Creeping flow

The solution for $\mathbf{u}_0$ and $p_0$ (creeping flow) is obtained by the method of §B.4.

Here the body force is

$$\mathbf{f}_0 = \mathbf{I}_0 - \mathbf{B}_0$$ (8.28)

$$= \mathbf{0} - \frac{T_0+Nm_0}{1+N}\mbox{\boldmath$\hat{\jmath}$}= -x\mbox{\boldmath$\hat{\jmath}$}$$ (8.29)

$$= -r\sin^2\theta\sin\phi\cos\phi\,\mbox{\boldmath$\hat{r}$} - r\sin\theta\cos\theta\sin\phi\cos\phi\,\mbox{\boldmath$\hat{\theta}$}$$

$$- r\sin\theta\cos^2\phi\,\mbox{\boldmath$\hat{\phi}$}.$$ (8.30)

The scalar defining its scaloidal part satisfies

$$\nabla^2\mbox{$\mathcal S$}[\mathbf{f}_0] = \mbox{\boldmath$\nabla$}\cdot \mathbf{f}_0 = 0$$ (8.31)

$$\mbox{\boldmath$\hat{r}$}\cdot\mbox{\boldmath$\nabla$}\mbox{$\mathcal S$}[\mathbf{f}_0] = \mbox{\boldmath$\hat{r}$}\cdot\mathbf{f}_0,\quad(r=\mbox{$\frac{1}{2}$})$$

$$= -\frac{1}{4}\sin^2\theta\sin\phi\cos\phi$$

$$= -\frac{1}{24}P_2^2(\cos\theta)\sin 2\phi.$$ (8.32)

This is

$$\mbox{$\mathcal S$}[\mathbf{f}_0] = -\frac{1}{4}r^2\sin^2\theta\sin 2\phi,$$ (8.33)

so that

$$\mathbf{f}_0^{(S)} = \mbox{\boldmath$\nabla$}\mbox{$\mathcal S$}[\mathbf{f}_0] = -\frac{1}{2}(r\sin^2\theta\sin 2\phi\,\mbox{\boldmath$\hat{r}$} + r\sin\theta\cos\theta\sin 2\phi\,\mbox{\boldmath$\hat{\theta}$}$$

$$+ r\sin\theta\cos 2\phi\,\mbox{\boldmath$\hat{\phi}$})$$ (8.34)

and

$$\mathbf{f}_0-\mathbf{f}_0^{(S)} = -\frac{1}{2}r\sin\theta\,\mbox{\boldmath$\hat{\phi}$},$$ (8.35)

which is obviously solenoidal, as expected.

Clearly

$$\mbox{$\mathcal P$}[\mathbf{f}_0-\mathbf{f}_0^{(S)}] = 0,$$ (8.36)

since

$$\mbox{\boldmath$\hat{r}$}\cdot(\mathbf{f}_0-\mathbf{f}_0^{(S)}) = 0.$$ (8.37)

The problem for the scalar defining the toroidal part of the body force is:

$${\mathcal L}^2\mbox{$\mathcal T$}[\mathbf{f}_0-\mathbf{f}_0^{(S)}] = -\mathbf{r}\cdot\mbox{\boldmath$\nabla$}\times(\mathbf{f}_0-\mathbf{f}_0^{(S)})$$ (8.38)

$$= r\cos\theta = rP_1^0(\cos\theta).$$ (8.39)

The solution is

$$\mbox{$\mathcal T$}[\mathbf{f}_0-\mathbf{f}_0^{(S)}] = -\frac{1}{2}r\cos\theta.$$ (8.40)

To illustrate the decomposition of the vector field here, the force-lines of $\mathbf{f}_0$, $\mathbf{f}_0^{(S)}$ and $\mathbf{f}_0^{(T)}$ in the plane $z=0$ are plotted in figure 8.3.

Figure 8.3: Decomposition of the vector field $\mathbf{f}_0$ into scaloidal, $\mathbf{f}_0^{(S)}$, and toroidal, $\mathbf{f}_0^{(T)}$, parts. The fields are represented by their force-lines in the plane $z=0$, the contours of $x^2/2$, $(x^2-y^2)/2$ and $(x^2+y^2)/4$.

Since, in this case, all three vector fields are plane and solenoidal, the force-lines can be represented as the contours of scalar functions. The functions are $x^2/2$, $(x^2-y^2)/4$ and $(x^2+y^2)/4$, respectively.

The problem for $\mbox{$\mathcal T$}[\mathbf{u}_0]$ is:

$$\nabla^2\mbox{$\mathcal T$}[\mathbf{u}_0] = \mbox{$\mathcal T$}[\mathbf{f}_0-\mathbf{f}_0^{(S)}] = -\frac{r}{2}P_1^0(\cos\theta)$$ (8.41)

$$\mbox{$\mathcal T$}[\mathbf{u}_0] = 0,\quad(r=1/2).$$ (8.42)

The solution is:

$$\mbox{$\mathcal T$}[\mathbf{u}_0] = \frac{r}{80}\left(1-4r^2\right)P_1^0(\cos\theta).$$ (8.43)

The other scalars, $\mbox{$\mathcal P$}[\mathbf{u}_0]$ and $s$, vanish.

Thus,

$$p_0 = -\mbox{$\mathcal S$}[\mathbf{f}_0] = \frac{r^2}{2}\sin^2\theta\sin\phi\cos\phi$$ (8.44)

$$\mathbf{u}_0 = \mbox{\boldmath$\nabla$}\times\mbox{$\mathcal T$}[\mathbf{u}_0]\mathbf{r} = \frac{r}{80}\left(1-4r^2\right)\sin\theta\mbox{\boldmath$\hat{\phi}$}.$$ (8.45)

The pressure is plotted in the plane of spanwise symmetry, $z=0$, in figure 8.4.

Figure 8.4: Zeroth order pressure, $p_0$, (8.46) in the plane $z=0$. The maxima are in the upper-right and lower-left quadrants. Contour levels at 0.01, 0.1(0.1)0.4, 0.6(0.1)0.9, 0.99 of range.

Since the velocity is purely toroidal, the pressure is due solely to the scaloidal part of the body force. The force-lines of $\mathbf{f}^{(S)}$ and the isobars, displayed in figures 8.3 and 8.4, are obviously related: they are mutually orthogonal.

The pressure is very simply expressed in terms of Cartesian coordinates:

$$p_0 = \frac{xy}{2}.$$ (8.46)

The fact that it is independent of $z$ suggests--consider the spanwise component of the equation of motion (2.54)--that $w_0=0$. This is the case:

$$\mathbf{u}_0 = \frac{4r^2-1}{80}(y\mbox{\boldmath$\hat{\imath}$}-x\mbox{\boldmath$\hat{\jmath}$}).$$ (8.47)

Notice that in the horizontal plane through the centre ($\sin\phi=y=0$) the velocity is purely vertical and identical to the fully developed flow in a cavity of circular horizontal section (7.65) except for a factor of $\frac{2}{5}$. Since $\mathbf{u}_0$ is independent of $\phi$, the contours of $v_n^{\circ}$ for $\mbox{$\mathcal S$}=1$ in figure 7.6 can also be interpreted as contours of the creeping speed in any plane passing through the $z$-axis of the sphere.