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For a locally uniform distribution of labelled connectors, a connector crosses a small surface when its midpoint lies in a slab of thickness $|\mathbf{R}\cdot \mathbf{n}|$. This gives the displayed unsigned count per unit area. Omitting the absolute value gives a signed count, or requires a restriction to one orientation. The signed spring force transmitted from the positive side to the negative side is $H\mathbf{R}\mathrm{sgn}(\mathbf{R}\cdot \mathbf{n})$, so multiplication by the unsigned count yields traction $nH⟨\mathbf{R}{\mathbf{R}}^T⟩\mathbf{n}$. Each labelled dumbbell is counted once.

The symmetric first moment of the traction exerted by a rigid inclusion, with the surface normal pointing into the inclusion. Its contraction with the background rate-of-strain tensor is the strain-related contribution to extra dissipation due to a rigid inclusion. A trace-free version is equivalent for an incompressible flow. This tensor is distinct from the axisymmetric force-dipole flow used as a far-field singularity.

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.

Write the supplied velocity in the radial formAt $r=a$, $f=1$ and $g=0$, giving the no-slip boundary condition $\mathbf{u}=\mathbf{U}$. At infinity $\mathbf{u}\to 0$ and the supplied gauge-fixed pressure tends to zero. These are the laboratory-frame conditions for a sphere translating through otherwise quiescent fluid.

The exterior Stokes equations can also be checked directly:andTo compute surface traction in translating-sphere Stokes flow, let $\mathbf{n}=\mathbf{x}/a$ point from the sphere into the fluid. At the surface, $f^{\prime }(a)=-3/(2a)$, $g^{\prime }(a)=3/(2a^3)$ and $g(a)=0$. HenceSubstitute into the Newtonian fluid stress tensor. Its viscous contribution to the surface traction is $3\mu [-\mathbf{U}+(\mathbf{U}\cdot \mathbf{n})\mathbf{n}]/(2a)$, so the normal part cancels the pressure contribution:The vector is constant over the entire sphere in the stated pressure gauge. Integrating it gives the fluid's drag force on the sphere,This is Stokes's law. On an exterior fluid domain the outward domain normal at its inner boundary is $-\mathbf{n}$; that distinction will matter in the reciprocal calculation.

Let $S_a$ be the sphere surface, ${\mathbf{n}}_s$ point from the sphere into the fluid, and writeBecause the prescribed exterior body forces are unchanged, they cancel on subtraction: the disturbance is a homogeneous Stokes flow in the exterior domain. On the sphere the no-slip boundary condition isallowing an arbitrary sphere rotation ${\mathbf{\omega }}_s$.

Use as auxiliary flow the translating-sphere solution of part b, with arbitrary test velocity $\mathbf{U}$. Its surface velocity is $\mathbf{U}$ and its surface traction is $-3\mu \mathbf{U}/(2a)$. Applying the Lorentz reciprocal theorem on the exterior domain gives, after cancelling the common inner-normal minus sign,The outer surface terms vanish for decaying disturbances: velocities are at most $O(r^{-1})$, stresses $O(r^{-2})$, and the integrated cross-work is $O(r^{-1})$.

There are no body forces inside the sphere in the original ambient problem. Thus the smooth ambient stress tensor has zero divergence there, andAlso ${\int }_{S_a}\mathbf{x}dS=0$, so the rotation term drops out of the left side. Consequently the reciprocal identity reduces toSince $\mathbf{U}$ is arbitrary, the result before imposing mechanical balance isFor a freely suspended sphere with negligible inertia and no applied external force, ${\mathbf{F}}_H=0$. Its sphere velocity as a surface average of ambient Stokes flow is thereforeThe implicit force-free condition is essential. More generally, an external force ${\mathbf{F}}_{\mathrm{ext}}$ gives $\mathbf{V}=⟨{\mathbf{u}}_{\infty }{⟩}_{S_a}+{\mathbf{F}}_{\mathrm{ext}}/(6\pi \mu a)$. For example, an externally pulled sphere in the zero ambient flow has nonzero velocity, contrary to an unrestricted surface-average-only claim. No torque-free assumption is needed for this translational identity, because the auxiliary surface traction is constant.

For antiplane shear, the displacement field is $\mathbf{u}=(0,0,w(x,y,t))$, independent of $z$. Its divergence is zero. Consequently the only possibly nonzero components of the stress tensor are the shear stressesThe $x$ and $y$ components of the Navier-Cauchy equation vanish identically. Its $z$ component gives $\rho w_{tt}={\partial }_x(\mu w_x)+{\partial }_y(\mu w_y)$. Each material has constant shear modulus and mass density, so the required wave equation and wave speeds are

At a perfectly bonded elastic interface, both the displacement field and the traction are continuous. Using the same normal ${\mathbf{e}}_y$ on both sides, the only nonzero traction component is ${\tau }_{zy}$. Thus the interface boundary conditions areContinuity of “stress” here means continuity of traction; it does not require equality of the full stress tensors. The rigid plate supplies the remaining boundary condition $w_1(x,0,t)=0$.

Substitution of the proposed normal mode into the wave equation gives the ordinary differential equationsA nonzero exterior displacement field has exponential decay at infinity only when $c<c_2$. Withits wave amplitude is $f_2(y)=D{e}^{-\ell (y-h)}$. At $c=c_2$ the exterior wave amplitude is affine and cannot decay; at $c>c_2$ it is oscillatory and cannot decay. Setting the exterior wave amplitude identically zero would impose both $f_1(h)=0$ and $f_1^{\prime }(h)=0$, forcing the entire normal mode to vanish.

We also need $c>c_1$. Indeed, if $c<c_1$, the rigid boundary condition gives $f_1(y)=A\sinh (qy)$ with $q=k\sqrt{1-c^2/c_1^2}>0$. At the interface,which contradicts continuity of traction because both values of the shear modulus are positive. At $c=c_1$, the layer wave amplitude is $Ay$ and the same contradiction follows from $f_1^{\prime }(h)/f_1(h)=1/h>0$.

Therefore $c_1<c<c_2$. Define the positive transverse wavenumberThe rigid boundary condition now gives $f_1(y)=A\sin (my)$. The interface boundary conditions giveNeither $\sin (mh)$ nor $\cos (mh)$ can vanish for a nontrivial normal mode with $m,\ell >0$. Dividing therefore gives the rigid-backed shear-wave dispersion relation:Conversely, positive $m,\ell$ satisfying these relations generate a nonzero normal mode through the wave amplitudes above. Its physical displacement field is the real part of the complex expression.

During increasing pressure, the elastic solution reaches the Tresca yield criterion with $t-q>0$. In the plastic shell of a material with perfect plasticity, this sign is retained andRadial force balance becomes $q^{\prime }-4{\tau }_Y/r=0$. Integrating from the cavity wall, where $q(a)=-P$, givesOutside the plastic shell, radial elastic compatibility and vanishing far-field stress again give $q=-C/r^3$ and $t=C/(2r^3)$. At the advancing plastic front $r=s$, the Tresca yield criterion gives $3C/(4s^3)={\tau }_Y$, so $C=P_c{s}^3$. Continuity of radial traction now givesThusThe complete stress field of this spherical elastoplastic cavity isAt $s$, the tangential stress also matches because $-P_c+2{\tau }_Y=P_c/2$. Outside $s$, the function in the Tresca yield criterion is ${\tau }_Y(s/r{)}^3<{\tau }_Y$, which verifies that the remaining material is elastic. As $P\downarrow P_c$, $s\downarrow a$ and the solution joins the entirely elastic regime. The radius $a$ is the fixed reference radius of the small-strain model used by the paper.

LetAt maximum pressure, the stress is the field from part (c) with $P=P_m$ and $s=s_m$. Under the assumed elastic unloading, previously accumulated plastic strain is frozen. Therefore it is the stress increment, rather than the total stress, that satisfies homogeneous linear elasticity and radial elastic compatibility. Applying the superposition principle, a pressure change $-\Delta$ produces the elastic incrementsThey have zero far-field stress and change the wall traction from $-P_m$ to $-P$. These increments are the same in the formerly plastic and formerly elastic regions because their elastic Lamé parameters agree. Adding them to the maximum-load field givesandThese formulae retain $s_m$: it records the plastic loading history and is not the instantaneous loading-front formula with $P$ substituted for $P_m$. If the entire return is elastic, setting $\Delta =P_m$ gives the residual stress after removal of the applied pressure.

To check admissibility, examine the signed difference of the principal stresses in the formerly plastic shell:It immediately moves into the interior of the Tresca yield criterion when unloading starts. It can next reach the opposite limit $D=+2{\tau }_Y$. For fixed $\Delta >0$, $D$ is largest at $r=a$ and stays above $-2{\tau }_Y$. At the wall,There is no earlier violation in the outer region. Indeed,so its magnitude is no larger than $|D(s_m)|$, which has already been controlled by the shell calculation. Hence the first reverse yielding occurs at the cavity wall, withIf $P_c<P_m<2P_c$, the stress stays within the yield surface throughout unloading to zero. If $P_m=2P_c$, it reaches reverse yield exactly at $P=0$; further unloading would produce reverse plastic deformation. If $P_m>2P_c$, reverse plastic deformation starts while the applied pressure is still positive. ThusIf “yield again while unloaded” includes first touching the yield surface at the final endpoint, the corresponding condition is $P_m\ge 8{\tau }_Y/3$. The displayed elastic unloading fields are physically valid only for $0\le \Delta \le \min (P_m,2P_c)$; beyond that interval a new reverse-plastic region must be solved for. Continuing the elastic formula to $P=0$ when $P_m>2P_c$ would violate the given Tresca yield criterion.

Given an admissible velocity field, recover the pressure gradient from the Stokes equation, $\nabla p=\mu {\nabla }^2\mathbf{v}$, and integrate it, fixing the additive constant by matching to the bath pressure near the open ends. The biharmonic equation ensures the compatibility of this integration. For a freely moving hinge, the laboratory velocity field also contains the unknown rigid translation and rotation. Solve the interior and exterior Stokes flows with these wall velocities, then impose zero total external force and torque on the complete apparatus if its motor supplies only internal actuation. The pinned solution alone cannot determine that free motion.

At a fixed angle, the actuation velocity scale is $U\sim L|\dot{\alpha }|$. The Stokes equation gives $\Delta p\sim \mu U/L\sim \mu |\dot{\alpha }|$. Integrating normal traction over a plate of length $L$ gives a force per unit width of scalewith a dimensionless signed coefficient depending on geometry and exterior matching. Normal viscous stress has the same dimensional scale and is included in $C$. The force reverses when the actuation reverses. A narrow gap can make $|C|$ large.

More specifically, the corrected local streamfunction gives, up to a reference pressure,Indeed, inserting $\mathbf{v}=(-\dot{\alpha }r{G}^{\prime }/2,\dot{\alpha }rG)$ into the vector Laplacian gives $p_r=-\mu \dot{\alpha }(G^{\prime \prime \prime }+4G^{\prime })/(2r)=\mu \dot{\alpha }K/r$ and $p_{\theta }=0$. For $\alpha <\pi /2$, opening produces reduced pressure toward the hinge when matched at $r_{\ast }\sim L$. The logarithmic singularity is integrable: ${\int }_0^L\log (L/r)dr=L$, consistent with the stated force scale. A real hinge regularizes the singularity; the local formula is not a full end-matched solution.

The scallop theorem applies to a single periodically actuated hinge in a stationary Newtonian fluid with negligible fluid and body inertia. At each shape, Linearity of Stokes flow makes its rigid velocity proportional to $\dot{\alpha }$. Closing retraces the sequence of shapes and reverses the corresponding motion, irrespective of different opening and closing rates. The hinge may move during each half-cycle, butA motor can generate instantaneous forces and consume energy through viscous dissipation without producing sustained swimming. Additional independently controlled shape changes, appreciable inertia, or a fluid with memory would require a different analysis.

The coefficient of the growing term is fixed by the far-field complex potential: $A=\beta e^{i\alpha }$. On $z=a{e}^{i\theta }$, the no-penetration boundary condition makes the cylinder a streamline, so the imaginary part of $A{a}^2{e}^{2i\theta }+B{a}^{-2}{e}^{-2i\theta }$ must be constant. Choosing the two terms to be complex conjugates makes that constant zero, givingThis is the zero-circulation solution, also obtained by the Milne-Thomson circle theorem. Impermeability and the leading far field alone permit an additional circulation term $-i\Gamma \log z/(2\pi )$; the requested algebraic form implicitly selects $\Gamma =0$. No no-slip boundary condition is imposed on an inviscid flow.

The streamfunction in polar coordinates isIt gives $v_r(a,\theta )=0$ andThe liquid still exerts normal pressure on each surface element even though tangential viscous traction is absent. For this steady flow with zero vorticity, the Bernoulli equation has one constant $C$ throughout the connected fluid region:There is no finite uniform far-field pressure to substitute for $C$, because this straining velocity field grows with radius. Taking $\mathbf{n}=(\cos \theta ,\sin \theta )$ outward from the cylinder, the fluid force per unit axial width isIndeed $p(\theta +\pi )=p(\theta )$ while $\mathbf{n}(\theta +\pi )=-\mathbf{n}(\theta )$: diametrically opposite elements cancel exactly. Zero resultant force does not mean zero local pressure or zero deformation tendency.

For an axisymmetric field without azimuthal motion, the divergence in spherical coordinates isSubstituting the given axisymmetric stream function representation givesThus incompressible flow is automatic.

The velocity components are laboratory-frame components resolved about the instantaneous bubble center. The center itself translates with $U\hat{\mathbf{z}}=U\cos \theta \hat{\mathbf{r}}-U\sin \theta \hat{\mathbf{\theta }}$. The kinematic boundary condition for a spherical interface givesFor negligible internal dynamic viscosity, continuity of tangential traction imposes zero tangential shear stress, rather than a prescribed tangential velocity. In terms of the strain-rate tensor,Equivalently, ${\partial }_r{v}_{\theta }(a)=v_{\theta }(a)/a+U\sin \theta /a$. Axisymmetry gives $v_{\varphi }=0$, and the far-field boundary condition is $\mathbf{v}\to 0$. There is no no-slip boundary condition $v_{\theta }=-U\sin \theta$ on a clean bubble; the liquid can slip tangentially relative to the translating interface.

These are the conditions of the prescribed-sphere clean-bubble Stokes drag model. A genuinely free bubble also requires a normal stress balance. If one additionally insists on spatially uniform gas pressure, while removing both surface tension and the variation of hydrostatic pressure, that extra condition is incompatible with the translating spherical solution in part (c). The approximation must therefore be understood as computing drag on an imposed spherical interface, or as retaining the normal balancing effects when checking a free bubble.

Differentiating the axisymmetric stream function givesThe normal kinematic boundary condition fixes $C=Ua/2$, soBoth velocity components vanish at infinity, and the axisymmetric stream function is regular at the polar axes. The tangential component of the strain-rate tensor vanishes everywhere:Thus the stress-free boundary condition is satisfied. The vorticity is ${\omega }_{\varphi }=2C\sin \theta /r^2=Ua\sin \theta /r^2$, and the supplied dipolar pressure is consistent with the Stokes equation.

Use the Newtonian fluid stress tensor $\mathbf{\sigma }=-p\mathbf{I}+2\eta \mathbf{e}$. At the bubble surface,The vertical force exerted by the liquid is the integral of the vertical traction:Since the integral is $2/3$, the drag is downward and has magnitudeThe constant pressure integrates to zero force. The pressure perturbation contributes $4\pi \eta aU/3$ and the normal viscous stress contributes $8\pi \eta aU/3$; there is no tangential traction contribution.

For the free-interface qualification in part (b), the calculated normal traction at $\theta =0$ and $\theta =\pi$ differs by $6\eta U/a$. It cannot equal a single uniform gas pressure if every restoring effect is discarded. Restoring the external-minus-internal variation of hydrostatic pressure supplies a term $({\rho }_w-{\rho }_a)ga\cos \theta$. It cancels the dynamic normal traction at the terminal velocity $U=({\rho }_w-{\rho }_a)g{a}^2/(3\eta )$, consistent with a spherical clean bubble without surface tension. This distinguishes the drag calculation from the extra conditions needed for a freely rising bubble.

Augustin-Louis Cauchy's importance lies in the interaction of foundations, methods and applications. He made the justification of limits, series and integrals a programme for teaching and research; developed complex analysis into a source of powerful general methods; and contributed to permutation groups, geometry and continuum mechanics. His work is best assessed through particular achievements and their limitations, rather than by treating every theorem now bearing his name as a modern statement he already possessed. His engineering training and subsequent mathematical career also make a rigid separation of pure and applied work historically unhelpful.^[ref]

The Cours d'analyse by Augustin-Louis Cauchy (1821) makes convergence a condition requiring argument. A series is studied through its partial sums; a limit describes a variable approaching a fixed value; an infinitesimal is a variable tending to zero, rather than simply a mysterious fixed quantity smaller than every positive number. The introduction explicitly challenges unrestricted transfers of rules from finite expressions to infinite ones. This is not an abandonment of calculation: it seeks to identify when calculations are valid.^[ref]

For example, the Cauchy criterion for series asks whether tails can all be made small:This is a modern formulation of the convergence requirement, with the completeness of the real numbers understood. Merely requiring $a_n\to 0$ is weaker. In the harmonic series, grouping the terms from $2^j+1$ to $2^{j+1}$ gives a block of at least $1/2$, so its tails cannot satisfy the criterion. In contrast, the root test compares a tail with a geometric series: if $\mathrm{limsup}|a_n{|}^{1/n}<q<1$, then eventually $|a_n|\le q^n$, which proves absolute convergence. Such arguments explain the usefulness of convergence tests instead of reducing Augustin-Louis Cauchy's contribution to a vocabulary of definitions.

His Résumé des leçons sur le calcul infinitésimal by Augustin-Louis Cauchy (1823) extends the programme to differentiation and integration. For a continuous function on a compact interval, an integral can be approached through sums of values times subinterval lengths. A modern justification uses uniform continuity: for a partition of mesh at most $\delta$, two choices of sample points give sums differing by at most $(b-a){\omega }_f(\delta )$, where the modulus of continuity ${\omega }_f(\delta )$ tends to zero. Refinement then establishes the common limit. This reconstruction makes clear why control of error, rather than a geometrical picture of area alone, is decisive. It does not attribute the later full Riemann integral theory for discontinuous functions to the 1823 text.^[ref]

The achievement has limits. The 1821 Sum theorem of Cauchy asserts continuity of sums of convergent series of continuous functions in terms that, under a pointwise reading, are insufficient. The original proof treats the tail as negligible without establishing the uniform control needed near the point concerned. A modern diagnostic is $f_n(x)=x^n$ on $[0,1]$: every $f_n$ is continuous, but its pointwise limit is zero for $x<1$ and one at $1$. Thus a proof must distinguish pointwise convergence from uniform convergence. Augustin-Louis Cauchy's use of variable and infinitesimal language deserves historical interpretation, but it does not license silently importing modern hypotheses into his printed theorem.^[ref]

In complex analysis, the Mémoire sur les intégrales définies, prises entre des limites imaginaires by Augustin-Louis Cauchy (1825) exemplifies a new productive setting for integration. A contour integral depends on a path in the complex plane; deforming that path and accounting for isolated singularities can evaluate otherwise difficult definite integrals. In modern form, Cauchy's integral theorem applies to a holomorphic function on a suitable domain and Cauchy integral formula states, for a circle enclosing $z$ and lying with its interior in that domain,The importance is structural: local values and derivatives are determined by data on a surrounding path. Historical proofs and hypotheses should be distinguished from later general forms of the theorem.^[ref]

A short modern consequence measures the power of this method. For an entire function bounded by $M$, the derivative version of Cauchy integral formula on a circle of radius $R$ gives $|f^{\prime }(z)|\le M/R$. Since $R$ is arbitrary, $f^{\prime }=0$, proving Liouville's theorem. If a nonconstant polynomial $P$ had no root, then $1/P$ would be an entire function, bounded outside a large disc by the leading term and inside the disc by compactness. Liouville's theorem would force it to be constant, a contradiction. This proves the fundamental theorem of algebra. The chain is an assessment of the reach of the methods, not a claim that Augustin-Louis Cauchy alone originated every named result.

Augustin-Louis Cauchy's algebra concerns concrete permutations as well as analytical formulas. His 1845 investigations developed the theory of substitutions; Cauchy theorem for groups says that a prime number dividing a finite group order occurs as an element order. One can appreciate its generality from a modern argument: rotate the coordinates of the tuples $(g_1,\dots ,g_p)$ satisfying $g_1\cdots g_p=e$. Rotation preserves this condition, since $g_2\cdots g_p{g}_1=g_1^{-1}(g_1\cdots g_p)g_1=e$. There are $|G{|}^{p-1}$ such tuples; nonfixed rotation orbits have $p$ elements, and fixed tuples correspond to $g^p=e$. When $p\mid |G|$, the number of fixed tuples is divisible by $p$, so besides $e$ there is an element of element order $p$. This is explanatory mathematics, not Augustin-Louis Cauchy's original proof. The dating itself needs care: Peter Michael Neumann's archival argument places the major burst of work in 1845 despite an 1844 date traditionally attached to a volume.^[ref]

Finally, Augustin-Louis Cauchy's work on elasticity shows the strength of making a local mathematical object out of physical interactions. Instead of describing only a force perpendicular to a surface, one studies the traction on a plane through a point, with its direction dependent on the plane's unit normal. In modern notation the relation is $\mathbf{t}(\mathbf{n})=\mathbf{\sigma }\mathbf{n}$, where the Cauchy stress tensor contains the necessary directional information. Together with balance of linear momentum, this yields equations for the internal motion of a body. The modern tensor notation is later explanatory language; the conceptual contribution is the reduction of the forces on arbitrary internal planes to a finite collection of local components.^[ref]

Augustin-Louis Cauchy's lasting contribution was a powerful research programme, not a finished foundation. His insistence on justified passage to a limit reshaped analysis, while complex integration, substitutions and local stresses opened methods whose value extended beyond their first applications. The inadequacy of a pointwise convergence hypothesis, the need to state domains in complex analysis, and the later construction of the real numbers show that nineteenth-century rigour remained a collective and revisable achievement. These qualifications strengthen an assessment of his influence by identifying what later mathematicians could extend and repair.

The substrate contact area is the square bottom face, so the total traction force magnitude is stress times area:By Newton's third law, the force exerted by the substrate on the biological cell has the same magnitude and opposite direction to the cell's force on the substrate.

First convert the dynamic viscosity to SI units:Using the stipulated effective-size version of Stokes's law with $a=20\times 10^{-6}\mathrm{m}$, the drag coefficient isThe speed is $v=10^{-7}\mathrm{m}{\mathrm{s}}^{-1}$, henceThe drag acts opposite the velocity. Its magnitude is about $1.06\times 10^5$ times smaller than the substrate force in (a); balancing surrounding-fluid drag alone therefore cannot account for the much larger substrate traction. The calculation uses the size parameter exactly as supplied. An exact drag calculation for a cube moving close to a wall would need geometry and boundary corrections beyond this stipulated approximation.

Here a prime denotes $d/d\sigma$. It is useful to retain terms through $O({ϵ}^2)$ in the normal mechanical equilibrium equation, even though only first-order shape corrections are sought. The undeformed elastic membrane is tensionless, so a first-order change of curvature contributes to normal force only at second order.

The geometric equations written in reference arc length areTheir first-order Taylor expansions giveand henceThe principal stretches and the first-order constitutive equation areBecause $t_s=t_{\phi }=t$ exactly, the tangential mechanical equilibrium equation gives ${\partial }_{s}t=0$. Thus both $t_1$ and $t_2$ are spatial constants: $t_1^{\prime }=t_2^{\prime }=0$.

Expand the geometric factors in the normal balance:ConsequentlyMatching the first-order coefficient gives the requested relationSince the prescribed pressure has no second-order term, and ${\theta }_1^{\prime }=-r_1^{\prime \prime }$, its second-order coefficient givesA closed boundary value problem for the three requested unknowns is thereforeThe first equation has order two, the second order one, and the last order one, so the four boundary conditions match the total order. The unknown constant $t_2$ is selected by the axial endpoint constraint, rather than by an additional specified traction.

For $p\ne 0$, an equivalent formulation eliminates $z_1$ and displays that constraint as an integral:After solving this system, recover $z_1$ by integration from $z_1(0)=0$. No second derivatives of the unspecified function $f$ are required: they enter the next-order shape corrections through the second-order constitutive equation, whereas the displayed normal balance already determines the leading shape. The case $p=0$ has the undeformed solution $r_1=z_1=t_2=0$.

The end displacement relative to the initial unit axial length isBy Hooke's law, the total resisting force supplied by the spring is $F_3=-K\delta$. To convert this resultant into the axial Cauchy stress tensor, divide by the current end cross-sectional area. The transverse total stretches are both ${\alpha }^{-1/2}$, so that area isUniform traction over that face consequently givesEquate this to the expression from part (ii):The extra factor $\alpha$ on the left is essential: the spring law specifies a total force, whereas the material constitutive equation specifies force per current area. The reference end area is one, so the same spring resultant can also be read as a nominal end traction.

Write $W_i=\partial W/\partial {\alpha }_i$, evaluated at $({\alpha }^{-1/2},{\alpha }^{-1/2},\alpha )$. Applying the chain rule to the reduced strain energy density givesMultiply by $\alpha$ and use ${\alpha }_1={\alpha }_2={\alpha }^{-1/2}$:The transverse symmetry specified in the hint has $W_1=W_2$. ThusThe lateral faces have zero traction, so their diagonal Cauchy stress tensor components satisfy $t_1=t_2=0$. The hydrostatic pressure enforcing elastic incompressibility is therefore $p={\alpha }_1{W}_1={\alpha }_2{W}_2$, and the axial component simplifies toThis elimination of the Lagrange multiplier $p$ is what makes the reduced strain energy density useful for the spring balance.

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.