Explicit Leray projection computations on ${\mathbb{T}}^3$: gradients, a mixed field, and a divergence-free mode

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: $\widehat{u}(\pm k)\cdot \overline{(I-P)v(\pm k)}=0$.