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When deformation is stopped, the rate-of-strain tensor and velocity gradient vanish, leaving $\mathsf{T}+\tau {\partial }_t\mathsf{T}=0$. HenceThe positive constant $\tau$ is the viscoelastic relaxation time. For slow weak deformation, all the transport and quadratic terms are smaller than $\mathsf{T}$, so $\mathsf{T}\simeq 2\mu \mathsf{E}$. Thus $\mu >0$ is the zero-rate dynamic viscosity, measured in stress times time. The corresponding short-time shear modulus of the Maxwell fluid response is $G_{\mathrm{el}}=\mu /\tau$. The source's $\alpha$ has units of time, since $\alpha \mathsf{TE}$ must have units of stress.

Let $g=\dot{\gamma }$ and use $G_{yx}=g$ for the source's velocity gradient convention. Seek the homogeneous stationary stress reached under steady simple shear flow. The material derivative of that stress tensor vanishes. In the $xy$ plane,The $xx$, $yy$ and $zz$ components of the constitutive equation give, respectively,Thus, putting $D=1+\alpha (2\tau -\alpha )g^2$ and using the shear stress calculated below,The out-of-plane shear components vanish on the homogeneous branch. The two normal-stress differences areThe ratio is defined at nonzero shear when $N_1\ne 0$; at $g=0$ both differences vanish, and the displayed constant is their limiting ratio. An arbitrary isotropic pressure in the total Cauchy stress tensor cancels from both differences; it is not an arbitrary addition to the constitutively defined extra stress $\mathsf{T}$ itself.

Take $\mathbf{R}={\mathbf{r}}_2-{\mathbf{r}}_1$ and label each dumbbell once. Locally assume its probability density and number density vary negligibly over a connector length. The oriented connector crossing count follows geometrically: a connector crosses a small plane element precisely when its midpoint lies within a slab of normal thickness $|\mathbf{R}\cdot \mathbf{n}|$. For configurations in $d^{3}R$, the unsigned crossing number per unit area is thereforeThe expression without the absolute value is an oriented crossing count, or an unsigned count restricted to $\mathbf{R}\cdot \mathbf{n}\ge 0$. It cannot be an unrestricted nonnegative number: a connector $\mathbf{R}=-R\mathbf{n}$ gives a negative printed expression despite physically intersecting the plane.

To obtain traction, keep the force orientation as well. For $\mathbf{R}\cdot \mathbf{n}>0$, the spring force on the bead below the plane exerted by the bead above it is $H\mathbf{R}$. For the opposite orientation it is $-H\mathbf{R}$. Multiplying that force by the unsigned crossing count in either case givesIntegrating over all labelled configurations yieldsThere is no extra factor of two: each dumbbell was labelled and counted once, and the orientation-dependent force has already supplied the sign. This is the local continuum stress approximation; resolving appreciable spatial variation across a finite connector would require a nonlocal stress construction.

Let $p_T=n{k}_{B}T$, $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$, and $M=I+\alpha \sigma$, so the single-bead mobility is $M/\zeta$. Assume an incompressible solvent, constant $n,H,T,\zeta$, negligible bead inertia, and a symmetric positive-definite $M$. The last requirement makes the resistance invertible and Brownian diffusion physically admissible. At a given position and time, $M$ depends on the moment $C$ but not on the individual configuration $\mathbf{R}$.

Use the source's gradient convention $K_{ij}={\partial }_i{u}_j$, and write $G=K^T$ for the component-first gradient. Then $\mathbf{u}({\mathbf{r}}_2)-\mathbf{u}({\mathbf{r}}_1)\simeq G\mathbf{R}$. The spring forces on the two beads are ${\mathbf{F}}_1=H\mathbf{R}$ and ${\mathbf{F}}_2=-H\mathbf{R}$. Their overdamped probability currents areHere the Brownian term written as $-k_{B}T\nabla \psi$ is a force weighted by probability density. Its corresponding conditional osmotic force is $-k_{B}T\nabla \log \psi$; treating the given term as an unweighted deterministic force per bead would not produce the stated Fokker–Planck equation.

With midpoint $\mathbf{x}=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$, the derivatives are ${\nabla }_{r_1}=\frac{1}{2}{\nabla }_x-{\nabla }_R$ and ${\nabla }_{r_2}=\frac{1}{2}{\nabla }_x+{\nabla }_R$. Subtracting the bead currents gives the configuration currentThe midpoint current is $\mathbf{u}\psi$ after neglecting the specified spatial Brownian diffusion. Configuration probability conservation therefore givesSince $G\mathbf{R}$ has components $R_j{\partial }_j{u}_i$, this agrees with the source's $\mathbf{R}\cdot \nabla \mathbf{u}$. The factor two in both spring relaxation and configuration diffusivity comes from relative motion of both beads.

The second moment $C$ is the unnormalized conformation tensor. Now multiply by $R_i{R}_j$ and integrate over configuration space. Assume sufficiently decaying tails so every boundary term at infinite separation vanishes. The affine current gives $GC+C{G}^T$; spring drift gives $-(2H/\zeta )(MC+CM)$; two integrations by parts of diffusion give $(4k_{B}T/\zeta )M$. HenceBecause $\sigma =nHC-p_{T}I$, $M$ is a polynomial in $C$ and commutes with it. ThusThis is the Hookean dumbbell conformation equation with anisotropic mobility, an exact second-moment closure within the stated model; no Gaussian approximation to $\psi$ was needed.

The upper-convected derivative of the identity is $I^{\nabla }=-(G+G^T)$. Applying it to the polymer stress therefore givesConsequently the coefficients and the closed stress equation arewhere ${\sigma }^{\nabla }=D_t\sigma -\sigma K-K^T\sigma$ in the printed convention. Here $\lambda$ is a polymer relaxation time, unrelated to the inverse-length parameter of question c; $\mu$ is the polymer viscosity scale. The coefficient $\alpha$ has inverse-stress units. In the dimensional mobility parameter in the Giesekus model, the dimensionless factor is $b=\alpha p_T$, not the dimensional $\alpha$ printed in this problem.

The upper-convected rate removes the affine stretching and rotation of material connectors. If $\dot{F}=GF$, then a tensor transported purely as $C=F{C}_0{F}^T$ has $C^{\nabla }=0$. Under a time-dependent rigid rotation $Q$, the rate transforms as $Q{C}^{\nabla }{Q}^T$: this is material frame indifference, not an ordinary componentwise time derivative. The resulting equation is the Giesekus model in these dimensional conventions.

The isotropic subtraction is fixed by zero polymer extra stress at thermal equilibrium. With no flow and positive mobility, the stationary distribution isso $\boxed{{\sigma }_{\mathrm{eq}}=0}$. The constitutive equation correspondingly has a zero-stress quiescent state and a Newtonian low-rate limit $\sigma \simeq \mu (K+K^T)$. Without subtracting exactly $p_{T}I$, the spring contribution would retain an equilibrium isotropic stress and the equation would acquire corresponding reference terms. Because mobility itself depends on $\sigma$, that reference cannot simply be shifted while retaining the same model. The separate fluid pressure enforcing incompressibility is not included in this polymer extra stress.

For steady uniaxial extension of the Giesekus model, the component-first velocity gradient is diagonal, $G=\mathrm{diag}(\gamma ,-\gamma /2,-\gamma /2)$. For a steady spatially uniform diagonal stress, the material derivative vanishes and the convected terms are $-2\gamma {\sigma }_{11}$ and $+\gamma {\sigma }_{22}$. The off-diagonal equations vanish identically, and symmetry permits equal transverse components. The remaining equations areSet $W=\lambda \gamma \ge 0$, $b=\alpha p_T$ and $s_i={\sigma }_{ii}/p_T$, with $\mu =\lambda p_T$. For $b>0$, the branches continuously connected to zero stress at zero rate areA sufficient nondegenerate range for physical solutions at every nonnegative rate is $0<b<1$. Indeed $M=(1-b)I+bA$ with $A=HC/(k_{B}T)$, so positive conformation implies positive mobility. The longitudinal quadratic equation is nonpositive at $s_1=0$ and has a nonnegative root of a polynomial. The transverse quadratic equation has values $b-1<0$ at $s_2=-1$ and $W\ge 0$ at $s_2=0$, so its larger root of a polynomial lies in $(-1,0]$. Thus the displayed branches satisfy $s_i>-1$, giving $A_i=1+s_i>0$. The other algebraic roots do not provide a physical branch connected to the equilibrium distribution. For constant positive diagonal $A$, the associated stationary configuration distribution is a Gaussian with covariance $C=(k_{B}T/H)A$.

For positive $\alpha$ on the physical branch, the stretching stress remains finite through $W=1/2$. At that rate,At large extension rate,Thus the stress continues to grow, but the uniaxial extensional viscosity ${\eta }_E=({\sigma }_{11}-{\sigma }_{22})/\gamma$ approaches the finite plateau $2\lambda /\alpha$ rather than developing a finite-rate pole. The transverse stress approaches $-p_T$ as its conformation tensor variance becomes small. At small rate the two stresses are $2\mu \gamma$ and $-\mu \gamma$, so ${\eta }_E\to 3\mu$.

When $\alpha =0$, the physical steady branch instead isThe stretching stress diverges as $W\to 1/2$ from below. Above that threshold its algebraic continuation has $1+s_1=1/(1-2W)<0$ and cannot be a conformation covariance. There is no physical finite steady stretching state. The transient equationgives linear growth at $W=1/2$ and exponential growth at larger rates, the Hookean coil-stretch failure of the upper-convected Maxwell limit.

Parameter qualification. Positivity of $\alpha$ alone, without a restriction on $b=\alpha n{k}_{B}T$, does not ensure real physical steady solutions for the entire tensor. For example $b=2$, $W=1$ makes the transverse equation $2s_2^2+2s_2+1=0$, which has no real root. Its diffusion/resistance interpretation is already inadmissible for some conformations. The physically positive-mobility regime stated above avoids that counterexample; the source's positive-$\alpha$ case must not be taken as a guarantee for arbitrary parameters.

The magnetic second-moment tensor is $M_{ij}=B_i{B}_j$. This is the magnetic dyadic called the Maxwell tensor in the paper; the isotropic part of the usual Maxwell stress tensor is already contained in the magnetohydrodynamic total pressure. The solenoidal magnetic-field constraint givesConsequently the magnetohydrodynamic momentum equation in part a becomes $D_t{u}_i=-{\partial }_{i}p+{\partial }_j{M}_{ij}$, where $D_t={\partial }_t+u_n{\partial }_n$ is the material derivative. Applying its product rule and the ideal magnetohydrodynamic induction equation givesThus the evolution uses only $\mathbf{u},M,p$, together with ${\partial }_i{u}_i=0$. To determine $p$, take the divergence of the magnetohydrodynamic momentum equation. Since ${\partial }_t{\partial }_i{u}_i=0$ and $u_j{\partial }_j{\partial }_i{u}_i=0$,This is the pressure Poisson equation for incompressible flow, with boundary conditions supplied by the imposed normal acceleration and normal stress, or by periodicity or decay at infinity as appropriate. For example, the normal component of the magnetohydrodynamic momentum equation requires $\mathbf{n}\cdot \nabla p=\mathbf{n}\cdot (\nabla \cdot M-D_t\mathbf{u})$ at a prescribed boundary. The pressure is determined up to a spatially constant function of time.

For equivalence to one resolved magnetic field, the initial $M$ must actually be $\mathbf{B}{\mathbf{B}}^T$ for a divergence-free vector field. In particular it is a symmetric positive semidefinite matrix of matrix rank at most one. The tensor equations also admit more general stresses; the averaged $M$ in part d need not have matrix rank one.

Average the equations in part b and make exactly the prescribed scale-separation approximation: the retained perturbations vary on lengths much greater than $\ell$, and perturbations at the tangle scale are omitted. Products of two retained perturbations can then be neglected. Writing $\mathbf{v}=\delta \mathbf{u}$, $m=\delta M$ and $\pi =\delta p$, the linearized equations areNo unmodelled small-scale response has been inserted into this closure.

Introduce the fluid displacement $\mathbf{\xi }$ by $\mathbf{v}={\partial }_t\mathbf{\xi }$ and choose $\nabla \cdot \mathbf{\xi }=0$. Integrating the magnetic second-moment tensor equation giveswhere $s$ is a time-independent integration tensor set by the initial stress. For a pure oscillatory normal mode, $s=0$. Its restoring force is therefore $v_A^2{\nabla }^2\mathbf{\xi }$: the isotropic magnetic tangle has the same transverse response as an incompressible elastic medium with shear modulus ${\rho }_0{v}_A^2$.

For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, with $\omega \ne 0$, incompressibility gives $\mathbf{k}\cdot \hat{\mathbf{v}}=0$. The tensor equation and magnetohydrodynamic momentum equation becomeDotting the second equation with $\mathbf{k}$ yields $\hat{\pi }=0$ for $k\ne 0$, since $k_i{k}_j{\hat{m}}_{ij}=0$. Substitution then givesThere are two independent transverse polarizations. The phase speed is $v_A$ in every direction, and the two signs of $\omega$ give group velocities $\pm v_A\hat{\mathbf{k}}$. These magnetoelastic waves in an isotropic magnetic tangle exist because a shear deformation distorts the second moment of the tangled magnetic field, producing a restoring magnetic tension even though its mean is zero.

For comparison, a uniform magnetic field with Alfvén velocity ${\mathbf{B}}_0$ gives the incompressible Alfvén wave relation ${\omega }^2=(\mathbf{k}\cdot {\mathbf{B}}_0{)}^2$. Its frequency depends on the direction of $\mathbf{k}$, and energy propagates along ${\mathbf{B}}_0$. Here the averaged restoring response is isotropic: there is no preferred mean-field direction. Both systems resist transverse deformation through magnetic tension, rather than through compressive sound.

The propagating branch is not the entire perturbation space. For example, $\mathbf{v}=0$, $m_{ij}=f(\mathbf{r}){\delta }_{ij}$ and $\pi =f(\mathbf{r})$ solve the linearized equations without propagation, for any small, slowly varying $f$. Such a static stress perturbation in magnetoelastic wave equations balances a pressure gradient. Thus the requested wave conclusion applies to the oscillatory branch; it must not exclude stationary initial stress perturbations.

Using the given linear momentum balance, multiply by the velocity and integrate over the fixed volume:This is integration by parts component by component, or the divergence theorem. The first term is boundary stress power. With component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the second is the double contraction $\Phi =\sigma :K$.

For a Newtonian fluid, $\sigma =-pI+2\mu e$, where $e=(K+K^T)/2$ is the rate-of-strain tensor and $\mu \ge 0$ is dynamic viscosity. Incompressibility removes the pressure contribution, and the symmetric tensor $e$ is orthogonal to the antisymmetric part of $K$. ThereforeThe displayed energy identity follows for the stated linear balance. A full advective momentum equation in a fixed control volume would also contribute an advective kinetic energy flux. The nonnegativity statement here is for a purely Newtonian viscous stress; a viscoelastic stress can exchange stored free energy with the flow.

For a steady axisymmetric flow ${\mathbf{u}}_0=r\Omega (r)\hat{\mathbf{\varphi }}$, incompressible flow is automatic. The only material acceleration is centripetal, $-r{\Omega }^2\hat{\mathbf{r}}$. The radial Euler equations therefore requireWith no viscosity there is no tangential shear stress to redistribute angular momentum. Any sufficiently regular $\Omega (r)$ can therefore be a steady inviscid profile, with the radial pressure adjusted by integrating this equation. This is a statement of equilibrium, not of stability: some such profiles are centrifugally unstable.

The Rankine-Hugoniot conditions follow by integrating conservation laws across a thin moving layer, then letting its macroscopic thickness tend to zero. For example, normal momentum conservation in one dimension has the conservative formFor a shock wave moving at speed $V$, integrate over a control interval whose endpoints move with the discontinuity. The Reynolds transport theorem giveswhere $[f]=f_R-f_L$ denotes the difference of limiting states. Mass conservation similarly gives $V[\rho ]=[\rho u]$. If $v=u-V$ is the normal velocity in the shock frame, combining the two relations yieldsThe same construction applied to conservation of energy, for a non-radiating normal shock wave with nonzero mass flux, gives $[h+v^2/2]=0$, where $h$ is the specific enthalpy. An equation of state closes these relations, and the physical compression branch has increasing entropy.

Finite dynamic viscosity replaces the ideal discontinuity by a thin smooth transition. A viscous stress contributes a derivative to the momentum flux inside the layer and converts directed kinetic energy into internal energy. If its gradients vanish in the uniform far states, the same limiting Rankine-Hugoniot conditions still hold there. Balancing nonlinear transport with viscous diffusion gives a thickness of order $\nu /|u_L-u_R|$ in the viscous Burgers equation, where $\nu$ is the kinematic viscosity.

The in-plane equilibrium equations are ${\tau }_{xx,x}+{\tau }_{xy,y}=0$ and ${\tau }_{xy,x}+{\tau }_{yy,y}=0$. On a simply connected patch, each divergence-free row has a streamfunction: write ${\tau }_{xx}={\varphi }_y$, ${\tau }_{xy}=-{\varphi }_x=-{\psi }_y$, ${\tau }_{yy}={\psi }_x$. Symmetry gives ${\varphi }_x={\psi }_y$, so another potential exists with $A_y=\varphi$, $A_x=\psi$. HenceIf two such Airy stress functions give the same stress, their difference has zero Hessian and is therefore an arbitrary affine function $a+bx+cy$ on a connected domain. This is the full nonuniqueness for a fixed stress. The existence argument is local, or global on a simply connected domain; domains that are multiply connected regions require the associated force and moment period conditions for a single-valued global potential. Equilibrium alone should not silently erase that topological qualification.

For plane strain put $s=e_{xx}+e_{yy}$. The constitutive equation gives ${\tau }_{xx}+{\tau }_{yy}=2(\lambda +\mu )s$. ConsequentlyThe displacement Saint-Venant compatibility equations $e_{xx,yy}+e_{yy,xx}-2e_{xy,xy}=0$ then becomesFor stable elastic moduli the coefficient is nonzero, so $\boxed{{\Delta }^{2}A=0}$. If $D$ is bounded and smooth and $A={\partial }_{n}A=0$ on its boundary, integrating twice by parts gives $0={\int }_{D}A{\Delta }^{2}AdV={\int }_D(\Delta A{)}^{2}dV$. Thus $A$ is harmonic with zero boundary value and $\boxed{A=0}$ by uniqueness of the Dirichlet problem. The normal-derivative condition is what removes the extra fourth-order boundary terms.

For the pressurized annulus, substituting the radial constitutive equations into equilibrium gives $u^{\prime \prime }+u^{\prime }/r-u/r^2=0$, so $u=Cr+D/r$. The stresses areUsing gauge stresses relative to ambient, the boundaries are ${\tau }_{rr}(a)=-P$, ${\tau }_{rr}(b)=0$. ThereforeandThe maximum occurs at the bore. Once the chosen yield criterion is reached, Hooke's law cannot describe the total strain alone: write total strain as elastic plus plastic strain and supplement the elastic stress law with a yield condition, a plastic flow rule and any required work hardening law. A plastic region starts near the bore and must be matched to the remaining elastic region. The exact conversion between a principal-stress difference and the quoted yield stress depends on whether the material criterion is Tresca or von Mises. This is the elastic stress concentration in a pressurized thick cylinder.

For antiplane displacement $u=(0,0,w)$ the only nonzero strain components are $e_{xz}=w_x/2$ and $e_{yz}=w_y/2$. Commutation of mixed partial derivatives gives the necessary local compatibility conditionThe supplied stress is symmetric, independent of $z$ and trace-free. Its first two equilibrium equations are automatic; its third givesaway from the origin. Antiplane shear has zero trace, so isotropic Hooke's law reduces to $\boxed{e=\tau /(2\mu )}$. HenceLocally this integrates to $w=\arg (x+iy)+C$, represented on a suitable branch by $\arctan (y/x)+C$. The strain condition is satisfied at every point of the punctured plane, but the displacement cannot be made globally single-valued there:A gradient of a globally single-valued smooth displacement would instead integrate to zero around any closed curve. Local compatibility is therefore not sufficient on this multiply connected region.

A physical construction is a Volterra screw dislocation: cut the cylinder along a radial half-plane ending on its axis, shift one cut face relative to the other parallel to the axis, and rejoin. The axial mismatch is the Burgers vector. In the given normalization its magnitude is $2\pi$; with a dimensional mismatch $b$ one uses $w=b\arg (x+iy)/(2\pi )$. An intact pristine solid cannot reach this state by a smooth single-valued elastic deformation alone. The axis is a dislocation core where the continuum elastic field is singular and must be regularized by microscopic structure or plastic deformation. On the punctured domain the differential condition holds locally; globally its nonzero circulation detects the defect. In a distributional description, that incompatibility is concentrated on the axis. This is local strain compatibility versus screw-dislocation circulation.

Take the trace of the constitutive relation at zero stress. It gives $0=(2\mu +3\lambda )e_{kk}-3c\alpha (T-T_0)$. With the observed stress-free volume strain this requiresThis is the bulk modulus. A stable solid resists uniform compression, so $\boxed{K>0}$; equivalently the elastic energy of a nonzero pure dilation is positive. The thermal strain is isotropic, $e_{ij}^{\mathrm{th}}=\alpha (T-T_0){\delta }_{ij}/3$, because $\alpha$ here is the volumetric, not linear, thermal expansion coefficient.

Let $\beta =K\alpha$. Using $e_{ij}=(u_{i,j}+u_{j,i})/2$, stress divergence isZero body force and equilibrium therefore giveFor an applied body force per unit volume $b$, equilibrium is $\nabla \cdot \tau +b=0$, so its right-hand side becomes $\beta \nabla T-b$. Consequentlycancels the effective thermal forcing in the displacement equation. Per unit mass the force is $\beta \nabla T/\rho$. This sign follows from the divergence of the thermal stress $-\beta (T-T_0)I$, which is $-\beta \nabla T$. Cancelling that divergence is distinct from removing the stress tensor itself: boundary tractions and thermal compatibility still matter.

For spatially uniform heating with the entire boundary fixed, $u=0$ is the unique equilibrium displacement in a stable elastic solid, andThe stress is compressive hydrostatic stress for a temperature rise. The fixed boundary supplies the tractions that prevent thermal expansion. If $T$ varies spatially and no compensating body force is imposed, fixing the boundary alone does not generally force the interior displacement to vanish: one must solve the displayed forced Navier-Cauchy equation with $u=0$ on the boundary, then recover the stress. The simple uniform hydrostatic expression should not be asserted for arbitrary temperature fields. This is uniform thermal stress in a fully constrained solid.

Choose zero energy in the unstrained state. The required strain energy density isIts variation is $\delta W=(\lambda \mathrm{tr}(e)I+2\mu e):\delta e$, giving the prescribed stress as the tensor gradient. This variational notation handles the symmetric off-diagonal components without a spurious factor of two.

Decompose $e=e^{\mathrm{dev}}+(\mathrm{tr}e)I/3$ into orthogonal deviatoric and spherical parts. ThenThe two squares vanish together only for $e=0$. Thus $\mu >0$ and $3\lambda +2\mu >0$ are sufficient for positive definiteness. They are necessary by testing a nonzero trace-free strain and a pure dilation separately. Equivalently both shear modulus $\mu$ and bulk modulus $K=\lambda +2\mu /3$ must be positive.

Multiply equilibrium by the displacement and integrate. Integration by parts and the traction condition giveThe stress is symmetric, so it contracts only with the symmetric part of $\nabla \mathbf{u}$, namely $e$. Hooke's law then gives $T:e=\lambda (\mathrm{tr}e{)}^2+2\mu e:e=2W$. HenceThe identity is the linear-elastic work balance: twice the stored quadratic energy equals the body-force and boundary-traction work paired with the final displacement.

Apply energy uniqueness for a linear elastic traction problem. Subtract two solutions with the same loading, writing their displacement difference as $\mathbf{v}$. Linearity makes its body force and boundary traction zero. The preceding energy identity therefore yields ${\int }_{B}2W(e(\mathbf{v}))dV=0$. Positive definiteness and continuity imply $e(\mathbf{v})=0$ everywhere. By the permitted zero-strain characterization,Thus the solution is unique modulo infinitesimal rigid translations and rotations, on each connected body. Conversely adding such a displacement changes neither strain nor stress, so these are exactly the unconstrained modes. This is conditional uniqueness; existence of a pure-traction solution additionally requires balanced resultant force and torque.

Use the torsion stress function for a simply connected bar. The only nonzero strain components are $e_{xz}=(w_x-\Omega y)/2$ and $e_{yz}=(w_y+\Omega x)/2$. ThusThe axial equilibrium equation is ${\partial }_x{\tau }_{xz}+{\partial }_y{\tau }_{yz}=0$. Since the section is simply connected, the planar divergence-free stress field has a stream function:Equality of the mixed warping derivatives gives $\boxed{\Delta \varphi =-2\mu \Omega }$. On the lateral boundary the axial traction is ${\varphi }_y{n}_x-{\varphi }_x{n}_y$, the tangential derivative of $\varphi$. Traction-free walls therefore make $\varphi$ constant on the connected boundary, and its arbitrary additive constant can be chosen so $\boxed{\varphi =0\text{ on }\partial D}$.

The axial torque isThe divergence theorem removes the boundary term because $\varphi =0$.

Tresca's yield criterion requires the maximum shear stress, half the largest principal-stress difference, not to exceed the material's shear yield stress ${\tau }_Y$. The stress eigenvalues here are $0,\pm \sqrt{{\tau }_{xz}^2+{\tau }_{yz}^2}$, so the maximum shear is $|\nabla \varphi |$ andIf yield is instead quoted as a uniaxial tensile stress, that parameter is $2{\tau }_Y$; the present convention is the source's shear parameter.

For the ellipse, the proposed quadratic vanishes on its boundary. Its Laplacian fixesThe corresponding warping is $w=\Omega (b^2-a^2)xy/(a^2+b^2)+\text{constant}$. Elliptic polar coordinates $x=ar\cos \vartheta$, $y=br\sin \vartheta$ give ${\int }_D\varphi =C\pi ab/2$, henceTo locate first yield, maximize $|\nabla \varphi |=2|C|\sqrt{x^2/a^4+y^2/b^4}$. It increases radially to the boundary. For $a>b$, the maximum there occurs at $(0,\pm b)$ and equals $2|C|/b$. ThereforeThe positive-twist value is the requested expression. The ends of the minor axis yield first, not the ends of the longer axis. This is elliptical-bar torsion and first yield.

Differentiating the displacement givesThe isotropic stress is diagonal, withThe two transverse traction-free conditions set the latter entries to zero. Hence $\lambda =2\nu (\lambda +\mu )$ andSubstitution shows $E=2\mu (1+\nu )$; solving for the two Lamé parameters givesTo determine the allowed interval without assuming signs not given, put $K=\lambda +2\mu /3>0$. Then $\nu =(3K-2\mu )/[2(3K+\mu )]$, soTherefore $\boxed{-1<\nu <1/2}$. The endpoints would require a degenerate elastic modulus, so they are excluded by the strict material inequalities.

For a surface with unit normal $\mathbf{n}$, traction is $t_i={\tau }_{ij}{n}_j$. Thus ${\tau }_{ij}$ is the $i$th force-per-area component on a plane whose positive normal is ${\mathbf{e}}_j$. Diagonal entries are normal stresses and off-diagonal entries are shear stresses, with tensile normal stress positive in this convention.

The symmetric strain has six independent components and so does the symmetric stress. A general linear map between these two six-dimensional spaces therefore hasEquivalently choose the minor symmetries $C_{ijkl}=C_{jikl}=C_{ijlk}$. Components antisymmetric in the strain indices have no observable effect and are omitted from this representative.

A further distinction is important: if a strain energy density exists, $C_{ijkl}={\partial }^{2}W/(\partial e_{ij}\partial e_{kl})$ has the major symmetry $C_{ijkl}=C_{klij}$. Then the six-by-six map is symmetric and has $6\cdot 7/2=21$ independent coefficients. Thus 36 follows from the two stated tensor symmetries alone; 21 additionally uses the energy/reciprocity assumption, as in the variational setting later in the paper. These are the minor and major symmetries of linear elasticity.

The perturbation preserves the prescribed displacement because $\mathbf{w}=0$ on the boundary. Strain varies as $e_{ij}(\mathbf{u}+\alpha \mathbf{w})=e_{ij}(\mathbf{u})+\alpha (w_{i,j}+w_{j,i})/2$. Taylor expansion of the strain energy density givesSymmetry of the stress makes its contraction with the two symmetric terms equal, henceA minimizer must have zero first variation for both signs of $\alpha$. Integrating by parts and using the zero boundary variation yields ${\int }_B(-{\partial }_j{\tau }_{ij}-\rho g_i)w_{i}dV=0$. The fundamental lemma of the calculus of variations, applied to arbitrary compactly supported variations, givesThis is the static momentum equation. Fixed-displacement data remove the surface variation; prescribed tractions would instead appear through boundary work in the potential energy.