Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2008 / B / 5

Near a prevertex of a Schwarz-Christoffel map $X_j$, a conformal map to an interior angle $(1-{\beta }_j)\pi$ has local formIts derivative therefore has the factor $(Z-X_j{)}^{-{\beta }_j}$. On each real interval between prevertex, the argument of the product of these factors is constant, so its integral traces a straight polygon side. Passing a prevertex of a Schwarz-Christoffel map changes the argument by the prescribed exterior angle ${\beta }_j\pi$. Thushas exactly the required straight sides and corner angles, with the branches chosen in the upper half-plane. Equivalently $f^{\prime \prime }/f^{\prime }$ has simple poles $-{\beta }_j/(Z-X_j)$; removing them leaves the constant normalization determined by behavior at infinity. The additive integration constant translates the polygon and $C$ rotates and scales it. This is the Schwarz-Christoffel formula.

The automorphisms of the upper half-plane are real Möbius transformations with positive determinant, modulo a common nonzero scalar. They have three real degrees of freedom, so three prevertex of a Schwarz-Christoffel map positions may be assigned independently. The remaining positions are constrained by the side-length ratios.

If $X_n=\infty$, omit its finite factor:Since ${\sum }_j{\beta }_j=2$, the derivative behaves like $Z^{{\beta }_n-2}$ at infinity; in the coordinate $U=1/Z$, this is exactly the required corner exponent there. Thus the missing angle is encoded by behavior at infinity.

The finite corners have interior angle $3\pi /2$, hence exterior-angle parameters $-1/2$. The channel end at $Z=0$ is logarithmic. Their Schwarz-Christoffel mapping is consequentlyChoose $R(Z)=\sqrt{Z^2-1}$ analytic in the upper half-plane with $R(Z)\sim Z$ at upper infinity. In particular $R(iy)=i\sqrt{1+y^2}$ for $y>0$. A direct differentiation givesHence $f=A+CH$ with $H=R+{\tan }^{-1}(1/R)$.

The inverse trigonometric functions here is the branch obtained by analytic continuation from $Z>1$ through the upper half-plane, not an independent Cauchy principal value evaluation at each point. Its endpoint values give $H(1)=\pi /2$, $H(-1)=-\pi /2$. Imposing $f(1)=1$ and $f(-1)=-1$ therefore gives $A=0$, $C=2/\pi$:Along the positive imaginary axis, $H$ is purely imaginary and tends to $-i\infty$ as $Z\to 0$; at upper infinity $H\sim Z$. The real intervals $(0,1)$ and $(-1,0)$ trace the channel walls $x=1$ and $x=-1$ downward, while the outer intervals trace the horizontal boundary. This verifies the stated end assignments as well as the constants. The map is the Schwarz-Christoffel map of a half-plane joined to a strip.

A uniform horizontal velocity potential in the upper $Z$-plane is a real constant times $Z$. Since $dz/dZ=(2/\pi )R/Z$, chooseup to an additive constant. Its complex velocity isAs $Z\to \infty$ this tends to one, giving physical velocity $(1,0)$. As $Z\to 0$, $R\to i$ and $Z/R\to 0$, giving zero velocity far down the channel. The imaginary part of $w$ is constant on the real $Z$-axis, so the physical boundary is a streamline and the impermeability condition holds. The usual complex velocity convention is $dw/dz=u-iv$.

To supply nonzero channel flux while letting the speed decay in the large upper region, take a logarithmic velocity potential:Use the logarithm analytic on the upper half-plane, with argument between zero and $\pi$. Its complex velocity isAt upper infinity it tends to zero. At the channel end $Z\to 0$, it tends to $1/i=-i$, which means $(u,v)=(0,1)$, as required. The streamfunction is constant on each real boundary interval, and differs by two between the two channel walls. That flux equals the channel width two times the unit upward speed, providing an independent normalization check.