 Solution

Use $-\pi <{\theta }_s<\pi$, with the crack along the negative radial axis. In the coefficient convention used here, the leading crack-tip expansion isThe square-root term is harmonic and has zero angular derivative on both crack faces. Regular contributions may include a constant and bounded-gradient terms under smooth loading.

If $E(s)$ is the minimized elastic-plus-load potential, the energy release rate is $\boxed{G(s)=-dE(s)/ds}$. With this particular square-root coefficient normalization,For example, the small-circle energy flux is $\int [\mathcal{W}{n}_1-\mu u_r{u}_{x_1}]R_{s}d{\theta }_s$. The singular term gives $\mathcal{W}=\mu k_s^2/(8R_s)$, $u_r=k_s\sin ({\theta }_s/2)/(2\sqrt{R_s})$ and $u_{x_1}=-k_s\sin ({\theta }_s/2)/(2\sqrt{R_s})$. Integration over $(-\pi ,\pi )$ produces the stated factor. Other stress-intensity normalizations change that factor.

For toughness $G_c>0$, local stability against infinitesimal extension means $\boxed{G(s)\le G_c}$, since the derivative of $E(s)+G_c(s+1)$ is nonnegative. Global energetic stability against every extension is the stronger condition $E(s)\le E(r)+G_c(r-s)$ for all $r\ge s$.

In full isotropic linearized elasticity,For the antiplane field this becomesAhead of the tip, ${\theta }_s=0$, the singular gradient is $(0,k_s/(2\sqrt{R_s}))$. ThusThis is stronger than the printed remainder $o(R_s^{-1})$. If $k_s$ denotes a nonnegative intensity, it has the printed leading coefficient; for a signed singular coefficient its absolute value is essential because the left side is a norm.