Kelvin solution for a point force

Example

Consider an infinite homogeneous isotropic linear elastic body filling $\R^3$, with shear modulus $\mu > 0$ and Poisson's ratio $\nu \in (-1, 1/2)$. A point force $F \in \R^3$ acts at the origin, so the body force is $F \delta_0$, where $\delta_0$ is the Dirac delta. The Kelvin solution is the displacement $$u_i(x) = \frac{1}{16 \pi \mu (1 - \nu)} \left( \frac{(3 - 4\nu) \, \delta_{ij}}{r} + \frac{x_i x_j}{r^3} \right) F_j, \qquad r = \abs{x} \ne 0,$$ with summation over $j$. It solves the Lamé equations $\mu \Delta u + (\lambda + \mu) \nabla (\nabla \cdot u) + F \delta_0 = 0$ in the sense of distributions, where $\lambda = 2\mu\nu/(1 - 2\nu)$ is the first Lamé parameter, and $u(x) \to 0$ as $\abs{x} \to \infty$.

Check away from the origin: let $\psi = \frac{F}{8\pi(1-\nu) r}$, a vector field whose components are harmonic on $\R^3 \setminus \{0\}$. Using $\nabla \frac{x \cdot F}{r} = \frac{F}{r} - \frac{(x \cdot F) x}{r^3}$, one finds $u = \frac{1}{2\mu} \big( 4(1-\nu) \psi - \nabla (x \cdot \psi) \big)$, a Papkovich–Neuber representation. Since $\Delta (x \cdot \psi) = 2 \nabla \cdot \psi$, this gives $\Delta u = -\frac{1}{\mu} \nabla (\nabla \cdot \psi)$ and $\nabla \cdot u = \frac{1 - 2\nu}{\mu} \nabla \cdot \psi$. With $\lambda + \mu = \frac{\mu}{1 - 2\nu}$, the sum $\mu \Delta u + (\lambda + \mu) \nabla (\nabla \cdot u)$ is $-\nabla (\nabla \cdot \psi) + \nabla (\nabla \cdot \psi) = 0$. The constant in front is fixed by requiring that the tractions on every sphere centred at $0$ balance the point force $F$. The stresses decay like $r^{-2}$.