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.