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1、Uniform Boundedness for aPhytoplankton-ToxicPhytoplankton-Zooplankton Model withCross-DiusionTian Wang西北师范大学研究生学位论文作者信息论文题目一类带毒素的浮游植物-浮游动物交错扩散模型解的一致有界性姓 名王 田学 号2010210784专业名称偏微分方程及其应用答辩日期2013.05.24联系电话15117288618E_通信地址(邮编)：备注：8Abstract1 1 ! 1 2 ! (Jiii191 3 ! (Jzy1131amuL35375a2i - 2i*.5. AUO Gaglia
2、rdo-Nirenberg .yT.N)35k.5.c: 2i- 2i.;*;N); k.5iAbstractIn this paper, the dissipativeness of a phytoplankton-toxic phytoplankton-zooplanktonmodel with cross-diusion is investigated. Using the energy estimates and Gagliardo-Nirenberg-type inequalities, the existence and uniform boundedness of global
3、solu-tions for the model are proved.Keywords: Phytoplankton-toxic; phytoplankton-zooplankton model; Cross-diusion; Global solution; Uniform boundedness;ii1 1 ! 3!9AY, 2i)gvkU, kf, U_Y6, 23Y). eka.2i): (1) , 2i, ), *; (2) , 2i, )32i. Cc5, kz2i- 2iX 4, 7, 14, 24, 28, 29, Banerjee Venturino 3 4 2i2i, 2
4、i2iO5. AO, du?Monod-Haldane UA, 2i2i, d, .= 22 1 b 12 1 dt K2 2 + 22,(1.1) i, Ki, i, ei (i = 1, 2), a, b, 2 , (t), 1(t) 2(t) OL2i, 2i92i+. 2i2i,2i, k, ,2i, 2iUO2i+. Monod-Haldane UA 26:2i, e, 2i;m2i, 2i2i. T.z 4.u(t) = A1( ), v(t) = B2( ), w(t) = D ( ), = Et,A = K1,B =2,D = e22,E =1e222 2,1 d11 dt =
5、 11 1 K a 12 11,d2 2 22 d e222 ,= e111 2 + 22dtEP t , KX (1.1)z: du dv v vw= r2v 1 buv K 1 + v2dt 1 + v2K2r1 = 1E, r2 = 2E, K = -20,2,(1.2)a =ae222, b =e222 2, c = 1DE =122,e = e11AE =e222 2, m = E =e222 2.X (1.2) kU6 : E0 = (0, 0, 0); E1 = (1, 0, 0); E2 =(0, K, 0); E3 = (u3, 0, w3), u3 =me 0, w3 =r
6、1c1 me 0, XJ m 0, v4 = r1Kbr2abK r1r2 0, XJ (ak r1)(b r2) 0;: E5 = (u, v, w) Jw/L5, O, XJr2Kr1ce,mb e(1.3)e, =cr1e mer2 bme mer2 bmae + bcer1r2K,(1.5)KX (1.2) 3: E5 = (u, v, w).2 dt = r1u(1 u) auv cuw, dt dwvw= euw mw,bK1e11K1PA1 = r1u r2vK+2v2w(1 + v2)2,A2 =r1u r2vK2v2w(1 + v2)2 abuv+ ceuw (1 v2)vw(1 + v2)3,A3 =1 v2w(1 + v2)2r1uv2 bcuv+ ew 1 + v2 cr2uvK+2cuv2w(1 + v2)2.Banerjee Venturino 3 4 yXe(:(1) E0 = (0, 0, 0) -;(2) X r2 b, e m , K E1 = (1, 0, 0) C-;(3) X r1 aK , K E2 = (0, K, 0) C-;(4) Xr2 abK,r1K(r2 b)(r1r2 abK )(r1r2 abK )2 + (b r2)2r12K 2= me, E4 =
