• $d\in\left\{1,\ldots,4\right\}$
  • $\Lambda\subseteq\mathbb R^d$ be bounded, nonempty and open
  • $\mathcal V:=\left\{\phi\in C_c^\infty(\Lambda,\mathbb R^d):\nabla\cdot\phi=0\right\}$, $V:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{H^1(\Lambda,\:\mathbb R^d)}}$ 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 $u,v,w\in V$
  • $\nu>0$
  • $f\in L^2(\Lambda,\mathbb R^d)$

I've found proofs showing that there is a $u\in V$ with $$\tilde{\mathfrak a}(u,v):=\nu\mathfrak a(u,v)+\mathfrak b(u,u,v)=\langle f,v\rangle_{L^2(\Lambda,\:\mathbb R^d)}=:\ell(v)\;\;\;\text{for all }v\in V\tag1$$ using the (Leray-)Schauder fixed-point theorem.

However, doesn't the existence (and even uniqueness) of such a $u$ simply follow from the Lax-Milgram theorem?

It's well-known that

  1. $\mathfrak a$ is a bounded and coercive bilinear form on $H_0^1(\Lambda,\mathbb R^d)$
  2. $\mathfrak b$ is a bounded trilinear form on $H_0^1(\Lambda,\mathbb R^d)$ with $$\mathfrak b(u,v,w)+\mathfrak b(u,w,v)=-\langle\left(\nabla\cdot u\right)v,w\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for all }u,v,w\in H_0^1(\Lambda,\mathbb R^d)\tag2$$

Since $$\tilde{\mathfrak a}(u,u)=\nu\mathfrak a(u,u)+\underbrace{\mathfrak b(u,u,u)}_{=\:0}=\nu\mathfrak a(u,u)\;\;\;\text{for all }u\in V\tag3\;,$$ $\tilde{\mathfrak a}$ is a bounded and coercive bilinear form on $V$. Moreover, $\ell$ is a bounded linear functional on $V$. So, there should be a unique $u\in V$ with $(1)$ by the Lax-Milgram theorem.

What am I missing?


What I was missing is that $$V\times V\ni(u,v)\mapsto\mathfrak b(u,u,v)$$ (and hence $\tilde{\mathfrak a}$) is not bilinear. It's obvious, but I missed that. However, we can use the Lax-Milgram theorem to show that for all $u\in V$ there is a unique $\Phi(u)\in V$ with $$\nu\mathfrak a(\Phi(u),v)+\mathfrak b(u,\Phi(u),v)=\ell(v)\tag4\;.$$ Now, the Schauder fixed-point theorem can be used to show that $\Phi$ has a fixed-point.


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