## Tag Archives: regularity

### One regularity of NS 3-D

$\displaystyle \begin{cases}u_t+u\cdot\nabla u-\nu\Delta u+\nabla p=0\\div u=0\\u(x,0)=u_0\end{cases}\text{ on }\mathbb{R}^3$

Thm: If ${u}$ solves NS 3-D and ${u\in C([0,T];L^3(\mathbb{R}^3))}$, then ${u}$ is regular.

Proof: Denote

$\displaystyle E=\int_{\mathbb{R}^3}|u_0|^2dx$

Note that ${\int_{\mathbb{R}^3}|u(x,t)|^2dx\leq E}$ for all ${t>0}$.

Fix ${M>0}$, which will be determined later, letting ${G_M=\{x|u(x,t)\leq M\}}$, by Chebyshev’s inequality

$\displaystyle M^2\left|G^c_M\right|\leq \int_{\mathbb{R}^3}|u(x,t)|^2dx\leq E$

$\displaystyle \left|G^c_M\right|\leq \frac{E}{M^2}.$

Since ${u\in C([0,T];L^3)}$, then ${u(\cdot,t)}$ is compact in ${L^3}$, in particular, ${u(\cdot,t)}$ is uniformly integrable, namely

$\displaystyle \sup_t\left|\int_B|u(x,t)|^3dx\right|\leq \varepsilon\text{ whenever }|B|\leq \delta(\varepsilon),t\in [0,T]$

Multiplying the NS equation by ${\Delta u}$, we get

$\displaystyle \frac{1}{2}\frac{d}{dt}\int |\nabla u|^2+\nu\int |\Delta u|^2dx\leq \int |u||\nabla u||\Delta u|$

$\displaystyle \leq \int_{G_M}|u||\nabla u||\Delta u|dx+\int_{G_M^c}|u||\nabla u||\Delta u|dx$

$\displaystyle \leq M\int |\nabla u||\Delta u|dx+\left(\int_{G_M^c} u^3dx\right)^{1/3}||\nabla u||_{L^6}||\Delta u||_{L^2}$

Choose ${M}$ such that ${\int_{G_M^c} u^3dx}$ small enough and use the Sobolev inequality ${||\nabla u||_{L^6}||\leq C||\Delta u||_{L^2}}$,

$\displaystyle \frac{1}{2}\frac{d}{dt}\int |\nabla u|^2+\frac{\nu}{2}\int |\Delta u|^2dx\leq C||\nabla u||_{L^2}||\Delta u||_{L^2}$

By Young’s inequality,

$\displaystyle \frac{1}{2}\frac{d}{dt}\int |\nabla u|^2\leq C||\nabla u||_{L^2}^2$

So ${\nabla u\in L^\infty([0,T]; L^2)}$. From the regularity theory, ${u}$ must be a regular solution. $\Box$

Remark: Follows from my note of lectures given by peter constantin.

### Counterexample of regularity of newtonian potential

Define the Newtonial potential in a bounded domain $\Omega$ by

$\displaystyle N(x)=\int_{\Omega}\Gamma(x-y)f(y)dy$

If $f$ is bounded and integrable on $\Omega$, then $N\in C^1(\mathbb{R}^n)$. And if $f$ is bounded and locally holder continuous in $\Omega$ then $N\in C^2(\Omega)$ and $\Delta N=f$. But when $f$ is only continuous, $N$ is not necessarily secondly differentiable. Here is a counterexample.

$\mathbf{Problem:}$ Let $P$ be a homogeneous harmonic polynomial of degree 2. Suppose $D^\alpha P\neq 0$ for some multi-index $\alpha=2$, for instance $P=x_1x_2, D_{12}P\neq 0$. Choose a cut-off function $\eta\in C^\infty_0(\{x||x|<2\})$ with $\eta=1$ when $|x|<1$. Denote $t_k=2^k$, and let $c_k\to 0$ as $k\to \infty$ and $\sum c_k$ divergent. Define

$\displaystyle f(x)=\sum\limits_0^\infty c_k\left(\Delta(\eta P)\right)(t_kx)$

Prove $f$ is a continuous function but $\Delta u=f$ has no $C^2$ solution near the origin.

$\mathbf{Proof:}$ If $x=0$, then $f=0$. If $x\neq 0$, then there exists only one $k_0$ such that $1\leq |t_{k_0}x|\leq 2$. Since $P$ is a harmonic polynomial and $\eta$ has compact support,

$f(x)=c_{k_0}\left(\Delta(\eta P)\right)(t_{k_0}x)\leq Const.c_{k_0}$

So $f\in C^\infty(\mathbb{R}^n\backslash \{0\})$. As $x\to 0$, $c_k\to 0$, so $f(x)\to 0$, which means $f$ is continuous at 0.

Let $\displaystyle w(x)=\sum\limits_{0}^\infty t^{-2}_kc_k(\eta P)(t_kx)$, then it is easy to prove $w$ is well defined and for $x\neq 0$ and the unique $k_0$ such that $1\leq |t_{k_0}x|\leq 2$,

$\displaystyle D^\alpha w(x)=\sum\limits_0^{k_0}c_k+c_{k_0}D^\alpha(\eta P)(t_{k_0}x)\quad (1)$

$\Delta w(x)=c_{k_0}\left(\Delta(\eta P)\right)(t_{k_0}x)=f(x) \quad (2)$

(1) means $w\in C^2(\mathbb{R}^n\backslash\{0\})$ but $w\not\in C^2(\mathbb{R}^n)$, otherwise $D^\alpha w(0)=\sum c_k$ which does not exist.

Suppose there exists $u$ is a $C^2$ solution at $B_{\epsilon}(0)$, (2) means that $\Delta (u-w)=0$ on $B_{\epsilon}\backslash\{0\}$. Since $u-w$ is bounded, by the removable singularity theorem,  $u-w$ is a harmonic function in $B_\epsilon (0)$, thus an analytic function. However this means $w=u-(u-w)$ is a $C^2$ function on $B_\epsilon (0)$. Contradiction.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Gilbarg Trudinger’s book. Chapter 4. Exercise 4.9

$\mathbf{Erratum:}$ $u-w$ is not bounded in general, because $w$ can not be proved bounded directly. But we have $w(x)=o(\log r)$. In fact suppose $2^{-l}\leq|x|< 2^{-l+1}$, some $l\geq 1$, then

$\displaystyle w(x)=\sum_{k=0}^{l-1} c_k+c_l\frac{(\eta P)(t_lx)}{t_l^2}=\sum_{k=0}^{l-1} c_k+c_l\eta (t_lx)P(x)$

Because $c_k\to 0$,

$\displaystyle \left|\frac{w(x)}{\log |x|}\right|\leq \frac{|\sum_{k=0}^{l-1} c_k+c_l\eta (t_lx)P(x)|}{l\log 2}\to 0\text{ as }l\to \infty$

So $u-w=o(\log r)$ as $r\to 0$. Then one can use the Removable singularity theorem(Bochner), or the conclusion of 3.7 in GT’s book.