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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.

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.