ISSN:0024-6115    

DOI:10.1112/plms/s3-17.2.342

出版商:Oxford University Press    

年代:2016

数据来源: WILEY

ISSN:0024-6115    

DOI:10.1112/plms/s3-17.2.355

出版商:Oxford University Press    

年代:2016

数据来源: WILEY

ISSN:0024-6115    

DOI:10.1112/plms/s3-17.2.381

出版商:Oxford University Press    

年代:2016

数据来源: WILEY

摘要:

Let ψ(Σ) denote the frontier of the square Σ (see § 3.1). A major part of the proof of the theorem rests on a tacit assumption, in §4.1, that if the condition(1) |ξ(t|⩽D3and |η(t)|⩽D3for allt⩾t0is not fulfilled then the corresponding curve γnecessarilyintersects ψ(Σ)in such a way that, from a certaintonwards,allthe intersection‐points can be enumerated one after another thus giving rise to the sequence {E8}. This assumption is incorrect, as has been pointed out to me by Professor H. Pachale. Indeed, if (1) does not hold, γ may well behave as described in §4.1, but there is also a possibility that, ast→ ∞, the representative point (ξ,η) on γ may move on the frontier ψ(Σ) itself for brief periods or that γ may intersect ψ(Σ) in a set of points with limit‐points on ψ(Σ), and in either case the special enumeration, referred to above, of all the points of intersection of γ with ψ(Σ) would be impossible.With the additional condition(2) |p(t)| ⩽A1<∞ for alltconsidered,onp(t), it is however possible to validate once again the entire content of §4.1, though with a somewhat different Σ; so that the boundedness theorem in the paper is valid subject to this further restriction (2) onp.Indeed letd1= max{D3,(A1+1)δ0−1},d2= max{D3,[A1+1+1+δ2d1]δ1−1};and letΣ*denote the rectangle|x| ⩽d1, |y| ⩽d2in the plane π. Here δ0, δ1are the constants in the hypotheses of the theorem, andD3is the constant given in §2.2. Our claim about the validity of the work in §4.1 under the present conditions rests upon the followingLEMMA.Assume that(1)holds and also that all the previous conditions on a, f, g, p are satisfied. Then every curve γ in the plane π necessarily satisfies one or other of the following:(I) |ξ(tverbar; ⩽d1and |η(t)| for allt⩾t0,(II)4as t increases to+∞the curveγintersects the frontier ofΣ*repeatedly in such a way that the intersection‐points can be enumerated one after another exactly as in Fig.1 (p. 108)but with the vertices Q1, Q2, Q3, Q4replaced by the points (−d1,d2), {d1,d2), {d1,−d2), (−d1, −d2)respectively.Proof of Lemma. Sinced1⩾D3andd2⩾D3it is clear from an argument in §3.2 that (ξ,η) cannot stay outside Σ*indefinitely. Therefore to prove the lemma it suffices to show that, under our present conditions, it is impossible to find an infinite sequence {tn} (n= 1,2,…) of values oft,with a finite limit‐point,t*say, such that any one of the following is satisfied:(3) |ξ(tn)| ⩽d1and η(tn) =d2(n=l,2,…),(4) |ξ(tn)| ⩽d1and η(tn) =−d2(n=l,2,…),(5) ξ(tn)=d1(n=l,2,…),(6) ξ(tn)= −d1(n=l,2,…),Our proof of this will be by contradiction. Suppose, on the contrary, that there is an infinite sequence {tn} with a finite limit‐pointt*, satisfying (3). Then, because of the continuity of ξ(t),η(t),|ξ(t*)| ⩽d1, ξ(t*)≡ η(t*)=d2.Also, from the fact thatη(t1)=η(t2)=…=η(tn)=…=d2there would be, by Rolle's theorem, an infinite sequence {τn}, witht*as a limit‐point, such that(7)ξ¨(τn)≡η˙τ(n)= 0 (n= 1,2,…);and from this, by virtue of the continuity ofξ¨(t), we derive the resultξ¨(t*)= 0.In view of (7) it is also clear, by applying Rolle's theorem to the sequence {ξ¨(τn)} and then using the continuity ofξ˙¨(t), thatξ˙¨(t*= 0.The substitution of these values,ξ˙(t*)=d2,ξ¨(t*)= 0 =ξ˙¨(t*),in the differential equation itself gives(8)f(ξ(t*))d2= −g(ξ(t*))+p(t*).Sincef⩾δ1, |p|⩽A1, |ξ(t*)|⩽d1,and since our hypotheses ongimply that|g(ξ)|⩽δ2|ξ|,it follows at once from (8) that(9)d2δ1⩽δ2d1+Aa.However,d2has been fixed to satisfyd2⩾δ1−1(A1+1 +δ2d1),so thatd2δ1−δ2d1⩾A1+1,and thus (9) cannot hold. Hence (3) is impossible. In the same way it can be shown that (4) cannot hold for any sequence {tn} with a finite limit point.It remains now to dispose of (5) and (6). Suppose, on the contrary, that there is an infinite sequence {tn}, with a finite limit‐pointt*, such that (5) holds. Then, in exactly the same way as before, we shall haveξ(t*)=d1,ξ˙(t*)=0=ξ¨(t*)=ξ˙¨(t*);and then, by substituting these values in the differential equation, one finds thatg(d1)=p(t*)or, since |g(d1)|⩾δ0d1and |p|⩽A1, thatδ0d1⩽A1But this is impossible since dx has been fixed so thatδ0d1⩾A1+1.Similarly it can be shown that (6) cannot hold; and the lemma is thereby proved.

ISSN:0024-6115    

DOI:10.1112/plms/s3-17.2.382-s

出版商:Oxford University Press    

年代:2016

数据来源: WILEY

ISSN:0024-6115    

DOI:10.1112/j.1460-244X.1967.tb00237.x

出版商:Oxford University Press    

年代:1967

数据来源: WILEY