An energy Navier–Stokes velocity on the square without endpoint-integrable pressure

The choices in Given exist by [F8]: take a bump centered at $1\mathrm{/}2$ with inner radius $1\mathrm{/}24$ and outer radius $1\mathrm{/}12$ for $a$, giving a nonzero function compactly supported in $(1\mathrm{/}3,2\mathrm{/}3)$. Restrict zero-centered bumps to the positive halfline with radii $(1\mathrm{/}32,1\mathrm{/}16)$ for $\eta$, $(1\mathrm{/}8,1\mathrm{/}4)$ for $\kappa$ and $\chi$, and $(1\mathrm{/}3,1)$ for $\rho$. They have every stated support, plateau and sign property. Put $a_n=2{\int }_0^{1}a(x)\sin (n\pi x)dx$ and define $\theta (x,y)={\sum }_{n\ge 1}{a}_n\sin (n\pi x)\frac{\sinh (n\pi (1-y))}{\sinh (n\pi )},\tilde{\varphi }(x,y)={\sum }_{n\ge 1}{a}_n\cos (n\pi x)\frac{\cosh (n\pi (1-y))}{\sinh (n\pi )}\mathrm{.}$ Repeated integration by parts makes $\mathrm{\mid }{a}_n\mathrm{\mid }\le C_N{n}^{-N}$ for every $N$, since $a$ vanishes near the endpoints. The hyperbolic factors and their derivatives of order $k$ are bounded by $C_k{n}^k$ uniformly on $[0,1]$. Thus every differentiated series converges absolutely uniformly, and both functions are smooth on the closed square and harmonic. The sine series equals $a$ at $y=0$: indeed the odd, $2$-periodic smooth extension of $a$ has its Fourier series equal to itself. To verify this last assertion directly, its rapid coefficients define a uniformly convergent Fourier series with the same coefficients; their difference has all Fourier coefficients zero. Convolution with the Fejer kernels, which are nonnegative unit-mass kernels concentrating at zero by their formula $N^{-1}\mathrm{\mid }{\sum }_{j=0}^{N-1}{e}^{ij\pi x}{\mathrm{\mid }}^2$, converges uniformly to a continuous periodic function by uniform continuity; it vanishes for a function whose coefficients all vanish. Hence the difference is zero. The other boundary values of $\theta$ are zero. Direct differentiation gives ${\tilde{\varphi }}_x={\theta }_y$ and ${\tilde{\varphi }}_y=-{\theta }_x$. Let $\varphi =\tilde{\varphi }-{\int }_D\tilde{\varphi }$, so $\varphi \in Q$, $\mathrm{\nabla }\varphi =\mathrm{rot}\theta$, and $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_2>0$ because $a\mathrm{\ne }0$ makes $\theta$ nonconstant.

Define $b=\theta -a\chi$. This is smooth and zero on all four sides. Its normal derivative equals ${\mathrm{\partial }}_n\theta$ everywhere on the boundary: at the bottom ${\chi }^{\mathrm{\prime }}=0$, at the top $\chi ={\chi }^{\mathrm{\prime }}=0$, and near the vertical sides $a=a^{\mathrm{\prime }}=0$. Take the function $R(t)=R_{\sqrt{t}}$ from [F2] and choose $T$ small enough for that construction, with $\sqrt{T}<1\mathrm{/}16$. Set ${\psi }_m=a(x)[\kappa (y)h(t,y)+M_0\chi (y)],\psi ={\psi }_m+m(t)\theta -m(t)bR(t),$ $u=\mathrm{rot}\psi ,p=-m^{\mathrm{\prime }}(t)\varphi ,r=-m(t)\mathrm{rot}(bR(t))\mathrm{.}$ All these fields are smooth for $t>0$. On the bottom ${\psi }_m=-am$, $\theta =a$, and $b=0$; on the other sides ${\psi }_m=\theta =b=0$. Thus $\psi =0$ on the boundary. The normal derivative of ${\psi }_m$ is zero on every side: on the bottom it is, up to its sign, $aw(t,0)=0$, and on all other sides the fixed cutoffs or $a,a^{\mathrm{\prime }}$ vanish. Hence ${\mathrm{\partial }}_n({\psi }_m+m\theta )=m{\mathrm{\partial }}_{n}b$, and [F2] makes ${\mathrm{\partial }}_n\psi =0$. Thus $u(t)\in V_{\sigma }$ exactly, with no-slip on every edge including compatible corner values.

Write $L(t)=\log (1\mathrm{/}t)$. By [F1] and [F2], $\mathrm{\parallel }r{\mathrm{\parallel }}_{\mathrm{\infty }}\le C\mathrm{\mid }m\mathrm{\mid },\mathrm{\parallel }r{\mathrm{\parallel }}_2\le C\mathrm{\mid }m\mathrm{\mid }{t}^{1\mathrm{/}4},\mathrm{\parallel }\mathrm{\nabla }r{\mathrm{\parallel }}_2\le C\mathrm{\mid }m\mathrm{\mid }{t}^{-1\mathrm{/}4},$ $\mathrm{\parallel }{r}_t{\mathrm{\parallel }}_2+\mathrm{\parallel }\mathrm{\Delta }r{\mathrm{\parallel }}_2\le C(\mathrm{\mid }{m}^{\mathrm{\prime }}\mathrm{\mid }{t}^{1\mathrm{/}4}+\mathrm{\mid }m\mathrm{\mid }{t}^{-3\mathrm{/}4})\le C{t}^{-1\mathrm{/}2}L(t{)}^{-\alpha }\mathrm{.}$ The last function is square integrable near zero because $2\alpha >1$. In particular $r\in L_t^{\mathrm{\infty }}{L}_x^2\cap L_t^2{H}_x^1$ and $r(t)\to 0$ strongly in $L^2$. The main stream rotation has a uniform $L^2$ bound from bounded $h$ and bounded $\mathrm{\parallel }w{\mathrm{\parallel }}_2$. Its gradient uses only bounded $h$, $w$ with fixed coefficients and $w_y$ with fixed coefficients, so the heat energy bound makes it $L_{t,x}^2$. The smooth field $m\mathrm{rot}\theta$ has the same required bounds. We obtain $u\in L^{\mathrm{\infty }}(0,T;L^2(D{)}^2)\cap L^2(0,T;V_{\sigma })\mathrm{.}$

The outer harmonic field cancels the entire rough gradient: $({\mathrm{\partial }}_t-\nu \mathrm{\Delta })(m\mathrm{rot}\theta )+\mathrm{\nabla }p=m^{\mathrm{\prime }}\mathrm{\nabla }\varphi -m^{\mathrm{\prime }}\mathrm{\nabla }\varphi =0$. The main stream has the explicit residual $({\mathrm{\partial }}_t-\nu \mathrm{\Delta }){\psi }_m=-\nu [a^{\mathrm{\prime }\mathrm{\prime }}\kappa h+2a{\kappa }^{\mathrm{\prime }}w+a{\kappa }^{\mathrm{\prime }\mathrm{\prime }}h]-\nu M_0[a^{\mathrm{\prime }\mathrm{\prime }}\chi +a{\chi }^{\mathrm{\prime }\mathrm{\prime }}]\mathrm{.}$ Its rotation contains only fixed coefficients times $h,w$, except for terms with ${\kappa }^{\mathrm{\prime }}{w}_y$. The latter are supported in $y\ge 1\mathrm{/}8$, a fixed positive distance from $\mathrm{supp}{w}_0\subset [0,1\mathrm{/}16]$, and are square integrable by the off-support Gaussian bound in [F1]. All other terms are uniformly $L^2$ bounded. Together with step 3.1 this proves $F_L=({\mathrm{\partial }}_t-\nu \mathrm{\Delta })u+\mathrm{\nabla }p\in L^2(0,T;L^2(D{)}^2)\mathrm{.}$

Let $h_0(y)=-{\int }_y^{\mathrm{\infty }}{w}_0(z)dz$, ${\psi }_0=a[\kappa h_0+M_0\chi ]$, and $u_0=\mathrm{rot}{\psi }_0$. Strong $L^2$ convergence of $w$, uniform convergence of $h$ on $[0,1]$, $m(t)\to 0$ and step 3.1 give $u(t)\to u_0$ strongly in $L^2$. The stream ${\psi }_0$ is $H^1$, supported away from the other three sides, and vanishes on the bottom because $h_0(0)=-M_0$. Here is a direct approximation proving ${\psi }_0\in H_0^1$, without relying on an unstated trace theorem. Near the bottom it equals $a(x){\int }_0^y{w}_0(z)dz$. If a cutoff removes $0<y<\epsilon$, the derivative-of-cutoff error is bounded in squared $L^2$ by $C{\epsilon }^{-2}{\int }_0^{2\epsilon }{\mid {\int }_0^y{w}_0(z)dz\mid }^{2}dy\le C{\int }_0^{2\epsilon }\mathrm{\mid }{w}_0(z){\mathrm{\mid }}^{2}dz⟶0.$ The inequality follows from $\mathrm{\mid }{\int }_0^y{w}_0{\mathrm{\mid }}^2\le y{\int }_0^y\mathrm{\mid }{w}_0{\mathrm{\mid }}^2$ and integrating $y\le 2\epsilon$. The other errors tend to zero by absolute continuity of the integrals of $\mathrm{\mid }{\psi }_0{\mathrm{\mid }}^2+\mathrm{\mid }\mathrm{\nabla }{\psi }_0{\mathrm{\mid }}^2$. The cut-off streams are smooth and compactly supported inside $D$, since $w_0$ is smooth away from zero. Their rotations belong to $V_{\sigma }$ and converge to $u_0$ in $L^2$, proving $u_0\in H_{\sigma }$ by its actual closure definition.

The normal component $u_2$ is uniformly bounded in space and time: $-{\mathrm{\partial }}_x{\psi }_m$ uses only bounded $h$ and fixed functions, the harmonic term has bounded amplitude $m$, and $r$ is uniformly bounded. Thus $u_2{\mathrm{\partial }}_{y}u\in L_{t,x}^2$ by step 3.1. The only unbounded part of $u_1$ is $a\kappa w$. Choose $J⋐(0,1)$ containing $\mathrm{supp}a$ and decrease $T$ so the last estimate in [F2] applies there. On $J\times (0,1)$, ${\mathrm{\partial }}_{x}r$ is uniformly bounded; the main and harmonic terms satisfy $\mathrm{\mid }{\mathrm{\partial }}_{x}u\mathrm{\mid }\le C(1+\mathrm{\mid }w\mathrm{\mid })$, so $\mathrm{\mid }{u}_1{\mathrm{\partial }}_{x}u\mathrm{\mid }\le C(1+\mathrm{\mid }w{\mathrm{\mid }}^2)$. This belongs to $L_{t,x}^2$ by $w\in L_{t,y}^4$. Outside $J$, the coefficient $a$ vanishes and $u_1$ is bounded, so $u_1{\mathrm{\partial }}_{x}u\in L_{t,x}^2$ there by the energy-gradient bound. Hence $N(u)=(u\cdot \mathrm{\nabla })u\in L_{t,x}^2$. Define $f=F_L+N(u)$. It belongs to $L_t^2{L}_x^2$, and the pointwise positive-time equation is $u_t-\nu \mathrm{\Delta }u+N(u)+\mathrm{\nabla }p=f$.

On $[\epsilon ,T]$, integrate the smooth equation from step 5.1 against $u$. The convection and pressure integrals vanish by divergence and no slip. Thus $\frac{1}{2}\mathrm{\parallel }u(t){\mathrm{\parallel }}_2^2+\nu {\int }_{\epsilon }^t\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2=\frac{1}{2}\mathrm{\parallel }u(\epsilon ){\mathrm{\parallel }}_2^2+{\int }_{\epsilon }^t(f,u)\mathrm{.}$ Let $\epsilon \downarrow 0$: strong initial convergence handles the endpoint and the energy and force bounds handle both integrals. This gives energy equality with initial value $u_0$. Similarly integrate the positive-time equation against a fixed solenoidal $V$ test and a smooth scalar time test, integrate by parts in time, and send $\epsilon \downarrow 0$. The pressure term vanishes; all remaining integrals converge by the established bounds. The resulting identity is precisely the solenoidal weak equation with initial value $u_0$. Positive-time smoothness and strong initial convergence give the required weak continuity on $[0,T]$. Thus $u$ is a solenoidal energy weak velocity of [F3].

The pressure has zero spatial mean and is in $L_t^{1}Q$, since $m^{\mathrm{\prime }}\in L^1$ and $\varphi \in Q$. It defines a time-distribution pressure for the full momentum equation: test the pointwise equation on compact positive-time intervals; its time-distribution identity on $(0,T)$ follows directly. Also $\mathrm{\parallel }p(t){\mathrm{\parallel }}_2=\mathrm{\mid }{m}^{\mathrm{\prime }}(t)\mathrm{\mid }\mathrm{\parallel }\varphi {\mathrm{\parallel }}_2$, so [F1] and $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_2>0$ imply $p\mathrm{\notin }{L}_t^{4\mathrm{/}3}Q$. If a second mean-zero $Q$-valued pressure distribution $q$ represented this same velocity and force, subtract their momentum equations. For each scalar time test $\zeta$, the $Q$ element $⟨q-p,\zeta ⟩$ has zero divergence pairing with every $V$ test. Pair with the bounded divergence right inverse of that very element, using [F4], to show its squared norm is zero. Thus $q=p$ as $Q$-valued distributions. In particular, if $q\in L_t^{4\mathrm{/}3}Q$, both are $L_t^{1}Q$ functions and equality of their distributions gives equality almost everywhere (test against a countable dense set in $Q$ and scalar tests), contradicting the nonintegrability just proved. Consequently this velocity does not meet the regular-pressure definition named in the Counterexample.

By Uniqueness and strong trace for two-dimensional solenoidal energy velocities, the constructed solenoidal energy velocity is the unique one for these two-dimensional square data. Every actual MINI subsequential limit supplied by [F5] is a solenoidal energy weak velocity with these same data, hence equals $u$. Steps 1.1–5.1 give all concrete properties and show that no mean-zero endpoint-integrable pressure exists for that limit. This refutes the universal regular-pressure assertion.