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Retain the factor $ϵ$ in the last temporal phase, as printed in the original PDF. Let $a=U/(2\gamma )$ and put $\eta =e^{ax}b(\xi )e^{st}$ with $s=-i{\omega }_0-iϵ{\omega }_1$. The drift-removing substitution for a Dirichlet Sturm-Liouville problem cancels the first derivative:With $\xi =\sqrt{ϵ}fx$, this becomesChoose the positive scale $f=(\nu /\gamma {)}^{1/4}$. To leave an order-one spectral parameter, cancel the numerator's order-one term. ThenThe assumed decaying solutions of the Hermite differential equation require $\lambda =2j+1$, $j=0,1,\dots$. They are proportional to $e^{-{\xi }^2/2}{H}_j(\xi )$, with physicists' Hermite polynomials. Therefore the global modes of a quadratically confined Ginzburg-Landau equation haveFor every fixed positive $ϵ$, multiplying the Hermite-Gaussian by $e^{Ux/(2\gamma )}$ still gives decay at both infinities. The strongest mode is $j=0$, so the whole-line system is globally unstable precisely whenEquality is a neutral ground mode. This spectrum is exact for the stated quadratic profile, even though the notation organizes it as an expansion in $ϵ$.

The homogeneous absolute-instability threshold in part (a) is $\mu =U^2/(4\gamma )$. The global threshold approaches that value as the profile varies increasingly slowly, but at finite $ϵ$ the quadratic confinement adds the loss $ϵ\sqrt{\gamma \nu }$. A locally absolutely unstable patch at the profile maximum is necessary but must be strong and broad enough to overcome that loss. Merely having positive local temporal growth ${\mu }_0>0$, which can be only convective wave-packet instability, is insufficient for a growing decaying global normal mode. This connects the local impulse classification to the discrete global spectrum.

Use the drift-removing substitution for a Dirichlet Sturm-Liouville problem, $y=e^{-x}v$. Direct differentiation gives $\mathcal{L}y=e^{-x}(-v^{\prime \prime }+v)$. The homogeneous boundary conditions become $v(0)=v(\pi )=0$, so the eigenvalues and eigenfunctions areThe self-adjoint weighted form isThus the weight and leading coefficient are both $e^{2x}$. The orthogonality of Sturm-Liouville eigenfunctions readsAfter multiplying the forcing equation by $e^x$, it becomes $-v^{\prime \prime }+v=x$. Expanding $x$ in the sine series of part (i) and dividing by $n^2+1$ givesEvery term satisfies the endpoint conditions. As a check, solving the transformed ODE directly gives $v=x-\pi \sinh x/\sinh \pi$, which has the same coefficients and forcing.

Use the drift-removing substitution for a Dirichlet Sturm-Liouville problem, $z(x)=e^{-2x}u(x)$. It gives $u^{\prime \prime }+\lambda u=0$ with $u(0)=u(1)=0$. Positive $\lambda =k^2$ requires $\sin k=0$; zero or negative $\lambda$ yields only the trivial solution under both endpoint conditions. HenceMultiplication of the original equation by $e^{4x}$ gives $-(e^{4x}{z}^{\prime }{)}^{\prime }-4e^{4x}z=\lambda e^{4x}z$. Thus the appropriate weighted inner product has weight $e^{4x}$, andMultiplying each eigenfunction by $\sqrt{2}$ gives weighted norm one.

Multiply the original equation by $p(x)=\exp ({\int }^{x}u(s)ds)>0$. Since $p^{\prime }=up$, it becomes $(p{y}^{\prime }{)}^{\prime }+qy+\lambda ry=0$ with $q=pv$, $r=pw>0$. This is Sturm-Liouville form. For two real eigenfunctions, multiply the equation for $y_n$ by $y_m$, subtract the equation for $y_m$ multiplied by $y_n$, and integrate. The Lagrange identity for a Sturm-Liouville operator givesEach separated homogeneous boundary condition makes the corresponding determinant zero, provided its pair of coefficients is not both zero. For distinct eigenvalues, this proves the orthogonality of Sturm-Liouville eigenfunctions.

For the constant drift, the drift-removing substitution for a Dirichlet Sturm-Liouville problem is $y=e^{-\alpha x}v$. It reduces the equation to $v^{\prime \prime }+\lambda v=0$ with $v(0)=v(\pi )=0$. Nonzero solutions occur exactly atThe zero and negative values give only the zero solution under both endpoint conditions. The self-adjoint expression is $(e^{2\alpha x}{y}^{\prime }{)}^{\prime }+{\alpha }^2{e}^{2\alpha x}y+\lambda e^{2\alpha x}y=0$. Hence the weight is $r=e^{2\alpha x}$, and for the displayed normalization,