Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2010 / B / 5

Orient the smooth contour so the plus side is to its left. Assume the density is Hölder continuous near the observation point and the integrals converge at any other endpoints. The Cauchy principal value isSubtract $f(t)$ from the numerator. The resulting quotient is locally integrable, so its limits from the two sides coincide. For the remaining constant density, straightening the contour gives a half-circle contribution $\pm i\pi f(t)$. Division by $2\pi i$ yields the Sokhotski–Plemelj formulaFor the requested real integral assume $0<\alpha <1$ and $-1<x<1$, ensuring integrable endpoints and an interior principal-value pole. Put $W(t)=(1-t{)}^{\alpha -1}/(1+t{)}^{\alpha }$. Choose $w(z)=(z-1{)}^{\alpha -1}/(z+1{)}^{\alpha }$ analytic off $[-1,1]$, positive for $z>1$, and $w(z)\sim 1/z$ at infinity. Its two boundary values are $w_{\pm }(t)=-e^{\pm i\pi \alpha }W(t)$.

The jump is $-2i\sin (\pi \alpha )W(t)$. The Cauchy representation for this decaying analytic function is therefore $w(z)=-\sin (\pi \alpha ){\int }_{-1}^{1}W(t)/(t-z)dt/\pi$. Taking the mean of its two boundary values removes the half-residue terms and gives

The complex potential of a two-dimensional potential flow maps the fluid region to the strip $0<\mathrm{Im}w<1$. Its lower boundary is the wall, with $B$ at $w=0$; upstream is $\mathrm{Re}w\to -\infty$ and downstream is $\mathrm{Re}w\to +\infty$. The free streamline is the upper strip boundary.

Write $q=w^{\prime }=u-iv$. The horizontal wall gives positive real velocities running from $q=1$ to the stagnation value $q=0$ at $B$. The vertical wall gives negative imaginary velocities from zero to $-i$. On the free surface $|q|=1$, with the angle turning continuously from zero to $-\pi /2$. Thus the hodograph plane domain is the fourth-quadrant quarter of the unit disk, bounded by those two radii and the unit-circle arc.

The following successive conformal maps straighten it. First $r=q^2$ maps to the lower half unit disk. Next $s=(r-1)/(r+1)$ maps to the third quadrant, since $\mathrm{Re}s=(|r{|}^2-1)/|r+1{|}^2<0$ and $\mathrm{Im}s=2\mathrm{Im}r/|r+1{|}^2<0$. Finally $s^2$ maps onto the upper half-plane. The three wall landmarks $q=1,0,-i$ go to $0,1,\infty$ respectively.

The strip map $e^{\pi w}$ also maps onto the upper half-plane, with these same landmarks $0,1,\infty$. The unique half-plane automorphism fixing all three is the identity. ThereforeThe free surface maps to the negative real axis in either construction, as required by its streamfunction value one. This completes the right-angle free-streamline hodograph map construction.