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Let $B$ denote the $n$-dimensional unit ball. Assume $u\in\bigcap_{1\le p<2} W_0^{1,p}(B)$ satisfies $$ \int a_{ij}\partial_j u \partial_iv =0,$$ for any $v\in C_c^{\infty}(B)$, where we assume $a_{ij}(x)$ are just uniformly elliptic, i.e. there exists $\lambda,\Lambda>0$ such that $$\lambda \lvert\xi\rvert^2\le a_{ij}(x) \xi_i\xi_j \le \Lambda \lvert\xi\rvert^2$$ for any $x\in B$. Under these conditions, can we show that $u$ is identically $0$?

If we just assume $u\in W^{1,1}_0(B)$, the answer is negative, see the post Uniqueness of solutions to elliptic PDE in $W^{1,1}$.

And when $a_{ij}=1$ ($i=j$), we can show $u\equiv 0$ by Poisson integral formula, while in general case this approach may not be generalized.

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Dec 5, 2022 at 12:56
  • $\begingroup$ Can you assume not and then approximate $u$ by functions in $W_0^{1, 2}$? $\endgroup$ Commented Dec 5, 2022 at 17:06
  • $\begingroup$ @DanielShapero Of course we can, but how to use that to prove $u=0$? $\endgroup$
    – Tian LAN
    Commented Dec 5, 2022 at 18:13
  • $\begingroup$ How you prove it for the Laplacian? Usually Poisson formula requires classical solutions. $\endgroup$ Commented Dec 5, 2022 at 21:27
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    $\begingroup$ @GiorgioMetafune because the Laplacian is so special that if $\Delta u=0$ in the distributional sense, then it must be a harmonic function. So for a fixed $x\in B$, we can apply the Poisson integral formula in $B_r$ for $r<1$, and then let $r\rightarrow 1$. Use the trace of $u$ on $\partial B_r$ converges to the trace on $\partial B$ in $L^1$. $\endgroup$
    – Tian LAN
    Commented Dec 5, 2022 at 23:34

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The answer is yes if and only if you have optimal elliptic regularity in $W^{-1,p'}(B)$ for some $p>2$ for the adjoint differential operator, i.e., the one given by the transpose matrix $A^\top$.

Consider your question from a functional-analytic point of view: You are essentially asking whether the (bounded linear) operator $$-\nabla \cdot A\nabla \colon W^{1,p}_0(B) \to W^{-1,p}(B), \qquad \langle -\nabla \cdot A \nabla u, v\rangle := \int_B A\nabla u \cdot \nabla v$$ is injective for some $p<2$, where of course $A = (a_{ij})$ and $W^{-1,p}(B)$ is the dual space of $W^{1,p'}_0(B)$.

For this it is sufficient (and in this case equivalent, since the Sobolev scale is reflexive, at least if we stay away from $p=1$) that the adjoint operator of $-\nabla \cdot A \nabla$ is surjective.

This adjoint operator happens to be $-\nabla \cdot A^\top \nabla \colon W^{1,p'}_0(B) \to W^{-1,p'}(B)$ which is still uniformly elliptic. Since $p' > 2$ and $B$ is bounded, we thus already know from Lax-Milgram that for every $f \in W^{-1,p'}(B)$ there is a unique $w \in W^{1,2}_0(B)$ such that $-\nabla \cdot A^\top \nabla w = f$.

So the question of surjectivity for $-\nabla \cdot A^\top \nabla$ as before becomes the question whether $w \in W^{1,p'}_0(B)$. But that is a very question of optimal elliptic regularity for $-\nabla \cdot A^\top\nabla$ in $W^{-1,p'}(B)$ for some $p' > 2$.

Such questions are classical and answered in many articles and textbooks. The property in question is true for the ball, and also for, say, (weak) Lipschitz domains. You can also have mixed boundary conditions within this class. (And also beyond, but that seems not to be the scope of this question.) I would like to point to the classical work [G89] which covers this question in some generality and also points to the historical development.


[G89] Gröger, Konrad, A $W^{1,p}$-estimate for solutions to mixed boundary value problems for second order elliptic differential equations, Math. Ann. 283, No. 4, 679-687 (1989). ZBL0646.35024.

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