# Incompressible Fluid Flow around sphere (Stokes)

Stokes solved this 1851. I have a question regarding the derivation. Following Batchelor the equations to be solved are \begin{align} \nabla \left( \frac{p - p_0}{\mu} \right) = \nabla^2 \vec{u} = -\nabla \times \vec{\omega} \\ \nabla \cdot \vec{u} = 0 \end{align} with boundary conditions $\vec{u} = \vec{U}$ at the surface of the sphere where U is the velocity of the sphere in z-direction. Also $\vec{u} \rightarrow 0, p-p_0 \rightarrow 0 \quad {\rm as} \quad r \rightarrow \infty$. The center of the sphere is instantaneous at the origin of the co-ordinate system. $\vec{\omega}=\nabla \times \vec{u}$ is the vorticity and $\mu$ the viscosity. Since $p-p_0$ solves the Laplace equation it can be written as a series in solid harmonics \begin{align} \frac{p-p_0}{\mu} = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} c_{lm} I_{l}^m \quad {\rm .} \end{align} The m-sum is absent because of the azimutal symmetry. He argues that $p-p_0 \propto \vec{U} \cdot \vec{r} \propto \cos \theta$ and therefore the only term $\neq 0$ is $l=1$. I cannot see why $p-p_0 \propto \cos \theta$ should be fulfilled though.

• Is it consistent with the differential equations and boundary conditions? – Chet Miller May 30 '18 at 11:15
• Is what consistent? – Diger May 30 '18 at 11:27
• The pressure dependence on the cosine of the angle. – Chet Miller May 30 '18 at 12:04

Stokes equation is linear in its dependent variables. Further its boundary condition is linearly dependent on translation velocity $\mathbf{U}$. This means that if $\{\mathbf{u},p\}$ is the solution of Stokes equation for translation velocity $\mathbf{U}$, then if the translation velocity is scaled to $\alpha\mathbf{U}$, the new solution solution corresponding to this new boundary condition is simply $\{\alpha\mathbf{u},\alpha p\}$. This suggests that both $\mathbf{u}$ and $p$ must be linear functions of $\mathbf{U}$.

In 3-dimensions, a solution of the Laplace equation is $1/r$, i.e. $\nabla^2(1/r)=0$, in which $r$ is the radial distance to a point in the flow field from the sphere center. Further solutions to the Laplace equation may be generated by taking successive gradients of $1/r$. Of course each new solution generated in this way is of higher tensorial rank than the previous one. The solutions are: $$\frac{1}{r},\nabla\frac{1}{r},\nabla\nabla\frac{1}{r},\ldots$$

Now the pressure field, which is a scalar field, must be linear in $\mathbf{U}$ while involving one (or more) of the solutions listed above. The only viable combination is $$p=\lambda \mathbf{U}\cdot\nabla\frac{1}{r}=-\lambda\frac{\mathbf{U}\cdot\mathbf{r}}{r^3}$$ in which $\lambda$ is a constant. This is how $p$ (or $p-p_0$, if your datum is not zero) is found to be proportional to $\cos\theta$, in which $\theta$ is the angle between the direction of motion and the position vector of a point in the flow field.

Reference: Guazzelli & Morris, A Physical Introduction to Suspension Dynamics, Cambridge University Press.

• Why is it necessary that p depends on $\vec{U}$ via scalar product and not just in terms of the absolute value? It would still scale with $\alpha$. – Diger May 30 '18 at 17:39
• @Diger Because the boundary condition involves the vector velocity and not its magnitude. If $p$ depended on the absolute magnitude $U$ then so would the flow velocity $\mathbf{u}$, via Stokes equation ($p~\alpha~U/r$ and so $\nabla p~\alpha~U\nabla (1/r)$). It would then not be possible to satisfy the boundary condition. – Deep May 31 '18 at 7:35
• Another way to see it: Consider 2 distinct cases, in which sphere moves with velocity $\mathbf{U}$ and $\mathbf{U}'$; both have same magnitude $U$ but different directions. The corresponding solutions are $\{\mathbf{u},p\}$ and $\{\mathbf{u}',p'\}$. Now consider a third case in which the sphere moves with velocity $\tilde{\mathbf{U}}=\mathbf{U}-\mathbf{U}'$, whose solution is $\{\tilde{\mathbf{u}},\tilde{p}\}$. If pressure depended on magnitude of translation velocity then we must have $p~\alpha~U,p'~\alpha~U,\tilde{p}~\alpha~|\mathbf{U}-\mathbf{U}'|$. $\ldots$ – Deep May 31 '18 at 9:55
• But since Stokes equations are linear we must also have $\tilde{\mathbf{u}}=\mathbf{u}-\mathbf{u}',\tilde{p}=p-p'$. But then $\mathbf{u}-\mathbf{u}'$ and $p-p'$ do not contain a term proportional to velocity magnitude. To avoid this contradiction we must suppose that the flow velocity and pressure do not depend on velocity magnitude. – Deep May 31 '18 at 9:55
• @Diger Batchelor has given in Chapter 3 of his book the derivation that constitutes my answer. – Deep May 31 '18 at 10:01