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Suppose we have a PDE $$\langle u', v \rangle + a(u,v) = 0$$ where $a:V\times V \to \mathbb{R}$ is a bounded symmetric bilinear form, then if $u_0 \in V$ then $u \in L^2(0,T; V)$ with $u' \in L^2(0,T;H)$. This works because we get a bound on $u_n'$ (Galerkin approximation) by writing $a(u_n, u_n')$ as $\frac{1}{2}\frac{d}{dt}a(u_n, u_n)$, integrating and using the boundedness condition.

If $a$ is not symmetric (eg. $a = \int_{\Omega}\nabla \cdot (\textbf{b}u)v)$$a = \int_{\Omega}\nabla \cdot (\textbf{b}u)v + \int_{\Omega}\nabla u \cdot \nabla v$), then we can't write it as a derivative alone. What can we do to get a uniform bound on $u_m'$ in $L^2(0,T;H)$ in this case? I know we can get a bound in $L^2(0,T;V^*)$ but I want it in a stronger space. Thanks.

Suppose we have a PDE $$\langle u', v \rangle + a(u,v) = 0$$ where $a:V\times V \to \mathbb{R}$ is a bounded symmetric bilinear form, then if $u_0 \in V$ then $u \in L^2(0,T; V)$ with $u' \in L^2(0,T;H)$. This works because we get a bound on $u_n'$ (Galerkin approximation) by writing $a(u_n, u_n')$ as $\frac{1}{2}\frac{d}{dt}a(u_n, u_n)$, integrating and using the boundedness condition.

If $a$ is not symmetric (eg. $a = \int_{\Omega}\nabla \cdot (\textbf{b}u)v)$, then we can't write it as a derivative alone. What can we do to get a uniform bound on $u_m'$ in $L^2(0,T;H)$ in this case? I know we can get a bound in $L^2(0,T;V^*)$ but I want it in a stronger space. Thanks.

Suppose we have a PDE $$\langle u', v \rangle + a(u,v) = 0$$ where $a:V\times V \to \mathbb{R}$ is a bounded symmetric bilinear form, then if $u_0 \in V$ then $u \in L^2(0,T; V)$ with $u' \in L^2(0,T;H)$. This works because we get a bound on $u_n'$ (Galerkin approximation) by writing $a(u_n, u_n')$ as $\frac{1}{2}\frac{d}{dt}a(u_n, u_n)$, integrating and using the boundedness condition.

If $a$ is not symmetric (eg. $a = \int_{\Omega}\nabla \cdot (\textbf{b}u)v + \int_{\Omega}\nabla u \cdot \nabla v$), then we can't write it as a derivative alone. What can we do to get a uniform bound on $u_m'$ in $L^2(0,T;H)$ in this case? I know we can get a bound in $L^2(0,T;V^*)$ but I want it in a stronger space. Thanks.

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  • 113
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Galerkin method for existence for PDE with nonsymmetric bilinear form

Suppose we have a PDE $$\langle u', v \rangle + a(u,v) = 0$$ where $a:V\times V \to \mathbb{R}$ is a bounded symmetric bilinear form, then if $u_0 \in V$ then $u \in L^2(0,T; V)$ with $u' \in L^2(0,T;H)$. This works because we get a bound on $u_n'$ (Galerkin approximation) by writing $a(u_n, u_n')$ as $\frac{1}{2}\frac{d}{dt}a(u_n, u_n)$, integrating and using the boundedness condition.

If $a$ is not symmetric (eg. $a = \int_{\Omega}\nabla \cdot (\textbf{b}u)v)$, then we can't write it as a derivative alone. What can we do to get a uniform bound on $u_m'$ in $L^2(0,T;H)$ in this case? I know we can get a bound in $L^2(0,T;V^*)$ but I want it in a stronger space. Thanks.