Uniqueness and strong trace for two-dimensional solenoidal energy velocities

For scalar $z\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$ the coordinate fundamental theorem gives $\mathrm{\mid }z(x,y){\mathrm{\mid }}^2\le 2{\int }_{\mathbb{R}}\mathrm{\mid }z{z}_x\mathrm{\mid }(s,y)ds=:A(y)$ and $\mathrm{\mid }z(x,y){\mathrm{\mid }}^2\le 2{\int }_{\mathbb{R}}\mathrm{\mid }z{z}_y\mathrm{\mid }(x,s)ds=:B(x)$. Thus $\mathrm{\mid }z{\mathrm{\mid }}^4\le A(y)B(x)$, and integration and Holder give $\mathrm{\parallel }z{\mathrm{\parallel }}_4^4\le 4\mathrm{\parallel }z{\mathrm{\parallel }}_2^2\mathrm{\parallel }{z}_x{\mathrm{\parallel }}_2\mathrm{\parallel }{z}_y{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }z{\mathrm{\parallel }}_2^2\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2^2$. Apply this to components of a vector and to differences of compact smooth approximants defining $H_0^1(D{)}^2$, extended by zero. The inequality makes the approximants Cauchy in $L^4$, so their limit is the same $L^2$ function; consequently $\mathrm{\parallel }z{\mathrm{\parallel }}_4^2\le C\mathrm{\parallel }z{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2$ on $H_0^1(D{)}^2$. Each energy velocity therefore belongs to $L_t^4{L}_x^4$.

With its $H^1$ inner product the complete space $H^1(D{)}^2$ is Hilbert. Its closed subspace $H_0^1(D{)}^2$ is Hilbert, as is $V_{\sigma }$, the kernel of bounded divergence therein. Poincare gives its equivalent gradient norm and [F2] gives reflexivity. Separability follows by mapping $H^1(D{)}^2$ isometrically, in its full norm, into the finite product of scalar $L^2$ spaces listing components and first derivatives. This product is separable by [F2]; every metric subspace of a separable metric space is separable: restrict a countable ambient base and, using Countable Choice, select a point from each nonempty restricted basic set. This proves separability of $V_{\sigma }$. The closure $H_{\sigma }$ is Hilbert as a closed $L^2$ subspace. Its inclusion of $V_{\sigma }$ is continuous by Poincare, injective and dense by definition. Thus the canonical pivot map $Jh(v)=(h,v{)}_2$ is bounded and injective and the triple has all hypotheses of [F3].

For a solenoidal energy velocity $u$, its convection functional on $V_{\sigma }$ is $N(u)(v)=-{\int }_D(u\otimes u):\mathrm{\nabla }v$, since the divergence and boundary terms vanish by approximation. Holder bounds its norm by $\mathrm{\parallel }u{\mathrm{\parallel }}_4^2$, which belongs to $L^2(0,T)$ by step 1.1. This is a strongly measurable dual-valued function: the map $L^4(D{)}^2\to V_{\sigma }^{\ast }$ is continuous and quadratic, and $u$ is strongly measurable in $L^4$ as a continuous image of its Bochner $V_{\sigma }$ representative. The viscous functional is a bounded linear image of $u\in L_t^2{V}_{\sigma }$, and the force functional is a bounded linear image of $f\in L_t^2{L}^2$. The fixed-test equation [F1] therefore says that the actual restricted derivative is the $L_t^2{V}_{\sigma }^{\ast }$ function $f-\nu (\mathrm{\nabla }u,\mathrm{\nabla }\cdot )-N(u)$. Apply [F3] to obtain a unique strong $C([0,T];H_{\sigma })$ representative and its energy chain rule. This representative agrees with the given weakly continuous one (fixed dense tests identify them), and so its initial value is $u_0$.

For two such velocities $u,\tilde{u}$ with the same data let $z=u-\tilde{u}$. Their difference has restricted derivative in $L_t^2{V}_{\sigma }^{\ast }$ by step 2.1, so [F3] permits pairing its equation with $z$ and differentiating its squared pivot norm. The trilinear cancellation leaves, with either velocity named $u$, $\frac{1}{2}\frac{d}{dt}\mathrm{\parallel }z{\mathrm{\parallel }}_2^2+\nu \mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2^2\le \mid {\int }_D(z\cdot \mathrm{\nabla })u\cdot z\mid \le C\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2\mathrm{\parallel }z{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2\le \frac{\nu }{2}\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2^2+C{\nu }^{-1}\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2\mathrm{\parallel }z{\mathrm{\parallel }}_2^2\mathrm{.}$ The first estimate uses the planar bound in step 1.1; the final one is $ab\le \nu a^2\mathrm{/}2+C{\nu }^{-1}{b}^2$. The initial difference is zero by step 2.1. Put $Y(t)=\mathrm{\parallel }z(t){\mathrm{\parallel }}_2^2$ and $B(t)=C{\nu }^{-1}\mathrm{\parallel }\mathrm{\nabla }u(t){\mathrm{\parallel }}_2^2$, increasing the constant to absorb the factor $2$. The chain rule makes $Y$ absolutely continuous and gives $Y(t)\le Y(s)+{\int }_s^{t}B(r)Y(r)dr$, with $B\ge 0$ in $L^1$. Choose $K>0$ with ${\int }_{\{B>K\}}B<1\mathrm{/}4$ by [F4], and choose $\delta >0$ with $K\delta <1\mathrm{/}4$. Partition $[0,T]$ into finitely many intervals of length at most $\delta$; each has ${\int }_{I}B<1\mathrm{/}2$. On the first interval $Y(0)=0$ and continuity gives ${\sup }_{I}Y\le ({\sup }_{I}Y){\int }_{I}B\le \frac{1}{2}{\sup }_{I}Y$, hence $Y=0$. Inductively the same argument applies on every next interval since its initial value is zero. Thus $z=0$ throughout $[0,T]$. This proves uniqueness and, together with the earlier steps, all stated regularity and trace assertions.