= Solution
{c}
A <dead-load traction> is prescribed <force> per unit reference area whose vector is independent of the deformation. It is imposed through $T_RN=t_R$ 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,
$$
T_R=DW(F)-p\,\operatorname{cof}F=DW(F)-pF^{-T},\qquad
T=DW(F)F^T-pI.
$$
The <pressure> field is determined by <equilibrium> and <boundary conditions>, rather than by the <strain energy density> alone.
For the diagonal <homogeneous deformation>, $F=\operatorname{diag}(v_1,v_2,v_3)$ and $v_1v_2v_3=1$. <Isotropy> makes $DW(F)$ diagonal here, with entries $\Phi_i$, so
$$
(T_R)_{ii}=\Phi_i-\frac p{v_i}.
$$
The six face conditions are exactly $T_Re_i=\tau e_i$. Thus $\Phi_i-p/v_i=\tau$ for each $i$, or
$$
\boxed{v_1\Phi_1-\tau v_1=v_2\Phi_2-\tau v_2
=v_3\Phi_3-\tau v_3=p,\qquad v_1v_2v_3=1.}
$$
Necessity follows from the <boundary conditions>. Conversely these equations give a constant <pressure> and constant <stress> satisfying those conditions, while $\operatorname{Div}T_R=0$ 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 $Q=v_1v_2+v_1v_3+v_2v_3$. <Elastic incompressibility> gives $v_i^{-1}=v_jv_k$, hence
$$
v_i\Phi_i-\tau v_i
=\mu(v_i^2-v_jv_k)-\tau v_i
=(\mu S-\tau)v_i-\mu Q.
$$
Equality 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 \b[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>, $S\geq3$, with equality only at the unit triple. Consequently $\tau\leq3\mu$ 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 set
$$
v_{2,3}=\frac{s-t\pm\sqrt{(s-t)^2-4/t}}2.
$$
At $t=1$ 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. Hence
$$
\boxed{\tau\leq3\mu:\text{ one triple};\qquad
\tau>3\mu:\text{ infinitely many triples}.}
$$
The 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 $S\geq3$ moves from $S=3$ to the whole level set $S=\tau/\mu$ when the load exceeds $3\mu$.
\Image[/past-exam-of-the-solid-mechanics-course-of-the-university-of-oxford/2007-cube-equilibrium-family.png]
{title=The positive stretch triples form a continuum above the critical <dead load>}
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