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The divergence is ${\partial }_x{u}_x+{\partial }_y{u}_y=1-1=0$, while the scalar vorticity is ${\partial }_x{u}_y-{\partial }_y{u}_x=0$. The flow is therefore incompressible flow and irrotational flow. Integrating the streamfunction equations givesThe time dependence must be retained: no time-independent function $\psi (x,y)$ represents this unsteady field for every $t$. At $t=0$ the streamlines are $xy=\mathrm{constant}$, with the coordinate axes as separatrices. Their arrows point towards increasing $|x|$ and decreasing $|y|$, as shown below.

A pathline solves the material ODEs, givingFor $x_0\ne 0$, elimination of time yieldsIf $x_0=0$, the pathline stays on the $y$ axis and the parametric expression remains valid; it cannot in general be written as a single-valued $y=f(x)$ along that vertical trajectory. For $x_0=y_0=1$, the path is $y=\log x-1+2/x$: it initially descends, reaches its minimum $(2,\log 2)$, then rises. Unlike a frozen streamline, it records the time-varying flow.

Irrotational flow permits $\mathbf{u}=\nabla \varphi$, and incompressible flow imposes $\nabla \cdot \mathbf{u}=0$, hence $\Delta \varphi =0$. In the laboratory frame, the no-penetration condition at the translating sphere is ${\partial }_r\varphi (a,\theta )=U\cos \theta$.

For separation of variables, take the angular factor $g(\theta )=\cos \theta$, which satisfies $\frac{1}{\sin \theta }(\sin \theta g^{\prime }{)}^{\prime }=-2g$. The radial equation iswith solutions $r$ and $r^{-2}$. Rest at infinity excludes the $r$ term. The normal boundary condition fixes the remaining coefficient, giving the potential flow around a translating sphereThere is no azimuthal velocity. With fluid density ${\rho }_f$, the kinetic energy isThe angular integral is $2\pi$, and the radial integral is $1/(3a^3)$, so $K_f={\rho }_f\pi a^3{U}^2/3$. Since $M=(4\pi a^3/3){\rho }_s$ and $k={\rho }_f/{\rho }_s$, $K_f=kM{U}^2/4$. Thus the added mass of a sphere is $kM/2$.

During a fall through $h$, gravity does work $Mgh$ and buoyancy does work $-kMgh$. Starting from rest, conservation of energy givesConsequently $U=\sqrt{4(1-k)gh/(2+k)}$. The mass cancels: the reduced speed comes from both buoyancy and the fluid's added mass.

For an incompressible flow, write the Newtonian fluid stress tensor as $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the rate-of-strain tensor. The Stokes equation gives ${\partial }_j{\sigma }_{ij}=0$. Symmetry of the Cauchy stress tensor therefore givesIntegrating and applying the divergence theorem expresses the viscous dissipation as boundary power:At a moving rigid body, the surface velocity is $U+\Omega \times (x-x_c)$. Its boundary-power contribution is $U\cdot F+\Omega \cdot G$, with the force and torque evaluated using the fluid's outward normal vector. Both resultants vanish for a force-free, torque-free inclusion. Thus the new viscous dissipation comes entirely from the unchanged outer boundary velocity. Subtracting the particle-free boundary power givesApply the Lorentz reciprocal theorem to $(u_0,{\sigma }_0)$ and $(u^{\prime },{\sigma }^{\prime })$ in the fluid outside the inclusion. On the outer boundary $u^{\prime }=0$, soWriting $n_p=-n$ for the particle's outward normal vector gives the required extra dissipation due to a rigid inclusion:This subtraction already accounts for the fluid volume displaced by the inclusion; it is not just the integral of the disturbance's local viscous dissipation.

For the sphere, the ambient rate-of-strain tensor $E$ is symmetric and trace free. The sphere in a uniform straining Stokes flow has zero translational and rotational velocity: inversion symmetry eliminates its force, and symmetry of $E$ eliminates its torque. To derive the disturbance, put $r=|x|$, $Q=x\cdot Ex$ and seekThe incompressible flow condition and Stokes equation reduce toA decaying family satisfying these equations is $f=A{r}^{-5}$, $g=B{r}^{-5}-5A{r}^{-7}/2$, $j=2B{r}^{-5}$. The no-slip boundary condition requires $f(a)=-1$ and $g(a)=0$, so $A=-a^5$ and $B=-5a^3/2$. ConsequentlyThe total velocity is zero at $r=a$, and the disturbance decays as $r^{-2}$. The pressure constant has been set to zero. These boundary and far-field conditions, together with Uniqueness of Stokes flow, establish the solution. There is no rotational background in a pure strain flow.

On the sphere $u^{\prime }=-u_0=-aE{n}_p$. Contracting the supplied Newtonian fluid stress tensor with $n_p$ makes its two terms proportional to $(n_p\cdot E{n}_p)n_p$ cancel, leaving $\sigma n_p=5\mu E{n}_p$. Hence the extra dissipation due to a rigid inclusion isHere ${\int }_{S^2}{n}_j{n}_{k}d\Omega =4\pi {\delta }_{jk}/3$. Taking the large outer boundary limit only after using the fixed-boundary power identity avoids replacing that prescribed boundary by an uncontrolled boundary at infinity.

If $N_v$ denotes the number of spheres per unit volume, $N_v(4\pi a^3/3)=\varphi$. ThusAdd this to the particle-free viscous dissipation density $2\mu E:E$. Comparing with $2{\mu }_{\mathrm{eff}}E:E$ gives the Einstein viscosity formula for a dilute suspension:to first order in $\varphi$. The hydrodynamic interactions neglected here contribute beyond that dilute order.