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The spin tensor $W$ is skew, and $\dot{Q}{Q}^T=W$ means $\dot{Q}=WQ$ and ${\dot{Q}}^T=-Q^{T}W$. Differentiate all factors to obtain the corotating-frame representation of a Jaumann derivative:Thus the Jaumann derivative is the ordinary derivative of the tensor components in this rotating frame.

Put $S={\sigma }^d$, $\Delta \mu ={\mu }_1-{\mu }_0$, $A=S-2{\mu }_{0}D$ and $\tilde{A}=Q^{T}AQ$, $\tilde{D}=Q^{T}DQ$. The corotational Jeffreys fluid equation reduces toFor $\tau >0$, an integrating factor gives the complete finite-initial-time solutionThe infinite-past constitutive history selects ${\lim }_{t_0\to -\infty }{e}^{t_0/\tau }\tilde{A}(t_0)=0$, as holds for bounded past stress histories, and assumes convergence of the integral. Taking that limit and rotating back provesA general initial stress would retain the displayed homogeneous term. Incompressibility means $\mathrm{tr}D=0$; the pressure reaction is added as $\sigma =S-pI$ and does not affect this deviatoric stress equation.

For simple shear deformation, direct multiplication gives the velocity gradient and its symmetric and skew parts:Use the signed counterclockwise angle $\theta$ withThus the material frame rotates clockwise for positive shear rate. This angle convention is necessary for the signs of the sine terms. Set $\delta =\theta (t)-\theta (s)$. With $H=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $R_{\delta }=Q(t)Q^T(s)$, multiplication yieldsSubstitution in the history formula provesThe suppressed third row and column are zero for the selected infinite-past history.

For constant shear rate $g=\dot{\gamma }$, $2\delta =-g(t-s)$. Put $u=t-s\ge 0$. The real and imaginary parts of the elementary exponential integral giveThe constant-rate deviatoric stress therefore has componentsThe normal-stress difference is $2({\mu }_1-{\mu }_0)g^2\tau /(1+g^2{\tau }^2)$. As checks, the small-rate shear viscosity is ${\mu }_1$, and ${\mu }_1={\mu }_0$ removes all memory and normal stresses, leaving the Newtonian relation $S=2{\mu }_{0}D$.

Let $w=v^{\prime }-v$ and $D(v)=\mathrm{sym}\nabla v$. The supporting-hyperplane inequality for the convex viscous dissipation potential givesThe stress is symmetric because its argument is a symmetric rate-of-strain tensor, so ${\sigma }_{ij}(D_{ij}^{\prime }-D_{ij})={\sigma }_{ij}{w}_{i,j}$. Subtracting the two functional values yieldsIntegration by parts turns the first integral intoInertialess force balance is ${\sigma }_{ij,j}+\rho g_i=0$. The variation vanishes on $S_v$, and the traction is $t_i^0$ on $S_t$. Thus both the volume and boundary contributions cancel exactly, proving the convex viscous potential minimum principleOnly convexity, not strict convexity, is required, so no uniqueness follows just from this argument. For incompressible flow, admissible trial velocities must also be divergence free. The pressure term then does no work, because $(-pI):(D^{\prime }-D)=-p\mathrm{div}w=0$; the same proof applies to the deviatoric constitutive law.

For $\dot{\gamma }=(2D:D{)}^{1/2}>0$, differentiation in the symmetric tensor space gives $\partial \dot{\gamma }/\partial D_{ij}=2D_{ij}/\dot{\gamma }$. Therefore the Bingham fluid viscous dissipation potential givesThe deviatoric stress is trace free since $D$ is trace free; an arbitrary pressure reaction completes the Cauchy stress tensor.

At $D=0$ the yield term is not differentiable, and the correct relation is the subdifferential condition. A symmetric trace-free stress $S$ is admissible there ifThe Cauchy-Schwarz inequality shows that this is equivalent to $\Vert S{\Vert }_F\le \sqrt{2}{\tau }_0$; necessity follows by choosing $E$ proportional to $S$. Hence the no-strain-rate restriction isThe quadratic term has zero derivative at zero and does not change this restriction. This is the unyielded part of the Bingham plastic law.

For flow along the plates, write $y=x_2$, $v=(u(y),0,0)$ and take $G=-p_{,1}>0$ as the pressure-drop magnitude. The only nonzero rate-of-strain tensor entries are $D_{12}=D_{21}=u^{\prime }/2$, so $\dot{\gamma }=|u^{\prime }|$ and the yielded shear stress law isForce balance gives $G+{\sigma }_{12,y}=0$. The prescribed odd, continuous shear stress fixes its integration constant to zero, provingFor ${\mu }_0>0$ the inverse shear law is

The original PDF also calls $u$ odd. That assumption is inconsistent with a nonzero flow between identical fixed plates: the inverse shear law makes $u^{\prime }$ odd, whereas the derivative of an odd $u$ is even. Both can hold only if $u^{\prime }=0$, and the no-slip wall values then give $u=0$. Thus with the literal odd-velocity requirement, there is no admissible flowing solution when $Gh>{\tau }_0$. The standard pressure-driven velocity is even; replacing that erroneous parity yields the intended profile below.

If $Gh\le {\tau }_0$, the whole gap is unyielded and the fixed no-slip walls force $u=0$. Otherwise let $y_0={\tau }_0/G<h$. The plug flow of a yield-stress fluid occupies $|y|\le y_0$. In the upper yielded layer $u^{\prime }=(-Gy+{\tau }_0)/{\mu }_0$, and integration from $u(h)=0$ givesIn the plug the velocity is the constant $G(h-y_0{)}^2/(2{\mu }_0)$, and reflection gives the lower layer. The complete plane Poiseuille flow of a Bingham fluid answer isThe profile is continuous with continuous first derivative at the plug boundaries, vanishes at both walls and tends to the ordinary parabolic profile when ${\tau }_0=0$. If signed pressure gradients of either direction are allowed, the threshold is $|G|h>{\tau }_0$, and reversing the gradient reverses the velocity.

Put $a={\sigma }_{13}$, $b={\sigma }_{23}$, $w=v_3$, and write $f_a=\partial f/\partial a$, $f_b=\partial f/\partial b$. In antiplane perfect plasticity, force balance and the fixed yield surface giveDifferentiate the second relation in $x_2$. Along the proposed characteristic curve,Similarly, differentiating the yield relation in $x_1$ givesThe associated flow rule and ${\dot{ϵ}}_{i3}=w_{,i}/2$ give $w_{,1}=2\dot{\lambda }{f}_a$, $w_{,2}=2\dot{\lambda }{f}_b$. ConsequentlyThus both shear stresses and the antiplane velocity are constant on each characteristic. Assume a regular yield gradient $(f_a,f_b)\ne 0$. Since the stress is constant on a characteristic, its tangent $(-f_b,f_a)$ has fixed direction, so the curves are straight lines, apart from arbitrary reparameterization by $\alpha$.

In a centred antiplane plastic fan those lines emanate from the crack tip. Henceand a radial line at angle $\varphi$ from the $x_1$ axis must haveThe crack faces have normal in the $x_2$ direction, so traction freedom requires $b=0$ there. The fan joins the adjacent constant crack-face stress region along the characteristic carrying that stress. Its boundary line therefore satisfieswith the direction interpreted directly when the denominator vanishes. The original PDF reverses this ratio. Its displayed ratio is $d{x}_1/d{x}_2$, whereas the tangent of an angle from the $x_1$ axis is $d{x}_2/d{x}_1$. The distinction cannot be removed while retaining both its stated characteristic direction and its angle convention.

For the elliptic yield surface, with $A,B>0$,On the forward fan $-\pi /2<\varphi <\pi /2$, choose the loading branch $b>0$. The radial tangent condition gives $a/A^2=-(b/B^2)\tan \varphi$. Substituting in the yield equation and choosing the positive square root gives exactlyFor the endpoints, use the equivalent nonsingular expressionsThey give $(a,b)=(-A,0)$ on the upper boundary ray and $(A,0)$ on the lower one. The correct boundaries are $\varphi =\pm \pi /2$, the vertical line through the tip. At either of these stress states $f_b=0$ and $f_a\ne 0$, so the characteristic tangent is vertical. The printed boundary ratio would instead give a horizontal line, providing a concrete counterexample to that intermediate assertion.

Extend the upper and lower rear sectors by the constant stresses $(-A,0)$ and $(A,0)$ respectively; these satisfy the traction-free crack faces and match the fan stresses continuously. The reversed loading has all stresses negated. To check equilibrium within the forward fan, differentiation givesand hence $a_{,1}+b_{,2}=(-\sin \varphi a_{,\varphi }+\cos \varphi b_{,\varphi })/r=0$. The yield equation is satisfied identically as well.

Finally $\nabla w=w^{\prime }(\varphi )e_{\varphi }/r$, while $(f_a,f_b)=2e_{\varphi }/H$. The associated flow rule reduces tofor this stress branch. Any such angular velocity field is constant along the rays and has constant values in the adjacent rear sectors. Its magnitude and angular variation require velocity or loading data beyond the specified traction-free faces; the local stress fan alone does not determine them.

At a sharp interface, continuity of velocity and traction gives, for unit incident velocity from the left,Thus $r_v=(Z_a-Z_b)/(Z_a+Z_b)$ and $t_v=2Z_a/(Z_a+Z_b)$. Normalizing the transmitted velocity by its elastic-wave energy flux gives $T_{ba}=\sqrt{Z_b/Z_a}{t}_v$, exactly the result above. Right incidence gives the reversed reflection sign and the same flux-normalized transmission. A stress reflection coefficient has the opposite sign to the velocity reflection coefficient; the convention matters.

Put $r=R_{aa}$ and $t=T_{ab}$. The incoming-to-outgoing matrix is $S_0=\left(\begin{array}{cc}r& t\\ t& -r\end{array}\right)$, with $r^2+t^2=1$. Therefore $S_0^{†}{S}_0=I$, proving equality of incoming and outgoing time-averaged flux for arbitrary coherent incident wave amplitudes, not just incidence from one side.

For all real frequencies the harmonic equations instead giveTheir right sides are equal; the diagonal terms $\pm i\omega /\beta$ have zero real contribution. ConsequentlyThe mean flux for real-part phasors is $(|{\varphi }_{+}{|}^2-|{\varphi }_{-}{|}^2)/2$. Evaluating it at the two endpoints proves exactly that total outgoing flux equals total incoming flux at every real frequency. This relies on real lossless moduli and mass density; absorption would change the identity.

Let $n$ point out of the drop, and use $-$ for the interior and $+$ for the exterior. The interfacial velocity is continuous. Define $\kappa ={\nabla }_s\cdot n$; for a sphere of radius $a$ it is $2/a$. With constant surface tension, the interfacial stress balance with variable surface tension reduces toUse the exterior-viscosity kernels to define $S[t]=\int J(y-x)t(x)d{S}_x$ and $D[u]=\int u(x)K(y-x)n(x)d{S}_x$. The interior viscosity is $\lambda \mu$, so its velocity kernel is $J/\lambda$, while its stress kernel $K$ is unchanged. The interior boundary equation isFor the decaying exterior flow, the inner boundary's fluid-domain normal is $-n$, soAdd these equations and use the traction jump. The required capillary boundary integral equation for a viscous drop isThe formula uses the drop-outward normal consistently in both integrals. If $\lambda =1$, the double-layer term cancels. A spherical drop of constant curvature has zero velocity: a constant Laplace-pressure jump is balanced without motion. A prescribed ambient flow would add its known far-boundary contribution; none is present for the surface-tension-driven flow here.

Use the Newtonian fluid stress tensor $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$. The Stokes equations for a point force on the fluid are $\nabla \cdot u=0$ and $\nabla \cdot \sigma +F\delta (x)=0$. Put $r=|x|$. Away from the origin, a harmonic vector potential proportional to $F/r$ in the Unscaled Papkovich–Neuber representation gives the rotationally covariant decaying solution. Specifically, take $\Phi =-F/(8\pi \mu r)$ and $\chi =0$. Differentiation yieldsThis satisfies the homogeneous Stokes equations for $r>0$. Its stress isTo verify the point-force normalization, on a sphere $r=\epsilon$ the traction is $\sigma n=-3(F\cdot n)n/(4\pi {\epsilon }^2)$. Since ${\int }_{S^2}{n}_i{n}_{j}d\Omega =4\pi {\delta }_{ij}/3$, its surface integral is $-F$. Thus integrating momentum balance over the small ball gives exactly the applied force $F$, fixing the normalization of the Stokeslet. Its radial velocity is $(F\cdot n)/(4\pi \mu r)$, soThere is no net volume or mass source at the singularity. The kernels have the parity $J(-x)=J(x)$ and $K(-x)=-K(x)$.

For two Stokes flows with the same viscosity and body forces $f^{(1)},f^{(2)}$, the Lorentz reciprocal theorem including body forces isIndeed, apply the divergence theorem to $u_i^{(1)}{\sigma }_{ij}^{(2)}-u_i^{(2)}{\sigma }_{ij}^{(1)}$. The cross-gradient terms cancel: incompressibility removes each pressure contraction and the remaining terms are both $2\mu e^{(1)}:e^{(2)}$. The two stress divergences give the body-force terms shown.

Choose flow 1 to be the desired body-force-free field and flow 2 to be a Stokeslet with arbitrary force vector $q$ at $y$. For $y\in V$, the body-force term is $-q\cdot u(y)$. Rearrange the reciprocal identity and use the parity of $J,K$; equality for every $q$ givesFor $y$ outside the volume there is no point force in $V$, so the same two integrals sum to zero. To obtain the smooth boundary value, exclude a small hemisphere about $y$, use continuity of $u$ to replace it by $u(y)$ on that hemisphere, and take the radius to zero. A full small sphere contributes $u(y)$; a hemisphere contributes half of it. Equivalently, the local solid-angle factor is $2\pi /(4\pi )$. The result is the boundary integral representation of Stokes flow with right side $\frac{1}{2}u(y)$ on a smooth boundary, and with the double layer interpreted by its boundary principal value. A nonsmooth corner would have a different solid-angle coefficient.

For body-force-free Stokes fields with the same viscosity,To prove this Lorentz reciprocal theorem, integrate the divergence of the difference of the two cross-work fluxes. Both stress divergences are zero; incompressibility and symmetry of the Newtonian stress reduce each cross-gradient contraction to $2\mu e^{(1)}:e^{(2)}$. These are equal, so the divergence integrates to zero. The argument applies to an exterior domain by first cutting it off and taking the outer radius to infinity; the decaying velocity and stress make the outer boundary contribution vanish.

Define the generalized force as force and torque exerted by the rigid body on the fluid. With a resting far field and no-slip boundary conditions, $u=U+\Omega \times x$ on the body, and linearity givesFor two such motions the reciprocal theorem becomesSince the six-component velocities can be chosen independently, $\boxed{\mathsf{R}={\mathsf{R}}^T}$. The ordinary work identity givesfor every nonzero rigid-body motion. If the integral were zero, $e=0$ would make the connected exterior flow an infinitesimal rigid motion. Its decay at infinity requires that motion to be zero, which is incompatible with a nonzero no-slip body velocity. Thus $\mathsf{R}$ is positive definite. Fluid-on-body forces have the opposite sign; their map is $-\mathsf{R}$, not a positive-definite resistance matrix.

Write $s_{\varphi }=\sin \varphi$ and $c_{\varphi }=\cos \varphi$. The helix has $|dX/d\theta |=b\sqrt{1+{\tan }^2\varphi }=b\sec \varphi$, soIts unit tangent is $X^{\prime }=c_{\varphi }{e}_{\theta }+s_{\varphi }{e}_z$, where $e_{\theta }=(-\sin \theta ,\cos \theta ,0)$. The given slender-body force density is the force per arc length exerted on the fluid. For pure axial translation, $V=U{e}_z$, it givesFor pure rotation, $V=b\Omega e_{\theta }$, it givesThe axial torque per unit arc length is $b{f}_{\theta }$. Integrating along the wire gives the axial resistance matrix of a slender helixThus translation gives $(F_z,G_z)=(\mathcal{A}U,\mathcal{B}U)$, and rotation gives $(\mathcal{B}\Omega ,\mathcal{D}\Omega )$. These are the requested axial components; the axial swimmer approximation does not require proving statements about the other components. The equality of the two off-diagonal coefficients checks reciprocity, andFor the organism, retain the specified entrainment approximation: the head moves with the local large-scale fluid translation, and its angular velocity $\Omega -\omega$ differs from the local fluid rotation $\Omega$ by $-\omega$. Put $D_0=8\pi \mu a^3$. Its torque on the fluid is therefore $-D_0\omega$, and its force is neglected. Neutral buoyancy and absence of external torque giveSolving,This is the entrained-head approximation for a helical microswimmer, which differs from treating the head as an isolated sphere in otherwise stationary fluid. The flagellum's angular speed is also $\Omega =D_0\omega (1+c_{\varphi }^2)/(CL{b}^2)$.

At fixed $b,L,a,\omega ,ϵ$, the speed is maximal at $\varphi =\pi /4$ and tends to zero at either limiting pitch. At fixed other parameters it decreases as $1/L$: the longer flagellum has more resistance at the fixed torque supplied in this model. Holding the number of turns fixed instead would make $L$ change with pitch, so that is a different comparison.

The motor exerts equal and opposite torques of magnitude $D_0\omega$ on flagellum and head, whose relative angular speed is $\omega$. Therefore its rate of working isEquivalently, the work delivered to the fluid is $G_{\mathrm{flag}}\Omega +G_{\mathrm{head}}(\Omega -\omega )=D_0\omega \Omega -D_0\omega (\Omega -\omega )$. Translation contributes no net work because the total force vanishes. This explains the independence from $U$ and $\Omega$, without assigning an isolated-fluid drag law to the entrained head.

The material derivative follows surfactant carried by the moving interface. The term $-C{\nabla }_s\cdot u_s$ is dilution or concentration by tangential area expansion or contraction. The term $-C(u\cdot n){\nabla }_s\cdot n$ is the additional area change produced by normal motion of a curved interface. The term $D_s{\nabla }_s^{2}C$ represents surface diffusion, while $-k(C-C_0)$ relaxes the concentration towards a reservoir value through exchange. All these terms refer to concentration per surface area, as in surfactant transport with exchange relaxation.

In the steady bubble frame, the spherical interface has $u\cdot n=0$. Write $C=C_0+C^{\prime }$ with uniform $C_0$ and regard both $u$ and $C^{\prime }$ as first order in the rising speed. Advection of $C^{\prime }$ and its product with surface dilation are second order. Hence the linear steady equation isSpherical symmetry and linearity leave only the tangential projection of the imposed translation vector: $u_s=A{I}_{s}U$. An azimuthal term $n\times U$ would change sign under reflection and is excluded for this non-chiral rising bubble. The degree-one forcing does not excite other spherical-harmonic degrees in the linear problem.

On $r=a$, differentiating $n=x/r$ givesEach component of $n$ is a degree-one spherical harmonic, soOne may also obtain the coefficient from $n\cdot n=1$: its surface Laplacian gives $n\cdot {\nabla }_s^{2}n=-|{\nabla }_{s}n{|}^2=-2/a^2$, while rotational symmetry makes the vector Laplacian parallel to $n$. Expanding $I_{s}U=U-n(U\cdot n)$ then yieldsThe candidate $C^{\prime }=BU\cdot n$ therefore solves the linear transport equation ifIts spherical mean is zero. This is the dipolar surfactant distribution. Assuming $k{a}^2+2D_s>0$, its consistency condition isDiffusion or exchange must restore uniform concentration faster than the interfacial flow redistributes it: this holds with a small surface Péclet number, fast exchange, or sufficiently immobilized surface velocity. These same conditions justify neglecting the perturbation's advection. Small capillary number and sufficiently small buoyancy-induced shape distortion are separate requirements for a spherical interface.

Take $+$ to mean the outer liquid, $-$ the bubble interior and $n$ outward from the bubble. The general interfacial stress balance with variable surface tension isFor $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$, its tangential part is $-{\nabla }_s\gamma ={\gamma }_{1}B{I}_{s}U/a$. Thus the requested form isThis positive parameter measures the strength of the restoring Marangoni stress relative to viscous stress, including the relaxation of concentration by diffusion and exchange.

The far velocity $-U$ is produced by the constant harmonic vector potential $\Phi =U$. The two independent decaying degree-one modes are the force-carrying vector potential $\alpha aU/r$ and the scalar dipole $\chi =\beta a^{3}U\cdot \nabla (1/r)$. They are harmonic outside the bubble and have precisely the symmetry of uniform translation. The Unscaled Papkovich–Neuber representation gives, at arbitrary radius,The velocity at $r=a$ agrees with the printed expression. However, the original PDF's traction formula has a factor-of-two error. Direct differentiation with $\sigma =-pI+\mu (\nabla u+\nabla u^T)$ givesnot the printed prefactor $12\mu /a$. This is traction of translating-sphere Papkovich–Neuber potentials. The $\alpha$ term follows immediately from the Stokeslet stress, and the scalar-potential part gives $6\mu \beta [U-3(U\cdot n)n]/a$; neither contribution contains the extra factor two.

With this consistent traction, no penetration requires $\beta -\alpha =1/2$, tangential velocity requires $A=-(1+\alpha +\beta )$, and tangential stress balance requires $\beta =A\lambda$. Solving,For comparison, taking the printed $12\mu /a$ literally would instead giveThose solve the supplied algebraic pair but not the Newtonian stress calculated from the supplied potentials. The factor-six result is the physically consistent translating surfactant-coated bubble solution.

As $\lambda \to 0$, $A\to -1/2$ and $\beta \to 0$: the clean bubble has a mobile interface and zero tangential traction. As $\lambda \to \infty$, $A\to 0$, $\alpha \to -3/4$ and $\beta \to -1/4$: the interface becomes immobilized and the exterior flow approaches that past a no-slip rigid sphere. The limiting tangential traction is $-3\mu I_{s}U/(2a)$, a useful check on the correct normalization.

The normal dynamic traction isapart from an arbitrary constant pressure. In the normal stress balance on a translating bubble, its degree-one part is balanced by the buoyant hydrostatic pressure difference and by the variation $-2{\gamma }_1{C}^{\prime }/a$ of normal capillary stress. A uniform internal gas pressure balances the constant Laplace-pressure term but cannot by itself balance this angular variation. If $U$ is upwards and $\Delta \rho$ is liquid minus bubble mass density, the degree-one balance can be writtenIt fixes $U=\Delta \rho g{a}^2(1+2\lambda )/[3\mu (1+3\lambda )]$, tending to the clean-bubble and rigid-sphere rise speeds. A degree-one radial displacement translates a sphere rather than changing its curvature, so a supposed dipolar shape distortion cannot replace this force balance.

Let $\mathbf{F}$ denote the applied force, and take the normal ${\mathbf{n}}_S$ outward from the solid. At leading order the Stokes drag law and the torque on a rotating sphere in Stokes flow giveThe exact no-slip boundary condition and the exact resultant-load conditions areThe stress condition specifies the total force and torque, not a pointwise prescribed traction.

For the boundary perturbation of a nearly spherical particle, evaluate the no-slip condition at $\mathbf{x}=(a+\epsilon f)\mathbf{n}$. Taylor expansion givesThere is no $f{\mathbf{\Omega }}_0\times \mathbf{n}$ term because ${\mathbf{\Omega }}_0=0$. Therefore

To expand the integral conditions correctly, use surface independence of Stokes force and torque integrals. The Stokes equation implies $\nabla \cdot \mathbf{\sigma }=0$, and symmetry of the stress tensor implies that the angular-momentum flux is divergence-free as well. Evaluate the exact force and torque on a fixed sphere of radius $R$ enclosing the perturbed particle. There are then no geometric terms: the $O(\epsilon )$ integrals of ${\mathbf{\sigma }}_1$ on that fixed sphere vanish because the external force and torque have no $O(\epsilon )$ corrections. Each perturbation field solves the homogeneous exterior Stokes equations, so the divergence theorem transfers those integrals back to $r=a$. HenceIf the integrals are expanded directly on the displaced surface, terms involving the displaced evaluation of ${\mathbf{\sigma }}_0$, the changed normal and the changed area do appear individually. Their sum vanishes by the same surface-independence identity. Thus their absence in the final equations is a cancellation, not permission to omit geometric terms independently.

Let the longitudinal variation scale be $\ell \gg h$. The stress-free boundary condition gives $\mu (u_z+w_x)=0$ at each face. Incompressibility implies $w=O(hu/\ell )$, so $w_x$ is smaller than a putative $u_z=O(u/h)$ by $O(h^2/{\ell }^2)$. The leading Stokes equation and the two shear-free conditions therefore make $u$ independent of $z$; its thickness-dependent correction is smaller-order. Symmetry gives $w=0$ at $z=0$, and incompressibility integrates toVertical momentum makes the leading normal stress independent of $z$. With the normal directed from liquid into gas, the upper-face curvature is $-h_{xx}/2$ to leading order. The Young–Laplace equation givesThe Newtonian fluid stress tensor also gives ${\sigma }_{zz}=-p+2\mu w_z=-p-2\mu u_x$. Hence $p=p_a-\gamma h_{xx}/2-2\mu u_x$ andThe factor four is the planar extension factor: both the longitudinal viscous stress and the pressure adjustment from transverse contraction contribute.