Past exam of the applied complex variables course of the University of Oxford / 2011 / 2

Let $\zeta =\zeta (z,t)$ be the inverse conformal map. With $Q$ real and nonzero, the function $Q\log |\zeta |$ is harmonic away from the source, vanishes on the moving boundary, and has the required logarithmic singularity. Indeed $F(\zeta ,t)=a_1(t)\zeta +O({\zeta }^2)$ with $a_1\ne 0$, so $\log |\zeta |=\log |z|-\log |a_1|+o(1)$ near the origin. Subtracting this candidate from the physical velocity potential leaves a bounded harmonic function near the source. Its singularity is removable, and the difference has zero boundary values. The maximum principle for harmonic functions makes it identically zero. ThereforeOnly the real part is needed, so the multivalued imaginary part of the logarithm causes no ambiguity in $\varphi$.

Differentiate $F(\zeta (z,t),t)=z$ while holding the physical point $z$ fixed:Also the local analytic complex potential of a two-dimensional potential flow has derivativeThe distinction between differentiating at fixed $z$ and fixed $\zeta$ is essential because the boundary map depends on time. Substituting into the boundary evolution law and multiplying by $|\zeta F_{\zeta }{|}^2/Q$ givesTaking the conjugate inside a real part leaves it unchanged, so the desired conformal map equation for planar Darcy growth isAs a normalization check, parameterize the boundary by $\zeta =e^{i\theta }$. Its outward normal vector times arclength is $\zeta F_{\zeta }d\theta$, so its area growth rate is ${\int }_0^{2\pi }\mathrm{Re}(F_t\overline{\zeta F_{\zeta }})d\theta =2\pi Q$, exactly the specified source flux. For an expanding circular map $F=R(t)\zeta$, the equation reduces to $R\dot{R}=Q$.

Orient the contour and call its left-hand side the plus side. At a smooth interior point $t$, take a local arclength parameter $\gamma (s)$ with $\gamma (0)=t$. The Cauchy principal value means the symmetric deletion limitThe two local pieces are deleted with the same arclength cutoff. For a sufficiently smooth regular contour, an equal-radius deletion about $t$ gives the same limit. The point is assumed to be away from endpoints and self-intersections.

To derive the Sokhotski–Plemelj formula, split the contour into a short smooth arc through $t$ and its complement. The complementary integral has the same limit from both sides. On the short arc subtract $f(z)$ from the numerator. Because $f$ is holomorphic nearby, $(f(\zeta )-f(z))/(\zeta -z)$ extends continuously as $\zeta$ approaches $z$; this part also has the same limit from either side. The remaining local term is $f(z)\int d\zeta /(\zeta -z)$. Straightening the oriented tangent shows its extra contribution is $+i\pi f(t)$ from the left and $-i\pi f(t)$ from the right: on a real segment this is the direct limit of ${\int }_{-\delta }^{\delta }du/(u\mp i\epsilon )$. Dividing by $2\pi i$ yieldsIn particular the jump is $w_{+}-w_{-}=f$, while the sum is the principal-value integral divided by $\pi i$. Smooth densities on a real interval satisfy the same formulas at every interior point, by the identical subtraction step with a Hölder-continuous remainder.

Now put $h=w/\tilde{w}$. On the interval, the sign reversal of the auxiliary boundary values givesThus the required Cauchy transform density isWhenever this density is integrable at the endpoints, taking $h=C[F]$ constructs a function with the required jump. Multiplication back gives $w_{+}+w_{-}=G$. The jump of the constructed $w$ isTo conclude that this jump solves the final singular equation, one must additionally justify $w=C[f]$. This reconstruction requires endpoint and infinity conditions; the sign reversal of $\tilde{w}$ alone does not ensure it. Here is a complete valid normalization. LetThe branch is holomorphic off $[0,1]$, with $S_{+}(x)=i\sqrt{x(1-x)}$ and $S_{-}=-S_{+}$. If $G$ extends smoothly to $[0,1]$, the density $F=G/{\tilde{w}}_{+}=i\sqrt{x(1-x)}G$ vanishes like a square root at each endpoint. Its Cauchy transform $h$ is bounded at the endpoints and is $O(z^{-1})$ at infinity. Therefore $w=h/S$ is at most inverse-square-root singular at each endpoint and is $O(z^{-2})$ at infinity. Its jump $f$ is integrable.

For Cauchy reconstruction from a jump across a finite interval, apply the Cauchy integral formula to $w$ on a large slit contour, and shrink its banks to $[0,1]$. The large circle contributes zero in the limit; each small endpoint circle contributes $O(\sqrt{\epsilon })$ and also vanishes. The two banks give the jump $f$, proving $w=C[f]$. The sum part of the Sokhotski–Plemelj formula now givesThis proves the intended finite Hilbert transform inversion with admissible endpoint behavior. Adding a multiple of $1/\sqrt{x(1-x)}$ to $f$ gives a homogeneous solution; equivalently one can add a constant multiple of $1/S$ to $w$, whose boundary sum is zero.

The omitted normalization is a real limitation of the printed claim. To see it explicitly, choose instead $\tilde{w}=S$ and $G=1$. This auxiliary function satisfies all the stated holomorphicity, nonvanishing and sign-reversal requirements. ButThe displayed integral identity follows from $\xi =(1+\cos \theta )/2$ and its continuation from real $z>1$. Hence the proposed jump formula gives $f=0$, which does not solve the equation with $G=1$. The failure is the nonzero constant at infinity: $w-C[f]=1/2$. Thus the final implication is valid with suitable endpoint and decay conditions, but is false for an unrestricted auxiliary function as printed. The valid reciprocal-square-root normalization above supplies an actual solution for $G$ smooth up to the two endpoints.