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Let

  • $d\in\mathbb N$
  • $\lambda^d$ denote the Lebesgue measure on $\mathbb R^d$
  • $\Lambda\subseteq\mathbb R^d$ be open
  • $\mathcal V:=\left\{v\in C_c^\infty(\Lambda)^d:\nabla\cdot v=0\right\}$, $$V:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{H^1(\Lambda,\:\mathbb R^d)}}$$ and $$H:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}=\overline V^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}\tag 1$$
  • $W^1((0,T),V):=\left\{u\in L^1_{\text{loc}}((0,T),V):u\text{ is weakly differentiable}\right\}$

I want to reformulate $$\left\{\begin{array}{rll}\displaystyle\frac{\partial u}{\partial t}-\nu\Delta u+\left(u\cdot\nabla\right)u+\frac1\rho\nabla p&=&f&&\text{in }[0,T]\times\Lambda\\ u(0,\;\cdot\;)&=&u_0&&\text{in }\Lambda\end{array}\right.\tag 2$$ as a semilinear evolution equation on $H$.

It's easy to see that $$(A_0u)v:=\sum_{i=1}^d\langle\nabla u_i,\nabla v_i\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }u,v\in V$$ is a bounded linear operator from $V$ to $V'$. Moreover, if $d\le 4$ and $\Lambda$ is bounded, then $$b(u,v,w):=\int_\Lambda(u\cdot\nabla)v\cdot w\:{\rm d}\lambda^d\;\;\;\text{for }u,v,w\in H_0^1(\Lambda,\mathbb R^d)$$ is a well-defined bounded trilinear form on $H_0^1(\Lambda,\mathbb R^d)$ and hence $$B(u,v):=\left.b(u,v,\;\cdot\;)\right|_V\;\;\;\text{for }u,v\in V$$ is a bounded bilinear operator from $V\times V$ to $V'$. Now, let $u$ be a classical solution of $(2)$ and $$\tilde u(t):=u(t,\;\cdot\;)\;\;\;\text{for }t\in(0,T)\;.$$ A straightforward calculation yields $\tilde u\in W^1((0,T),V)$ with $$\tilde u'(t)+\nu A_0\tilde u(t)+B(\tilde u(t),\tilde u(t))=0\;\;\;\text{for all }t\in[0,T].\tag 3$$

Note that $(3)$ is an equation in $V'$ and hence any solution $\tilde u\in W^1((0,T),V)$ of $(3)$ corresponds to the notion of a weak solution of $(2)$. Since I'm interested in a mild solution of $(2)$, I somehow need to reformulate $(2)$ as an equation in $H$.

It's well-known that $$A:=\left.A_0\right|_{\mathcal D(A)}$$ with $D(A):=V\cap H^2(\Lambda,\mathbb R^d)$ is a densely-defined linear operator on $H$. Moreover, $-A$ is the generator of a contraction $C_0$-semigroup $S$ on $H$.

Now, I've often seen that people call $$v(t)=S(t)v_0+\int_0^tS(t-s)B(v(s),v(s))\:{\rm ds}\;\;\;\text{for all }t\in[0,T]\tag 4$$ the mild equation corresponding to $(2)$.

However, I expect that a solution $v$ of $(4)$ is $H$-valued! As stated above, if $v(t)\in\mathcal D(A)$, then $A_0v(t)\in H$ for all $t\in[0,T]$, but the $B(v(t),v(t))$ are still $V'$-valued.

Is this simply ruled out by $V'\cong V$ or has $(4)$ to be understood in a special sense?

In order to prevent any confusion: My problem with $(4)$ is that $A_0v(t)+B(v(t),v(t))\not\in H$, cause $B(v(t),v(t))\not\in H$, even when $v(t)\in\mathcal D(A)$, for any $t\in[0,T]$.

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  • $\begingroup$ Have you really used the additional $H^2$ regularity of $v(t) \in D(A)$? Shouldn't this be (by far) enough for $B$ to give rise to a continuous linear form on $H$? $\endgroup$
    – Hannes
    Commented Dec 7, 2016 at 9:27
  • $\begingroup$ @Hannes I need a continuous linear operator on $H$, not a continuous linear form. $\endgroup$
    – 0xbadf00d
    Commented Dec 7, 2016 at 10:47
  • $\begingroup$ Yes, that's what I meant. Sorry, I shouldn't have called it form in this context. $\endgroup$
    – Hannes
    Commented Dec 7, 2016 at 12:08
  • $\begingroup$ @Hannes Meanwhile, I've seen that some authors define $\tilde B(u,v):=\operatorname P_H\left[(u\cdot\nabla)v\right]$ even for $u,v\in V$, where $\operatorname P_H$ is the orthogonal projection from $L^2(\Lambda,\mathbb R^d)$ onto $H$; see, for example, here on page 2. But I don't see that $(u\cdot\nabla)v\in L^2(\Lambda,\mathbb R^d)$ for all $u,v\in V$. I've just asked for this on MSE. $\endgroup$
    – 0xbadf00d
    Commented Dec 7, 2016 at 13:55
  • $\begingroup$ I think you need $u,v$ in $D(A)$, both in the paper regarding the projection, and here, and use the additional $H^2$ regularity for elements from $D(A)$. $\endgroup$
    – Hannes
    Commented Dec 7, 2016 at 15:08

1 Answer 1

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You are right, $(4)$ has to be understood in the special sense that the semigroup $S(t)$ (and the (Leray?) $L^2$-projection operator to $H$, as well) extend to a wider space, that of "distributional derivatives of $L^1$ functions". Then $(4)$ makes sense as soon as $v(t)\in H$ (with some measurability in $t$).

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  • $\begingroup$ Please take note of my last three comments below the question. I'm only interested in mild solutions of $(2)$; which should be easier than finding a weak solution. From my understanding, $(3)$ is the wrong equation to consider, unless one is searching for weak solutions. However, I can't find any good reference which is considering mild solutions of $(2)$ (and hence of $(6)$). $\endgroup$
    – 0xbadf00d
    Commented Dec 7, 2016 at 16:54

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