Mixing
evolution equation satisfied by projection
$begingroup$
Consider the linear parabolic problem (as in Evans)
$$begin{cases}
partial_t u+Lu=f : mbox{in} :Omega times (0,T]\
u=0 : mbox{on} :partial Omega times (0,T]\
u(cdot,0)=g : mbox{in} :Omega
end{cases}$$
where $L$ is the operator given by
$$Lu=-mathrm{div}(A(x,t)nabla u)+b(x,t)cdotnabla u+c(x,t)u$$
which we assume to be parabolic with bounded coefficients (and $A$ symmetric).
A weak solution is given by a function $u in L^2(0,T;H^1_0(Omega))$ with $u'in L^2(0,T;H^{-1}(Omega))$ that satisfies the initial condition and the following equation:
$$forall 0 < t leq T,:forall v in H^1_0(Omega),: langle u'(t), v rangle + a(t;u(t),v)=(f(t),v)_{L^2}$$
(here $a$ is the usual bilinear form, depending on time, associated to $L$)
Consider a Hilbert basis $(w_i)$ of $L^2(Omega)$ formed by the eigenfunction of the Laplace operator, so that
$$w_i in H^1_0(Omega),:int_Omega w_i w_j = delta_{ij},:int_Omega nabla w_icdot nabla w_j =||nabla w_i||^2_{L^2} delta_{ij}$$
Set $V_m=overline{mathrm{span}}({ w_i| 1 leq i leq m})$ and denote by $P_m$ the projection on $V_m$.
If $u$ is a weak solution, is it true that $u_m=P_m circ u$ is a solution of the following approximated problem on $V_m$?
$$forall 0 < t leq T,:forall v in V_m,: (u_m'(t), v)_{L^2} + a(t;u_m(t),v)=(f(t),v)_{L^2}$$
I have been able to prove the following. If $u(t)=sum_{i=1}^infty y_i(t) w_i$, then $y_i in H^1(0,T)$ so that $u_m in H^1(0,T;V_m)$ with $u_m(t)=sum_{i=1}^m y_i(t)w_i$ and $u'_m(t)=sum_{i=1}^m y'_i(t)w_i$. Then one can see that for all $v in V_m$
$$langle u'(t),v rangle=(u'_m(t),v)_{L^2}$$
$$ int_Omega A(cdot,t)nabla u(t) cdot nabla v=int_Omega A(cdot,t)nabla u_m(t) cdot nabla v$$
$$ int_Omega c(cdot,t)u(t)v=int_Omega c(cdot,t)u_m(t)v$$
I don't know if is true that it works also for the $b$ term, that is if for $v in V_m$
$$int_Omega b(cdot,t)cdot v nabla u(t)=int_Omega b(cdot,t) cdot v nabla u_m(t)$$
I think it would be sufficient to know that $int_Omega w_j nabla w_i=0$ for all $i neq j$, but I don't know if it is true and (eventually) how to prove it.
pde parabolic-pde bochner-spaces
asked Jan 15 at 8:30
LouisLouis
1247
$endgroup$
add a comment |
$begingroup$
Consider the linear parabolic problem (as in Evans)
$$begin{cases}
partial_t u+Lu=f : mbox{in} :Omega times (0,T]\
u=0 : mbox{on} :partial Omega times (0,T]\
u(cdot,0)=g : mbox{in} :Omega
end{cases}$$
where $L$ is the operator given by
$$Lu=-mathrm{div}(A(x,t)nabla u)+b(x,t)cdotnabla u+c(x,t)u$$
which we assume to be parabolic with bounded coefficients (and $A$ symmetric).
A weak solution is given by a function $u in L^2(0,T;H^1_0(Omega))$ with $u'in L^2(0,T;H^{-1}(Omega))$ that satisfies the initial condition and the following equation:
$$forall 0 < t leq T,:forall v in H^1_0(Omega),: langle u'(t), v rangle + a(t;u(t),v)=(f(t),v)_{L^2}$$
(here $a$ is the usual bilinear form, depending on time, associated to $L$)
Consider a Hilbert basis $(w_i)$ of $L^2(Omega)$ formed by the eigenfunction of the Laplace operator, so that
$$w_i in H^1_0(Omega),:int_Omega w_i w_j = delta_{ij},:int_Omega nabla w_icdot nabla w_j =||nabla w_i||^2_{L^2} delta_{ij}$$
Set $V_m=overline{mathrm{span}}({ w_i| 1 leq i leq m})$ and denote by $P_m$ the projection on $V_m$.
If $u$ is a weak solution, is it true that $u_m=P_m circ u$ is a solution of the following approximated problem on $V_m$?
$$forall 0 < t leq T,:forall v in V_m,: (u_m'(t), v)_{L^2} + a(t;u_m(t),v)=(f(t),v)_{L^2}$$
I have been able to prove the following. If $u(t)=sum_{i=1}^infty y_i(t) w_i$, then $y_i in H^1(0,T)$ so that $u_m in H^1(0,T;V_m)$ with $u_m(t)=sum_{i=1}^m y_i(t)w_i$ and $u'_m(t)=sum_{i=1}^m y'_i(t)w_i$. Then one can see that for all $v in V_m$
$$langle u'(t),v rangle=(u'_m(t),v)_{L^2}$$
$$ int_Omega A(cdot,t)nabla u(t) cdot nabla v=int_Omega A(cdot,t)nabla u_m(t) cdot nabla v$$
$$ int_Omega c(cdot,t)u(t)v=int_Omega c(cdot,t)u_m(t)v$$
I don't know if is true that it works also for the $b$ term, that is if for $v in V_m$
$$int_Omega b(cdot,t)cdot v nabla u(t)=int_Omega b(cdot,t) cdot v nabla u_m(t)$$
I think it would be sufficient to know that $int_Omega w_j nabla w_i=0$ for all $i neq j$, but I don't know if it is true and (eventually) how to prove it.
pde parabolic-pde bochner-spaces
asked Jan 15 at 8:30
LouisLouis
1247
$endgroup$
add a comment |
$begingroup$
Consider the linear parabolic problem (as in Evans)
$$begin{cases}
partial_t u+Lu=f : mbox{in} :Omega times (0,T]\
u=0 : mbox{on} :partial Omega times (0,T]\
u(cdot,0)=g : mbox{in} :Omega
end{cases}$$
where $L$ is the operator given by
$$Lu=-mathrm{div}(A(x,t)nabla u)+b(x,t)cdotnabla u+c(x,t)u$$
which we assume to be parabolic with bounded coefficients (and $A$ symmetric).
A weak solution is given by a function $u in L^2(0,T;H^1_0(Omega))$ with $u'in L^2(0,T;H^{-1}(Omega))$ that satisfies the initial condition and the following equation:
$$forall 0 < t leq T,:forall v in H^1_0(Omega),: langle u'(t), v rangle + a(t;u(t),v)=(f(t),v)_{L^2}$$
(here $a$ is the usual bilinear form, depending on time, associated to $L$)
Consider a Hilbert basis $(w_i)$ of $L^2(Omega)$ formed by the eigenfunction of the Laplace operator, so that
$$w_i in H^1_0(Omega),:int_Omega w_i w_j = delta_{ij},:int_Omega nabla w_icdot nabla w_j =||nabla w_i||^2_{L^2} delta_{ij}$$
Set $V_m=overline{mathrm{span}}({ w_i| 1 leq i leq m})$ and denote by $P_m$ the projection on $V_m$.
If $u$ is a weak solution, is it true that $u_m=P_m circ u$ is a solution of the following approximated problem on $V_m$?
$$forall 0 < t leq T,:forall v in V_m,: (u_m'(t), v)_{L^2} + a(t;u_m(t),v)=(f(t),v)_{L^2}$$
I have been able to prove the following. If $u(t)=sum_{i=1}^infty y_i(t) w_i$, then $y_i in H^1(0,T)$ so that $u_m in H^1(0,T;V_m)$ with $u_m(t)=sum_{i=1}^m y_i(t)w_i$ and $u'_m(t)=sum_{i=1}^m y'_i(t)w_i$. Then one can see that for all $v in V_m$
$$langle u'(t),v rangle=(u'_m(t),v)_{L^2}$$
$$ int_Omega A(cdot,t)nabla u(t) cdot nabla v=int_Omega A(cdot,t)nabla u_m(t) cdot nabla v$$
$$ int_Omega c(cdot,t)u(t)v=int_Omega c(cdot,t)u_m(t)v$$
I don't know if is true that it works also for the $b$ term, that is if for $v in V_m$
$$int_Omega b(cdot,t)cdot v nabla u(t)=int_Omega b(cdot,t) cdot v nabla u_m(t)$$
I think it would be sufficient to know that $int_Omega w_j nabla w_i=0$ for all $i neq j$, but I don't know if it is true and (eventually) how to prove it.
pde parabolic-pde bochner-spaces
asked Jan 15 at 8:30
LouisLouis
1247
$endgroup$
Consider the linear parabolic problem (as in Evans)
$$begin{cases}
partial_t u+Lu=f : mbox{in} :Omega times (0,T]\
u=0 : mbox{on} :partial Omega times (0,T]\
u(cdot,0)=g : mbox{in} :Omega
end{cases}$$
where $L$ is the operator given by
$$Lu=-mathrm{div}(A(x,t)nabla u)+b(x,t)cdotnabla u+c(x,t)u$$
which we assume to be parabolic with bounded coefficients (and $A$ symmetric).
A weak solution is given by a function $u in L^2(0,T;H^1_0(Omega))$ with $u'in L^2(0,T;H^{-1}(Omega))$ that satisfies the initial condition and the following equation:
$$forall 0 < t leq T,:forall v in H^1_0(Omega),: langle u'(t), v rangle + a(t;u(t),v)=(f(t),v)_{L^2}$$
(here $a$ is the usual bilinear form, depending on time, associated to $L$)
Consider a Hilbert basis $(w_i)$ of $L^2(Omega)$ formed by the eigenfunction of the Laplace operator, so that
$$w_i in H^1_0(Omega),:int_Omega w_i w_j = delta_{ij},:int_Omega nabla w_icdot nabla w_j =||nabla w_i||^2_{L^2} delta_{ij}$$
Set $V_m=overline{mathrm{span}}({ w_i| 1 leq i leq m})$ and denote by $P_m$ the projection on $V_m$.
If $u$ is a weak solution, is it true that $u_m=P_m circ u$ is a solution of the following approximated problem on $V_m$?
$$forall 0 < t leq T,:forall v in V_m,: (u_m'(t), v)_{L^2} + a(t;u_m(t),v)=(f(t),v)_{L^2}$$
I have been able to prove the following. If $u(t)=sum_{i=1}^infty y_i(t) w_i$, then $y_i in H^1(0,T)$ so that $u_m in H^1(0,T;V_m)$ with $u_m(t)=sum_{i=1}^m y_i(t)w_i$ and $u'_m(t)=sum_{i=1}^m y'_i(t)w_i$. Then one can see that for all $v in V_m$
$$langle u'(t),v rangle=(u'_m(t),v)_{L^2}$$
$$ int_Omega A(cdot,t)nabla u(t) cdot nabla v=int_Omega A(cdot,t)nabla u_m(t) cdot nabla v$$
$$ int_Omega c(cdot,t)u(t)v=int_Omega c(cdot,t)u_m(t)v$$
I don't know if is true that it works also for the $b$ term, that is if for $v in V_m$
$$int_Omega b(cdot,t)cdot v nabla u(t)=int_Omega b(cdot,t) cdot v nabla u_m(t)$$
I think it would be sufficient to know that $int_Omega w_j nabla w_i=0$ for all $i neq j$, but I don't know if it is true and (eventually) how to prove it.
pde parabolic-pde bochner-spaces
pde parabolic-pde bochner-spaces
asked Jan 15 at 8:30
LouisLouis
1247
asked Jan 15 at 8:30
LouisLouis
1247
asked Jan 15 at 8:30
LouisLouis
1247
asked Jan 15 at 8:30
LouisLouis
1247
asked Jan 15 at 8:30
LouisLouis
1247
1247
add a comment |
add a comment |
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