Past exam of the mathematical mechanical biology course of the University of Oxford / 2023 / 3 (Constrained growth)

Before contact with both walls, the elastic rod can grow freely, so its elastic stretch ratio is $\alpha =1$ and its length is its stress-free grown length,The first contact occurs when $\ell (t_c)=L_1$. For the intended growth case $c>0$,The condition $L_1>L_0$ ensures $t_c>0$. If $c=0$, contact never occurs; negative $c$ describes shrinkage rather than the stated continuing growth.

Write $\lambda =L_1/L_0>1$. The one-dimensional multiplicative decomposition of the deformation gradient reads $\ell /L_0=\alpha \gamma$. Before contact, $\alpha =1$; afterwards, the imposed fixed length givesSubstituting in the linear constitutive equation yieldsHere negative axial stress means compression. The two branches agree at $t_c$ because $1+c{t}_c=\lambda$. At late times,Thus the law permits arbitrarily severe elastic shortening at a finite compressive stress. It is a small-strain approximation, appropriate near $\alpha =1$, and is not a credible material law as $\alpha \to 0$. The one-dimensional constraint rules out buckling in this particular model; the finite limiting stress is a defect of extrapolating the linear constitutive equation, even with that constraint retained.

The kinematic relation is unchanged: $\alpha =1$ before contact and $\alpha =\lambda /(1+ct)$ afterwards. Therefore the nonlinear constitutive equation givesThe first term in the post-contact expression is $O(t^{-2})$. Hence its asymptotic expansion isIn this model the compressive axial stress diverges in magnitude, because the $-{\alpha }^{-1}$ term penalizes vanishing elastic stretch ratio without bound.

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.

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.

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.

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.