Nonlinear Analysis 42 (2000) 651 – 661
www.elsevier.nl/locate/na
On global stability of third-order nonlinear dierential equations Chuanxi Qian ∗ Department of Mathematics and Statistics, Mississippi State University, Mississippi State, MS 39762, USA Received 2 August 1997; accepted 26 August 1998
Keywords: Nonlinear dierential equation; Trivial solution; Global stability; Lyapunov’s method
1. Introduction It is well known that the stability is a very important problem in the theory and applications of dierential equations. So far, the most eective method to study the stability of nonlinear dierential equations is still the Lyapunov’s direct method. For the second-order nonlinear dierential equations, many stability results have been established by using this method. However, the results about the stability of nonlinear dierential equations whose orders are more than two are relatively scarce. This is perhaps due to the diculty of constructing proper Lyapunov functions for higher-order nonlinear dierential equations. We refer to [1–9] and the references cited there for this topic. In 1970, Barbashin [1] discussed the global stability of the following third-order nonlinear dierential equation: x000 + (x; x0 )x00 + f(x; x0 ) = 0;
(1.1)
where ; f;
x
and fx ∈ C(R × R; R)
∗
Tel.: +601-325-3414; fax: +601-325-0005. E-mail address:
[email protected] (C. Qian)
0362-546X/00/$ - see front matter ? 2000 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 9 ) 0 0 1 2 0 - 0
(1.2)
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C. Qian / Nonlinear Analysis 42 (2000) 651 – 661
and the following result was proved: Theorem A (Barbashin [1]). Assume that (i) f(x; R y 0)x¿0 for x 6= 0; (ii) 0 f(0; v) dv¿0 for y 6= 0; (iii) y x (x; y) ≤ 0 for y 6= 0 and there is a positive number B such that (iv) (x; y) ≥ B; Ry (v) B[f(x; y) − f(x; 0)]y¿y 0 fx (x; v) dv for y 6= 0; Ry Rx (vi) 4B 0 f(u; 0) du 0 [f(x; v) − f(x; 0)] dv − y2 f2 (x; 0)¿0 for xy 6= 0; Rx Ry (vii) lim|x|→∞ {B 0 f(u; 0) du + 0 f(x; v) dv} = ∞ for all ÿxed y: Then the trivial solution of Eq. (1:1) is globally asymptotically stable. Theoretically, this is a very interesting result since (1.1) is a rather general third-order nonlinear dierential equation. For example, many third-order dierential equations which have been discussed in [7] are special cases of Eq. (1.1), and some known results can be obtained by using this theorem. However, it is not easy to apply Theorem A to these special cases to obtain new or better results since Theorem A has some hypotheses which are not necessary for the stability of many nonlinear equations. Our aim in this paper is to further study the global stability of Eq. (1.1). In the next section, we establish a criterion for the stability of Eq. (1.1), which extends and improves Theorem A. Then, in Section 3, we apply our result to some special cases of Eq. (1.1) which have been discussed in the literature to obtain some better global stability results. In the following discussion, we always assume (1.2) holds without further mention. 2. Main result Our main result in this section is the following theorem. Theorem. Assume that (1) xf(x; R y 0)¿0 for x 6= 0; (2) 0 f(0; v) dv¿0 for y 6= 0; Rx (3) lim|x|→∞ 0 f(u; 0) du = ∞ and there is a positive number B such that (4) (x; y) ≥ B; Ry Ry (5) B[f(x; y) − f(x; 0) − 0 x (x; v)v dv]y ≥ y 0 fx (x; v) dv; Ry Ry (6) B[f(x; y) − f(x; 0) − 0 x (x; v)v dv]y + (x; y)¿ 0 fx (x; v) dv + B for y 6= 0; Ry Ry Rx (8) 4B 0 f(u; 0) du{ 0 [f(x; v) − f(x; 0)] dv + B 0 [ (x; v) − B]v dv}¿y2 f2 (x; 0) for xy 6= 0: Then the trivial solution of Eq. (1:1) is globally asymptotically stable. Proof. Clearly, Eq. (1.1) is equivalent to the system x0 = y;
y0 = z;
z 0 = −f(x; y) − (x; y)z
(2.1)
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653
and (0; 0; 0) is a solution of (2.1). Now, consider the Lyapunov function V (x; y; z) = BF(x) + yf(x; 0) + (x; y) Z y (z + By)2 +B ( (x; v) − B)v dv; + 2 0 where
Z F(x) =
0
x
Z f(u; 0) du
Observe that 0 (x; y; z) = y V(2:1)
and
(x; y) =
y
0
Z B f(x; 0) − f(x; y) +
0
(2.2)
[f(x; v) − f(x; 0)] dv:
y
x (x; v)v dv
Z +
y
0
+ [B − (x; y)]z 2 :
fx (x; v) dv (2.3)
0 (x; y; z) ≤ 0. From the hypotheses, we see that V(2:1) In the following, we are going to show that V is a positive-de nite function, every 0 (x; y; z)=0} does positive semi-trajectory of (2.1) is bounded and the set {(x; y; z): V(2:1) not contain the whole trajectory of any solution of (2.1) except the trivial solution (0; 0; 0). First we show that V is a positive-de nite function. Observe that when x = 0, Z y (z + By)2 +B ( (0; v) − B)v dv: (2.4) V (0; y; z) = (0; y) + 2 0
Hence, in view of (2) and (4), it is easy to see that V (0; y; z) is a positive-de nite function of y and z. Next, assume that x 6= 0. Then F(x)¿0 and so V (x; y; z) can be written in the form #2 " p yf(x; 0) (z + By)2 BF(x) + p + V (x; y; z) = 2 2 BF(x) Z y y2 f2 (x; 0) : ( (x; v) − B)v dv − + (x; y) + B 4BF(x) 0
(2.5)
If y = 0, then V (x; 0; z) = BF(x) +
z2 ≥ BF(x)¿0; 2
if y 6= 0, then V (x; y; z) ≥ (x; y) + B
Z 0
y
( (x; v) − B)v dv −
y2 f2 (x; 0) ; 4BF(x)
which, in view of (7), implies that V (x; y; z)¿0. Hence, V is a positive-de nite function.
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Next, we show that the positive semi-trajectory of every solution of (2.1) is bounded. To this end, consider the set D(r; M; N ) = {(x; y; z) | V (x; y; z)¡r 2 ; x2 ¡M 2 ; y2 ¡N 2 }; where r is any xed positive number and M =M (r) and N =N (r) are positive constants which satisfy BF(±M )¿r 2 and
(2.6)
√
2r ; (2.7) B respectively. The existence of M is guaranteed by (3). From the expression of V (x; y; z) we see that every point (x; y; z) ∈ D satis es the condition N¿
(z + By)2 ≤ 2r 2 ;
(2.8)
which implies that z is bounded since y is bounded. Hence, D is a bounded set. In the following, we show that any positive semi-trajectory + of (2.1) setting o from a 0 (x; y; z) ≤ 0, we see that if point in D cannot go out of D. By noting the fact that V(2:1) + goes out of D, it must go through either 1 , or 2 , or 3 , or 4 where 1 ; 2 ; 3 and 4 are the segments of the planes x = M; x = −M; y = N and y = −N on the boundary @D of D, respectively. We claim that this is impossible also. First we show that + cannot go through 1 , or 2 . Observe that when (x; y; z) ∈ 1 ∪ 2 ; x = M or −M and satis es p yf(x; 0) (2.9) ≤ r: BF(x) + p 2 BF(x) Hence, by noting (2.6), we see that yf(x; 0)|(x;y; z)∈1 ∪2 ¡0: Then it follows that f(x; 0) |(x;y; z)∈1 ∪2 ¡0; x which, in view of (1), implies that xy|(x;y; z)∈1 ∪2 ¡0. Therefore, xy
dx2 |(x;y; z)∈1 ∪2 = 2xy¡0: dt From this fact, we see that the direction eld of (2.1) enters D on 1 and 2 and so + cannot go through 1 or 2 . Next, we show that + cannot go through 3 or 4 either. In fact, from (2.7) and (2.8) we see that if there is a moment T such that |y(T )| = N , then z(T )y(T )¡0. By noting y0 (T ) = z(T ) it follows that y0 (T )y(T )¡0 which implies that the direction eld of (2.1) enters D on 3 and 4 . Hence, + cannot go through 3 or 4 . Therefore, any positive semi-trajectory of (2.1) setting o from a point in D cannot go out of D and so is bounded. Since for any bounded region in the space, we can choose a sucient large r such that ⊂ D, it follows that every positive semi-trajectory is bounded.
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Finally, we show that the set 0 (x; y; z) = 0} = {(x; y; z): V(2:1)
does not contain the whole trajectory of any solution of (2.1) except the trivial solution (0; 0; 0). Assume that (x(t); y(t); z(t)) is a solution of (2.1) which satis es 0 (x(t); y(t); z(t)) ≡ 0 V(2:1)
for all t:
(2.10)
We claim that y(t) ≡ 0. Otherwise, by the continuity of y(t), there are T1 and T2 with T1 ¡T2 such that y(t) 6= 0 for any t ∈ (T1 ; T2 ). Then, in view of (4), (5) and (2.3), it follows from (2.10) that Z y(t) B[f(x(t); 0) − f(x(t); y(t)) + x (x; v)|x=x(t) v dv] 0
Z +
y(t) 0
fx (x; v)|x=x(t) dv ≡ 0
for t ∈ (T1 ; T2 ):
Hence, by noting (6), we nd that (x(t); y(t)) − B¿0
for t ∈ (T1 ; T2 )
and so (2.10) yields z(t) ≡ 0 for t ∈ (T1 ; T2 ). Then it follows from (2.1) that y(t) ≡ c
and
x(t) = ct + d
for t ∈ (T1 ; T2 );
where c is a nonzero constant and d is a constant. By using continuation theorem of solutions, it is easy to see that x(t) = ct + d; y(t) ≡ c and z(t) ≡ 0
for t ∈ (−∞; ∞):
However, it has been shown above that the positive trajectory of every solution of (2.1) is bounded. Hence c = 0 which is a contradiction and so y(t) ≡ 0 for t ∈ (−∞; ∞). Then, it follows from (2.1) that x(t) ≡ d; z(t) ≡ 0
for t ∈ (−∞; ∞)
and
f(d; 0) = 0;
where d is a constant, which, in view of (1), implies that d = 0 and so x(t) ≡ 0 for t ∈ (−∞; ∞). Therefore, the set does not contain the whole trajectory of any solution of (2.1) except the trivial solution (0; 0; 0). Hence, by the Barbashin–Krasovskii theorem (see, for example [4]) the trivial solution of (2.1) is globally asymptotically stable. The proof is complete. Remark 1. Clearly, our theorem is an improvement and extension of Theorem A. In particular, from our theorem we see that (iii) assumed in Theorem A is not necessary, and (vii) can be replaced by a better condition (3) for the global stability of the trivial solution of Eq. (1.1). 3. Some special cases In this section, we apply our theorem to some special cases of Eq. (1.1) to extend and improve some well-known global stability results in the literature.
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First, consider the third-order dierential equation x000 + (x; x0 )x00 + (x0 ) + g(x) = 0;
(3.1)
where ;
x
∈ C[R × R; R]
and
; g0 ∈ C[R; R]:
The global stability of Eq. (3.1) has been studied by Ezeilo [2] and Ogurtsov [5] and the following results have been established by them. Theorem B (Ezeilo [2]). Assume that (i) g(x)=x ≥ c0 ¿0 for x 6= 0; g0 (x) ≤ c; (ii) (y)=y ≥ b¿0 for y 6= 0; (iii) (x; y) ≥ a¿c=b; (iv) y x (x; y) ≤ 0 for y 6= 0; where a; b; c and c0 are positive constants. Then the trivial solution of Eq. (3:1) is globally asymptotically stable. Theorem C (Ogurtsov [5]). Assume that (i) xg(x)¿0 for x 6= 0; (ii) [a(y) − g0 (x)y]y¿0 for y 6= 0; (iii) (x; y) ≥ a¿0; Ry R x y x (x; y) ≤ 0; (iv) U (x; y) = a 0 g(u) du + yg(x) + 0 (v) dv positive-de nite and limx2 +y2 →∞ U (x; y) = ∞, where a is a positive constant. Then the trivial solution of Eq. (3:1) is global asymptotically stable. However, by noting that Eq. (3.1) is a special case of Eq. (1.1) with f(x; y) = g(x) + (y) and by applying our theorem, we have the following result immediately. Corollary 1. Assume that (1) xg(x)¿0 for x 6= 0; Ry (2) 0 (v) dv¿0 for y 6= 0; Rx (3) lim|x|→∞ 0 g(u) du = ∞ and there is a positive number B such that (4) (x; y) ≥ B; Ry (5) B[(y) − 0 x (x; v)v dv]y ≥ g0 (x)y2 ; Ry (6) (x; y) + B[(y) − 0 x (x; v)v dv]y¿B + g0 (x)y2 for y 6= 0; Ry Ry Rx (7) 4B 0 g(u) du{ 0 (v) dv + B 0 [ (x; v) − B]v dv}¿g2 (x)y2 for xy 6= 0: Then the trivial solution of Eq. (3:1) is globally asymptotically stable. We claim that both Theorems B and C are direct consequences of Corollary 1. Let us consider Theorem B rst. Suppose all the hypotheses in Theorem B hold. It suces to show that all the hypotheses in Theorem 2 hold also. Clearly, (i) and (ii)
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implyR(1)–(3). Take B = c=b. Then (iii) implies (4). By noting (ii) and (iv), we see y that ( 0 x (x; v)v dv)y ≤ 0 and Z y (x; v)v dv y ≥ B(y)y ≥ Bb y2 = cy2 ≥ g0 (x)y2 B (y) − x 0
and so (5) holds. Then, (6) follows from (5) and the fact see that g0 (x)g(x) ≥ cg(x)
for x¿0 and g0 (x)g(x) ≤ cg(x)
and so it follows that Z x g(u) du ≥ 12 g2 (x)¿0 0
(x; y)¿B. From (i) we
for x¡0
for x 6= 0:
In addition, observe that (ii) implies that Z y (v) dv ≥ 12 by2 : 0
Hence,
Z
4B
0
x
Z g(u) du Z
¿4B
0
x
y
0
g(u) du
≥ g2 (x)y2
Z (v) dv + B Z 0
y
0
y
[ (x; v) − B]v dv
(v) dv
for xy 6= 0
and so (7) holds. Therefore, all the hypotheses in Corollary 1 hold. Next, assume that all the hypotheses in Theorem C hold. We show that all the hypotheses in Theorem 2 hold also. Clearly, (1) – (3) follow from (i) and (iv) immediately. Now, take B = a and so (iii) implies (4). Then by noting (ii) and (iii) we see that Z y 0 2 for y 6= 0; B (y) − x (x; v)v dv y ≥ B(y)y¿g (x)y 0
which implies (5) and (6) hold. Clearly, (ii) yields Z y (v) dv¿ 12 g0 (x)y2 for y 6= 0: a 0
Hence, by noting (i) also, Z y (v) dv¿ 12 g(x)g0 (x)y2 ag(x) 0
and
Z ag(x)
0
y
(v) dv¡ 12 g(x)g0 (x)y2
for x¿0 and y 6= 0
for x¡0 and y 6= 0:
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C. Qian / Nonlinear Analysis 42 (2000) 651 – 661
Then it follows that Z Z x g(u) du a 0
y
0
(v) dv¿ 14 g2 (x)y2
for xy 6= 0;
which implies (7) holds. Hence, all the hypotheses in Corollary 1 hold. Example 1. Consider the equation x000 + [(sin x)x0 + (x0 )2 + 2]x00 + (x0 )3 + x0 +
x = 0: 1 + x2
(3.2)
Eq. (3.2) is in the form of (3.1) with (x; y) = (sin x)y + y2 + 2;
g(x) =
x 1 + x2
(y) = y3 + y:
and
Take B = 1. Observe that Z y 3 3 1 (y) − x (x; v)v dv y = [y + y − 3 (cos x)y ]y 0
¿ y2 and
Z 4
x 0
Z g(u)du
0
y
1 − x2 = y2 g0 (x) (1 + x2 )2
(v) dv = 2 ln(1 + x2 ) ¿
1 4 4y
+ 12 y2
x2 y2 = g2 (x)y2 (1 + x2 )2
for y 6= 0
for xy 6= 0:
Then it is easy to check all the hypotheses in Corollary 1 are satis ed and so the trivial solution of Eq. (3.2) is globally asymptotically stable. However, both Theorems B and C cannot be applied here. For example, observe that y x (x; y) = (cos x)y2 ; which does not satisfy the condition y x (x; y) ≤ 0 assumed in both Theorems B and C. When g(x) = cx and (y) = by, where b and c are positive constants, Eq. (3.1) reduces to x000 + (x; x0 )x00 + bx0 + cx = 0:
(3.3)
Shimanov [8] has established the following stability theorem for this equation: Theorem (i) b (ii) y Then the
D (Shimanov [8]). Assume that (x; y)¿c; x (x; y) ≤ 0; trivial solution of Eq. (3:3) is globally asymptotically stable.
Here by using Corollary 1 and by taking B = c=b we have the following result which is an improvement of Theorem D.
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Corollary 2. Assume that (1) b R(x; y) ≥ c; y (2) y 0 x (x; v)v dv ≤ 0; Ry (3) (x; y) − y 0 x (x; v)v dv¿c=b for y 6= 0; Ry (4) 0 [ (x; v) − bc ]v dv¿0 for y 6= 0: Then the trivial solution of Eq. (3:3) is globally asymptotically stable. Next, consider the third-order dierential equation x000 + h(x0 )x00 + (x0 ) + g(x) = 0;
(3.4)
where h; ; g
and
g0 ∈ C[R; R]:
Eq. (3.4) is a special case of Eq. (3.1) with 1 we can establish the following result.
(x; y) = h(y) and so by using Corollary
Corollary 3. Assume that (1) xg(x)¿0 Rfor x 6= 0; x (2) lim|x|→∞ 0 g(x) dx = ∞ and there is a positive number B such that (3) h(y) ≥ B; (4) B(y)y ≥ g0 (x)y2 (5) h(y) + B(y)y¿B + g0 (x)y2 for y 6= 0: Then the trivial solution of Eq. (3:4) is globally asymptotically stable. Proof. Clearly, it suces to show that (2) and (7) in Corollary 1 hold. From (1) we see that there is an x0 6= 0 such that g0 (x0 )¿0. Hence, it follows from (4) that B(y)y ≥ g0 (x0 )y2 ¿0
for y 6= 0
which implies that (2) in Corollary 1 holds. In view of (3), (4) and (5), we see that B(y) + B2 (h(y) − B)y¿g0 (x)y
for y¿0
B(y) + B2 (h(y) − B)y¡g0 (x)y
for y¡0:
and Then it follows that Z y Z (v) dv + B B 0
y
0
[h(y) − B]y dy ¿ 12 g0 (x)y2
By noting (1), we see that Z y Z (v) dv + B Bg(x) 0
¿ 12 g0 (x)g(x)y2
0
y
[h(y) − B]y dy
for x¿0 and y 6= 0
for y 6= 0:
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C. Qian / Nonlinear Analysis 42 (2000) 651 – 661
and
Z Bg(x)
0
y
Z (v) dv + B
¡ 12 g0 (x)g(x)y2 Hence, B
Z 0
x
Z g(u) du
0
y
0
y
[h(y) − B]y dy
for x¡0 and y 6= 0: Z (v) dv + B
0
y
[h(y) − B]y dy ¿ 14 g2 (x)y2
for xy 6= 0;
which implies that (7) in Corollary 1 holds. The proof is complete. Corollary 3 improves several known results in the literature. For the lack of the space, we give only one example below. For a survey, one can see [7]. Consider the following third-order dierential equation: x000 + h(x0 )x00 + (x0 )x0 + k(x)x = 0;
(3.5)
where h; ; k and kx ∈ C[R; R]: The global stability of Eq. (3.5) has been studied by Goldwyn and Narendra [3] and the following result has been established. Theorem E (Goldwyn and NarendraR[3]). Assume that x (i) k(x)¿0 for x 6= 0; lim|x|→∞ 0 k(u)u du = ∞: (ii) inf y {(y)} = 0 ¿0; 0 h0 − K0 ¿0 where h0 = inf y {h(y)} and K0 = supx {(k(x) x)0 }: Then the trivial solution of Eq. (3:5) is globally asymptotically stable. However, by using Corollary 3 we have the following result. Corollary 4. Assume that Rx (1) k(x)¿0 for x 6= 0; lim|x|→∞ 0 k(u)u du = ∞ and there is a positive number B such that (2) h(y) ≥ B; (3) B(y) ≥ (xk(x))0 ; (4) h(y) + B(y)¿B + (xk(x))0 : Then the trivial solution of Eq. (3:5) is globally asymptotically stable. By taking B = K0 =0 , it is easy to see that Theorem E is a direct consequence of Corollary 4. While the following example shows that Corollary 4 has a wider range of applications. Example 2. Consider the following equation: 1 x00 + x0 + x = 0: x000 + 1 + 1 + (x0 )2
(3.6)
C. Qian / Nonlinear Analysis 42 (2000) 651 – 661
661
Eq. (3.6) is in the form of (3.5) with h(y) = 1 +
1 ; 1 + y2
(y) ≡ 1
and
k(x) ≡ 1:
By taking B = 1, it is easy to check all the hypotheses of Corollary 4 are satis ed and so the trivial solution of Eq. (3.6) is globally asymptotically stable. However, the equation does not satisfy (ii) in Theorem E. References [1] E.A. Barbashin, Lyapunov Functions, Izdat.‘Nauka’, Moscow, 1970. [2] J.O.C. Ezeilo, On the stability of solutions of certain dierential equations of the third order, Quart. J. Math. Oxford Ser. 11 (1960) 64–69. [3] M. Goldwyn, S. Narendra, Stability of Certain Nonlinear Dierential equations using the second method of Lyapunov, Craft Lab. Harvard Univ., Cambridge, MA, 1963, pp. 1–14. [4] N.N. Krasovskii, Stability of Motion, Stanford University Press, Stanford, CA, 1963. [5] A.I. Ogurtsov, On the stability of the solutions of some nonlinear dierential equations of the third and fourth order, Izv. Vyssh, Uchebn. Zaved. Matematika 10 (1959) 200–209. [6] Y. Qin, M. Wang, L. Wang, Theory and Applications of Stability of Motion, Academic Press, Beijing, 1981. [7] R. Reissig, G. Sansone, R. Conti, Nonlinear Dierential Equations of Higher Order, Noordho Inter. Pub., Leyden, 1974. [8] S.N. Shimanov, On the stability of the solution of a nonlinear equation of the third order, Prikl. Mat. Mekh. 17 (1953) 369–372. [9] L. Wang, M. Wang, Analysis of construction of Lyapunov functions of third-order nonlinear systems, Acta Math. Appl. Sinica 6 (1983) 309–323.