一般网站 要 加入 友情链接吗,中国网站开发语言,品牌营销推广要怎么做,常用网站代码波动方程初值问题能量不等式的证明
Gronwall 不等式
若非负函数 G ( τ ) G(\tau) G(τ) 在 [ 0 , T ] [0,T] [0,T] 上连续可微#xff0c; G ( 0 ) 0 G(0)0 G(0)0#xff0c;且对 τ ∈ [ 0 , T ] \tau\in[0,T] τ∈[0,T]满足 d G ( τ ) d τ ≤ C G ( τ ) F ( τ …波动方程初值问题能量不等式的证明
Gronwall 不等式
若非负函数 G ( τ ) G(\tau) G(τ) 在 [ 0 , T ] [0,T] [0,T] 上连续可微 G ( 0 ) 0 G(0)0 G(0)0且对 τ ∈ [ 0 , T ] \tau\in[0,T] τ∈[0,T]满足 d G ( τ ) d τ ≤ C G ( τ ) F ( τ ) \frac{dG(\tau)}{d\tau}\leq CG(\tau)F(\tau) dτdG(τ)≤CG(τ)F(τ) 其中 C C C 为常数且 C 0 C0 C0 F ( τ ) F(\tau) F(τ) 是 [ 0 , T ] [0,T] [0,T] 上不减的非负可积函数
那么有 d G ( τ ) d τ ≤ e C τ F ( τ ) \frac{dG(\tau)}{d\tau}\leq e^{C\tau}F(\tau) dτdG(τ)≤eCτF(τ) G ( τ ) ≤ C − 1 ( e C τ − 1 ) F ( τ ) G(\tau)\leq C^{-1}(e^{C\tau}-1)F(\tau) G(τ)≤C−1(eCτ−1)F(τ)
常用的替换技巧 u t u t t 1 2 ⋅ 2 u t ⋅ u t t 1 2 ( u t 2 ) t u_{t}u_{tt}\frac{1}{2}\cdot 2u_{t}\cdot u_{tt}\frac{1}{2}(u_{t}^2)_t ututt21⋅2ut⋅utt21(ut2)t u u t 1 2 ⋅ 2 u ⋅ u t 1 2 ( u 2 ) t uu_t\frac{1}{2}\cdot 2u\cdot u_t \frac{1}{2}(u^2)_t uut21⋅2u⋅ut21(u2)t u u x x ( u u x ) x − u x 2 u x 2 u u x x − u x 2 uu_{xx}(uu_x)_x-u_x^2u_x^2uu_{xx}-u_x^2 uuxx(uux)x−ux2ux2uuxx−ux2 ∵ ( u t u x ) x u t x u x u t u x x \because(u_{t}u_{x})_xu_{tx}u_{x}u_{t}u_{xx} ∵(utux)xutxuxutuxx ∴ u t u x x ( u t u x ) x − u t x u x \therefore u_{t}u_{xx}(u_{t}u_{x})_x-u_{tx}u_{x} ∴utuxx(utux)x−utxux u u u 二阶导函数连续时 u x t u t x u_{xt}u_{tx} uxtutx ∴ u t x u x 1 2 ( u x 2 ) t 1 2 ⋅ 2 u x ⋅ u x t u x u t x \therefore u_{tx}u_{x} \frac{1}{2}(u_{x}^2)_t\frac{1}{2}\cdot 2u_{x}\cdot u_{xt}u_{x}u_{tx} ∴utxux21(ux2)t21⋅2ux⋅uxtuxutx ∴ u t u x x ( u t u x ) x − 1 2 ( u x 2 ) t \therefore u_{t}u_{xx}(u_{t}u_{x})_x-\frac{1}{2}(u_{x}^2)_t ∴utuxx(utux)x−21(ux2)t
能量不等式证明过程
规定 x x x 增长的方向为正方向而 G r e e n Green Green 公式曲线积分时组成梯形区域 K τ K_{\tau} Kτ 边界 ∂ K τ \partial K_{\tau} ∂Kτ 的四条线段按照右手法则只有 Ω 0 \Omega_0 Ω0 这条线段的方向是与右手法则一致所以有 ∂ K τ Ω 0 ∪ ( − Γ τ 2 ) ∪ ( − Ω τ ) ∪ ( − Γ τ 1 ) \partial K_{\tau} \Omega_0\cup(-\Gamma_{\tau_{2}})\cup(-\Omega_{\tau})\cup(-\Gamma_{\tau_{1}}) ∂KτΩ0∪(−Γτ2)∪(−Ωτ)∪(−Γτ1) 此外在 ∂ K τ \partial K_{\tau} ∂Kτ 计算曲线积分时关于 x , t x,t x,t 都积分但是这两条边 Ω 0 , Ω τ \Omega_0,\Omega_{\tau} Ω0,Ωτ 只在 x x x 方向有增长而在 t t t 方向无增长那么这两条线段在计算曲线积分时关于 d t dt dt 的积分项为 0 0 0
待证明的两个不等式 ∫ Ω τ [ u t 2 ( x , τ ) a 2 u x 2 ( x , τ ) ] d x ≤ M [ ∫ Ω 0 ( ψ 2 a 2 φ x 2 ) d x ∫ K τ f 2 ( x , t ) d x d t ] \int_{\Omega_\tau} \left[u_t^2(x, \tau) a^2 u_x^2(x, \tau)\right] dx \leq M \left[ \int_{\Omega_0} (\psi^2 a^2 \varphi_x^2) dx \int_{K_\tau} f^2(x,t) dxdt \right] ∫Ωτ[ut2(x,τ)a2ux2(x,τ)]dx≤M[∫Ω0(ψ2a2φx2)dx∫Kτf2(x,t)dxdt] ∫ K τ [ u t 2 ( x , t ) a 2 u x 2 ( x , t ) ] d x d t ≤ M [ ∫ Ω 0 ( ψ 2 a 2 φ x 2 ) d x ∫ K τ f 2 ( x , t ) d x d t ] \int_{K_\tau} \left[u_t^2(x,t) a^2 u_x^2(x,t)\right] dxdt \leq M \left[ \int_{\Omega_0} (\psi^2 a^2 \varphi_x^2) dx \int_{K_\tau} f^2(x,t) dxdt \right] ∫Kτ[ut2(x,t)a2ux2(x,t)]dxdt≤M[∫Ω0(ψ2a2φx2)dx∫Kτf2(x,t)dxdt]
在波动方程 ∂ 2 u ∂ t 2 − a 2 ∂ 2 u ∂ x 2 f \frac{\partial^2 u}{\partial t^2}-a^2\frac{\partial^2 u}{\partial x^2}f ∂t2∂2u−a2∂x2∂2uf 两端同乘以 ∂ u ∂ t \frac{\partial u}{\partial t} ∂t∂u 并在区域 K τ K_\tau Kτ 上积分并按照前面的常用替换得 ∬ K τ u t u t t − a 2 u t u x x d x d t ∬ K τ u f d x d t \iint_{K_\tau}u_{t}u_{tt}-a^2u_{t}u_{xx}dxdt\iint_{K_\tau}ufdxdt ∬Kτututt−a2utuxxdxdt∬Kτufdxdt$ ∬ K τ 1 2 ( u t 2 a 2 u x 2 ) t − a 2 ( u t u x ) x d x d t ∬ K τ u f d x d t \iint_{K_\tau}\frac{1}{2}(u_{t}^2a^2u_{x}^2)_t-a^2(u_tu_{x})_{x}dxdt\iint_{K_\tau}ufdxdt ∬Kτ21(ut2a2ux2)t−a2(utux)xdxdt∬Kτufdxdt
应用下面这个 G r e e n Green Green 公式将上面等式左边替换 ∬ Ω ∂ P ∂ t ∂ Q ∂ x d σ ∮ ∂ Ω − P d x Q d t \iint_{\Omega}\frac{\partial P}{\partial t}\frac{\partial Q}{\partial x}d\sigma\oint_{\partial\Omega}-PdxQdt ∬Ω∂t∂P∂x∂Qdσ∮∂Ω−PdxQdt其中 P 1 2 ( u t 2 a 2 u x 2 ) P\frac{1}{2}(u_{t}^2a^2u_{x}^2) P21(ut2a2ux2) Q − a 2 ( u t u x ) Q-a^2(u_tu_{x}) Q−a2(utux)得闭合曲线积分 ∮ ∂ K τ [ − 1 2 ( u t 2 a 2 u x 2 ) d x − a 2 ( u t u x ) d t ] \oint_{\partial{K_{\tau}}}[-\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx-a^2(u_tu_{x})dt] ∮∂Kτ[−21(ut2a2ux2)dx−a2(utux)dt] − ∮ ∂ K τ a 2 ( u t u x ) d t 1 2 ( u t 2 a 2 u x 2 ) d x -\oint_{\partial{K_{\tau}}}a^2(u_tu_{x})dt\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx −∮∂Kτa2(utux)dt21(ut2a2ux2)dx
记 a 2 ( u t u x ) d t 1 2 ( u t 2 a 2 u x 2 ) d x □ a^2(u_tu_{x})dt\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx \square a2(utux)dt21(ut2a2ux2)dx□ ∵ ∂ K τ Ω 0 ∪ ( − Γ τ 2 ) ∪ ( − Ω τ ) ∪ ( − Γ τ 1 ) \because \partial K_{\tau} \Omega_0\cup(-\Gamma_{\tau_{2}})\cup(-\Omega_{\tau})\cup(-\Gamma_{\tau_{1}}) ∵∂KτΩ0∪(−Γτ2)∪(−Ωτ)∪(−Γτ1) ∴ − ∮ ∂ K τ a 2 ( u t u x ) d t 1 2 ( u t 2 a 2 u x 2 ) d x \therefore -\oint_{\partial{K_{\tau}}}a^2(u_tu_{x})dt\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx ∴−∮∂Kτa2(utux)dt21(ut2a2ux2)dx − [ ∫ Ω 0 □ − ∫ Ω τ □ − ∫ Γ τ 1 □ − ∫ Γ τ 2 □ ] -[\int_{\Omega_0}\square-\int_{\Omega_{\tau}}\square-\int_{\Gamma_{\tau_{1}}}\square-\int_{\Gamma_{\tau_{2}}}\square] −[∫Ω0□−∫Ωτ□−∫Γτ1□−∫Γτ2□] Ω 0 , Ω τ \Omega_0,\Omega_{\tau} Ω0,Ωτ 只在 x x x 方向有增长而在 t t t 方向无增长那么这两条线段在计算曲线积分时关于 d t dt dt 的积分项为 0 0 0可得 − ∮ ∂ K τ a 2 ( u t u x ) d t 1 2 ( u t 2 a 2 u x 2 ) d x -\oint_{\partial{K_{\tau}}}a^2(u_tu_{x})dt\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx −∮∂Kτa2(utux)dt21(ut2a2ux2)dx ∫ Γ τ 1 ∪ Γ τ 2 a 2 ( u t u x ) d t 1 2 ( u t 2 a 2 u x 2 ) d x − ∫ Ω 0 1 2 ( u t 2 a 2 u x 2 ) d x ∫ Ω τ 1 2 ( u t 2 a 2 u x 2 ) d x \int_{\Gamma_{\tau_{1}}\cup\Gamma_{\tau_{2}}}a^2(u_tu_{x})dt\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx-\int_{\Omega_0}\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx\int_{\Omega_\tau}\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx ∫Γτ1∪Γτ2a2(utux)dt21(ut2a2ux2)dx−∫Ω021(ut2a2ux2)dx∫Ωτ21(ut2a2ux2)dx
记上面等式的三项分别为 J 1 , J 2 , J 3 J_1,J_2,J_3 J1,J2,J3即 J 1 ∫ Γ τ 1 ∪ Γ τ 2 a 2 ( u t u x ) d t 1 2 ( u t 2 a 2 u x 2 ) d x J_1\int_{\Gamma_{\tau_{1}}\cup\Gamma_{\tau_{2}}}a^2(u_tu_{x})dt\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx J1∫Γτ1∪Γτ2a2(utux)dt21(ut2a2ux2)dx J 2 − ∫ Ω 0 1 2 ( u t 2 a 2 u x 2 ) d x J_2-\int_{\Omega_0}\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx J2−∫Ω021(ut2a2ux2)dx J 3 ∫ Ω τ 1 2 ( u t 2 a 2 u x 2 ) d x J_3\int_{\Omega_\tau}\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx J3∫Ωτ21(ut2a2ux2)dx
应用下面 d x dx dx 和 d t dt dt 的关系将 J 1 J_1 J1 统一为只关于 d t dt dt 的积分 Γ τ 1 : d x a d t \Gamma_{\tau_{1}}:dxadt Γτ1:dxadt Γ τ 2 : d x − a d t \Gamma_{\tau_{2}}:dx-adt Γτ2:dx−adt
即 J 1 ∫ Γ τ 1 a 2 ( u t u x ) d t 1 2 a ( u t 2 a 2 u x 2 ) d t ∫ Γ τ 2 a 2 ( u t u x ) d t − 1 2 a ( u t 2 a 2 u x 2 ) d t J_1\int_{\Gamma_{\tau_{1}}}a^2(u_tu_{x})dt\frac{1}{2}a(u_{t}^2a^2u_{x}^2)dt\int_{\Gamma_{\tau_{2}}}a^2(u_tu_{x})dt-\frac{1}{2}a(u_{t}^2a^2u_{x}^2)dt J1∫Γτ1a2(utux)dt21a(ut2a2ux2)dt∫Γτ2a2(utux)dt−21a(ut2a2ux2)dt ∫ Γ τ 1 a 2 ( 2 a u t u x u t 2 a 2 u x 2 ) d t ∫ Γ τ 1 a 2 ( 2 a u t u x − u t 2 − a 2 u x 2 ) d t \int_{\Gamma_{\tau_{1}}}\frac{a}{2}(2au_tu_{x}u_{t}^2a^2u_{x}^2)dt\int_{\Gamma_{\tau_{1}}}\frac{a}{2}(2au_tu_{x}-u_{t}^2-a^2u_{x}^2)dt ∫Γτ12a(2autuxut2a2ux2)dt∫Γτ12a(2autux−ut2−a2ux2)dt
利用完全平方公式得 J 1 ∫ Γ τ 1 a 2 ( u t a u x ) 2 d t − ∫ Γ τ 2 a 2 ( u t − a u x ) 2 d t J_1\int_{\Gamma_{\tau_{1}}}\frac{a}{2}(u_tau_x)^2dt-\int_{\Gamma_{\tau_{2}}}\frac{a}{2}(u_t-au_x)^2dt J1∫Γτ12a(utaux)2dt−∫Γτ22a(ut−aux)2dt
因为在 Γ τ 1 \Gamma_{\tau_{1}} Γτ1 线段上规定的正方向是 t t t 增长的方向而 Γ τ 2 \Gamma_{\tau_{2}} Γτ2 上相反所以 t t t 的积分限分别是 0 0 0 到 τ \tau τ 和 τ \tau τ 到 0 0 0 也即 J 1 ∫ 0 τ a 2 ( u t a u x ) 2 d t − ∫ τ 0 a 2 ( u t − a u x ) 2 d t ≥ 0 J_1\int_{0}^{\tau}\frac{a}{2}(u_tau_x)^2dt-\int_{\tau}^0\frac{a}{2}(u_t-au_x)^2dt\geq0 J1∫0τ2a(utaux)2dt−∫τ02a(ut−aux)2dt≥0
第一项平方项且积分下限小于上限故其积分非负而第二项是平方项在积分下限大于上限加上前面的负号故也是非负所以 J 1 ≥ 0 J_1\geq0 J1≥0
把 J 1 , J 2 , J 3 J_1,J_2,J_3 J1,J2,J3 写回来根据 J 1 ≥ 0 J_1\geq0 J1≥0 构造不等关系 J 1 J 2 J 3 ∬ K τ u t f d x d t J_1J_2J_3\iint_{K_{\tau}}u_{t}fdxdt J1J2J3∬Kτutfdxdt ∵ J 1 ≥ 0 \because J_1\geq0 ∵J1≥0 ∴ J 3 ≤ J 1 J 3 J 1 J 2 J 3 − J 2 ∬ K τ u t f d x d t − J 2 \therefore J_3\leq J_1J_3J_1J_2J_3-J_2\iint_{K_{\tau}}u_{t}fdxdt-J_2 ∴J3≤J1J3J1J2J3−J2∬Kτutfdxdt−J2 也即 ∫ Ω τ 1 2 ( u t 2 a 2 u x 2 ) d x ≤ ∬ K τ u t f d x d t ∫ Ω 0 1 2 ( u t 2 a 2 u x 2 ) d x \int_{\Omega_\tau}\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx\leq \iint_{K_{\tau}}u_{t}fdxdt\int_{\Omega_0}\frac{1}{2}(u_{t}^2a^2u_{x}^2)dx ∫Ωτ21(ut2a2ux2)dx≤∬Kτutfdxdt∫Ω021(ut2a2ux2)dx 两边同乘 2 2 2 且 Ω 0 \Omega_0 Ω0 是 t 0 t0 t0 的线段而 t 0 t0 t0 时 u t ( x , 0 ) ψ , u x ( x , 0 ) φ x u_t(x,0)\psi,u_x(x,0)\varphi_{x} ut(x,0)ψ,ux(x,0)φx即 ∫ Ω τ ( u t 2 a 2 u x 2 ) d x ≤ ∬ K τ 2 u t f d x d t ∫ Ω 0 ( ψ 2 a 2 φ x 2 ) d x \int_{\Omega_\tau}(u_{t}^2a^2u_{x}^2)dx\leq \iint_{K_{\tau}}2u_{t}fdxdt\int_{\Omega_0}(\psi^2a^2\varphi_{x}^2)dx ∫Ωτ(ut2a2ux2)dx≤∬Kτ2utfdxdt∫Ω0(ψ2a2φx2)dx
对上面不等式右边 ∬ K τ 2 u t f d x d t \iint_{K_{\tau}}2u_{t}fdxdt ∬Kτ2utfdxdt 这一项应用 C a u c h y Cauchy Cauchy 不等式 2 a b ≤ a 2 b 2 2ab\leq a^2b^2 2ab≤a2b2 得 ∫ Ω τ ( u t 2 a 2 u x 2 ) d x ≤ ∬ K τ u t 2 f 2 d x d t ∫ Ω 0 ( ψ 2 a 2 φ x 2 ) d x \int_{\Omega_\tau}(u_{t}^2a^2u_{x}^2)dx\leq \iint_{K_{\tau}}u_{t}^2f^2dxdt\int_{\Omega_0}(\psi^2a^2\varphi_{x}^2)dx ∫Ωτ(ut2a2ux2)dx≤∬Kτut2f2dxdt∫Ω0(ψ2a2φx2)dx
此时对比要证明的不等式 ∫ Ω τ ( u t 2 a 2 u x 2 ) d x ≤ ∬ K τ f 2 d x d t ∫ Ω 0 ( ψ 2 a 2 φ x 2 ) d x \int_{\Omega_\tau}(u_{t}^2a^2u_{x}^2)dx\leq \iint_{K_{\tau}}f^2dxdt\int_{\Omega_0}(\psi^2a^2\varphi_{x}^2)dx ∫Ωτ(ut2a2ux2)dx≤∬Kτf2dxdt∫Ω0(ψ2a2φx2)dx 发现只是右端多了一项 ∬ K τ u t 2 d x d t \iint_{K_{\tau}}u_{t}^2dxdt ∬Kτut2dxdt
利用 G r o n w a l l Gronwall Gronwall 不等式将该项消去
令 G ( τ ) ∬ K τ ( u t 2 a 2 u x 2 ) d x d t ∫ 0 τ ∫ x 0 − a ( t 0 − t ) x 0 a ( t 0 − t ) ( u t 2 a 2 u x 2 ) d x d t G(\tau)\iint_{K_{\tau}}(u_{t}^2a^2u_{x}^2)dxdt\int_{0}^{\tau}\int_{x_0-a(t_0-t)}^{x_0a(t_0-t)}(u_{t}^2a^2u_{x}^2)dxdt G(τ)∬Kτ(ut2a2ux2)dxdt∫0τ∫x0−a(t0−t)x0a(t0−t)(ut2a2ux2)dxdt
则 d G ( τ ) d τ ∫ Ω τ ( u t 2 a 2 u x 2 ) d x \frac{dG(\tau)}{d\tau}\int_{\Omega_\tau}(u_{t}^2a^2u_{x}^2)dx dτdG(τ)∫Ωτ(ut2a2ux2)dx 是待证明不等式的左端
令 F ( τ ) ∬ K τ f 2 d x d t ∫ Ω 0 ( ψ 2 a 2 φ x 2 ) d x F(\tau)\iint_{K_{\tau}}f^2dxdt\int_{\Omega_0}(\psi^2a^2\varphi_{x}^2)dx F(τ)∬Kτf2dxdt∫Ω0(ψ2a2φx2)dx
给 ∫ Ω τ ( u t 2 a 2 u x 2 ) d x ≤ ∬ K τ u t 2 f 2 d x d t ∫ Ω 0 ( ψ 2 a 2 φ x 2 ) d x \int_{\Omega_\tau}(u_{t}^2a^2u_{x}^2)dx\leq \iint_{K_{\tau}}u_{t}^2f^2dxdt\int_{\Omega_0}(\psi^2a^2\varphi_{x}^2)dx ∫Ωτ(ut2a2ux2)dx≤∬Kτut2f2dxdt∫Ω0(ψ2a2φx2)dx 右边添加 ∬ K τ a 2 u x 2 d x d t \iint_{K_{\tau}}a^2u_{x}^2dxdt ∬Kτa2ux2dxdt则 ∫ Ω τ ( u t 2 a 2 u x 2 ) d x ≤ ∬ K τ u t 2 a 2 u x 2 f 2 d x d t ∫ Ω 0 ( ψ 2 a 2 φ x 2 ) d x \int_{\Omega_\tau}(u_{t}^2a^2u_{x}^2)dx\leq \iint_{K_{\tau}}u_{t}^2a^2u_{x}^2f^2dxdt\int_{\Omega_0}(\psi^2a^2\varphi_{x}^2)dx ∫Ωτ(ut2a2ux2)dx≤∬Kτut2a2ux2f2dxdt∫Ω0(ψ2a2φx2)dx
那么此时该不等式满足 G r o n w a l l Gronwall Gronwall 不等式的前提条件 d G ( τ ) d τ ≤ C G ( τ ) F ( τ ) \frac{dG(\tau)}{d\tau}\leq CG(\tau)F(\tau) dτdG(τ)≤CG(τ)F(τ) 此时 C 1 C1 C1那么由 G r o n w a l l Gronwall Gronwall 不等式的两个结论可分别得到如下两个能量不等式其中 M e τ Me^\tau Meτ ∫ Ω τ [ u t 2 ( x , τ ) a 2 u x 2 ( x , τ ) ] d x ≤ M [ ∫ Ω 0 ( ψ 2 a 2 φ x 2 ) d x ∫ K τ f 2 ( x , t ) d x d t ] \int_{\Omega_\tau} \left[u_t^2(x, \tau) a^2 u_x^2(x, \tau)\right] dx \leq M \left[ \int_{\Omega_0} (\psi^2 a^2 \varphi_x^2) dx \int_{K_\tau} f^2(x,t) dxdt \right] ∫Ωτ[ut2(x,τ)a2ux2(x,τ)]dx≤M[∫Ω0(ψ2a2φx2)dx∫Kτf2(x,t)dxdt] ∫ K τ [ u t 2 ( x , t ) a 2 u x 2 ( x , t ) ] d x d t ≤ M [ ∫ Ω 0 ( ψ 2 a 2 φ x 2 ) d x ∫ K τ f 2 ( x , t ) d x d t ] \int_{K_\tau} \left[u_t^2(x,t) a^2 u_x^2(x,t)\right] dxdt \leq M \left[ \int_{\Omega_0} (\psi^2 a^2 \varphi_x^2) dx \int_{K_\tau} f^2(x,t) dxdt \right] ∫Kτ[ut2(x,t)a2ux2(x,t)]dxdt≤M[∫Ω0(ψ2a2φx2)dx∫Kτf2(x,t)dxdt]
波动方程混合问题能量不等式的证明
待证明的两个不等式 ∫ 0 l [ u t 2 ( x , τ ) a 2 u x 2 ( x , τ ) ] d x ≤ M [ ∫ 0 l ( ψ 2 a 2 φ x 2 ) d x ∫ Q τ f 2 d x d t ] \int_{0}^{l} \left[u_t^2(x, \tau) a^2 u_x^2(x, \tau)\right] dx \leq M \left[ \int_{0}^{l} (\psi^2 a^2 \varphi_x^2) dx \int_{Q_\tau} f^2 dxdt \right] ∫0l[ut2(x,τ)a2ux2(x,τ)]dx≤M[∫0l(ψ2a2φx2)dx∫Qτf2dxdt] ∬ Q τ ( u t 2 a 2 u x 2 ) d x d t ≤ M [ ∫ 0 l ( ψ 2 a 2 φ x 2 ) d x ∫ Q τ f 2 d x d t ] \iint_{Q_\tau} \left(u_t^2 a^2 u_x^2\right) dxdt \leq M \left[ \int_{0}^{l} (\psi^2 a^2 \varphi_x^2) dx \int_{Q_\tau} f^2 dxdt \right] ∬Qτ(ut2a2ux2)dxdt≤M[∫0l(ψ2a2φx2)dx∫Qτf2dxdt] 二维区域的情况下这里可以不用 G r e e n Green Green 公式将面积分转为线积分可以直接按照二重积分展开。
在波动方程 u t t − a 2 u x x f u_{tt}-a^2u_{xx}f utt−a2uxxf 两端同乘 u t u_t ut 并在区域 Q τ ( 0 , l ) × ( 0 , τ ) Q_\tau(0,l)\times(0,\tau) Qτ(0,l)×(0,τ) 上积分 ∬ Q τ u t u t t − a 2 u t u x x d x d t ∬ Q τ u t f d x d t \iint_{Q_\tau}u_{t}u_{tt}-a^2u_{t}u_{xx}dxdt\iint_{Q_\tau}u_tf dxdt ∬Qτututt−a2utuxxdxdt∬Qτutfdxdt
利用常用的替换技巧 u t u t t 1 2 ⋅ 2 u t ⋅ u t t 1 2 ( u t 2 ) t u_{t}u_{tt}\frac{1}{2}\cdot 2u_{t}\cdot u_{tt}\frac{1}{2}(u_{t}^2)_t ututt21⋅2ut⋅utt21(ut2)t u t u x x ( u t u x ) x − 1 2 ( u x 2 ) t u_{t}u_{xx}(u_{t}u_{x})_x-\frac{1}{2}(u_{x}^2)_t utuxx(utux)x−21(ux2)t得 ∬ Q τ 1 2 ( u t 2 ) t 1 2 a 2 ( u x 2 ) t − ( a 2 u t u x ) x d x d t \iint_{Q_\tau}\frac{1}{2}(u_{t}^2)_t\frac{1}{2}a^2(u_{x}^2)_t-(a^2u_tu_x)_xdxdt ∬Qτ21(ut2)t21a2(ux2)t−(a2utux)xdxdt这里用二重积分展开 ∬ Q τ 1 2 ( u t 2 ) t 1 2 a 2 ( u x 2 ) t − ( a 2 u t u x ) x d x d t \iint_{Q_\tau}\frac{1}{2}(u_{t}^2)_t\frac{1}{2}a^2(u_{x}^2)_t-(a^2u_tu_x)_xdxdt ∬Qτ21(ut2)t21a2(ux2)t−(a2utux)xdxdt ∫ 0 τ ∫ 0 l [ ( 1 2 u t 2 1 2 a 2 u x 2 ) t − ( a 2 u t u x ) x ] d x d t \int_{0}^{\tau}\int_{0}^{l}[(\frac{1}{2}u_{t}^2\frac{1}{2}a^2u_{x}^2)_t-(a^2u_tu_x)_x]dxdt ∫0τ∫0l[(21ut221a2ux2)t−(a2utux)x]dxdt ∫ 0 l 1 2 u t 2 ( x , τ ) 1 2 a 2 u x 2 ( x , τ ) d x − ∫ 0 l 1 2 u t 2 ( x , 0 ) 1 2 a 2 u x 2 ( x , 0 ) d x − ∫ 0 τ a 2 u t ( l , t ) u x ( l , t ) d t ∫ 0 τ a 2 u t ( 0 , t ) u x ( 0 , t ) d t \int_{0}^{l}\frac{1}{2}u_{t}^{2}(x,\tau)\frac{1}{2}a^2u_{x}^{2}(x,\tau)dx-\int_{0}^{l}\frac{1}{2}u_{t}^{2}(x,0)\frac{1}{2}a^2u_{x}^{2}(x,0)dx-\int_{0}^{\tau}a^2u_{t}(l,t)u_{x}(l,t)dt\int_{0}^{\tau}a^2u_{t}(0,t)u_{x}(0,t)dt ∫0l21ut2(x,τ)21a2ux2(x,τ)dx−∫0l21ut2(x,0)21a2ux2(x,0)dx−∫0τa2ut(l,t)ux(l,t)dt∫0τa2ut(0,t)ux(0,t)dt
根据初始条件和齐次边界条件替换上面表达式
初始条件 u t ( x , 0 ) ψ ( x ) u_t(x,0)\psi(x) ut(x,0)ψ(x) u x ( x , 0 ) φ x ( x ) u_x(x,0)\varphi_x(x) ux(x,0)φx(x) 边界条件 u ( 0 , t ) 0 u(0,t)0 u(0,t)0 u ( l , t ) 0 u(l,t)0 u(l,t)0即杆的两端是固定的也即速度为 0 0 0 u t ( l , t ) u t ( 0 , t ) 0 u_t(l,t)u_t(0,t)0 ut(l,t)ut(0,t)0 带入 1. 1. 1. 得到的等式并两边同乘 2 2 2 得 ∫ 0 l u t 2 ( x , τ ) a 2 u x 2 ( x , τ ) d x ∫ 0 l ψ 2 a 2 φ x 2 d x 2 ∫ 0 τ ∫ 0 l u t f d x d t \int_{0}^{l}u_{t}^{2}(x,\tau)a^2u_{x}^{2}(x,\tau)dx\int_{0}^{l}\psi^2a^2\varphi_{x}^{2}dx2\int_{0}^{\tau}\int_{0}^{l}u_tfdxdt ∫0lut2(x,τ)a2ux2(x,τ)dx∫0lψ2a2φx2dx2∫0τ∫0lutfdxdt 对等式右边应用 C a u c h y Cauchy Cauchy 不等式以及添加非负项 ∫ 0 τ ∫ 0 l a 2 u x 2 d x d t \int_{0}^{\tau}\int_{0}^{l}a^2u_{x}^2dxdt ∫0τ∫0la2ux2dxdt 得 ∫ 0 l u t 2 ( x , τ ) a 2 u x 2 ( x , τ ) d x ≤ ∫ 0 l ψ 2 a 2 φ x 2 d x ∫ 0 τ ∫ 0 l f 2 d x d t ∫ 0 τ ∫ 0 l u t 2 a 2 u x 2 d x d t \int_{0}^{l}u_{t}^{2}(x,\tau)a^2u_{x}^{2}(x,\tau)dx\leq \int_{0}^{l}\psi^2a^2\varphi_{x}^{2}dx\int_{0}^{\tau}\int_{0}^{l}f^2dxdt\int_{0}^{\tau}\int_{0}^{l}u_{t}^2a^2u_{x}^2dxdt ∫0lut2(x,τ)a2ux2(x,τ)dx≤∫0lψ2a2φx2dx∫0τ∫0lf2dxdt∫0τ∫0lut2a2ux2dxdt
应用 G r o n w a l l Gronwall Gronwall 不等式
令 G ( τ ) ∫ 0 τ ∫ 0 l u t 2 a 2 u x 2 d x d t G(\tau) \int_{0}^{\tau}\int_{0}^{l}u_{t}^2a^2u_{x}^2dxdt G(τ)∫0τ∫0lut2a2ux2dxdt F ( τ ) ∫ 0 l ψ 2 a 2 φ x 2 d x ∫ 0 τ ∫ 0 l f 2 d x d t F(\tau)\int_{0}^{l}\psi^2a^2\varphi_{x}^{2}dx\int_{0}^{\tau}\int_{0}^{l}f^2dxdt F(τ)∫0lψ2a2φx2dx∫0τ∫0lf2dxdt
那么 d G ( τ ) d τ ∫ 0 l u t 2 ( x , τ ) a 2 u x 2 ( x , τ ) d x \frac{dG(\tau)}{d\tau}\int_{0}^{l}u_{t}^{2}(x,\tau)a^2u_{x}^{2}(x,\tau)dx dτdG(τ)∫0lut2(x,τ)a2ux2(x,τ)dx
将 G ( τ ) , F ( τ ) , d G ( τ ) d τ G(\tau),F(\tau),\frac{dG(\tau)}{d\tau} G(τ),F(τ),dτdG(τ) 带入 G r o n w a l l Gronwall Gronwall 不等式的两个结论即得到待证不等式其中 C 1 C1 C1
