ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu
ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS OF THE NONLINEAR DIFFERENTIAL EQUATION t2u00= un
MENG-RONG LI, HSIN-YU YAO, YU-TSO LI
Abstract. In this article we study properties of positive solutions of the or-dinary differential equation t2u00= unfor 1 < n ∈ N, we obtain conditions for
their blow-up in finite time, and some properties for global solutions. Equa-tions containing more general nonlinear terms are also considered.
1. Introduction
Some interesting results on the blow-up, blow-up rates, and estimates for the life-span of solutions of the Emden-Fowler equation and the semi-linear wave equation u + f (u) = 0 have been obtained, as shown in the references.
Here we wish to study the Emden-Fowler type wave equation, i.e. solutions, independent of the space variable x, of the equation t2utt− ∆u = un for n > 1.
The existence and uniqueness of local solutions of the initial-value problem t2u00= un, 1 < n ∈ N,
u(1) = u0, u0(1) = u1,
(1.1) follow by standard arguments. Considering the transformation t = es, u(t) = v(s), we have t2u00(t) = −v
s(s) + vss(s), v(s)n = −vs(s) + vss(s) and v(0) = u(1) = u0; vs(0) = u0(1) = u1, the problem (1.1) can be transformed into
vss(s) − vs(s) = vn(s), v(0) = u0, vs(0) = u1.
(1.2) Thus, the existence of local solutions u for (1.1) in (1, T ) is equivalent to the existence of local solutions v for (1.2) in (0, ln T ). In this article, we give estimates for the life-span T∗ of positive solutions u of (1.1) in three different cases. The main results are as follows:
(a) u1= 0, u0> 0: T∗≤ ek1, k1:= s0+ 2(n+3) 8− 2 n−1v(s0) 1−n 2 , ε ∈ (0, 1); (b) u1> 0, u0> 0: (i) E(0) ≥ 0, T∗≤ ek2, k 2:= n−12 q n+1 2 u 1−n 2 0 ; (ii) E(0) < 0, T∗≤ ek3, k 3:= n−12 uu01; (c) u1< 0, u0∈ (0, (−u1) 1 n): u(t) ≤ (u0− u1− un 0) + (u1+ un0)t − u n 0ln t.
2000 Mathematics Subject Classification. 34A34, 34C11, 34C60.
Key words and phrases. Nonlinear differential equation; Emden-Fowler equation; blow-up rate. c
2013 Texas State University - San Marcos.
Submitted October 5, 2013. Published November 20, 2013.
where E(0) is defined in the next section and s0 is given by (3.3).
In Section 6, we replace the nonlinear term vn by a more general increasing function f (v).
2. Notation and some lemmas For a given function v, we use the following functions
a(s) := v(s)2, E(0) := u21− 2 n + 1u n+1 0 , J (s) := a(s) −n−1 4 ,
where u0and u1 are the given initial conditions.
By an easy calculation we can obtain the following two Lemmas; we shall omit the proof of the first lemma.
Lemma 2.1. Suppose that v ∈ C2[0, T ] is the solution of (1.2), then E(s) = vs(s)2− 2 Z s 0 vs(r)2dr − 2 n + 1v(s) n+1= E(0), (2.1) (n + 3)vs(s)2= (n + 1)E(0) + a00(s) − a0(s) + 2(n + 1) Z s 0 vs(r)2dr, (2.2) J00(s) = n 2− 1 4 J (s) n+3 n−1(E(0) − a 0(s) n + 1+ 2 Z s 0 vs(r)2dr), (2.3) J0(s)2= J0(0)2+(n − 1) 2 4 E(0)(J (s) 2(n+1) n−1 − J (0) 2(n+1) n−1 ) +(n − 1) 2 2 J (s) 2(n+1) n−1 Z s 0 vs(r)2dr. (2.4)
Lemma 2.2. For u0> 0, the positive solution v of (1.2) satisfies:
(i) If u1≥ 0, then vs(s) > 0 for all s > 0 (2.5) (ii) If u1< 0, u0∈ (0, (−u1)
1
n), then vs(s) < 0 for all s > 0. (2.6)
Proof. (i) Since vss(0) = u1+ un0 > 0, we know that vss(s) > 0 in [0, s1) and vs(s) is increasing in [0, s1) for some s1 > 0. Moreover, since v and vs are increasing in [0, s1), vss(s1) = vs(s1) + v(s1)n > vs(0) + v(0)n > 0 for all s ∈ [0, s1) and vs(s1) > vs(s) > 0 for all s ∈ [0, s1), we know that there exists a positive number s2 > 0, such that vs(s) > 0 for all s ∈ [0, s1+ s2). Continuing this process, we obtain vs(s) > 0 for all s > 0, for which the solution exists.
(ii) Since vss(0) = vs(0) + v(0)n = u1+ un0 < 0, there exists a positive number s1 > 0 such that vss(s) < 0 in [0, s1), vs(s) is decreasing in [0, s1); therefore, vs(s) < vs(0) = u1< 0 for all s ∈ [0, s1) and v(s) is decreasing in [0, s1). Moreover, since v and vs are decreasing in [0, s1), vss(s) = vs(s) + v(s)n < vs(0) + v(0)n < 0 for all s ∈ [0, s1) and vs(s1) < vs(s) < 0 for all s ∈ [0, s1), we know that there exists a positive number s2 > 0, such that vs(s) < 0 for all s ∈ [0, s1+ s2). Continuing this process, we obtain vs(s) < 0 for all s > 0 in the interval of existence.
3. Life-span of positive solutions of (1.1) when u1= 0, u0> 0 In this section we want to estimate the life-span of a positive solution u of (1.1) if u1 = 0, u0 > 0. Here the life-span T∗ of u means that u is the solution of equation (1.1) and u exists only in [0, T∗) so that the problem (1.1) possesses a positive solution u ∈ C2[0, T∗).
Theorem 3.1. For u1 = 0, u0 > 0, the positive solution u of (1.1) blows up in finite time; that is, there exists T∗< ∞ so that
u(t)−1→ 0 as t → T∗.
Proof. By (2.5), we know that vs(s) > 0, a0(s) > 0 for all s > 0 provided that u1= 0, u0> 0. By Lemma 2.1, a00(s) − a0(s) = 2(vs(s)2+ v(s)n+1), (a0(s)e−s)0 = e−s(a00(s) − a0(s)) = 2e−s(vs(s)2+ v(s)n+1), a0(s)e−s= 2 Z s 0 e−r(vs(r)2+ v(r)n+1)dr ≥ 4 Z s 0 e−rvs(r)v(r) n+1 2 dr,
and a0(0) = 0, hence we have a0(s)e−s≥ 8 n + 3(v(r) (n+3)/2e−r|s r=0+ Z s 0 v(r)(n+3)/2e−rdr) = 8 n + 3(v(s) (n+3)/2e−s− v(0)(n+3)/2) + 8 n + 3 Z s 0 v(r)(n+3)/2e−rdr. Since a0(s) > 0 for all s > 0, v is increasing on (0, ∞) and
a0(s)e−s≥ 8 n + 3(v(s) (n+3)/2e−s− v(0)(n+3)/2) + 8 n + 3 Z s 0 v(0)(n+3)/2e−rdr = 8 n + 3(v(s) (n+3)/2e−s− v(0)(n+3)/2) + 8 n + 3v(0) (n+3)/2(1 − e−s), a0(s) ≥ 8 n + 3(v(s) (n+3)/2− v(0)(n+3)/2) = 8 n + 3(v(s) (n+3)/2− u(n+3)/2 0 ). (3.1) Using u1= 0 and integrating (1.2), we obtain
vs(s) = v(s) − u0+ Z s 0 v(r)ndr, vs(s) ≥ v(s) − u0+ Z s 0 v(0)ndr = v(s) − u0+ un0s, (e−sv(s))s= e−s(vs(s) − v(s)) ≥ e−s(un0s − u0), a0(s) ≥ 8 n + 3(v(s) (n+3)/2− v(0)(n+3)/2) = 8 n + 3(v(s) (n+3)/2− u(n+3)/2 0 ). (3.2)
According to (3.2), and since v0(s) > 0, v(s)(n+3)/2≥ (u
0+ un0(es− 1 − s))(n+3)/2, for all ∈ (0, 1), we obtain
v(s)(n+3)/2≥ (u0+ un0(e s− 1 − s))(n+3)/2, v(s)(n+3)/2− 8u(n+3)/20 ≥ (u0+ un0(e s− 1 − s))(n+3)/2− 8u(n+3)/2 0 ≥ (u(n+3)/20 + u n(n+3) 2 0 (e s− 1 − s)(n+3)/2) − 8un+32 0 = ( − 8)u(n+3)/20 + u n(n+3) 2 0 (e s − 1 − s)(n+3)/2. Now, we want to find a number s0> 0 such that
es0− s 0= 1 + 8 − u n+3 2 (1−n) 0 2/(n+3) . (3.3)
This means that there exists a number s0 > 0 satisfying (3.3) with ∈ (0, 1) such that
v(s)(n+3)/2− 8u(n+3)/20 ≥ 0 for all s ≥ s0. From (3.1), it follows that
a0(s) ≥ 8 n + 3v(s) (n+3)/2− 8 n + 3u (n+3)/2 0 = 8 − n + 3v(s) (n+3)/2+v(s)(n+3)/2− 8u (n+3)/2 0 n + 3 ≥ 8 − n + 3v(s) (n+3)/2, for all s ≥ s 0. For all s ≥ s0, ∈ (0, 1), we obtain that
2v(s)vs(s) ≥ 8 − n + 3v(s) (n+3)/2, v(s)−n+12 vs(s) ≥ 8 − 2(n + 3), 2 1 − n(v(s) 1−n 2 )s≥ 8 − 2(n + 3) and hence (v(s)1−n2 )s≤ 8 − 2(n + 3) 1 − n 2 . Integrating the above inequality, we conclude that
v(s)1−n2 ≤ v(s0) 1−n 2 − 8 − 2(n + 3) n − 1 2 (s − s0). Thus, there exists a number
s∗1≤ s0+ 2(n + 3) 8 − 2 n − 1v(s0) 1−n 2 =: k1
such that v(s)−1 → 0 for s → s∗
1, that is, u(t)−1 → 0 as t → e
k1, which implies
that the life-span T∗ of a positive solution u is finite and T∗≤ ek1.
4. Life-span of positive solutions of (1.1) when u1> 0, u0> 0 In this section we estimate the life-span of a positive solution u of (1.1) whenever u1> 0, u0> 0.
Theorem 4.1. For u1 > 0, u0 > 0, the positive solution u of (1.1) blows up in finite time; that is, there exists a number T∗< ∞ so that
u(t)−1→ 0 as t → T∗.
Proof. We separate the proof into two parts depending on whether E(0) ≥ 0 or E(0) < 0.
(i) Assume that E(0) ≥ 0. By (2.1) and (2.5) we have vs(s)2− 2 n + 1v(s) n+1 ≥ E(0), vs(s)2≥ 2 n + 1v(s) n+1+ E(0), v s(s) ≥ r 2 n + 1v(s) n+1+ E(0).
Since E(0) ≥ 0, we obtain vs(s) ≥ r 2 n + 1v(s) n+1 2 , v(s)−n+12 · vs(s) ≥ r 2 n + 1, (v(s)1−n2 )s≤1 − n 2 r 2 n + 1. Integrating the above inequality, we obtain
v(s)1−n2 ≤ u 1−n 2 0 + 1 − n 2 r 2 n + 1s. Thus, there exists
s∗2≤ 2 n − 1 r n + 1 2 u 1−n 2 0 =: k2 such that v(s)−1→ 0 for s → s∗
2; that is, u(t)−1→ 0 as t → ek2, which means that the life-span T∗ of a positive solution u is finite and T∗≤ ek2.
(ii ) Assume that E(0) < 0. From (2.1) and (2.5) we obtain that J0(s) = −n−1
4 a(s) −n+3
4 a0(s), a0(s) > 0, vs(s) > 0 for all s > 0 and
J0(s) = −n − 1 2 s 2 n + 1+ E(0)a(s) −n+1 2 + 2a(s)− n+1 2 Z s 0 vs(r)2dr ≤ −n − 1 2 r 2 n + 1+ E(0)a(s) −n+1 2 , J (s) ≤ J (0) −n − 1 2 Z s 0 r 2 n + 1+ E(0)a(r) −n+1 2 dr.
Since E(0) < 0 and a0(s) > 0 for all s > 0, J (s) ≤ J (0) −n − 1 2 Z s 0 r 2 n + 1+ E(0)a(0) −n+1 2 dr = a(0)−n−14 −n − 1 2 r 2 n + 1+ E(0)a(0) −n+1 2 s.
Thus, there exists a number s∗3 ≤ 2 n − 1a(0) −n−1 4 ( 2 n + 1+ E(0)a(0) −n+1 2 )−12 =: k3
such that J (s∗3) = 0 and a(s)−1 → 0 for s → s∗
3; that is, u(t)−1 → 0 as t → ek3. This means that the life-span T∗ of u is finite and T∗≤ ek3.
5. Life-span of positive solutions of (1.1) when u1< 0
Finally, we estimate the life-span of a positive solution u of (1.1) when u1< 0. Theorem 5.1. For u1< 0, u0∈ (0, (−u1)
1
n) we have
u(t) ≤ (u0− u1− un0) + (u1+ un0)t − u n 0ln t,
and in particular, if E(0) ≥ 0, we have u(t) ≤ (u 1−n 2 0 + n − 1 2 r 2 n + 1ln t) 2 1−n.
Proof. (i) By (1.2) and integrating this equation with respect to s, we get vs(s) = (u1− u0) + v(s) +R
s 0 v(r)
ndr. By (2.6), we have that v is decreasing and
vs(s) ≤ (u1− u0) + v(s) + Z s 0 v(0)ndr = (u1− u0) + v(s) + un0s, e−sv(s) − u0≤ (u1− u0) Z s 0 e−rdr + un0 Z s 0 re−rdr = (u1− u0)(1 − e−s) + un0(−se −s− e−s+ 1); that is, u(t) ≤ (u0− u1) + u1t + un0(t − 1 − ln t) = (u0− u1− un0) + (u1+ un0)t − u n 0ln t. (ii) If E(0) ≥ 0, by (2.1), we have
vs(s)2− 2 n + 1v(s) n+1= E(0) + 2 Z s 0 vs(r)2dr ≥ E(0), vs(s)2≥ E(0) + 2 n + 1v(s) n+1≥ 2 n + 1v(s) n+1. By (2.6), we obtain that −vs(s) ≥ q 2 n+1v(s) n+1 2 , 2 n−1(v(s) 1−n 2 )s≥ q 2 n+1 and r 2 n + 1s ≤ 2 n − 1(v(s) 1−n 2 − v(0) 1−n 2 ), v(s)1−n2 ≥ (u 1−n 2 0 + n − 1 2 r 2 n + 1s). Then, we know that
v(s) ≤ (u 1−n 2 0 + n − 1 2 r 2 n + 1s) 2 1−n, for all s ≥ 0; that is, u(t) ≤ (u 1−n 2 0 + n − 1 2 r 2 n + 1ln t) 2 1−n for all t ≥ 1. 6. A generalization of Theorem 4.1
In this section we want to extent the blow-up result for the following generaliza-tion of (1.2),
vss(s) − vs(s) = f (v), v(0) = v0, vs(0) = v1,
(6.1) where f is an increasing continuous function with f (0) = 0. We have the following result.
Theorem 6.1. Suppose that f is an increasing function with f (0) = 0 and suppose v is a positive solution of (6.1). If F (v) :=R0vf (r)dr, then
¯
E(s) := vs(s)2− 2 Z s
0
vs(r)2dr − 2F (v(s)) (6.2) is constant. Furthermore, if there exists a positive constant k such that F (s) ≥ ksp+1, p > 1 for all s ≥ 0, and v
1> 0, then the life span of v is finite.
Proof. By an argument similar to that used in proving (2.1), we easily obtain that ¯
E(s) is a constant. Since f is increasing, we have
vf (v) = (v − 0) · (f (v) − f (0)) ≥ 0 for v ≥ 0, thus (v2)s− 2v2(s) ≥ 2v0(v1− v0) + 2 Z s 0 v2s(r)dr. (6.3)
By (6.2) and (6.3), we have ¯E(s) = v2
1− 2F (v0) := ¯E, and v2(s) ≥ v0v1e2s− v0(v1− v0), (6.4) (v2)s− 2v2(s) ≥ 2v0(v1− v0) + 2 Z s 0 v2s(r)dr = 2v0(v1− v0) + 2 Z s 0 ( ¯E + 2F (v(r)) + 2 Z r 0 vs(η)2dη)dr ≥ 2v0(v1− v0) + 2 ¯Es + 4k Z s 0 vp+1(r)dr ≥ 2v0(v1− v0) + 2 ¯Es + 4ks1−p( Z s 0 v2(r)dr)p+12 . (6.5) LetRs 0 v 2(r)dr := b(s), e−sb(s) = B(s). Then b(s)00− 2b(s)0 ≥ 2v0(v1− v0) + 2 ¯Es + 4ks1−pb(s) p+1 2 for s > 0 and by (6.5), we have (e−s(b(s)0− b(s)))0 = e−s(b(s)00− 2b(s)0+ b(s)) ≥ 2v0(v1− v0) + 2 ¯Es + 4ks1−pe−sb p+1 2 + e−sb(s) = 2v0(v1− v0) + 2 ¯Es + 4ks1−p(e−sb(s)) p+1 2 e p−1 2 s+ e−sb(s) ≥ 0, (6.6) (e−sb(s))00= (e−s(b(s)0− b(s)))0 ≥ 2v0(v1− v0) + 2 ¯Es + 4k( es s2) p+1 2 (e−sb(s)) p+1 2 + e−sb(s) ≥ 2v0(v1− v0) + 2 ¯Es + 22− p+1 2 k(e−sb(s)) p+1 2 + e−sb(s), B(s)00≥ 2v0(v1− v0) + 2 ¯Es + 22− p+1 2 kB(s) p+1 2 + B(s). (6.7)
From (6.4) it follows that b(s) ≥ v0v1 2 (e 2s − 1) − v0(v1− v0)s, B(s) ≥ v0v1 2 (e s− e−s) − v 0(v1− v0)se−s,
2v0(v1− v0) + 2 ¯Es + B(s)
2 ≥ 0, s ≥ s0 for some s0> 0. Therefore,
B(s)00≥B(s) 2 ≥ v0v1 5 e s, s ≥ s 0, B0(s) ≥ v0v1 5 (e s − es0) + B0(s 0) > 0, s ≥ s1 for some s1> s0. By (6.7), for all s ≥ s1, ((B(s)0)2)0= 2B(s)0B(s)00 ≥ 22−p−1 2 kB(s) p+1 2 B(s)0 = 22−p−12 k 2 p + 3(B(s) p+3 2 )0, (B0)2− B0(s1)2≥ 23−p−12 k p + 3 (B p+3 2 − B(s1) p+3 2 ), (B0)2≥ 2 3−p−12 k p + 3 (B p+3 2 − B(s1) p+3 2 ) + B0(s1)2 = 2 3−p−12 k 2(p + 3)B p+3 2 + 23− p−1 2 k 2(p + 3)B p+3 2 − 2B(s1) p+3 2 + B0(s1)2, B0≥ 2 7−p 4 √ p + 3B p+3 4 for s ≥ s2,
for some s2> s1; hence, for s ≥ s2, 4 1 − p(B 1−p 4 )0 = B p+3 −4B0(s) ≥ 2 7−p 4 √ p + 3, B(s)1−p4 ≤ B(s2) 1−p 4 −p − 1 4 27−p4 √ p + 3(s − s2) for all s ≥ s2> 0.
Thus B(s) blows up at a finite s∗. Since b(s) = esB(s), b(s) also blows up at s∗. Further, since v2(s) = b0(s) ≥ 2b(s), v(s) blows up at s∗, as well.
Acknowledgements. Thanks are due to Professors Tai-Ping Liu, Ton Yang and Shih-Shien Yu for their continuous encouragement; to the anonymous referee for his/her helpful comments; and to Professor K. Schmitt for his comments and sug-gestions on Theorem 6.1. The authors want to thank Metta Education, Grand Hall and Auria Solar for their financial assistance.
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Meng-Rong Li
Department of Mathematical Sciences, National Chengchi University, Taipei, Taiwan E-mail address: liwei@math.nccu.edu.tw
Hsin-Yu Yao
Department of Mathematical Sciences, National Chengchi University, Taipei, Taiwan E-mail address: diadia0914@gmail.com
Yu-Tso Li
Department of Aerospace and Systems Engineering, Feng Chia University, Taichung, Taiwan
