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The opening of an internal cavity under tensile loading in a solid. For spherically symmetric deformation of an isotropic hyperelastic material with incompressibility, and an empty cavity without surface tension, radial static elastic equilibrium gives the displayed critical signed external pressure, where $h(\lambda )=W({\lambda }^{-2},\lambda ,\lambda )$ and positive $P$ denotes compression. Indeed, the principal stress formula for incompressible hyperelasticity gives $d{t}_r/d\lambda =h^{\prime }(\lambda )/(1-{\lambda }^3)$. Integration between the free inner surface and the outer surface, followed by the vanishing-initial-hole and vanishing-current-cavity limits, gives the formula. For a neo-Hookean solid with shear modulus $\mu >0$, the tensile magnitude is $-P_{\mathrm{crit}}=5\mu /2$.

The radial unit vector in spherical coordinates is ${\mathbf{e}}_R=\mathbf{X}/R$, so $\mathbf{X}\otimes \mathbf{X}/R^2={\mathbf{e}}_R\otimes {\mathbf{e}}_R$. The identity matrix is the sum of the three basis outer products. Substitution in the given deformation gradient givesThe reference and current radial directions coincide for this spherically symmetric deformation. The positive principal stretches are therefore ${\lambda }_r=f^{\prime }(R)$ and ${\lambda }_{\theta }={\lambda }_{\varphi }=f(R)/R$. Incompressibility imposesIntegrating and using $f(A)=a$ givesIn particular $b^3-a^3=B^3-A^3$, expressing conservation of shell volume. Writing $\lambda =r/R$ also gives ${\lambda }_r={\lambda }^{-2}$ and ${\lambda }_{\theta }={\lambda }_{\varphi }=\lambda$.

Let $W_r,W_{\theta },W_{\varphi }$ denote derivatives of the strain energy density with respect to its three principal stretches. The principal stress formula for incompressible hyperelasticity isIt follows by applying $\mathbf{T}=\mathbf{F}({\partial }_{\mathbf{F}}W{)}^T-p1$ in principal axes. Equal tangential principal stretches and material isotropy give $W_{\theta }=W_{\varphi }$ at $({\lambda }^{-2},\lambda ,\lambda )$. Differentiating the reduced strain energy density by the chain rule givesThe unknown pressure cancels in this difference. Substituting in part (i), and then using the supplied derivative of $\lambda$, givesConsequently the desired ordinary differential equation isFor completeness, the supplied derivative follows from $R^3=r^3-(a^3-A^3)$: $dR/dr=r^2/R^2={\lambda }^2$, and differentiating $\lambda =r/R$ gives $(1-{\lambda }^3)/R$. At the undeformed value $\lambda =1$ the division must be interpreted by a limit; the spherically symmetric deformation with $a=A$ has $\lambda \equiv 1$ and should be handled directly by the radial equilibrium equation.