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Let $0<a<b$ and impose zero Dirichlet boundary conditions at $a,b$. With the convention $L_{x}G(x,s)=\delta (x-s)$ for the Dirac delta distribution, the Green's function is continuous and its first derivative has jump one. For $n\ne 0$ choose $\ell (x)=x^{-n}-a^{-n}$ and $r(x)=x^{-n}-b^{-n}$. Their Wronskian is $n(a^{-n}-b^{-n})s^{-n-1}$, soFor $n=0$, take the continuous limit or use $\ell =\log (x/a)$ and $r=\log (x/b)$ to obtainThe solution is $u(x)={\int }_a^{b}G(x,s)f(s)ds$. These kernels use ordinary measure $ds$. The factor $s^{n+1}$ comes from the nonconstant Wronskian and must not be discarded by confusing the original operator with its multiplied divergence form $(x^{n+1}{u}^{\prime }{)}^{\prime }$.

For the advection-diffusion equation $u_t=u_{xx}+\alpha u_x$ on $x>0$, the spectral divergence form uses $\omega =k^2-i\alpha k$ and factor $e^{-ikx+\omega t}$. With negative-exponential Half-range Fourier transforms, integration gives the displayed global relation for $\mathrm{Im}k\le 0$. The boundary transform $G_1$ refers to $u_x(0,t)$, the negative of the outward normal derivative.

Put $w(k)=-i(k^3+k)$ and define the spectral fluxWriting $E=e^{-ikx+w(k)t}$, division of the proposed divergence form by $E$ gives $q_t+wq+X_x-ikX$. Direct differentiation yieldsThe last term cancels $wq$, so the divergence form is exactly the normalized partial differential equation. This calculation fixes both the sign of the dispersion relation and the sign of the boundary flux used below.

Use the spectral parameter for a linear boundary value problem $k\in \mathbb{C}$ and define the dispersion relation $\omega (k)=k^2-i\alpha k$. Direct differentiation gives the divergence formIndeed the coefficient of $u$ left over after expansion is $\omega -k^2+i\alpha k=0$, and the remaining factor is $u_t-u_{xx}-\alpha u_x$. Thus this is equivalent to the advection-diffusion equation for every $k$.

Introduce the Half-range Fourier transforms and finite-time spectral boundary transformswhere $g_1(t)=u_x(0,t)$ is the unknown normal derivative with the positive-$x$ convention. The outward normal at zero instead gives $-g_1$. The spatial transforms are analytic for $\mathrm{Im}k<0$ and continuous on the real axis under the stated decay assumptions. Integrating the divergence form on $(0,\infty )\times (0,t)$ gives the global relationThe sign follows from the lower spatial endpoint: the integrated spatial derivative is minus its value at zero. This sign will determine the boundary-forcing term in the solution.

Let $\Omega$ be a bounded piecewise smooth domain, $f=u{|}_{\partial \Omega }$ and $q={\partial }_{n}u$, with $n$ the outward normal vector. Integrating the divergence form from part (i) and using the divergence theorem gives two global relations for a linear boundary value problem:These conjugate global relations for the modified Helmholtz equation are identities between the prescribed Dirichlet boundary condition and its unknown normal derivative; neither is a pointwise boundary equation. Both hold for every nonzero boundary spectral parameter under the usual trace regularity assumptions.

If $u$ is real, its two boundary traces $f,q$ are real. Since ${\tilde{v}}_{\lambda }=\overline{v_{\overline{\lambda }}}$ and the outward normal vector is real,Thus apply complex conjugation to the first global relation for a linear boundary value problem, and replace its parameter by $\overline{\lambda }$, to get the second. For complex $u$, conjugation would also replace its traces by their conjugates, so that argument requires the stated reality assumption. There is also an algebraic redundancy in this particular parametrization: ${\tilde{v}}_{\lambda }=v_{1/\lambda }$, hence $\tilde{\mathcal{G}}(\lambda )=\mathcal{G}(1/\lambda )$ even for complex data. This does not change the requested conjugation identity.

For a possibly unstable normal mode, $c_i\ne 0$ ensures $V=U-c$ never vanishes, so choose one continuous branch of $V^a$. Substitute $\hat{w}=V^{a}q$ into the Taylor–Goldstein equation and collect terms:Multiplication by $V^{2a}$ puts its first two terms in divergence form:Multiply by the complex conjugate $\overline{q}$ and apply integration by parts. Since $q(\pm L)=0$, the boundary term vanishes and the power-transformed Taylor–Goldstein energy identity isThe weight is $V^{2a}$, not $|V{|}^{2a}$: preserving its complex phase is essential for the subsequent stability proofs.

The homogeneous linear differential equation can be written in divergence form:One integration gives $u^{\prime }=C{x}^{-n-1}$. Therefore the two cases areIn each case the displayed functions have linear independence throughout the interval. Imposing both homogeneous Dirichlet boundary conditions forces $A=B=0$: for $n\ne 0$, $2^n\ne 1$, and for $n=0$, $\log 2\ne 0$. This also establishes uniqueness for the inhomogeneous boundary value problem.