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For the convention $e^{i\omega t}$, complex viscosity is the ratio of oscillatory shear stress amplitude to shear rate amplitude. A causal relaxation modulus gives the displayed transform. Its associated complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, where $G^{\prime }$ is the storage modulus and $G^{\prime \prime }$ the loss modulus.

Linearizing the differential pom-pom model about its relaxed state leaves no first-order stretch perturbation in incompressible flow. Its anisotropic orientation obeys $\dot{a}+a/{\tau }_1=A_1$, producing the displayed relaxation modulus. In a slow-flow expansion, $A=I+{\tau }_1{A}_1+{\tau }_1^2(2A_1^2-A_2)+O(3)$. Trace normalization and the first stretch correction affect only isotropic stress at this order, giving the displayed second-order fluid coefficients. Linear small-amplitude response allows arbitrary frequency, whereas this local second-order expansion also requires slow time dependence.

A linear viscoelastic fluid has causal stress response linear in the deformation history, often expressed as ${\sigma }^d(t)=2{\int }_0^{\infty }G(s)E(t-s)ds$. The approximation requires small deformation and rotation over the relevant memory times. High frequency is allowed at sufficiently small amplitude. An integrable decaying relaxation modulus gives a finite zero-shear viscosity; an instantaneous solvent term may be written separately.

The constitutive convolution is the response of a linear viscoelastic fluid close to its relaxed state. Deformations and rotations during the material's memory time must be small, so the stress is linear in the deformation history and evaluating that history at a fixed position is consistent with linearization about rest. For an isotropic incompressible flow, the deviatoric stress is proportional to the history of the rate-of-strain tensor; an arbitrary isotropic pressure is separate. Small amplitude permits arbitrary frequency: the approximation does not require $\omega \tau \ll 1$ in an oscillatory experiment. Large accumulated deformation, substantial advection of memory, and nonlinear normal-stress differences are outside this formula's scope.

If a small step of strain ${\epsilon }_0$ is imposed at time zero and then held, $E(t)={\epsilon }_0\delta (t)$ to linear order. The convolution consequently gives ${\sigma }^{\prime }(t)=2G(t){\epsilon }_0$ for $t>0$. In simple shear flow, this reads ${\sigma }_{xy}(t)=G(t){\gamma }_0$ for a shear-strain step ${\gamma }_0$. Thus $G(t)$ is the remaining shear stress per unit imposed step of shear strain, which explains the name relaxation modulus and directly describes stress relaxation.

Use the convention $E(t)=\mathrm{Re}[\hat{E}{e}^{i\omega t}]$. Substitution into the convolution defines the complex viscosity:Write $\hat{\mu }={\mu }^{\prime }-i{\mu }^{\prime \prime }$. Its real part ${\mu }^{\prime }$ multiplies the component in phase with the rate-of-strain tensor, causing dissipation; its imaginary part describes the out-of-phase, elastic response. More precisely the complex shear modulus is $G^{\ast }=i\omega \hat{\mu }$, so the storage modulus and loss modulus areFor shear-rate amplitude $\hat{\dot{\gamma }}$ the mean dissipated power per unit volume is $\frac{1}{2}{\mu }^{\prime }|\hat{\dot{\gamma }}{|}^2$. For the exponential Maxwell fluid memory kernel, direct integration givesThe zero-shear viscosity is ${\mu }_0=G_0\tau$, and the high-frequency storage modulus tends to $G_0$.

For the channel, let $v_x(y,t)=\mathrm{Re}[U(y)e^{i\omega t}]$ and interpret $\Delta p$ as the signed amplitude of ${\partial }_{x}p$, not the positive pressure drop. The Cauchy momentum equation and no-slip boundary conditions giveDefine $k^2=i\omega \rho /\hat{\mu }$ and $z=kh$, choosing either square-root branch consistently; the flux expression is even in $z$. The symmetric solution isIts integral across the channel gives the oscillatory channel flux of a linear Maxwell fluid:This includes inertia while using the linear constitutive response. The pressure-gradient amplitude must be small enough that the resulting deformation stays in that response regime.

As $\omega \to 0$, $\tanh z/z=1-z^2/3+2z^4/15+\cdots$. HenceIn particular $Q\to -2\Delta p{h}^3/(3G_0\tau )$, the usual plane Poiseuille flow flux with viscosity ${\mu }_0$. The fluid has enough time to relax and behaves as a Newtonian fluid at leading order.

For fixed positive $h,\tau ,\rho ,G_0$, let $c=\sqrt{G_0/\rho }$. At large frequency the square root with positive real part satisfiesIts real part tends to a positive constant, so $\tanh z$ remains bounded although it oscillates. Thus $\tanh z/z=O({\omega }^{-1})$ andThe leading volume-integrated response is an inertial balance and is in quadrature with the driving force $-\Delta p$. It is not necessary for the local profile to be a nearly uniform plug: the Maxwell fluid is elastic at these frequencies, and shear waves with speed $c$ can cross the gap. The oscillatory part of $U$ largely cancels on integration. Small attenuation allows resonance-like enhancements at finite frequencies; the leading flux limit still holds for fixed nonzero relaxation and gap parameters. A Newtonian thin viscous-layer argument would not justify the Maxwell profile here.

Finally take the different ordering $\omega \tau \gg 1$ but $\rho {\omega }^2{h}^2/G_0\ll 1$. Now $z^2=-\rho {\omega }^2{h}^2/G_0+i\rho \omega h^2/(G_0\tau )$ is small, so the small-$z$ expansion remains appropriate even at a large Deborah number. The result isHere inertia is negligible throughout the narrow gap. Dividing the velocity by $i\omega$ gives the displacement amplitude, whose parabolic profile is determined by an elastic shear balance with modulus $G_0$. The velocity therefore grows like frequency for a fixed displacement response. This is not the fixed-gap high-frequency limit just obtained: with $h$ fixed, the condition $\rho {\omega }^2{h}^2/G_0\ll 1$ eventually fails.

A simple fluid reduces to a linear viscoelastic fluid when its deformation and rotation over the relevant memory interval are small enough to linearize the constitutive history about an isotropic relaxed state. Small amplitude is essential; small frequency is not. For example, arbitrarily high-frequency oscillation can still be linear if its strain amplitude is sufficiently small. The relaxation modulus $G(t)$ is the shear-stress response to a unit small step of shear strain.

Choose the harmonic convention $e^{i\omega t}$. If the imposed shear rate is $\mathrm{Re}(\hat{\dot{\gamma }}{e}^{i\omega t})$, the corresponding shear stress is $\mathrm{Re}({\eta }^{\ast }(\omega )\hat{\dot{\gamma }}{e}^{i\omega t})$. Substitution into the constitutive convolution gives the complex viscosity:For a nondecaying elastic part this formula is understood through the limit from a positive Laplace damping parameter. The complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, so $G^{\prime }$ is the storage modulus, $G^{\prime \prime }$ the loss modulus, and ${\eta }^{\ast }=G^{\prime \prime }/\omega -i{G}^{\prime }/\omega$ in this convention.

Creep is the progressive deformation produced by a held stress. For a small step $S_{0}H(t)$ of shear stress, define the creep compliance $J$ by $\gamma (t)=S_{0}J(t)$ from a relaxed state. Writing hats for Laplace transforms, the same constitutive law gives $\hat{S}=p\hat{G}\hat{\gamma }$. Since $\hat{S}=S_0/p$, this yieldsHere $H$ is the Heaviside step function. Inversion determines the full creep response, including any instantaneous elastic strain jump. Equivalently, the causal relaxation/creep convolution includes that jump through ${\int }_{0^{-}}^{t}G(t-s)dJ(s)=1$ for $t>0$.

For the Maxwell fluid, $\hat{G}(p)=G_0/(p+1/\tau )$. Thus the two material responses areThe creep response is an elastic jump followed by constant-rate viscous deformation. The zero-frequency shear viscosity is $\eta =G_0\tau$.

For the wall problem, take $v=u(y,t)e_x$ and denote the shear component of stress by $S$. The incompressibility constraint is satisfied, and the advective part of the material derivative vanishes because no field depends on $x$. With no imposed longitudinal pressure gradient, momentum conservation and the linear Maxwell law giveThe initial conditions are $u=S=0$ in $y>0$; the no-slip boundary condition is $u(0,t)=UH(t)$, with decay as $y\to \infty$. Transforming these equations givesChoose the analytic square-root branch in $\mathrm{Re}p>0$ with $\mathrm{Re}q>0$, so that the spatially increasing solution is excluded. The Maxwell start-up shear layer consequently hasThe Bromwich inversion formula now givesThe vertical Bromwich contour is directed upward and lies to the right of all singularities, including the origin and the branch points at $0$ and $-1/\tau$. A cut joining these branch points is a possible continuation into the left half-plane; inversion only requires the stated right-half-plane branch.

The Newtonian limit holds $\eta =G_0\tau$ fixed while $\tau \to 0$, so $G_0\to \infty$. Then $q(p)\to \sqrt{\rho p/\eta }$. With $\nu =\eta /\rho$, the inverse Laplace transform gives the complementary error function profileThis is diffusive momentum penetration in a Newtonian fluid, with thickness of order $\sqrt{\nu t}$.

The elastic limit instead holds $G_0$ fixed while $\tau \to \infty$. Write $c=\sqrt{G_0/\rho }$. On the chosen branch, $q(p)\to p/c$, so the Laplace transform time-shift rule yieldsAn elastic shear front travels at speed $c$: the material ahead is at rest, while that behind moves at the wall speed. Indeed the limiting displacement is $d(y,t)=U(t-y/c)H(t-y/c)$, whose shear strain is $d_y=-(U/c)H(t-y/c)$. A nonzero pointwise velocity gradient at the front is interpreted distributionally, or by smoothing the ideal wall start.

The use of linear viscoelasticity is conditional, rather than automatically justified for arbitrary $U$. The material derivative simplifying to a time derivative does not remove the nonlinear constitutive effects of finite shear strain. In this Maxwell problem the rapid elastic response supplies the natural strain scale $U/c=U\sqrt{\rho /G_0}$, which must be small for linearization through start-up. Away from the front, the strain accumulated over an effective memory interval must likewise be small. In the late diffusive regime its characteristic size is $\tau U/\sqrt{\nu t}$, which decreases with time. Large absolute wall displacement $Ut$ alone is not a disqualification for a fluid with finite memory; it is relative deformation during that memory that matters. In the elastic limit, where memory does not decay, the explicit strain behind the front confirms the condition $U/c\ll 1$. Without such a small-strain restriction, the computed field is a formal linear response and need not describe an arbitrary simple fluid.

At rest the tensor equation relaxes $A$ toward $I$ and the stretch equation relaxes $\lambda$ toward one. ThusThe equilibrium nonpressure stress $GI/3$ is isotropic and can be absorbed into the arbitrary pressure in incompressible flow.

For the linear response, put $A=I+a+O({\epsilon }^2)$, $\lambda =1+\ell +O({\epsilon }^2)$ and take the velocity gradient to be $O(\epsilon )$. The first-order tensor equation isIts matrix trace has no forcing, because incompressibility gives $\mathrm{tr}K=0$. Starting relaxed, $\mathrm{tr}a=0$. The first-order stretch equation is $D_t\ell +\ell /{\tau }_2=(I/3):K=0$, hence $\ell =0$. Consequently $B=I/3+a/3+O({\epsilon }^2)$, and its anisotropic stress is $Ga/3$. From the relaxed remote past,The linear and second-order response of the differential pom-pom model therefore hasThe stretch time ${\tau }_2$ does not enter the linear shear relaxation modulus.

For the second-order fluid constants, additionally make the slow-flow expansion: count $K$ as first order and $D_t$ as raising the order by one. IteratinggivesUsing the Rivlin-Ericksen tensor convention $A_2=D_t{A}_1+A_{1}K+K^T{A}_1$, the identityreduces this to $A=I+{\tau }_1{A}_1+{\tau }_1^2(2A_1^2-A_2)+O(3)$. Since $\mathrm{tr}{A}_2=\mathrm{tr}{A}_1^2$, normalization givesStretch first changes at second order. Its equation gives $\lambda -1={\tau }_1{\tau }_2\mathrm{tr}(A_1^2)/6+O(3)$. Multiplication by ${\lambda }^2$ therefore adds only an isotropic stress at second order; multiplying its second-order increment by the anisotropic first-order term would be third order. Absorb these isotropic corrections into $p_{\ast }$ to obtainComparison with the viscometric functions parametrization yieldsEquivalently, in $\sigma =-pI+{\mu }_0{A}_1+{\alpha }_1{A}_2+{\alpha }_2{A}_1^2$, the constants are ${\alpha }_1=-G{\tau }_1^2/3$ and ${\alpha }_2=2G{\tau }_1^2/3$. The general small-amplitude linear response above permits fast oscillation; the local second-order fluid expansion also requires the slow-flow ordering just used.

For a linear viscoelastic fluid, causal superposition givesThe relaxation modulus $G(s)$ is the stress response to a unit small step of shear strain. The equation assumes an isotropic reference state and deformations and rotations small enough over the significant memory times that nonlinear transport and strain products can be neglected. Small-amplitude motion can have high frequency; linearity does not require low frequency by itself. The infinite-past integral encodes the preceding history, rather than assuming that stress vanishes whenever the present rate vanishes.

For a small constant shear rate maintained from the remote past, the rate can be taken outside the integral. Therefore the zero-shear viscosity isprovided the kernel is integrable, with any instantaneous contribution counted fully. A nondecaying solid-like modulus would not yield a finite fluid viscosity.

To linearize the stated Oldroyd-B model, put $A=I+C$ and absorb the isotropic $G_{0}I$ into pressure. The products of the small deformation tensor C and velocity gradient are second order. Using the paper's derivative-index-first gradient convention therefore givesAn integrating factor, with the prescribed preceding history, gives $C(t)=2{\int }_0^{\infty }{e}^{-s/\tau }E(t-s)ds$. Thus the linear nonpressure stress isThe regular relaxation tail is $G_0{e}^{-s/\tau }$, while the solvent is instantaneous. Equivalently, the instantaneous solvent term in a relaxation modulus can be represented byHere the causal endpoint delta is defined by ${\int }_0^{\infty }{\delta }_{+}(s)f(s)ds=f(0)$. If instead a symmetric Dirac delta is assigned half its mass at the endpoint, the same solvent term is written $2{\mu }_0\delta (s)$. Stating the convention prevents a factor-of-two error in the viscosity; the explicit stress formula above has no such ambiguity.

In a cone-and-plate rheometer, the small-angle gap is $\alpha r$ and the local wall speed is $\Omega r$, so the shear rate is independent of radius to leading order: $g(t)=\Omega (t)/\alpha$. Write $s(t)$ for shear stress and $P(t)$ for its polymer part. The linear constitutive response isThe uniform shear stress gives torqueFor the initial steady torque T, put $s_{+}=3T/(2\pi a^3)$ and ${\mu }_p=G_0\tau$, $\mu ={\mu }_0+{\mu }_p$. The steady polymer stress is $P_{+}={\mu }_p{g}_{+}$, with $g_{+}=s_{+}/\mu$. Hence the initial angular velocity is

For $t>0$, the total stress is $s=-s_{+}$, but the polymer stress is continuous at the torque reversal. Since ${\mu }_0>0$, a finite jump in shear rate is possible even though P cannot jump. Its immediate value isEliminating P from the constant-stress equations for $t>0$ givesThe decay time is the Retardation time of an Oldroyd fluid ${\lambda }_2={\mu }_0\tau /({\mu }_0+{\mu }_p)$. Solving this first-order equation with the actual initial value yields the complete torque reversal of a linear Oldroyd fluid:In particular, $\Omega (\infty )=-{\Omega }_{+}$ but $\Omega (0^{+})$ is more negative. The stored polymer stress initially retains its old positive sign, requiring a larger negative solvent shear stress to attain the new negative total torque. This produces the backward-speed overshoot and its subsequent relaxation. Apparatus inertia would smooth the instantaneous jump, but is neglected in the question. The torque must be sufficiently small for the entire transient, including its enhanced initial rate, to remain in the linear viscoelastic regime; small pre-reversal rate alone is not sufficient when $G_0\tau /{\mu }_0$ is large.

The affine linear PTT fluid considered here uses a symmetric dimensionless tensor $A$ with $\stackrel{\nabla }{A}+(1+\alpha \mathrm{tr}A)A/\tau =2E$ and polymer stress $G_{0}A$. Here $\stackrel{\nabla }{A}$ is the upper-convected derivative, $G_0,\tau >0$ and the nonlinear relaxation parameter $\alpha$ is usually positive. Linearizing about rest gives the Maxwell fluid relaxation modulus $G(s)=G_0{e}^{-s/\tau }$. The PTT family also admits other relaxation functions and nonaffine derivatives; their nonlinear predictions need not equal those of this specified version.

Stress relaxation is decay of stored stress after a deformation is imposed and then held fixed. It probes a material's memory through its relaxation modulus. An instantaneous generalized Newtonian fluid has no stored shear stress once the shear rate is zero and cannot describe a nontrivial relaxation tail.

For an integrable relaxation modulus, a small constant shear rate sustained from the remote past gives stress equal to that rate times the displayed integral, including any instantaneous solvent mass. For the linear Oldroyd-B model it is ${\mu }_0+G_0\tau$. A nondecaying elastic modulus would make this integral infinite rather than describe a fluid with finite steady viscosity.