本文考虑Kirchhoff方程
| \[\left\{ {\begin{array}{*{20}{l}} { - (a + b\int {_\Omega |\nabla u{|^2}\Delta u} = f(u),}&{x \in \Omega ,}\\ {u(x) = 0,}&{x \in \partial \Omega } \end{array}} \right.\] | (1) |
变号解的存在性,其中Ω是R2中有界光滑区域,a,b>0是常数. 非线性项f满足下面条件:
(f1) 对每个β>0,存在一个正数Cβ,使得|f(t)|,|f′(t)|≤Cβexp(β t2),t∈R；
(f2)$\mathop {\lim }\limits_{t \to 0} \frac{{f(t)}}{t} = 0$；
(f3)$\frac{{f(t)}}{{|t{|^3}}}$在区间(0,+∞)和 (-∞,0)上递增；
(f4)$\mathop {\lim }\limits_{t \to \infty } \frac{{F(t)}}{{{t^4}}} = + \infty ,F(t) = \int_\Omega {tf(s){\rm{ds}}} \cdot $
近年来,Kirchhoff方程
| \[\left\{ {\begin{array}{*{20}{l}} { - (a + b\int {_\Omega |\nabla u{|^2}\Delta u} + V(x)u = f(x,u),}&{x \in \Omega ,}\\ {u(x) = 0,}&{x \in \partial \Omega ,} \end{array}} \right.\] | (2) |
已经被国内外许多作者进行了深入研究,并且得到许多重要结果[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],其中Ω是RN中的区域,V(·):Ω→R,f∈ C(Ω×R,R),a,b>0是常数. 许多作者研究了Kirchhoff方程的基态解、正解、多重解等问题. 文献[5]用Nehair流形及紧性原理证明了问题(2)基态解的存在性. 其他有关Kirchhoff方程的研究可参看文献[12, 13, 14, 15, 16].然而,研究Kirchhoff方程变号解的结果还不多,文献[17]研究了三维情形下Kirchhoff方程变号解的存在性,他们的非线性项都是次临界增长,也就是|f(x,t)|≤ C(1+|t|p),p∈(1,5).受此启发,本文研究带有指数增长的Kirchhoff方程变号解的存在性.
记X=H01(Ω) 是通常意义下的Sobolev空间,其范数和内积分别为
| \[(u,\upsilon ) = \int {_\Omega \nabla u \cdot \nabla \upsilon ,||u|| = } {(u,u)^{1/2}}.\] |
空间Lp(Ω)是通常的Lebesgue空间,其范数记为|·|p,1≤p＜∞.C和Ck代表不同的正常数.u+(x)=max｛0,u(x)｝,u-(x)=min>｛0,u(x)｝.R+:=[0,+∞).
问题(1)对应的能量泛函为
| \[I(u) = \frac{a}{2}||u|{|^2} + \frac{b}{4}||u|{|^4} - \int {_\Omega F(u),u \in X.} \] |
显然I∈C1(X,R),且∀u,v∈X,
| \[\left\langle {{I^\prime }(u),} \right.\left. \upsilon \right\rangle = a\int {_\Omega \nabla u \cdot \nabla \upsilon + } b\int {_\Omega |\nabla u{|^2}} \int {\nabla u \cdot \nabla \upsilon - \int {_\Omega f(u)\upsilon .} } \] |
定义M=｛u∈X:u±≠0,〈I′(u),u+〉=〈I′(u),u-〉=0｝. m=min｛I(u):u∈M｝.
定理1 假设 (f1)~(f4)成立,那么方程(1)有一个最小能量变号解.
1 预备知识引理1 假设f满足(f1)~(f4),则对每个u≠0,u∈ X,有
| \[\begin{array}{l} \left( {\rm{i}} \right)\mathop {\lim }\limits_{t \to 0} \int_\Omega {\frac{{f(tu)u}}{t}} = 0.\\ \left( {{\rm{ii}}} \right)\mathop {\lim }\limits_{{\rm{|}}t{\rm{|}} \to \infty } \int_\Omega {\frac{{f(tu)u}}{{{t^3}}}} = + \infty .\\ \left( {{\rm{iii}}} \right)\mathop {\lim }\limits_{{\rm{|}}t{\rm{|}} \to \infty } \int_\Omega {\frac{{F(tu)}}{{{t^4}}}} = + \infty . \end{array}\] |
证明 (ⅰ) 由(f1)，(f2)知,∀β,ε>0,固定的p≥1,都存在Cβ,ε,p，使得
| \[|f(t)| \le \varepsilon |t| + {C_{\beta ,\varepsilon ,p}}|t{|^{p - 1}}\exp (\beta {t^2}),t \in {\rm{R}},\] | (3) |
| \[|F(t)| \le \frac{\varepsilon }{2}|t{|^2} + {C_{\beta ,\varepsilon ,p}}|t{|^p}\exp (\beta {t^2}),t \in {\rm{R}}{\rm{.}}\] | (4) |
由文献[18],存在常数C只与Ω有关,使得当u∈X\｛0｝,α‖u‖2≤4π时,
| \[\int_\Omega {\exp (a|u(x){|^2}{\rm{)d}}x = } \int_\Omega {\exp (a||u|{|^2}{\rm{)}}} \left[ {u(x)/||u|{|^2}} \right]{\rm{d}}x \le C.\] | (5) |
取β‖u‖2≤2π,p>2.由式(3)及(5)知,
| \[\int_\Omega {f(u)u| \le } \varepsilon \int_\Omega {{u^2}} + {C_{\beta ,\varepsilon ,p}}\int_\Omega {|u{|^p}} \exp (\beta {u^2}) \le C\varepsilon ||u|{|^2} + C{C_{\beta ,\varepsilon ,p}}|u{|^p}.\] | (6) |
由式(6)知
| \[\begin{array}{l} |\int_\Omega {f(tu)tu| \le } C\varepsilon {t^2}||u|{|^2} + C{C_{\beta ,\varepsilon ,p}}{t^p}||u|{|^p}.\\ |\int_\Omega {\frac{{f(tu)u}}{t}| \le } C\varepsilon ||u|{|^2} + C{C_{\beta ,\varepsilon ,p}}{t^{p - 2}}||u|{|^p}. \end{array}\] |
故由ε的任意性及p>2可得
| \[\mathop {\lim }\limits_{t \to 0} \int_\Omega {\frac{{f(tu)u}}{t}} = 0.\] |
(ⅱ) 由(f2)～(f4)易知,∀M>0,∃C>0,使得
f(t)t≥Mt4-Ct2,t∈R.
故可得
f(tu)tu≥Mt4u4-Ct2u2,t∈R.
上式同除t4并积分后取极限得
| \[\mathop {\lim }\limits_{{\rm{|}}t{\rm{|}} \to \infty } \int_\Omega {\frac{{f(tu)u}}{{{t^3}}}} \ge M|u|_4^4.\] |
由M的任意性可得
| \[\mathop {\lim }\limits_{{\rm{|}}t{\rm{|}} \to \infty } \int_\Omega {\frac{{f(tu)u}}{{{t^3}}}} = + \infty .\] |
(ⅲ)与(ⅱ)的证明类似.
引理2[19] 假设f满足条件(f1)~(f4),序列 ｛un｝⊂X,使得${u_n} \to u$,则
| \[\int_\Omega {f({u_n}){u_n} \to } \int_\Omega {f(u)u,} \int_\Omega F ({u_n}) \to \int_\Omega F (u).\] |
类似于文献[17]有下面几个引理,但值得注意的是引理3的证明不同于文献[16],本文用Brouwer不动点定理证明M的非空性.
引理3 假设(f1)~(f4)成立,则∀u∈ X,u±≠0,存在唯一正数对(su,tu),使得suu++tuu-∈M.
证明 对给定的u∈X,u±≠0,定义泛函Φu(s,t)=I(su++tu-),(s,t)∈R+×R+. 直接计算可得
| \[\nabla {\Phi _u}(s,t) = \left( {\frac{{\partial {\Phi _u}}}{{\partial s}}(s,t)\frac{{\partial {\Phi _u}}}{{\partial t}}(s,t)} \right) = \left( {\left\langle {{I^\prime }(s{u^ + } + t{u^ - }),\left. {{u^ + }} \right\rangle ,\left\langle {{I^\prime }(s{u^ + } + t{u^ - })\left. {{u^ - }} \right\rangle } \right.} \right.} \right).\] |
故su++tu-∈M当且仅当(s,t)是Φu的一个临界点.下证Φu存在临界点. 首先证明对任意给定的t0∈R+,都存在唯一的s0,使得$\frac{\partial }{{\partial s}}{\Phi _u}({s_0},{t_0})$.
事实上,由Φu的定义知
| \[\frac{\partial }{{\partial s}}{\Phi _u}(s,{t_0}) = s\left( {a||{u^ + }|{|^2} + b{s^2}||{u^ + }|{|^4} + bt_0^2||{u^ + }|{|^2}||{u^ - }|{|^2} - \int {_\Omega \frac{{f(s{u^ + }){u^ + }}}{s}} } \right),\] | (7) |
| \[\frac{\partial }{{\partial s}}{\Phi _u}(s,{t_0}) = {s^3}\left( {\frac{a}{{{s^2}}}||{u^ + }|{|^2} + b||{u^ + }|{|^4} + \frac{{bt_0^2}}{{{s^2}}}||{u^ + }|{|^2}||{u^ - }|{|^2} - \int {_\Omega \frac{{f(s{u^ + }){u^ + }}}{{{s^3}}}} } \right).\] | (8) |
由式(7)及引理1(i)知,当s>0充分小时,$\frac{\partial }{{\partial s}}{\Phi _u}(s,{t_0}) > 0$;由式(8)及引理1(ⅱ)知,当s>0充分大时,$\frac{\partial }{{\partial s}}{\Phi _u}(s,{t_0}) ＜ 0$. 因此,∃s0>0,使得$\frac{\partial }{{\partial s}}{\Phi _u}(s,{t_0}) = 0$.下证s0是唯一的.若不然,假设存在s1,s2,不妨设 0＜s1＜s2,使得$\frac{\partial }{{\partial s}}{\Phi _u}({s_1},{t_0}) = \frac{\partial }{{\partial s}}{\Phi _u}({s_2},{t_0}) = 0$.则由式(8)知
| \[\left( {\frac{1}{{s_1^2}} - \frac{1}{{s_2^2}}} \right)a||{u^ + }|{|^2} + \left( {\frac{1}{{s_1^2}} - \frac{1}{{s_2^2}}} \right)bt_0^2||{u^ + }|{|^2}||{u^ - }|{|^2} = \int {_\Omega \left( {\frac{{f({s_1}{u^ + }){u^ + }}}{{{{({s_1})}^3}}} - \frac{{f({s_1}{u^ + }){u^ + }}}{{{{({s_2})}^3}}}} \right)} .\] |
由(f3)得矛盾,故存在唯一的s0>0,使得$\frac{\partial }{{\partial s}}{\Phi _u}({s_0},{t_0}) = 0$.
定义函数φ1(t):=s(t),t∈R+,其中s(t)为上面得到的s且 s(t0)=s0. 由定义知$\frac{\partial }{{\partial s}}{\Phi _u}({\varphi _1}(t),t) = 0$,容易证明,φ1是连续的有正下界，且当 t充分大时,φ1(t)≤ t. 类似地,对每个s∈R+,可以定义函数φ2(s),使得$\frac{{\partial {\Phi _u}}}{{\partial s}}(s,{\varphi _2}(s)) = 0$,s∈ R+,φ2是连续的有正下界，且当s充分大时,φ2(s)≤s.
下面用Brouwer不动点定理证明Φu存在临界点.由φ1,φ2的性质知存在C1>0,使得φ1(t)≤t,t>C1;φ2(s)≤ s,s>C1. 令C2=max｛maxt∈[0,C1]φ1(t),maxs∈[0,C1]φ2(s)｝,C=max｛C1,C2｝.定义H:[0,C]×[0,C]→R+×R+,H(s,t)=(φ1(t),φ2(s)).由定义可知H(s,t)∈[0,C]×[0,C].注意到H是连续的,故由Brouwer不动点定理知∃(s′,t′)∈[0,C]×[0,C],使得(φ1(t′),φ2(s′))=(s′,t′). 通过φ1,φ2的定义知
| \[\frac{{\partial {\Phi _u}}}{{\partial s}}\left( {s\prime ,t\prime } \right) = \frac{{\partial {\Phi _u}}}{{\partial t}}\left( {s\prime ,t\prime } \right) = 0,\] |
因此,(s′,t′)是Φu的临界点. 下证临界点是唯一的.
显然,v∈M时,(1,1) 是Φv的临界点.首先证明(1,1)是Φv的唯一临界点. 假设$\left( {\tilde s,\tilde t} \right)$也是Φv的临界点,不妨设$o \le \tilde t \le \tilde s$ ，则有
| \[a\tilde s||{\upsilon ^ + }|{|^2} + b{{\tilde t}^3}||{\upsilon ^ + }|{|^4} + b\tilde s{{\tilde t}^2}||{\upsilon ^ + }|{|^2}||{\upsilon ^ - }|{|^2} = {\smallint _\Omega }f\left( {\tilde s{\upsilon ^ + }} \right){\upsilon ^ + }.\] | (9) |
| \[a\tilde t||{\upsilon ^ - }|{|^2} + b{{\tilde t}^3}||{\upsilon ^ - }|{|^4} + b\tilde t{{\tilde s}^2}||{\upsilon ^ + }|{|^2}||{\upsilon ^ - }|{|^2} = {\smallint _\Omega }f\left( {\tilde t{\upsilon ^ - }} \right){\upsilon ^ - }.\] | (10) |
由式(9)知
| \[\left( {\frac{1}{{{{\tilde s}^2}}} - 1} \right)a||{\upsilon ^ + }|{|^2} \ge \int {_\Omega } \left( {\frac{{f\left( {\tilde s{\upsilon ^ + }} \right)}}{{{{\left( {\tilde s{\upsilon ^ + }} \right)}^3}}} - \frac{{f\left( {{\upsilon ^ + }} \right)}}{{{{\left( {{\upsilon ^ + }} \right)}^3}}}} \right){\left( {{\upsilon ^ + }} \right)^4}.\] | (11) |
由式(11)及(f3)知$o \le \tilde t \le \tilde s \le 1$. 类似地,由式(10)及(f3)知,$\tilde t \ge 1$.因此,$\tilde s = \tilde t = 1$,即(1,1) 是Φv 的唯一临界点. 若u∈X,u≠0,(s1,t1),(s2,t2)都是Φu的临界点,则v1=s1u++t1u-∈M,v2=s2u++t2u-∈M.
| \[{\upsilon ^2} = \left( {\frac{{{s_2}}}{{{s_1}}}} \right){s_1}{u^ + } = \left( {\frac{{{t_2}}}{{{t_1}}}} \right){t_1}{u^ - } = \left( {\frac{{{s_2}}}{{{s_1}}}} \right)\upsilon _1^ + + \left( {\frac{{{t_2}}}{{{t_1}}}} \right){\upsilon _1} - \in M.\] |
由Φv,v∈M临界点的唯一性知,s2s1=t2t1=1. 因此,Φu(u∈ X,u±≠0)的临界点是唯一的.
引理4[13] 假设u∈X,u±≠0,则(su,tu)是Φu(s,t)的唯一的最大值点,其中(su,tu)由引理3得到.
引理5[13] 假设(f1)~(f4)成立,且u∈X,u±≠0,〈I′(u),u+〉≤0,〈I′(u),u-〉≤0,则0<su,tu≤1,其中(su,tu)由引理3得到.
引理6[13] 假设(f1)～(f4)成立,则 m>0可达,即∃u∈M,使得m=I(u).
2 定理1的证明证明 证明引理6得到的极小化点u就是问题(1)的变号解. 用形变引理证明I′(u)=0.
令λ=min｛|u+|2,|u-|2｝,由嵌入定理知|u|2≤S‖u‖,其中S为嵌入常数.
反证法. 假设I′(u)≠0,则∃r>0,α>0,使得‖v-u‖≤r时,‖I′(v)‖≥α.
令δ∈(0,min｛λ/(2S),r/3｝),θ∈(0,min｛1/2,δ/(2‖u‖)｝),D=(1-θ,1+θ)×(1-θ,1+θ),(s,t)=su++tu-,(s,t)∈D.由上面的定义及引理4可得,$\tilde m: = \mathop {\max I}\limits_{\partial D} \circ \phi < m.$
取$\varepsilon : = \min \left\{ {\left( {m - \tilde m} \right)/2,a\delta /8} \right\},S = \left\{ {\upsilon \in X,||\upsilon - u|| \le \delta } \right\}.$则由假设知
| \[||{I^\prime }(\upsilon )|| \ge 8\varepsilon /\delta ,\upsilon \in {I^{ - 1}}\left( {\left[ {m - 2\varepsilon ,m + 2\varepsilon } \right]} \right) \cap {S_{2\delta .}}\] | (12) |
应用形变引理[20],则存在泛函η∈([0,1]×X,X),使得
(a) $u \notin {I^{ - 1}}\left( {\left[ {m - 2\varepsilon ,m + 2\varepsilon } \right]} \right) \cap {S_{2\delta .}}$时,η(1,u)=u.
(b) $\eta \left( {1,{I^{m + \varepsilon }} \cap S} \right) \subset {I^{m - \varepsilon }}.$
(c) $||\eta \left( {1,u} \right) - u|| \le \delta ,u \in X.$
首先证明
| \[\mathop {\max I}\limits_{(s,t) \in \bar D} \left( {\eta \left( {1,\phi (s,t)} \right)} \right) < m.\] | (13) |
由引理4知,I(φ(s,t)))≤m<m+ε,即φ(s,t)∈Im+ε. 由于‖φ(s,t)-u‖2=‖su++tu--u+-u-‖2≤2((s-1)2‖u+‖2+(t-1)2‖u-‖2)≤2 θ2‖u‖2<δ2,则φ(s,t)∈ S. 故由(b)知,I(η(1,φ(s,t)))<m-ε.因此式(13)成立.
下证
| \[\eta \left( {1,\phi (D)} \right) \cap M \ne \emptyset .\] |
定义γ(s,t):=η(1,φ(s,t)),
Ψ1(s,t)=(〈I′(φ(s,t)),su+〉),〈I′(φ(s,t)),su-〉)=(P(s,t),Q(s,t)),
Ψ2(s,t)=(〈I(γ(s,t)),(γ(s,t))+〉,〈I(γ(s,t)),(γ(s,t))-〉).
直接计算得
| \[\begin{array}{*{20}{l}} {\frac{{\partial P(s,t)}}{{\partial s}}{|_{(1,1)}}}&{ = a||{u^ + }|{|^2} + 3b||{u^ + }|{|^4} + b||{u^ + }|{|^2}||{u^ - }|{|^2} - \int {_\Omega } {f^\prime }({u^ + }){{({u^ + })}^2} \le }\\ {}&{a||{u^ + }|{|^2} + 3b||{u^ + }|{|^4} + b||{u^ + }|{|^2}||{u^ - }|{|^2} - 3\int {_\Omega } {f^\prime }({u^ + }){u^ + } = }\\ {}&{ - 2a||{u^ + }|{|^2} - 2b||{u^ + }|{|^2}||{u^ - }|{|^2},} \end{array}\] |
| \[\begin{array}{*{20}{l}} {\frac{{\partial Q(s,t)}}{{\partial s}}{|_{(1,1)}}}&{ \le - 2a||{u^ - }|{|^2} - 2b||{u^ + }|{|^2}||{u^ - }|{|^2},\frac{{\partial P}}{{\partial t}}{{\left| {_{(1,1)} = \frac{{\partial Q}}{{\partial s}}} \right|}_{(1,1)}} = 2b||{u^ + }|{|^2}||{u^ - }|{|^2},}\\ {}&{{J_{{\psi _1}}}(1,1) = \frac{{\partial Q}}{{\partial s}}{{\left| {_{(1,1)} = \frac{{\partial P}}{{\partial t}}} \right|}_{(1,1)}}{{\left. { - \frac{{\partial Q}}{{\partial s}}} \right|}_{(1,1)}}{{\left. { = \frac{{\partial Q}}{{\partial s}}} \right|}_{(1,1)}} > 0.} \end{array}\] |
因为Ψ1是C1的且(1,1)是一个孤立零点,故
deg(ψ1,D,0)=ind(ψ1,(1,1))=sgnJΨ1(1,1)=1.
由于$\tilde m < m - \varepsilon $,故由(a)知ψ(s,t)=γ(s,t),(s,t)∈$\partial $D. 因此,由Brouwer度的边界值性质知,deg(Ψ1,D,0)=deg(Ψ2,D,0)=1.即∃(s0,t0)∈D,使得Ψ2(s0,t0)=0. 下证(γ(s0,t0))±≠0.|(ψ(s0,t0))+|2=s0|u+|2≥λ/2,|(ψ(s0,t0))-|2=t0|u-|2≥λ/2.由(c)知|γ(s0,t0)-ψ(s0,t0)|2≤S‖γ(s0,t0)-ψ(s0,t0)‖≤Sδ,这表明|(γ(s0,t0))±-(ψ(s0,t0))±|2≤|γ(s0,t0)-ψ(s0,t0)|2≤Sδ. 因此,|(γ(s0,t0))±|2≥|(ψ(s0,t0))±|2-Sδ>0. 故(γ(s0,t0))±≠0. 因此,γ(s0,t0)∈M,这与式(13)矛盾. 综上所述,I′(u)=0,即u是问题(1)的一个变号解,且是能量最小的变号解.证毕.
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