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For real $C$, this analytic homogeneous field has real boundary values on both faces of the negative-axis crack, so it adds no face shear traction. Ahead of the tip it adds stress $-\mu C/\sqrt{x}$. Its stress tends to zero at infinity, but its primitive grows like $\sqrt{z}$, producing unbounded remote displacement and infinite total elastic energy. Thus merely requiring vanishing far stress does not uniquely fix the face-loaded crack solution. Excluding an added remote stress-intensity field, for example through the localized-load bounded-displacement far condition, removes this homogeneous ambiguity. Ordinary inverse-square-root tip stress remains allowed and has finite energy in any bounded neighbourhood of the tip.

The hydrodynamic torque on a body about a reference point is $T={\int }_{S}r\times (\sigma \cdot n)dS$, with $r$ measured from that point and $n$ pointing from the body into the fluid. The stress includes pressure and viscous contributions, and $\sigma \cdot n$ is the fluid traction on the body. Changing the reference point by $d$ changes the torque by $-d\times F$, where $F={\int }_S\sigma \cdot ndS$ is the fluid force. A torque-free body has zero resultant torque, including any external torque.

A hyperelastic material derives its elastic stress from a strain energy density. In reference-volume units its first Piola-Kirchhoff stress tensor is the derivative of that density with respect to the deformation gradient.

The two minor symmetries of the elastic stiffness tensor make it a map between six-dimensional spaces of symmetric strain and stress. An elastic energy gives major symmetry, so the matrix of this map is symmetric and has 21 independent entries. For a symmetry operation acting on strains with parity spaces of dimensions $m$ and $6-m$, invariance eliminates the mixed block and leaves $m(m+1)/2+(6-m)(7-m)/2$ independent coefficients.

Using the extensional screening length of a slumping column, an initially uniform column of length $\Lambda$ obeys $W_{ZZ}=1+W$, $W(0)=0$ and $W_Z(\Lambda )=0$. Thus $W=\cosh (\Lambda -Z)/\cosh \Lambda -1$ and its initial area-fraction growth is ${\varphi }_T={\varphi }_0\sinh (\Lambda -Z)/\cosh \Lambda$. For $\Lambda \gg 1$, the velocity is $e^{-Z}-1$ outside exponentially small end corrections; for $\Lambda \ll 1$, it is $Z^2/2-\Lambda Z$. The first limit balances excess weight with annular hydraulic drag, the second with extensional viscous stress.

The Kirchhoff tensor is the Cauchy stress tensor multiplied by the local volume ratio. Its stress power is per reference volume. Pushing forward the second Piola-Kirchhoff stress tensor gives the displayed formula.

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.

Apply the divergence theorem to ${\partial }_k(x_i{\sigma }_{jk})$ and use momentum balance ${\sigma }_{jk,k}=\rho (a_j-b_j)$ and stress symmetry to obtain the displayed mean stress identity. Applying the same theorem to the infinitesimal strain tensor gives $V{\overline{e}}_{ij}=\frac{1}{2}{\int }_S(u_i{n}_j+u_j{n}_i)dS$. The acceleration term cannot be dropped during a transient.

Let a convex strain energy density have derivative equal to the actual stress tensor. Under mixed displacement and traction data, any equilibrated stress ${\sigma }_0$ with the prescribed body force and boundary tractions represents the external work. For any compatible trial strain ${\epsilon }^{\prime }$, the convex supporting-plane inequality gives $W({\epsilon }^{\prime })-W(\epsilon )\ge \sigma :({\epsilon }^{\prime }-\epsilon )$. The integral of $(\sigma -{\sigma }_0):({\epsilon }^{\prime }-\epsilon )$ vanishes by integration by parts. Thus the actual strain minimizes $\int (W-{\sigma }_0:\epsilon )$.

For an incompressible neo-Hookean solid, a convenient strain energy density is $W=\mu (\mathrm{tr}C-3)/2$, where $C$ is the right Cauchy-Green deformation tensor. An additive constant in $W$ does not affect the stress. The shear modulus $\mu$ fixes the small-strain stiffness.

A Newtonian fluid has viscous stress proportional to the local rate-of-strain tensor, with coefficients independent of that rate. For an isotropic fluid these coefficients are viscosities. Its Newtonian fluid stress tensor specifies the corresponding constitutive relation; the material and its stress tensor are distinct concepts.

For isotropic incompressible flow of a Newtonian fluid, the stress tensor is the displayed expression, with pressure $p$, dynamic viscosity $\mu$, velocity $\mathbf{u}$, and identity matrix $I$. In Couette flow, its off-diagonal component is the shear stress.

Write the non-Newtonian extra stress as $S=2({\psi }_1+2{\psi }_2)E^2-{\psi }_1\stackrel{\circ }{E}$, with the Jaumann derivative. For a steady Stokes flow, $\nabla \times \nabla \cdot \stackrel{\circ }{E}=0$. If ${\psi }_1+2{\psi }_2=0$, the extra force is a gradient and only alters pressure. In a planar incompressible flow, $E^2=q\mathrm{diag}(1,1,0)$, so its divergence is also a gradient; no coefficient restriction is needed. A Newtonian velocity with prescribed boundary velocities consequently solves the second-order model with an altered pressure. On the regular weak-flow branch, Uniqueness of Stokes flow makes each velocity correction vanish. This does not establish uniqueness of arbitrary large nonlinear branches or the absence of higher-order effects in a real material.

Here the convention places the material index first: $P_{Ii}$. Its stress power per reference volume is $P_{Ii}{\dot{F}}_{iI}$. It is the transpose of the first Piola-Kirchhoff stress tensor; some texts instead call that first tensor itself nominal stress.

For gap $h\ll \ell$ and speed $U$, shear stress scales as $\mu U/h$ and viscoelastic normal stresses as ${\psi }_i{U}^2/h^2$. Their force contributions scale respectively as $\mu U/h^2$ and ${\psi }_i{U}^2/(h^2\ell )$. Thus the displayed criterion permits neglecting the normal-stress corrections in the leading lubrication approximation. It is weaker than requiring the normal stresses themselves to be smaller than the shear stress. Rapid time dependence can impose an additional material-time criterion.

Use material coordinates $X$ and write the placement as $y=X+u$. The strain energy density $W$ in the printed convention is per unit reference mass. It is convenient to use the conventional first Piola-Kirchhoff stress tensor $P=s^{\mathsf{T}}$, soAssume the undeformed reference state is stress-free. With $H=A-I$ and infinitesimal strain tensor $e=(H+H^{\mathsf{T}})/2$, material isotropy and material frame indifference give the quadratic expansionConsequently the Lamé parameters from hyperelastic energy Hessian areIndeed, differentiating the quadratic expression twice givesthe elastic stiffness tensor of linear elasticity. A prestressed reference state requires the corresponding initial-stress corrections; the displayed formula concerns the natural state.

For a plane elastic wave, $u=u(X_1,t)$, put $r=e_1+u_{,X_1}$ and $\Psi (r)=W([r,e_2,e_3])$. Balance of momentum without body force becomesThis is a material-coordinate equation: the reference density cancels between inertia and nominal stress. If $C{a}_{\alpha }=c_{\alpha }^2{a}_{\alpha }$, the first-order system has characteristic speeds $\pm c_{\alpha }$ and right eigenvectors $(a_{\alpha },\mp c_{\alpha }{a}_{\alpha })$. Along a characteristic curve with $d{X}_1/dt=s=\pm c_{\alpha }$, its differential compatibility relation isThis follows by multiplying the system by the left eigenvector $(-s{a}_{\alpha }^{\mathsf{T}},a_{\alpha }^{\mathsf{T}})$. It does not assert the existence of globally integrable Riemann invariants for an arbitrary three-component constitutive law.

There is a necessary qualification to the finite-amplitude transverse request. Set $u=(0,w,0)$ and $q=w_{,X_1}$, so $r=(1,q,0)$. The longitudinal momentum equation requiresThus arbitrary nonuniform pure transverse profiles are possible only if the compatibility condition for pure nonlinear transverse waves holds:The third component vanishes by the reflection symmetry implied by isotropy and objectivity. The shear equation then closes:

General compressible hyperelastic material assumptions alone do not imply the required compatibility. For a concrete counterexample, take the Saint Venant–Kirchhoff hyperelastic energyFor $A=[(a,q,0{)}^{\mathsf{T}},e_2,e_3]$, this becomesAt $a=1$, $P_{11}=(\lambda +2\mu )q^2/2$. A varying shear therefore produces a longitudinal force $(\lambda +2\mu )q{q}_{,X_1}$ and cannot preserve $u_1=0$. Infinitesimal transverse waves do exist, because this coupling is second order, but a general pure finite-amplitude transverse wave does not.

The compatible nonlinear shear energy class is nonempty and can be genuinely nonlinear. Define $J=\det A$, $S=\mathrm{tr}(A^{\mathsf{T}}A)-2J$, and, for $h\ge 0$, takeThis is objective and isotropic. Near $A=I$, $S-1=2\mathrm{tr}(e^2)-(\mathrm{tr}e{)}^2+O(|H{|}^3)$, so it has the specified Lamé parameters. On the plane deformation above, $S-1=(a-1{)}^2+q^2$ and $J=a$. Hence $P_{11}=0$ at $a=1$, whileFor positive $\lambda ,\mu$, the relevant incremental plane moduli are positive. This supplies an explicit stable pure transverse family and, when $h>0$, amplitude-dependent wave speed.

For any compatible family, letThe equations $q_t=v_{,X_1}$ and $v_t=c(q{)}^2{q}_{,X_1}$ give the Riemann invariantsConsider the initially undisturbed half-space $X_1>0$. A right-going simple wave has the incoming invariant $v+H=0$, so $v=-H(q)$. Denote the prescribed boundary nominal stress by $T(t)=P_{21}(0,t)=\tau (q_b(t))$. The physical boundary traction, whose outward normal is $-e_1$, is $-T(t)e_2$. Invert the monotone law $\tau$ to obtain $q_b$. The wave state emitted at boundary time $s$ is transported alongAhead of the leading front $X_1=c(0)t$ the state is zero. Before characteristic crossing, these equations determine the unique emission time $s$ for each disturbed point. The displacement itself is also determined, rather than just its derivatives:Differentiating the implicit emission relation verifies $w_{,X_1}=q_b(s)$ and $w_t=-H(q_b(s))$, so this is the complete pre-shock simple wave solution.

Write $c_b(s)=c(q_b(s))$. The boundary-generated characteristic curves lose invertibility whenThus the characteristic crossing in a boundary-generated elastic simple wave criterion isHere ${\dot{c}}_b=c^{\prime }(q_b)T^{\prime }/({\rho }_0{c}_b^2)$. A smoothly increasing boundary stress therefore steepens into a shock wave only when later states travel faster: $c^{\prime }(q_b)T^{\prime }>0$. Increasing stress alone is insufficient. For the compatible example with $h>0$ and $q_b>0$, $c^{\prime }(q_b)=3h{q}_b/({\rho }_0{c}_b)>0$, so a nonzero increasing ramp produces finite-time crossing. For $h=0$ all rays are parallel and no such shock forms. This criterion locates the first gradient singularity; subsequent discontinuous evolution additionally requires the jump conditions and an admissibility condition.

For a general vector simple wave, replace the pure-shear line by an integral curveSubstitution gives $a_t+c_{\alpha }(a)a_{,X_1}=0$, with the same emission construction and crossing criterion. Boundary tractions must lie on this one-family curve. Generic finite-amplitude quasi-transverse waves may contain longitudinal displacement; arbitrary vector boundary forcing need not generate a single simple wave.