Past exam of the mathematical mechanical biology course of the University of Oxford / 2023 (C5.9, A13386W1)

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.

The material frame of a rod obeys the director kinematics of a rodIn the stress-free reference configuration, the constitutive equation gives $u_1=0$, $u_2=\hat{u}$, and $u_3=0$, since all three stiffnesses are positive. Thus $\mathbf{u}=\hat{u}{\mathbf{d}}_2$. Its cross product with ${\mathbf{d}}_2$ vanishes, soConsequently this director is a fixed vector in space, even though the other two directors rotate along the elastic rod.

The right-handed material frame of a rod has ${\mathbf{d}}_2\times {\mathbf{d}}_1=-{\mathbf{d}}_3$ and ${\mathbf{d}}_2\times {\mathbf{d}}_3={\mathbf{d}}_1$. Hence the director kinematics of a rod reduce toSolving these constant-coefficient ordinary differential equations with the given initial frame givesThe inextensible rod has ${\mathbf{r}}^{\prime }={\mathbf{d}}_3$. An integration using $\mathbf{r}(0)=0$ yields, for $\hat{u}\ne 0$,Indeed,so the centreline is a circular arc of radius $1/|\hat{u}|$ in the plane perpendicular to ${\mathbf{e}}_2$. The continuous $\hat{u}\to 0$ limit is ${\mathbf{d}}_1={\mathbf{e}}_1$, ${\mathbf{d}}_3={\mathbf{e}}_3$, and $\mathbf{r}=S{\mathbf{e}}_3$.

The second moment of area about an axis weights each element of cross-sectional area by its squared perpendicular distance from that axis. The ellipse extends farther along ${\mathbf{d}}_2$ than along ${\mathbf{d}}_1$, so its material lies, on balance, farther from the ${\mathbf{d}}_1$ axis. ThereforeWith zero intrinsic curvature of an elastic filament, the respective bending moments required by the constitutive equation are $E{I}_1\kappa {\mathbf{d}}_1$ and $E{I}_2\kappa {\mathbf{d}}_2$. Thus their magnitudes satisfyfor $\kappa \ne 0$: bending about the minor axis is harder. Here $E$ is Young's modulus, and $E{I}_i$ is the corresponding bending stiffness.

Zero applied end force and the force balance ${\mathbf{n}}^{\prime }=0$ imply $\mathbf{n}(S)=0$ everywhere. The moment balance then reduces to ${\mathbf{m}}^{\prime }=0$.

Choose an untwisted straight configuration: all three directors are constant, ${\mathbf{r}}^{\prime }={\mathbf{d}}_3$, and $\mathbf{u}=0$. The constitutive equation giveswhich is constant in space and therefore satisfies the moment balance. The required end bending moment is consequentlyIts sign opposes the positive intrinsic curvature of an elastic filament. If the initial position and frame from part (ii) are retained, this solution is simply $\mathbf{r}(S)=S{\mathbf{e}}_3$ with ${\mathbf{d}}_i={\mathbf{e}}_i$; the clamp supplies the opposite reaction moment at the other end.

Because the two elastic rods each have thickness $a$, their centrelines are separated by $a$ along the common director ${\mathbf{d}}_1$. The bonded geometry isDifferentiate with respect to reference arc length. The first rod is an inextensible rod, and the printed director kinematics of a rod giveFor the physically admissible orientation-preserving configuration $1-au>0$, the rods have the same unit tangent ${\mathbf{d}}_3^{+}={\mathbf{d}}_3$ and the same planar material frame of a rod. ThereforeIn particular, the second rod is shortened on the inside of a bend with $u>0$.

The bending strain of a rod is defined here by differentiation with respect to reference arc length. Since the reference labels match,The distinction from spatial curvature matters: the second centreline hasThus $u/(1-au)$ is its signed spatial curvature, rather than the $u^{+}$ appearing in the given elastic energy.

The first term is the bending energy of rod 1, whose preferred bending strain of a rod is zero. The second is the bending energy of rod 2 relative to its preferred intrinsic curvature of an elastic filament $\hat{u}$. The third is its stretching energy, with stretching stiffness $k{B}_2$ and stress-free stretch ratio ${\alpha }^{+}=1$. Each term is an energy per unit reference arc length before integration.

Substitute $u^{+}=u$ and ${\alpha }^{+}-1=-au$. The resulting quadratic function of $u$ has densityThere are no derivatives of $u$ or imposed end constraints in this minimum potential energy principle in elasticity. Sincethe unique pointwise minimizer is also the constant global minimizer:For example, completing the square giveswhich also proves uniqueness, up to equality almost everywhere, without merely assuming a uniform curvature.

With $k$, $a$ and $\hat{u}$ fixed, the two limits areIn the first limit, rod 1 dominates the bending energy and suppresses the bend. In the second, rod 1 offers negligible bending resistance, but following rod 2's preferred bending strain of a rod still shortens rod 2 through the bonded geometry. The competition between its bending energy and stretching energy leaves $u_{\ast }<\hat{u}$ when $\hat{u}>0$ and $k{a}^2>0$. Only when that stretching penalty also vanishes does $u_{\ast }\to \hat{u}$.

These formulae describe an admissible bonded elastic rod provided $1-a{u}_{\ast }>0$. If parameters instead force $a{u}_{\ast }\ge 1$, the unconstrained quadratic minimizer is outside the kinematic regime of the model; the given energy does not itself enforce positive stretch ratio.

Use the cylindrical coordinate system basis ${\mathbf{e}}_r,{\mathbf{e}}_{\phi },{\mathbf{e}}_z$. The meridional unit tangent vector and outward unit normal vector consistent with the given geometry areAn surface of revolution has $\theta$ independent of $\phi$. ConsequentlyApply the product rule to the vector force balance:Equating the three independent components to zero, and using $r_s=\cos \theta$, givesThus, away from the rotation axis where $r>0$, the scalar mechanical equilibrium equations areThe two geometric factors ${\theta }_s$ and $\sin \theta /r$ are the signed meridional and azimuthal principal curvatures in this convention. The normal equation is the Young–Laplace equation generalized to unequal principal membrane tensions.

Take $s=\sigma$ up to an irrelevant additive constant, $r=a$ constant, $z=z_0-\sigma$, and $\theta =\pi /2$. These satisfy the geometric relations and give the principal stretchesThe constitutive equations therefore produce constant membrane tensionsThe tangential mechanical equilibrium equations hold automatically, while the normal equation reduces toEvery positive root of this algebraic equation provides a cylindrical solution with no axial stretch ratio change. There is no further constraint on $t_s$ in the interior equations when end boundary conditions are discarded. In particular, a closed-end pressure balance would impose additional axial loading data that are absent here. Existence of a positive root for an arbitrary specified $P$ depends on the given constitutive equation.

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 $r_1=z_1=t_2=0$.

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.