GENERAL BOUNDEDNESS THEOREMS TO SOME SECOND ORDER NONLINEAR DIFFERENTIAL EQUATION WITH INTEGRABLE FORCING TERM

In this note we present a boundedness theorem to the equation x"+ c(t,x,x’)+ a(t)b(x)= e(t)where e(t) is a continuous absolutely integrable function over the nonnegative real line. We then extend the result to the equation x" + c(t, x, :r’) + a(t, x) e(t). The first theorem provides the motivation for the second theorem. Also, an example illustrating the theory is then given. KEY WOROS AND PHRASES. Integrable forcing term, bounded, nonlinear differential equation. 1990 AMS SUBJECT CLASSIFICATION CODE. 34C11.


INTRODUCTION.
In this article we shall discuss using standard methods the boundedness properties of a second order nonlinear differential equation with integrable forcing term, i.e. the equation, " + (,, ') + ()b() () (.l)Our purpose here is to simplify some of the previous proofs to this well-known equation as well as extending some of the previous results.For example, we are replacing the condition c(t, x, y)y > 0 for y :/: 0 with c(t, x, y)y _> 0 and letting a(t) be non-increasing (see and [2] for details, especially [2] for its excellent bibliography of previous work).Also, as in [2] we shall not need to make use of any Liapunov function.Finally, the result will be of such a nature that it covers the ease when no damping factor appears, i.e. it covers the equation, x" + a()b(x) e() (1.2) Later we shall briefly mention how this result carries over to the more general nonlinear equation, " + (, , ') + (, ) () (.3)However, this ease requires a more delicate discussion.We now state and prove the boundedness theorem.Without loss ofgenerality, we shall assume t _> 0.
PROOF.By standard existence theory, there is a solution to (1) which exists on [0, T) for some T > O for any initial conditions x(0) and x'(0).Multiply equation ( 1) by z and perform an integration by parts on the last term from 0 to < T in order to obtain, x'(t)/2 + c(, z(s), x'(s))z'(s)d + a(t)B(x(t)) a' (s)B(x(s))ds '(o/ + (/:'(/a _< z'(o// + I(l'(lld (.l Now if x(t) becomes unbounded then we must have that all terms on the LHS of (2.1) become positive from our hypotheses.By the mean value theorem, equation (2.l) may be rewritten as, x'(t)2/2 + c(a,x(s),x'(s))ds + a(t)B(x(t)) a'(a)B(x(s))ds Now from (2.2) we see that if' Ixl approaches oo then so must Ix'(t)l.Otherwise, the LHS of' (2.2) becomes unbounded while the RHS stays bounded which is impossible.Also, as la (t)l approaches oo so must Iz()l.Now on any compact subinterval choose t where z(t) is a maximum.Integrate equation (1.1) as before from 0 to t and divide by :d (t) (assume x(t) > 0, a similar argument works for z (t) < 0 only the inequality is reversed) in order to obtain, x'(t)/2 + 1/x'(t) ( fot C(S, x(s), x'())d + a(t)B (x(t) fot a' (s)B(x() )d) Now if x'(t) approaches oo then the LHS of (2.3) becomes unbounded while the RHS of (2.3) stays bounded which is a contradiction.Thus, Izl and I'1 must stay bounded on [0, T).A standard argument ([3, pp.17-18]) now permits the solution to be emended on all of [0, oo).
As for equation (1.3) we may multiply it by x' and integrate as before obtaining the following, '(t)/ + (,(),e())'()a + .' (t, )d _' dd Oa(s, .4) f0 We see here that as long as a(t, u)du oo uniformly in t and xa(t, x) <_ 0 then we may use the same argument as in our first theorem.We now state this final result.
TIIEOREM IL Given equation (1.3).Suppose c(t,x,y) is continuous on [0,oo) x R x R, c(t, x, y)y > O, a(t, x) continuous on [0, oo) x R with x a(t, x) < O. Furthermore, if fo:att u)du oo uniformly in t and e(.) is continuous on [0, oo) with fo le(t)ldt < oo, then all solutions to equation (1.3) as well as their derivatives are bounded as oo.
EXAMPLE.Consider the nonlinear differential equation, x" + cx2'-lx + bx 2n-exp( t) (2.5) where t > 0, c, b are positive and m, n are positive integers.By Theorem we see that all solution to equation (2.5) are bounded.[3]