= Solution
Choose the extending direction as $x$ and write $s=\dot\gamma>0$. The <uniaxial extensional flow>, satisfying <incompressibility>, has
$$
K=\operatorname{diag}(s,-s/2,-s/2).
$$
Starting from rest, symmetry preserves $A=\operatorname{diag}(a,b,b)$. The <differential pom-pom model> tensor equation becomes
$$
\dot a=(2s-1/\tau_1)a+1/\tau_1,\qquad
\dot b=-(s+1/\tau_1)b+1/\tau_1,\qquad a(0)=b(0)=1.
$$
Put $x=s\tau_1$. For $x\ne1/2$, the solutions are
$$
a(t)=\frac{1-2x e^{(2x-1)t/\tau_1}}{1-2x},\qquad
b(t)=\frac{1+x e^{-(1+x)t/\tau_1}}{1+x}.
$$
If $x<1/2$, both transients decay, giving
$$
\boxed{A_\infty=\operatorname{diag}\left(\frac1{1-2x},\frac1{1+x},\frac1{1+x}\right).}
$$
If $x>1/2$, the axial component is instead $a=(2x e^{(2x-1)t/\tau_1}-1)/(2x-1)$ and grows exponentially without bound, while $b\to1/(1+x)$. At the boundary $x=1/2$, which lies between the two stated regimes, $a=1+t/\tau_1$ grows linearly and $b\to2/3$.
Normalization makes the molecular orientation finite even when the unnormalized <conformation tensor> diverges. For $0\le x<1/2$, direct division by $a+2b$ gives
$$
\boxed{B_\infty=\operatorname{diag}\left(\frac{1+x}{3(1-x)},\frac{1-2x}{3(1-x)},\frac{1-2x}{3(1-x)}\right).}
$$
For $x>1/2$, the divergent axial component dominates the <matrix trace>, so
$$
\boxed{B_\infty=\operatorname{diag}(1,0,0).}
$$
The same normalized limit holds at $x=1/2$, agreeing continuously with the limit of the subcritical expression. The limiting orientations have nonnegative <eigenvalues> and unit <matrix trace>, as required for the normalized molecular orientation.
Now set $S(t)=B(t):K=s(a-b)/(a+2b)$. Its limiting value is
$$
S_\infty(s)=\begin{cases}
s^2\tau_1/(1-s\tau_1),&s\tau_1<1/2,\\
s,&s\tau_1\ge1/2.
\end{cases}
$$
The <scalar> stretch equation is $\dot\lambda=[S(t)-1/\tau_2]\lambda+1/\tau_2$. If $S_\infty<1/\tau_2$, its coefficient is eventually bounded above by a negative constant. The <variation of constants> formula gives a bounded solution, converging to
$$
\lambda_\infty=\frac1{1-\tau_2S_\infty}.
$$
If $S_\infty>1/\tau_2$, the coefficient is eventually positive, and the positive stretch grows without bound. At equality a bounded stretch would imply $\dot\lambda\to1/\tau_2>0$, which contradicts boundedness. Thus equality is not an additional bounded steady regime.
The function $S_\infty(s)$ is continuous and strictly increasing from zero to infinity. Consequently there is a unique critical extension rate defined by
$$
\tau_2 S_\infty(\dot\gamma_c)=1,
$$
and \b[the molecular stretch remains bounded precisely for $\dot\gamma<\dot\gamma_c$]. This is distinct from the tensor threshold $\dot\gamma\tau_1=1/2$. If $\tau_2<2\tau_1$, for example, the normalized orientation can remain well defined and the stretch remain bounded for some rates above the tensor threshold; divergence of $A$ alone does not prove divergence of the physical <stress>.
In the requested doubly subcritical regime, write $D=1-s\tau_1-s^2\tau_1\tau_2>0$. The <steady uniaxial extension of the differential pom-pom model> has
$$
\lambda_\infty=\frac{1-s\tau_1}{D},\qquad
B_{xx,\infty}-B_{yy,\infty}=\frac{s\tau_1}{1-s\tau_1}.
$$
The isotropic <pressure> cancels from the tensile <normal-stress difference>. Hence the <extensional viscosity> is
$$
\eta_E=\frac{\sigma_{xx}-\sigma_{yy}}s
=\frac{G\lambda_\infty^2}{s}(B_{xx,\infty}-B_{yy,\infty}),
$$
or
$$
\boxed{\eta_E(s)=\frac{G\tau_1(1-s\tau_1)}{(1-s\tau_1-s^2\tau_1\tau_2)^2},\qquad
s\tau_1<\frac12,\quad s<\dot\gamma_c.}
$$
Its zero-rate limit is $G\tau_1=3\mu_0$, recovering the <Trouton ratio> of three. The restrictions exclude both loss of a finite tensor steady state and loss of a finite stretch steady state; neither denominator may be continued through its singularity as a physical steady constitutive response.
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