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.

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Comments

  • J B  On October 10, 2012 at 4:40 am

    I think the equation 1 might be off. With w as you defined, we have that
    $w = t_0^{-2}c_0P(t_0x) + \cdots + t_{k_0}c_{k_0}P(t_{k_0}x)$. When you take derivatives, the $P$ term shouldn’t go away.

    • Sun's World  On October 11, 2012 at 12:44 pm

      No. It should go away. Cuz p is haromic polynomial of degree 2. Take P=x_1x_2 and \alpha=(1,1) for example.

    • Sun's World  On June 27, 2013 at 8:59 pm

      You are right. I have changed the expression (1). Thank you very much.

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