2
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Let

  • $d\in\left\{1,\ldots,4\right\}$
  • $\Lambda\subseteq\mathbb R^d$ be a bounded Lipschitz domain
  • $\mathcal V:=\left\{\phi\in C_c^\infty(\Lambda,\mathbb R^d):\nabla\cdot\phi=0\right\}$, $V:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{H^1(\Lambda,\:\mathbb R^d)}}$ and $H:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}$
  • $\mathfrak a(u,v):=\sum_{i=1}^d\langle\nabla u_i,\nabla v_i\rangle_{L^2}$ for $u,v\in V$
  • $\mathfrak b(u,v,w):=\langle((u\cdot\nabla)v,w\rangle_{L^2}$ for $u,v,w\in V$
  • $\nu>0$
  • $f\in L^2(\Lambda,\mathbb R^d)$
  • $u^0\in V$
  • $h>0$

Now, let $$\eta_u(v):=\frac1h\langle u-u^0,v\rangle_H+\nu\mathfrak a(u,v)+\mathfrak b(u,u,v)-\langle f,v\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }v\in V$$ for $u\in V$. It's easy to see that $$\left|\mathfrak a(u,v)\right|\le C_1\left\|u\right\|_H\left\|v\right\|_H\;\;\;\text{for all }u,v\in V\tag1$$ and $$\left|\mathfrak b(u,u,v)\right|\le C_2\left\|u\right\|_V^2\left\|v\right\|_V\;\;\;\text{for all }u,v\in V\tag2$$ for some $C_1,C_2\ge 0$. Thus, $$\left|\eta_u(v)\right|\le\left(\left(\frac1h+\nu C_1\right)\left\|u\right\|_H+C_2\left\|u\right\|_V^2+\left\|f\right\|_{L^2(\Lambda,\:\mathbb R^d)}+\frac1h\left\|u_0\right\|_H\right)\left\|v\right\|_V\;\;\;\text{for all }v\in V\tag3\;,$$ i.e. $$\eta_u\in V'\;,\tag4$$ for all $u\in V$. By $(4)$ and Riesz' representation theorem, for all $u\in V$ there is a unique $\Phi(u)\in V$ with $$\langle\Phi(u),v\rangle_V=\eta_u(v)\;\;\;\text{for all }v\in V\tag5\;.$$

How can we show that $V\ni u\mapsto\Phi(u)$ is continuous?

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  • $\begingroup$ Maybe, continuity of $V\ni u\mapsto\Phi(u)$ follows from continuity of $\eta_u$ with respect to $u$? $\endgroup$ – VorKir Mar 3 '17 at 4:35

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