一類非線性梁組的解的存在性

一類非線性梁組的解的存在性

sim環(huán)繞考慮了在邊界條件下，b.0，1.2.0的特殊情況下，方程解的存在性，android證明了在齊次邊界條件下，當(dāng)1.2.0時，方程弱解的存在性和唯一性，而to傅里葉證明了非線性邊界條件下方程u的存在性和能量衰減的估計。該工作證明了當(dāng)=1時（1）到（5）的總體解的存在性?！+c¨v+η1˙u+αu(4)-(a+b∫l0(u(1))2dx)u(2)=f1(x,t),(1)c¨u+γ¨u+η2¨v+δv(4)-β0¨v(2)=f2(x,t).(2)式中,α,δ,γ,η1,η2,a,b,β0,c是常數(shù),且c2>γ.邊界條件:u(0,t)=u(1)(0,t)=u(2)(l,t)=0,(3)u(3)(l,t)-(a+b∫l0(u(1))2dx)u(1)(l,t)=f(u(l,t))+g(˙u(l,t)),v(0,t)=v(l,t)=v(2)(0,t)=v(2)(l,t)=0;(4)初始條件:u(x,0)=u0(x),v(x,0)=v0(x),˙u(x,0)=u1(x),˙v(x,0)=v1(x).(5)1v20,lv0.t記V1={u∈Η2(0?l)|u(0,t)=u(1)(0?t)=0},V2={v∈Η2(0,l)|v(0,t)=v(l,t)=0},W1={u∈V1∩Η4(0?l)|u(2)(l,t)=0},W2={v∈V2∩Η4(0?l)|v(2)(0,t)=v(2)(l,t)=0}.2計算步驟1.2.2,5.2回2.2,2,4.2回2.2定理:給定函數(shù)u0(x)∈W1,v0(x)∈W2,u1(x)∈V1,v1(x)∈V2,對任意T>0在下面條件下1)f1(x,t),f2(x,t)∈L2(0,T;L2(0,l));2)f,g:R→R為連續(xù)可導(dǎo)函數(shù),且f(s)s-2Λf(s)≥0??s∈R,其中Λf(s)=∫s0f(z)dz;3)g(0)=0,(g(r)-g(s))(r-s)≥ρ|r-s|2,?r,s∈R,其中ρ>0為常數(shù),存在定義在[0,T]上有意義的函數(shù){u(x,t),v(x,t)}滿足:a.u(x,t)∈L∞(0,T;V1),v(x,t)∈L∞(0,T;V2),b.˙u(x,t),˙v(x,t)∈L∞(0?Τ;H2),c.對所有的w1∈V1,w2∈V2,{u(x,t),v(x,t)}是下面方程組在D′(0,T)的解(¨u?w1)+c(¨v?w2)+η1(˙u?w1)+w1(l)[f(u(l,t))+g(˙u(l,t))]+(a+b∫l0(u(1))2dx)(u(1)?w(1)1)+(u(2),w(2)1)=(f1(x,t),w1),c(¨u?w1)+γ(¨v?w2)+η2(˙v?w2)+β0(v(1)?w(1)2)+δ(v(2)?w(2)2)=(f2(x,t),w2),d.u(x,0)=u0(x)?˙u(x,0)=u1(x)?˙v(x,0)=v1(x),v(x,0)=v0(x).證明:Stepl近似解我們考慮:a)～d)對應(yīng)的變形問題,即求u(x,t)∈L∞(0,T;V1),v(x,t)∈L∞(0,T;V2)使對一切wi∈Vi(i=1,2)有3)中二式成立。這可用Galerkin逼近法得到,設(shè){w1j},{w2j}為V1,V2中的完備正交基,且{u0?u1}∈span{w11?w12}?{v0?v1}∈span{w21?w22}.對于一切m∈N,令wim=span{wi1,wi2,??wim},(i=1,2).我們來求函數(shù)um(t)=m∑j=1gmj(t)w1j?vm(t)=m∑j=1hmj(t)w2j.使對?w1j∈w1m,?w2j∈w2m,um(t),vm(t)滿足逼近方程(¨um(t),w1)+c(¨vm(t),w2)+η1(˙um(t),w1)+w1(l)[f(um(l,t)+g(˙um(l,t))]+(a+b∫10(u(1)m)2dx))(u(1)m(t),w(1)1)+(u(2)m(t),w(2)1)=(f1(x,t),w1),(6)c(¨um(t),w1)+γ(¨vm(t),w2)+η2(˙vm(t),w2)+β0(v(1)m(t),w(1)2)+δ(v(2)m(t),w(2)2)=(f2(x?t),w2),(7)及初始條件:um(x,0)=u0m(x),vm(x,0)=v0m(x),u˙m(x,0)=u1m(x),v˙m(x,0)=v1m(x);且u0m→u0?v0m→v0在Η4中強(qiáng)收斂,u1m→u1?v1m→v1在Η1中強(qiáng)收斂.(8)注意到,式(6)～(8)事實上是變量t的m×m常微分方程組,由常微分方程理論知:存在tm>0,使得存在系統(tǒng)(6)～(8)的解:{um(x,t),vm(x,t)}Step2先驗估計估計1:在(6),(7)式中取w1=2u˙m(t),w2=2v˙m(t)并從0到t積分且相加后,根據(jù)(8)的收斂性及條件2),3)及fΛ(um(l,t))的性質(zhì),依據(jù)許瓦茲不等式及放縮原理我們有:(1+c)|u˙m(t)|2+a|um(1)(t)|2+|um(2)(t)|2+(c+γ)|v˙m(t)|2+β0|vm(1)(t)|2+δ|vm(2)(t)|2≤c+c∫0l[|u˙m(t)|2+|v˙m(t)|2+|um(1)(t)|2+|um(2)(t)|2+|vm(1)(t)|2+|vm(2)(t)|2]ds.據(jù)Gronwall不等式得|u˙m(t)|,|v˙m(t)|,|um(2)(t)|,|um(1)(t)|?|um(2)(t)|?|vm(1)(t)|＜c.其中c與m無關(guān)。估計2:現(xiàn)在我們來得出u¨(0)?v¨(0)在L2中模的估計式,在(6),(7)式中取w1=u¨m(0),w2=v¨m(0)及t=0并相加后且根據(jù)(8)的收斂性有(1+c)|u¨m(0)|2+(c+γ)|v¨m(0)|2≤(η1|u1|-|u0(4)|+(a+b|u0(1)|2)|u0(2)|+|f1(x,0)|)|u¨m(0)|+(η2|v1|-δ|v0(4)(0)|+β0|v0(2)|+|f2(x,0)|)|v¨m(0)|.因為η1|u1|-|u0(4)|+(a+b|u0(1)|2)|u0(2)|+|f1(x,0)|)|≤d1,η2|v1|-δ|v0(4)|+β0|v0(2)|+|f2(x,0)|)≤d2,所以有|u¨m(0)|2+c+γc+1|v¨m(0)|2-d1c+1|u¨m(0)|-d2c+1|v¨m(0)|≤0.則根據(jù)配方法我們有|u¨m(0)|≤c,|v¨m(0)|≤c(d1,d2為常數(shù)).估計3:現(xiàn)在我們將推導(dǎo)u¨?v¨在L2中模的估計式,在(6)(7)都分別取t=t+ξ,t=t,并將(7)所得兩式相減(8)所得兩式相減,且令w1=u˙m(t+ξ)-u˙m(t),w2=v˙m(t+ξ)-v˙m(t).再將相減所得二式相加得12ddtΦm(t,ξ)+η1|u˙m(t+ξ)-u˙m(t)|2+η2|v˙m(t+ξ)-v˙m(t)|2+A+B+Ι1=(f1(x,t+ξ)-f1(x,t),u˙m(t+ξ)-u˙m(t))+(f2(x,t+ξ)-f2(x,t),v˙m(t+ξ)-v˙m(t)).其中:Ι1=b|um(1)(t+ξ)|2[(um(1)(t+ξ)?u˙m(1)(t+ξ)-u˙m(1)(t))]-b|um(1)(t)|2[(um(1)(t),u˙m(1)(t+ξ)-u˙m(1)(t))],A=[f(um(l,t+ξ))-f(um(l,t))]×[u˙m(l,t+ξ)-u˙m(l,t)],B=[g(u˙m(l,t+ξ))-g(u˙m(l,t))]×[u˙m(l,t+ξ)-u˙m(l,t)],Φm(t,ξ)=(1+c)|u˙m(t+ξ)-u˙m(t)|2+a|um(1)(t+ξ)-um(1)(t)|2+|um(2)(t+ξ)-um2(t)|2+(c+γ)|v˙m(t+ξ)-v˙m(t)|2+β0|vm(1)(t+ξ)|-vm(1)(t)|2+δ|vm(2)(t+ξ)-vm2(t)|2.通過估計I1可得:|Ι1|≤c2|um(2)(t+ξ)-um(2)(t)|2+|u˙m(t+ξ)-u˙m(t)|2+ρ2|u˙m(l,t+ξ)-u˙m(l,t)|2;通過估計A得A≤c3|um(2)(t+ξ)-um(2)(t)|2+ρ2|u˙m(l,t+ξ)-u˙m(l,t)|2;通過估計B得B≥ρ|u˙m(l,t+ξ)-u˙m(l,t)|2.所以ddtΦm(t,ξ)≤2Μ1Φm(t,ξ).根據(jù)Gronwall不等式可得:Φm(t?ξ)≤Φm(0,ξ)exp(ΜΤ)??t∈[0?Τ]上述不等式除以ξ2,并令ξ→0根據(jù)導(dǎo)數(shù)的意義得:(1+c)|u¨m(t)|2+a|u˙m(1)(t)|2+|u˙m(2)(t)|2+(γ+c)|v¨m(t)|2+β0|v˙m(1)(t)|2+δ|v˙m(2)(t)|2≤(1+c)|u¨m(0)|2+a|u˙1(1)|2+|u˙1(2)|2+(γ+c)|v¨m(0)|2+β0|v˙1(1)|2+δ|v˙1(2)(0)|2.由估計1,估計2可得僅依賴于T的常數(shù)M2使得,對?m∈Ν?t∈[0?Τ](1+c)|u¨m(t)|2+a|u˙m(1)(t)|2+|u˙m(2)(t)|2+(γ+c)|v¨m(t)|2+β0|v˙m(1)(t)|2+δ|v˙m(2)(t)|2≤Μ2.所以|u¨m(t)|,|u˙m(1)(t)|,|u˙m(2)(t)|,|v¨m(t)|,|v˙m(1)(t)||v¨m(2)(t)|≤c.Step3收