# Global Existence and Boundedness of Solutions to a Second-Order Nonlinear Differential System

Authors: Changjian Wu; Shushuang Hao; Changwei Xu
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We investigate the global existence and boundedness of solutions to a second-order nonlinear differential system.


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## Body

## 1. Introduction

In this paper, we study the nonlinear system(1.1)x′=1a(x)[c(y)-b(x)],y′=-a(x)[h(x)-e(t)],
where a:R→(0,∞), b, c,  h:R→R, and e:R→R are  continuous.As a particular case of (1.1) we have well-known  Lie~nard equation as follows:
(1.2)x′′+f(x)x′+h(x)=e(t).
with a(x)=1, b(x)=∫0xf(s)ds, c(x)=x, x∈R and the second-order nonlinear differential equation as follows:
(1.3)x′′+(f(x)+g(x)x′)x′+h(x)=e(t)
for  a(x)=exp(∫0xg(s)ds),  b(x)=∫0xa(s)f(s)ds,  c(x)=x,  x∈R.System (1.1) can be regarded as a mathematical model for many phenomena in applied sciences (theory of feedback electronic circuits, motion of  a mass-spring system). It has been investigated by several authors, compare [1–4] and the citations therein.The purpose of this paper is to present new results on the global existence and boundedness of solutions for the system (1.1). The obtained results improve the recent results in [1, 5]. Our paper is divided into two section. In Section 2, we prove the global existence of solutions for (1.1). In Section 3, we get some new results on boundedness of solutions for the system (1.1).


## 2. Global Existence

In this section, we will present new results on the global existence of solutions to system (1.1) under general conditions on the nonlinearities.Let us first define(2.1)C(y)=∫0yc(s)ds,H(x)=∫0xa2(s)h(s)ds.
Then, we have the following.Theorem 2.1.
Assume that(i)
there exists someK≥0, such that
(2.2)sgn(x)H(x)+K≥0,x∈R,sgn(y)C(y)+K≥0,y∈R,(ii)
there exist someN≥0 and Q>0, such that
(2.3)|H(x)|<Q,|x|>N,|C(y)|<Q,|y|>N,(iii)
lim|y|→∞sgn(y)C(y)=Q, lim|x|→∞[1/(Q-sgn(x)H(x))+sgn(x)b(x)]=∞,(iv)
there exist two positive functionsμ,ω∈C([0,K+Q),(0,∞)) such that
(2.4)a(x)|c(y)|≤min{μ(sgn(x)H(x)+K)+ω(sgn(x)H(x)+K),μ(sgn(y)C(y)+K)+ω(sgn(y)C(y)+K)},|x|>N,|y|>N,(v)
sgn(x)a(x)b(x)h(x)≥-[μ(sgn(x)H(x)+K)+ω(sgn(x)H(x)+K)], |x|>N and |h(x)|≤M<∞,x∈R.
If(2.5)∫0K+Qdsμ(s)+ω(s)=∞,
then every solution of  (1.1)  exists globally.Proof.
Due thata:R→(0,∞), b, c, h:R→R and e:R→R are  continuous, by Peano’s Existence Theorem [6], we have that the system (1.1) with any initial data (x0,y0) possesses a solution (x(t),y(t))  on [0,T) for some maximal T>0. If T<∞, one has
(2.6)limt→T(|x(t)|+|y(t)|)=∞.
First, assume that limt→T|y(t)|=∞.
Sincey(t) is continuous, there exists 0≤T0<T such that
(2.7)|y(t)|>N,t∈[T0,T).
Take V1(t,x,y)=sgn(y)C(y)+K,   t∈R+,  x,y∈R. Differentiating V1(t,x,y) with respect to t along solution (x(t),y(t))  of  (1.1), we have
(2.8)dV1dt=sgn(y)[-a(x)c(y)h(x)+a(x)c(y)e(t)]≤(|h(x)|+|e(t)|)a(x)|c(y)|≤(M+|e(t)|)[μ(sgn(y)C(y)+K)+ω(sgn(y)C(y)+K)],(|h(x)|+|e(t)|)a(x)|c(y)|t∈[T0,T).
Since  0≤sgn(y(t))C(y(t))+K<Q+K,  t∈[T0,T), we obtain
(2.9)dV1(t)μ(V1(t))+ω(V1(t))≤(M+|e(t)|)dt,t∈[T0,T).
We denote that  V1(t)=V1(t,x(t),y(t)).
Sincelim|y|→∞sgn(y)C(y)=Q,  ∫0K+Q(ds/(μ(s)+ω(s)))=∞,  y(t),  C(y) are continuous, there exists T0≤t1<t2<T such that
(2.10)∫V1(t1)V1(t2)dsμ(s)+ω(s)>M∫0Tdt+∫0T|e(t)|dt.
Integrating (2.9) on [t1,t2] with respect to t and using the above relation, we obtain the following contradiction:
(2.11)M∫0Tdt+∫0T|e(t)|dt<∫V1(t1)V1(t2)dsμ(s)+ω(s)=∫t1t2dV1(t)μ(V1(t))+ω(V1(t))≤∫t1t2(M+|e(t)|)dt≤M∫0Tdt+∫0T|e(t)|dt.
Thus, there exists an M>0 such that
(2.12)|y(t)|≤M,t∈[0,T).
Second, by the result above, we have limt→T|x(t)|=∞.
Iflim|x|→∞(1/(Q-sgn(x)H(x)))=∞, that is, lim|x|→∞sgn(x)H(x)=Q, we set
(2.13)V2(t,x,y)=sgn(x)H(x)+K,t∈R+,x,y∈R.
Since x(t) is continuous, there exists 0≤T1<T such that
(2.14)|x(t)|>N,t∈[T1,T).
Differentiating V2(t,x,y) with respect to talong solution (x(t),y(t)) of  (1.1), we have
(2.15)dV2dt=sgn(x)[a(x)c(y)h(x)-a(x)b(x)h(x)]≤(M+1)[μ(sgn(x)H(x)+K)+ω(sgn(x)H(x)+K)],t∈[T1,T).
Since  0≤sgn(x(t))H(x(t))+K<Q+K, t∈[T1,T), we obtain
(2.16)dV2(t)μ(V2(t))+ω(V2(t))≤(M+1)dt,t∈[T1,T).
We denote  V2(t)=V2(t,x(t),y(t)).
Sincelim|x|→∞sgn(x)H(x)=Q, ∫0K+Q(ds/(μ(s)+ω(s)))=∞,  x(t),H(x) are continuous, there exists T1≤t3<t4<T such that
(2.17)∫V2(t3)V2(t4)dsμ(s)+ω(s)>(M+1)∫0Tdt.
Integrating (2.16) on [t3,t4] with respect to t and using the above relation, we obtain the contradiction as follows:
(2.18)(M+1)∫0Tdt<∫V2(t3)V2(t4)dsμ(s)+ω(s)=∫t3t4dV2(t)μ(V2(t))+ω(V2(t))≤∫t3t4(M+1)dt≤(M+1)∫0Tdt.
So consider lim|x|→∞(1/(Q-sgn(x)H(x)))<∞.
By (iii), we havelim|x|→∞sgn(x)b(x)=∞.
Set(2.19)W(t,x,y)=x,t∈R+,x,y∈R.
Then, along solutions to (1.1) we have
(2.20)dWdt=1a(x)[c(y)-b(x)].
If  limt→Tx(t)=∞, we deduce that there exist x1 and x2 such that x0<x1<x2 and
(2.21)dWdt<0,x1≤x≤x2,|y|≤M.
Then, by the continuity of the solution, there exist 0<t1<t2<T such that x(t1)=x1, x(t2)=x2. Integrating (2.21) on [t1,t2], we have
(2.22)W(t1,x(t1),y(t1))=x1>x2=W(t2,x(t2),y(t2)).
This contradicts x1<x2. Hence x(t) is bounded from above.
Similarly, iflimt→Tx(t)=-∞, we can obtain a contradiction by setting W(t,x,y)=-x. Thus, it follows that x(t) is also bounded from above. This forces T=∞ and completes the proof of Theorem 2.1.Example 2.2.
Consider the following nonlinear system:(2.23)x′=1+x21+y2,y′=-11+x2+t21+x2.
Set a(x)=1/1+x2, b(x)=0, c(y)=1/(1+y2), h(x)=1, e(t)=t2. Then we have
(2.24)C(y)=arctany,H(x)=arctanx.
Take K=N=0,  Q=π/2, and  μ(θ)=ω(θ)=cosθ/2,  θ∈[0,π/2). Note that
(2.25)a(x)|c(y)|=11+x2·11+y2≤min{11+x2,11+y2}=min{μ(sgn(x)arctanx+K)+ω(sgn(x)arctanx+K),μ(sgn(y)arctany+K)+ω(sgn(y)arctany+K)}sgn(x)a(x)b(x)h(x)=0,|h(x)|=1 and
(2.26)∫0π/2dθμ(θ)+ω(θ)=∫0π/2dθcosθ≥12∫01dx1-x=∞.
Applying Theorem 2.1, we know that every solution of (2.23) exists globally. Observe that the theorem and corollary in [5] cannot be used in the present case.Theorem 2.3.
Assume that
(i) there exists someK≥0, such that
(2.27)H(x)+K≥0,x∈R,C(y)+K≥0,y∈R,
(ii) there exist someN≥0 and Q>0, such that
(2.28)|H(x)|<Q,|x|>N,|C(y)|<Q,|y|>N,
(iii)lim|y|→∞C(y)=Q, lim|x|→∞[1/(Q-H(x))+sgn(x)b(x)]=∞,
(iv) there exist two positive functionsμ,ω∈C([0,K+Q),(0,∞)) such that
(2.29)a(x)|c(y)|≤min{μ(H(x)+K)+ω(H(x)+K),μ(C(y)+K)+ω(C(y)+K)},|x|>N,|y|>N,
(v)a(x)b(x)h(x)≥-[μ(H(x)+K)+ω(H(x)+K)], |x|>N and |h(x)|≤M<∞, x∈R.
If(2.30)∫0K+Qdsμ(s)+ω(s)=∞,
then every solution of  (1.1)  exists globally.Proof.
The proof of Theorem2.3 is similar to that of Theorem 2.1, so we omit it.Example 2.4.
Consider the following nonlinear system:(2.31)x′=2y(1+y2)2+2yln(1+x2)(1+y2)2,y′=-2x(1+x2)[1+ln(1+x2)]+t31+ln(1+x2).
Seta(x)=1/(1+ln(1+x2)),  b(x)=0,  c(y)=2y/(1+y2)2,  h(x)=2x/(1+x2), e(t)=t3. Then we have C(y)=1-(1/(1+y2)), H(x)=1-(1/(1+ln(1+x2))). Take K=N=0,  Q=1 and μ(t)=ω(t)=(1-t)/2,  t∈[0,1). Note that
(2.32)a(x)|c(y)|=11+ln(1+x2)·2|y|(1+y2)2≤min{11+ln(1+x2),11+y2}=min{μ(H(x)+K)+ω(H(x)+K),μ(C(y)+K)+ω(C(y)+K)}.a(x)b(x)h(x)=0,|h(x)|=2|x|/(1+x2)≤1 and
(2.33)∫01dsμ(s)+ω(s)=∫01ds1-s=∞.
Applying Theorem 2.3, we know that every solution of (2.31) exists globally. Observe that the theorem and corollary in [5] cannot be uses in the present case.


## 3. Boundedness

In this section, we will present some results on the boundedness of solutions to (1.1) under general conditions on the nonlinearities.Theorem 3.1.
Assume that
(i) there exist functionsf1,  f2∈C(R+,R) such that
(3.1)f1(t)≤y′(t)=-a(x)[h(x)-e(t)]≤f2(t),x∈R,t∈R+,
and|∫0∞f1(t)dt|<∞,|∫0∞f2(t)dt|<∞,
(ii)lim|x|→∞sgn(x)b(x)=∞.
If the solution(x(t),y(t)) of  (1.1)  exists globally, then (x(t),y(t)) is bounded.Proof.
By (i), we have(3.2)f1(t)≤y′(t)≤f2(t),x∈R,t∈R+.
Integrating (3.2) on [0,t] with respect to t, we have
(3.3)∫0tf1(s)ds≤y(t)-y0≤∫0tf2(s)ds.
Since |∫0∞f1(t)dt|<∞,  |∫0∞f2(t)dt|<∞. Thus, there exists a Y>0 such that
(3.4)|y(t)|≤Y,t≥0.
Set(3.5)W(t,x,y)=-x,t∈R+,x,y∈R.
Then, along solutions to (1.1), we have
(3.6)dWdt=-1a(x)[c(y)-b(x)].
If  limt→∞x(t)=-∞, we deduce that there exist x1 and x2 such that x(0)>x1>x2 and
(3.7)dWdt<0,x2≤x≤x1,|y|≤Y.
Then, by the continuity of the solution, we have that there exist 0<t1<t2<∞ such that x(t1)=x1 and x(t2)=x2. Integrating (3.7) on [t1,t2], we get
(3.8)W(t1,x(t1),y(t1))=-x1>-x2=W(t2,x(t2),y(t2)).
This contradicts x1>x2. Hence x(t) is bounded from below.
Similarly, iflimt→Tx(t)=∞, we can obtain a contradiction by setting W(t,x,y)=x. Thus, it follows that x(t) is also bounded from above. This completes the proof of Theorem 3.1.Example 3.2.
Consider the following nonlinear system:(3.9)x′=1+x21+y2-x1+x2,y′=11+x2·11+t2.
Set a(x)=1/1+x2,  b(x)=x,  c(y)=1/(1+y2),  h(x)=1,  e(t)=1+(1/(1+t2)). Then we have C(y)=arctany and  H(x)=arctanx. Take K=N=0,  Q=π/2 and μ(θ)=ω(θ)=cosθ/2, θ∈[0,π/2). Applying Theorem 2.1, we know that every solution of (3.9) exists globally.
Takef1(t)=0, f2(t)=1/(1+t2), we have
(3.10)f1(t)=0≤y′(t)≤f2(t)=11+t2,x∈R,t∈R+
and |∫0∞f2(t)dt|=π/2<∞, lim|x|→∞sgn(x)b(x)=lim|x|→∞xsgn(x)=∞. Applying Theorem 3.1, we know that every solution of (3.9) is bounded.Theorem 3.3.
Assume that
(i) there exists someK≥0, such that
(3.11)sgn(x)H(x)+K≥0,x∈R,sgn(y)C(y)+K≥0,y∈R,
(ii) there exist someN≥0 and Q>0, such that
(3.12)|H(x)|<Q,|x|>N,|C(y)|<Q,|y|>N,
(iii)lim|y|→∞sgn(y)C(y)=Q, lim|x|→∞[1/(Q-sgn(x)H(x))+sgn(x)b(x)]=∞,
(iv) there exist two positive functionsμ,ω∈C([0,K+Q),(0,∞)) such that
(3.13)a(x)|c(y)|≤min{μ(sgn(x)H(x)+K)+ω(sgn(x)H(x)+K),μ(sgn(y)C(y)+K)+ω(sgn(y)C(y)+K)},|x|>N,|y|>N,
(v)sgn(x)a(x)b(x)h(x)≥0, |x|>N and |h(x)|≤M<∞, x∈R,
(vi)E=∫0∞|e(t)|dt<∞, and ∫0K+Q(ds/(μ(s)+ω(s)))=∞.
If there existsg(x) such that
(3.14)0<a(x)|c(y)-b(x)|≤g(x),x,y∈R
and G=∫-∞∞g(x)|h(x)|dx<∞, then every solution of  (1.1)  is bounded.Proof.
Let(x(t),y(t)) be a solution to (1.1) with initial data (x0,y0). By Theorem 2.1, we have that (x(t),y(t)) exists globally. If (x(t),y(t)) is unbounded, we have
(3.15)limt→∞|x(t)|=∞orlimt→∞|y(t)|=∞.
First, assume thatlimt→T|y(t)|=∞.
Sincey(t) is continuous, there exists  T1≥0 such that
(3.16)|y(t)|>N,t∈[T1,∞).
Take V1(t,x,y)=sgn(y)C(y)+K, t∈R+, x,y∈R. Differentiating V1(t,x,y) with respect to t along solution (x(t),y(t)) of  (1.1), we have
(3.17)dV1dt=sgn(y)[-a(x)c(y)h(x)+a(x)c(y)e(t)]≤(|h(x)|+|e(t)|)a(x)|c(y)|≤(|h(x)|+|e(t)|)[μ(sgn(y)C(y)+K)+ω(sgn(y)C(y)+K)],t∈[T1,∞).
Since 0≤sgn(y(t))C(y(t))+K<Q+K, t∈[T1,∞), we obtain
(3.18)dV1(t)μ(V1(t))+ω(V1(t))≤(|h[x(t)]|+|e(t)|)dt,t∈[T1,∞).
We denote V1(t)=V1(t,x(t),y(t)).
By (vi), there existsT1≤t1<t2<∞ such that
(3.19)∫V1(t1)V1(t2)dsμ(s)+ω(s)>G+E.
Integrating (3.18) on [t1,t2] with respect to t and using the above relation, we obtain the contradiction as follows:
(3.20)G+E<∫V1(t1)V1(t2)dsμ(s)+ω(s)=∫V1(t1)V1(t2)dV1(t)μ(V1(t))+ω(V1(t))≤∫t1t2|h(x(t))|dt+∫t1t2|e(t)|dt≤∫x(t1)x(t2)|h(x)|dxx′(t)+E≤∫-∞∞g(x)|h(x)|dx+E=G+E.
Thus, there exists a Y>0 such that
(3.21)|y(t)|≤Y,t∈[0,∞).
Second assume thatlimt→T|x(t)|=∞.
Iflim|x|→∞(1/(Q-sgn(x)H(x)))=∞, that is, lim|x|→∞sgn(x)H(x)=Q, we set
(3.22)V2(t,x,y)=sgn(x)H(x)+K,t∈R+,x,y∈R.
Since x(t) is continuous, there exists 0≤T2<∞ such that
(3.23)|x(t)|>N,t∈[T2,∞).
Differentiating V2(t,x,y) with respect to talong solution (x(t),y(t)) of  (1.1), we have
(3.24)dV2dt=sgn(x)[a(x)c(y)h(x)-a(x)b(x)h(x)]≤|h(x)|[μ(sgn(x)H(x)+K)+ω(sgn(x)H(x)+K)],t∈[T2,∞).
Since 0≤sgn(x(t))H(x(t))+K<Q+K, t∈[T2,∞), we obtain
(3.25)dV2(t)μ(V2(t))+ω(V2(t))≤|h[x(t)]|dt,t∈[T2,∞).
We denote that V2(t)=V2(t,x(t),y(t)).
By (vi), there existsT2≤t3<t4<T such that
(3.26)∫V2(t3)V2(t4)dsμ(s)+ω(s)>G.
Integrating (3.25) on [t3,t4] with respect to t and using the above relation, we obtain the contradiction as follows:
(3.27)G<∫V2(t3)V2(t4)dsμ(s)+ω(s)=∫t3t4dV2(t)μ(V2(t))+ω(V2(t))≤∫t3t4|h(x(t))|dt=∫x(t3)x(t4)|h(x)|dxx′(t)≤∫-∞∞g(x)|h(x)|dx=G.
So consider lim|x|→∞(1/((Q-sgn(x)H(x))))<∞.
By (iii), we havelim|x|→∞sgn(x)b(x)=∞. The proof of this condition is similar to that of Theorem 3.1, so we omit it. Thus, it follows that x(t) is also bounded from above. Then every solution of  (1.1)  is bounded. This completes the proof of Theorem 3.3.Theorem 3.4.
Assume that
(i) there exists someK≥0, such that
(3.28)H(x)+K≥0,x∈R,C(y)+K≥0,y∈R,
(ii) there exist someN≥0 and Q>0, such that
(3.29)|H(x)|<Q,|x|>N,|C(y)|<Q,|y|>N,
(iii)lim|y|→∞C(y)=Q,  lim|x|→∞[1/(Q-H(x))+sgn(x)b(x)]=∞,
(iv)   there exist two positive functionsμ,ω∈C([0,K+Q),(0,∞)) such that
(3.30)a(x)|c(y)|≤min{μ(H(x)+K)+ω(H(x)+K),μ(C(y)+K)+μ(C(y)+K)+ω(C(y)+K)},|x|>N,|y|>N,
(v)sgn(x)a(x)b(x)h(x)≥0, |x|>N and |h(x)|≤M<∞, x∈R,
(vi)E=∫0∞|e(t)|dt<∞, and ∫0K+Q(ds/(μ(s)+ω(s)))=∞.
If there existsg(x) such that
(3.31)0<a(x)|c(y)-b(x)|≤g(x),x,y∈R
and G=∫-∞∞g(x)|h(x)|dx<∞, then every solution of  (1.1)  is bounded.Proof.
The proof of Theorem3.4 is similar to that of Theorem 3.3, so we omit it.



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*Source: 603783-2012-10-03.xml*