Plane strain

In plane strain, the deformation map has $x_1=x_1(X_1,X_2)$, $x_2=x_2(X_1,X_2)$ and $x_3=X_3$. The out-of-plane stretch is one and out-of-plane shear vanishes. This kinematic condition need not make the out-of-plane stress vanish.

For an isotropic elastic annulus $a<r<b$ with internal excess pressure $P$ and zero outer gauge pressure, radial equilibrium gives $u=Cr+D/r$. Its stress tensor has ${\tau }_{rr}=2(\lambda +\mu )C-2\mu D/r^2$ and ${\tau }_{\theta \theta }=2(\lambda +\mu )C+2\mu D/r^2$. Enforcing ${\tau }_{rr}(a)=-P$, ${\tau }_{rr}(b)=0$ gives $2\mu D=P{a}^2{b}^2/(b^2-a^2)$, so the maximum difference occurs at the bore and equals the displayed expression. Once an appropriate yield criterion is reached, stress must be related to elastic strain while total strain includes a plastic contribution. The elastic formula alone cannot describe further irreversible deformation.

The displayed scalar stress potential automatically satisfies the two in-plane static momentum equations without body force, because the mixed third derivatives cancel. For homogeneous isotropic linear elasticity, strain compatibility further makes $A$ biharmonic. Since stresses depend only on second derivatives, $A$ is unique only up to $ax+by+c$ on a connected domain. Conversely a function with all second derivatives zero is affine, proving that this is exactly the gauge freedom. It does not license adding an arbitrary biharmonic function, which generally changes stress.

For plane strain in $y>0$ under surface normal traction $-P(x)$ and zero surface shear, use the forward Fourier transform phase $e^{ikx}$. At $k\ne 0$, biharmonicity gives $({\partial }_y^2-k^2{)}^2\hat{A}=0$, whose decaying part is $(C+Dy)e^{-|k|y}$. Boundary traction fixes $C=\hat{P}/k^2$, $D=|k|C$, proving the displayed formula. Constitutive inversion and a decaying displacement mode give $2\mu \hat{v}=[y+2(1-\nu )/|k|]\hat{P}{e}^{-|k|y}$. Multiplication by $-ik$ produces an unambiguous displacement-gradient convolution kernel.

The zero mode is separate: a nonzero resultant load $\hat{P}(0)$ conflicts with an ordinary fully decaying displacement normalization in two dimensions and leads to a logarithmic reference issue. Balanced loading or a distributional/reference-displacement treatment is required before extending the transformed formulas globally. Dividing by $k^2$ and assigning a finite value at $k=0$ is invalid.