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Relative to the kinetic-energy inner product, integration by parts reverses incompressible base transport, transposes the velocity-gradient coupling, and preserves viscous diffusion. With reverse time $\tau =-t$, the adjoint base state is evaluated at $-\tau$. The Leray-Helmholtz projection or adjoint pressure enforces divergence-free adjoint velocity. The curl of a cross product and vorticity cross-product identity give the equivalent form ${\partial }_{\tau }v=\Omega \times v-\nabla \times (U\times v)-\nabla \pi +{\mathrm{Re}}^{-1}\Delta v$.

Using $r=(x,y,z)$ and constant $a$, the divergence isFor the curl, the curl of a cross product identity givesThusFor a quick sign check, taking $a=e_z$ gives $r\times a=(y,-x,0)$, whose curl is $(0,0,-2)$.

In Einstein summation convention, the curl of a cross product has componentsThe second line uses the contraction of two Levi-Civita symbols; the last uses the product rule. These components prove the curl of a cross product identity, including both divergence and directional-derivative terms.

Now apply that identity to $m\times v$. On the orientable surface, $m=n$, $m\cdot v=0$, and $|m|=1$. Differentiating $m\cdot m=1$ gives $m_i{\partial }_j{m}_i=0$. Taking a dot product with $m$ in the curl of a cross product formula therefore gives, on $S$,Only tangency on the surface is needed; no claim is made that the extension $v$ is tangent away from it. The right side is the surface divergence of the tangent vector field.

With the boundary oriented according to Stokes theorem, cyclic invariance of the scalar triple product giveswhere $u$ is the outward conormal of a surface boundary. Thus Stokes theorem provesThis is the surface divergence theorem for a tangent field.

For the disc in the $z=0$ plane, take $n=(0,0,1)$ and $v=\mathbf{r}=(x,y,z)$. The field is tangent on the disc, and ${\partial }_j{v}_i={\delta }_{ij}$. The contraction is $3-|n{|}^2=2$, so the surface integral is $2\pi R^2$. On the positively oriented circular boundary, $u=(\cos \theta ,\sin \theta ,0)$ and $v=Ru$, so the boundary flux is $R(2\pi R)=2\pi R^2$. Both sides agree.

Use the Levi-Civita symbol and sum over repeated indices. The contraction of two Levi-Civita symbols and the product rule giveReading the four terms as divergence and directional derivatives proves the curl of a cross product identity

In Cartesian coordinates, independent coordinate variables satisfywhere ${\delta }_{ij}$ is the Kronecker delta. Applying the product rule and chain rule to $u_i=r^{\alpha }{x}_i$ gives the Jacobian matrixThis tensor is symmetric in $i,j$. Hence its contraction with the antisymmetric Levi-Civita symbol vanishes:Similarly, $v_i={ϵ}_{ijk}{k}_j{u}_k$ and the constancy of $\mathbf{k}$ giveThe trace of the Jacobian matrix iswhich vanishes for $\alpha =-3$. The curl of a cross product identity, with $\mathbf{k}$ constant, now givesFor $\alpha =-3$,soAll identities for this singular case hold on $r>0$. In particular, zero divergence here is a pointwise assertion away from the origin, not an assertion that the flux through a sphere enclosing the singularity is zero.

Use the Levi-Civita symbol and summation over repeated indices. Contracting the two symbols givesThus the curl of a cross product is

Use the kinetic-energy inner product $⟨v,u⟩=\int v^{\ast }\cdot udV$, with perturbation energy $\Vert u{\Vert }^2/2$. On divergence-free fields the primal generator, before the pressure projection, isThe base state is also divergence-free. The periodic boundary terms in $x,z$ cancel, and decay at $|y|\to \infty$ removes the remaining boundary terms. Integration by parts changes $-\mathbf{U}\cdot \nabla$ to $+\mathbf{U}\cdot \nabla$; the Laplacian is self-adjoint under these conditions. Writing $J_{ij}={\partial }_j{U}_i$, the shear term $-Ju$ has adjoint operator $-J^{T}v$. A pressure gradient pairs to zero with a divergence-free field. Hencewhere the last term enforces the divergence-free projection.

To conserve $⟨v(t),u_p(t)⟩$ over the finite interval, the physical-time adjoint obeys $-{\partial }_{t}v=A^{†}(t)v$. Put $\tau =-t$ and $u_d(\tau )=v(-\tau )$; then ${\partial }_{\tau }{u}_d=A^{†}(-\tau )u_d$. This reverses the interval from $t=T$ towards $t=0$, or $\tau =-T$ towards $\tau =0$.

For $\mathbf{\Omega }=\nabla \times \mathbf{U}$, the vorticity cross-product identity and the curl of a cross producthold when both fields are divergence-free. Their difference equals the transport and shear terms in the adjoint operator. Therefore the adjoint linearized Navier-Stokes evolution isThe Leray-Helmholtz projection determines the adjoint pressure. The time-dependent base state is evaluated at $-\tau$; no extra time derivative of that base state appears in the spatial adjoint operator.

Write ${\partial }_i=\partial /\partial x_i$ and use $\times$ for the paper's wedge notation, which denotes the cross product. The given expression for the curl and the product rule giveApply the vector triple product identity $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})=\mathbf{b}(\mathbf{a}\cdot \mathbf{c})-\mathbf{c}(\mathbf{a}\cdot \mathbf{b})$ separately to the two terms:Summing over the Cartesian basis vectors identifies the two directional derivatives and the two divergences:Here, for example, $((\mathbf{w}\cdot \nabla )\mathbf{v}{)}_j={\sum }_i{w}_i{\partial }_i{v}_j$. This proves the curl of a cross product directly from the formula supplied.

For constant $m$, the curl of a cross product identity reduces toCommuting mixed second partial derivatives gives $(m\cdot \nabla )\nabla \varphi =\nabla (m\cdot \nabla \varphi )$. Hence the requested sum equals $m\Delta \varphi$, which is zero because $\varphi$ is harmonic. The argument assumes the indicated second derivatives are continuous on the domain.

For the curl of a cross product, the Cartesian contraction isThe product rule then givesFor the fluid equation, replace the advective term by $\mathbf{w}\times \mathbf{u}+\nabla (|\mathbf{u}{|}^2/2)$ using the preceding identity. Take its curl. Both gradient terms have zero curl, while $\nabla \cdot \mathbf{w}=\nabla \cdot (\nabla \times \mathbf{u})=0$ and incompressibility gives $\nabla \cdot \mathbf{u}=0$. The vorticity equation is consequentlyThe left side is the material derivative of vorticity; the right is vortex stretching. It follows here because the given force per mass is a gradient, in addition to the flow being inviscid and incompressible.

For a differentiable vector field in Cartesian coordinates, the divergence and curl arewith summation over repeated indices. In components the curl is $({\partial }_y{F}_z-{\partial }_z{F}_y,{\partial }_z{F}_x-{\partial }_x{F}_z,{\partial }_x{F}_y-{\partial }_y{F}_x)$. The contraction of two Levi-Civita symbols and the product rule giveRecognizing the directional derivatives gives the curl of a cross product identity

In Cartesian coordinates, use the Levi-Civita symbol and the Einstein summation convention to expandThe contraction of the two symbols uses the Kronecker delta, and the final step is the product rule. Thus the convective acceleration identity isPut $\mathbf{\omega }=\nabla \times \mathbf{u}$. The curl of a gradient vanishes, and the curl of a cross product identity givesSince the divergence of a curl is zero, taking the curl of the first identity yieldsTwice continuous differentiability suffices for the mixed-derivative identities used here.

For a twice continuously differentiable vector field in Cartesian coordinates, the curl and divergence arePut $\mathbf{\omega }=\nabla \times \mathbf{v}$. Using the Levi-Civita symbol, the Kronecker delta contraction and the product rule,The last term vanishes because the divergence of a curl is zero: the symmetric mixed derivatives contract against an antisymmetric symbol. Thus the curl of a cross product identity becomes

For orthogonal coordinates $q_i$ with position vector $\mathbf{r}(q_1,q_2,q_3)$, the scale factors of orthogonal coordinates and local unit vectors areOrthogonality gives $d{s}^2={\sum }_i{h}_i^{2}d{q}_i^2$, and right-handedness fixes ${\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3$. In cylindrical coordinates, $\mathbf{r}=(\rho \cos \varphi ,\rho \sin \varphi ,z)$, soThe radial vector points away from the axis; the azimuthal vector is tangent in the direction of increasing $\varphi$; ${\mathbf{e}}_z$ is the Cartesian positive $z$ direction. The azimuthal basis is undefined on the coordinate axis itself.

For the axial velocity profile, the cylindrical coordinates formula givesThe associated cross product isThis field has only a radial component depending on $\rho$, so its curl is zero. Also $\nabla \cdot \mathbf{u}={\partial }_z{u}_z=0$. The identity above therefore yieldsIn fact both directional derivatives vanish: the first differentiates $u_z{\mathbf{e}}_z$ only in the azimuthal direction, and the second differentiates $\mathbf{\omega }$ only in the $z$ direction. All physical vector fields extend smoothly through $\rho =0$, despite the coordinate-basis singularity.

Write the curl and cross product with the Levi-Civita symbol, then use its contraction identity:Expanding with the product rule givesRecognize the divergence in the first and third terms, and the directional derivatives in the other two. The curl of a cross product identity is thereforeThe derivatives act on both vector fields. No assumption of vanishing divergence or spatially constant fields has been made.

Work in a Cartesian coordinate system, write ${\partial }_j=\partial /\partial x_j$, and use the Einstein summation convention. For a twice continuously differentiable vector field, the contraction of two Levi-Civita symbols givesHere the Kronecker deltas select components, and equality of mixed partial derivatives gives the curl of the curl identityFor the second identity, the same contraction followed by the product rule yieldsThus the curl of a cross product isThe directional derivatives in the last two terms act on the vector field immediately to their right.

Use the Levi-Civita symbol and the contraction of two Levi-Civita symbols in Cartesian coordinates. With repeated indices summed,The last line is the product rule. Reading its four terms in vector notation proves the curl of a cross product identity:

Use Einstein summation convention, writing ${\partial }_j=\partial /\partial x_j$. The contraction of two Levi-Civita symbols givesThe last equality is the product rule. Recognizing the divergences and directional derivatives proves the curl of a cross product identity:

For a smooth tangent vector field $v$ on a compact orientable surface with smooth boundary and outward conormal of a surface boundary, the integral of its surface divergence is its boundary flux. Extend the unit normal vector to a unit field $m$. The curl of a cross product gives $m\cdot \nabla \times (m\times v)=\nabla \cdot v-m_i{m}_j{\partial }_j{v}_i$ on the surface: $m\cdot v=0$ there and $m\cdot {\partial }_{j}m=0$ by unit length. Apply Stokes theorem and use $(m\times v)\cdot d\mathbf{r}=v\cdot (d\mathbf{r}\times m)$ to obtain the formula.

For differentiable vector fields, the displayed identity relates vorticity to the antisymmetric part of the velocity gradient. Together with the curl of a cross product, it converts the adjoint linearized Navier-Stokes evolution between component and rotational forms.