Nonuniqueness of Leray–Hopf solutions to the forced fractional Navier–Stokes Equations in three dimensions, up to the J. L. Lions exponent

If $\Omega_{n}\in D(\boldsymbol{L}_{\alpha,\beta})$ converges to $\Omega$ and $\boldsymbol{L}_{\alpha,\beta}\Omega\in L^{2}$ converges to $W$, then $W=\boldsymbol{L}_{\alpha,\beta}\Omega$ in the sense of distributions. Since $\boldsymbol{L}$ is closed, in order to show that $\boldsymbol{L}_{\alpha,\beta}$ is closed, we only need to show that each $Z\in L^{2}$ corresponds to a unique solution $\Omega\in H^{2\alpha}$ to the equation $(-\boldsymbol{L}_{\alpha,\beta}+1+\beta\boldsymbol{L})\Omega=Z$, i.e.

$$-\frac{1}{2\alpha}\xi\cdot\nabla\Omega+(-\Delta)^{\alpha}\Omega=Z.$$

We deal with the unbounded term here and in the sequel by using the fact that transforming back to physical variables by $\xi=x/t^{1/2\alpha}$ converts $\xi\cdot\nabla_{\xi}$ into a time derivative. Specifically, we consider the functions

$\displaystyle h(x,t)=\Omega(x/t^{1/2\alpha}),\quad g(x,t)=\frac{1}{t}Z(x/t^{1/2\alpha}).$

(2.1)

It is easy to check that $h$ solves the fractional heat equation

$$\partial_{t}h+(-\Delta)^{\alpha}h=g,$$

and attains zero initial data in $L^{2}$ since $\alpha>0$:

$$\|h\|_{L^{2}}=t^{3/4\alpha}\|\Omega\|_{L^{2}}\xrightarrow[t\to 0]{}0.$$

Since $\|g\|_{L^{2}}=t^{-1+3/4\alpha}\|Z\|_{L^{2}}$ is integrable, we obtain via the theory of the fractional heat equation that $\Omega=h(\cdot,1)\in H^{2\alpha}$. Indeed, firstly, $h(\cdot,1/2)$ is $L^{2}$ by Lemma 2.1. Then for times $t\geq 1/2$, by writing

$$h(x,t)=\textup{e}^{-t(-\Delta)^{\alpha}}h(x,1/2)+\mathcal{G}_{0}(g(\cdot,t+1/2))(x,t-1/2),$$

since $g(\cdot,t+1/2)$ is not singular at $t=0$, Lemma 2.1 shows that $(-\Delta)^{\alpha}h\in C([1/2,1];L^{2})$, so we obtain the claimed result. Hence, $\boldsymbol{L}_{\alpha,\beta}$ is closed. ∎

We want to use that the Laplace transform of the semigroup is the resolvent (see (2.6) below), so let $\Omega^{\beta}$ be the solution to the initial value problem $(\partial_{\tau}-\beta^{-1}\boldsymbol{D}_{\alpha}-\boldsymbol{M}-\SS)\Omega=0$, i.e.

$$\left\{\begin{aligned} \partial_{\tau}\Omega^{\beta}-\frac{1}{\beta}\Big{(}\frac{3}{4\alpha}+\frac{1}{2\alpha}\xi\cdot\nabla-(-\Delta)^{\alpha}\Big{)}\Omega^{\beta}+\overline{U}\cdot\nabla\Omega^{\beta}&=\SS\Omega^{\beta},\\ \Omega^{\beta}|_{\tau=0}&=\Omega_{0},\end{aligned}\right.$$

and let $\Omega$ solve the $\beta\to\infty$ limit equation

$$\left\{\begin{aligned} \partial_{\tau}\Omega+\overline{U}\cdot\nabla\Omega&=\SS\Omega,\\ \Omega|_{\tau=0}&=\Omega_{0},\end{aligned}\right.$$

For smooth, compactly supported data, $\Omega$ exists and is smooth. $\Omega^{\beta}$ can be shown to exist as a strong solution in $C^{0}H^{2\alpha}$ by again undoing the similarity variable transform which removes the term $\xi\cdot\nabla$ to get a more standard fractional heat equation with drift, similarly to (2.1). Specifically, one sets $h^{\beta}(x,t)=\Omega^{\beta}(xt^{-1/2\alpha},\beta\log t)$ and $f^{\beta}(x,t)=\beta t^{-1}\SS\Omega^{\beta}(xt^{-1/2\alpha},\beta\log t)$ to get

$$\partial_{t}h^{\beta}-\left(\frac{3}{4\alpha}-(-\Delta)^{\alpha}\right)h^{\beta}+\beta\overline{u}\cdot\nabla h^{\beta}=f^{\beta}.$$

Notably, $\overline{u}$ is smooth in both space and time, as the initial data is prescribed at $t=1$, so well-posedness is easily proven first for $f^{\beta}=0$, then for $f^{\beta}\neq 0$ with Duhamel’s formula.

From the energy inequality for $\Omega^{\beta}$, and from the Lagrangian formulation for $\Omega$, we have the estimates

$\displaystyle\|\Omega\|_{L^{2}}^{2}+\|\Omega^{\beta}\|_{L^{2}}^{2}\lesssim\textup{e}^{\tau\smash{\|\SS\|_{L^{2}_{{\textup{aps}}}\to L^{2}}}}\|\Omega_{0}\|^{2}_{L^{2}}=e^{\tau\mu}\|\Omega_{0}\|^{2}_{L^{2}}.$

(2.3)

For the limit, we study the equation for the difference $\tilde{\Omega}\coloneq\Omega^{\beta}-\Omega$,

$$(\partial_{\tau}-\beta^{-1}\boldsymbol{D}_{\alpha}-\boldsymbol{M}-\SS)\tilde{\Omega}=\beta^{-1}\boldsymbol{D}_{\alpha}\Omega,$$

initially for $\Omega_{0}\in C^{\infty}_{c}$. Testing against $\tilde{\Omega}$ in $L^{2}$, and using Young’s inequality ($ab\leq a^{2}/2+b^{2}/2$) gives

$$\frac{\mathop{}\!\mathrm{d}}{\mathop{}\!\mathrm{d}\tau}\|\tilde{\Omega}\|_{L^{2}}^{2}+\frac{1}{\beta}\|(-\Delta)^{\alpha/2}\tilde{\Omega}\|_{L^{2}}^{2}\leq\mu\|\tilde{\Omega}\|_{L^{2}}^{2}+\frac{1}{\beta}\langle\boldsymbol{D}_{\alpha}\Omega,\tilde{\Omega}\rangle_{H^{-\alpha},H^{\alpha}}\\ \leq\mu\|\tilde{\Omega}\|_{L^{2}}^{2}+\frac{1}{2\beta}\|\boldsymbol{D}_{\alpha}\Omega\|_{H^{-\alpha}}^{2}+\frac{1}{2\beta}\Big{(}\|\tilde{\Omega}\|_{L^{2}}^{2}+\|(-\Delta)^{\alpha/2}\tilde{\Omega}\|_{L^{2}}^{2}\Big{)}.$$

(2.4)

Absorbing $\frac{1}{2\beta}\|(-\Delta)^{\alpha/2}\tilde{\Omega}\|_{L^{2}}^{2}$ into the left-hand side, integrating in $0\leq\tau\leq T$ and using that $\tilde{\Omega}|_{\tau=0}=0$, we obtain for each fixed $T$ that

$\displaystyle\|\tilde{\Omega}\|_{L^{\infty}(0,T;L^{2})}^{2}\lesssim_{T}\beta^{-1}\sup_{t\leq T}\|\boldsymbol{D}_{\alpha}\Omega(\cdot,t)\|_{H^{-\alpha}}^{2}\xrightarrow[\beta\to\infty]{}0.$

(2.5)

Recall that for a densely defined operator $\boldsymbol{A}$ with sufficient growth bounds on the semigroup $e^{t\boldsymbol{A}}$, the resolvent $R(\lambda,\boldsymbol{A})$ can be written as the Laplace transform

$\displaystyle R(\lambda,\boldsymbol{A})\Omega_{0}=\int_{0}^{\infty}\textup{e}^{-t(\lambda-\boldsymbol{A})}\Omega_{0}\mathop{}\!\mathrm{d}t.$

(2.6)

In our case, this formula and the growth bound (2.3) shows the boundedness (2.2). It follows that for all $\lambda$ with $\operatorname{Re}\lambda>\mu$,

	$\displaystyle\|R(\lambda,\beta^{-1}\boldsymbol{D}_{\alpha}+\boldsymbol{M}+\SS)\Omega_{0}-R(\lambda,\boldsymbol{M}+\SS)\Omega_{0}\|_{L^{2}}$
	$\displaystyle\leq\int_{0}^{\infty}\textup{e}^{-\operatorname{Re}\lambda t}\|\textup{e}^{t(\beta^{-1}\boldsymbol{D}_{\alpha}+\boldsymbol{M}+\SS)}\Omega_{0}-\textup{e}^{t(\boldsymbol{M}+\SS)}\Omega_{0}\|\mathop{}\!\mathrm{d}t$
	$\displaystyle\leq\|\tilde{\Omega}\|_{L^{\infty}(0,T;L^{2})}\int_{0}^{T}\textup{e}^{-\operatorname{Re}\lambda t}\mathop{}\!\mathrm{d}t+\int_{T}^{\infty}\textup{e}^{-\operatorname{Re}\lambda t}(\|\Omega(t)\|_{L^{2}}+\|\Omega^{\beta}(t)\|_{L^{2}})\mathop{}\!\mathrm{d}t$
	$\displaystyle\leq\frac{1}{\operatorname{Re}\lambda}\|\tilde{\Omega}\|_{L^{\infty}(0,T;L^{2})}e^{-\operatorname{Re}\lambda T}+\frac{1}{\operatorname{Re}\lambda-\mu}\|\Omega_{0}\|_{L^{2}}e^{-(\operatorname{Re}\lambda-\mu)T}.$

The strong convergence now follows from (2.5) for all compactly supported data, by first taking $T\gg 1$ and then $\beta\to\infty$. We can then cover the case $\Omega_{0}\in L^{2}$ by density. ∎

To prove this, we let $\gamma\subset\rho(\boldsymbol{L})\cap\{\operatorname{Re}\lambda>\mu\}$ be a sufficiently small circle around the unstable eigenvalue $\lambda_{\infty}$ which encloses no other part of $\sigma(\boldsymbol{L})$. We need to show that we can define the spectral projection operator

$\displaystyle\operatorname{Pr}_{\alpha,\beta}\coloneq\frac{1}{2\pi i}\int_{\gamma}R(\lambda,\boldsymbol{L}^{\prime}_{\alpha,\beta})\mathop{}\!\mathrm{d}\lambda.$

(2.7)

and show it converges in norm to the corresponding spectral projection for $\boldsymbol{L}=\boldsymbol{M}+\SS+\boldsymbol{K}$,

$\displaystyle\operatorname{Pr}\coloneq\frac{1}{2\pi i}\int_{\gamma}R(\lambda,\boldsymbol{L})\mathop{}\!\mathrm{d}\lambda.$

(2.8)

From Lemma 1.1, $\operatorname{Pr}$ is non-trivial, so $\operatorname{Pr}_{\alpha,\beta}$ is nontrivial for sufficiently large $\beta$. This means that $\gamma$ encloses an eigenvalue of $\boldsymbol{L}_{\alpha,\beta}$ with finite multiplicity, which proves the Lemma.

So we just need to justify (2.7). Consider the identity

$$\lambda-\boldsymbol{L}^{\prime}_{\alpha,\beta}=(\lambda-\beta^{-1}\boldsymbol{D}_{\alpha}-\boldsymbol{M}-\SS)\big{(}I-R(\lambda,\beta^{-1}\boldsymbol{D}_{\alpha}+\boldsymbol{M}+\SS)\boldsymbol{K}\big{)},$$

which makes sense for $\operatorname{Re}\lambda>\mu$ by (2.2). As $R(\lambda,\beta^{-1}\boldsymbol{D}_{\alpha}+\boldsymbol{M}+\SS)\to R(\lambda,\boldsymbol{M}+\SS)$ in the strong topology and $\boldsymbol{K}$ is compact, we have the norm convergence

$$R(\lambda,\beta^{-1}\boldsymbol{D}_{\alpha}+\boldsymbol{M}+\SS)\boldsymbol{K}\xrightarrow[\beta\to\infty]{}R(\lambda,\boldsymbol{M}+\SS)\boldsymbol{K},$$

locally uniformly in $\lambda$. In addition, for $\lambda\in\rho(\boldsymbol{L})\cap\{\operatorname{Re}\lambda>\mu\}\subset\rho(\boldsymbol{M}+\SS)$, the identity

$$\lambda-\boldsymbol{L}=\lambda-(\boldsymbol{M}+\SS+\boldsymbol{K})=(\lambda-\boldsymbol{M}+\SS)(I-R(\lambda,\boldsymbol{M}+\SS)\boldsymbol{K})$$

shows that $I-R(\lambda,\boldsymbol{M}+\SS)\boldsymbol{K}$ is invertible. As the set of invertible operators is open, it follows that $I-R(\lambda,\beta^{-1}\boldsymbol{D}_{\alpha}+\boldsymbol{M}+\SS)\boldsymbol{K}$ is invertible for $\beta\gg 1$, locally uniformly in $\lambda$. Hence, $R(\lambda,\boldsymbol{L}^{\prime}_{\alpha,\beta})$ for $\beta\gg 1$ makes sense on the compact set $\gamma$, which verifies (2.7). This finishes the proof. ∎

Algebraically, this is true because the operators in velocity and vorticity formulation are linked by conjugation with the curl operator,

$$\nabla\times\boldsymbol{T}_{\!\alpha,\beta}V=\boldsymbol{L}_{\alpha,\beta}\nabla\times V.$$

Hence, $\boldsymbol{T}_{\!\alpha,\beta}$ has the same unstable eigenvalue as $\boldsymbol{L}_{\alpha,\beta}$, and if $\zeta$ is the eigenfunction for $\boldsymbol{L}_{\alpha,\beta}$, then $\eta=\operatorname{BS}\zeta$ is formally the eigenfunction for $\boldsymbol{T}_{\!\alpha,\beta}$. The only detail to check is that $\eta\in D(\boldsymbol{T}_{\!\alpha,\beta})$, and the conditions defining $D(\boldsymbol{T}_{\!\alpha,\beta})$ are all satisfied save for potentially $\eta\in L^{2}$. As $\operatorname{BS}$ is bounded from $L^{2}$ to $L^{6}$, and from $L^{1}$ to $L^{3/2}$, it suffices by interpolation to show that $\zeta\in L^{1}$.

As an eigenfunction of $\boldsymbol{L}_{\alpha,\beta}$, $\zeta$ satisfies the equation

$$\lambda_{\alpha,\beta}\zeta-\Big{(}1+\frac{1}{2\alpha}\xi\cdot\nabla-(-\Delta)^{\alpha}\Big{)}\zeta=\boldsymbol{L}\zeta.$$

We again undo the similarity variables by setting $\xi=x/t^{1/2\alpha}$ and

$$h(x,t)\coloneq t^{\lambda_{\alpha,\beta}-1}\zeta\Big{(}\frac{x}{t^{1/2\alpha}}\Big{)},\quad g(x,t)\coloneq t^{\lambda_{\alpha,\beta}-2}(\boldsymbol{L}\zeta)\Big{(}\frac{x}{t^{1/2\alpha}}\Big{)}.$$

The corresponding equation is

$$\partial_{t}h+(-\Delta)^{\alpha}h=g,\quad h|_{t=0}=0,$$

where the initial data is attained in $L^{2}$ once $\operatorname{Re}\lambda_{\alpha,\beta}>1-\frac{3}{4\alpha}$, using $\beta\gg 1$. Observe that $\boldsymbol{L}\zeta\in L^{2}$ is compactly supported. It follows from the integral formulation of the Duhamel-type operator $\mathcal{G}_{0}$ that $g\in C^{0}([0,T];L^{1})$ for each $T$ (again using $\beta\gg 1$). Lemma 2.1 shows that $h$ is also $C^{0}L^{1}$, so in particular $\zeta=h(\cdot,1)\in L^{1}$, as needed. ∎