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Let

  • $d\in\left\{1,\ldots,4\right\}$
  • $\lambda$ denote the Lebesgue measure on $\mathbb R^d$
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open
  • $\Delta t>0$
  • $\nu>0$

Note that the norm induced by $$\mathfrak a(u,v):=\sum_{i=1}^d\langle\nabla u_i,\nabla v_i\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }u,v\in H^1(\Lambda,\mathbb R^d)$$ is equivalent to $\left\|\;\cdot\;\right\|_{H^1(\Lambda,\:\mathbb R^d)}$ on $H_0^1(\Lambda,\mathbb R^d)$ and hence we will assume that the latter space is equipped with that norm. Moreover, $$\mathfrak c(u,v,w):=\int_\Lambda(u\cdot\nabla)v\cdot w)\:{\rm d}\lambda\;\;\;\text{for }u,v,w\in H_0^1(\Lambda,\mathbb R^d)$$ is a bounded trilinear form with $$\mathfrak c(u,v,w)+\mathfrak c(u,w,v)=-\langle(\nabla\cdot u)v,w\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for all }u,v,w\in H_0^1(\Lambda,\mathbb R^d)\tag1\;.$$

Now, let $$\tilde{\mathfrak a}(u,v):=\langle u,v\rangle_{L^2(\Lambda,\:\mathbb R^d)}+\Delta t\nu\mathfrak a(u,v)+\Delta t\mathfrak c(u^0,u,v)\;\;\;\text{for }u,v\in H_0^1(\Lambda,\mathbb R^d)$$ for some $u^0\in H_0^1(\Lambda,\mathbb R^d)$. How can we show that $$\tilde{\mathfrak a}(u,u)>0\;\;\;\text{for all }u\in H_0^1(\Lambda,\mathbb R^d)\tag2\;;$$ at least for $\Delta$ sufficiently small?

Since $$\tilde{\mathfrak a}(u,u)=\left\|u\right\|_{L^2(\Lambda,\:\mathbb R^d)}+\Delta t\nu\left\|u\right\|_{H_0^1(\Lambda,\:\mathbb R^d)}-\frac{\Delta t}2\langle(\nabla\cdot u^0)u,u\rangle_{L^2(\Lambda,\:\mathbb R^d)}\tag3$$ for all $u\in H_0^1(\Lambda,\mathbb R^d$, $(2)$ would hold, for any $\Delta t$, if $\nabla\cdot u^0=0$. How can we show $(2)$ for arbitrary $u^0$?

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  • $\begingroup$ Why is $\mathfrak c$ well-defined for arbitrary $u,v,w \in H_0^1(\Lambda,\mathbb R^d)$? $\endgroup$ – Michał Miśkiewicz Apr 22 '17 at 17:17
  • $\begingroup$ @MichałMiśkiewicz because $H^1_0$ is continuously embedded in $L^2$ for $d\le 4$. $\endgroup$ – daw Apr 22 '17 at 18:13
  • $\begingroup$ I doubt that the claim is true without further assumptions (i.e., $d<4$, $u^0\in W^{1,\infty})$. For the given regularity, the term $c(u^0,u,u)$ is at the edge of being well-defined. It seems to be impossible to make any use of the addition $L^2$-term in $\tilde a$. $\endgroup$ – daw Apr 22 '17 at 18:19

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