Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2007 / B / 7

The integral construction requires differentiation under the integral and vanishing boundary terms. With those assumptions, integration by parts givesThe standard construction therefore sets $g^{\prime }={\zeta }^{2}g$, giving $g=C{e}^{{\zeta }^3/3}$. Convergent contours run between rays where $\Re {\zeta }^3<0$: convenient central rays have arguments $-\pi /3$, $\pi /3$ and $\pi$. Two choices are the oriented pathsThe cubic exponential dominates the linear one at both ends. To verify independence directly, put $A={\int }_0^{\infty }{e}^{-r^3/3}dr>0$ and $B={\int }_0^{\infty }r{e}^{-r^3/3}dr>0$. At $z=0$ the value and derivative of the integral along rays ${\theta }_1,{\theta }_2$ are $A(e^{i{\theta }_2}-e^{i{\theta }_1})$ and $B(e^{2i{\theta }_2}-e^{2i{\theta }_1})$. For the two contours above their derivative-to-value ratios are respectively $B/A$ and $(B/A)(e^{i\pi /3}-1)$, which differ. Hence their Wronskian is nonzero.

For the second equation multiply by $z$ before integrating by parts:Setting the residual weight to zero gives $g=C(1+{\zeta }^2{)}^{-1/2}$. One contour is the segment from $-i$ to $i$, with the square root positive at $\zeta =0$. Its integral is a nonzero multiple of $J_0(z)$: substituting $\zeta =it$ gives $i{\int }_{-1}^1{e}^{izt}(1-t^2{)}^{-1/2}dt$, an entire function nonzero at zero.

A second contour starts at $i$ and follows $\zeta =i-t$, $0\le t<\infty$, choosing a continuous branch along the path. For $\Re z>0$ it converges at infinity; at the branch endpoint its integrable singularity is $t^{-1/2}$. In both cases the boundary term contains $\sqrt{1+{\zeta }^2}$ and vanishes at branch endpoints. The second integral has a nonzero logarithmic divergence as $z\to 0$ through positive real values, because its large-$t$ integrand is a nonzero multiple of $e^{-zt}/t$. It is therefore independent of the entire first solution; analytic continuation supplies solutions elsewhere away from the equation's singular point.

Strictly, nonzero $w$ and an arbitrary contour alone do not prove the printed necessity for $g$. Necessity also needs injectivity of the chosen Laplace representation, while the vanishing-integrand argument above is the usual sufficient construction. For a concrete counterexample to unconditional necessity in the Bessel case, take a positively oriented circle $|\zeta |=R>1$, choose the branch $(1+{\zeta }^2{)}^{-1/2}\sim 1/\zeta$ outside the cut $[-i,i]$, and add one to that weight. The added entire integrand integrates to zero. Expanding the original weight on this circle gives integral $2\pi i{J}_0(z)$, which is nonzero and solves the equation, although the modified weight is not a constant multiple of the printed one. The standard weights and contours above still give the requested independent solutions.

Letbe the root with positive real and positive imaginary parts. More explicitly, with $r_j=\sqrt{{\kappa }_j^4+{ϵ}^2}$,Decay at each infinity and continuity at zero force $y=A{e}^{-i{q}_{0}x}$ for $x<0$ and $y=A{e}^{i{q}_{1}x}$ for $x>0$. The derivative jump is $i(q_1+q_0)A=1$, soThe two roots lie in the first quadrant, so their sum cannot vanish.

Integrating the differential equation twice by parts on each half-line givesTheir sum is the stated identity with right side one. Equivalently the global distributional equation is $y^{\prime \prime }+{\kappa }^{2}y={\delta }_0$, since the derivative has jump one.

The direct half-transforms areHere ${\hat{y}}_{+}$ is analytic above its lower-half-plane pole $-q_1$, while ${\hat{y}}_{-}$ is analytic below its upper-half-plane pole $q_0$.

To recover the same result from the transform identity alone, use Wiener-Hopf factorization:Multiply by $(q_1+k)/(q_0+k)$. The coefficient of ${\hat{y}}_{-}$ becomes $(q_0-k)/(q_1-k)$, analytic and zero-free in the lower half-plane, while the coefficient of ${\hat{y}}_{+}$ is analytic and zero-free in the upper half-plane. Split the right side asThe first term is analytic above and the second below. Equating the separated functions and using their decay at infinity fixes the entire remainder to zero, yielding the boxed ${\hat{y}}_{+}$, with the branches fixed before factorization.