Derivation of the differential equation: $E^2(E^2\psi)=0$

The derivation of the differential equation $E^2(E^2\psi)=0$ is part of the steps in deriving Stoke’s law.

We have, in the previous section, derived the Navier-Stokes equation:

$\rho\frac{D\textbf{\textit{u}}}{Dt}=-\nabla p+\rho g+\mu \nabla^2\textbf{\textit{u}}\; \; \; \; \; \; \; (32)$

The terms $\rho \frac{D\textbf{\textit{u}}}{Dt}$ and $\rho g$ are attributed to inertia forces, while the terms $\nabla p$ and $\mu \nabla^2\textbf{\textit{u}}$ are due to hydrostatic forces and viscous forces respectively. For a very viscous fluid, the viscosity of the fluid is very high but fluid velocities are very low. Therefore, the non-inertia forces dominate the inertia forces and eq37 is reduced to:

$\mu \nabla^2\textbf{\textit{u}}=\nabla p\; \; \; \; \; \; \; (33)$

Taking the curl on both sides of eq33,

$\nabla \times \mu \nabla^2\textbf{\textit{u}}=\nabla \times \nabla p$

Since the curl of the gradient of a function is zero and the viscosity of an incompressible fluid is constant,

$\nabla\times\mu \nabla^2\textbf{\textit{u}}=0\; \; \; \; \Rightarrow \; \; \; \; \nabla\times\nabla^2\textbf{\textit{u}}=0\; \; \; \; \; \; \; (33a)$

Substituting eq14 where $\nabla\cdot\textbf{\textit{u}}=0$ in the vector identity $\nabla^2\textbf{\textit{u}}=\nabla\left ( \nabla\cdot \textbf{\textit{u}}\right )-\nabla \times \left ( \nabla \times \textbf{\textit{u}}\right )$ , we get:

$\nabla^2\textbf{\textit{u}}=-\nabla\times\left ( \nabla\times \textbf{\textit{u}}\right )\; \; \; \; \ (33b)$

Substitute eq33b in eq33a

$\nabla \times \nabla \times \left ( \nabla \times \textbf{\textit{u}}\right )=0\; \; \; \; \; \; \; (34)$

Assuming that the fluid is flowing in the $e_\phi$ direction, eq3 becomes $\textbf{\textit{u}}=u_r\hat{r}+u_\theta \hat{\theta}$ . Noting that the curl of a function in spherical coordinates is:

$\nabla\times\textbf{\textit{u}}=\frac{1}{rsin\theta}\left [ \frac{\partial }{\partial \theta}\left ( u_\phi sin\theta \right )-\frac{\partial }{\partial \phi}u_\theta \right ]\hat{r}$

$+\frac{1}{r}\left [\frac{1}{sin\theta} \frac{\partial }{\partial \phi} u_r-\frac{\partial }{\partial r}\left (ru_\phi \right ) \right ]\hat{\theta}+\frac{1}{r}\left [ \frac{\partial }{\partial r}\left (r u_\theta \right )-\frac{\partial }{\partial \theta}u_r \right ]\hat{\phi}$

and substituting eq15 and eq16 in $\textbf{\textit{u}}=u_r\hat{r}+u_\theta \hat{\theta}$ , we get:

$\textbf{\textit{u}}=\frac{1}{r^2sin\theta}\frac{\partial \psi}{\partial \theta}\hat{r}-\frac{1}{rsin\theta}\frac{\partial \psi}{\partial r}\hat{\theta}=\nabla\times \frac{\psi}{rsin\theta}\hat{\phi}\; \; \; \; \; \; \; (35)$

Question

Show that $\nabla\times\frac{\psi}{rsin\theta}\hat{\phi}=\frac{1}{r^2sin\theta}\frac{\partial \psi}{\partial \theta}\hat{r}-\frac{1}{rsin\theta}\frac{\partial \psi}{\partial r}\hat{\theta}$ .

Answer

Using the curl of a function in spherical coordinates,

$\nabla\times\left ( 0\hat{r}+0\hat{\theta}+\frac{\psi}{rsin\theta}\hat{\phi} \right )=\frac{1}{rsin\theta}\left [ \frac{\partial }{\partial \theta}\left ( \frac{\psi}{rsin\theta} sin\theta\right )-\frac{\partial }{\partial \phi}0 \right ]\hat{r}$

$+\frac{1}{r}\left [ \frac{1}{sin\theta}\frac{\partial }{\partial \phi}0-\frac{\partial }{\partial r}\left ( \frac{r\psi}{rsin\theta} \right ) \right ]\hat{\theta}\: +\frac{1}{r}\left [ \frac{\partial }{\partial r}r0-\frac{\partial }{\partial \theta}0 \right ]\hat{\phi}\:$

$\nabla\times\frac{\psi}{rsin\theta}\hat{\phi}=\frac{1}{r^2sin\theta}\frac{\partial \psi}{\partial \theta}\hat{r}-\frac{1}{rsin\theta}\frac{\partial \psi}{\partial r}\hat{\theta}$

Substituting eq35 in eq34 and working out the algebra, we have:

$\nabla\times\nabla\times\left ( -\frac{1}{rsin\theta}E^2\psi\hat{\phi} \right )=0\; \; \; \; \; \; \; (35a)$

where $E^2=\frac{\partial ^2}{\partial r^2}+\frac{sin\theta}{r^2}\frac{\partial }{\partial \theta}\left ( \frac{1}{sin\theta}\frac{\partial }{\partial \theta} \right )$ .