Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2008 / B (Applied Complex Variables)

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.

The lower wall has streamfunction zero. The streamline through the lip and along the upper fluid free boundary has streamfunction equal to the downstream volumetric flow rate, namely $a$ because the downstream speed is one and its depth is $a$. The complex potential of a two-dimensional potential flow therefore fillsThe chosen velocity potential origin gives $w(D)=ia$. The remote reservoir endpoints $A,E$ have velocity potential tending to $-\infty$; the downstream endpoints $B,C$ have velocity potential tending to $+\infty$.

Put $q=dw/dz=u-iv$. On the floor its values run from zero to one along the positive real axis. Above the plate the flow goes left toward the lip, so the plate velocities run along the negative real axis from zero at $E$ to $-1$ at $D$. On the fluid free boundary Bernoulli's equation and its constant pressure give $|q|=1$. Its downward velocity has $v<0$, hence $\mathrm{Im}q>0$. Consequently the hodograph plane domain isthe upper unit semicircle. Its marked boundary values are $q(A)=q(E)=0$, $q(D)=-1$, and $q(B)=q(C)=1$.

For the velocity potential strip, $e^{\pi w/a}$ maps to the upper half-plane. The real Möbius transformationpreserves that half-plane and sends its marked endpoints to $-1,\infty ,0$ in the order $A/E,D,B/C$.

For the hodograph plane, first put $r=(q-1)/(q+1)$. Its image is the second quadrant, bounded by the negative real ray and positive imaginary ray. Squaring takes that quadrant to the lower half-plane, and negating restores the upper half-plane. Thushas exactly the same three marked values. Uniqueness of a conformal map to a half-plane with three assigned boundary values makes these maps equal. Cancel the minus signs to obtainThe sign in the intermediate map is important for identifying the upper, rather than lower, half-plane.

On the fluid free boundary write $q=e^{-i\theta }$, with $-\pi <\theta <0$: the lip has $q=-1$ and the downstream limit has $q=1$. ThenSubstitution into the relation from part (ii) givesThe surface lies on the top of the velocity potential strip, so its logarithm has imaginary part $\pi$, givingFinally $dz/dw=1/q=e^{i\theta }$, and thereforeThe source's typeset $\cot \theta /2$ is interpreted as $\cot (\theta /2)$, as the derivation requires.

Integrate real and imaginary parts from the lip value $z(-\pi )=i$:The identities $\sin s\cot (s/2)=1+\cos s$ andgive the explicit parametrizationAs $\theta \uparrow 0$, $x\to +\infty$ and $y\to 1-a$. This must be the downstream height $a$, soThe surface initially moves slightly left of the lip before turning right. Its leftmost point occurs at $\theta =-\pi /2$, with $x_{\min }=-(1-\log 2)/(2\pi )$. This is the half-contraction free streamline at a horizontal canal lip.

Orient $\Gamma$ and label its left side plus and its right side minus. Assume the density is Hölder continuity near the point in question, or has sufficient regularity for the boundary limits below. The source does not explicitly state this regularity assumption. Parametrize locally by oriented arclength $\gamma (s)$, with $\gamma (0)=t$. Define the Cauchy principal value by removing equal arclengths on the two sides:Subtract $f(t)$ from the numerator near $t$. The Hölder continuity remainder is locally bounded in absolute value by a constant times $|s{|}^{\nu -1}$ for some $\nu >0$, so its integral has the same limit from the two sides. The constant part contributes half the residue, with sign determined by the side of the small semicircle. This proves the Sokhotski–Plemelj theoremIn particular $f=w_{+}-w_{-}$ and the Cauchy principal value term with coefficient $1/(\pi i)$ equals $w_{+}+w_{-}$. Substituting these two identities into the singular integral equation gives

Choose a canonical homogeneous solution $\tilde{w}$ whose boundary values do not vanish, and account for any permitted zeros or poles when forming a quotient. The assumptions $a\pm b\ne 0$ alone do not ensure that an arbitrary homogeneous solution has this property. Its jump relation isDividing the inhomogeneous relation by $(a+b){\tilde{w}}_{+}$ therefore givesThe Cauchy integral solves this additive jump. More generally every solution in the chosen analytic class haswhere $H$ has no jump across $\Gamma$ and obeys the applicable growth, endpoint and divisor conditions. Setting $H=0$ gives exactly the intermediate formula printed in the source.

Let $C\rho =(2\pi i{)}^{-1}\mathrm{PV}{\int }_{\Gamma }\rho (\zeta )/(\zeta -t)d\zeta$. Applying Sokhotski–Plemelj theorem to the displayed solution and subtracting boundary values yieldsUseThe normalized particular solution is consequentlyand the full solution isThus the printed expression is a particular solution; homogeneous terms need separate normalization. To see why “general” cannot hold without qualification, take the positively oriented unit circle, $a=(1+z)/2$, $b=(1-z)/2$, and $c=0$. Then $a+b=1$ and $a-b=z$ are nonzero on the contour. The analytic pair ${\tilde{w}}_{+}=1$, ${\tilde{w}}_{-}=1/z$ satisfies the homogeneous jump and decays on the exterior side. Its nonzero density $f=1-1/z$ has these very Cauchy boundary values and solves the original homogeneous equation, whereas the source formula with $c=0$ gives zero. This is homogeneous freedom in a scalar singular integral equation; endpoint and index conditions determine which jump-free additions are permitted.

The two holomorphic function functions agree throughout their overlap region strip, so they define a single holomorphic function on the union of their domains, which is the whole complex plane. The resulting entire function is bounded outside a sufficiently large disk by the assumptions at infinity. It is bounded on the disk by continuity and compactness. Liouville theorem therefore makes it constant:Agreement on an open overlap region is what permits analytic continuation; equality of unspecified boundary values alone would need additional hypotheses.

First state the normalization needed for the displayed logarithmic factorization: choose a continuous holomorphic function branch $L=\log F$ in the strip that tends to zero at both horizontal ends and has sufficient decay for its Cauchy integrals. In particular the index of the kernel along a horizontal line must be zero.

Choose $\alpha <c_{+}<c_{-}<\beta$ and let ${\Gamma }_{\pm }=\{\xi +i{c}_{\pm }:\xi \in \mathbb{R}\}$, both oriented from left to right. PutThe first is analytic above the lower line and the second below the upper line. On the region between the lines, closing a rectangle gives $C_{+}L-C_{-}L=L$: the lower horizontal side is traversed forward and the upper side backward. HenceThe equation $F{w}_{+}=w_{-}$ becomes $F_{+}{w}_{+}=F_{-}{w}_{-}$ on the overlap region. With the canonical bounded factors normalized at infinity, these two products glue to a bounded entire function. Part (a) givesThis is Wiener-Hopf factorization with the quotient convention used in this question.

The source hypotheses require a qualification: nonvanishing in the strip and $F\to 1$ do not by themselves force index zero or a logarithm tending to zero at both ends. For exampleis analytic and nonzero for $|\mathrm{Im}z|<1/2$, tends to one at both ends, but winds once around zero along the real line. A logarithm continued along that line changes by $2\pi i$, so it cannot have the stated normalization at both ends. One must first separate the index factor, or allow appropriately subtracted factors and a correspondingly different common-function growth class. The displayed constant conclusion follows under the canonical zero-index assumptions above; it is not an unconditional consequence of the two printed properties of $F$. This is the issue measured by the Index of a Wiener-Hopf kernel.

Use the same normalized factors and contours as in part (b), and put $H=G{F}_{-}$. Multiplying the inhomogeneous equation by $F_{-}$ givesThe additive Cauchy decomposition has $C_{+}H-C_{-}H=H$ on the overlap region. Subtract it from this equation:Under the same convergence and boundedness assumptions, these expressions are analytic on complementary half-planes, agree on their overlap region, and glue to a bounded entire function. ThusIf the decay is too weak for an unsubtracted Cauchy integral, the subtraction and resulting common-function growth condition must be specified. Merely asserting pointwise $G\to 0$ is not a substitute for the needed integral transform convergence.

For $\mathrm{Im}k>0$, put $s=-ik$, so $\mathrm{Re}s>0$. The Laplace transform for the Gamma function, analytically continued from positive real $s$, givesChoose $\arg k\in (0,\pi )$ in the upper half-plane. Since $\arg (-ik)=\arg k-\pi /2$,This integral is absolutely convergent throughout the upper half-plane. On the nonzero real axis it converges conditionally by Dirichlet test for improper integrals, and its value is the upper boundary limit of the same branch. In particular negative real $k$ has argument $\pi$, not $-\pi$ in this formula.

At $k=0$ the integral diverges at infinity; in the lower half-plane its exponential function factor grows and the ordinary improper integral diverges. The origin itself causes no local-integrability problem for $x^{-1/4}$. A useful continuation of the chosen power uses a branch cut descending from $k=0$ with argument in $(-\pi /2,3\pi /2)$.

Use the Fourier transform convention $\overline{U}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{ikx}dx$, initially in a class where the integral transforms and subsequent limits are defined. Transforming Laplace's equation gives ${\overline{U}}_{yy}-k^2\overline{U}=0$ on real $k$. Decay for $y\to \infty$ selectsThe boundary trace is the sum of its negative-half-line unknown and the given positive-half-line datum. The normal derivative is zero on the negative side. ThereforeFor the $e^{ikx}$ convention, negative support gives a integral transform analytic for $\mathrm{Im}k<0$, and positive support gives one analytic for $\mathrm{Im}k>0$, subject to the appropriate endpoint interpretation. The normal trace in this particular problem requires a finite-part interpretation, detailed below.

For the Wiener-Hopf method, regularize the real-axis kernel byTake the branch cut of $a_{+}$ vertically downward from $-iϵ$ and that of $a_{-}$ vertically upward from $iϵ$. Their product on the real axis is the positive square root $\sqrt{k^2+{ϵ}^2}$. The branch of $H_{+}$ has its cut downward from zero. Dividing the regularized equation by $a_{+}$ givesThe forcing $P$ is analytic in $0<\mathrm{Im}k<ϵ$, between its two cuts. Decompose $P=P_{+}-P_{-}$ by Cauchy integrals on two horizontal lines within that strip. Separation giveswhen the endpoint growth class makes the common entire remainder bounded. For $\mathrm{Im}k<0$ one can take the minus Cauchy integral on $\mathrm{Im}\hat{k}=\delta$, with $0<\delta <ϵ$, obtainingThe integrand is $O(|\hat{k}{|}^{-5/4})$ at infinity, so the Cauchy integral converges there. In the limit choose $\delta =\delta (ϵ)$, for example $\delta =ϵ/2$; a fixed positive $\delta$ cannot remain between zero and $ϵ$. With $\sqrt{k}$ denoting the limiting lower-analytic branch and with the prescribed limiting interpretation,The numerator square root has a cut going upward from $iϵ$; the power ${\hat{k}}^{-3/4}$ has a cut going downward from zero. The contour passes between them. The pole $\hat{k}=k$ is below this contour when $k$ is in the lower integral transform domain. These branch choices are essential, rather than independent principal powers of the factors.

There is an important mathematical limitation in the source's ordinary integral definition of $\bar{g}$. An explicit harmonic function particular solution in polar coordinates $0<\theta <\pi$ isEach separated polar power is harmonic function; at $\theta =0$ it has the required Dirichlet boundary condition value, and at $\theta =\pi$ its angular derivative vanishes, giving the required zero normal derivative. It also decays at infinity. Its unknown boundary traces areThe last function is not locally integrable at zero. Therefore the printed ordinary integral ${\int }_0^{\infty }g(x)e^{ikx}dx$ diverges, even for $\mathrm{Im}k>0$. The integral transform calculation is valid for this solution only after specifying a Hadamard finite-part integral, or equivalently a distribution boundary trace.

For the canonical homogeneous extension of this power, the two transformed traces areThe first is analytic below and the second above. The second formula follows from $\Gamma (3/4)=(-1/4)\Gamma (-1/4)$ and the finite-part continuation of the Laplace transform. Substituting their real-axis boundary values directly verifies the boxed integral transform relation for both signs of $k$, so the qualification can be checked without relying only on formal factorization.

There is also homogeneous corner freedom:satisfies zero Dirichlet boundary condition data on the positive axis, zero Neumann boundary condition data on the negative axis and decay at infinity. Its negative trace has integral transform $C\Gamma (1/2)(ik{)}^{-1/2}$, corresponding to the term $A/\sqrt{k}$. Higher singular modes $r^{-(n+1/2)}\sin ((n+1/2)\theta )$ would also satisfy the homogeneous boundary conditions for integers $n\ge 1$, but are excluded if the negative trace is required to be locally integrable. Thus the one-constant formula belongs to a specified corner-growth class. The original boundary conditions alone, without that class, do not assert uniqueness. These facts form the finite-part normal trace for mixed quarter-power boundary data.