在本节中,我们将介绍一些将在整个工作中使用的材料
$$H_{\Gamma_{0}}^{1}(\Omega)=\bigl\{u\在H^{1{(\欧米茄)\vert u=0\mbox{on}\Gamma_{0}\bigr\}$$
并捐赠\(H_{\Gamma_{0}}^{1}(\Omega)\)希尔伯特结构由\(H^{1}(\Omega)\).我们有\(H_{\Gamma_{0}}^{1}(\Omega)\)是希尔伯特空间。为了简单起见,我们表示\(\Vert\cdot\Vert_{p}=\Vert\cdot\Vert_{L^{p}(\Omega)}\),\(\Vert\cdot\Vert_{p,\Gamma}=\Vert\cdot\Vert_{L^{p}(\Gamma)}\),\(1).
我们提出了证明主要结果所需的一些假设和初步准备。
我们做出以下假设:
-
(H1):
-
\(克:\mathbb{右}_{+}\rightarrow\mathbb{右}_{+}\)是一个可微函数,使得
$$1-\int_{0}^{\infty}g(s)\,ds=l>0,\qquad g(t)\geq0,\q quad g'(t)\ leq0,\ quad \ for all t\geq0$$
(7)
-
(H2):
-
对于非线性项,我们有
$$开始{aligned}&2<p\leq\frac{2(n-1)}{n-2}\quad\mbox{if}n\geq3\quad\\mbox{和}\quad p>2\quad\mbox{if{n=1,2,\end{aligned}$$
(8)
$$\开始{aligned}&2<\rho\leq\frac{2}{n-2}\quad\mbox{if}n\geq3\quad\\mbox{和}\quad\rho>0\quad_mbox{if}n=1,2。\结束{对齐}$$
(9)
-
(H3):
-
\(\Phi:\mathbb{R}\rightarrow\mathbb{R}\)是单调的、连续的,并且存在正常数\(m{q}\)和\(M_{q}\)这样的话
$$m_{q}\vert s\vert^{q}\leq\Phi(s)s\leq m_{qneneneep \vert s \ vert^},\quad\对于所有s\在\mathbb{R}中$$
(10)
-
(H4):
-
功能(f),q个,小时本质上是有界的
$$f(x)>0,\qquad q(x)>0\quad\mbox{和}\quad h(x)大于0,\quad_对于\Gamma_{1}中的所有x$$
我们在没有证明的情况下陈述了一种局部存在,这种局部存在可以通过组合[4,5].
假设(H1)-(H4)成立,\(u{0}\在H_{\Gamma{0}}^{1}(\Omega)中),\(u_{1}\在L^{2}(\Omega)中\)、和\(L^{2}(Gamma{1})中的y_{0}).然后是问题(1)–(6)承认局部解决方案薄弱\((u,y)\)这样,对一些人来说\(T>0\),
在L^{\infty}\bigl([0,T);H_{\Gamma_{0}}^{1}(\Omega)\bigr)中的$$开始{对齐}&u,在L^}\infty}\bigle([0,T);L^{2}(\ Omega Gamma_{1}\bigr),在L^{2}\bigl([0,T)$$
为了获得全局不存在的结果,我们需要以下引理。
引理2.1
假设(H1)–(H4)持有.让
\(u(t)\)
解决问题(1)–(6).然后能量函数
\(E(t)\)
问题的数量(1)–(6)没有增加.此外,以下能量不等式成立:
哪里
和
引理2.2
假设(H1)–(H4)持有.让
\(u(t)\)
是问题的解决方案(1)–(6).此外,假设
$$E(0)<E_{1}=\biggl(\frac{1}{2}-\frac}1}{p}\biggr)B_{1}^{-\frac{2p}{p-2}}$$
和
$$\Vert\nabla u_{0}\Vert\geq B_{1}^{-\frac{p}{p-2}}$$
哪里
\(B_{1}=B/l^{\frac{1}{2}})
和
B类
是Sobolev嵌入的最佳常数
\(H_{0}^{1}(\Omega)\hookrightarrow L^{p}(\欧米茄)\).然后存在一个常数
\(β>B_{1}^{-\frac{p}{p-2}}\)
这样的话
$$\biggl(1-\int_{0}^{t}g(s)\,ds\biggr)\bigl\Vert\nabla u(t)\bigr\Vert_{2}\geq\beta^{2},\quad\forall t>0$$
(14)
和
$$\bigl\Vert u(t)\bigr\Vert_{p}\geq B_{1}\beta,\quad\对于所有t>0$$
(15)
证明
发件人(6)和嵌入定理,我们有
哪里\(\xi=((1-\int_{0}^{t}g(s)\,ds)\Vert\nabla u(t)\Vert^{2})^{\frac{1}{2}\)。很容易看出\(G(\xi)\)取其最大值\(\ xi=\ xi ^{*}=B_{1}^{-\frac{p}{p-2}}\),将严格增加\(0<\xi<\xi^{*}\),严格减少\(\xi>\xi^{*}\),\(G(\xi)\右箭头-\infty\)作为\(\xi\rightarrow\infty\)、和
$$G\bigl(\xi^{*}\bigr)=\biggl(\frac{1}{2}-\frac}1}{p}\biggr)B_{1}^{-\frac{2p}{p-2}}=E_{1}$$
自\(E(0)<E_{1}\),存在\(\beta>\xi^{*}\)这样的话\(G(β)=E(0)).设置\(\xi_{0}=\Vert\nabla u(0)\Vert\),由(16),我们看到了
$$G(\xi_{0})\leq E(0)=G(\beta)$$
这意味着
$$\Vert\nabla u_{0}\Vert=\xi_{0}>\测试版$$
为了证明(14)相反,我们认为
$$\biggl(\biggl(1-\int_{0}^{t}g(s)\,ds\biggr)\bigl\Vert\nabla u(t)\bigr\Vert^{2}\biggr)^{\frac{1}{2}}<\beta$$
对一些人来说\(t=t{0}>0\).通过连续性\((1-\int_{0}^{t}g(s)\,ds)\Vert\nabla u(t)\Vert^{2}\),我们可以选择\(t_{0}\)这样的话
$$\beta>\biggl(\biggl(1-\int_{0}^{t_{0}g(s)\,ds\biggr)\bigl\Vert\nabla u(t_{0{)\bigr\Vert^{2}\bigger)^{\frac{1}{2}>\xi^{*}$$
然后它从(16)那个
这与引理相矛盾2.1.因此(14)已被证明。现在我们要证明(15). 发件人(12), (13), (14)、和引理2.1,我们推断
因此,引理的证明2.2已完成。□
定理2.1
让
\(2<m<p),\(2 \leq q<p)
并假设(H1)–(H4)持有.假设
\(\rho<p-2),\(0<\varepsilon_{0}<\frac{p}{2}-1\),和
$$\int_{0}^{\infty}g$$
(17)
都很满意,那么就不存在问题的全局解决方案(1)–(6)如果
$$E(0)<\bigl(1-C M_{q}\frac{\lambda ^{q}}{q}\frac{2}{p-2}-\frac{1}{p(p-2)}\frac{1-l}{l}\biggr)\bigl(\frac{1}{2}-\frac{1}{p}\biggr)B_{1}^{-\frac{2 p}{p-2}}$$
(18)
和
$$\Vert\nabla u_{0}\Vert>B_{1}^{-\frac{p}{p-2}}$$
(19)
证明
假设解决方案\(u(t)\)第页,共页(1)–(6)是全球性的。我们设置了
其中常量\(E_{2}\ in(E(0),E_{1})\)应稍后选择。按引理2.1,函数\(H(t)\)正在增加。那么,对于\(t \geq s \geq0),
因此,从(14)我们得到
$$\开始{对齐}[b]H(t)&\leqE_{2}-\frac{1}{2}\biggl(1-\int_{0}^{t}g(s)\,ds\biggr)\bigl\Vert\nabla u(t)\bigr\Vert^{2}+\frac}1}{p}\bigl\ Vert_u(t{2}\biggl(1-\int_{0}^{t}g(s)\,ds\biggr)\bigl\Vert\nabla u(t)\bigr\Vert^{2}+\frac{1}{p}\bigl\ Vert u(t\bigr\Vert_{p}^{p}\\&\leq E_{1}-\frac{1}{2}B_{1}^{-\frac{2p}{p-2}}+\frac}1}{p}\bigl\Vert u(t)\bigr\ Vert_{p}^{p}\\&=\biggl \压裂{2p}{p-2}}-\压裂{1}{2}B{1}^{-\frac{2p{{p-2{}}+\frac}1}{p}\bigl\Vertu(t)\bigr\Vert_{p}^{p}\\&\leq\frac[1}{p{2}}+\frac{1}\bigr\Vert_{p}^{p}。\结束{对齐}$$
(22)
现在,我们定义
$$开始{对齐}[b]L(t)&=H^{1-\sigma}(t)+\frac{\varepsilon}{\rho+1}\int_{\Omega}\bigl\vert u{t}(t)\bigr\vert^{\rho}u{t{(t y^{2}(t)\,d\Gamma-\varepsilon\int_{\Gamma_{1}}H(x)u(t)y(t)$$
(23)
其中常量\(0<西格玛<1),\(\varepsilon>0\)应稍后选择。
求…的导数(23),使用(7)–(10)和引理2.1,我们有
$$开始{对齐}L'(t)&=(1-\sigma)H^{-\sigma}(t)H'(t u(t)\,dx\\&\四{}-\varepsilon\int_{\Gamma{1}}H(x)f(x)y(t)y_{t}(t),d\Gamma-\varepsilon\int_{\Gamma_{1}}h(x)u_{t}(t)y(t)\,d\Gamma\\&\quad{}-\varesilon\int_}\Gamma_1}}h(x)u(t)y_{tneneneep(t)\垂直u_{t}(t)\bigr\Vert_{\rho+2}^{\rho+2}\\&\quad{}+\varepsilon\int_{\Omega}u(t)\biggl[\Delta u(t)-\int_{0}^{t}g(t-s)\Delta u(s)\,ds-\bigl\vert u_{t}(t x)f(x)y(t)y{t}(t)y(t)\,d\Gamma\\&&\quad{}-\varepsilon\int_{\Gamma_{1}}h(x)u(t)y_{t}(t)\,d\Gamma\\&&=(1-\sigma)h^{-\sigma}(t)h'(t)+\frac{\varepsilon}{\rho+1}\bigl\Vert u_{t}(t)\bigr\Vert_{\rho+2}^{\rho+2}\\&&\quad{}-\varepsilon\int_{\ Omega}\bigl\Vert\nabla u(t)\bigr\Vert^{2}\,dx+\varepsilon\int_{\Omega}\nabla u(t)\int_{0}^{t}g(t-s)\nabla u(s)\,ds\,dx\\&quad{}-\varepsilon\int_}\Omega}\bigl\vert u_{t}(t)\ bigr\vert^{m-2}u_{t}(t)u(t epsilon\int_{\Gamma_{1}}u(t)\biggl(\frac{\partialu(t分形{部分u(s)}{部分nu}\,ds\biggr)\,d\Gamma\\&\quad{}-\varepsilon\int_{\Gamma_{1}}h(x)y(t)\bigl(f(x)y_{t}(t)+u_{t{(t t)\,d\Gamma\\&=(1-\sigma)h^{-\sigma}(t)h'(t)+\frac{\varepsilon}{\rho+1}\bigl\Vert u_{t}(t)\bigr\Vert _{\rho+2}^{\rho+2}-\varepsilon\bigl\ Vert\nabla u(t)\ bigr\Vert^{2}\\&\ quad{}+\varepsilon\int_{\Omega}\nabla u(t)\nint_{0}^{t}g(t-s)\nabla u(s)\,ds\,dx-\varepsi lon\int_{\Omega}\ bigl\Vert u{t}(t)\bigr\Vert ^{m-2}u{tneneneep(t)u(t)\,dx\\&\四{}+\varepsilon\bigl\Vertu(t)\bigr\Vert_{p}^{p}-\varepsilon\int_{\Gamma_{1}}\Phi\bigl(u_{t}(t)\ bigr)u(t),d\Gamma+\varepsiolon\int_{\Gamma_{1{}h(x)q(x)y^{2}(t)\,d\ Gamma。\结束{对齐}$$
(24)
利用霍尔德和杨氏不等式\(\varepsilon_{1}\)(\(0<\varepsilon_{1}<1\)),我们获得
因此,从(24)和(25),我们到达
因此,从(11), (12), (20),以及(26),我们推断
从这个关系和使用
由此可见
$$开始{对齐}L'(t)&\geq(1-\sigma)H^{-\sigma}n(1-\varepsilon_{1})pH+\varepsilon\biggl(\frac{1}{\rho+1}+\frac{p(1-\varepsilon_{1})}{\rro+2}\biggr)\bigl\Vert u_t}(t)\bigr\Vert_{\rho2}^{\rho+2}\\&\quad{}+\varepsi lon\varepsion_{1{\bigl\ Vert u(t)\ bigr\Vert_{p}^{p}-\varepsilon\int_{\Omega}\bigl\Vert u_{t}(t)\bigr\Vert^{m-2}u_{t}(t)u(t)\,dx\\&\quad{}-\varepsilon\int_{\Gamma_{1}}\Phi\bigl(u_{t}(t)\biger)u(t)裂缝{(1-\varepsilon_{1})p}{2}-1\biggr)\biggl(1-\int_{0}^{t}g(s)\,ds\biggr)\bigl\Vert\nabla u(t)\bigr\Vert^{2}\\&&\quad{}-\frac{\varepsilon}{2 p(1-\varepsilon\{1})}\int_{0}^{t}g(s)\,ds\bigl\Vert\nabla u(t)\bigr\Vert^{2}\\&&\geq(1-\sigma)H^{-\sigma}(t)\bigl\Vert u_{t}(t)\bigr\Vert_{m}^{m}+pH(t)\varepsilon(1-\varepsilon\{1})\\&&\quad{}+\varepsilon\biggl)\biggl(1-\int_{0}^{t}g(s)\,ds\biggr)\bigl\Vert\nabla u(t)\bigr\Vert^{2}\\&\quad{}-\frac{\varepsilon}{2p(1-\varepsilon_{1})}\int_{0}^{t}g(s)\,ds\bigl\Vert\nabla u(t)\bigr\Vert^{2}-\varepsilon(1-\verepsilon_1})pE_{2}+\varepsi lon\varepssilon_{1}\bigl\ Vert u(t \Omega}\bigl\Vert u_{t}(t)\bigr\Vert ^{m-2}u_{t}(t)u(t)\,d\Gamma-\varepsilon\int_{\Gamma_{1}}\Phi\bigl(u_{t}(t)\biger)u(t)(压裂{(1-\varepsilon_{1})p}{2}+1\biggr)\int{\Gamma{1}}H(x)q(x)y^{2}(t)\,d\Gamma。\结束{对齐}$$
(28)
根据Hölder和Young不等式,条件\(m<p\), (22),和嵌入定理(\(L^{p}(\Omega)\hookrightarrow L^{m}(\欧米茄)\)),我们获得
$$开始{对齐}[b]\int_{\Omega}\bigl\vert u_{t}bigl\vert u(t)\bigr\vert ^{m}\,dx\biggr)^{\frac{1}{m}}\\&\leq\bigl\vert u{t}(t)\ bigr\vert{m}^{m-1}\bigl\Vert_u(t)\bigr\Vert_{m}\\&\leq C\bigl\ Vert_u{t}C{p}{m}}\bigl\垂直u(t)\bigr\垂直_}\bigl(\varepsilon_{1}\bigl\Vertu(t)\bigr\Vert_{p}^{p}+C(\varesilon_{1{)\bigl\ Vert_u{t}(t)\ bigr\Vert_{m}^{m}\bigr)\\&\leq CH(t)^{frac{1}{p}-\frac{1}}{m}}\big)\bigr\Vert_{p}^{p}+C(\varepsilon_{2})\bigl\Vert_u{t}(t)\birgr\Vert_{m}^{m}\bigr),\end{aligned}$$
(29)
哪里C类是一个通用的正常量,可能会在不同的行之间变化,并且\(\varepsilon_{2}>\varepsilon_{1}p^{1/p-1/m}\).
我们在这里选择
$$0<\sigma<\min\biggl$$
(30)
并采取\(α=\压裂{m-p}{pm}+\西格玛=-(压裂{1}{m}-\压裂{1{p})+\西格玛<0\)。然后是属性(21)函数的\(H(t)\)向大家展示
$$H(t)^{\frac{1}{p}-\frac{1}{m}}=H(t)^{-\sigma}H(t)^{\alpha}\leq H(t)^{-\sigma}H(0)^{\alpha}$$
因此从不平等(30)它如下
$$\int_{\Omega}\bigl\vert u_{t}(t)\bigr\vert ^{m-2}u_{t}(t)u(t)\,d\Gamma\leq CH(t)^{-\sigma}H(0)^{\alpha}\bigr t)\bigr\vert_{m}^{m}\bigr)$$
(31)
此外,来自(10),很明显
$$\int_{\Gamma_{1}}\Phi\bigl(u_{t}(t$$
(32)
以及下面的杨氏不等式
$$XY\leq\frac{\lambda ^{\gamma}X^{\gamma}+\frac{\lambda ^{-\beta}Y^{\beta}}$$
\(X,Y\geq0\),\(\lambda>0),\(伽马射线,β射线{右}_{+}\)这样的话\(压裂{1}{\gamma}+\frac{1}}{\beta}=1\),然后从(10),我们得到
$$开始{对齐}[b]\int_{\Gamma_{1}}\Phi\bigl(u_{t}(t)\bigr)u(t)}{q}\bigl\Vertu(t)\bigr\vert_{q,\Gamma_{1}}^{q}+M_{q}\分形{q-1}{q}\lambda^{-\frac{q}{q-1}}\bigl\Vertu{t}(t)\bigr\Vert_{q,\Gamma{1}}^{q}。\结束{对齐}$$
(33)
因此,从(28)和(31)–(33),我们推断
$$\boot{aligned}[b]L'(t)&\geq(1-\西格玛)H^{-\西格玛}(t)\bigl\Vert u_{t}(t)\bigr\Vert{m}^{m}+\varepsilon\biggl[\frac{1}{\rho+1}+\frac{p(1-\varepsilon\{1})}{\rho+1}\biggr]\bigl\Vert u_{t}(t)\bigr\Vert+2}+\varepsilon(1-\varepsilon\{1})pH(t)\hspace{-20pt}\&&\quad{}+\varepsilon\biggl[\biggl(\frac{(1-\varepsilon_{1})p}{2}-1\biggr{2}空格{-20pt}\\&\四{}-\varepsilon\bigl\Vert u(t)\bigr\Vert_{p}^{p}\\&\quad{}-\varepsilon CH(t)^{-\sigma}H(0)^{\alpha}\bigl(\varepsilon_{2}\bigr\ Vert u r)\\&\四{}+(1-\sigma)H^{-\sigma}(t)m_{q}\bigl\Vertu{t}(t)\bigr\Vert_{q,\Gamma{1}}^{q}\\&\quad{}-\varepsilon M_{q}\frac{\lambda^{q}{q}\bigl\Vert u(t)\bigr\Vert_{q,\Gamma_{1}}^{q{-\varesilon M_{q}\frac{q-1}{q{\lambda ^{-\frac}{q-1{}\bigl\ Vert u{t}(t)\ bigr\Vert_{q 1}^{q}\\&\四{}+\varepsilon\biggl(\frac{(1-\varepsilon_{1})p}{2}+1\biggr)\int_{Gamma{1}}h(x)q(x)y^{2}(t)\,d\Gamma\\&=H^{-\sigma}(t)\bigl[1-\simma-\varepsilon C H^{\alpha}(0)C(\varepsilon_{2})\bigr]\bigl\Vertu{t}\biggr]\bigl\Vert u_{t}(t)\bigr\Vert_{\rho+2}^{\rho+2}\hspace{-20pt}\\&\quad{}+瓦雷普西隆(1-\varepsilon_{1})pH(t)-\varepsi隆(1-\ varepsilen_{1{)pE_{2}\\&\quad{}+\varepsilon\biggl[\biggl(\frac{(1-\verepsilon_1})p}{2}-1\biggr)\biggal(1-\int_{0}^{t}g(s)\,ds\biggro)-\frac}{1}{2p(1-\varepsilon_{1})}\int_{0}^{t}g(s)\,ds\biggr]\bigl\Vert\nabla u(t)\bigr\Vert^{2}\hsspace{-20pt}\\&\quad{}+\varepsilon\bigl[\varepsilon_{1}-\varepsilon_{2}C H^{-\sigma}(t)H^{\alpha}}^{q}\\&\quad{}+\biggl[(1-\sigma)H^{-\sigma}(t)M_{q}-\varepsilon M_{q{\frac{q-1}{q}\lambda^{-\frac{q}{q-1}}\biggr]\bigl\Vertu{t}(t)\bigr\Vert_{q,\Gamma{1}}^{q}\\&\quade{}+\varepsilon\biggl,d\Gamma,\quad\表示所有t\geq t_{0}。\结束{对齐}$$
(34)
我们还使用了嵌入定理。让我们回顾一下不平等(C类表示一般的正常数)
$$\bigl\Vertu(t)\bigr\Vert_{q,\Gamma_{1}}\leq C\bigl\ VertU(t)\ bigr\Vert_{H^{s}(\Omega)}$$
哪里\(问题1)和\(0\leq s<1),\(s\geq\压裂{N}{2}-\压裂{N-1}{q}>0\)以及插值和庞加莱不等式(参见[11])
$$\bigl\Vert-u(t)\bigr\Vert_{H^{s}(\Omega)}\leq C\bigl\ Vert-u$$
如果\(s{2}{q}\),再次使用Young不等式,我们得到
$$\bigl\Vert_u(t)\bigr\Vert_{q,\Gamma_{1}}^{q}\leq C\bigl[\bigl(\bigl\ Vert-u(t}}\bigr]$$
对于\(\frac{1}{\mu}+\frac{1}{\theta}=1\)。在这里我们选择\(θ=frac{2}{qs}\)得到\(\mu=\frac{2}{2-qs}\)因此,前面的不等式
$$\bigl\Vert u(t)\bigr\Vert _{q,\Gamma_{1}}^{q}\leq C\biggl[\bigl(\bigl\Vert u(t)\bigr\Vert _{p}^{p}\bigr)^{\frac{2 q(1-s)}{(2-qs)p}}+\biggl(1-\int_{0}^{t}g(s)\,ds\biggr)\bigl\Vert\nabla u(t)\bigr\Vert ^{2}\biggr]$$
(35)
现在,选择秒这样的话
$$0<s\leq\frac{2(p-q)}{q(p-2)}$$
我们得到
$$\frac{2q(1-s)}{(2-qs)p}\leq1$$
(36)
曾经的不平等(36)如果满足,我们使用经典代数不等式
$$\chi^{\nu}\leq(\chi+1)\leq\biggl(1+\frac{1}{w}\biggr)(\chi+w),\quad\forall \chi\geq0,0<\nu\leq1,w\geq0$$
(37)
具有\(\chi=\垂直u(t)\Vert_{p}^{p}\),\(d=1+\压裂{1}{H(0)}\),\(w=H(0)\)、和\(nu=\压裂{2q(1-s)}{(2-qs)p}\)得到以下估计:
$$\bigl(\bigl\Vertu(t)\bigr\Vert_{p}^{p}\bigr)^{\frac{2q(1-s)}{(2-qs)p}}\leq d\bigl,\quad\用于所有t\geq0$$
(38)
发件人(35)和(38),我们有
$$\bigl\Vert_u(t)\bigr\Vert_{q,\Gamma_{1}}^{q}\leq C\biggl(\bigl \Vert_u(t)\ bigr\ Vert_{p}^{p}+\biggl(1-\int_{0}^{t}g(s)\,ds\biggr)\bigl\ Vert\nabla u(t$$
(39)
插入估算(39)到(34)和使用(14),我们到达
$$开始{对齐}L'(t)&\geq H^{-\sigma}(t)\bigl[1-\simma-\varepsilon C H^{\alpha}(0)C(\varepsilon_{2})\bigr]\bigl\Vert u_{t}n_{1})p}{\rho+2}\biggr]\bigl\Vertu{t}(t)\bigr\Vert_{\rho2}^{\rho+2}+\varepsilon(1-\varepsilon_{1})pH(t)\\&\quad{}+\varebsilon\biggl[\biggl(\frac{p(1-\valepsilon_{1')}{2}-1\biggr)\biggal(1-\int_{0}^t}g(s)\,ds\bigger)-\frac}{1}{2p(1-\ varepsilen_{1{)}\int_0}^{t}g(s)\,ds\biggr]\\&\quad{}\times\bigl\Vert\nabla u(t)\bigr\Vert^{2}-\varepsilon(1-\varepsilon_{1})pE_{2}+\varepsilon\bigl[\varepsilon_{1}-\varepsilon_{2}C H^{-\sigma}(t)H^{\alpha}(0)\bigr]\bigl\Vertu(t顶点{p}^{p}+\biggl(1-\int_{0}^{t}g(s)\,ds\biggr)\bigl\Vert\nabla u(t)\bigr\Vert^{2}+2H(t)\大gr]\\&\quad{}+\biggl[(1-\sigma)H^{-\sigma}(t)m_{q}-\varepsilon m_{q{q}\frac{q-1}{1}\lambda^{\frac}{q-1{}}\biggr]\bigl\垂直u{t}gl[\frac{(1-\varepsilon_{1})p}{2}+1\biggr]\int_{\Gamma{1}}H(x)q(x)y^{2}(t)\,d\Gamma\\&=H^{-\sigma}(t)\bigl[1-\sigma-\varepsilon C H^{\alpha}(0)C(\varepsilon_{2})\bigr]\bigl\Vertu{t}(t)\bigr\Vert_{\rho+2}^{\rho2}\hspace{-20pt}\\&\quad{}+\varepsilon\biggl[(1-\varepsilon_{1})p-2 C M_{q}\frac{\lambda^{q}}{q}\biggr]H(t)-\varepsilon(1-\varepsilon_{1}(1-\int_{0}^{t}g(s)\,ds\biggr)\\&\quad{}-\frac{1}{2(1-\varepsilon{1})p}\int_}0}^}t}g\bigl\Vert\nabla u(t)\bigr\Vert^{2}\\&\quad{}+\varepsilon\biggl[\varepsilon_{1}-\varepsilon_{2}C H^{-\sigma}(t)H^{\alpha}(0)-C M_{q}\frac{\lambda^{q}{q}{q}\ biggr]\bigl\Vert u(t{}+\biggl[(1-\sigma)H^{-\sigma}(t)M_{q}-\varepsilon M_{q{frac{q-1}{q}\lambda^{-\frac{q}{q-1-}}\biggr]\bigl\Vertu(t)\bigr\Vert_{q,\Gamma_{1}}^{q}\\&\quad{}+\varepsilon\biggl[\frac{(1-\varepsilon_{1{)p}{2}+1\biggr]\int_{Gamma_}1}}h(x)q(x)y^{2}(t)\,d\Gamma\\&\geq h^{-\sigma}(t)\bigl[1-\sigma-\varepsilon C h^{\alpha}(0)C(\varepsilon_{2})\bigr]\bigl\Vert u_{t}(t)\biger\Vert _{m}^{m}+\varepsilon\biggl[\frac{1}{\rho+1}+\frac{(1-\varepsilon_{1})p}{\rro+2}\biggr]\bigl\Vertu{t}(t)\bigr\Vert_{\rho2}^{\rho+2}\hspace{-20pt}\\&\quad{}+\varebsilon\biggl[(1-\valepsilon_{1\)p-2C M_{q}\frac{\lambda^{q}}{q}\biggr]H(t)-\varepsilon{1-\int_{0}^{t}g(s)\,ds}\biggl[\biggl(\frac{(1-\varepsilon_{1})p}{2}-CM_{q}\frac}\lambda^{q}}{q}-1\biggr)l-\frac[1}{2(1-\verepsilon_2})p}(1-l)\biggr]\beta^{2}\\&\quadr{}+\varepsi silon\biggl[\varepsilon_{1}-\varepsilon_{2}C H^{-\sigma}(t)H^{\alpha}(0)-C M_{q}\frac{\lambda^{q}}{q}\ biggr]\bigl\Vert u(t)\bigr\Vert _{p}^{p}\\&&\quad{}+\biggl[(1-\sigma)H^{-\sigma}(t)m_{q}-\varepsilon m_{q}\frac{q-1}{q}\lambda ^{-\frac{q}{q-1}}\biggr]\bigl\Vert u(t)\bigr\Vert _{q,\Gamma_{1}}^{q}\\&&\quad{}+\varepsilon \biggl[\frac{(1-\varepsilon{1})p}{2}+1\biggr]\int_{\Gamma_{1}}H(x)q(x)y^{2}(t)\,d\伽玛。\结束{对齐}$$
(40)
自
我们有
$$\biggl(\frac{p}{2}-1-C M_{q}\frac}\lambda^{q}}{q}\biggr)\biggl(1-\int_{0}^{infty}g(s)\,ds\bigger)-\frac[1}{2p}\int_}0}^}{infty}g$$
很容易看出存在\(\varepsilon_{1}^{*}>0\)和\(T_{0}>0\)这样,对于\(0<\varepsilon_{1}<\varesilon_{1{^{*}:=1-\frac{2(1+\varepsilon_{0})}{p}\),\(0<\varepsilon_{0}<\frac{p}{2}-1\)、和\(t>t_{0}\),
$$开始{聚集}\frac{[([frac{(1-\varepsilon_{1})p}{2}-CM_{q}\frac{\lambda^{q}}{q}-1)l-\frac}{1}{2(1-\verepsilon_1})p}(1-l)]\beta^{2}{1-\int_{0}^{t}g(1-\varepsilon_{1})p}{2}-CM_{q}\frac{\lambda^{q}{q}-1)l-\frac}{2(1-\verepsilon_1})p}(1-l)}{1-\int_{0}^{\infty}g(s)\,ds}B_{1}^{-\frac{2p}{p-2}}。\结束{聚集}$$
现在,我们可以选择\(\varepsilon_{1}>0\)足够小且\(E_{2}\ in(E(0),E_{1})\)足够近\(E(0)\)这样的话
$$\frac{(1-\varepsilon_{1})p}{2}-CM_{q}\frac{\lambda^{q}}{q}-1)l-\frac{1}{2(1-\varepsilon_{1')p}(1-l)}{1-\int_{0}^{infty}g(s)(1-\varepsilon_{1})E_{2}>0$$
(41)
自从
$$\开始{对齐}E(0)&<E_{2}<\biggl(\frac{1}{2}-C M_{q}\frac{\lambda ^{q}}{q}\frac{1}{p}-\frac{1}{2 p ^{2}}\ frac{1-l}{l}\biggr)B_{q}\frac{2}{p-2}-\frac{1}{p(p-2)}\frac{1-l}{l}\biggr)\biggl(\frac{1}{2}-\frac{1}{p}\biggr)B_{1}^{-\frac{2 p}{p-2}}<\biggl(\frac{1}{2}-\frac}1}{p}\biggr)B_{1}^{-\frac{2p}{p-2{}。\结束{对齐}$$
发件人(40)和(41),我们到达
$$开始{对齐}[b]L'(t)&\geq H^{-\sigma}(t)\bigl[1-\simma-\varepsilon C H^{\alpha}(0)C(\varepsilon_{2})\bigr]\bigl\Vert u_{t}p}{\rho+2}\biggr]\bigl\Vertu{t}(t)\bigr\Vert_{\rho2}^{\rho+2}\hspace{-20pt}\\&\ quad{}+\varepsilon\biggl[(1-\varepsilon_{1})p-2 C M_{q}\frac{\lambda^{q}}{q}\biggr]H(t)\\&\quad{}+\varepsi lon\bigl[\varepssilon_{1'-\varepsion_{2}C H^{-\sigma}(t)H^{\alpha}(0)-C M_{q}\frac{\lampda^{q}}}}{q}\biggr]\bigl\Vertu(t)\bigr\Vert_{p}^{p}\\&\quad{}+\biggl[(1-\sigma)H^{-\sigma}(t)M_{q}-\varepsilon M_{q}\压裂{q-1}{q}\lambda^{-\frac{q}{q-1{}\biggr]\bigl\Vertu(t)\bigr\Vert_{q,\Gamma_{1}}^q}\\&\quad}h(x)q(x)y^{2}(t)\,d\伽马。\结束{对齐}$$
(42)
此时,对于\(\varepsilon_{2}C H^{-\sigma}(t)H^{\alpha}(0)<\varepsilon_{1}<\min\{1,\varepsilon_{2]p^{1/m-1/p}\}\),我们可以λ足够小以至于
$$开始{聚集}(1-\varepsilon_{1})p-2 C M_{q}\frac{\lambda^{q}}{q}>0,\\varepsilen_{1{-\varepsilon_{2}C H^{-\sigma}(t)H^{\alpha}(0)-C M_{q}\frac{\lampda^{q}}{q}>0。\结束{聚集}$$
我们再次选择ε足够小以至于
$$开始{聚集}1-\sigma-\varepsilon C H^{\alpha}(0)C(\varepsilon_{2})>0,\\(1-\simma)H^{-\sigma}(t)m_{q}-\varepsi lon m_{q{\frac{q-1}{q}\lambda^{-\frac{q}{q-1{}>0。\结束{聚集}$$
然后从(42)存在一个正常数\(K_{1}>0\)这样,以下不等式成立:
$$L'(t)\geq\varepsilon K_{1}\biggl(H(t)+\bigl\Vertu(t)\ bigr\Vert_{\rho+2}^{\rho+2}+\bigr\vertu(t \,d\Gamma\biggr)$$
(43)
另一方面,根据定义(23)从那以后\(f,h>0),我们有
$$L(t)\leq H^{1-\sigma}(t)+\frac{\varepsilon}{\rho+1}\int_{\Omega}\bigl\vert u_{t}(t)\bigr\vert^{\rho}u_{t}$$
因此,上述估计导致
$$开始{对齐}[b]L^{\frac{1}{1-\sigma}}(t)和\leq C(\varepsilon,\sigma,\rho)\biggl[H(t)+\biggl(\int_{\Omega}\bigl\vert u{t}\quad{}+\biggl(\int_{\Gamma{1}}H(x)q(x)y(t)\,d\Gamma\biggr)^{\frac{1}{1-\sigma}}\biggr]。\结束{对齐}$$
(44)
我们现在估计(参见[17])
$$开始{对齐}\biggl(\int_{\Omega}\bigl\vert u_{t}(t)\bigr\vert^{\rho}u_{t}(t)u(t)\,dx\biggr)bigl\Vertu(t)\bigr\vert_{p}^{\frac{1}{1-\sigma}}\\&\leq C\bigl\vertu{t}(t)\ bigr\vert_{\rho+2}^{\ frac{\rho+1}{1-\sigma}\mu}\bigl\Vertu(t)\bigr\Vert_{p}^{\frac{1}{1-\sigma}\theta},\end{aligned}$$
哪里\(\frac{1}{\mu}+\frac{1}{\theta}=1\)。选择\(\mu=\frac{(1-\sigma)(\rho+2)}{\rho+1}>1\),那么
$$\frac{\theta}{1-\sigma}=\frac{\rho+2}{(1-\sigma)(\rho2)-(\rho+1)}$$
发件人(30),我们知道
$$\frac{\theta}{1-\sigma}<p$$
(45)
然后从(22)我们推断
$$\开始{对齐}[b]\bigl\Vertu(t)\bigr\Vert_{p}^{\frac{\theta}{1-\sigma}}&=\bigl\ Vertu(t)\ bigr\ Vert_{p}^{p-(p-\frac}\theta}{1-\sigma})}=\bigr\ vertu(t)\bigr\Vert_{p{p}}^{-k}\\&\leq C\bigl\Vertu(t)\bigr\Vert_{p}^{p}H(0)^{-\frac{k}{p}},\end{aligned}$$
(46)
哪里\(k=p-\frac{\theta}{1-\sigma})是一个正常数。因此,从(46)我们获得
$$\biggl(\int_{\Omega}\bigl\vert u_{t}^{p}H(0)^{-\frac{k}{p}}$$
(47)
另一方面,使用与中相同的方法[13],我们获得
$$开始{aligned}\biggl\vert\int_{\Gamma_{1}}h(x)u(t)y(t)\,d\Gamma\biggr\vert&=\biggl \vert\int(x)q(x)}{q(x{1}{2}}\Vertq\vert_{L^{infty}}^{\frac{1}}{2{}}{q_{0}}\biggl(\int_{\Gamma{1}h(x)q(x)y^{2}(t)\,d\Gamma\biggr)^{\frac{1}{2}}\biggl(int_{\Gamma{1}}u^{2{(t。\结束{对齐}$$
使用嵌入\(L^{p}(\Gamma_{1})\hookrightarrow L^{2}(\ Gamma_})\)和杨氏不平等,我们得到
$$开始{aligned}\biggl\vert\int_{\Gamma_{1}}h(x)u(t)y(t)}^{\frac{1}{2}}\Vertq\vert_{L^{\infty}}^{\frac{1}[2}}{q_{0}}\biggl(\int_{\Gamma{1}}h(x)q(x)y^{2}(t)\,d\Gamma\biggr。\结束{对齐}$$
因此,存在一个正常数\(\波浪号{C}(C)_{1} =\波浪线{C}(C)_{1} (\Vert h\Vert_{L^{infty}},\Vert q\Vert_{L^},q_{0},\sigma)\)这样的话
$$\开始{aligned}&\biggl(\int_{\Gamma_{1}}h(x)u(t)y(t)\,d\Gamma\biggr)^{\frac{1}{1-\sigma}}\\&\quad\leq\ tilde{C}(C)_{1} \biggl(\int_{\Gamma_{1}}h(x)q(x)y^{2}(t)\,d\Gamma\biggr)。\结束{对齐}$$
将杨氏不等式应用于上述不等式的右侧,存在一个正常数,也表示为\(\波浪号{C}(C)_{2}\),使得
$$\开始{对齐}[b]&\biggl(\int_{\Gamma_{1}}h(x)u(t)y(t)\,d\Gamma\biggr)^{\frac{1}{1-\sigma}}\\&\quad\leq\tilde{C}(C)_{2} \biggl(int_{\Gamma_{1}}h(x)q(x)y^{2}(t)\,d\Gamma\biggr)$$
(48)
对于\(压裂{1}{\tau}+\frac{1}}{\theta}=1\).我们接受\(θ=2(1-σ)),因此\(τ=frac{2(1-\sigma)}{1-2\sigma}\)得到
$$开始{对齐}[b]&\biggl(\int_{\Gamma{1}}h(x)u(t)y(t)\,d\Gamma\biggr)-\sigma)}}+\int_{\Gamma_{1}}h(x)q(x)y^{2}(t)\,d\Gamma\biggr]。\结束{对齐}$$
(49)
通过使用(30)和代数不等式(37)带有\(\chi=\int_{\Gamma_{1}}\vert u(t)\vert ^{p}\,d\Gamma\),\(d=1+\压裂{1}{H(0)}\),\(w=H(0)\)、和\(nu=\frac{2}{p(1-2\sigma)}\),条件(30)上的σ确保\(0<\nu<1\),我们得到
$$\chi^{\nu}\leq d\bigl(\chi+H(0)\bigr)\leq 2\bigl.(\chi+H(t)\biger)$$
因此,从(49)存在一个正常数C̃这样,对所有人来说\(t \geq0),
$$开始{对齐}[b]&\biggl(\int_{\Gamma_{1}}h(x)u(t)y(t)\,d\Gamma\biggr)(x)q(x)y^{2}(t),d\Gamma\biggr)。\结束{对齐}$$
(50)
因此,从(39)和(50)我们得到
$$开始{对齐}[b]&\biggl(\int_{\Gamma_{1}}h(x)u(t)y(t)\,d\Gamma\biggr)^{\frac{1}{1-\sigma}}\\&\quad\leq C\biggl(h(t)+\bigl\Vert u(t)\bigl\Vert\nabla u(t)\bigr\Vert^{2}+\int_{\Gamma{1}}h(x)q(x)y^{2{(t)\,d\Gamma\biggr)\\&\quad\leq C\biggl(H(t)+\bigl\Vert u(t)\bigr\Vert_{p}^{p}+\bigle\Vert\nabla u(t$$
(51)
哪里C类是一个正常数。因此,从(44), (47),以及(51),我们到达
$$开始{对齐}[b]L^{\frac{1}{1-\sigma}}(t)和\leq\上划线{C}\biggl[H(t)+\biggl(\int_{\Omega}\bigl\vert u_{t}(t)\bigr\vert^{\rho}u_{t}ma_{1}}H(x)q(x)y(t)\,d\Gamma\biggr)^{frac{1}{1-\sigma}}\biggr]\\&\leq\上划线{C}\bigl[H(t)+\bigl\Vert u_{t}(t)\bigr\Vert _{\rho+2}^{\rho+2}+\bigl\ Vert u(t)\ bigr\Vert _{p}^{p}+\bigl\Vert\nabla u(t$$
(52)
哪里C̅是一个常数,取决于ε,σ,ρ,C̃,C类因此,结合(43)和(52),对一些人来说\(\xi>0\),我们得到
$$L'(t)\geq\xi L^{\frac{1}{1-\sigma}}(t),\quad\forall t\geq0$$
对于ε足够小,存在一些常数\(T_{1}\)这样的话
$$开始{对齐}L(T_{1})&=H^{1-\sigma}\int_{\Gamma_{1}}H(x)f(x)y^{2}(T_{1}),d\Gamma\\&>0。\结束{对齐}$$
因此我们得到
$$L'(t)\geq\xi L^{\frac{1}{1-\sigma}}(t)>0,\quad\对于所有t\geqT_{1}$$
(53)
以下内容的简单集成(53)超过\((T_{1},T)\)产量
$$L^{\压裂{\西格玛}{1-\西格马}}(t)\geq\frac{1}{L^{压裂{-\sigma}{1-\sigma}}$$
因此\(L(t)\)在有限时间内爆炸
$$T^{*}\leq\frac{1-\sigma}{xi\sigma L^{frac{\sigma}{1-\sigma}}(T_{1})}$$
因此,定理的证明2.1已完成。□