Marcinkiewicz Estimates for Solutions of Some Elliptic Problems with Singular Data

Abstract

In this paper we prove regularity result for solutions of the boundary value problem

$$ \left\{ \begin{array}{cl} -{{\,\textrm{div}\,}}(M(x)\,\nabla u) + u = -{{\,\textrm{div}\,}}(u\,E(x)) + f(x)\,, &{} \text{ in }\,\, \Omega , \\ u = 0\,, &{} \text{ on }\,\,\partial \Omega , \end{array} \right. $$

with the vector field E(x) and the function f(x) belonging to some Marcinkiewicz spaces.

Appendix

As stated in the Introduction, we prove here a Real Analysis lemma which will be used in the proof of Theorem 2.1.

Lemma A.1

Let $\{u_{n}\}$ be a sequence bounded in $L^{1}(\Omega )$: $\Vert {u_{n}}\Vert _{1} \le R$. Let $0< \theta < 1$, $k_{0} > 0$, $Q > 0$, and suppose that

$$\begin{aligned} g_{n}(k) \le Q\,|A_{n}(k)|^{1 - \theta }\,, \qquad \forall \, k \ge k_{0}\,, \end{aligned}$$

(1.1)

where

$$ g_{n}(k) = \displaystyle \int _{\Omega }\,|G_{k}(u_{n})|\,. $$

Then there exist $k_{1} = k_{1}(Q,R)\ge k_{0}$ and $C = C(\theta )$ such that

$$\begin{aligned} |A_{n}(k)| \le C\,\frac{Q^{\frac{1}{\theta }}}{k^{\frac{1}{\theta }}}\,, \qquad \forall \, k \ge k_{1}\,. \end{aligned}$$

(1.2)

Proof

We recall the classical result

$$ |A_{n}(k)| = -g'_n(k)\,, $$

whose simple proof uses Cavalieri’s principle as follows:

$$ \begin{array}{rcl} g_{n}(k) &{} = &{} \displaystyle \int _{\Omega }\,|G_{k}(u_{n})| = \displaystyle \int _0^{+\infty }|\{|G_k(u_{n})|\ge t\}|\,dt \\ &{} = &{} \displaystyle \int _0^{+\infty }|\{|u_{n}|\ge t+k\}|\,dt = \displaystyle \int _k^{+\infty }|A_n(s)|\,ds\,. \end{array} $$

Therefore, (A.1) can be rewritten as

$$ g_{n}(k) \le Q\,(-g'_n(k))^{1 - \theta } \quad \iff \quad Q^{\frac{1}{1-\theta }}\,g'_n(k) \le -(g_{n}(k))^{\frac{1}{1-\theta }}\,. $$

Dividing by $(g_{n}(k))^{\frac{1}{1-\theta }}$, we thus have

$$ Q^{\frac{1}{1-\theta }}\,\frac{g'_n(k)}{(g_{n}(k))^{\frac{1}{1-\theta }}} \le -1\,. $$

Integrating between $k_{0}$ and k, we have

$$ Q^{\frac{1}{1-\theta }}\,\int _{k_{0}}^{k}\,\frac{g_{n}'(h)}{(g_{n}(h))^{\frac{1}{1-\theta }}}\,dh \le k_{0} - k\,, $$

so that

$$ -Q^{\frac{1}{1-\theta }}\,\frac{1-\theta }{\theta }\,s^{-\frac{\theta }{1 - \theta }}\Big |_{g_{n}(k_{0})}^{g_{n}(k)} \le k_{0} - k\,. $$

From this inequality it follows that

$$ Q^{\frac{1}{1-\theta }}\,\frac{1-\theta }{\theta }[(g_{n}(k_{0}))^{-\frac{\theta }{1 - \theta }} - (g_{n}(k))^{-\frac{\theta }{1 - \theta }}] \le k_{0} - k\,, $$

which implies, after some straightforward algebraic passages, that

$$\begin{aligned} g_{n}(k) \le \frac{Q^{\frac{1}{\theta }}}{[Q^{\frac{1}{1-\theta }}\,(g_{n}(k_{0}))^{-\frac{\theta }{1-\theta }} + \frac{\theta }{1-\theta }\,(k - k_{0})]^{\frac{1 - \theta }{\theta }}}\,. \end{aligned}$$

(1.3)

We now remark that

$$ g_{n}(k_{0}) \le g_{n}(0) = \displaystyle \int _{\Omega }|u_{n}| \le R\,, $$

so that

$$ Q^{\frac{1}{1-\theta }}\,(g_{n}(k_{0}))^{-\frac{\theta }{1-\theta }} + \frac{\theta }{1-\theta }\,(k - k_{0}) \ge Q^{\frac{1}{1-\theta }}\,R^{-\frac{\theta }{1-\theta }} + \frac{\theta }{1-\theta }\,(k - k_{0}) \ge \frac{\theta }{2(1-\theta )}\,k\,, $$

if $k \ge \tilde{k} = \tilde{k}(Q, R, k_{0})$. Thus, from (A.3), it follows that

$$\begin{aligned} g_{n}(k) \le \bigg [ \frac{2(1-\theta )}{\theta }\bigg ]^{\frac{1-\theta }{\theta }}\,Q^{\frac{1}{\theta }}\,\frac{1}{k^{\frac{1-\theta }{\theta }}}\,, \qquad \forall \, k \ge \tilde{k}\,. \end{aligned}$$

(1.4)

We now remark that

$$ g_{n}(k) = \displaystyle \int _{\Omega }|G_{k}(u_{n})| = \displaystyle \int _{A_{n}(k)}|G_{k}(u_{n})| \ge \int _{A_n(2k)}\,|G_{k}(u_{n})| \ge k\,|A_n(2k)|\,, $$

so that from (A.4) it follows that

$$ k\,|A_n(2k)| \le \bigg [ \frac{2(1-\theta )}{\theta }\bigg ]^{\frac{1-\theta }{\theta }}\,Q^{\frac{1}{\theta }}\,\frac{1}{k^{\frac{1-\theta }{\theta }}}\,, \qquad \forall \, k \ge \tilde{k}\,. $$

Dividing by k we have therefore proved that

$$ |A_n(2k)| \le \bigg [ \frac{2(1-\theta )}{\theta }\bigg ]^{\frac{1-\theta }{\theta }}\,\frac{Q^{\frac{1}{\theta }}}{k^{\frac{1}{\theta }}}\,, \qquad \forall \, k \ge \tilde{k}\,. $$

Defining $k_{1} = 2\tilde{k}$, and

$$ C(\theta ) = 2^{\frac{1}{\theta }}\,\bigg [ \frac{2(1-\theta )}{\theta }\bigg ]^{\frac{1-\theta }{\theta }}\,, $$

we thus have proved that

$$ |A_n(2k)| \le C(\theta )\,\frac{Q^{\frac{1}{\theta }}}{(2k)^{\frac{1}{\theta }}}\,, \qquad \forall \,\;k: \ 2k \ge k_{1}\,, $$

which is (A.2).$\square$