Krylov’s Approach of Recovering Pressure from Velocity
The mathematician Nicolai Vladimirovich Krylov wrote a book Lectures on Elliptic and Parabolic Equations in Sobolev Space and this article is meant to give a complete solution to the three exercises at the end of Section 9, Chapter 1 of this book.
Let me first introduce the background. Let $u(x)$ be the velocity of some incompressible fluid at the point $x\in \mathbb{R}^d$. We also assume $d\geq 3$. It is well-known that the pressure $p(x)$ satisfies the following PDE
\[-\Delta p= \partial_{ij}(u_iu_j),\](we use $u_i$ to denote the $i^{th}$ component of $u$ and the Einstein notation) and can be recovered from $u$ by using the following singular integral formula:
\[\begin{align*} p(x)&=-\frac{\delta_{i j}}{d} u_i(x) u_j(x)+ \text{p.v.}\int_{\mathbb{R}^d} K_{i j}(y) u_i(x-y) u_j(x-y) dy\\ &=-\frac{\delta_{i j}}{d} u_i(x) u_j(x)+\lim_{r\downarrow 0} \int_{|y|\geq r} K_{ij}(y) u_i(x-y) u_j(x-y) dy \hspace{2mm}(1), \end{align*}\]where $\alpha_d$ is the volume of the unit ball in $\mathbb{R}^d$ and
\[K_{i j}(y)=\frac{dy_iy_j-\delta_{ij}|y|^2}{d\alpha_d|y|^{d+2}}.\]Krylov does not mention fluid mechanics in Section 9, Chapter 1, but he talks about how to derive the formula above and to prove an estimate for $p$. More specifically, we can use his method to show that if $u\in C_0^2$ (we use 0 in subscript to mean that $u$ is compactly supported), then
\[\begin{align*} p(x)=&\Big(\int_{\mathbb{R}^d}\frac{y_i}{d\alpha_d|y|^d}\partial_j\zeta(y)dy\Big)u_i(x)u_j(x)\\ &+\lim_{r\downarrow 0} \int_{|y|\geq r} K_{ij}(y)u_i(x-y)u_j(x-y)dy, \end{align*}\]where $\zeta\in C_0^\infty$ is a radially symmetric function such that $\zeta(0)=1$. Hence, to obtain Formula (1), we just need to solve the following exercise.
Exercise 1 :
\[N_{ij}:=\int_{\mathbb{R}^d}\frac{y_i}{d\alpha_d|y|^d}\partial_j\zeta(y)dy=-\frac{\delta_{ij}}{d}.\]Proof : Define a function $g:[0,\infty)\rightarrow \mathbb{R}$ by using the following formula:
\[g(r)=\zeta((r,0,\cdots, 0)).\]Clearly $g\in C_0^\infty$, $\zeta(x)=g(\lvert x\rvert)$, and we have
\[\begin{align*} N_{ij}&=\int_{\mathbb{R}^d}\frac{y_i}{d\alpha_d|y|^d}\partial_j\zeta(y)dy\\ &=\int_{\mathbb{R}^d}\frac{y_iy_j}{d\alpha_d|y|^d}g'(|y|)dy. \end{align*}\]From this formula, we can see that $N_{ii}$ is independent of $i$. Therefore, we have
\[\begin{align*} dN_{ii}=\sum_{i=1}^dN_{ii}&=\sum_{i=1}^d\int_{\mathbb{R}^d}\frac{y_i}{d\alpha_d|y|^d}\partial_i\zeta(y)dy. \end{align*}\]Krylov also shows in this section that
\[\int_{\mathbb{R}^d}\frac{y_i}{d\alpha_d|y|^d}\partial_i\zeta(y)dy=\partial_{ii}v(0),\]where $v$ satisfies $-\Delta v=\zeta$. We then can deduce that
\[\begin{align*} dN_{ii}&=\Delta v(0)=-\zeta(0)=-1,\\ N_{ii}&=-\frac{1}{d}. \end{align*}\]If $i\ne j$, we can use a change of variable $y_i\rightarrow -y_i$ to see that $N_{ij}=-N_{ij}$, which implies that $N_{ij}=0$. $\square$
Now we want to obtain an estimate of the $L^q$ norm of $p$ in terms of the the $L^{2q}$ norm of $u$ (at least for some $q\in (1,\infty)$). The theory of singular integral operators (see Grafakos’ Classical Fourier Analysis) implies that for each pair $(i,j)$ and every $q\in (1,\infty)$,
\[\begin{align*} \|\lim_{r\downarrow 0} \int_{|y|\geq r} K(y)f(x-y)dy|\|_{L^q}&\lesssim \|u_iu_j\|_{L^q}\leq \|u\|_{L^{2q}}^2. \end{align*}\]Here’s a rough outline of the proof that is applicable for any reasonably nice functions (e.g. $C_0^\infty$) $K$ and $f\in L^q\cap C_0^\infty$. We first prove this inequality for $q=2$. It is easy to see that we have the following pointwise convergence:
\[\begin{align*} &\lim_{r\downarrow 0} \int_{|y|\geq r} K(y)f(x-y)dy\\&=\lim_{r\downarrow 0} \int_{r\leq |y|\leq \frac{1}{r}} K(y)f(x-y)dy \end{align*}\]as $K(y)f(x-y)$ is integrable. By Fatou’s Lemma and Plancherel theorem, we have
\[\begin{align*} &\|\lim_{r\downarrow 0} \int_{|y|\geq r} K(y)f(x-y)dy\|_{L^2}\\&\leq \lim_{r\downarrow 0} \|\int_{r\leq |y|\leq \frac{1}{r}} K(y)f(x-y)dy\|_{L^2}\\ &=\lim_{r\downarrow 0} \|\mathcal{F}(K\chi_{\{r\leq |y|\leq \frac{1}{r}\}})\mathcal{F}(f)\|_{L^2}\\ &\leq \sup_{r>0}\|\mathcal{F}(K\chi_{\{r\leq |y|\leq \frac{1}{r}\}})\|_{L^\infty}\|f\|_{L^2}. \end{align*}\]As a result, we will obtain the desired $L^2$ boundedness if we have a uniform estimate of $\parallel\mathcal{F}(K\chi_{{r\leq \lvert y\rvert\leq \frac{1}{r}}})\parallel_{L^\infty}$, which we omit the details. Once we have the $L^2$ boundedness, we can prove a weak type (1,1) estimate by using the Calderón–Zygmund Decomposition. We can then finally obtain $L^q$ boundedness for all $q\in (1,\infty)$ by using the Marcinkiewicz Interpolation Theorem.
In the case that $K=K_{ij}$ and $i\ne j$, we don’t need to use Fatou’s Lemma because we have the $L^2$ convergence.
Exercise 2 : Let $u\in C_0^\infty$ and $f=-\Delta u$. Show that $K_{ij}\chi_{r<\lvert y\rvert<\frac{1}{r}}*f$ converges to $\partial_{ij} u$ in the $L^2$ sense.
Proof : Since $i\ne j$, $K_{ij}$ is antisymmetric in $y^i$, $K_{ij}\zeta$ has vanishing integral over any compact annulus, which implies the following pointwise convergence:
\[\begin{align*} \partial_{ij} u(x)&=\lim_{r\downarrow 0} \int_{|y|\geq r} K_{ij}(y)f(x-y)dy\\ &=\lim_{r\downarrow 0} \int_{r\leq |y|\leq \frac{1}{r}} K_{ij}(y)f(x-y)dy\\ &=\lim_{r\downarrow 0} \int_{r\leq |y|\leq \frac{1}{r}} K_{ij}(y)(f(x-y)-f(x)\zeta(y))dy\\ \end{align*}\]If we can find a dominating function of
\[\Big|\int_{r\leq \lvert|y|\leq \frac{1}{r}} K_{ij}(y)(f(x-y)-f(x)\zeta(y))dy\Big|^2,\]then we can apply DCT to obtain the $L^2$ convergence.
Let $R>0$ be large enough so that $\zeta(x)=f(x)=0$ for any $\lvert x\rvert>R$. We consider two cases.
Case 1: Suppose $\lvert x\rvert\leq R+1$. For any $\lvert y\rvert >R+\lvert x\rvert$, we have
\[\begin{align*} |x-y|&\geq |y|-|x|>R. \end{align*}\]Hence, $f(x-y)-f(x)\zeta(y)=0$ for any $\lvert y\rvert >R+\lvert x\rvert$. We can then use the mean value inequality to prove that for all $r$ that is small enough,
\[\begin{align*} &\Big|\int_{r\leq |y|\leq \frac{1}{r}} K_{ij}(y)(f(x-y)-f(x)\zeta(y))dy\Big|\\ &\leq \int_{r\leq |y|\leq R+|x|} |K_{ij}(y)||f(x-y)-f(x)+f(x)(1-\zeta(y))|dy\\ &\lesssim \int_{|y|\leq R+|x|}\frac{(\|\nabla f\|_{L^\infty}+\|f\|_{L^\infty}\|\nabla \zeta\|_{L^\infty})|y|}{|y|^d}dy\\ &\lesssim\int_{|y|\leq R+|x|}\frac{1}{|y|^{d-1}}dy\\ &\lesssim R+|x|\\ &\leq 2R+1. \end{align*}\]Case 2: Suppose $\lvert x\rvert>R+1$. In this case, $f(x)=0$, and for any $\lvert y\rvert<\lvert x\rvert-R$, we have
\[\begin{align*} |x-y|&\geq |x|-|y|>R. \end{align*}\]Hence, $f(x-y)=0$ for any $\lvert y\rvert<\lvert x\rvert-R$. For all $r$ that is small enough, we have
\[\begin{align*} &\Big|\int_{r\leq |y|\leq \frac{1}{r}} K_{ij}(y)f(x-y)dy\Big|\\ &\lesssim \int_{|x|-R\leq |y|} \frac{|f(x-y)|}{|y|^d}dy\\ &\leq \frac{1}{(|x|-R)^d}\int_{|x|-R\leq |y|} |f(x-y)|dy\\ &\leq \frac{\|f\|_{L^1}|x|^d}{(|x|-R)^d}\frac{1}{|x|^d}\\ &\lesssim \frac{1}{|x|^d}. \end{align*}\]We have the last inequality because $\lvert x\rvert/(\lvert x\rvert -R)$ is a a bounded function over the region $\lvert x\rvert>R+1$.
Clearly, the function $g$ defined by $g(x)=2R+1$ for $\lvert x\rvert\leq R+1$ and $g(x)=\lvert x\rvert^{-d}$ for $\lvert x\rvert>R+1$ is in $L^2$, so $g(x)^2$ is a suitable domaining function of $K_{ij}\chi_{r<\lvert y\rvert<\frac{1}{r}}*f$. $\square$
Here’s another way of deriving the $L^2$ estimate for the singular integral operators defined by the kernel $K_{ij}$. The general idea of this approach can be summarized by one word: pertubation. We first consider $\lambda-\Delta$ and prove an estimate for this operator, where $\lambda>0$ , and then obtain an estimate for $-\Delta$ by taking $\lambda\rightarrow 0$.
Lemma : For all $u\in C^\infty\cap W^{2,2}=C^\infty\cap H^2$,
\[\begin{align*} \lambda^2\|u\|^2_{L^2}+2\lambda\sum_{i}\|\partial_iu\|_{L^2}^2+\sum_{i,j}\|\partial_{ij}u\|_{L^2}^2&=\|\lambda u-\Delta u\|^2_{L^2}\\ \|\partial_{ij}u\|_{L^2}&\leq \|\lambda u-\Delta u\|_{L^2}\\ &\leq \lambda \|u\|_{L^2}+\|\Delta u\|_{L^2} \end{align*}\]Proof : Use Fourier transform. See Lemma 1 in Krylov’s book. $\square$
We are now ready to solve Exercise 3.
Exercise 3 : Let $f\in C_0^\infty$. Define $u$ by
\[u(x)=\int_{\mathbb{R}^d}\frac{1}{d(d-2)\alpha_d|y|^{d-2}}f(x-y)dy.\]Show that $u\in C^\infty\cap H^2$ and
\[\|\partial_{ij}u\|_{L^2}\leq \|\Delta u\|_{L^2}.\]Conclude that the singular integral operator defind by $K_{ij}$ is bounded from $L^2$ to $L^2$.
Proof : $u\in C^\infty$ because $f\in C^\infty$ and one can commute $\partial_i$ with $\int$. Moreover, we can apply Young’s convolution inequality to deduce that
\[\begin{align*} \partial_i u(x)&=-\int_{\mathbb{R}^d}\frac{y_i}{d\alpha_d|y|^d}f(x-y)dy\\ \|\partial_i u(x)\|_{L^2}&\lesssim \|\frac{y_i}{|y|^d}\chi_{|y|\leq R}\|_{L^1}\|f\|_{L^2}<\infty \end{align*}\]where $R>0$ is a large enough number so that the support of $f$ is contained in the We can similarly prove an estimate for second-order derivatives. Therefoere, $u\in H^2$, which allows us to use Lemma 1 to derive the following inequality:
\[\begin{align*} \forall \lambda>0, \|\partial_{ij}u\|_{L^2}&\leq \lambda \|u\|_{L^2}+\|\Delta u\|_{L^2},\\ \|\partial_{ij}u\|_{L^2}&\leq \|\Delta u\|_{L^2},\\ \|\lim_{r\downarrow 0} \int_{|y|\geq r} K_{ij}(y)f(x-y)dy\|_{L^2}&\leq \|f\|_{L^2}. \end{align*}\]$\square$