= Solution
For a differentiable <crack> path, the local quasistatic Griffith laws with constant positive toughness are
$$
\boxed{\dot\ell(t)\geq0,\qquad
G(\ell(t),t)\leq G_c,\qquad
\dot\ell(t)[G_c-G(\ell(t),t)]=0.}
$$
They express irreversibility, local stability, and energy balance during growth. A strictly subcritical release rate forces the <crack> to stop.
Fix a <crack> parameter $s$ and let $u(s,t)$ denote the equilibrium <displacement> with the prescribed time-dependent loads. The scalar equilibrium problem is linear and has a unique solution for its fixed Dirichlet data. Both body force and boundary <displacement> scale by $t$, so
$$
\boxed{u(s,t)=t\,u(s,1).}
$$
The minimized potential energy is quadratic under this scaling: the strain term scales by $t^2$ and so does the load term $-f(t)u(s,t)$. Hence $E(s,t)=t^2E(s,1)$ and
$$
\boxed{G(s,t)=t^2G(s,1).}
$$
Equivalently the <crack>-tip coefficient scales by $t$ and its squared energy-release formula gives the same result.
Put $s_0=\ell(0)$. Finiteness and <continuity> of $G(s,1)$ make it bounded by some finite $M\geq0$ on a neighborhood of $s_0$. <Continuity> of $\ell$ supplies a positive interval on which $\ell(t)$ stays in that neighborhood. Shorten it to $[0,t_0]$ so that $t_0^2M<G_c$. Then $G(\ell(t),t)\leq t_0^2M<G_c$, and complementarity forces $\dot\ell(t)=0$ throughout. Therefore
$$
\boxed{\ell(t)=\ell(0)\quad(0\leq t\leq t_0).}
$$
No differentiating an unknown <crack> path or assuming its length already constant is needed; its <continuity> first gives the local release-rate bound.
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