= Solution
{c}
For the <neo-Hookean solid>, the axial <Cauchy stress tensor> under the unloaded lateral <boundary conditions> is
$$
t_3=\mu(\alpha^2-\alpha^{-1}).
$$
With $K,\mu>0$, the spring balance is equivalent, after multiplication by $\alpha>0$, to the <cubic equation>
$$
\boxed{(K\gamma+\mu)\alpha^3-K\alpha^2-\mu=0.}
$$
For a fixed $\gamma\ge1$, write its left side as $h(\alpha)$. It obeys $h(0)=-\mu<0$ and tends to $+\infty$ as $\alpha\to\infty$. Moreover
$$
h'(\alpha)=\alpha\big[3(K\gamma+\mu)\alpha-2K\big].
$$
Thus $h$ first decreases from a negative value and then increases strictly, crossing zero exactly once on $\alpha>0$. For $\gamma>1$,
$$
h(1)=K(\gamma-1)>0,\qquad
h(1/\gamma)=\mu(\gamma^{-3}-1)<0,
$$
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 $\gamma$:
$$
\gamma=\frac1\alpha+\frac\mu K(\alpha^{-3}-1).
$$
This is a strictly decreasing function of $\alpha>0$, since its <derivative> is
$$
-\alpha^{-2}-\frac{3\mu}{K}\alpha^{-4}<0.
$$
It tends to $+\infty$ only as $\alpha\to0^+$. Consequently unlimited growth forces $\alpha\to0^+$, and
$$
\boxed{t_3=\mu(\alpha^2-\alpha^{-1})\to-\infty.}
$$
More precisely, the <dominant balance> is $K\gamma\alpha^3\sim\mu$, giving
$$
\boxed{\alpha\sim\left(\frac\mu{K\gamma}\right)^{1/3},\qquad
t_3\sim-\mu^{2/3}(K\gamma)^{1/3}.}
$$
The 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:
$$
\ell=\alpha\gamma\sim\left(\frac\mu K\right)^{1/3}\gamma^{2/3},\qquad
\mathcal A=\alpha^{-1}\sim\left(\frac K\mu\right)^{1/3}\gamma^{1/3},
$$
whose product is the total volume $\gamma$, as required by <elastic incompressibility> and the imposed <growth tensor>.
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