A hybrid asymptotic-numerical method for calculating drag coefficients in 2-D low Reynolds number flows

Abstract

Steady-state incompressible low-Reynolds-number fluid flow past a cylindrical body in an unbounded two-dimensional domain is a singular perturbation problem involving an infinite logarithmic expansion in the Reynolds number ${\displaystyle \varepsilon }$ as ${\displaystyle \varepsilon }\rightarrow 0$. The central difficulty with applying a conventional matched asymptotic approach to this problem is that only the first few terms in the infinite logarithmic expansion of the drag coefficient and of the flow field can be calculated analytically. To overcome this difficulty, a hybrid asymptotic-numerical method that incorporates all logarithmic correction terms is implemented for three low-Reynolds-number flow problems. In particular, for a nanocylinder of circular cross section with surface roughness, modeled by a Navier boundary condition involving a sliplength parameter, a hybrid asymptotic-numerical method is formulated and implemented to determine an approximation to the drag coefficient that is accurate to all powers of ${-1/\log {\displaystyle \varepsilon }}$. A similar analysis is done to determine a corresponding approximation of the drag coefficient for a porous cylinder, where the flow inside the cylinder is modeled by the Brinkman equation. For both the nano- and porous-cylinder problems, the hybrid asymptotic-numerical method is extended to calculate the first transcendentally small correction term to the Stokes flow near a body. This term, which governs weak upstream/downstream asymmetry in the Stokes flow, is extrapolated to finite ${\displaystyle \varepsilon }$ to predict the formation of any eddies near the body. Finally, the hybrid method is used to determine the drag coefficient, valid to within all logarithmic terms, for two identical cylinders of circular cross section in tandem alignment with the free stream. An extension of the theoretical framework to more general slow viscous flow problems is discussed.

Appendices

Appendix 1: Drag coefficient

In [30], Imai gives an expression for the drag force in terms of an arbitrary closed circular contour of radius $r_0$ surrounding the circular cross section of a cylinder. Converting this expression to polar coordinates, and defining the Reynolds number as ${\displaystyle \varepsilon }\equiv \text{ Re }\equiv {U_{\infty } L\rho _\mathrm{{f}}/\mu }$, we can express the drag coefficient, $C_\mathrm{{D}}$, as

$$\begin{aligned} C_\mathrm{{D}}&= \frac{r_0}{2}\int \limits _{0}^{2\pi } \left[ \cos \theta \left( \frac{\partial \psi }{\partial r}\right) ^{2} - \frac{\cos \theta }{r^{2}}\left( \frac{\partial \psi }{\partial \theta } \right) ^{2}-\frac{2\sin \theta }{r}\frac{\partial \psi }{\partial r}\frac{\partial \psi }{\partial \theta }\right] {\Bigg \vert }_{r=r_0} \,\mathrm{{d}}\theta \nonumber \\&\quad -r_0\int \limits _{0}^{2\pi }(\sin \theta )\,\omega \frac{\partial \psi }{\partial \theta } {\Big \vert }_{r=r_0} \, \mathrm{{d}}\theta +\frac{r_0^{2}}{{\displaystyle \varepsilon }}\int \limits _{0}^{2 \pi }(\sin \theta )\frac{\partial \omega }{\partial r}{\Big \vert }_{r=r_0} \, \mathrm{{d}}\theta - \frac{r_0}{{\displaystyle \varepsilon }}\int \limits _{0}^{2\pi }(\sin \theta ) \, \omega {\vert }_{r=r_0} \,\mathrm{{d}}\theta . \end{aligned}$$

(7.12)

Here $\psi$ satisfies (2.1)–(2.3), and the vorticity $\omega$ is $\omega = - \Delta \psi$.

To evaluate the integrals in this expression for $C_\mathrm{{D}}$ when ${\displaystyle \varepsilon }\rightarrow 0$, we use the far-field behavior

$$\begin{aligned} \psi \sim S \left( r\log {r} - \frac{r}{2} + \frac{1}{2r}\right) \sin \theta , \end{aligned}$$

(7.13)

valid for $r\rightarrow \infty$, of the Stokes solution. By symmetry, only the last two integrals in the expression for $C_\mathrm{{D}}$ are nonzero, and consequently

$$\begin{aligned} C_\mathrm{{D}}\sim \frac{r_0^2}{{\displaystyle \varepsilon }}\int \limits _0^{2\pi } (\sin \theta )\, \frac{\partial \omega }{\partial r} {\Big \vert }_{r=r_0}\, \mathrm{{d}}\theta - \frac{r_0}{{\displaystyle \varepsilon }}\int \limits _0^{2\pi } (\sin \theta ) \omega \vert _{r=r_0}\, \mathrm{{d}}\theta . \end{aligned}$$

(7.14)

Since $\omega =-{2Sr^{-1}}\sin \theta$, (7.14) becomes $C_\mathrm{{D}}\sim 4\pi {\displaystyle \varepsilon }^{-1}S$, as given in (2.25).

Appendix 2: Stream function for flow past a porous cylinder

In this appendix we derive (5.1)–(5.6) for the stream function for flow past a porous cylinder. Inside the cylinder, we assume that the flow is governed by the Brinkman equation

$$\begin{aligned} \nabla P^i= -\frac{\mu }{k} \mathbf{v}^i + \hat{\mu } \Delta \mathbf{v}^i. \end{aligned}$$

(7.15)

Here $k$ is the permeability of the porous cylinder, $\mu$ is the dynamic viscosity of the fluid, $\hat{\mu }$ is the effective viscosity of the porous medium, $P^i$ is the pressure, and $\mathbf{v^i=(v_r^i,v_{\theta }^i)}$ is the fluid velocity in the porous medium. The components of the fluid velocity in terms of stream function can be written as $\mathbf{v}^i=(v_r^i,v_{\theta }^i)= (\partial \psi ^i/(r \partial \theta ), -\partial \psi ^i/ \partial r)$.

In general, the effective viscosity $\hat{\mu }$ is not expected to be the same as the viscosity of the fluid $\mu$. A derivation of the the dependency of the effective viscosity and permeability on the solid volume fraction in a porous medium through a renormalization of the Stokes equation was obtained in [31]. There, the authors showed that the Brinkman equation is valid in the limit of dilute porous media. In this limit, the effective viscosity, $\hat{\mu }$, can be well approximated by the fluid viscosity, $\mu$. Hence, in our derivation below, we assume that $\hat{\mu } =\mu$.

We adopt the following scalings for Eq. (7.15):

$$\begin{aligned} \bar{P^i}= \frac{P^i}{\mu U_{\infty }/L},\quad \bar{\mathbf{v}^\mathbf{i}}= \frac{\mathbf{v}^\mathbf{i}}{U_{\infty }},\quad \bar{r}= \frac{r}{L}. \end{aligned}$$

Upon dropping the overbar, we write

$$\begin{aligned} \nabla P^i = -\chi ^2 \mathbf{v^i} + \Delta \mathbf{v^i}. \end{aligned}$$

(7.16)

Note that in the case where $\hat{\mu } \ne \mu$, the preceding equation can be replaced by $1/m\nabla P^i= -(\chi ^2/m)\mathbf{v^i} + \Delta \mathbf{v^i}$, where $m=\hat{\mu }/\mu$. Therefore, the scaling, governing equations, and boundary conditions will not change. In (7.16), the pressure can be eliminated by taking the curl of the Brinkman equation. Then we obtain (5.2) upon substituting the velocity components in terms of the stream function. Outside the porous cylinder, the flow is governed by (5.1). Modeling the porous medium by the Brinkman equation rather than Darcy’s law allows us to satisfy the continuity of velocity and stresses at the interface between the fluid region and porous cylinder. The validity of the interfacial boundary conditions is discussed in [32]. We may write the boundary conditions at the interface as follows:

Continuity of velocity components

$$\begin{aligned} v_r^i&= v_r^e \quad \Rightarrow \quad \frac{\partial \psi ^i}{\partial \theta }= \frac{\partial \psi ^e}{\partial \theta } \Rightarrow \psi ^i=\psi ^e \quad \text{ on } \ r=1,\\ v_{\theta }^i&= v_{\theta }^e \quad \Rightarrow \quad \frac{\partial \psi ^i}{\partial r}= \frac{\partial \psi ^e}{\partial r} \quad \text{ on } \ r=1; \end{aligned}$$

Continuity of traction

$$\begin{aligned}&\textit{Normal stress balance:}\; [\![ n\cdot T\cdot n]\!]=0\;\text{ across } \quad r=1,\\&\textit{Tangential stress balance:}\; [\![n\cdot T\cdot t]\!]=0\;\text{ across } \quad r=1, \end{aligned}$$

where $[\![A]\!]=[A]^i-[A]^e$ denotes the jump in $A$ across the surface of the porous cylinder. Here $T$ is the traction tensor, and $n=e_r$ and $t=e_{\theta }$ are the unit normal and tangential vectors to the surface of the porous cylinder, respectively. We may write the traction tensor as follows:

$$\begin{aligned} T= \left( \begin{array}{cc} \tau _{rr}-P &{}\quad \tau _{r \theta } \\ \tau _{r \theta } &{}\quad \tau _{\theta \theta }-P \end{array}\right) =\left( \begin{array}{cc} 2\frac{\partial v_r}{\partial r} -P &{}\quad \frac{1}{r}\frac{\partial v_r }{\partial \theta }+r \frac{\partial }{\partial r} (\frac{v_{\theta }}{r})\\ \frac{1}{r}\frac{\partial v_r}{\partial \theta } +r \frac{\partial }{\partial r} (\frac{v_{\theta }}{r}) &{}\quad \frac{2}{r}\frac{\partial v_{\theta }}{\partial \theta } +\frac{2v_r}{r}-P \end{array}\right) . \end{aligned}$$

Upon writing the tangential and normal stress balance in terms of the stream function, we apply the continuity of velocity across the interface $r=1$ to obtain

$$\begin{aligned}&\Big [\!\Big [\tau _{r\theta }\Big ]\!\Big ] = 0 \quad \Rightarrow \quad \Big [\!\Big [-\frac{1}{r^2}\frac{\partial ^2 \psi }{\partial \theta ^2}-\frac{1}{r}\frac{\partial \psi }{\partial r}+\frac{\partial ^2 \psi }{\partial r^2}\Big ]\!\Big ]=0 \quad \Rightarrow \quad \Big [\!\Big [{\partial ^2 \psi }/{\partial r^2}\Big ]\!\Big ]=0\\&\Big [\!\Big [\tau _{rr}-P\Big ]\!\Big ] = 0 \quad \Rightarrow \quad \Big [\! \Big [\frac{2}{r^2}\frac{\partial \psi }{\partial \theta }-\frac{2}{r} \frac{\partial ^2 \psi }{\partial r\partial \theta }-P\Big ]\!\Big ]=0 \quad \Rightarrow \quad \Big [\!\Big [P\Big ]\!\Big ]=0 \quad \Rightarrow \quad \Big [\!\Big [\frac{\partial P}{\partial \theta }\Big ]\!\Big ]=0. \end{aligned}$$

Then, upon using the Brinkman and Navier–Stokes equations inside and outside the porous cylinder, we obtain

$$\begin{aligned} -\frac{1}{r} \frac{\partial P^i}{\partial \theta }=-(\nabla ^2-\chi ^2) v_{\theta }^i=-\frac{1}{r}\frac{\partial }{\partial r}\left( r \frac{\partial v_{\theta }^i}{\partial r}\right) -\frac{1}{r^2} \frac{\partial ^2v_{\theta }^i}{\partial \theta ^2}-\frac{2}{r^2} \frac{\partial v_r^i}{\partial \theta }+\frac{v_{\theta }^i}{r^2}+ \chi ^2 v_{\theta }^i \end{aligned}$$

and

$$\begin{aligned} -\frac{1}{r} \frac{\partial P^e}{\partial \theta }=-\frac{1}{r} \frac{\partial }{\partial r}\left( r\frac{\partial v_{\theta }^e}{\partial r}\right) -\frac{1}{r^2}\frac{\partial ^2 v_{\theta }^e}{\partial \theta ^2}-\frac{2}{r^2}\frac{\partial v_r^e}{\partial \theta }+\frac{v_{\theta }^e}{r^2}\,. \end{aligned}$$

Upon equating these two expressions and then applying the continuity of velocity and tangential stress, we obtain that

$$\begin{aligned} -\frac{1}{r}\frac{\partial }{\partial r}\left( r\frac{\partial v_{\theta }^i}{\partial r}\right) +\chi ^2 v_{\theta }^i=-\frac{1}{r}\frac{\partial }{\partial r}\left( r\frac{\partial v_{\theta }^e}{\partial r}\right) \quad \Rightarrow \quad \frac{\partial ^3 \psi ^e}{\partial r^3}-\frac{\partial ^3 \psi ^i}{\partial r^3}=-\chi ^2\frac{\partial \psi ^i}{\partial r}\quad \text{ on }\; r=1. \end{aligned}$$

This completes the derivation of (5.1)–(5.6).

Appendix 3: Solution to Stokes problem for flow past two cylinders

In this appendix we outline the derivation of (6.12)–(6.16) using the formulation of [22]. Let $\mathbf{y}=(y_1,y_2)^\mathrm{{t}}=r(\cos \theta ,\sin \theta )$, and introduce the bipolar coordinate system $(\xi ,\eta )$ defined by

$$\begin{aligned} y_1 = r \cos \theta = \frac{c\sinh \xi }{\cosh \xi -\cos \eta }, \quad y_2 = r \sin \theta = \frac{c\sin \eta }{\cosh \xi -\cos \eta }. \end{aligned}$$

(7.17)

Then lines of constant $\xi$ map to the circles $(y_1-c\coth \xi )^2+y_2^2 = c^2 \text{ csch }^{2}\xi$. Therefore, if we choose $\alpha$ and $c$ by

$$\begin{aligned} c \coth \alpha ={d/2}, \quad c \, \text{ csch }\alpha = 1, \end{aligned}$$

(7.18)

so that

$$\begin{aligned} \alpha = \log \left( \frac{d}{2} + \sqrt{ \frac{d^2}{4}-1}\,\right) , \quad c= \sqrt{ \frac{d^2}{4} -1 }, \end{aligned}$$

(7.19)

it follows that the region ${\mathbb {R}}^2 \backslash D$ in (6.6) maps one to one to the rectangle $\varOmega \equiv \lbrace {(\xi ,\eta ) \, \vert \,\,|\xi |\le \alpha , \, -\pi <\eta \le \pi \rbrace }$, and that $(\xi ,\eta )\rightarrow \mathbf{0}$ corresponds to $r\rightarrow \infty$. For $r\rightarrow \infty$, we readily obtain the following local behavior of the mapping (7.17):

$$\begin{aligned} \xi ^2 + \eta ^2 \sim \frac{4c^2}{r^2}, \quad \xi \sim \frac{2c \cos \theta }{r}, \quad \eta \sim \frac{2c \sin \theta }{r}. \end{aligned}$$

(7.20)

In terms of these bipolar coordinates, (6.6)–(6.9) is transformed into a problem for $\varPhi (\xi ,\eta )$ given by (cf. [33])

$$\begin{aligned} \left[ \partial _{\xi \xi \xi \xi } + 2\partial _{\xi \xi \eta \eta } + \partial _{\eta \eta \eta \eta } - 2\partial _{\xi \xi } +2\partial _{\eta \eta } + 1\right] \varPhi = 0, \qquad (\xi ,\eta )\in \varOmega , \end{aligned}$$

(7.21)

$$\begin{aligned} \varPhi =\varPhi _{\xi } = 0, \quad \text{ on } \quad \xi =\pm \alpha ; \quad \varPhi \quad \text{ is } \,\, 2\pi \,\, \text{ periodic } \text{ in } \eta , \end{aligned}$$

(7.22)

with $\varPhi$ having an appropriate singularity behavior as $(\xi ,\eta )\rightarrow \mathbf{0}$. In particular, $\varPhi \rightarrow 0$ but is not $C^1$ as $(\xi ,\eta )\rightarrow \mathbf{0}$. In terms of $\varPhi$, the solution $\psi _c$ to (6.6)–(6.9) is

$$\begin{aligned} \psi _c(y_1,y_2)=\psi \left[ \xi (y_1,y_2),\eta (y_1,y_2)\right] , \quad \text{ where } \; \psi (\xi ,\eta )= f \varPhi (\xi ,\eta ), \; f\equiv \frac{c}{\cosh \xi -\cos \eta }. \end{aligned}$$

(7.23)

The appropriate singularity behavior of $\varPhi$ as $(\xi ,\eta )\rightarrow \mathbf{0}$ must be such that, in terms of polar coordinates, it yields $\psi \sim r\log {r}\sin \theta$ as $r\rightarrow \infty$. Since $f\sim {2c/(\xi ^2+\eta ^2)}= {\mathcal O}(r^2)$ as $(\xi ,\eta )\rightarrow \mathbf{0}$ or, equivalently, as $r\rightarrow \infty$, from (7.23) we conclude that $\psi$ has the required far-field behavior provided that

$$\begin{aligned} \varPhi \rightarrow 0, \quad \text{ as } \quad (\xi ,\eta )\rightarrow \mathbf{0} \, \quad \text{ with } \text{ local } \text{ behavior } \quad \varPhi \sim -\frac{1}{2}\eta \log (\xi ^2 + \eta ^2). \end{aligned}$$

(7.24)

To verify this claim, we calculate for $(\xi ,\eta )\rightarrow \mathbf{0}$ using (7.20) that

$$\begin{aligned} \psi =f \varPhi \sim \frac{2c}{\xi ^2+\eta ^2}\left( -\frac{\eta }{2} \log (\xi ^2+\eta ^2)\right) \sim \frac{2c r^2}{4c^2}\left( -\frac{1}{2} \frac{2c\sin \theta }{r} \log \left( \frac{4c^2}{r^2}\right) \right) \sim r\log {r}\sin \theta \end{aligned}$$

(7.25)

as $r\rightarrow \infty$. Therefore, problem (6.6)–(6.9) is transformed into a problem of finding a solution to (7.21)–(7.22) satisfying the local behavior in (7.24) as $(\xi ,\eta )\rightarrow \mathbf{0}$.

Next, we construct an exact solution of (7.21) with the local behavior (7.24). To determine such a solution, we recall in terms of polar coordinates that $\Delta ^2 \left[ \frac{r^2}{2}\log {r}\right] =0$. Upon differentiating with respect to $y_2$, we obtain that $\Delta ^2[ y_2\log {r} + {y_2/2}]=0$, so that $\Delta ^2[y_2\log {r}]=0$. To convert this exact biharmonic solution having the correct behavior at infinity to bipolar coordinates, we first calculate $\log {r}$ from (7.17) as

$$\begin{aligned} \log {r} = \log \left( \frac{c\left( \sinh ^2\xi +\sin ^2\eta \right) ^{1/2}}{{\cosh \xi -\cos \eta }}\right) = -\frac{1}{2}\log \left( \cosh \xi - \cos \eta \right) + \frac{1}{2}\log \left( \cosh \xi + \cos \eta \right) + \log {c}. \end{aligned}$$

(7.26)

Then, upon calculating $y_2\log {r}$ from using (7.17) for $y_2$, it follows that an exact solution to (7.21) with the local behavior (7.24) is

$$\begin{aligned} \tilde{\varPhi }_0 = -\frac{1}{2}\sin \eta \log \left( \cosh \xi - \cos \eta \right) + \frac{1}{2}\sin \eta \log \left( \cosh \xi + \cos \eta \right) . \end{aligned}$$

(7.27)

The second term in (7.27) is regular as $(\xi ,\eta )\rightarrow \mathbf{0}$ but is not $C^{1}$ at $(\xi ,\eta )=(0,\pm \pi )$. Moreover, by expanding this term in a Fourier sine series in $\eta$ to generate a separation-of-variables type of solution, it is readily verified that it satisfies (7.21). As such, an exact solution to (7.21), which has the local behavior (7.24) as $(\xi ,\eta )\rightarrow \mathbf{0}$ but is otherwise smooth in $\varOmega$, is simply the first term in (7.27), which we label by

$$\begin{aligned} \varPhi _0\equiv -\frac{1}{2}\sin \eta \log \left( \cosh \xi -\cos \eta \right) . \end{aligned}$$

(7.28)

We then must add to $\varPhi _0$ a smooth solution $\varPhi _s$, constructed using the separation of variables applied to (7.21), in order to satisfy the boundary conditions in (7.22) at $\xi =\pm \alpha$. Since $\varPhi _0$ is even in $\xi$, we need only choose $\varPhi _s$ to be an even function of $\xi$, with $\varPhi _s=-\varPhi _0$ and $\partial _\xi \varPhi _s=-\partial _\xi \varPhi _0$ on $\xi =\alpha$.

To expand $\varPhi _0$ as a Fourier sine series in $\eta$, we first define $z=\mathrm{{e}}^{-\xi +\mathrm{{i}}\eta }$ and derive the identity

$$\begin{aligned} \sum _{m=1}^{\infty }\frac{\mathrm{{e}}^{-m\xi }}{m}\cos (m\eta )=\mathrm{Re}\left[ \sum _{m=1}^{\infty } \frac{z^m}{m}\right] =-\log |1-z|= \frac{\xi }{2}-\frac{\log {2}}{2} - \frac{1}{2} \log \left( \cosh \xi -\cos \eta \right) . \end{aligned}$$

(7.29)

From this identity, and using trigonometric addition formulas, we obtain for $\xi >0$ that $\varPhi _0$ has the Fourier sine series

$$\begin{aligned} \varPhi _0&= -\frac{\sin \eta }{2}\left[ \xi -\log {2} -2\sum _{m=1}^{\infty } \frac{\mathrm{{e}}^{-m\xi }}{m} \cos (m\eta )\right] ,\nonumber \\&=-\frac{1}{2}\left( \xi -\log {2} + \frac{1}{2}\mathrm{{e}}^{-2\xi }\right) \sin (\eta ) + \frac{1}{2} \sum _{k=2}^{\infty } \left( \frac{\mathrm{{e}}^{-(k-1)\xi }}{k-1} - \frac{\mathrm{{e}}^{-(k+1)\xi }}{k+1} \right) \sin (k\eta ). \end{aligned}$$

(7.30)

Then, since the Fourier sine series separation of variables solution $\varPhi _s$ for (7.21), which is even in $\xi$, is a linear combination of $\cosh [(k+1)\xi ]\sin (k\eta )$ and $\cosh [(k-1)\xi )]\sin (k\eta )$ for $k\ge 1$, it follows that the solution $\varPhi =\varPhi _0+\varPhi _s$ to (7.21)–(7.23) with (7.24) has the form

$$\begin{aligned} \varPhi = h_1(\xi ) \sin (\eta ) + \sum _{k=1}^{\infty } h_{k}(\xi )\sin (k\eta ), \end{aligned}$$

(7.31)

where

$$\begin{aligned} h_1(\xi ) \equiv -\frac{\xi }{2} + \frac{\ln {2}}{2} - \frac{\mathrm{{e}}^{-2\xi }}{4} - \frac{a_1}{2}\cosh (2\xi ) - \frac{b_1}{2}, \end{aligned}$$

(7.32)

$$\begin{aligned} h_k(\xi ) \equiv \frac{1}{2}\left( \frac{\mathrm{{e}}^{-(k-1)\xi }}{k-1} - \frac{\mathrm{{e}}^{-(k+1)\xi }}{k+1} \right) - \frac{a_k}{2}\cosh [(k+1)\xi ] - \frac{b_k}{2} \cosh [(k-1)\xi ], \quad k\ge 2. \end{aligned}$$

(7.33)

Upon setting $h_k=h_k^{\prime }=0$ at $\xi =\alpha$ for $k\ge 1$, we obtain that $a_k$ and $b_k$ for $k\ge 1$ satisfy the linear system

$$\begin{aligned} a_1\cosh (2\xi ) + b_1=- \alpha + \ln {2} - \frac{1}{2}\mathrm{{e}}^{-2\alpha }, \quad 2a_1 \sinh (2\alpha )=-1 + \mathrm{{e}}^{-2\alpha },\end{aligned}$$

(7.34)

$$\begin{aligned} a_k \cosh \left[ (k+1)\alpha \right] + b_k \cosh \left[ (k-1)\alpha \right] = \frac{\mathrm{{e}}^{-(k-1)\alpha }}{k-1}- \frac{\mathrm{{e}}^{-(k+1)\alpha }}{k+1}, \quad k\ge 2,\end{aligned}$$

(7.35)

$$\begin{aligned} a_k (k+1) \sinh \left[ (k+1)\alpha \right] + b_k (k-1)\sinh \left[ (k-1) \alpha \right] = \mathrm{{e}}^{-(k+1)\alpha } - \mathrm{{e}}^{-(k-1)\alpha }, \quad k\ge 2. \end{aligned}$$

(7.36)

The solution of this system is given in (6.12)–(6.16).

Finally, we calculate the correction to the local behavior of $\varPhi$ as $(\xi ,\eta )\rightarrow \mathbf{0}$. We obtain from (7.31)–(7.33) that

$$\begin{aligned} \varPhi \sim -\frac{\eta }{2}\log \left( \frac{\xi ^2+\eta ^2}{2}\right) - \frac{\eta }{2} \sum _{m=1}^{\infty } m (a_m + b_m) \quad \text{ as }\ (\xi ,\eta )\rightarrow \mathbf{0}. \end{aligned}$$

(7.37)

Recalling that $\psi =f\varPhi$, we use the local behavior (7.20) to obtain that $\psi _c$ has the form (6.9), where $q$ is defined in (6.12). This completes the derivation of (6.12)–(6.16).