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A Vapore-Free Vacuum Seal(en)(2s) by Prund A. H.

By Prund A. H.

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Theorem 7. Assume that ð116Þ g is nonincreasing and that there exists a nondecreasing function YACðRþ 0 ; RÞ; with Yð0Þ ¼ 0; such that f ðr; uÞ XYðuÞ gðuÞ for all rX0 and all If lim sAðsÞ ¼ N s-N u > 0: ð117Þ R. Filippucci / J. Differential Equations 188 (2003) 353–389 385 and Y satisfies the condition Z N H À1 Z t YðtÞ dt À1 dtoN ð118Þ then inequality (105) does not admit any entire positive solution. Proof. Assume by contradiction that (105) admits a positive entire solution u : Rn -Rþ : Since AðsÞps2 AðsÞ for s > 0 by (C1) and gu ðuÞp0 for all uX0; by (105) and (117), we have divðAðjrujÞruÞX gu ðuÞ f ðjxj; uÞ ½AðjrujÞ À jruj2 Aðjrujފ þ gðuÞ gðuÞ X YðuÞ: ð119Þ Consider now the initial value problem ðrnÀ1 Aðjv0 jÞv0 Þ0 ¼ rnÀ1 YðvÞ; vð0Þ ¼ aAð0; uð0ÞÞ; v0 ð0Þ ¼ 0: ð120Þ By Theorem 1 the initial value problem (120) admits a local solution v : ½0; RÞ-Rþ ; where R > 0: Without loss of generality, we may assume that ½0; RÞ is the maximal interval of existence of v: By Theorem 5 we have that RoN: As noted in the proof of Theorem 5 it results that v0 ðrÞ > 0 for rAð0; RÞ: Hence either lim vðrÞ ¼ N or r-RÀ lim v0 ðrÞ ¼ N: r-RÀ Case 1: lim vðrÞ ¼ N: r-RÀ In this case, we can take R1 Að0; RÞ so that vðR1 ÞXmaxfuðxÞ : jxj ¼ R1 g: ð121Þ Define B1 ¼ fxARn : jxjoR1 g: Then the function v ¼ vðjxjÞ is such that divðAðjrvjÞrvÞ ¼ YðvÞ in B1 and we have that vXu on @B1 : Hence, by Theorem 6 applied with O ¼ B1 ; it results that upv in B1 and this is a contradiction since vð0Þ ¼ aouð0Þ: Case 2: lim v0 ðrÞ ¼ N: r-RÀ 386 R.

Filippucci / J. À1 Z t Z N À1 1 H f ðtÞ dt dt ¼ N; n vð0Þ vð1Þ which contradicts (109). Hence (106) does not admit any global positive solution. & In the sequel, we will also need the weak comparison principle which is due to Pucci, Serrin and Zou (see [15]), which we state only when the domain OCRn is bounded. Theorem 6 (Pucci et al. [15, Lemma 3]). Let u and v be respective solutions of the differential inequalities: divðAðjrujÞruÞ À YðuÞX0; uX0; ð114Þ divðAðjrvjÞrvÞ À YðvÞp0; vX0 ð115Þ in a bounded domain O of Rn ; nX2: Assume that the function YACðRþ 0 ; RÞ is such that Yð0Þ ¼ 0; Y is nondecreasing in ½0; dÞ; 0odpN: % with uod in O and vXu on @O; then vXu in D.

If lim sAðsÞ ¼ N s-N and Z N H À1 Z À1 t f ðtÞ dt dtoN; ð107Þ where HðsÞ ¼ s2 AðsÞ À AðsÞ; s > 0; ð108Þ then the initial value problem (106) does not admit any global positive solution. Remark. Theorem 5 was proved by Naito and Usami [6] under stronger regularity assumptions on the operator A and by assuming condition (5) in place of (107). Note that when (107) fails then (106) admits global positive solutions, as shown by Naito and Usami. Thus (107) is a necessary and sufficient condition for the nonexistence of global solutions of (106).

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