Work in the body frame and approximate the narrow annulus locally by a planar gap. Let $y=0$ be the cylindrical wall, moving at $-U$ in this frame, and $y=h(x)$ the stationary cell surface. The red blood cell is represented by the given compliant body; the surrounding fluid is modeled as incompressible and Newtonian. Assume $h\ll a$, small local gap slope, and negligible inertia. The lubrication equations are $p_y=0$ and $\mu w_{yy}=p_x$, with no-slip conditions $w(0)=-U$, $w(h)=0$. Integration givesMass conservation makes the body-frame flux constant. With the sign convention of the problem,so the lubrication model of a compliant translating plug givesThe circumferential multiplier is $2\pi a$ to leading order, giving the relative volume flux $-2\pi aQ$.

Let $p_d$ be the pressure in front of the cell. Match the film to the reservoirs by $p(L)=p_d$ and $p(-L)=p_d+\Delta p$. Use $h=a-r$ and symmetric end matching for the parabolic-gap reduction. The elastic law givesFor $r_0(-L)=r_0(L)$ this gives $h(-L)-h(L)=\Delta p/\alpha$. If the outer unstressed shape is not symmetric at the matching ends, the general relation instead includes $h(-L)-h(L)=\Delta p/\alpha -[r_0(-L)-r_0(L)]$. The proposed reduced end equation assumes the symmetric matching used here. The gap must remain positive. To determine the unknown flux and translating speed one also needs zero net axial force on the freely moving body, and a specified absolute downstream pressure relative to $p_0$, not merely a pressure difference. If an external force pulls the body, it must replace the force-free condition. The stated elasticity law fixes $p_0$; treating it as an unknown internal pressure would require a further cell-volume constraint.

The force-free condition is most easily obtained from a control volume containing both body and surrounding film. The pressure force is $\pi a^2\Delta p$, and the resisting wall traction is $2\pi a\mu {\int }_{-L}^L{w}_y(0)dx$. The local velocity solution gives $w_y(0)=4U/h-6Q/h^2$, soThis is the force-free pressure drop of an annular lubrication plug. It includes the film pressure and shear contributions without separately approximating the deformed body's surface normal.

Put $s=2Q/U>0$, $\ell =\sqrt{s/\kappa }$, $h=sH$ and $x=\ell X$. The local parabolic elastic law is $p=p_0+\alpha [r_{00}-a-\kappa x^2/2+h]$, whence $p_x=\alpha \sqrt{\kappa s}(H_X-X)$. Substitution into the local lubrication equation givesThe derivative is with respect to the scaled coordinate $X$; a derivative with respect to dimensional $x$ would be inconsistent with this nondimensional equation.

Let $C=s/a$ and $\tilde{L}=L/\ell$. The elastic end condition gives $\Delta p=\alpha s[H(-\tilde{L})-H(\tilde{L})]$. Scaling the force integral and eliminating $\Delta p$ then givesThis completes the requested formulation and both dimensionless equations. The local parabolic approximation is used in the region contributing most strongly to the lubrication integrals; matching to the outer shape and reservoirs supplies the stated end conditions.

Use mass conservation and conservative momentum conservation, with viscous stress ${\sigma }_{ij}$ defined so that its divergence is the viscous force:Differentiate the first equation in time and eliminate the momentum time derivative using the second. This givesSet ${\rho }^{\prime }=\rho -{\rho }_0$ and $p^{\prime }=p-p_0$, where the reference values and $c_0$ are constant. Subtract $c_0^2{\nabla }^2{\rho }^{\prime }$ to obtain the exact Lighthill acoustic analogyThe Lighthill stress tensor includes nonlinear momentum transport, departure from the reference linear pressure-density relation, and viscous stress. For an inviscid fluid the last term is absent. The derivation is an exact rearrangement of the conservation equations; treating the tensor as a known localized source is a subsequent approximation.

Represent the leading real pattern by $A(T)e^{i{k}_{c}x}+\overline{A}(T)e^{-i{k}_{c}x}$. Before forcing, continuous translation symmetry acts as $A\mapsto e^{i\theta }A$, and spatial reflection acts as $A\mapsto \overline{A}$. Equivariance excludes quadratic terms and permits a linear term and $|A{|}^{2}A$ at cubic order. Symmetry under reflection makes their coefficients real. A supercritical bifurcation has a saturating cubic coefficient; rescaling amplitude and time makes it $-1$. The imperfection has a Fourier component at the same wavenumber, so its resonant projection contributes an additive real coefficient. On the spatially uniform wave envelope subspace the amplitude equation is therefore $A_T=\mu A-|A{|}^{2}A+\alpha$. Allowing modulations would additionally produce the spatial terms of the real Ginzburg–Landau equation.

For a forcing pattern moving in the positive $x$ direction, its laboratory-frame coefficient is $\alpha e^{-i\Omega T}$, where $\Omega$ is proportional to $k_{c}c$ after converting to slow time. The comoving amplitude is $B=e^{i\Omega T}A$, since the fast carrier becomes $e^{i{k}_c(x-ct)}$. Differentiation givesA constant phase choice makes $\alpha >0$, and complex conjugation reverses the sign of $\Omega$, so take $\Lambda =\Omega >0$. Put $B={\alpha }^{1/3}a$ and $\tau ={\alpha }^{2/3}T$. The rescaled linear coefficients are $\mu /{\alpha }^{2/3}$ and $\Lambda /{\alpha }^{2/3}$, and the forcing becomes one. Below the rescaled variables are renamed $A,T,\mu ,\Lambda$, giving the directly forced cubic amplitude equation.

Take $M^2>0$ initially to avoid an infrared divergence. A Schwinger parameterization and the Euclidean Gaussian integral give, in the convergent range $0<\mathrm{Re}d<4$,Integration over $t$ supplies the Gamma function. Its analytic continuation defines the answer in dimensional regularization:Use $d=4-ϵ$ near four dimensions and $d=6-ϵ$ near six dimensions. The Gamma function recurrence implies $\Gamma (ϵ/2)=2/ϵ+O(1)$ and $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$. ThereforeBoth are simple poles. Equivalently their residues as functions of $d$ are $-2/(4\pi {)}^2$ at $d=4$ and $2M^2/(4\pi {)}^3$ at $d=6$. The negative pole near six dimensions is a feature of analytic continuation; it is not a negative value of the original convergent positive integral. A massless zero-momentum integral is scaleless and cannot be used indiscriminately to extract these ultraviolet poles, because dimensional regularization then combines ultraviolet and infrared contributions.

For the following Feynman diagrams, specify the Euclidean action and additive counterterms explicitly so their signs are unambiguous. The four-dimensional interaction is ${\mu }^{ϵ}\lambda {\varphi }^4/4!$, the six-dimensional interaction is ${\mu }^{ϵ/2}g{\varphi }^3/3!$, and the common free inverse scalar propagator is $p^2+m^2$. Here $ϵ=4-d$ or $6-d$, respectively. The factors of $\mu$ give dimensionless renormalized couplings and change no leading pole residues. The cubic Euclidean theory is treated as a formal perturbation expansion about its massive quadratic scalar field theory; its real cubic potential is not bounded below.

A phase diagram divides a space of control parameters into equilibrium thermodynamic phases, marking phase coexistence boundaries and thermodynamic critical points. Its dimension counts independently variable controls, not spatial dimensions. An explicit three-dimensional example uses the controls $(r,u,h)$ of a sextic even Landau potential:Here $M$ is a scalar order parameter, $h$ its conjugate field, and $r$ a thermal control. The equilibrium magnetization minimizes $V$ globally. At $h=0$, $u>0$, the boundary $r=0$ is a continuous phase transition. At $h=0$, $u<0$, the boundary $r=3u^2/(16v)$ is a first-order phase transition between $M=0$ and $M=\pm \sqrt{-3u/(4v)}$. The two boundaries meet at the tricritical point $r=u=h=0$. A fixed strictly positive quartic coupling alone cannot produce this tricritical point; the stabilizing sextic term is essential when the quartic coefficient is tuned through zero.

The full phase diagram also contains the $h=0$ phase coexistence sheet between positive and negative ordered phases. Crossing it changes the sign of $M$ discontinuously. For $u<0$, the three-phase line $r=3u^2/(16v)$ is the junction of this sheet and two tricritical wings, on which two unequal same-sign equilibrium magnetizations coexist at nonzero $h$. The tricritical wing critical edges are lines of ordinary thermodynamic critical points; they terminate the first-order sheets and meet at the tricritical point. Away from these boundaries, varying $h$ produces a smooth crossover.

For an explicit construction of the tricritical wings, write their two coexisting order parameters as $M_1=s-d$ and $M_2=s+d$, with $s>0$ and $0<d\le s$. Equality of their Landau free energies and first derivatives givesIndeed, for these controls the tricritical wing coexistence factorization isThe last factor is $(M+2s{)}^2+s^2-d^2\ge 0$, proving that both specified minima are global. The reflected wing has negative $h$ and negative $M$. Letting $d\to 0$ gives the tricritical wing critical edge $u=-10v{s}^2/3$, $r=5v{s}^4$, $h=8v{s}^5/3$; there $V^{\prime \prime }(s)=V^{\prime \prime \prime }(s)=0$ and $V^{\prime \prime \prime \prime }(s)=40v{s}^2>0$. Letting $d=s$ recovers the three-phase line. This supplies the first-order sheets and their continuous boundary curves, rather than only a zero-field section.

For a density matrix $\rho$ on a finite-dimensional Hilbert space, its Von Neumann entropy in bits iswhere ${\lambda }_j$ are its eigenvalues and $0{\log }_{2}0=0$. It is the Shannon entropy of the spectrum, is zero for a pure state, and is invariant under unitary conjugation.

The positive state-price density alternative characterizes one-period arbitrage. Interpret $A_0$ as the vector of current asset prices and $A_1$ as their terminal payments. A fixed vector of holdings produces initial cost $x^{\top }{A}_0$ and terminal payment $x^{\top }{A}_1$. The first alternative is a nonpositive-cost investment with a nonnegative payment and strict improvement either initially or on an event of positive probability. A strictly positive state-price density prices every such payment and excludes both forms of arbitrage, as proved below.

If there is a traded bank account costing one and paying a deterministic $R>0$, the density satisfies $1=R\mathbb{E}\nu$. Hence $dQ/dP=R\nu$ defines an equivalent martingale measure andConversely this pricing relation under an equivalent martingale measure gives $\nu =R^{-1}dQ/dP$. Without such an account, $\nu$ is a pricing density, not necessarily a probability density or even an integrable random variable by itself; the stated requirement is integrability of $\nu A_1$. This distinction keeps the abstract alternative valid without an unstated asset paying one.