For each fixed $n$, the next term must be negligible relative to its predecessor:The ratios are understood where their denominators are nonzero near the limiting point, or equivalently through the little o notation definition. For example $1,ϵ,{ϵ}^2,\dots$ form an asymptotic sequence.

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.

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.

The potential plane is the strip $0<\psi <1$. The wall maps to its lower boundary, with upstream $\varphi \to -\infty$, the corner at $w=0$, and downstream $\varphi \to +\infty$. The free surface maps to its upper boundary.

Write $q=w^{\prime }=u-iv$. Far upstream $q=1$; far downstream the flow points vertically down and $q=i$. On the two wall rays, $q$ runs along the positive real and positive imaginary axes. The reentrant fluid corner has angle $3\pi /2$, so its potential behaves locally as $(z-B{)}^{2/3}$ and its speed diverges like $|z-B{|}^{-1/3}$. Thus the corner corresponds to $q=\infty$, not $q=0$. The free surface has $|q|=1$, giving the hodograph domainIt is the exterior of a quarter unit disc in the first quadrant.

Now $s=q^2$ maps this domain to the part of the upper half-plane outside the unit disc. The maptakes it to the first quadrant: its real and imaginary parts are positive because $|s|>1$ and $\Im s>0$. Finally $\eta =t^2$ maps that quadrant to the upper half-plane. The corresponding strip map is $\eta =e^{\pi w}$. Both give upstream $\eta =0$, corner $\eta =1$, and downstream $\eta =\infty$, so the three boundary normalizations fix the remaining real Möbius freedom. ConsequentlyOn the free surface $q=e^{i\theta }$, $t=i\tan \theta$, so $\eta$ is negative and $\Im w=1$ as required. On the wall $\eta >0$, giving $\Im w=0$.

Conformality implies $f^{\prime }(0)\ne 0$, so $a_1\ne 0$. Apply the stated Green's theorem with $F=z^n\bar{z}$:Here $\bar{\zeta }={\zeta }^{-1}$ on the unit circle. The lowest power in $f^n{f}^{\prime }$ is ${\zeta }^n$, while the most negative power in the conjugate polynomial is ${\zeta }^{-N}$. For $n\ge N$ their product has no ${\zeta }^{-1}$ term. The residue theorem gives $M_n=0$ for all $n\ge N$.

When $N=2$, the residue for $n=0$ is $|a_1{|}^2+2|a_2{|}^2$. For $n=1$, expandOnly $a_1^2\zeta {\bar{a}}_2{\zeta }^{-2}$ contributes. ThereforeThe first is the area; the second records its complex first moment, and need not be real.

The affine competitor $1-x$ has finite energy and the energy is nonnegative. A minimizing sequence therefore has bounded $\int u_j^{\prime 2}$. Its fixed value at zero gives $|u_j(x)|\le 1+\Vert u_j^{\prime }{\Vert }_2$, so it is also bounded in $L^2$ and hence in $W^{1,2}$. The estimategives uniform equicontinuity. Extract a weakly convergent subsequence in $W^{1,2}$ and, by the Arzelà-Ascoli theorem, a uniformly convergent subsequence to the same limit $u$. The endpoint values are retained. Uniform convergence on a common bounded range gives $\int u_j^{2008}\to \int u^{2008}$, while the derivative norm is weakly lower semicontinuous. Consequentlyso the global minimum is attained in the stated $W^{1,2}$ class. In fact the gradient term is strictly convex on this fixed-endpoint affine class and $t^{2008}$ is convex, so the minimizer is unique as well.

With coefficients evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$, put $p=f_{vv}$, $r=f_{uv}$ and $s=f_{uu}$. The second variation isThe Jacobi equation in divergence form is $-(p{z}^{\prime }+rz{)}^{\prime }+r{z}^{\prime }+sz=0$; if $r$ is differentiable it is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$. A conjugate point $c>a$ admits a nonzero Jacobi solution vanishing at $a,c$.

There are two positivity conventions worth distinguishing. Strict pointwise positivity means $q(\varphi )>0$ for each nonzero fixed-endpoint variation. Coercive positivity means $q(\varphi )\ge c\int {\varphi }^{\prime 2}$ with one $c>0$ for all such variations; the endpoint Poincaré estimate makes this equivalent to a positive lower bound in the full $H^1$ norm. Under the regular Legendre hypothesis $p>0$, the necessary and sufficient Jacobi criterion is no conjugate point in $(a,b]$; in that regular case it gives both strict and coercive positivity. Equivalently the Jacobi equation has a solution positive on $[a,b]$, and the square identitygives positivity and a uniform derivative lower bound by the positive minimum of $z$ and boundedness of $z,z^{\prime }$. Conversely a conjugate Jacobi field, extended by zero beyond its second zero, is a nonzero Lipschitz variation with $q=0$, contradicting either positivity convention.

Coercivity itself forces $p>0$: local high-frequency variations make the leading coefficient at least the coercivity constant. Mere strict positivity need not force it, as $q(\varphi )=\int {\varphi }^2$ shows. If that weaker convention is intended without regularity, a complete criterion is $p\ge 0$, no conjugate point for every regularized form $q_{\eta }=q+\eta \int {\varphi }^{\prime 2}$ with $\eta >0$, and no nonzero zero-endpoint Lipschitz weak solution of the original Jacobi equation. The regularized criteria imply $q_{\eta }\ge 0$, hence $q\ge 0$ by passage to the limit. For a nonnegative quadratic form, $q(\varphi )=0$ forces all mixed terms with test variations to vanish, giving exactly the weak Jacobi kernel; excluding it gives strict positivity. Necessity follows in reverse. This specifies the degenerate case without imposing an unstated strict Legendre condition.

For weak local minimality, sufficient conditions are the Euler-Lagrange equation and a coercively positive second variation with the stated smoothness. Taylor's remainder is bounded by $\epsilon \Vert \varphi {\Vert }_{H^1}^2$ for small $W^{1,\infty }$ perturbations, so the positive quadratic term dominates it. These are the classical weak minimum sufficient conditions.