几类带有p-Laplacian算子微分方程边值问题解存在性.doc

Aa‘kp ? Laplacian?f??·§?ˉK  ) ? ? ) 6 ?: o  ·35 +  ? ‰ ; ’: A^ê? ? ? · ·: ??5??9A^  ·  “: ü ? r ¤ ? m: 2013 c 4  ? -H“‰??a??  ??  M5`2  x-(2: d?¤J  a???5Aa‘kp?Laplacian? f??·§?ˉK) ·356, ′ 3 “· e, 3-H “‰????a?? ?m?á?1??ó?¤ ¤J. ? ?￥?52ü? ?·1|?2uL? ??¤J. é ? ??ó?‰?-? z ?ú8N, t?3?￥?2( ·a52. (2 {?(Jò d ?ú.  ??\?:  F?: -H“‰??a??  ??|^?  ?  5Aa‘kp ? Laplacian?f??·§?ˉK)  ·356X 3-H“‰????a?? ?m, 3 “· e ¤ a ?? ??. ?? ??¤J8-H“‰??¤k, ?? ??SN? ±ù|ü ??uL.  )-H“‰? ?u ·!|^? ?? 5?, ó?? 3?·kü€ x ?? E?úf? , #N?? ú/ . ? -H“‰??, ?±?^K?ù|E??? ·??, ?±úm uL?? ü?ü?SN.  ??\?: “\?:  F?: F?: -H“‰??a??  ??  Aa‘kp ? Laplacian?f??·§?ˉK) ·35 á ? ‘X‰???uD, X8‘kp ? Laplacian?f ??5ˉK?F?ú?? 2 ·5, ??5???¤·y“ê?￥ -???··??, p ? Laplacian?f?? 9A^′??5??￥ ??-??|, ?ùUéD )og,.￥ ??? g ,y–é IS ê?.úg,‰?. -à. ??5??·§?ˉK uA^ ê?, ?n?, ??? ??A^?‰￥, ′8c??5??9A^￥??··1 +·??. ?|^In?ú??:n?, ?? Aa?é9è??ˉK) ·35 9??5, ?r (J. ?aSN ì?(JA^ ?A ??5??·§ ?ˉK, ??·±enù: ? # 1?ù 3 ù￥, ·?·????mt?kp ? Laplacian?fóà??·§  ? ? ?  ?u|t=tk = Ik(u(tk)), (k = 1, 2, 3, · · · , n),  (1.1.1) ? ?  u(+∞) = βu(0), u (+∞) = αu (0)  )  ·35, ù￥?p(s) =| s |p?2 s, p 1, β 1, 0 α 1, J = [0, +∞), 0 t1 t2 t3 · · · tn +∞, J = J\{t1, t2, t3, · · · , tn}, Ik ∈ C[R+, R+]. u(+∞) = limt→+∞ u(t), u (+∞) = limt→+∞ u (t) ≥ 0, R+ = [0, +∞), ?u|t=tkL??êu(t)3t = tk?a Y, ùpu(t+k ), u(t?k )?OL?t = tk? m4·??4·. u(t)3t = tk??Y, 3t = tk???Y…u(t+k )·3, k = 1, 2, 3, · · · , n. ·?é???mt p ? Laplacian? fóà??·§ ) ·35Dm??, |^I. ?? ??:?n ? ) · 35, ?‰???~f?·ì?(? A^. 1 ù 3 ù￥,·?·??é: p ? Laplacian?fóà??·§  ? ? ?  ?u|t=tk = Ik(u(tk)), (k = 1, 2, 3, · · · , n),  (2.1.1) ? ?  u(1) = u(0) = 0  i -H“‰??a??  ??  )  ·35, ù￥?p(s) =| s |p?2 s, p 1, g(z)3R \ {0}?K?Y  , 3z = 0?? é, f(t, z, r)3[0, 1] × R × Rt?K?Y. J = [0, 1], J+ = (0, 1), 0 t1 t2 t3 · · · tn 1, J+ = J+\{t1, t2, t3, · · · , tn}, Jk = (tk?1, tk), k = 1, 2, 3, · · · , n. Ik ∈ C[R+, R+], ?u|t=tkL