Weak Formulation of Navier-Stokes
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I was looking into the Navier-Stokes Weak formulation
Let $fin L^2(Omega_T)$, $u_0 in H(Omega)=lbrace uin L^2(Omega):text{div }u=0text{ in }Omega;ucdot n|_{partial Omega}=0rbrace$.
A measurable function $u:Omega_Trightarrow mathbb{R}^N$, $N=2,3,$ is said to be a weak solution to Navier-Stokes equations in $Omega_T$ if
$u in V_T equiv L^2(0,T;H^1_0)cap L^{infty}(0,T;H)$
and $u$ satisfies $$int_0^{infty}Big(big(u,frac{partialvarphi}{partial t}big)-nu(nabla u,nablavarphi)-(ucdotnabla u,varphi)Big)text{ d}t=-int_0^{infty}(f,varphi)text{ d}t-(u_0,varphi(0))$$
$$forall varphiin mathcal{D}_T=lbrace varphi in C_0^{T}(Omega_T):; text{div}varphi(x,t)=0 text{ in } Omega_Trbrace.$$
Then I've found this lemma, said to be useful
Let $u$ be a weak solution in $Omega_T$. Then $u$ can be redefined on a
set of zero Lebesgue measure in such a way that $u(t) in L^2(Omega)$ $forall tin[0,T)$ and satisfies
begin{align*}
int_s^tBig(big(u,frac{partialvarphi}{partial t}big)-nu(nabla u,nablavarphi)-(ucdotnabla u,varphi)Big)text{ d}tau\
=-int_s^{t}(f,varphi)text{ d}tau+(u(t),varphi(t))-(v(s),varphi(s)),\forall sin[0,t],t<T,text{ and }forall varphiin mathcal{D}_T.
end{align*}
Then there are some theorems about equivalency of this formulations.
But I don't really think I can understand well the main idea, so obviously we use smaller integration interval but how does this contribute to futher results.
I would be really grateful for some explanation, thanks!
fluid-dynamics
asked Dec 19 '18 at 19:13
user396656user396656
232
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add a comment |
$begingroup$
I was looking into the Navier-Stokes Weak formulation
Let $fin L^2(Omega_T)$, $u_0 in H(Omega)=lbrace uin L^2(Omega):text{div }u=0text{ in }Omega;ucdot n|_{partial Omega}=0rbrace$.
A measurable function $u:Omega_Trightarrow mathbb{R}^N$, $N=2,3,$ is said to be a weak solution to Navier-Stokes equations in $Omega_T$ if
$u in V_T equiv L^2(0,T;H^1_0)cap L^{infty}(0,T;H)$
and $u$ satisfies $$int_0^{infty}Big(big(u,frac{partialvarphi}{partial t}big)-nu(nabla u,nablavarphi)-(ucdotnabla u,varphi)Big)text{ d}t=-int_0^{infty}(f,varphi)text{ d}t-(u_0,varphi(0))$$
$$forall varphiin mathcal{D}_T=lbrace varphi in C_0^{T}(Omega_T):; text{div}varphi(x,t)=0 text{ in } Omega_Trbrace.$$
Then I've found this lemma, said to be useful
Let $u$ be a weak solution in $Omega_T$. Then $u$ can be redefined on a
set of zero Lebesgue measure in such a way that $u(t) in L^2(Omega)$ $forall tin[0,T)$ and satisfies
begin{align*}
int_s^tBig(big(u,frac{partialvarphi}{partial t}big)-nu(nabla u,nablavarphi)-(ucdotnabla u,varphi)Big)text{ d}tau\
=-int_s^{t}(f,varphi)text{ d}tau+(u(t),varphi(t))-(v(s),varphi(s)),\forall sin[0,t],t<T,text{ and }forall varphiin mathcal{D}_T.
end{align*}
Then there are some theorems about equivalency of this formulations.
But I don't really think I can understand well the main idea, so obviously we use smaller integration interval but how does this contribute to futher results.
I would be really grateful for some explanation, thanks!
fluid-dynamics
asked Dec 19 '18 at 19:13
user396656user396656
232
$endgroup$
add a comment |
$begingroup$
I was looking into the Navier-Stokes Weak formulation
Let $fin L^2(Omega_T)$, $u_0 in H(Omega)=lbrace uin L^2(Omega):text{div }u=0text{ in }Omega;ucdot n|_{partial Omega}=0rbrace$.
A measurable function $u:Omega_Trightarrow mathbb{R}^N$, $N=2,3,$ is said to be a weak solution to Navier-Stokes equations in $Omega_T$ if
$u in V_T equiv L^2(0,T;H^1_0)cap L^{infty}(0,T;H)$
and $u$ satisfies $$int_0^{infty}Big(big(u,frac{partialvarphi}{partial t}big)-nu(nabla u,nablavarphi)-(ucdotnabla u,varphi)Big)text{ d}t=-int_0^{infty}(f,varphi)text{ d}t-(u_0,varphi(0))$$
$$forall varphiin mathcal{D}_T=lbrace varphi in C_0^{T}(Omega_T):; text{div}varphi(x,t)=0 text{ in } Omega_Trbrace.$$
Then I've found this lemma, said to be useful
Let $u$ be a weak solution in $Omega_T$. Then $u$ can be redefined on a
set of zero Lebesgue measure in such a way that $u(t) in L^2(Omega)$ $forall tin[0,T)$ and satisfies
begin{align*}
int_s^tBig(big(u,frac{partialvarphi}{partial t}big)-nu(nabla u,nablavarphi)-(ucdotnabla u,varphi)Big)text{ d}tau\
=-int_s^{t}(f,varphi)text{ d}tau+(u(t),varphi(t))-(v(s),varphi(s)),\forall sin[0,t],t<T,text{ and }forall varphiin mathcal{D}_T.
end{align*}
Then there are some theorems about equivalency of this formulations.
But I don't really think I can understand well the main idea, so obviously we use smaller integration interval but how does this contribute to futher results.
I would be really grateful for some explanation, thanks!
fluid-dynamics
asked Dec 19 '18 at 19:13
user396656user396656
232
$endgroup$
I was looking into the Navier-Stokes Weak formulation
Let $fin L^2(Omega_T)$, $u_0 in H(Omega)=lbrace uin L^2(Omega):text{div }u=0text{ in }Omega;ucdot n|_{partial Omega}=0rbrace$.
A measurable function $u:Omega_Trightarrow mathbb{R}^N$, $N=2,3,$ is said to be a weak solution to Navier-Stokes equations in $Omega_T$ if
$u in V_T equiv L^2(0,T;H^1_0)cap L^{infty}(0,T;H)$
and $u$ satisfies $$int_0^{infty}Big(big(u,frac{partialvarphi}{partial t}big)-nu(nabla u,nablavarphi)-(ucdotnabla u,varphi)Big)text{ d}t=-int_0^{infty}(f,varphi)text{ d}t-(u_0,varphi(0))$$
$$forall varphiin mathcal{D}_T=lbrace varphi in C_0^{T}(Omega_T):; text{div}varphi(x,t)=0 text{ in } Omega_Trbrace.$$
Then I've found this lemma, said to be useful
Let $u$ be a weak solution in $Omega_T$. Then $u$ can be redefined on a
set of zero Lebesgue measure in such a way that $u(t) in L^2(Omega)$ $forall tin[0,T)$ and satisfies
begin{align*}
int_s^tBig(big(u,frac{partialvarphi}{partial t}big)-nu(nabla u,nablavarphi)-(ucdotnabla u,varphi)Big)text{ d}tau\
=-int_s^{t}(f,varphi)text{ d}tau+(u(t),varphi(t))-(v(s),varphi(s)),\forall sin[0,t],t<T,text{ and }forall varphiin mathcal{D}_T.
end{align*}
Then there are some theorems about equivalency of this formulations.
But I don't really think I can understand well the main idea, so obviously we use smaller integration interval but how does this contribute to futher results.
I would be really grateful for some explanation, thanks!
fluid-dynamics
fluid-dynamics
asked Dec 19 '18 at 19:13
user396656user396656
232
asked Dec 19 '18 at 19:13
user396656user396656
232
asked Dec 19 '18 at 19:13
user396656user396656
232
asked Dec 19 '18 at 19:13
user396656user396656
232
asked Dec 19 '18 at 19:13
user396656user396656
232
232
add a comment |
add a comment |
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