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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.

Choose a fixed vertical datum so the physical free surface is at $H+\zeta$, with $H$ constant; its depth above the bottom is $h=H+\zeta -b$. The gradient of the constant datum does not enter the pressure force. With $D_H={\partial }_t+u{\partial }_x+v{\partial }_y$, the rotating shallow water equations areThe fluid is hydrostatic with depth-independent horizontal velocity at this order. The pressure-gradient force uses the free-surface slope, not the layer-depth slope: $\nabla \zeta =\nabla h+\nabla b$.

Let $q=v_x-u_y$ be relative vertical vorticity, $q_a=f+q$ the absolute vorticity, and $\delta =u_x+v_y$. Taking the horizontal curl of the two momentum equations and expanding the advective derivatives givesMass conservation is $D_{H}h=-h\delta$. The two stretching factors cancel, proving the exact shallow-water potential vorticity conservation lawThe denominator is $h$, the actual column thickness that changes under horizontal convergence; an arbitrary surface elevation $\zeta$ does not obey that column mass equation. There is no $w{\partial }_z$ term because the reduced layer quantities have no vertical dependence. The vertical motion has already supplied the layer-thickness equation, rather than remaining an independent coordinate of this model.

Now take $H=h_{00}$ and assume $|q|\ll |f|$, $|\zeta |,|b|\ll H$. To first order,For a small Rossby number, leading geostrophic flow satisfies $fv=g{\zeta }_x$ and $fu=-g{\zeta }_y$. ThusFor $f\ne 0$, the barotropic deformation radius is $L_R=\sqrt{gH}/|f|$. Multiplication of the first-order PV by $H$ givesThis is the required shallow-water quasi-geostrophic potential vorticity, including the bottom contribution.

Write $S=fϵ/H$ and $L=L_R$. The initial state has $\psi =0$, so ${\mathcal{Q}}_i=f+(f/H)b$. The specified homogenization sets ${\mathcal{Q}}_f=f$ in $|y|<a$ and leaves it unchanged outside. Hence the requested change isFor $f>0$ this is a descending straight segment from $+Sa$ to $-Sa$, with zero exterior values and jumps at the two edges. Endpoint values do not affect the inversion. It is the anomaly, not the full final PV profile, that appears as the source below.

For the zonal mean, $v=0$, $u=-{\psi }^{\prime }(y)$, and potential-vorticity inversion requireswith decaying disturbance at both infinities. The odd source makes the unique decaying solution odd. A particular interior solution is $S{L}^{2}y$, so writeBoth $\psi$ and ${\psi }^{\prime }$ must be continuous at $y=\pm a$: a jump would introduce a delta function or its derivative into a source that has only finite jumps. Matching at $a$ givesand thereforeDifferentiate the interior streamfunction to obtainThe exterior profiles needed for the sketches are $u=(D/L)e^{-(|y|-a)/L}$ and $\zeta =(f/g)\mathrm{sgn}(y)D{e}^{-(|y|-a)/L}$.

For $L\gg a$, put $A=a/L$, $Y=y/L$, so $|Y|\le A$. Uniform Taylor expansion givesConsequentlyFor $f>0$ the leading interior current is westward and approximately parabolic.

For $L\ll a$ and away from the edges, $u\simeq -S{L}^2$ and $\zeta =(f/g)\psi \simeq ϵy$: the surface follows the bottom slope centrally. Near $|y|=a$, the exponentially thin corrections give positive edge jets of width $L$ and peak scale $SLa/2$ when $f>0$. Those jets decay in the exterior. The elevation is odd, has opposite-sign extrema just inside the edges, and decays to zero outside; at the positive edge $\zeta (a)\simeq ϵ(a-L)/2$. The following sketch uses the full matched solution with $L/a=0.08$. For negative $f$, the current and PV-change signs reverse, while this elevation profile does not.

Here write $\sigma$ for the buoyancy perturbation about the resting constant-$N$ stratification; the total buoyancy also includes its background gradient $N^2$. Use kinematic perturbation pressure $p$. The weak-forcing Linearized Boussinesq equations areWeakness means that perturbation advection is small compared with the retained linear terms. With the specified streamfunction convention, the transverse vorticity isDifferentiate the horizontal momentum equation in $z$ and subtract the $x$ derivative of vertical momentum. The pressure derivatives cancel, givingThe second equation uses ${\sigma }_t=N^2{\psi }_x$ and is the forced internal-wave equation.

For free waves at frequency $|\omega |\ll N$, the internal gravity wave dispersion relation gives $|k|/|m|\simeq |\omega |/N\ll 1$. Thus the horizontal response scale is much larger than the vertical scale. Vertical acceleration is small in the vertical momentum equation, leaving the hydrostatic approximation $p_z=\sigma$. Correspondingly the transverse vorticity is predominantly $u_z$: neglect $-w_x={\psi }_{xx}$ relative to ${\psi }_{zz}$ in the inertial vorticity term. The restoring term $N^2{\psi }_{xx}$ must still be retained, since it balances the small time derivatives.

For $m\ne 0$ put $\psi =e^{imz}P(x,t)$, $\sigma =e^{imz}B(x,t)$ and interpret physical fields as real parts. The reduced equation isOne can solve it while obtaining the buoyancy simultaneously. Define $U=imP$ and $H=iB/(cm)$. The first-order hydrostatic equations becomeHence $R_{+}=U+H$ and $R_{-}=U-H$ obeyApply the method of characteristics with zero fields in the past. Withthe solution is $R_{\pm }=I_{\pm }$. Consequently the hydrostatic response to a localized horizontal force isThese are exactly the required two characteristic integrals, derived with their relative sign fixed by the buoyancy equation. Quiet-past initial conditions exclude additional freely propagating solutions.

For the consistency check, set $\tau =s-t$, so both integrals run over the fixed range $(-\infty ,0)$. Differentiation at fixed $x,z$ now givesThis change of variable is important: it includes the time dependence of both the sampled forcing and the moving integration geometry without omitting a term. Each straight path intersects $|x|\le a$ for a time interval of length at most $2a/c$. If $L=\sup |f_{xt}|$, thenThe characteristic size of the hydrostatic buoyancy and pressure-gradient balance is $|m|a{f}_{\max }$, as follows from the analogous bound on $B$ in its integral formula. Thus the omitted acceleration has relative scale at most a constant timesTogether with the low-frequency assumption, this verifies the requested hydrostatic consistency. It is a comparison of characteristic magnitudes, rather than division by a buoyancy value at a node where both terms may vanish. The capital $N$ here is the buoyancy frequency, as printed in the PDF.

For $t\ge 0$, only $0\le s\le t$ contributes. If $|x|>ct+a$, every point $x\mp c(s-t)$ lies outside the source interval, so both integrals vanish:This finite propagation cone belongs to the reduced hydrostatic wave equation.

The hypotheses do not require an even forcing, so no single symmetric graph follows for every permitted $f$. For an illustrative smooth even nonnegative profile, $U$ is even and $H$ is odd. At late time and fixed $x$, if $A_f={\int }_{-a}^{a}f(r,\infty )dr$, thenThus the streamfunction develops a broad nearly level interior response with outgoing fronts, while the buoyancy changes sign across the source and approaches opposite interior plateaus to its left and right. Both vanish beyond the fronts. For $m>0$, a vertical phase with $\sin (mz)=1$ gives real fields $\psi =U/m$ and $\sigma =cmH$. The figure plots those profiles up to positive normalization factors, for an even compact forcing increasing slowly from zero.

Let ${\rho }_0$ be a constant reference density and define total buoyancy acceleration by $\sigma =-g(\rho -{\rho }_0)/{\rho }_0$. Absorb the reference hydrostatic pressure into a kinematic pressure $p$. The ideal nonrotating Boussinesq equations areFor a resting density profile ${\rho }_b(z)$, the background satisfies $p_b^{\prime }(z)={\sigma }_b(z)$ and the squared buoyancy frequency isStable stratification has $N^2>0$. The Boussinesq approximation requires fractional density variations to be small over the domain and parcel excursions, so density can be replaced by ${\rho }_0$ in inertia while its small variation is retained in the gravitational force. For an approximately uniform density gradient over height $H$, this requires $N^{2}H/g\ll 1$. The stipulated incompressibility alone does not justify neglecting density variation in inertia.

Put $b=\sigma -{\sigma }_b$ and $p^{\prime }=p-p_b$. To linear order the two-dimensional internal gravity wave equations areThe exact impermeability condition on $z=\eta (x,t)$ is the kinematic boundary conditionTaylor expansion about $z=0$ permits $w(x,0,t)={\eta }_t$ at leading order if the boundary slope and its displacement relative to the vertical wave scale are small: $|kϵ|\ll 1$ and $|mϵ|\ll 1$. The second condition controls both the omitted evaluation correction $\eta w_z$ and the product $u{\eta }_x$, since continuity gives $u\sim (m/k)w$. Hence for the harmonic forcing,The propagating calculation assumes $k\ne 0$ and $0<|\omega |<N$; a time-independent boundary does no oscillatory work on fluid initially at rest. If $k=0$ but $\omega \ne 0$, incompressibility forces the vertical velocity to be independent of height, so the unbounded column has no finite-energy radiating internal-wave solution. The propagating wave-generation problem therefore needs the stated nonzero horizontal wavenumber.

For a plane mode with phase $\theta =kx+mz-\omega t$, let $U,W,P,B$ be its complex amplitudes. The linear equations giveSubstituting into $-i\omega W=-imP+B$ gives the dispersion relationNo waves enter from above. The appropriate radiation condition therefore selects upward vertical group velocity, not upward phase propagation:The boundary-forced internal gravity wave satisfying the boundary condition isThe displacement is $ϵ\sin \theta$, consistent with $b=-N^2$ times the displacement. For a right-moving boundary choose $k>0$, $\omega >0$ without loss of generality. Then $m<0$, and the constant-phase lines of an internal gravity wave have slope $dz/dx=-k/m>0$: the crests rise toward the right. Their acute angle with the horizontal isThe group velocity is along these crests in the upward-right direction, whereas the wavevector and phase propagation point downward-right.

The physical pressure perturbation is ${\rho }_0{p}^{\prime }$. Pressure work at the lower boundary, to second order in amplitude, is its period-mean pressure times vertical boundary velocity. Static hydrostatic terms average to zero over a cycle. ThereforeAs an independent energy check, the wave's mean kinetic-plus-buoyancy energy density is ${\rho }_0{ϵ}^2{N}^2/2$, and multiplication by $c_{gz}$ gives exactly this upward energy flux.

For the exact horizontal mean-momentum equation, incompressibility writes the horizontal momentum equation asAverage over a horizontal period. Require the pressure and velocity fluxes to be periodic, or more generally their horizontal derivative averages to vanish; in particular there must be no imposed mean pressure gradient. ThenHorizontal averaging also makes ${\partial }_z\overline{w}=0$; the zero mean boundary displacement rate gives $\overline{w}=0$. These conditions exclude an independent mean mass flux or horizontal force. The mean equation applies in the fluid, with flattening of the small moving boundary understood at the order used above.

For slow switch-on, define $A(z,t)=ϵ(t-z/c_{gz})$, using $ϵ(s)=0$ for $s<0$. The outgoing internal-wave envelope radiation condition is $A_t+c_{gz}{A}_z=0$. Replace $ϵ$ in the oscillatory fields by $A$. A modulation time $T$ must satisfy $|\omega |T\gg 1$ and $|m|c_{gz}T\gg 1$, with a frequency bandwidth sufficiently narrow to linearize the dispersion relation. The exact boundary derivative also contains $\dot{ϵ}\sin (kx-\omega t)$; it is smaller by the slow-modulation parameter and is beyond this leading envelope approximation.

To leading slow-modulation order, the wave momentum flux isSince ${\partial }_z{A}^2=-c_{gz}^{-1}{\partial }_t{A}^2$, integrating the mean-momentum equation from rest gives the internal-wave momentum depositionThis is second order in wave amplitude, with smaller relative slow-modulation corrections. Mean-flow feedback on the leading wave is higher order. The mean current is created as the wave envelope arrives; an already uniform wavetrain has no local flux divergence.

Finally move horizontally with speed $c_p=\omega /k$. The initial uniform relative flow is $-c_p$, with infinite total energy in the unbounded column. Its finite mean-energy change is insteadAt finite time the envelope is absent sufficiently far above, so integrating the momentum equation givesThis is the boundary work and mean-flow energy of an internal wave relation. In the moving frame the boundary pattern is stationary to leading modulation order. It extracts energy from the incident mean flow and converts it to outgoing internal-wave energy; in the laboratory frame the same energy is supplied as boundary work.

In infinitesimal elasticity and plasticity in continuum mechanics, decompose the symmetric strain as $e=e^e+e^p$. For incompressible isotropic elasticity, ${\sigma }^{\prime }=2G{e}^e$, with $G$ the shear modulus and both strains traceless. The associated flow rule for the Von Mises yield criterion has ${\dot{e}}^p=\Lambda {\sigma }^{\prime }$ after absorbing a positive factor into $\Lambda \ge 0$. Plane strain has $e_{33}=0$, soIf ${\sigma }_{33}^{\prime }=0$ initially, it remains zero by this homogeneous scalar differential equation. Then $e_{33}^e=0$ and ${\dot{e}}_{33}^p=0$, so $e_{33}^p=0$ is consistent with zero initial plastic component. This proves consistency, not a claim that arbitrary prestressed plane-strain histories must have zero separate components. The out-of-plane shear components can consistently be zero by the same argument.

Write $p=-\mathrm{tr}\sigma /3$, so ${\sigma }_{33}=-p$. The proposed in-plane parameterization hasIts squared contraction is $2k^2({\sin }^{2}2\varphi +{\cos }^{2}2\varphi )=2k^2$, proving the yield condition identically. Its principal deviatoric stresses are $k,-k,0$. Define the two slip line families by tangent unit vectorsThus $\alpha$ lines have tangent angle $\varphi$ and $\beta$ lines have tangent angle $\varphi +\pi /2$. Their normal extensions from the associated plastic strain rate vanish, while their shear stresses have maximum magnitude $k$; they are not the principal stress directions.

In the absence of body force, differentiate the stress parameterization in the two equilibrium equations. One obtainsThe symmetric matrix multiplying $2k\nabla \varphi$ has eigenvectors $a,b$ with eigenvalues $+1,-1$. Taking its contraction with these vectors provesHence the Hencky stress relations areConstants can differ between lines, and these signs depend on the specified negative shear-stress convention.

Place the notch tip at the origin, with the opening towards negative $x_1$ and tensile loading in the transverse direction. For a usual reentrant notch, $0\le \gamma \le \pi /2$. The upper face has tangent angle ${\vartheta }_f=\pi -\gamma$. Choose its tangent $t=(\cos {\vartheta }_f,\sin {\vartheta }_f)$ directed away from the tip and outward material normal $n=(-\sin {\vartheta }_f,\cos {\vartheta }_f)$ pointing into the notch. Direct rotation of the stress gives the normal and tangential traction componentsTake the notch faces traction-free, as intended by the notch problem. Then the cosine vanishes. Select the tensile branch $\sin (2\varphi -2{\vartheta }_f)=1$, for which the face-tangential normal stress is $2k$. Thus the upper face has $\boxed{p=-k,{\varphi }_f=\pi /4-\gamma (\mathrm{mod}\pi )}$. Choosing the opposite branch would describe compression rather than the tensile field.

Construct the centred slip-line fan at a traction-free V-notch to connect this surface state to the forward symmetry axis. Near the upper face there is a constant-stress region. It joins a fan bounded by raysInside the fan choose $\beta$ lines radial and $\alpha$ lines circular about the tip, so their angle parameter is $\varphi =\vartheta -\pi /2$. Along a circular $\alpha$ line the first invariant givesThe second invariant is automatically constant along each radial $\beta$ line because $p,\varphi$ are independent of radius. Thus this is an actual equilibrium fan, not just an assumed transport of a surface value. At ${\vartheta }_2$ it matches the free-face constant state. At ${\vartheta }_1$ it matches a forward constant-stress region withReflect the upper field across the symmetry axis to obtain the lower field. The forward region has zero shear, respects this symmetry, and contains the positive $x_1$ axis. There $\sin 2\varphi =-1$, and henceThis value holds in the fully plastic notch-tip region described by the field. It is not a uniform stress assertion for the entire axis of an arbitrary finite specimen, including any elastic or far-field region.

Differentiate the pulled-back Kirchhoff stress tensor $T=F^T\tau F$, using the velocity gradient convention $L=\dot{F}{F}^{-1}$:Thus the lower-convected stress pull-back givesFor an explicit objective time derivative check, superpose $x^{\ast }=Q(t)x+c(t)$, with $Q^{T}Q=I$ and $\Omega =\dot{Q}{Q}^T$ skew. Then ${\tau }^{\ast }=Q\tau Q^T$ and $L^{\ast }=QL{Q}^T+\Omega$, whereasThe additional terms in $(L^{\ast }{)}^T{\tau }^{\ast }+{\tau }^{\ast }{L}^{\ast }$ are $-\Omega {\tau }^{\ast }+{\tau }^{\ast }\Omega$. They cancel the extra material-derivative terms, proving $\boxed{{\mathcal{D}}_l{\tau }^{\ast }=Q({\mathcal{D}}_l\tau )Q^T}$. The translation has no effect on a stress or velocity gradient.

For the incompressible Oldroyd-A model, avoid using the same symbol for a stress tensor and its relaxation time by referring to $\tau >0$ below only as the scalar relaxation time. Define $S={\sigma }^d-2{\mu }_{r}D$. The constitutive equation becomesBy the pull-back identity, $A=F^{T}SF$ satisfiesAn integrating factor givesPush forward by $F(t{)}^{-T}$ and $F(t{)}^{-1}$. For a convergent, fading remote-past history with the transported homogeneous term tending to zero as $t_0\to -\infty$, this proves the fading-memory representation of the Oldroyd-A model:For arbitrary finite-time initial data retain $e^{-(t-t_0)/\tau }F(t{)}^{-T}A(t_0)F(t{)}^{-1}$. Thus the printed integral-only expression incorporates an initial-memory condition; it is not an unconditional replacement for all initial-value solutions.

For time-dependent simple shear,Put $\Delta \gamma =\gamma (s)-\gamma (t)$. The relative deformation is $F(s)F(t{)}^{-1}=\left(\begin{array}{cc}1& \Delta \gamma \\ 0& 1\end{array}\right)$, so the entire transported integrand simplifies toConsequently the explicit nonpressure stresses areSymmetry gives ${\sigma }_{21}^d={\sigma }_{12}^d$, and the unshown third-direction nonpressure components vanish. The superscript $d$ here labels the paper's nonpressure constitutive stress; it need not have zero trace. Subtracting $pI$ supplies the total Cauchy stress tensor.

For constant shear rate $s_0$, use $\gamma (s)-\gamma (t)=-s_0(t-s)$ and the integrals ${\int }_0^{\infty }{e}^{-a/\tau }da=\tau$ and ${\int }_0^{\infty }a{e}^{-a/\tau }da={\tau }^2$. This gives the steady simple shear of an Oldroyd-A fluid resultand thereforeThe incompressibility pressure cancels from this normal-stress difference. With the usual $\mu \ge {\mu }_r\ge 0$, it is nonnegative, while the second normal-stress difference equals its negative.

For fixed nominal boundary tractions and dead body forces, use the total potentialThe dead-load terms are linear in the deformation map. Let $x(\epsilon )$ be a constraint-preserving perturbation path through equilibrium, with $H=\delta F$ and $K={\delta }^{2}F$. Differentiating $\varphi (F)=0$ twice gives ${\varphi }_{,F}:H=0$ and ${\varphi }_{,F}:K=-{\varphi }_{,FF}[H,H]$. Equilibrium has nominal stress tensor $P_{Ii}=W_{,F_{iI}}+q{\varphi }_{,F_{iI}}$. Its virtual-work relation, applied to ${\delta }^{2}x$, gives the otherwise remaining first-derivative contribution ${\Pi }^{\prime }[{\delta }^{2}x]=-\int q{\varphi }_{,F}:KdX$. Therefore the constrained second variation isFor a time-dependent linearized perturbation $y$, the incremental reaction stress contains $\delta q{\varphi }_{,F}$, but its power vanishes because ${\varphi }_{,F}:\nabla \dot{y}=0$. Multiply the incremental momentum equation by $\dot{y}$, integrate by parts, and use zero incremental dead-load traction or the prescribed displacement condition. One obtainsPositive $\mathcal{Q}$ is therefore the standard constrained second-variation stability in elasticity criterion: perturbations cannot grow in this quadratic energy norm. A statement in a specified displacement norm also uses the usual coercivity and boundary conditions removing uncontrolled rigid-motion modes; translations with $\delta F=0$ are not tested by the displayed condition. This is a small-perturbation energy argument, not a claim of global nonlinear stability.

For the cube use incompressibility in the form $\varphi =\det F-1$. A neo-Hookean solid has $W_{,F}=\mu F$, and hence the spatial-index-first transpose of nominal stress is $\mu F+q{F}^{-T}$. On the diagonal family, equality to $TI$ givesThe first relation immediately gives $\boxed{q=T\lambda -\mu {\lambda }^2}$. Eliminating $q$ and factoring yieldsThus $\boxed{\lambda =1\text{or}T/\mu =\lambda +{\lambda }^{-1/2}}$.