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Use the conventional Laplace operator $\Delta ={\partial }_{x_1}^2+{\partial }_{x_2}^2$. The fundamental solution of the Laplace equation in two dimensions is $G(x)=(2\pi {)}^{-1}\log |x|$, with $\Delta G={\delta }_0$. To check its sign, integrate the outward normal derivative around a circle: ${\int }_{\partial B_{\epsilon }}{\partial }_n\log |x|ds=2\pi$. Applying integration by parts against a smooth compactly supported test function and shrinking the circle gives the distributional identity.

The printed stream function has $\varphi =-G\ast \omega$. For each multi-index $\beta$ with $|\beta |\le 2$, transfer derivatives to the compactly supported $C^2$ function:The logarithmic singularity is locally integrable. On a bounded region in $x$, the convolution uses one fixed bounded region in the integration variable; splitting off a small disk around the singularity proves continuity of these derivatives. Thus $\varphi \in C^2$, and convolution of the distributional identity givesThe asserted plus sign is false for the printed kernel and the conventional Laplacian. Any nonzero smooth compactly supported $\omega$ is a counterexample. To obtain $\Delta \varphi =\omega$, replace the kernel's minus sign by a plus sign, or explicitly adopt the negative Laplace operator. Below, retain the printed kernel and velocity convention. With that convention, $\mathrm{curl}u=\Delta \varphi =-\omega$, so $\omega$ is the negative of conventional scalar vorticity. This sign does not change the transport or flow arguments.

Take $\alpha >0$; a real disturbance with negative wavenumber is recovered by complex conjugation. To obtain the pressure equation for a shear-flow normal mode, apply $i\alpha$ to the streamwise momentum equation and differentiate the normal momentum equation. The mass conservation equation cancels the viscous divergence and the terms proportional to $U-c$, leaving $2i\alpha U^{\prime }v={\alpha }^{2}p-p^{\prime \prime }$. Thus the fluid pressure equation and wall value areThe wall value follows from streamwise momentum with no-slip boundary condition. Normal momentum also gives $p^{\prime }(0)=v^{\prime \prime }(0)/R$.

In the lower viscous boundary layer, mass conservation gives $v=O(\alpha \delta )$ if $u=O(1)$. Since $U\sim y$, convection and shear have size $\alpha \delta$ when $c=O(\delta )$; matching these with transverse viscous diffusion gives $R^{-1}{\delta }^{-2}\sim \alpha \delta$. The pressure gradient balance then gives $p=O(\delta )$. Thus the lower scalings areThe neglected streamwise viscous term is smaller by $(\alpha \delta {)}^2$. In these variables the leading equations areDifferentiating the second and using the first yields the Airy reduction $\boxed{u_{YYY}=i(Y-C)u_Y}$.

Put $q=e^{i\pi /6}$, so $q^3=i$. The bounded matching branch of the Airy function gives $u_Y=A\mathrm{Ai}(q(Y-C))$; the other independent Airy solution grows along the ray and is excluded. Integrating with the two wall conditions giveswhile the wall momentum equation gives the constant fluid pressureDefine the convergent Airy displacement integralsThe Airy function equation and its decaying derivative imply $I_1-C{I}_0=iq{\mathrm{Ai}}^{\prime }(-qC)$. Consequently, as $Y\to \infty$,Both differentiated lower momentum and the constant of integration are satisfied; retaining that constant is essential for the final matching.

For the middle region introduce a perturbation stream function $\Phi$ with $u={\Phi }^{\prime }$ and $v=-i\alpha \Phi$. The leading inviscid flow streamwise equation isFor a shear flow, this is the integrated form of the Rayleigh equation for inviscid shear flow. The leading matched solution for fixed $y=O(1)$ is $\Phi \sim BU$, so $u\sim B{U}^{\prime }$ and $v\sim -i\alpha BU$. In the overlap with the lower viscous boundary layer, the first constant correction iswhich agrees with the integrated lower solution. For a regular smooth profile, the normal momentum equation gives $p^{\prime }\sim -{\alpha }^2(U-c)\Phi$. Thus fluid pressure variation across a fixed-width middle region is $O({\alpha }^{2}B)$; in the final distinguished scaling this is smaller than its $O(\delta )$ value, so $p\sim \delta P_0$ throughout that region. Since $U\to 1$, the middle velocity tends to $\boxed{v\sim -i\alpha B}$, which does not decay and requires another region.

The fluid pressure equation becomes $p^{\prime \prime }-{\alpha }^{2}p=0$ when $U\sim 1$. Therefore the upper scale is $\Delta =1/\alpha$, with $z=\alpha y$. The inviscid decaying upper solution isMatching velocity as $z\downarrow 0$ gives $P_{\infty }\sim \alpha B$; matching fluid pressure to the middle region gives $\delta P_0\sim \alpha B$. This is three-layer long-wave shear-flow matching, expressed as a matched asymptotic expansion.

Under $\alpha =k{R}^{-1/4}$ with fixed positive $k$, $\delta =k^{-1/3}{R}^{-1/4}$, and $\alpha /\delta =k^{4/3}$. Eliminating the nonzero amplitude $A$ gives the Airy wall-layer matching determinant and dispersion relationEquivalently, with $\mathcal{T}(C)={\int }_{-qC}^{\infty e^{i\pi /6}}\mathrm{Ai}(\zeta )d\zeta =q{I}_0(C)$,The determinant form does not divide by a possibly vanishing $I_0$. The wavespeed is $c=\delta C$; temporal growth has sign $\mathrm{Im}C$ for positive $\alpha$.

There is a genuine conflict in the printed assumptions: fixed-$k$ $\alpha =k{R}^{-1/4}$ cannot satisfy $R^{-1/7}\ll \alpha$ as $R\to \infty$. For order-one lower matching amplitudes, constant leading middle fluid pressure requires ${\alpha }^2\ll \delta$, equivalently $\alpha \ll R^{-1/7}$; this is the reverse inequality and is satisfied by the final distinguished scaling. Thinness also requires $\alpha R\gg 1$. The dispersion relation above is therefore derived under the final distinguished limit, for which the complete consistent window is $R^{-1}\ll \alpha \ll R^{-1/7}$. Both printed limits cannot be imposed simultaneously.

The divergence and scalar vorticity areThus the flow is irrotational but not incompressible on the plane. A velocity potential satisfying $\mathbf{u}=\nabla \varphi$ isIndeed its two partial derivatives are the two velocity components. A usual two-dimensional stream function would satisfy $u_x={\psi }_y$, $u_y=-{\psi }_x$ and hence have identically zero divergence. No such stream function exists on an open region where this divergence is not identically zero; the line $y=0$ is not a two-dimensional flow domain.

The temporal integration by parts terms areThe terminal states are free and the objective has no explicit terminal-velocity dependence. ThereforeThe factor two follows from the displayed objective without a $1/2$ prefactor. The initial scalar field perturbation is fixed, so $\delta \theta (0)=0$ and there is no independently prescribed adjoint-scalar condition at $t=0$. Its value there is obtained by backward integration. Independent variation of the initial velocity state gives ${\mathbf{u}}^{†}(0)={\mathbf{u}}_0^{†}$.

Variation of the candidate ${\mathbf{u}}_0$ in the energy-augmented functional gives the fixed-energy initial-condition optimality condition. For the usual divergence-free no-slip control space $H$, write $P_H$ for its orthogonal projection, the Leray-Helmholtz projection in the usual incompressible velocity space. This is the standard formal control condition when pressure is determined by incompressible flow and normal momentum. If the independently imposed direct pressure flux is retained as an additional constraint, the admissible control variations must also satisfy its compatibility conditions along the trajectory; they need not fill the usual space $H$. The standard projection formula alone does not prove stationarity for that more restricted problem. ThenFor the natural solenoidal adjoint space, $P_H{\mathbf{u}}^{†}(0)={\mathbf{u}}^{†}(0)$. Equivalently, its component tangent to the sphere in a normed vector space of fixed kinetic energy vanishes:Here $E_0>0$ is needed for the sphere in a normed vector space to be regular and for division by $2E_0$. The multiplier is real and can have either sign; normalizing the initial adjoint with a prescribed positive sign is not a general necessary condition for minimization. If $E_0=0$, the only feasible initial velocity is zero and this tangent formula for the sphere in a normed vector space is inapplicable. First-order stationarity alone also allows maxima or saddles; a local minimum requires the appropriate nonnegative constrained second variation.

The spatial boundary terms vanish with periodic adjoints in $x,z$, homogeneous no-slip boundary conditions ${\mathbf{u}}^{†}=0$ at $y=\pm 1$, and ${\partial }_y{\theta }^{†}=0$ there. To see this, velocity variations vanish at the wall while their normal derivatives need not, so the viscous boundary term forces the adjoint velocity to vanish. Scalar variations have zero normal derivative but free values, forcing the adjoint scalar's normal derivative to vanish. Pressure variation gives adjoint incompressible flow; no independent terminal or initial datum is assigned to $p^{†}$. Its additive time-dependent constant may be fixed by a zero spatial mean.

In particular, a homogeneous Neumann condition for the direct pressure does not by variational transposition require ${\partial }_y{p}^{†}=0$. For a smooth adjoint solution, the wall-normal adjoint momentum equation instead suppliesat the flat walls, because the direct scalar normal derivative is zero there. A zero adjoint-pressure derivative would be an additional compatibility choice, valid only when this right-hand side vanishes.

There are also literal direct-data compatibility qualifications in the printed setup. The initial total scalar field is $\overline{\theta }=-\mathrm{erf}(30y)$, whose wall derivative is $-60e^{-900}/\sqrt{\pi }\ne 0$. It is extraordinarily small but mathematically not zero. Thus exact initial scalar data and exact zero wall flux are incompatible for a classical solution smooth at $t=0$. One may use a parabolic mild solution of an abstract Cauchy problem with the boundary condition enforced for $t>0$, or replace the initial profile by an exactly compatible smooth one; these are conventions, not an equality $e^{-900}=0$. The reference scalar field is also not a stationary profile of the diffusion equation at finite $Pe$: the full $\Delta \Theta$ term must be retained.

The pressure compatibility at a no-slip wall is similarly important. Direct normal momentum gives ${\partial }_{y}p=\nu \Delta u_y-R{i}_B\theta$. The separately imposed zero pressure derivative requires this right side to vanish. It is not automatic for arbitrary fixed-energy initial velocities. For example, the divergence-free no-slip field generated by the periodic stream function $S=A(1-y^2{)}^2\sin (\alpha x)$ with $\alpha =\pi /L_x>0$ has $u_x={\partial }_{y}S$, $u_y=-{\partial }_{x}S$ and $\Delta u_y{|}_{y=\pm 1}=-8A\alpha \cos (\alpha x)\ne 0$, while $\theta (0)=0$. With $A\ne 0$ adjusted to any positive energy, no classical solution smooth up to the initial wall can satisfy that extra pressure condition. The displayed adjoints are the formal necessary equations along admissible smooth trajectories; a well-posed physical formulation normally determines pressure from incompressible flow and normal momentum, rather than imposing independent homogeneous pressure flux for every control.

For Darcy-Bénard convection, the dimensionless Darcy law is $\mathbf{u}=-\nabla p+R\theta {\mathbf{e}}_z$, with incompressibility and a heat equation about $T_0=1-z$. Here $R$ is a Darcy thermal-time Rayleigh number, with permeability replacing the squared-depth dependence in the fluid-layer Rayleigh number. The stated stream function convention gives $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$, so $w={\psi }_x$. The vertical-plane curl of Darcy law gives ${\nabla }^2\psi =R{\theta }_x$; omitting the quadratic perturbation advection from the heat equation gives ${\theta }_t={\psi }_x+{\nabla }^2\theta$.

On the impermeable connected square boundary, zero normal velocity makes $\psi$ constant along the boundary; choose that constant as zero. The prescribed conductive boundary temperature fixes $\theta =0$ on all four sides. Thus the perfect-conductor wording is interpreted through the supplied homogeneous perturbation data: the side temperatures retain the background vertical profile, rather than imposing a different isothermal side state.

At marginal stability put $q=\sqrt{R}$ and $P=\psi +iq\theta$. The two coupled equations implyThe gauge transformation for conducting-square Darcy onset $P=e^{-iq(x-1/2)/2}F$ removes the first derivative:The square's Dirichlet Laplacian eigenfunctions are $\sin (m\pi x)\sin (n\pi z)$, with eigenvalues of $-{\nabla }^2$ equal to ${\pi }^2(m^2+n^2)$, $m,n\ge 1$. The first marginal value is thereforeIt is genuinely the first instability threshold, not just an isolated neutral value. Eliminating $\psi$ gives the temperature linear operator $L_R={\Delta }_D+R{\partial }_x{\Delta }_D^{-1}{\partial }_x$, where ${\Delta }_D$ is the Dirichlet realization of an elliptic operator for the Laplacian. Integration by parts shows the second term is self-adjoint and nonnegative, since its quadratic form is $-⟨{\theta }_x,{\Delta }_D^{-1}{\theta }_x⟩\ge 0$. Thus the spectrum is real, starts negative at $R=0$, and its leading eigenvalue cannot cross into growth without a zero eigenvalue. The smallest zero value is the one found above.

Let $S=\sin \pi x\sin \pi z$, $\xi =x-1/2$ and $a=q_c/2=\sqrt{2}\pi$. Multiplying the lowest real $F=S$ by independent real and imaginary constants gives two real physical eigenfunctions:The subscripts indicate even and odd parity about $x=1/2$. Because $S$ is even in $\xi$, the first has even $\psi$ and odd $\theta$, while the second reverses those parities. They are linearly independent and satisfy all four Dirichlet boundary conditions. The lowest complex $F$ eigenfunction is simple over the complex numbers, but its arbitrary complex multiplier supplies two independent real pairs; this explains the physical degeneracy.

The unit normal pointing out of the fluid at its lower boundary is $-{\mathbf{e}}_y$. Since impermeability gives $v(x,0)=0$, also $v_x(x,0)=0$, and the tangential viscous traction exerted on the fluid is $t_x=-\mu u_y(x,0)$. The cilia do positive work at rate $t_{x}u=\rho P$, soFor positive downstream velocity this requires $u_y<0$, as expected when the wall-driven flow slows away from the cilia. The wall velocity is not prescribed independently; the driving condition prescribes power.

In a laminar boundary layer, momentum diffuses over a thickness $\delta$ during the downstream advection time $x/U$, giving ${\delta }^2\sim \nu x/U$. The wall power condition gives $U^2/\delta \sim P/\nu$. With these balances the downstream Reynolds number grows and $\delta /x\propto x^{-3/5}$ tends to zero. This explains the thin layer for large $x$ within the laminar approximation. Pressure is uniform across it and has no imposed streamwise gradient because the outer fluid is at rest. The two-dimensional boundary-layer equations are

Choose the scales to satisfy the balances exactly:The stream function $\psi =U\delta f(\eta )$, $\eta =y/\delta$, enforces continuity equation and givesThus the advective term is $U^2[\frac{1}{5}{f}^{\prime 2}-\frac{3}{5}f{f}^{\prime \prime }]/x$, while $\nu u_{yy}=\nu U{f}^{\prime \prime \prime }/{\delta }^2=U^2{f}^{\prime \prime \prime }/x$. HenceImpermeability, imposed wall power and matching to resting outer fluid giveA regular decaying profile also has $f^{\prime \prime }\to 0$ and finite positive $f_{\infty }$; no arbitrary additional wall value for $f^{\prime }$ should be imposed.

The volume flux per unit span is $Q={\int }_0^{\infty }udy=U\delta f_{\infty }\propto x^{3/5}$, so continuity equation gives $v_{\infty }=-dQ/dx<0$: outer fluid is entrained toward the wall. In the physical positive decaying similarity profile, $f^{\prime }>0$ and $f^{\prime \prime }<0$. To justify monotonicity, $f^{\prime \prime }(0)<0$ by the power condition; if $f^{\prime \prime }$ first crossed zero with $f^{\prime }>0$, the equation would give $f^{\prime \prime \prime }=f^{\prime 2}/5>0$, after which $f^{\prime \prime }>0$ cannot cross back and $f^{\prime }$ could not decay to zero. Therefore $f(\eta )={\int }_0^{\eta }{f}^{\prime }(s)ds\ge \eta f^{\prime }(\eta )$, and the velocity formula gives $2\eta f^{\prime }-3f<0$ for $\eta >0$. Consequently the transverse velocity is negative throughout that physical layer, and zero at the wall. This pointwise conclusion uses the positive decaying branch; the flux argument already establishes entrainment of the exterior.

The velocity field is incompressible becauseWith the convention $u_x={\psi }_y$ and $u_y=-{\psi }_x$, a stream function isThe vorticity is vertical:For a steady inviscid incompressible flow, the vorticity equation is $(\mathbf{u}\cdot \nabla )\mathbf{\omega }=(\mathbf{\omega }\cdot \nabla )\mathbf{u}$. Here the right side is zero because the flow is independent of $z$, while the left side is $2(\mathbf{u}\cdot \nabla \psi ){\mathbf{e}}_z=0$ because velocity is tangent to the level sets of $\psi$. Thus the equation is satisfied.

The streamlines are the level curves $\sin x\sin y=\text{constant}$. The lines $x=\pi$ and $y=\pi$, together with the boundary, are separatrices; each of the four cells contains closed nested curves around a centre at $(\pi /2,\pi /2)$, $(3\pi /2,\pi /2)$, $(\pi /2,3\pi /2)$, or $(3\pi /2,3\pi /2)$.

With the conventiona stream function isup to an additive constant. Differentiation recovers both velocity components.

For a two-dimensional incompressible flow, the stream function conventionis satisfied by

With zero imposed pressure gradient, the two-dimensional boundary-layer equations areIntroduce the stream function byThe boundary conditions areThe additive constant in $\psi$ may be chosen so that $\psi (x,0)=0$.

The steady incompressible Stokes flow equations areTaking the curl eliminates the pressure gradient and commutes with the Laplacian, so for the vorticity $\mathbf{\omega }=\nabla \times \mathbf{u}$,This is the vorticity part of Harmonic pressure and vorticity in Stokes flow.

For a two-dimensional incompressible flow, choose the stream function conventionThenApplying the harmonic-vorticity equation givesthe Biharmonic stream function for planar Stokes flow equation.

Define the residual-circulation stream function byThis satisfies residual mass conservation identically. LetThe momentum equation is

Differentiate geostrophic balance vertically and hydrostatic pressure meridionally to obtain thermal-wind balanceAfter a time derivative, the transformed density equation givesOn the other hand, the $z$ derivative of momentum givesEliminating ${\overline{u}}_{tz}$ yields the Eliassen equation for residual circulation