= Solution
With no net body force, the tensile resultant is constant along the thread: $3\mu Aw_z=F(t)$. In material coordinates $z_0$, volume conservation gives $D_tA=-F/(3\mu)$. Thus
$$
\boxed{A=A_0(z_0)-\Delta(t),\qquad
\Delta(t)=\frac1{3\mu}\int_0^tF(t')\,dt'.}
$$
The material Jacobian satisfies $A\,\partial z/\partial z_0=A_0$. Integrating with $z(0,t)=0$ gives
$$
\boxed{z=z_0+\int_0^{z_0}\frac{\Delta\,dz'_0}{A_0(z'_0)-\Delta}.}
$$
This remains valid while $\Delta<\min A_0$. The even profile now specified ensures symmetric endpoints $\pm L(t)$.
For $A_0=C(1+k^2z_0^2)$ and $k\ne0$, direct integration yields
$$
L=L_0+\frac{\Delta}{|k|\sqrt{C(C-\Delta)}}
\tan^{-1}\left(|k|L_0\sqrt{\frac C{C-\Delta}}\right).
$$
For $k=0$, instead $L=CL_0/(C-\Delta)$. A constant tensile $F>0$ gives $\Delta=Ft/(3\mu)$ and \b[the finite blow-up time]
$$
\boxed{t^*=\frac{3\mu C}{F}.}
$$
Put $\alpha=F/(3\mu)$ and $\delta=C-\Delta=\alpha(t^*-t)$. For fixed nonzero $k$, the arctangent tends to $\pi/2$, whereas for the uniform profile the whole length has the same thinning. Consequently \b[the <localized-neck and uniform-thread stretching exponents> are]
$$
\boxed{
\begin{aligned}
k\ne0:\quad&\beta=-\tfrac12,\qquad
B=\frac{\pi}{2|k|}\sqrt{\frac{3\mu C}{F}},\\
k=0:\quad&\beta=-1,\qquad B=\frac{3\mu CL_0}{F}.
\end{aligned}}
$$
A nonuniform thread's divergent integral is concentrated in a material interval $|z_0|\sim\sqrt{\delta/C}/|k|$ near its quadratic minimum. Its width shrinks like $\delta^{1/2}$ while the Jacobian grows like $\delta^{-1}$, yielding $\delta^{-1/2}$. The remote ends no longer affect this leading coefficient, explaining independence from $L_0$. For $k=0$ every element contributes the same $\delta^{-1}$ stretch. The limits $k\to0$ and $t\to t^*$ do not commute.
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