论文标题
线性Navier-Stokes方程的解决方案的估计值
Estimates of solutions to the linear Navier-Stokes equation
论文作者
论文摘要
The linear Navier-Stokes equations in three dimensions are given by: $u_{it}(x,t)-ρ\triangle u_i(x,t)-p_{x_i}(x,t)=$ $w_i(x,t)$ , $div \textbf{u}(x,t)=0,i=1,2,3$ with initial conditions: $ \ textbf {u} | _ {（t = 0）\ bigcup \partialΩ} = 0 $。 dirichlet问题的绿色函数$ \ textbf {u} | _ {（t = 0）\ bigcup \partialΩ} = 0 $的方程$ u_ u_ {it}（x，x，x，x，x，t）-ρ\ triangle u_i（x，x，x，x，x，x，t）= f_i（x，t）= f_i（x，x，x，x，t） $ g（x，t;ξ，τ）= z（x，t;ξ，τ）+v（x，t;ξ，τ）。$其中$ z（x，x，t;ξ，τ）= \ frac {1} {1} {8π^{3/2} {3/2}（3/2}（t-t- t- t- t-τ） e^{ - \ frac {（x_1-ξ_1）^2+（x_2-ξ_2）^2+（x_3-ξ_3）^2} {4（t-τ）}} $是该方程的基本解决方案，$ v（x，x，x，t; t;ξ，τ）$是$（x，t；函数$ g（x，t;ξ，τ）$的构建在书中[1 p.106]。通过绿色函数，我们将Navier-Stokes方程呈现为： $ u_i（x，t）= \ int_0^t \int_Ω\ big（z（x，t;ξ，τ）+v（x，x，t;ξ，τ）\ big）\ fr ac {dp（ξ，τ）} {dξ}dξDτ+\ int_0^t t \int_Ωg（x，t; t;ξ，τ）但是$ div \ textbf {u}（x，x，t）= \ sum_1^3 \ frac {du_i（x，x，t）} {dx_i} = 0。 x_i} = - \ frac {d z（x，t;ξ，τ）} {d耳}，$，用于定义未知压力p（x，t）的定义，我们将收到积分方程。从这个积分方程式中，我们定义了压力的显式表达：$ p（x，t）= - \ frac {d} {dt} {dt} \ triangle^{ - 1} \ ast \ ast \ int_0^t \ int_0^t \int_Ω\ sum_1 \ sum_1^3 \ frac {dg（x，t;ξ，τ）} {dx_i} w_i（ξ，τ）dξdDτ+ρ+ρ\ cdot \ cd \ int_0^t \int_Ω\ sum_1^3 \ sum_1^3 \ frac {dg（x，x，x，x，t; t;ξ，τ）} $ y this dx_i} {dx_i} w_i（def）以下估计值：$ \ int_0^t \ sum_1^3 \ big \ | \ frac {\ partial p（x，x，x，τ）} {\ partial x_i} \ big | ___________ {l_2（ω）}^2 D D τ<c \ cdot \ int_0^t s sum_1^3 \ | w_i（x，τ）\ | _ | _ {l_2（ω）}^2dτ$ holds。
The linear Navier-Stokes equations in three dimensions are given by: $u_{it}(x,t)-ρ\triangle u_i(x,t)-p_{x_i}(x,t)=$ $w_i(x,t)$ , $div \textbf{u}(x,t)=0,i=1,2,3$ with initial conditions: $\textbf{u}|_{(t=0)\bigcup\partialΩ}=0$. The Green function to the Dirichlet problem $\textbf{u}|_{(t=0)\bigcup\partialΩ}=0$ of the equation $u_{it}(x,t)-ρ\triangle u_i(x,t)=f_i(x,t)$ present as: $G(x,t;ξ,τ)=Z(x,t;ξ,τ)+V(x,t;ξ,τ).$ Where $Z(x,t;ξ,τ)=\frac{1}{8π^{3/2}(t-τ)^{3/2}}\cdot e^{-\frac{(x_1-ξ_1)^2+(x_2-ξ_2)^2+(x_3-ξ_3)^2}{4(t-τ)}}$ is the fundamental solution to this equation and $V(x,t;ξ,τ)$ is the smooth function of variables $(x,t;ξ,τ)$. The construction of the function $G(x,t;ξ,τ)$ is resulted in the book [1 p.106]. By the Green function we present the Navier-Stokes equation as: $u_i(x,t)=\int_0^t\int_Ω\Big(Z(x,t;ξ,τ)+V(x,t;ξ,τ)\Big)\frac{dp(ξ,τ)}{dξ}dξdτ+\int_0^t\int_ΩG(x,t;ξ,τ)w_i(ξ,τ)dξdτ$. But $div \textbf{u}(x,t)=\sum_1^3 \frac{du_i(x,t)}{dx_i}=0.$ Using these equations and the following properties of the fundamental function: $Z(x,t;ξ,τ)$: $\frac{dZ(x,t;ξ,τ)}{d x_i}=-\frac{d Z(x,t; ξ,τ)}{d ξ_i},$ for the definition of the unknown pressure p(x,t) we shall receive the integral equation. From this integral equation we define the explicit expression of the pressure: $p(x,t)=-\frac{d}{dt}\triangle^{-1}\ast\int_0^t\int_Ω\sum_1^3 \frac{dG(x,t;ξ,τ)}{dx_i}w_i(ξ,τ)dξdτ+ρ\cdot\int_0^t\int_Ω\sum_1^3\frac{dG(x,t;ξ,τ)}{dx_i}w_i(ξ,τ)dξdτ.$ By this formula the following estimate: $\int_0^t\sum_1^3\Big\|\frac{\partial p(x,τ)}{\partial x_i}\Big\|_{L_2(Ω)}^2 d τ<c\cdot\int_0^t\sum_1^3\|w_i(x,τ)\|_{L_2(Ω)}^2 dτ$ holds.
