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Feb 6, 2020 at 13:12 history edited YCor
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Dec 7, 2016 at 16:50 comment added 0xbadf00d @Hannes So, my last question is: Is there any reason why I shouldn't consider $(5)$ if I'm interested in a mild solution of $(2)$? And is there any reference which considers $(5)$?
Dec 7, 2016 at 16:50 comment added 0xbadf00d @Hannes (2) So, if we equip $W:=H\cap H^2(Λ,ℝ^d)$ and define $$\tilde L(u):=L(u,u)\;\;\;\text{for }u∈W\;,$$ then $A$ would be a densely-defined operator on $W$ and it would make sense to consider the equation $$u'(t)+νAu(t)+\tilde L(u(t))=0\;\;\;\text{for all }t∈[0,T]\;.\tag 6$$ Since each classical solution of $(2)$ is obviously an solution of $(5)$, I don't understand why $(3)$ is mostly considered in the literature instead of $(5)$. The only thing I could imagine is that most authors search for weak solutions of $(2)$ and, obviously, $(3)$ is the natural equation for this kind of solution.
Dec 7, 2016 at 16:50 comment added 0xbadf00d @Hannes (1) In order to prevent any misunderstanding: Let's define $$L(u,v):=(u⋅\nabla)v\;\;\;\text{for }u∈ H_0^1(Λ,ℝ^d)\text{ and }v∈ H_0^2(Λ,ℝ^d)$$ and assume that $d≤4$ and $Λ$ is bounded and open. Then, $$u_j,\frac{∂ v_i}{∂ x_j}∈H_0^1(Λ)⊆L^4(Λ)\tag 5$$ by the Sobolev inequalities and hence $B(u,v)∈L^2(Λ,ℝ^d)$ for all $u∈ H_0^1(Λ,ℝ^d)$ and $v∈H_0^2(Λ,ℝ^d)$.
Dec 7, 2016 at 16:49 answer added Jean Duchon timeline score: 2
Dec 7, 2016 at 15:08 comment added Hannes 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 $D(A)$.
Dec 7, 2016 at 13:55 comment added 0xbadf00d @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.
Dec 7, 2016 at 12:08 comment added Hannes Yes, that's what I meant. Sorry, I shouldn't have called it form in this context.
Dec 7, 2016 at 10:47 comment added 0xbadf00d @Hannes I need a continuous linear operator on $H$, not a continuous linear form.
Dec 7, 2016 at 9:27 comment added Hannes 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 $H$?
Dec 6, 2016 at 17:03 history asked 0xbadf00d CC BY-SA 3.0