Conformal Maps

As we saw in Section 6.7, conformal maps are extremely useful in the theory of two-dimensional, irrotational, incompressible flows. It turns out that such maps also have applications to the theory of axisymmetric, irrotational, incompressible flows. Consider the general coordinate transformation

$$z+ {\rm i}\,\varpi = f(\xi+{\rm i}\,\eta),$$ (7.79)

where $f$ is an analytic function. The Cauchy-Riemann relations (see Section 6.3) yield

$$\frac{\partial z}{\partial\xi} =\frac{\partial\varpi}{\partial\eta},$$ (7.80)

$$\frac{\partial \varpi}{\partial\xi} =-\frac{\partial z}{\partial\eta}.$$ (7.81)

It follows, from the previous two expressions, that $\nabla\xi\cdot\nabla\eta=0$ . In other words, $\xi$ and $\eta$ are orthogonal coordinates in the meridian plane. [Incidentally, we are assuming that $(\xi,\,\eta,\,\varpi)$ are a right-handed set of coordinates.] Furthermore, it can also be shown from the Cauchy-Riemann relations that

$$h_\xi=h_\eta= \left[\left(\frac{\partial z}{\partial\xi}\right)^2... ...}\right)^2+ \left(\frac{\partial \varpi}{\partial\eta}\right)^2\right]^{\,1/2},$$ (7.82)

where $h_\xi=\vert\nabla\xi\vert^{\,-1}$ and $h_\eta=\vert\nabla\eta\vert^{\,-1}$ . Writing the flow velocity in terms of a velocity potential, so that ${\bf v}=-\nabla\phi$ , or, alternatively, in terms of a Stokes stream function, so that ${\bf v} = \nabla\varphi\times\nabla\psi$ , we get

$$v_\xi =-\frac{1}{h_\xi}\,\frac{\partial\phi}{\partial\xi} =-\frac{1}{\varpi\,h_\eta}\,\frac{\partial\psi}{\partial \eta},$$ (7.83)

$$v_\eta = -\frac{1}{h_\eta}\,\frac{\partial\phi}{\partial\eta} =\frac{1}{\varpi\,h_\xi}\,\frac{\partial\psi}{\partial \xi}.$$ (7.84)

Of course, writing the velocity field in terms of a Stokes stream function ensures that the field is incompressible, which also implies that $\nabla^{\,2}\phi=0$ . The additional requirement that the field be irrotational yields $\omega_\varphi=0$ . Making use of the analysis of Appendix C, this requirement reduces to

$$\frac{\partial(h_\eta\,v_\eta)}{\partial\xi} -\frac{\partial(h_\xi\,v_\xi)}{\partial \eta}=0,$$ (7.85)

or

$$\frac{\partial}{\partial\xi}\left(\frac{1}{\varpi}\,\frac{\partia... ...partial\eta}\left(\frac{1}{\varpi}\,\frac{\partial\psi}{\partial\eta}\right)=0.$$ (7.86)

Figure 7.5: An axisymmetric solid body moving through an incompressible irrotational fluid.

Let $\xi=\xi_0$ represent the surface of an axisymmetric solid body moving with velocity ${\bf V} =V\,{\bf e}_z$ through an incompressible irrotational fluid that is at rest a long way from the body. Let the fluid occupy the region $\xi_0<\xi<\infty$ , where $\xi\rightarrow\infty$ far from the body. (See Figure 7.5.) Let $\eta$ be an angular coordinate such that $\eta=0$ on the positive $z$ -axis, and $\eta=\pi$ on the negative $z$ -axis. The fact that the fluid is at rest at infinity implies that $\psi$ asymptotes to a constant a long way from the body. Without loss of generality, we can chose this constant to be zero. Thus, one constraint on the system is that

$$\psi(\xi\rightarrow\infty,\eta)= 0.$$ (7.87)

The appropriate constraint at the surface of the body is that

$$v_\xi = {\bf V}\cdot{\bf e}_\xi,$$ (7.88)

where ${\bf e}_\xi=\nabla\xi/\vert\nabla\xi\vert$ . However, we can write ${\bf V} = \nabla\varphi\times\nabla\psi_0$ , where $\psi_0(\varpi,z)= -(1/2)\,V\,\varpi^{\,2}$ . (See Section 7.6.) Hence, from Equation (7.83), the previous constraint becomes

$$\frac{\partial\psi}{\partial\eta} = \frac{\partial\psi_0}{\partial\eta}$$ (7.89)

when $\xi=\xi_0$ . Integrating, making use of the constraint (7.87) (which implies that $\psi=0$ on the $z$ -axis, where $\psi$ is constant, by symmetry), we obtain

$$\psi(\xi_0,\eta)= -\frac{1}{2}\,V\,\varpi^{\,2}.$$ (7.90)

We can also set the velocity potential, $\phi$ , to zero at infinity, and on the $z$ -axis.

The total kinetic energy of the fluid surrounding the moving body is

$$K = \frac{1}{2}\,\rho\int v^{\,2}\,dV = \frac{1}{2}\,\rho\int(\nabla\phi)^2\,dV = \frac{1}{2}\,\rho\int\nabla\cdot(\phi\,\nabla\phi)\,dV,$$ (7.91)

where we have made use of the fact that $\nabla^{\,2}\phi=0$ . Here, $\rho$ is the fluid mass density, and $d V$ is an element of the volume obtained by rotating the area $ABCDEFA$ , shown in Figure 7.5, about the $z$ -axis. Making use of the divergence theorem, we obtain

$$K=\frac{1}{2}\,\rho\oint \phi\,\nabla\phi\cdot{\bf n}\,2\pi\,\var... ...{ABC}\phi\,\frac{1}{h_\xi}\,\frac{\partial\phi}{\partial\xi}\,2\pi\,\varpi\,ds.$$ (7.92)

where $ds$ is an element of the curve $ABCDEFA$ , and ${\bf n}$ is an outward pointing, unit, normal vector to the area $ABCDEFA$ . Here, we have made use of the fact that the velocity potential is zero at infinity (i.e., on $DEF$ ), and also on the $z$ -axis (i.e., on $CD$ and $FA$ ). On the curve $ABC$ , we can write $ds= h_\eta\,d\eta$ . Furthermore, it follows from Equations (7.82) and (7.83) that $h_\xi=h_\eta$ , and $\partial\phi/\partial\xi = \varpi^{\,-1}\,\partial\psi/\partial\eta$ . Thus,

$$K= -\pi\,\rho\left.\int_{0}^\pi \phi\,\frac{\partial\psi}{\partial\eta}\,d\eta\right\vert _{\xi=\xi_0},$$ (7.93)

or

$$K = -\pi\,\rho\left.\int_{\eta=0}^{\eta=\pi} \phi\,d\psi\right\vert _{\xi=\xi_0}.$$ (7.94)

As a simple example, consider the conformal map

$$z+{\rm i}\,\varpi = c\,\exp\,(\xi+{\rm i}\,\eta),$$ (7.95)

where $c$ is real and positive. It follows that

$$z = c\,{\rm e}^{\,\xi}\,\cos\eta,$$ (7.96)

$$\varpi = c\,{\rm e}^{\,\xi}\,\sin\eta,$$ (7.97)

which implies that

$$z^{\,2}+\varpi^{\,2} = r^{\,2},$$ (7.98)

where

$$r(\xi)=c\,{\rm e}^{\,\xi}.$$ (7.99)

Thus, the constant-$\xi$ surfaces are concentric spheres of radius $r(\xi)$ . If we set

$$a = c\,{\rm e}^{\,\xi_0}$$ (7.100)

then the problem reduces to that of a sphere, of radius $a$ , moving through a fluid that is at rest at infinity. This problem was solved, via different methods, in Section 7.10. The constraints (7.87) and (7.90) yield

$$\psi(\xi\rightarrow\infty,\eta) = 0,$$ (7.101)

$$\psi(\xi_0,\eta) = -\frac{1}{2}\,V\,c^2\,{\rm e}^{\,2\,\xi_0}\,\sin^2\eta,$$ (7.102)

where use has been made of Equation (7.97). This suggests that we can write

$$\psi(\xi,\eta)=F(\xi)\,\sin^2\eta.$$ (7.103)

Substitution into the governing equation, (7.86), gives

$$\frac{d}{d\xi}\left({\rm e}^{-\xi}\,\frac{dF}{d\xi}\right) -2\,{\rm e}^{-\xi}\,F = 0,$$ (7.104)

whose most general solution is

$$F(\xi)=A\,{\rm e}^{\,2\,\xi} + B\,{\rm e}^{-\xi}.$$ (7.105)

The constraints (7.101) and (7.102) yield

$$A = 0,$$ (7.106)

$$B =-\frac{1}{2}\,V\,c^{\,2}\,{\rm e}^{\,3\,\xi_0},$$ (7.107)

respectively. Thus, we obtain

$$\psi(\xi,\eta) = -\frac{1}{2}\,V\,a^{\,3}\,\frac{\sin^2\eta}{c\,{\rm e}^{\,\xi}}.$$ (7.108)

Now, from Equations (7.84) and (7.97),

$$\frac{\partial\phi}{\partial\eta} = -\frac{1}{\varpi}\,\frac{\par... ...artial\xi} = -\frac{1}{2}\,V\,a^{\,3}\,\frac{\sin\eta}{(c\,{\rm e}^{\,\xi})^2},$$ (7.109)

which can be integrated to give

$$\phi(\xi,\eta)= \frac{1}{2}\,V\,a^{\,3}\,\frac{\cos\eta}{(c\,{\rm e}^{\,\xi})^2}.$$ (7.110)

Note that the previous expression is formally the same as expression (7.63), as long as we make the identifications $V\rightarrow V_z(t)$ , $c\,{\rm e}^{\,\xi}\rightarrow r$ , and $\eta\rightarrow\theta$ .

On the surface of the sphere, $\xi=\xi_0$ , we obtain

$$\psi(\xi_0,\eta) = -\frac{1}{2}\,V\,a^{\,2}\,\sin^2\eta,$$ (7.111)

$$\phi(\xi_0,\eta) = \frac{1}{2}\,V\,a\,\cos\eta.$$ (7.112)

Thus,

$$K= -\pi\,\rho\left.\int_{\eta=0}^{\eta=\pi} \phi\,d\psi\right\ver... ...rho\,V^{\,2}\int_{-1}^1 \mu^{\,2}\,d\mu= \frac{\pi}{3}\,a^{\,3}\,\rho\,V^{\,2}.$$ (7.113)

As is clear from the analysis of Section 7.10, the sphere's added mass can be written

$$m_{\rm added} = \frac{K}{(1/2)\,V^{\,2}} = \frac{2\pi}{3}\,\,a^{\,3}\,\rho.$$ (7.114)

Hence, we arrive at the standard result that the added mass is half the displaced mass [i.e., half of $(4\pi/3)\,a^{\,3}\,\rho$ ].