非线性Klein-Gordon-Maxwell方程如式(1)
$ \left\{ \begin{array}{l} - \Delta u + \left[ {V\left( x \right) - {{\left( {\omega + \phi } \right)}^2}} \right]u = \lambda f\left( u \right), \;\;\;\;\;x \in {R^3}\\ - \Delta \phi + {u^2}\phi = - \omega {u^2}, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in {R^3} \end{array} \right. $ | (1) |
其中ω>0为相位,λ∈R为参数。此类方程最先被文献[1-2]引入,用于描述在三维空间中非线性Klein-Gordon方程与静电场相互作用产生的孤立波问题。其中场函数u和电磁位势ϕ为未知变量,非线性项f(u)用来模拟多个粒子的作用或外部非线性项的干扰。
$ \left\{ \begin{array}{l} - \Delta u + \left[ {{m^2} - {{\left( {\omega + \phi } \right)}^2}} \right]u = {\left| u \right|^{p - 2}}u, \;\;\;\;\;x \in {R^3}\\ - \Delta \phi + {u^2}\phi = - \omega {u^2}, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in {R^3} \end{array} \right. $ | (2) |
并得出|m|>|ω|和4 < p < 6时,方程(2)有无穷多个解。D’Aprile等[3]发现当p≤2或p≥6且m≥w>0时,方程(2)的解不存在。Cassani[4]发现当4 < p < 6或p=4且λ充分大时,方程(3)至少有一个径向对称解。
$ \left\{ \begin{array}{l} - \Delta u + \left[ {{m^2} - {{\left( {\omega + \phi } \right)}^2}} \right]u = \\ \;\;\;\;\;\lambda {\left| u \right|^{p - 2}}u + {\left| u \right|^{{2^ * } - 2}}u, \;\;\;\;\;\;\;\;\;\;x \in {R^3}\\ - \Delta \phi + {u^2}\phi = - \omega {u^2}, \;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in {R^3} \end{array} \right. $ | (3) |
近年来,带有位势函数的问题引起了人们的关注,形如式(4)
$ \left\{ \begin{array}{l} - \Delta u + \left[ {V\left( x \right) - {{\left( {\omega + \phi } \right)}^2}} \right]u = f\left( u \right), \;\;\;\;\;x \in {R^3}\\ - \Delta \phi + {u^2}\phi = - \omega {u^2}, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in {R^3} \end{array} \right. $ | (4) |
Carriao等[5]证明了方程(4)在V(x)为周期位势且非线性项临界增长时,方程有基态解。Jing等[6]证明了当V(x)为衰减位势时,方程有无穷多个非平凡解。
本文通过变分方法研究Klein-Gordon-Maxwell方程在径向对称位势下,且方程的非线性项f(u)只在零点附近有定义时方程的解的存在性,并得到解关于参数λ的依赖性。
1 定理的提出首先对非线性项f(u)及势函数V(x)假设如下条件:
① 存在δ0>0,使得f(u)∈C[-δ0, δ0];
②
③ 存在μ∈[4, 6)和δ1>0,使得0 < |u| < δ1时,有0 < μF(u)≤uf(u),其中F(u)=∫0uf(t)dt;
④ V(x)∈C(R3, R)且V(x)是径向对称函数;
⑤
条件①~③只是f(u)在零点附近的条件,其中条件③为Ambrosetti-Rabinowitz条件[7]在零点附近的局部形式,在无穷远处对f(u)不加任何条件的限制。
其次,为了对本文的结果作准确的说明,给出如下定义:令Hr1(R3)、Dr1, 2(R3)分别为H1(R3)和D1, 2(R3)={u∈L6(R3):|∇u|∈L2(R3)}的径向对称函数空间;H1(R3)的范数记为‖u‖H12=∫R3(|∇u|2+u2)dx,D1, 2(R3)的范数记为
定理1 若条件①~⑤成立,则存在参数Λ>0,使得λ>Λ时,方程(1)至少存在一个非平凡的解(u, ϕu)∈Hr1(R3)×Dr1, 2(R3),且当λ→∞时,‖u‖Hr1→0,‖ϕu‖D→0。
由于f(u)只在零点附近有定义,其条件较弱,不能直接应用变分法。为了克服此困难,参考Costad等[8]的方法修正并扩张f(u)为新的
观察条件②的隐含条件易知,存在δ2>0及C1、C2>0,使得|u| < δ2时有
$ \left\{ \begin{array}{l} F\left( u \right) \ge {C_1}{\left| u \right|^p}\\ F\left( u \right) \le {C_2}{\left| u \right|^p} \end{array} \right. $ | (5) |
令ρ(t)∈C1(R, [0, 1])是一个满足tρ'(t)≤0的截断函数,且
$ \rho \left( t \right) = \left\{ \begin{array}{l} 1, \;\;\;\;\;\left| t \right| < \delta \\ 0, \;\;\;\;\;\left| t \right| > 2\delta \end{array} \right. $ |
式中
$ \tilde F\left( u \right) = \rho \left( u \right)F\left( u \right) + \left( {1 - \rho \left( u \right)} \right){C_2}{\left| u \right|^p} $ | (6) |
引理1[8] 由
由式(6)可将式(1)修正为式(7)
$ \left\{ \begin{array}{l} - \Delta u + \left[ {V\left( x \right) - {{\left( {\omega + \phi } \right)}^2}} \right]u = \lambda \tilde f\left( u \right), \;\;\;\;\;x \in {R^3}\\ - \Delta \phi + {u^2}\phi = - \omega {u^2}, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in {R^3} \end{array} \right. $ | (7) |
式(7)所对应的泛函为
$ \begin{array}{l} G\left( {u, \phi } \right) = \frac{1}{2}\int_{{R^3}} {\left( {{{\left| {\nabla u} \right|}^2} + V\left( x \right){u^2}} \right){\rm{d}}x} - \frac{1}{2}\int_{{R^3}} {} \\ {\left| {\nabla \phi } \right|^2}{\rm{d}}x - \frac{1}{2}\int_{{R^3}} {{{\left( {\omega + \phi } \right)}^2}{u^2}{\rm{d}}x} - \lambda \int_{{R^3}} {\tilde F\left( u \right){\rm{d}}x} \end{array} $ | (8) |
引理2[2] 对每个u∈Hr1(R3),都存在唯一的ϕu∈Dr1, 2(R3),且满足-Δϕ+u2ϕ=-ωu2。
由引理2知,ϕu为方程-Δϕ+u2ϕ=-ωu2的弱解。在-Δϕ+u2ϕ=-ωu2两边同乘ϕu,再进行积分得
$ \int_{{R^3}} {{{\left| {\nabla {\phi _u}} \right|}^2}{\rm{d}}x} + \int_{{R^3}} {{{\left| u \right|}^2}\phi _u^2{\rm{d}}x} = - \omega \int_{{R^3}} {{u^2}{\phi _u}{\rm{d}}x} $ | (9) |
将式(9)带入式(8),得到新的泛函G(u, ϕu),记为
$ \begin{array}{l} \tilde I\left( u \right) = \frac{1}{2}\int_{{R^3}} {\left( {{{\left| {\nabla u} \right|}^2} + V\left( x \right){u^2}} \right){\rm{d}}x} - \frac{1}{2}\int_{{R^3}} {\left( {{\omega ^2} + } \right.} \\ \left. {\omega {\phi _u}} \right){u^2}{\rm{d}}x - \lambda \int_{{R^3}} {\tilde F\left( u \right){\rm{d}}x} \end{array} $ | (10) |
进一步可得式(11)
$ \begin{array}{l} \left\langle {\tilde I'\left( {{u_n}} \right), {u_n}} \right\rangle = \int_{{R^3}} {\left\{ {{{\left| {\nabla {u_n}} \right|}^2} + \left[ {V\left( x \right) - \left( {{\omega ^2} + } \right.} \right.} \right.} \\ \left. {\left. {\left. {\omega {\phi _u}} \right)} \right]u_n^2} \right\}{\rm{d}}x - \lambda \int_{{R^3}} {\tilde f\left( {{u_n}} \right){u_n}{\rm{d}}x} \end{array} $ | (11) |
引理3[9] ϕu有以下性质:
1) ϕu≤0;
2) ϕu≥-ω在集合{x|u(x)≠0}上成立;
3) 存在常数C0>0, 使得‖ϕu‖D≤C0‖u‖H12成立。
证明 由式(7)中的-Δϕ+u2ϕ=-ωu2两端同乘ϕ+=max{ϕu, 0}≥0,得0≤∫R3|∇ϕ+|2+u2ϕ+dx=-ω∫R3u2ϕ+dx≤0,故ϕ+≡0,即ϕu≤0。
固定u∈Hr1(R3),在-Δϕ+u2ϕ=-ωu2两端同乘(ω+ϕu)-≡-min {ω+ϕu, 0},当ω>0时,有-∫ϕu < -ω(Dϕu)2dx=∫ϕu < -ω(ω+ϕu)2u2dx,故(ω+ϕu)-≡0。
即ω+ϕu≥0,进一步有ϕu≥-ω。
性质3)显而易见是成立的。引理3得证。
引理4[2] 泛函G(u, ϕ)∈C1(Hr(R3)×Dr1, 2(R3), R),且它的临界点为方程(7)的解。
引理5[9] 命题①、②等价:
①(u, ϕu)∈Hr1(R3)×Dr1, 2(R3)是G(u, ϕ)的临界点;
② u是
由引理4及引理5可知,若已求得
定义1(P-S条件[7]) 设I∈C1(E, R),如果{I(uk)}在R上有界,且当k→∞时,I'(uk)→0在E*中的每一个序列{uk}(uk∈E, k=1, 2, …)于E中都是列紧集,则称泛函I满足P-S条件。
引理6(山路定理[7]) 设E是Banach空间,I∈C1(E, R)且满足:①I(0)=0时,存在ρ0>0,使得I∂Bρ0(0)≥α>0;②存在外部结点e∈E\Bρ0(0),使得I(e)≤0。令Γ是E中联结0与e的道路集合,即Γ={g∈C([0, 1], E)|g(0)=0, g(1)=e},再记
引理7(Sobolev嵌入定理[7]) 设Ω是RN中的有界区域,则有H01(Ω)↺↺Lp(Ω),其中
引理8(Sobolev紧嵌入定理[10]) 当N≥2时,有Hr1(RN)↺↺Lr(RN) (r∈(2, 2*))。
引理9(Ni's不等式[11]) 设N≥2,x≠0,则存在
引理10 设{un}⊂Hr1(R3)是泛函
证明 由{un}⊂Hr1(R3)是泛函
由式(10)、(11)得
$ \begin{array}{l} p\tilde I\left( {{u_n}} \right) - \left\langle {\tilde I'\left( {{u_n}} \right), {u_n}} \right\rangle = \left( {\frac{p}{2} - 1} \right)\int_{{R^3}} {\left( {{{\left| {\nabla {u_n}} \right|}^2} + } \right.} \\ \left. {\left( {V\left( x \right) - {\omega ^2}} \right)u_n^2} \right){\rm{d}}x - \omega \left( {\frac{p}{2} - 1} \right)\int_{{R^3}} {u_n^2{\phi _u}{\rm{d}}x} - \\ \lambda \int_{{R^3}} {\left( {p\tilde F\left( {{u_n}} \right) - \tilde f\left( {{u_n}} \right){u_n}} \right){\rm{d}}x} \end{array} $ |
再由条件③、⑤、p∈[4, 6)及引理3,有
$ \begin{array}{l} p\tilde I\left( {{u_n}} \right) - \left\langle {\tilde I'\left( {{u_n}} \right), {u_n}} \right\rangle \ge \left( {\frac{p}{2} - 1} \right)\int_{{R^3}} {\left\{ {{{\left| {\nabla {u_n}} \right|}^2} + } \right.} \\ \left. {\left[ {V\left( x \right) - {\omega ^2}} \right]u_n^2} \right\}{\rm{d}}x \ge {{\hat C}_1}\left\| {{u_n}} \right\|_{H_r^1}^2 \end{array} $ | (12) |
又由假设得
$ \begin{array}{l} p\tilde I\left( {{u_n}} \right) - \left\langle {\tilde I'\left( {{u_n}} \right), {u_n}} \right\rangle \le p\left| {\tilde I\left( {{u_n}} \right)} \right| + \left| {\left\langle {\tilde I'\left( {{u_n}} \right), } \right.} \right.\\ \left. {\left. {{u_n}} \right\rangle } \right| \le pM + \varepsilon {\left\| {{u_n}} \right\|_{H_r^1}} \end{array} $ | (13) |
式中ε为任意小的数。最后由式(12)、(13)得
故知{un}在Hr1(R3)中有界。引理10得证。
引理11
证明 因为
由引理10知,{un}在Hr1(R3)中有界,则通过对{un}进行选子列可知,存在u∈Hr1(R3)使得在Hr1(R3)中有un⇀u;继而由引理8得,在Lr(R3)中有un→u(2 < r < 6),即式(14)成立
$ \begin{array}{l} \left\langle {\tilde I'\left( {{u_n}} \right) - \tilde I'\left( u \right), {u_n} - u} \right\rangle = \int_{{R^3}} {\left( {{{\left| {\nabla \left( {{u_n} - u} \right)} \right|}^2} + } \right.} \\ \left. {\left( {V\left( x \right) - {\omega ^2}} \right){{\left( {{u_n} - u} \right)}^2}} \right){\rm{d}}x - \omega \int_{{R^3}} {\left( {{\phi _{{u_n}}}{u_n} - {\phi _u}u} \right)\left( {{u_n} - } \right.} \\ \left. u \right){\rm{d}}x - \lambda \int_{{R^3}} {\left( {\tilde f\left( {{u_n}} \right) - \tilde f\left( u \right)} \right)\left( {{u_n} - u} \right){\rm{d}}x} = \\ {{\bar C}_2}\left\| {{u_n} - u} \right\|_{H_r^1}^2 - \omega \int_{{R^3}} {\left( {{\phi _{{u_n}}}{u_n} - {\phi _u}u} \right)\left( {{u_n} - u} \right){\rm{d}}x} - \\ \lambda \int_{{R^3}} {\left( {\tilde f\left( {{u_n}} \right) - \tilde f\left( u \right)} \right)\left( {{u_n} - u} \right){\rm{d}}x} \end{array} $ | (14) |
再由Hölder不等式及引理8得式(15)、(16)
$ \begin{array}{l} \int_{{R^3}} {\left( {{\phi _{{u_n}}}{u_n} - {\phi _u}u} \right)\left( {{u_n} - u} \right){\rm{d}}x} = \int_{{R^3}} {{\phi _{{u_n}}}} \\ {\left( {{u_n} - u} \right)^2}{\rm{d}}x + \int_{{R^3}} {\left( {{\phi _{{u_n}}} - {\phi _u}} \right)u\left( {{u_n} - u} \right){\rm{d}}x} \le {\left\| {{\phi _{{u_n}}}} \right\|_6}\\ \left\| {{u_n} - u} \right\|_{\frac{{12}}{5}}^2 + {\left\| {{\phi _{{u_n}}} - {\phi _u}} \right\|_6}{\left\| u \right\|_{\frac{{12}}{5}}}{\left\| {{u_n} - u} \right\|_{\frac{{12}}{5}}} \to 0 \end{array} $ | (15) |
$ \begin{array}{l} \int_{{R^3}} {\left( {\tilde f\left( {{u_n}} \right) - \tilde f\left( u \right)} \right)\left( {{u_n} - u} \right){\rm{d}}x} \le \int_{{R^3}} {\left( {\left| {\tilde f\left( {{u_n}} \right)} \right| + } \right.} \\ \left. {\left( {\left| {\tilde f\left( u \right)} \right|} \right)} \right)\left| {{u_n} - u} \right|{\rm{d}}x \le \int_{{R^3}} {{C_{13}}\left( {{{\left| {{u_n}} \right|}^{p - 1}} + {{\left| u \right|}^{p - 1}}} \right)} \\ \left| {{u_n} - u} \right|{\rm{d}}x \le {{\bar C}_{13}}\left( {\left\| {{u_n}} \right\|_p^{p - 1}{{\left\| {{u_n} - u} \right\|}_p} + \left\| u \right\|_p^{p - 1}} \right.\\ \left. {{{\left\| {{u_n} - u} \right\|}_p}} \right) \to 0 \end{array} $ | (16) |
式中4≤p < 6。
最后将式(15)、(16)带入式(14)得
$ o\left( 1 \right) = \left\| {{u_n} - u} \right\|_{H_r^1}^2 + o\left( 1 \right) $ |
故在Hr1(R3)中,有un→u成立。引理11得证。
引理12 泛函
证明 首先,由式(10)得
$ \begin{array}{l} \tilde I\left( u \right) \ge {C_3}\left\| u \right\|_{H_r^1}^2 - \lambda \int_{{R^3}} {\tilde F\left( u \right){\rm{d}}x} \ge {C_3}\left\| u \right\|_{H_r^1}^2 - \\ \lambda {C_1}\left\| u \right\|_p^p \end{array} $ | (17) |
由引理8得
$ {\left\| u \right\|_p} \le {{\tilde C}_1}{\left\| u \right\|_{H_r^1}} $ | (18) |
将式(18)带入式(17)得
$ \tilde I\left( u \right) \ge {C_4}\left\| u \right\|_{H_r^1}^2 - \lambda {{\tilde C}_2}\left\| u \right\|_{H_r^1}^p $ | (19) |
那么,由式(19)可知,存在参数α、ρ0,使得
其次,取定u0∈Hr1(R3),且u0≠0,则有
$ \begin{array}{l} \tilde I\left( {t{u_0}} \right) \le {C_5}{t^2}\left\| {{u_0}} \right\|_{H_r^1}^2 - \lambda {C_6}{t^p}\left\| {{u_0}} \right\|_p^p \to - \infty \\ \left( {t \to + \infty } \right) \end{array} $ |
故可取充分大的t>0,使得e=tu0满足‖e‖Hr1>ρ0且I(e)≤0。引理12得证。
综上,由引理11知
引理13 假设u∈Hr1(R3)为
证明 由式(11)易知
$ \left\langle {\tilde I'\left( u \right), u} \right\rangle = 0 $ | (20) |
所以由式(10)、(20)及引理1可得
$ \begin{array}{l} \left( {\frac{1}{2} - \frac{1}{\theta }} \right)\int_{{R^3}} {\left( {{{\left| {\nabla u} \right|}^2} + \left( {V\left( x \right) - {\omega ^2}} \right){u^2}} \right){\rm{d}}x} - \\ \left( {\frac{1}{2} - \frac{1}{\theta }} \right)\int_{{R^3}} {\omega {\phi _u}{u^2}{\rm{d}}x} - \lambda \int_{{R^3}} {\left( {\tilde F\left( u \right) - \frac{1}{\theta }\tilde f\left( u \right)u} \right)} \\ {\rm{d}}x = \tilde I\left( u \right) \end{array} $ | (21) |
由式(21)可得
引理14 定义
证明 由引理1得
$ \left\{ {\begin{array}{*{20}{c}} \begin{array}{l} \tilde F\left( u \right) \ge F\left( u \right) \ge {C_1}{\left| u \right|^p}, \\ \tilde F\left( u \right) = {C_2}{\left| u \right|^p}, \end{array}&\begin{array}{l} \left| u \right| \le \delta \\ \left| u \right| > 2\delta \end{array} \end{array}} \right. $ |
定义新的泛函
$ \begin{array}{l} J\left( u \right) = \frac{1}{2}\int_{{R^3}} {\left( {{{\left| {\nabla u} \right|}^2} + V\left( x \right){u^2}} \right){\rm{d}}x} - \frac{1}{2}\int_{{R^3}} {\left( {{\omega ^2} + } \right.} \\ \left. {\omega {\phi _u}} \right){u^2}{\rm{d}}x - \lambda \int_{{R^3}} {{{\bar C}_1}{{\left| u \right|}^p}{\rm{d}}x} \end{array} $ | (22) |
式中C1=min {C1, C2}。由式(5)、(10)、(22)得
$ \tilde I\left( u \right) \le J\left( u \right) $ | (23) |
那么,由式(23)可得
$ \begin{array}{l} c = \mathop {\inf }\limits_{u > 0} \mathop {\sup }\limits_{0 \le t < \infty } \tilde I\left( {tu} \right) \le \mathop {\inf }\limits_{0 < u \le 2\delta } \mathop {\sup }\limits_{0 \le t < \infty } \tilde I\left( {tu} \right) \le \mathop {\inf }\limits_{0 < u \le 2\delta } \\ \mathop {\sup }\limits_{0 \le t < \infty } J\left( {tu} \right) \end{array} $ | (24) |
又因
$ \begin{array}{l} J\left( {tu} \right) \le \frac{{{t^2}}}{2}\int_{{R^3}} {\left( {{{\left| {\nabla u} \right|}^2} + \left( {V\left( x \right) - {\omega ^2}} \right){u^2}} \right){\rm{d}}x} - \\ \lambda {t^p}\int_{{R^3}} {{{\bar C}_1}{{\left| u \right|}^p}{\rm{d}}x} = \frac{{{t^2}}}{2}\left\| u \right\|_{H_r^1}^2 - \lambda {t^p}{{\bar C}_1}\left\| u \right\|_p^p = {C_8}\frac{{{t^2}}}{2} - \\ {C_9}\lambda {t^p} \le {C_ * }{\lambda ^{ - \frac{2}{{p - 2}}}} \end{array} $ | (25) |
式(25)中C8、C9、C*为常数。故由式(24)、(25)得
引理15 设u∈Hr1(R3)为泛函
证明 由引理13、14、
$ \begin{array}{l} {\left\| u \right\|_D} \le \hat C{\left| x \right|^{ - \frac{1}{2}}}{\left\| {\nabla u} \right\|_2} \le \hat C{\left| x \right|^{ - \frac{1}{2}}}{\left\| u \right\|_{H_r^1}} \le \hat C\\ {\left| x \right|^{ - \frac{1}{2}}}{\left( {\tilde C{C_ * }} \right)^{\frac{1}{2}}}{\lambda ^{ - \frac{1}{{p - 2}}}} = {C_{10}}{\left| x \right|^{ - \frac{1}{2}}}{\lambda ^{ - \frac{1}{{p - 2}}}} \end{array} $ | (26) |
式(26)中,x≠0,
综上,证得‖u‖∞ < δ,即(u, ϕu)∈Hr1(R3)×Dr1, 2(R3)为方程(1)的解;再由引理3及引理13、14、15得,当λ→∞时,‖u‖Hr1→0,‖ϕu‖D→0。
定理1得证。证毕。
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