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Volume preservation by the elastic accommodation in morphoelasticity. If $\mathbf{F}=\mathbf{AG}$, elastic incompressibility imposes $\det \mathbf{A}=1$, and hence $\det \mathbf{F}=\det \mathbf{G}$. Total volume can increase through growth while the elastic response remains incompressible. A hydrostatic pressure acts as a Lagrange multiplier enforcing this constraint in the Cauchy stress tensor.

Fix a positive shear modulus $\mu$ and let the other Lamé parameter tend to infinity in $S_{\lambda }=\lambda \mathrm{tr}(e_{\lambda })I+2\mu e_{\lambda }$. If strain and stress converge to finite limits and differentiation commutes with the limit, taking the trace shows $\mathrm{tr}{e}_{\lambda }=O({\lambda }^{-1})$. Thus $\mathrm{div}u=0$ and $S=-pI+2\mu e$ for $p=-\lim \lambda \mathrm{tr}{e}_{\lambda }$. Force balance becomes the displayed Stokes equation, with an elastic shear modulus in place of fluid dynamic viscosity. The vanishing volumetric strain is the linearization of elastic incompressibility $\det Dy=1$.

The deformation gradient is the derivative of the deformation map with respect to the reference configuration coordinates, at fixed time:Its determinant $J=\det F$ is the local ratio of current to reference volume: $dv=JdV$ for an orientation-preserving deformation in continuum mechanics. Thus elastic incompressibility is the pointwise constraint $\boxed{J=1}$, rather than a requirement that each individual principal stretch be one.

For the axisymmetric deformation map, take $\theta =\Theta$, $z=Z$, and $r=r(R)$. A reference cylindrical shell of unit axial length has volume $2\pi RdR$, and the corresponding current shell has volume $2\pi rdr$. Elastic incompressibility therefore gives $rdr=RdR$.

Integrating from the inner surface, where $r(A)=a$, gives $r^2-a^2=R^2-A^2$. The positive radius is consequentlyThe outer surface is not an additional independent unknown: $\boxed{b=\sqrt{B^2+a^2-A^2}}$, and hence $h=b-a$.

Use orthonormal cylindrical bases in the reference configuration and current configuration. The deformation gradient maps a radial increment $dR$ to $r^{\prime }dR$, an azimuthal arc $Rd\Theta$ to $rd\Theta$, and an axial increment $dZ$ to itself. ThereforeDifferentiating the radius relation gives $r^{\prime }=R/r$. Its three principal stretches areTheir product is one, as required by elastic incompressibility. The symbols called “strains” in the source are these stretch ratios; engineering strain would instead subtract one. The axial stretch is one because this is plane strain.

The multiplier enforcing elastic incompressibility contributes an arbitrary hydrostatic stress. Denote it by $q(r)$ to distinguish it from the applied pressure $P$. For the neo-Hookean solid, the Cauchy stress tensor is $\tau =-qI+\mu F{F}^T$. Thus its principal components areThe axial component is $t_z=-q+\mu$, despite the unit axial stretch. The quantity $q$ is a Lagrange multiplier determined together with equilibrium and the traction conditions; it is not obtained by differentiating the reduced stretch energy while ignoring its constraint.

In particular the stress difference eliminates this multiplier:

Solutions to all three questions and all sixteen terminal parts of the mathematical mechanical biology course of the University of Oxford examination, sat on 7 June 2023. Source: original released PDF, inclusive bundle pages 335–338; paper pages 1–4. In Question 1(b), bending strain of a rod is measured per reference arc length, exactly as in the printed kinematic identities. In Question 3(b), elastic incompressibility constrains the elastic part of the deformation gradient; growth can still increase the total volume.

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.

In morphoelasticity, the deformation gradient factors as $\mathbf{F}=\mathbf{AG}$, where $\mathbf{G}$ accounts for local stress-free growth and $\mathbf{A}$ accounts for the subsequent elastic accommodation. Elastic incompressibility means that the latter preserves volume:It does not forbid volume increase by growth: here $\det \mathbf{F}=\det \mathbf{G}=\gamma$.

For the homogeneous diagonal deformation, the two unloaded transverse directions are equivalent. Taking the transversely symmetric branch implicit in the paper's hint gives ${\alpha }_1={\alpha }_2>0$. Set ${\alpha }_3=\alpha >0$. The determinant constraint is thensoEquality of the transverse stretches uses that symmetry assumption; elastic incompressibility alone would only give ${\alpha }_1{\alpha }_2{\alpha }_3=1$. The total deformation gradient and the actual axial length arebecause the initial cuboid is a unit cube.

For the neo-Hookean solid, the axial Cauchy stress tensor under the unloaded lateral boundary conditions isWith $K,\mu >0$, the spring balance is equivalent, after multiplication by $\alpha >0$, to the cubic equationFor a fixed $\gamma \ge 1$, write its left side as $h(\alpha )$. It obeys $h(0)=-\mu <0$ and tends to $+\infty$ as $\alpha \to \infty$. MoreoverThus $h$ first decreases from a negative value and then increases strictly, crossing zero exactly once on $\alpha >0$. For $\gamma >1$,so the unique equilibrium has $1/\gamma <\alpha <1$. The tissue is elastically compressed, while its actual length $\alpha \gamma$ exceeds one and compresses the spring.

To determine the large-growth limit without presuming a power law, solve for $\gamma$:This is a strictly decreasing function of $\alpha >0$, since its derivative isIt tends to $+\infty$ only as $\alpha \to 0^{+}$. Consequently unlimited growth forces $\alpha \to 0^{+}$, andMore precisely, the dominant balance is $K\gamma {\alpha }^3\sim \mu$, givingThe surrounding spring softens the divergence relative to rigid confinement but does not bound the compressive stress. Both axial extension and lateral expansion help accommodate growth:whose product is the total volume $\gamma$, as required by elastic incompressibility and the imposed growth tensor.

A dead-load traction is prescribed force per unit reference area whose vector is independent of the deformation. It is imposed through $T_{R}N=t_R$ on the reference boundary. It does not rotate or rescale with the current surface as a follower load would.

Elastic incompressibility imposes $\det F=1$ and adds a scalar Lagrange multiplier to the constitutive law. With the pressure sign convention used here,The pressure field is determined by equilibrium and boundary conditions, rather than by the strain energy density alone.

For the diagonal homogeneous deformation, $F=\mathrm{diag}(v_1,v_2,v_3)$ and $v_1{v}_2{v}_3=1$. Isotropy makes $DW(F)$ diagonal here, with entries ${\Phi }_i$, soThe six face conditions are exactly $T_R{e}_i=\tau e_i$. Thus ${\Phi }_i-p/v_i=\tau$ for each $i$, orNecessity follows from the boundary conditions. Conversely these equations give a constant pressure and constant stress satisfying those conditions, while $\mathrm{Div}{T}_R=0$ because the stress is constant and there is no body force. This proves sufficiency as well.

For the particular energy, ${\Phi }_i=\mu (v_i-v_i^{-2})$. Put $S=v_1+v_2+v_3$ and $Q=v_1{v}_2+v_1{v}_3+v_2{v}_3$. Elastic incompressibility gives $v_i^{-1}=v_j{v}_k$, henceEquality for every pair is therefore equivalent to $(\mu S-\tau )(v_i-v_j)=0$. If all stretches are equal, their positive product one gives $v_1=v_2=v_3=1$, with $p=-\tau$. Otherwise $S=\tau /\mu$. Conversely any positive triple with this sum and product one solves the equations, with $p=-\mu Q$. Thus the complete solutions within the stated homogeneous diagonal ansatz are the unit triple and the stated sum-product family.

By the arithmetic-geometric mean inequality, $S\ge 3$, with equality only at the unit triple. Consequently $\tau \le 3\mu$ admits only that triple, including equality, when the two descriptions coincide.

To prove infinitely many for $s=\tau /\mu >3$, choose $v_1=t$ near one and setAt $t=1$ the discriminant is $(s-1{)}^2-4>0$; it stays positive on an interval about one, and both roots remain positive. Every such $t$ gives a different admissible triple. HenceThe degeneracy has an energy explanation: under the product constraint the strain energy density is $\mu S^2/2$, and the total potential under dead loading is $\mu S^2/2-\tau S$. It depends on the three stretches only through their sum. Its minimum over attainable $S\ge 3$ moves from $S=3$ to the whole level set $S=\tau /\mu$ when the load exceeds $3\mu$.

Taking traces of the constitutive equation givesThe assumed finite limits of stress and strain make the right side bounded. Therefore $\mathrm{tr}{e}_{\lambda }=O({\lambda }^{-1})$, and the limiting strain satisfies $\mathrm{tr}e=0$. The permitted interchange of derivatives and limits gives $\mathrm{tr}e=\mathrm{div}u$. DefineThen the limit stress is $S=-pI+2\mu e$. Passing to the limit in force balance and using $2\mathrm{Div}e=\Delta u+\nabla \mathrm{div}u$ givesThe pressure acts as the Lagrange multiplier enforcing the volume constraint. The differential equations determine it up to an additive constant, while a specified absolute limit stress can fix that constant.

Nonlinear elastic incompressibility is $\det Dy=1$. Expanding the determinant for $Dy=I+Du$ gives $\det (I+Du)=1+\mathrm{div}u+O(|Du{|}^2)$, so its first-order form is precisely $\mathrm{div}u=0$. The volumetric energy term $\lambda (\mathrm{div}u{)}^2/2$ becomes increasingly expensive as $\lambda$ grows, while the fixed positive $\mu$ leaves shear strain possible. Thus the infinite-bulk-stiffness limit is the incompressible limit of isotropic linear elasticity, with the same equilibrium operator as the steady Stokes equations.