How to get the following explicit expression of the pressure term of Navier-Stokes equations?

Question

In 1.3.4 of 'the Mathematical Analysis of the Incompressible Eluer and Navier-Stokes Equations, Jacob Bedrossian & Vlad Vicol', Theorem 1.13 says that

Assume $u\in L^{2q}(\mathbb{R}^d)$ for some $q\in (1,\infty).$ Then the equation$-\Delta p=\partial_{ij}(u_iu_j)$ has a unique solution $p\in L^q(\mathbb{R}^d),$ which is given explicitly as $$p(x)=-\frac{\delta_{ij}}{d}u_i(x)u_j(x)+p.v.\int_{\mathbb{R}^d}K_{ij}u_i(x-y)u_j(x-y)dy,$$ where $$K_{ij}=\frac{y_iy_j-\frac{\delta_{ij}}{d}|y|^2}{\alpha_d|y|^{d+2}},$$ where $\delta_{ij}$ is Kronecker symbol, $\alpha_d$ is the volume of the unit ball in $\mathbb{R}^d,$ and we use Einstein summation convention here.

The author's proof is quite confused for me. I have the following question about his proof,

1. He first assume that $u\in C_{c}^2,$ and he says $$p(x)=\lim_{\epsilon\to 0}\int_{|y|>\epsilon}\mathcal{N}(y)\partial_i\partial_j(u_iu_j(x-y))dy$$ if $u\in C_{c}^2,$ where $\mathcal{N}$ is the Newtonian potential.

In my opinion, if $u\in C_{c}^2, \partial_{ij}(u_iu_j)$ doesn't belong to $C^2_{c},$ and from Evans's book ' Partial differential Equation', I know if $f\in C^2_c,$ we can write $$u(x)=\int_{\mathbb{R}^n}\mathcal{N}(y)f(x-y)dy$$ as the expilicit expression of Poisson's equation$-\Delta u=f.$

but for the equation$-\Delta p=\partial_{ij}(u_iu_j)$, if $u\in C^2_c,$ $\partial_{ij}(u_iu_j)$ doesn't belong to $C^2_c,$ so why the author can write the expression of $p(x)$ like that?

2. The author says 'The general $L^{2q}$ case follows by approximation using the Calderon-Zygmund theorem. I can't understand it, how can I get the details?

Answer 1

1. To obtain the explicit solution to $-\Delta u = f$, the assumption that $f\in C_c^2$ in Evans is a technical assumption and is not needed. For example, for any locally integrable $f$ the convolution representation solves $-\Delta u = f$ in the distributional sense; this follows from the fact that $-\Delta \mathcal{N} = \delta_0$ in the sense of distributions (this is a more or less straightforward calculation, similar to the one done in Evans), so that $-\Delta(\mathcal{N}*f) = (-\Delta\mathcal{N})*f = f$.

2. The Calderon-Zygmund theorem says that for a suitable kernel $K$, $\|K*f\|_p\leq C\|f\|_p$ for $1<p<\infty$. There are various characterizations of such kernels, but one common set of assumptions is

   1. $K$ is smooth away from the origin and $|\nabla K(x)|\leq C\frac{1}{|x|^{d+1}}$
   2. $K$ has the representation $K(x) = \Omega(x)|x|^{-d}$ where $\Omega$ is homogeneous of degree zero and $$\int_{|x|=1}\Omega(x) dS(x)=0$$

You can find this theorem in any harmonic analysis text, like Stein's Singular Integrals. The claim is then immediate:

$$\|p\|_q \leq \frac1d\||u|^2\|_q + C\sum_{i,j}\|u_iu_j\|_q \leq C'\|u\|_{2q}$$

Answer 2

From harmonic analysis, the Calderon-Zygmund operator $ T = (-\Delta)^{-1}\partial_{i}\partial_j$ is bounded $L^p\to L^p$ for any $p\in(1,\infty)$. What this means is that for any $v\in L^p$, $Tv\in L^p$ with the bound $$ \|Tv\|_{L^p} \le C \|u\|_{L^p}$$ for some $C$ that is independent of $v$. If you are following this for the first time, maybe it is good to outline the approximation argument. One usually proves the following slightly different statement first: for all "nice" functions $u$ (Schwartz, or $C^\infty_c$, or in your specific case, $u\in C^2_c$ is enough), $$ \|Tv\|_{L^p} \le C \|v\|_{L^p}$$ This inequality is then used to define $Tv$ for $v\in L^p$: Let $v_n$ be a sequence of "nice" functions, with $v_n\to v$ in $L^p$. Then we define $Tv$ as the $L^p$-limit of $T v_n$. The inequality above tells us that $Tv_n \in L^p$ for each $n$, and the convergence is guaranteed by the linearity of $T$: $$ \|Tv_n - T v_m \|_{L^p} = \| T(v_n - v_m)\|_{L^p} \le C \|v_n - v_m\|_{L^p} \to 0.$$

Your case is precisely the above, as for $u\in C^2_c$, we also have $v=u_i u_j \in C^2_c$.