= Solution
Expanding the matrix <determinant> from part (b) gives
$$
\det M=ik(\alpha-\alpha^2)\Delta(k)=-\sqrt3k\Delta(k),\qquad
\Delta(k)=e^{-ikL}+\alpha e^{-i\alpha kL}+\alpha^2e^{-i\alpha^2kL}.
$$
The <finite-interval Airy spectral determinant> obeys $\Delta(\alpha k)=\alpha^2\Delta(k)$. At the origin,
$$
\Delta(k)=-\frac32L^2k^2+O(k^5).
$$
This zero comes from coalescence of the three rotated equations; it is removable in the underlying boundary system. Indeed, $v'''=0$ with $v(0)=v(L)=v'(L)=0$ has only the zero quadratic solution.
For $k\ne0$, a zero of $\Delta$ is equivalent to the existence of a nonzero combination $v(x)=\sum_{j=0}^2c_je^{i\alpha^jkx}$ satisfying the homogeneous endpoint conditions. This can be checked by expanding the determinant of those three endpoint equations. Such a function satisfies $v'''=\omega(k)v$, so it is an <eigenfunction> of $A$ with <eigenvalue> $-\omega(k)$. <Integration by parts> yields
$$
-\operatorname{Re}\omega(k)\,\|v\|_2^2=\operatorname{Re}\langle v,Av\rangle=-\frac12|v'(0)|^2,
$$
so every nonzero zero has $\operatorname{Re}\omega(k)\geq0$. None lies on the boundary $\operatorname{Re}\omega=0$: by a cube-root rotation it suffices to take $K=kL$ real. With $u=\sqrt3K/2$, the determinant equation becomes
$$
e^{-3iK/2}=\cosh u-i\sqrt3\sinh u.
$$
The right-hand side has squared modulus $1+4\sinh^2u>1$ for $K\ne0$, whereas the left-hand side has modulus one. Therefore \b[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 $k=-is/L$, $s>0$. Then
$$
\Delta(-is/L)=e^{-s}-2e^{s/2}\cos(\sqrt3s/2+\pi/3).
$$
Thus zeros occur when
$$
e^{-3s/2}=2\cos(\sqrt3s/2+\pi/3).
$$
Between 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>. \b[The valid conclusion is absence of poles in the reconstruction sectors, not absence of operator eigenvalues.]
Back to article page
