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Use the Papkovich–Neuber representation in the conventionIt gives $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=-\mu \nabla (\nabla \cdot \mathbf{\Phi })=\nabla p$. Additive pressure constants are immaterial.

Let $q=\mathbf{x}\cdot \mathbf{Ex}$, with $\mathbf{E}$ symmetric and traceless. The harmonic vector $\mathbf{Ex}/r^3$ is a derivative of $1/r$. The scalar $q/r^5$ is a decaying degree-two harmonic because $q$ is a harmonic homogeneous quadratic. These Papkovich potentials for a strained sphere have precisely the strain symmetry and decay required for the disturbance, so trySubstitution givesAt $r=a$, require ${\mathbf{u}}^{\prime }=-\mathbf{Ex}$. The independent vector structures give $B=a^5$ and $A=-5a^3/3$. The disturbance for a sphere in a uniform straining Stokes flow isIt decays at infinity and satisfies no slip exactly at the sphere. Its leading far field is a stresslet, of velocity scale $E{a}^3/r^2$.

A force-free sphere in a linear background translates with ${\mathbf{U}}_0$, by Faxén's first law, and a couple-free sphere rotates with the background rigid-body rotation $\mathbf{\Omega }$, by Faxén's rotational law. Uniform translation and solid-body rotation can then be matched without disturbance. The remaining relative boundary velocity is exactly $-\mathbf{Ex}$. The Linearity of Stokes flow and Uniqueness of Stokes flow therefore give the same disturbance. The pure strain produces no resultant force or couple, by its symmetric degree-two angular structure.