Past exam of the applied complex variables course of the University of Oxford / 2011 / 3 / b

Write $Y_{\pm }={\overline{y}}_{\pm }$, $y_0=y(0)$ and $d_{\pm }=y^{\prime }(0\pm )$. In a common convergence strip, two integrations by parts on each half-line giveThe boundary terms at infinity vanish by exponential decay; the ordinary differential equations also give decay of the required first derivatives. Substituting the two equations for $y^{\prime \prime }$ yieldsContinuity makes the $y_0$ terms cancel on addition, while the unit derivative jump givesMultiply by $-1/((k-a)(k+b))$ and factor the quadratics. Since $\mathrm{Im}(a+b)>0$, their sum is nonzero, and partial fractions giveThis is the Wiener-Hopf equation for the two half-line Fourier transforms. The plus and minus subscripts refer to support in physical space; their holomorphic domains in transform space are determined next.

Let $\eta =\mathrm{Im}k$. On $x>0$, the modulus of the Fourier factor is $e^{-\eta x}$, so the given decay bound makes $Y_{+}$ converge absolutely when $\gamma +\eta >0$. On $x<0$, write $x=-r$; the bound becomes $e^{-(\gamma -\eta )r}$, so $Y_{-}$ converges absolutely when $\eta <\gamma$. Uniform domination on compact subsets permits differentiation with respect to $k$, provingThus the half-planes guaranteed directly by the given common decay bound have $\alpha =-\gamma$ and $\beta =\gamma$, and the equation holds on their overlapThis is a statement about the domains guaranteed by that bound, not necessarily their largest possible domains. The exact solution below decays at different rates on the two half-lines and gives the sharper convergence half-planes $\mathrm{Im}k>-\mathrm{Im}a$ and $\mathrm{Im}k<\mathrm{Im}b$, with maximal common strip $-\mathrm{Im}a<\mathrm{Im}k<\mathrm{Im}b$.

The constant-coefficient equations and decay imply that only the modes $e^{iax}$ on the positive half-line and $e^{-ibx}$ on the negative half-line can occur; the opposite modes grow. Continuity and the nonzero derivative jump then imply nonzero amplitudes, so any common decay exponent satisfies $0<\gamma \le \mathrm{Im}a<\mathrm{Im}b$. In particular $-b$ lies below the plus holomorphic half-plane, and $a$ lies outside the minus half-plane. This locates the poles needed for the transform calculation, without yet computing the amplitudes.

Rearrange the equation from part (i) asThe left side is holomorphic in the plus half-plane, and the right side in the minus half-plane. Their equality on the overlap glues them into one entire function. The assumed decay of $Y_{\pm }$, together with the rational factors tending to one and the subtracted fractions tending to zero, makes this entire function tend to zero at infinity. It is therefore identically zero by Liouville's theorem. This is the additive splitting step of the Wiener-Hopf method. Solving the two zero identities givesTheir apparent cancellation formulas extend holomorphically wherever a cancelled factor vanishes; the only actual poles are at $-a$ and $b$.

With the Fourier convention in use, inversion has factor $e^{-ikx}$ and normalization $1/(2\pi )$. For $x>0$, close the contour in the lower half-plane clockwise, collecting the pole at $-a$. For $x<0$, close in the upper half-plane counterclockwise, collecting the pole at $b$. The residue theorem givesEquivalently, each displayed transform follows immediately by integrating the corresponding decaying exponential on its half-line, so Fourier uniqueness gives the same inversion without a contour estimate. Both values at zero are $1/(i(a+b))$, and the derivative jump is $(a+b)/(a+b)=1$, directly checking the interface conditions.

The moduli are a nonzero constant times $e^{-\mathrm{Im}ax}$ for $x>0$ and $e^{-\mathrm{Im}b|x|}$ for $x<0$. Hence the largest common exponential decay rate isStrictly, the stated big-$O$ bound is also true for every smaller positive $\gamma$; it does not specify a unique exponent unless the sharp rate is intended. The two transform convergence half-planes are exactly the sharper domains stated in part (ii), and the poles on their boundaries prevent extending either half-line integral domain any farther.