Cm Math Fluid — Examples

Part (a): the coefficients of $\phi$ are $\widehat{\phi }(0,0,0)=0$ and, for the cosine term, $\widehat{\phi }(\pm e_1)=\frac{1}{2}$; for the sine term, $\widehat{\phi }(e_2+e_3)=\frac{1}{2i}$ and $\widehat{\phi }(-e_2-e_3)=-\frac{1}{2i}$. Hence $\widehat{v}(k)=2\pi ik\widehat{\phi }(k)$ is a vector parallel to $k$ at every frequency (zero at frequency $0$), and the orthogonal projection ${\pi }_k$ kills it: for $k\mathrm{\ne }0$, ${\pi }_k(2\pi ik\widehat{\phi }(k))=2\pi i\widehat{\phi }(k)(1-\frac{k\cdot k}{\mathrm{\mid }k{\mathrm{\mid }}_2^2})k=0$. Thus $Pv=0$, and $v=Pv+\mathrm{\nabla }\phi$ is the Hodge decomposition with zero divergence-free part.

Part (b): the coefficients of $u$ are supported on $k=\pm e_2$ (first component $1\mathrm{/}(2i)$ and $-1\mathrm{/}(2i)$) and $k=\pm e_3$ (third component $1\mathrm{/}2$ and $1\mathrm{/}2$). For $k=e_2$: $\widehat{u}(e_2)=(1\mathrm{/}(2i),0,0)$ and $k\cdot \widehat{u}(e_2)=0$, so ${\pi }_k\widehat{u}=\widehat{u}$; for $k=-e_2$ likewise. For $k=e_3$: $\widehat{u}(e_3)=(0,0,1\mathrm{/}2)$ and $k\cdot \widehat{u}(e_3)=1\mathrm{/}2$, so ${\pi }_k\widehat{u}=\widehat{u}-\frac{1}{2}\cdot \frac{1}{1}(0,0,1)=(0,0,0)$; for $k=-e_3$: $\widehat{u}=(0,0,1\mathrm{/}2)$, $k\cdot \widehat{u}=-1\mathrm{/}2$ and ${\pi }_k\widehat{u}=(0,0,1\mathrm{/}2)-\frac{-1\mathrm{/}2}{1}(0,0,-1)=(0,0,0)$. Hence $Pu=(\sin (2\pi x_2),0,0)$; moreover $\mathrm{\nabla }f$ for $f=\frac{1}{2\pi }\sin (2\pi x_3)$ has coefficients $2\pi ik\widehat{f}(k)$ equal to $(0,0,1\mathrm{/}2)$ at $k=e_3$ and $(0,0,1\mathrm{/}2)$ at $k=-e_3$, which is exactly $u-Pu$, so $u-Pu=\mathrm{\nabla }f$ is a gradient field.

Part (b), orthogonality and norms: since $\sin (2\pi x_2)$ and $\cos (2\pi x_3)$ are functions of different coordinates, $⟨Pu,u-Pu⟩={\int }_{{\mathbb{T}}^3}\sin (2\pi x_2)\cdot 0+0\cdot 0+0\cdot \cos (2\pi x_3)dx=0$; and ${\int }_0^1{\sin }^2(2\pi s)ds={\int }_0^1{\cos }^2(2\pi s)ds=\frac{1}{2}$, so $\mathrm{\parallel }Pu{\mathrm{\parallel }}_{L^2}^2=\frac{1}{2}$, $\mathrm{\parallel }u-Pu{\mathrm{\parallel }}_{L^2}^2=\frac{1}{2}$, and $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2=1$ (the mixed terms vanish by separation of variables). This verifies the orthogonal decomposition identity of Periodic vector Sobolev spaces on the three-torus, the weight family, and the Leray projection on this example.

Part (c): for each $k^{\mathrm{\prime }}\in {\mathbb{Z}}^3$, $\widehat{u}(k^{\mathrm{\prime }})=\frac{1}{2i}A$ if $k^{\mathrm{\prime }}=k$, $-\frac{1}{2i}A$ if $k^{\mathrm{\prime }}=-k$, and $0$ otherwise; since $A\cdot k=0$, $k\cdot \widehat{u}(k)=0$ and ${\pi }_k\widehat{u}(k)=\widehat{u}(k)$ (and likewise at $-k$), so $Pu=u$. The divergence has coefficients $2\pi i(k^{\mathrm{\prime }}\cdot \widehat{u}(k^{\mathrm{\prime }}))=0$ at $k^{\mathrm{\prime }}=\pm k$, so $\mathrm{div}u=0$; and for any $v\in H^0$ the pairing $⟨u,v⟩={\sum }_{k^{\mathrm{\prime }}}\widehat{u}(k^{\mathrm{\prime }})\cdot \overline{\widehat{v}(k^{\mathrm{\prime }})}$ only involves $k^{\mathrm{\prime }}=\pm k$ where $\widehat{u}\perp k^{\mathrm{\prime }}$, so it equals the pairing of $u$ with $Pv$ because $(I-P)v$ has coefficients parallel to $k^{\mathrm{\prime }}$ there: $\widehat{u}(\pm k)\cdot \overline{(I-P)v(\pm k)}=0$.

Part (a), nonlinear term: for $u_0=A\sin (2\pi k\cdot x)$ one has ${\mathrm{\partial }}_j{u}_0=2\pi k_{j}A\cos (2\pi k\cdot x)$, so $(u_0\cdot \mathrm{\nabla })u_0=2\pi (A\cdot k)\cos (2\pi k\cdot x)A\sin (2\pi k\cdot x)=0$ because $A\cdot k=0$; and $\mathrm{div}{u}_0=2\pi (A\cdot k)\cos (2\pi k\cdot x)=0$.

Part (a), norms: $\widehat{u}(t,\pm k)=\pm \frac{1}{2i}c(t)A$ are the only nonzero coefficients, so $\mathrm{\parallel }u(t){\mathrm{\parallel }}_{H^m}=\mathrm{\mid }c(t)\mathrm{\mid }\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{H^m}=e^{-4{\pi }^2\nu \mathrm{\mid }k{\mathrm{\mid }}_2^{2}t}\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{H^m}$ for every $m$; in particular the solution exists for all $t\ge 0$.

Part (c), divergence and Laplacian: ${\mathrm{\partial }}_1{u}_1^0=2\pi \cos {\theta }_1\cos {\theta }_2$, ${\mathrm{\partial }}_2{u}_2^0=-2\pi \cos {\theta }_1\cos {\theta }_2$, so $\mathrm{div}{u}^0=0$; and ${\mathrm{\partial }}_1^2{u}_1^0=-(2\pi {)}^2{u}_1^0$, ${\mathrm{\partial }}_2^2{u}_1^0=-(2\pi {)}^2{u}_1^0$, likewise for the second component, so $\mathrm{\Delta }{u}^0=-8{\pi }^2{u}^0$.

Part (c), nonlinear term and pressure: $(u^0\cdot \mathrm{\nabla })u_1^0=u_1^0{\mathrm{\partial }}_1{u}_1^0+u_2^0{\mathrm{\partial }}_2{u}_1^0=2\pi \sin {\theta }_1\cos {\theta }_2\cos {\theta }_1\cos {\theta }_2+2\pi \cos {\theta }_1\sin {\theta }_2\sin {\theta }_1\sin {\theta }_2=2\pi \sin {\theta }_1\cos {\theta }_1=\pi \sin 2{\theta }_1$, while $-{\mathrm{\partial }}_1{p}^0=-\frac{1}{4}(-2\cdot 2\pi \sin 2{\theta }_1)=\pi \sin 2{\theta }_1$ (because ${\mathrm{\partial }}_1\cos 2{\theta }_1=-2\cdot 2\pi \sin 2{\theta }_1$); similarly $(u^0\cdot \mathrm{\nabla })u_2^0=\pi \sin 2{\theta }_2=-{\mathrm{\partial }}_2{p}^0$. Hence $(u^0\cdot \mathrm{\nabla })u^0=-\mathrm{\nabla }{p}^0$.

Part (a), equation: with $c(t):=e^{-4{\pi }^2\nu \mathrm{\mid }k{\mathrm{\mid }}_2^{2}t}$ one has ${\mathrm{\partial }}_{t}u={\mathrm{\partial }}_{t}cA\sin (2\pi k\cdot x)$ and $\nu \mathrm{\Delta }u=c\nu (-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}_2^2)A\sin (2\pi k\cdot x)={\mathrm{\partial }}_{t}u$; since $(u(t)\cdot \mathrm{\nabla })u(t)=c(t{)}^2(u_0\cdot \mathrm{\nabla })u_0=0$ by step 1.1 and $P$ is linear with $P0=0$, the projected equation holds; the primitive equation holds with $p=0$ by Reconstruction of the periodic pressure from the velocity: $\mathrm{\nabla }p=-(I-P)((u\cdot \mathrm{\nabla })u)$ and $\mathrm{\Delta }p=-{\sum }_{i,j}{\mathrm{\partial }}_i{\mathrm{\partial }}_j(u_i{u}_j)$ (whose defining formula gives $p=0$ when $(u\cdot \mathrm{\nabla })u=0$). The Galerkin reduction is the one-dimensional ODE $c^{\mathrm{\prime }}=-4{\pi }^2\nu \mathrm{\mid }k{\mathrm{\mid }}_2^{2}c$, since the mode $k$ is not mixed by the convolution.

Part (c), the equations: with $u(t)=e^{-8{\pi }^2\nu t}{u}^0$ and $p(t)=e^{-16{\pi }^2\nu t}{p}^0$, one computes ${\mathrm{\partial }}_{t}u=-8{\pi }^2\nu u$, $\nu \mathrm{\Delta }u=\nu e^{-8{\pi }^2\nu t}(-8{\pi }^2{u}^0)=-8{\pi }^2\nu u$ (so ${\mathrm{\partial }}_{t}u-\nu \mathrm{\Delta }u=0$), and $(u\cdot \mathrm{\nabla })u=e^{-16{\pi }^2\nu t}(u^0\cdot \mathrm{\nabla })u^0=-e^{-16{\pi }^2\nu t}\mathrm{\nabla }{p}^0=-\mathrm{\nabla }p$; therefore ${\mathrm{\partial }}_{t}u+(u\cdot \mathrm{\nabla })u=-\mathrm{\nabla }p+\nu \mathrm{\Delta }u$. Since $(u\cdot \mathrm{\nabla })u$ is a gradient, its $P$-image vanishes, and the projected equation ${\mathrm{\partial }}_{t}u=\nu \mathrm{\Delta }u-P((u\cdot \mathrm{\nabla })u)=\nu \mathrm{\Delta }u$ holds; this also shows the solution is global.

For $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}=L$ the weight satisfies $2\pi L\le w_k\le cL$ with $c:=\sqrt{1+12{\pi }^2}$, and the shell has exactly $(2L+1{)}^3-(2L-1{)}^3=24L^2+2$ points. Hence for $L\ge 2$, with ${\mathrm{\ell }}_L:=\log (2\pi L)$ one has two-sided bounds $c^{-3\mathrm{/}2}{L}^{-3\mathrm{/}2}\mathrm{/}\log (cL)\le \mathrm{\mid }\widehat{u}(k){\mathrm{\mid }}_2\le (2\pi L{)}^{-3\mathrm{/}2}\mathrm{/}{\mathrm{\ell }}_L$ on that shell, so shell summation (multiplicity $24L^2+2$) gives $c_1{\sum }_{L\ge 2}{L}^{1\mathrm{/}2}\mathrm{/}\log L\le {\sum }_k\mathrm{\mid }\widehat{u}(k){\mathrm{\mid }}_2\le c_2{\sum }_{L\ge 2}{L}^{1\mathrm{/}2}\mathrm{/}\log L$ and $c_3{\sum }_{L\ge 2}1\mathrm{/}(L{\log }^{2}L)\le {\sum }_k\mathrm{\mid }\widehat{u}(k){\mathrm{\mid }}_2^2\le c_4{\sum }_{L\ge 2}1\mathrm{/}(L{\log }^{2}L)$ for positive constants $c_i$.

By [F1] the series ${\sum }_{L\ge 2}1\mathrm{/}(L{\log }^{2}L)$ converges (its integral test function $t^{-1}(\log t{)}^{-2}$ has finite integral from $2$ to $\mathrm{\infty }$, the antiderivative being $-1\mathrm{/}\log t$) while ${\sum }_{L\ge 2}{L}^{1\mathrm{/}2}\mathrm{/}\log L$ diverges (its summands do not tend to $0$); by [F2] this means $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^0}^2={\sum }_k\mathrm{\mid }\widehat{u}(k){\mathrm{\mid }}_2^2<+\mathrm{\infty }$ and ${\sum }_k\mathrm{\mid }\widehat{u}(k){\mathrm{\mid }}_2=+\mathrm{\infty }$.

The family is real-symmetric by construction, so by [F3] it is the coefficient family of a real field $u\in H^0$; by step 2.1 the hypothesis of the false statement is satisfied but its conclusion fails. Hence the false statement is refuted.