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With no net body force, the tensile resultant is constant along the thread: $3\mu A{w}_z=F(t)$. In material coordinates $z_0$, volume conservation gives $D_{t}A=-F/(3\mu )$. ThusThe material Jacobian satisfies $A\partial z/\partial z_0=A_0$. Integrating with $z(0,t)=0$ givesThis remains valid while $\Delta <\min A_0$. The even profile now specified ensures symmetric endpoints $\pm L(t)$.

For $A_0=C(1+k^2{z}_0^2)$ and $k\ne 0$, direct integration yieldsFor $k=0$, instead $L=C{L}_0/(C-\Delta )$. A constant tensile $F>0$ gives $\Delta =Ft/(3\mu )$ and the finite blow-up timePut $\alpha =F/(3\mu )$ and $\delta =C-\Delta =\alpha (t^{\ast }-t)$. For fixed nonzero $k$, the arctangent tends to $\pi /2$, whereas for the uniform profile the whole length has the same thinning. Consequently the localized-neck and uniform-thread stretching exponents areA 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 $k\to 0$ and $t\to t^{\ast }$ do not commute.