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Sep 18, 2018 at 16:55 comment added Willie Wong Without the $\langle B, B\rangle$ term the situation is worse. You don't even need $\lambda$ sufficiently small. If it has the "correct" sign it breaks the coercivity.
Sep 18, 2018 at 16:29 comment added Cogicero Thanks again. This is clear now. I have accepted the answer, but please I have one more related question. Does it change anything if the $\langle B,B \rangle$ term is not on the left i.e. the left hand is simply $(D(x),D(v))_{L^2} + (∇⋅x,∇⋅v)_{L^2}$ (that's what I get when I approach the problem a different way)? It seems to me, that it would mean $\langle A,B \rangle$ is on the left, which isn't necessarily non-negative.
Sep 18, 2018 at 15:38 vote accept Cogicero
Sep 18, 2018 at 13:59 comment added Willie Wong Essentially, you have some vector $A$ (which I use as a stand in for $\nabla x$) and some vector $B$ (which I use as a stand in for $\nabla v$). Your left hand side is basically $\langle A,B\rangle + \langle B, B\rangle$. Let $B' = \lambda B$. As long as $A$ and $B$ is not orthogonal, then you have $$ \langle A, B'\rangle + \langle B', B'\rangle = \lambda (\langle A, B\rangle + \lambda \langle B, B\rangle) $$ so for any $\lambda$ between $0$ and $-\langle A,B\rangle / \langle B, B\rangle$ the quantity is negative.
Sep 18, 2018 at 5:36 comment added Cogicero Also, I am sorry I see how the left is linear in $x$ so I think that infers linearity in $v$. However I can't see how the left hand side is negative. Could you please expand on that? Thanks!
Sep 17, 2018 at 23:02 comment added Cogicero Thanks for your useful answer! I am inclined to accept this right away and move on, but a friend just suggested I could try integrating by parts on the first term on the left. Does this look like it makes sense? I'm clutching at straws here. Thanks!
Sep 17, 2018 at 20:28 history answered Willie Wong CC BY-SA 4.0