A partial uniqueness result and an asymptotically sharp nonuniqueness result for the Zhikov problem on the torus

Abstract

We consider the stationary diffusion equation $-\mathrm {div} (\nabla u + bu )=f$ in d-dimensional torus $\mathbb {T}^d$, where $f\in H^{-1}$ is a given forcing and $b\in L^p$ is a divergence-free drift. Zhikov (Funkts Anal Prilozhen, 38(3):15–28, 2004) considered this equation in the case of a bounded, Lipschitz domain $\Omega \subset \mathbb {R}^d$, and proved existence of solutions for $b\in L^{2d/(d+2)}$, uniqueness for $b\in L^2$, and has provided a point-singularity counterexample that shows nonuniqueness for $b\in L^{3/2-}$ and $d=3,4,5$. We apply a duality method and a DiPerna–Lions-type estimate to show uniqueness of the solutions constructed by Zhikov for $b\in W^{1,1}$. We use a Nash iteration to demonstrate sharpness of this result, and also show that solutions in $H^1\cap L^{p/(p-1)}$ are flexible for $b\in L^p$, $p\in [1,2(d-1)/(d+1))$; namely we show that the set of $b\in L^p$ for which nonuniqueness in the class $H^1\cap L^{p/(p-1)}$ occurs is dense in the divergence-free subspace of $L^p$.

Appendix: Proof of the maximum principle (5)

Here we show that, if $f\in L^\infty$ and $b\in L^2$ satisfies the divergence-free condition (1), then any solution $u\in L^\infty \cap \dot{H}^1$ to $-\mathrm{{div\ }}(\nabla u + bu ) =f$ (in the sense of (3)) satisfies

$$\begin{aligned} \Vert u \Vert _{\infty } \le C_n \Vert f \Vert _\infty , \end{aligned}$$

(35)

where $C_n>0$ depends only on the dimension d. Note that, since the constant does not depend on b, this proves (35) for all $u\in \dot{H}^1$ (i.e. we can obtain $u\in L^\infty$, rather than assume it), which is (5), as desired. Indeed, one can approximate any $b\in L^{2d/(d+2)}$ in the $L^{2d/(d+2)}$ norm by a $C^\infty$ function and also approximate f in $H^{-1}$ by a smooth function, which gives a unique and smooth solution the approximate system. The limiting procedure described below (6) then gives an approximation solution u with (35). Since for $b\in L^2$ the uniqueness of solutions to (3) holds (recall (8)), we obtain (5).

We first recall the inequality

$$\begin{aligned} \Vert g \Vert _2^2 \le \varepsilon \Vert \nabla g \Vert ^2_2 + C_n \varepsilon ^{-\frac{n}{2}} \Vert g \Vert _1^2, \end{aligned}$$

(36)

valid for every $g\in H^1 (\mathbb {T}^d)$, $\varepsilon \in (0,1)$. Indeed the Gagliardo-Nirenberg-Sobolev inequality gives that

$$\begin{aligned} \Vert g \Vert ^2_2 \le C \Vert \nabla g \Vert _2^{\frac{2d}{d+2}} \Vert g \Vert _1^{\frac{4}{d+2}} \le \varepsilon \Vert \nabla g \Vert _2^2 + C_d \varepsilon ^{-\frac{d}{2}} \Vert g \Vert _1^2 \end{aligned}$$

if $\int g =0$, where we used Young’s inequality in the second step. Thus the claim follows for such g. If $\int g \ne 0$ then

$$\begin{aligned} \begin{aligned} \Vert g \Vert _2^2&\le C \left\| g- \int g \right\| _2^2 + C\left| \int g \right| ^2 \\&\le \varepsilon \Vert \nabla g \Vert _2^2 + C_n \varepsilon ^{-\frac{n}{2}} \left\| g - \int g\right\| _1^2 + C\Vert g \Vert _1^2 \\&\le \varepsilon \Vert \nabla g \Vert _2^2 + C_n \varepsilon ^{-\frac{n}{2}} \left\| g \right\| _1^2 , \end{aligned} \end{aligned}$$

where we used the triangle inequality and the assumption that $\varepsilon <1$ in the last step.

We can now prove (35). Applying (36) with $g:={u^{2^{k-1}}}$ gives

$$\begin{aligned} \int u^{2^k} \le \varepsilon \int \left| \nabla u^{2^{k-1}} \right| ^2 + C\varepsilon ^{-\frac{n}{2}} \left( \int u^{2^{k-1}}\right) ^2, \end{aligned}$$

(37)

where $C>1$ is a generic constant, which depends on d only. The value of C may change from line to line in the following calculation.

Note that, since we assume $b\in L^2$, we can test (3) with $u^{2^k-1}$. Indeed, one can, for example, take $\phi :=(u^{2^k-1})_\varepsilon$ (the mollification of u, recall (12)) and take the limit $\varepsilon \rightarrow 0$. We obtain

$$\begin{aligned} \frac{2^k -1}{2^{2k-2}} \int \left| \nabla u^{2^{k-1}} \right| ^2 = \int \nabla u \cdot \nabla \left( u^{2^k-1} \right) = \int f u^{2^k-1} \le \frac{1}{2} \int f^{2^k} + \frac{1}{2} \int u^{2^k}, \end{aligned}$$

(38)

where we used the fact that $\int bu \cdot \nabla \left( u^{2^k-1} \right) =(1-2^{-k}) \int b \cdot \nabla \left( u^{2^k} \right) =0$, due to the divergence-free assumption on b and the assumed regularity $u\in L^\infty \cap H^1$, $b\in L^2$. Applying this inequality in (37) gives

$$\begin{aligned} \int u^{2^k} \le \varepsilon \frac{2^{2k-3}}{2^k-1} \int f^{2^k} + \varepsilon \frac{2^{2k-3}}{2^k-1} \int u^{2^k} + C \varepsilon ^{-\frac{n}{2}} \left( \int u^{2^{k-1}} \right) ^2. \end{aligned}$$

Taking $\varepsilon :=2^{-k}$ and noting that

$$\begin{aligned} 2^{-k} \frac{2^{2k-3}}{2^k-1} \le \frac{1}{4} \qquad \text { for } k\ge 1, \end{aligned}$$

we obtain

$$\begin{aligned} \int u^{2^k} \le \frac{1}{4} \int f^{2^k} + \frac{1}{4} \int u^{2^k} + C 2^{\frac{kn}{2}} \left( \int u^{2^{k-1}} \right) ^2 , \end{aligned}$$

and so

$$\begin{aligned} \int u^{2^k} \le \int f^{2^k} + C 2^{\frac{kn}{2}} \left( \int u^{2^{k-1}} \right) ^2 \le \Vert f\Vert _{\infty }^{2^k} + C 2^{\frac{kn}{2}} \Vert u \Vert _{2^{k-1}}^{2^k} \end{aligned}$$

for $k\ge 1$. Taking both sides to power $2^{-k}$ gives

$$\begin{aligned} \Vert u \Vert _{2^k} \le \left( \Vert f \Vert _{\infty }^{2^k} + C 2^{\frac{kn}{2}} \Vert u \Vert _{2^{k-1}}^{2^k} \right) ^{2^{-k}} \end{aligned}$$

for $k\ge 1$. Noting that the above inequality holds trivially when the left-hand side is replaced by $\Vert f \Vert _\infty$, we set

$$\begin{aligned} m_k :=\max \left\{ \Vert f \Vert _\infty , \Vert u\Vert _{2^k} \right\} \end{aligned}$$

and obtain that

$$\begin{aligned} m_k \le \left( 1+C 2^{\frac{kn}{2}} \right) ^{2^{-k}} m_{k-1} \le \left( C 2^{\frac{kn}{2}} \right) ^{2^{-k}} m_{k-1} \end{aligned}$$

for $k\ge 1$. (Recall that the value of $C>1$ may change from line to line.) Since

$$\begin{aligned}&\prod _{k=1}^\infty \left( C 2^{\frac{kn}{2}} \right) ^{2^{-k}} = \exp \left( \log \left( \prod _{k=1}^\infty \left( C 2^{\frac{kn}{2}} \right) ^{2^{-k}} \right) \right) \\&\quad = \exp \left( \log C\sum _{k=1}^\infty 2^{-k} + \log \left( 2^{\frac{n}{2}} \right) \sum _{k=1}^\infty k2^{-k} \right) \le C, \end{aligned}$$

we see that

$$\begin{aligned} m_k \le C m_0\qquad \text { for } k\ge 1. \end{aligned}$$

In particular $\Vert u \Vert _{2^k } \le C m_0$ for $k\ge 1$, which implies that $u \in L^\infty$ with

$$\begin{aligned} \Vert u \Vert _\infty \le C m_0. \end{aligned}$$

(39)

On the other hand the Poincaré inequality and (38) applied with $k=1$ gives

$$\begin{aligned} \Vert u \Vert _2^2 \le C \Vert \nabla u \Vert _2^2 =C \int fu \le \frac{1}{2} \Vert u \Vert _2^2 + C \Vert f \Vert _\infty ^2, \end{aligned}$$

which, after absorbing the first term on the right-hand side, implies that $m_0 \le C \Vert f \Vert _\infty$. Applying this in (39) gives (35), as required.