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Let $\varepsilon > 0$ and consider the (complex) Sturm–Liouville differential operator on $[0,1]$ given by $$ \mathcal{L}_\varepsilon f(x) = \varepsilon^2 f''(x) + i V(x) f(x), $$ with Neumann boundary conditions $f'(0) = f'(1) = 0$; where, $$ V(x) = \begin{cases} 1, & \text{if } 0\leq x <1/2, \\ -1, & \text{if } 1/2\leq x \leq 1, \end{cases} $$ and $i = \sqrt{-1}$.

Classical Sturm–Liouville/PDE's theory ensures that $\mathcal{L}_\varepsilon$ has a compact resolvent, implying that $\mathcal L_\varepsilon$ has countable spectrum $\sigma(\mathcal{L}_\varepsilon) = \{\lambda_i(\varepsilon)\}_{i \in \mathbb{N}}$. Furthermore, it is straightforward to verify that $$ \lambda_j(\varepsilon) \in \{a + ib \mid a < 0 \text{ and } b \in [-1,1]\} \quad \text{for all } j \in \mathbb{N}. $$

My question: Define $$ L(\varepsilon) := \sup\{\mathrm{Re}(\lambda_i(\varepsilon)) \mid i \in \mathbb{N}\} \leq 0. $$ Is it possible to show that $L(\varepsilon) \to 0$ as $\varepsilon \to 0$ ?

I have performed some numerical simulations, which suggest that the above statement is true. However, I have not been able to prove it.

By direct computation, we can analytically solve the eigenvalue problem $$ \mathcal{L}_\varepsilon y = \lambda y, \quad y'(0) = y'(1) = 0. $$ Such solutions are given as $$ y(x)= \begin{cases} y_0 \cosh \left(\frac{x \sqrt{\lambda - i}}{2 \varepsilon}\right), & \text{if } x \in [0, 1/2], \\[10pt] \frac{y_0}{2\sqrt{\lambda +i}} \left(\sqrt{\lambda + i} - \sqrt{\lambda - i}\right) \cosh \left(\frac{\sqrt{\lambda - i} + \sqrt{\lambda + i} - 2 \sqrt{\lambda + i} x}{2 \varepsilon}\right) + \left(\sqrt{\lambda - i} + \sqrt{\lambda + i}\right) \cosh \left(\frac{\sqrt{\lambda - i} - \sqrt{\lambda + i} + 2 \sqrt{\lambda + i} x}{2 \varepsilon}\right), & \text{if } x \in (1/2, 1], \end{cases} $$ provided that $\lambda$ satisfies the equation $$ \sinh \left(\frac{\sqrt{\lambda - i}}{2 \varepsilon}\right) \cosh \left(\frac{\sqrt{\lambda + i}}{2 \varepsilon}\right) \sqrt{\lambda - i} + \sinh \left(\frac{\sqrt{\lambda + i}}{2 \varepsilon}\right) \cosh \left(\frac{\sqrt{\lambda - i}}{2 \varepsilon}\right) \sqrt{\lambda + i} = 0. $$

Unfortunately, the above equation is tricky to analyse. I have also consulted the literature but could only find useful spectral bounds for self-adjoint operators, which do not apply in this case. Does anyone have ideas or references that might help prove that $L(\varepsilon) \to 0$ as $\varepsilon \to 0$ ?

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  • $\begingroup$ Your boundary condition only involves 0 and 1, so the operator acts on functions defined on $[0,1]$. Then what role is played by $V(x)=-1$ for $x<0$ in the definition of potential $V$? Or this is a misprint? $\endgroup$
    – Alexandre Eremenko
    Commented Nov 6 at 12:10
  • $\begingroup$ Thank you for pointing this out. It was a typo, sorry for that. I have changed it $\endgroup$
    – Matheus Manzatto
    Commented Nov 6 at 12:13
  • $\begingroup$ Can you also give a hint how "it is straightforward to verify that the eigenvalues are in the left half-plane"? $\endgroup$
    – Alexandre Eremenko
    Commented Nov 6 at 12:16
  • $\begingroup$ Sure, take $y:[0,1]\to \mathbb C$ such that $\mathcal L_\varepsilon y = \lambda y$ and $\|y\|_{L^2} = 1$. It follows that $$\lambda =\int_0^1 \lambda y(x) \overline{y}(x) \mathrm{d} x =\int_0^1 \mathcal L_\varepsilon y(x) \overline{y}(x) \mathrm{d} x= \varepsilon \int_0^1 y''(x)\overline{y}(x) \mathrm{d}x + i \int_0^1 V(x) y(x) \overline{y}(x) \mathrm{d}x,$$ Hence, $$\lambda = -\varepsilon^2 \int_0^1 y'(x)\overline{y'}(x) \mathrm{d}x + i \int_0^1 V(x) y(x) \overline{y}(x) \mathrm{d}x$$ So $\lambda = -\varepsilon^2 \|y'\|_{L^2}^2 + i \int V(x) \|y\|^2(x) \mathrm{d}x$ $\endgroup$
    – Matheus Manzatto
    Commented Nov 6 at 12:27
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    $\begingroup$ I would suggest trying the ansatz $\lambda=\pm i+\epsilon\mu$ and looking at the asymptotics for small $\epsilon$. $\endgroup$
    – Michael Renardy
    Commented Nov 6 at 21:58

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This operator has been studied in the paper by Miloslav Znojil, PT-symmetric square well, Phys. Lett. A, 285 (2001) 7-10, and then by the same and many other authors. However I did not find out whether anyone addressed your question about real parts of eigenvalues.

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    $\begingroup$ But $\sqrt{\lambda-i}$ is not a conjugate to $\sqrt{\lambda+i}$ if $\lambda$ is not real? $\endgroup$
    – Aleksei Kulikov
    Commented Nov 6 at 18:01
  • $\begingroup$ You are right:-) $\endgroup$
    – Alexandre Eremenko
    Commented Nov 6 at 18:04
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    $\begingroup$ (I only wrote it because I committed the same mistake and was myself halfway through writing an answer:)) $\endgroup$
    – Aleksei Kulikov
    Commented Nov 6 at 18:05

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