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The rotated global relations form a square system for the three missing boundary traces. Solve this system and substitute the resulting finite-time spectral boundary transforms into the integral representation. The pieces containing $Q(k_j,T)$ have zero contour contribution by analyticity, rotation of the relevant sectors and the Cauchy integral theorem. The remaining expression involves only $Q_0,F_0,G_0,G_1$. The finite-interval Airy spectral determinant has no nonremovable zeros in the contour domains, as established below, so this elimination introduces no extra spectral data there. Differentiating the representation, and using temporal inversion and compatibility at the corners, verifies the partial differential equation, initial value and the three prescribed boundary values.

For a direct Hadamard well-posedness check, the homogeneous difference satisfiesThis proves uniqueness and continuous dependence for homogeneous boundary data. More generally subtract a smooth quadratic boundary lift $b(x,t)$ satisfying the three prescribed endpoint conditions. The remaining problem has homogeneous boundary conditions and forcing $-b_t-b_{xxx}$. Its semigroup generator is $A=-d^3/d{x}^3$ on $H^3(0,L)$ with $v(0)=v(L)=v^{\prime }(L)=0$. It is a dissipative operator; for every real $\zeta >0$ the equation $(\zeta -A)v=h$ has a unique solution, since the homogeneous third-order boundary system is injective by the same energy identity and therefore invertible. Thus $A$ is maximal dissipative and generates a contraction semigroup. This gives existence and continuous dependence on the lifted data and forcing. Smooth compatible data therefore define a well-posed problem. This operator argument and the spectral elimination describe the same evolution.

Expanding the matrix determinant from part (b) givesThe finite-interval Airy spectral determinant obeys $\Delta (\alpha k)={\alpha }^2\Delta (k)$. At the origin,This zero comes from coalescence of the three rotated equations; it is removable in the underlying boundary system. Indeed, $v^{\prime \prime \prime }=0$ with $v(0)=v(L)=v^{\prime }(L)=0$ has only the zero quadratic solution.

For $k\ne 0$, a zero of $\Delta$ is equivalent to the existence of a nonzero combination $v(x)={\sum }_{j=0}^2{c}_j{e}^{i{\alpha }^{j}kx}$ satisfying the homogeneous endpoint conditions. This can be checked by expanding the determinant of those three endpoint equations. Such a function satisfies $v^{\prime \prime \prime }=\omega (k)v$, so it is an eigenfunction of $A$ with eigenvalue $-\omega (k)$. Integration by parts yieldsso every nonzero zero has $\mathrm{Re}\omega (k)\ge 0$. None lies on the boundary $\mathrm{Re}\omega =0$: by a cube-root rotation it suffices to take $K=kL$ real. With $u=\sqrt{3}K/2$, the determinant equation becomesThe right-hand side has squared modulus $1+4{\sinh }^{2}u>1$ for $K\ne 0$, whereas the left-hand side has modulus one. Therefore there are no nonremovable spectral poles in $\overline{D}$, and the contour representation has no discrete residue contribution from that domain.

Taken literally as a claim about the operator's discrete spectrum, however, the printed assertion is false. To see this constructively, put $k=-is/L$, $s>0$. ThenThus zeros occur whenBetween successive positive cosine maxima and adjacent negative minima the difference changes sign. There are consequently infinitely many positive roots, producing genuine negative eigenvalues $-s^3/L^3$ of $A$. The third-order resolvent maps into $H^3(0,L)$, whose inclusion into $L^2(0,L)$ is compact, so these are indeed discrete spectrum of an operator with compact resolvent. The valid conclusion is absence of poles in the reconstruction sectors, not absence of operator eigenvalues.