¿ 0; (3) f(t; x; y) ≥ !(x) + c y2p for t ∈ [a; b] and large x2 + y2 , x ¿ 0; or f(t; x; y) ≤ −!(x) − c y2p for t ∈ [a; b] and large x2 + y2 ; x ¡ 0; where p ¿ =( + 1); !(x) – positive valued continuous function, c ¿ 0. Then the S-property is not fulÿlled for Eq. (1). The proof is similar to that of Corollary 4.1.

Appendix We will show that equations of the type (1) can have nonextendable oscillatory solutions. Consider the equation x00 + k 2 |x| sign x = |x0 |( −1)=( +1) · x0 ;

(59)

where ¿ 1 and   2 1 : k ¿

+ 1 ( + 1)2 2

(60)

Note that the power (2 )=( + 1) in the right-hand side of Eq. (59) is lying in (1; 2). Rewrite Eq. (59) as x0 = y; y0 = |y|( −1)=( +1) · y − k 2 |x| sign x:

(61)

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

531

Let (x(t); y(t)) be a solution of Eq. (61). Consider the quadratic (with respect to variables |x|( +1)=2 sign x and y) form V =

y2 k2 |x| +1 − A|x|( +1)=2 sign x · y + ;

+1 2

1 ¿ A ¿ 0:

(62)

Using the inequality − A|x|( +1|)=2 sign x · y ¡ 12 A|x| +1 + 12 y2 ;

(63)

one can write Eq. (62) in the form   2 A k + |x| +1 + y2 : V ≤

+1 2

(64)

The full derivative for V with respect to t is

+ 1 −1 2 dV = V 0 = k 2 |x| sign x · y − A |x| 2 y dt 2   −A|x|( +1)=2 sign x |y|( −1)=( +1) y − k 2 |x| sign x  

−1 y − k 2 |x| sign x +y |y|

+1

(65)

or

+ 1 ( −1)=2 2 |x| y 2 −A |x|( +1)=2 sign x|y|( −1)=( +1) y + |y|( −1)=( +1) y2 :

V 0 = A k 2 |x|(3 +1)=2 − A

(66)

Holder inequality implies that |x|( −1)2 · y2 ≤


p 1
q 1 |x|( −1)=2 + y2 ; p q

1 1 + = 1; p q

p ¿ 1; q ¿ 1:

For p=

3 + 1 ;

−1

q=

3 + 1 ; 2( + 1)

we get the inequality |x|( −1)=2 · y2 ≤

1 (3 +1)2 1 (3 +1)=( +1) + |y| : |x| p q

(67)

Analogically, ( +1)=2
p1
q1 1 1 |x| sign x · |y|( −1)( +1) y ≤ + |x|( +1)=2 |y|2 =( +1) : p1 q1 For p1 = (3 + 1)=( + 1) and q1 = (3 + 1)=2 one gets ( +1)=2 1 1 |x| sign x · |y|( −1) =( + 1)y ≤ |x|(3 +1)2 + |y|(3 +1)=( +1) : p1 q1

(68)

532

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

Inequalities (66)–(68) imply that V 0 ≥ A1 |x|(3 +1)=2 + B1 |y|(3 +1)=( +1) ; where



A1 = A

k2 −

+1 1 − 2p p1

(69) 

 ;

B1 =

1−

A( + 1) A − 2q q1

 :

Choose k 2 so large and A so close to zero that A1 ¿ 0 and B1 ¿ 0. One obtains then from Eq. (69) that V0 ≥

A1 |x|(3 +1)=2 + B1 |y|(3 +1)=( +1) · V r; Vr

where r = (3 + 1)=(2( + 1)). Using the elementary inequality   |a + b|r ≤ 2r |a|r + |b|r ; which holds for r ≥ 1 and inequality (64), one gets the estimation for V r   2 r A k + |x| +1 + y2 Vr ≤

+1 2 " # r A k2 r (3 +1)=2 (3 +1)=( +1) + ≤2 |x| + |y| :

+1 2

(70)

(71)

(72)

Inequalities (70) and (72) together imply V0 ≥

A1 |x|(3 +1)=2 + B1 |y|(3 +1)=( +1) · V r; A2 |x|(3 +1)=2 + B2 |y|(3 +1)=( +1)

where A2 = 2r



A k2 +

+1 2

r ;

(73)

B2 = 2r :

Inequality (73) implies that V0 ≥ cVr

(74)

for some c ¿ 0. It follows from Eq. (74) that V (t0 ) V (t) ≥  1=(r−1) : r−1 · c(r − 1)(t − t0 ) 1 − V (t0 ) Hence for some positive constant L V (t) ≥

L ( − t)1=(r−1)

(75)

Yu.A. Klokov, F.Zh. Sadyrbaev / Nonlinear Analysis 39 (2000) 519 – 533

533

and the existence of a solution which does not extend to a nite interval [a; b] is obvious. This solution must be of oscillatory type since the growth rate of the righthand side of Eq. (59) is less than 2 and blow-up behaviour is prohibited by the Nagumo–Bernstein theorem ([3, Ch. I]). Remark Using the same type of argument, more complicated technically, one can show that the equation x00 + k 2 |x| sign x = |x0 | · x0 ; where 0 ¡  ¡ ( − 1)=( + 1), also allows for nonextendable solutions. References [1] A. Bahri, H. Berestycki, Forced vibrations of superquadratic Hamiltonian systems, Acta Math. 152 (1984) 143–197. [2] A. Bahri, H. Berestycki, Existence of forced oscillations for some nonlinear dierential equations, Comm. Pure Appl. Math. 37 (1984) 403–442. [3] S. Bernfeld, V. Lakshmikantham, An Introduction to Nonlinear Boundary Value Problems, Academic Press, New York, 1974. [4] A. Capietto, M. Henrard, J. Mawhin, F. Zanolin, A continuation approach to some forced superlinear Sturm–Liouville boundary value problems, Topological Meth. Nonlinear Anal. 3 (1994) 81–100. [5] A. Capietto, J. Mawhin, F. Zanolin, A continuation approach to superlinear periodic boundary value problems, J. Dierential Equations 88 (1990) 347–395.  [6] H. Ehrmann, Uber die Existenz der Losungen von Randwertaufgaben bei gewohnlichen nichtlinearen Dierentialgleichungen zweiter Ordnung, Math. Ann. 134 (1957) 167–194.   [7] S. Fucik, V. Lovicar, Periodic solutions of the equation x00 + g(x) = p; Casopis Pest. Mat. 100 (1975) 160–175. [8] S. Hastings, Boundary value problems in one dierential equation with a discontinuity, J. Dierential Equations 1 (1965) 346–369. [9] M.A. Krasnoselskii, A.I. Perov, A.I. Povolockii, P.P. Zabreiko, Plane Vector Fields, Academic. Press, New York, 1966. [10] Yu.A. Klokov, F. Zh. Sadyrbaev, On the number of solutions of boundary value problems with nonlinear asymptotics, Dierencial’nye uravnenija (Dierential Equations) 34 (4) (1998) 471–479. (in Russian). [11] A. Lazer, Solvability of nonlinear equations and boundary value problems, Bull. Amer. Math. Soc. 8 (1983) 482–487. [12] J. Mawhin, W. Omana, A priori bounds and existence of positive solutions for some Sturm–Liouville superlinear boundary value problems, Preprint – Rapport No. 155 (1989), Seminaire Mathematique, Louvain–La–Neuve Math. Inst., p. 12. [13] G.R. Morris, An in nite class of periodic solutions of x00 + 2x3 = p(t); Proc. Cambridge Phil. Soc. 61 (1965) 157–164. [14] B. Ruf, S. Solimini, On a class of superlinear Sturm–Liouville problems with arbitrarily many solutions, SIAM J. Math. Anal. 17 (4) (1986) 761–771. [15] B. Ruf, P.N. Srikanth, Multiplicity results for ODEs crossing all but a nite number of eigenvalues, Nonlinear Anal. TMA 10 (2) (1986) 157–163. [16] M. Struwe, Multiple solutions of anticoercive boundary value problems for a class of ordinary dierential equations of second order, J. Dierential Equations 37 (1980) 285–295.