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Constructive Determination of the Spectral Invariants

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Multidimensional Periodic Schrödinger Operator
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Abstract

This chapter describes the constructive determination of the spectral invariants explicitly expressed with respect to the Fourier coefficients of the potential by using the Bloch eigenvalues as input data. At the same time, it gives a rich set of invariants that is enough to determine the potential q. This chapter consists of five sections. First section is the introduction and preliminary facts where we discuss the related papers, describe briefly the scheme of this chapter, and recall the results of Chap. 3 which are used essentially in this chapter. In Sect. 4.2, we develop the asymptotic formulas obtained in Chap. 3 and write the first and second terms of the asymptotic formulas for the Bloch eigenvalues in the explicit form. In Sect. 4.3, we investigate the derivatives of the band functions \(\Lambda _{n}\) with respect to the quasimomentum. In Sect. 4.4, using the results of the previous sections, we determine constructively a family of spectral invariants of this operator from the given Bloch eigenvalues. Some of these invariants generalize the well-known invariants, and others are entirely new. The new invariants are explicitly expressed by Fourier coefficients of the potential which present the possibility of determining the potential constructively by using the Bloch eigenvalues as input data in the next chapter. Section 4.5 is the Appendix, where we give some estimations and calculations of previous sections.

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References

  1. Eskin, G., Ralston, J., Trubowitz, E.: On isospectral periodic potential in Rn. Commun. Pure Appl. Math. 37, 647 (1984)

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  2. Feldman, J., Knorrer, H., Trubowitz, E.: The perturbatively unstable spectrum of the periodic Schrödinger operator. Comment. Math. Helv. 66, 557–579 (1991)

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Correspondence to Oktay Veliev .

4.5 Appendices

4.5 Appendices

4.1.1 Appendix 1. The Proof of (4.2.28)

Here, we estimate the conjugate \(\overline{C_{1}(j^{^{\prime }},\lambda _{j,\beta })}\) of \(C_{1}(j^{^{\prime }},\lambda _{j,\beta }),\) namely, we prove that

$$\begin{aligned} \sum _{(j_{1},\beta _{1})\in Q(\rho ^{\alpha },9r)}\dfrac{\overline{A(j^{^{\prime }},\beta ,j^{^{\prime }}+j_{1,}\beta +\beta _{1}})\overline{A(j^{^{\prime }} +j_{1,}\beta +\beta _{1},j,\beta })}{\lambda _{j,\beta }-\lambda _{j^{^{\prime } }+j_{1},\beta +\beta _{1}}}=O(\rho ^{-2a}r^{2}), \end{aligned}$$
(4.5.1)

[see (4.2.26)], where

$$ Q(\rho ^{\alpha },9r)=\{(j_{1},\beta _{1}):\mid j_{1}\delta \mid<9r,0<\mid \beta _{1}\mid<\rho ^{\alpha }\},j\in S_{1}(\rho ),\mid j^{^{\prime }}\delta \mid <r,r=O(\rho ^{\frac{1}{2}\alpha _{2}}). $$

The conditions on indices \(j^{^{\prime }}\), \(j_{1},\) j and (4.2.20) imply that

$$ \mu _{j^{^{\prime }}+j_{1}}=O(r^{2}),\mu _{j}=O(r^{2}). $$

These with \(\beta \notin V_{\beta _{1}}^{\delta }(\rho ^{a})))\), where \(\beta _{1}\in \Gamma _{\delta }(p\rho ^{\alpha }),\) [see (4.2.9)] give

$$\begin{aligned} \lambda _{j,\beta }-\lambda _{j^{^{\prime }}+j_{1},\beta +\beta _{1}}=-2\langle \beta ,\beta _{1}\rangle +O(r^{2}), \mid \langle \beta ,\beta _{1}\rangle \mid \,>\frac{1}{3}\rho ^{a}. \end{aligned}$$
(4.5.2)

Using this, (4.2.15), and (4.5.1), we get

$$\begin{aligned} \overline{C_{1}(j^{^{\prime }},\lambda _{j,\beta })}=\sum _{\beta _{1}} \dfrac{C^{^{\prime }}}{-2\langle \beta ,\beta _{1}\rangle }+O(\rho ^{-2a} r^{2}), \end{aligned}$$
(4.5.3)

where

$$ C^{^{\prime }}=\sum _{j_{1}}\overline{A(j^{^{\prime }},\beta ,j^{^{\prime }} +j_{1,}\beta +\beta _{1}})\overline{A(j^{^{\prime }}+j_{1,}\beta +\beta _{1},j,\beta }). $$

In Chap. 3, we proved that [see (3.3.21), (3.3.7), Lemma 3.3.3]

$$\begin{aligned} \overline{A(j^{^{\prime }},\beta ,j^{^{\prime }}+j_{1,}\beta +\beta _{1}} )=\sum \limits _{n_{1}:(n_{1},\beta _{1})\in \Gamma ^{^{\prime }}(\rho ^{\alpha } )}c(n_{1},\beta _{1})a(n_{1},\beta _{1},j^{^{\prime }},\beta ,j^{^{\prime }} +j_{1,}\beta +\beta _{1}), \end{aligned}$$
(4.5.4)
$$ \overline{A(j^{^{\prime }}+j_{1,}\beta +\beta _{1},j,\beta })=\sum \limits _{n_{2} :(n_{2},-\beta _{1})\in \Gamma ^{^{\prime }}(\rho ^{\alpha })}c(n_{2},-\beta _{1})a(n_{2},-\beta _{1},j^{^{\prime }}+j_{1,}\beta +\beta _{1},j,\beta ), $$
$$ \Gamma ^{^{\prime }}(\rho ^{\alpha })=\{(n_{1},\beta _{1}):\beta _{1}\in \Gamma _{\delta }\backslash 0,n_{1}\in \mathbb {Z},\beta _{1}+(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\delta \in \Gamma (\rho ^{\alpha })\}, $$
$$\begin{aligned} c(n_{1},\beta _{1})=q_{\gamma _{1}},\gamma _{1}=\beta _{1}+(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\delta \in \Gamma (\rho ^{\alpha }), \end{aligned}$$
(4.5.5)
$$ a(n_{1},\beta _{1},j^{^{\prime }},\beta ,j^{^{\prime }}+j_{1,}\beta +\beta _{1})=(\mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\zeta }\varphi _{j^{^{\prime }},v(\beta )}(\zeta ),\varphi _{j^{^{\prime }}+j_{1} ,v(\beta +\beta _{1})}(\zeta )), $$
$$ a(n_{2},-\beta _{1},j^{^{\prime }}+j_{1,}\beta +\beta _{1},j,\beta )=(\mathrm{e}^{i(n_{2} -(2\pi )^{-1}\langle -\beta _{1},\delta ^{*}\rangle )\zeta }\varphi _{j^{^{\prime }}+j_{1},v(\beta +\beta _{1})}(\zeta ),\varphi _{j,v(\beta )}(\zeta )) $$
$$\begin{aligned} =(\varphi _{j^{^{\prime }}+j_{1},v(\beta +\beta _{1})}(\zeta ),\mathrm{e}^{-i(n_{2} -(2\pi )^{-1}\langle -\beta _{1},\delta ^{*}\rangle )\zeta }\varphi _{j,v(\beta )}(\zeta )) \end{aligned}$$
(4.5.6)
$$ =\overline{(\mathrm{e}^{-i(n_{2}-(2\pi )^{-1}\langle -\beta _{1},\delta ^{*}\rangle \zeta }\varphi _{j,v(\beta )}(\zeta ),\varphi _{j^{^{\prime }}+j_{1},v(\beta +\beta _{1})}(\zeta )),} $$

where \(\delta ^{*}\) is the element of \(\Omega \) satisfying \(\langle \delta ^{*},\delta \rangle =2\pi .\)

Now, to estimate the right-hand side of (4.5.3) we prove that

$$\begin{aligned} \sum _{j_{1}}a(n_{1},\beta _{1},j^{^{\prime }},\beta ,j^{^{\prime }}+j_{1,} \beta +\beta _{1})a(n_{2},-\beta _{1},j^{^{\prime }}+j_{1},\beta +\beta _{1} ,j,\beta ) \end{aligned}$$
(4.5.7)
$$ =a(n_{1}+n_{2},0,j^{^{\prime }},\beta ,j,\beta )+O(\rho ^{-p\alpha }). $$

By definition, we have

$$ a(n_{1}+n_{2},0,j^{^{\prime }},\beta ,j,\beta )=(\mathrm{e}^{i(n_{1}+n_{2})\zeta } \varphi _{j^{^{\prime }},v(\beta )}(\zeta ),\varphi _{j,v(\beta )}(\zeta ))= $$
$$ (\mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\zeta } \varphi _{j^{^{\prime }},v(\beta )}(\zeta ),\mathrm{e}^{-i(n_{2}-(2\pi )^{-1}\langle -\beta _{1},\delta ^{*}\rangle )\zeta }\varphi _{j,v(\beta )}(\zeta )). $$

This, (4.5.6), and the following formulas

$$ \mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\zeta } \varphi _{j^{^{\prime }},v(\beta )}(\zeta )= $$
$$ \sum _{\mid j_{1}\delta \mid <9r}a(n_{1},\beta _{1},j^{^{\prime }},\beta ,j^{^{\prime }}+j_{1,}\beta +\beta _{1})\varphi _{j^{^{\prime }}+j_{1} ,v(\beta +\beta _{1})}(\zeta )+O(\rho ^{-p\alpha }), $$
$$ \mathrm{e}^{-i(n_{2}-(2\pi )^{-1}\langle -\beta _{1},\delta ^{*}\rangle )\zeta } \varphi _{j,v(\beta )}(\zeta )= $$
$$ \sum _{\mid j_{1}\delta \mid <9r}\overline{a(n_{2},-\beta _{1},j^{^{\prime }} ,\beta ,j^{^{\prime }}+j_{1,}\beta +\beta _{1})}\varphi _{j^{^{\prime }} +j_{1},v(\beta +\beta _{1})}+O(\rho ^{-p\alpha }) $$

and

$$\begin{aligned} \sum \limits _{j_{_{1}}}\mid a(n_{1},\beta _{1},j^{^{\prime }},\beta ,j^{^{\prime } }+j_{1,}\beta +\beta _{1})\mid =O(1) \end{aligned}$$
(4.5.8)

[see (3.3.16), (3.3.17) of Chap. 3] give the proof of (4.5.7). Now from (4.5.7), (4.5.4), and (4.5.3), we obtain

$$ C^{^{\prime }}=\sum \limits _{n_{1}} \sum \limits _{n_{2}}(c(n_{1},\beta _{1})c(n_{2},-\beta _{1})a(n_{1}+n_{2},0,j^{^{\prime }},\beta ,j,\beta )+O(\rho ^{-p\alpha })), $$
$$ \overline{C_{1}(j^{^{\prime }},\lambda _{j,\beta })}=\sum _{\beta _{1}} \sum \limits _{n_{1}} \sum \limits _{n_{2}}C_{1}^{^{\prime }}(\beta _{1},n_{1},n_{2})+O(\rho ^{-2a}r^{2}), $$

where

$$ C_{1}^{^{\prime }}(\beta _{1},n_{1},n_{2})=\dfrac{c(n_{1},\beta _{1} )c(n_{2},-\beta _{1})a(n_{1}+n_{2},0,j^{^{\prime }},\beta ,j,\beta )}{-2\langle \beta ,\beta _{1}\rangle }. $$

It is clear that

$$\begin{aligned} C_{1}^{^{\prime }}(\beta _{1},n_{1},n_{2})+C_{1}^{^{\prime }}(-\beta _{1} ,n_{2},n_{1})=0. \end{aligned}$$
(4.5.9)

Therefore

$$ \overline{C_{1}(j^{^{\prime }},\lambda _{j,\beta })}=O(\rho ^{-2a}r^{2}). $$

4.1.2 Appendix 2. The Proof of (4.2.35)

Arguing as in the proof of (4.2.27), we see that

$$ C_{2}(\Lambda _{j,\beta })=C_{2}(\lambda _{j,\beta })+O(\rho ^{-3a}) $$

and by (4.5.4)

$$\begin{aligned} \overline{C_{2}(\lambda _{j,\beta })}&=\sum _{\beta _{1},\beta _{2}} \Bigg (\sum \limits _{n_{1},n_{2},n_{3}}\Bigg (\sum _{j_{1},j_{2}}\dfrac{c(n_{1},\beta _{1})c(n_{2},\beta _{2})c(n_{3},-\beta _{1}-\beta _{2})}{(\lambda _{j,\beta }-\lambda _{j(1),\beta (1)})(\lambda _{j,\beta }-\lambda _{j(2)_,\beta (2)} )}a(n_{1},\beta _{1},j,\beta ,j(1)_,\beta (1))\times \\&a(n_{2},\beta _{2},j(1)_,\beta (1),j(2)_,\beta (2))a(n_{3},-\beta _{1}-\beta _{2},j(2),\beta (2),j,\beta )\Bigg )\Bigg ), \end{aligned}$$

where

$$ (j_{1},\beta _{1})\in Q(\rho ^{\alpha },9r_{1}),(j_{2},\beta _{2})\in Q(\rho ^{\alpha },90r_{1}),j\in S_{1},\beta _{1}+\beta _{2}\ne 0. $$

Applying (4.5.7) two times and using (4.5.8), we get

$$\begin{aligned} \sum _{j_{1}}a(n_{1},\beta _{1},j,\beta ,j(1)_,\beta (1))&\left( \sum _{j_{2}} a(n_{2},\beta _{2},j(1)_,\beta (1),j(2)_,\beta (2))a(n_{3},-\beta _{1}\right. \\&\left. \quad -\beta _{2},j(2)_,\beta (2),j,\beta )\right) \end{aligned}$$
$$ =\sum _{j_{1}}a(n_{1},\beta _{1},j,\beta ,j(1)_,\beta (1))(a(n_{2}+n_{3} ,-\beta _{1},j(1)_,\beta (1),j,\beta )+O(\rho ^{-p\alpha })) $$
$$ \quad =a(n_{1}+n_{2}+n_{3},0,j,\beta ,j,\beta )+O(\rho ^{-p\alpha }). $$

Using this in the above expression for \(C_{2}(\lambda _{j,\beta })\) and taking into account that

$$ \lambda _{j,\beta }-\lambda _{j(1)_,\beta (1)}=-2\langle \beta ,\beta _{1} \rangle +O(\rho ^{2\alpha _{1}}),\mid \langle \beta ,\beta _{1}\rangle \mid \,>\frac{1}{3}\rho ^{a} $$

and

$$ \lambda _{j,\beta }-\lambda _{j(2)_,\beta (2)}=-2\langle \beta ,\beta _{1} +\beta _{2}\rangle +O(\rho ^{2\alpha _{1}}),\mid \langle \beta ,\beta _{1}+\beta _{2}\rangle \mid \,>\frac{1}{3}\rho ^{a} $$

which can be proved as (4.5.2), we have

$$ C_{2}(\lambda _{j,\beta })=O(\rho ^{-3a+2\alpha _{1}})+ $$
$$ \sum _{\beta _{1},\beta _{2}} \sum \limits _{n_{1},n_{2},n_{3}}\frac{c(n_{1},\beta _{1})c(n_{2},\beta _{2})c(n_{3},-\beta _{1}-\beta _{2} )a(n_{1}+n_{2}+n_{3},0,j,\beta ,j,\beta )}{4\langle \beta ,\beta _{1}\rangle \langle \beta ,\beta _{1}+\beta _{2}\rangle }. $$

Grouping the terms with the equal multiplicands

$$\begin{aligned}&c(n_{1},\beta _{1})c(n_{2},\beta _{2})c(n_{3},-\beta _{1}-\beta _{2}),\\&c(n_{2},\beta _{2})c(n_{1},\beta _{1})c(n_{3},-\beta _{1}-\beta _{2}),\\&c(n_{1},\beta _{1})c(n_{3},-\beta _{1}-\beta _{2})c(n_{2},\beta _{2}),\\&c(n_{2},\beta _{2})c(n_{3},-\beta _{1}-\beta _{2})c(n_{1},\beta _{1}),\\&c(n_{3},-\beta _{1}-\beta _{2})c(n_{1},\beta _{1})c(n_{2},\beta _{2}),\text { }\\&c(n_{3},-\beta _{1}-\beta _{2})c(n_{2},\beta _{2})c(n_{1},\beta _{1}) \end{aligned}$$

and using the obvious equality

$$ \frac{1}{\langle \beta ,\beta _{1}\rangle \langle \beta ,\beta _{1}+\beta _{2}\rangle }+\frac{1}{\langle \beta ,\beta _{2}\rangle \langle \beta ,\beta _{2}+\beta _{1}\rangle }+\frac{1}{\langle \beta ,\beta _{1}\rangle \langle \beta ,-\beta _{2}\rangle }\,\,+ $$
$$ \frac{1}{\langle \beta ,\beta _{2}\rangle \langle \beta -,\beta _{1}\rangle }+\frac{1}{\langle \beta ,-\beta _{1}-\beta _{2}\rangle \langle \beta ,-\beta _{2}\rangle }+\frac{1}{\langle \beta ,-\beta _{1}-\beta _{2}\rangle \langle \beta ,-\beta _{1}\rangle }=0, $$

we see that

$$ C_{2}(\lambda _{j,\beta })=O(\rho ^{-3a+2\alpha _{1}}). $$

4.1.3 Appendix 3. The Proof of (4.2.34)

By (4.2.27) we have

$$ C_{1}(\Lambda _{j,\beta })=C_{1}(\lambda _{j,\beta })+O(\rho ^{-3a}). $$

Therefore, we need to prove that

$$ \overline{C_{1}(\lambda _{j,\beta })}=\frac{1}{4}\int _{F}\left| f_{\delta ,\beta +\tau }(x)\right| ^{2}\left| \varphi _{j,v}^{\delta }(\langle \delta ,x\rangle )\right| ^{2}\text {dx}+O(\rho ^{-3a+2\alpha _{1}}), $$

where

$$ \overline{C_{1}(\lambda _{j,\beta })}\equiv \sum _{\beta _{1}}\sum _{j_{1} }\dfrac{\overline{A(j,\beta ,j+j_{1,}\beta +\beta _{1}})\overline{A(j+j_{1,} \beta +\beta _{1},j,\beta })}{\lambda _{j,\beta }-\lambda _{j+j_{1},\beta +\beta _{1} }}, $$
$$ (j_{1},\beta _{1})\in Q(\rho ^{\alpha },9r_{1}),j\in S_{1}, $$

and by (4.5.4)

$$ \overline{C_{1}(\lambda _{j,\beta })}=\sum _{\beta _{1}} \sum \limits _{n_{1}:(n_{1},\beta _{1})\in \Gamma ^{^{\prime }}(\rho ^{\alpha })}\text { }\sum \limits _{n_{2}:(n_{2},-\beta _{1})\in \Gamma ^{^{\prime }}(\rho ^{\alpha } )} \sum _{j_{1}}\frac{c(n_{1},\beta _{1})c(n_{2},-\beta _{1})}{\lambda _{j,\beta }-\lambda _{j+j_{1},\beta +\beta _{1}}}\times $$
$$ a(n_{1},\beta _{1},j,\beta ,j+j_{1,}\beta +\beta _{1})a(n_{2},-\beta _{1} ,j+j_{1},\beta +\beta _{1},j,\beta ). $$

Replacing \(\lambda _{j,\beta }-\lambda _{j+j_{1},\beta +\beta _{1}}\) by

$$ -(2\langle \beta +\tau ,\beta _{1}\rangle +\mid \beta _{1}\mid ^{2}+\mu _{j+j_{1} }(v(\beta +\beta _{1}))-\mu _{j}(v(\beta ))) $$

and using (4.5.7) for \(j^{^{\prime }}=j,\) we have

$$ \overline{C_{1}(j,\lambda _{j,\beta })}=\sum _{\beta _{1}} \sum \limits _{n_{1}} \sum \limits _{n_{2}}\frac{c(n_{1},\beta _{1} )c(n_{2},-\beta _{1})a(n_{1}+n_{2},0,j,\beta ,j,\beta )}{-2\langle \beta +\tau ,\beta _{1}\rangle }+ $$
$$ \sum _{\beta _{1}} \sum \limits _{n_{1}} \sum \limits _{n_{2}}\text { }\sum _{j_{1}}\frac{c(n_{1},\beta _{1})c(n_{2},-\beta _{1})a(n_{1},\beta _{1},j,\beta ,j+j_{1,}\beta +\beta _{1})}{2\langle \beta +\tau ,\beta _{1} \rangle (2\langle \beta +\tau ,\beta _{1}\rangle +\mid \beta _{1}\mid ^{2}+\mu _{j+j_{1}}-\mu _{j})}\times $$
$$ a(n_{2},-\beta _{1},j+j_{1},\beta +\beta _{1},j,\beta )(\mid \beta _{1}\mid ^{2} +\mu _{j+j_{1}}(v(\beta +\beta _{1}))-\mu _{j}(v(\beta ))). $$

The formula (4.5.9) shows that the first summation of the right-hand side of this equality is zero. Thus we need to estimate the second sum. For this, we use the following relation

$$ \mu _{j+j_{1}}(v(\beta +\beta _{1}))a(n_{1},\beta _{1},j,\beta ,j+j_{1},\beta +\beta _{1})=(\mathrm{e}^{i(n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi })\zeta }\varphi _{j,v(\beta )}(\zeta ),T_{v}\varphi _{j+j_{1},v(\beta +\beta _{1} )}(\zeta )) $$
$$ =(T_{v}(\mathrm{e}^{i(n_{1}-(2\pi )^{-1}(\beta _{1},\delta ^{*}\rangle )\zeta } \varphi _{j,v(\beta )}(\zeta )),\varphi _{j+j_{1},v(\beta +\beta _{1})}(\zeta )) $$
$$ =(\mid n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle \mid ^{2} \mid \delta \mid ^{2}+\mu _{j}(v))(\mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1} ,\delta ^{*}\rangle )\zeta }\varphi _{j,v(\beta )}(\zeta ),\varphi _{j+j_{1} ,v(\beta +\beta _{1})}(\zeta )) $$
$$ -2i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\mid \delta \mid ^{2}(\mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\zeta } \varphi _{j,v(\beta )}^{^{\prime }}(\zeta )),\varphi _{j+j_{1},v(\beta +\beta _{1} )}(\zeta )). $$

Using this, (4.5.7), and the formula

$$ \sum _{j_{1}}((\mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\zeta }\varphi _{j,v(\beta )}^{^{\prime }}(\zeta )),\varphi _{j+j_{1},v(\beta +\beta _{1})}(\zeta ))a(n_{2},-\beta _{1},j+j_{1},\beta +\beta _{1},j,\beta ) $$
$$ =((\mathrm{e}^{i(n_{1}+n_{2})\zeta }\varphi _{j,v(\beta )}^{^{\prime }}(\zeta )),\varphi _{j,v(\beta )}(\zeta ))+O(\rho ^{-p\alpha }) $$

which can be proved as (4.5.7), we obtain

$$\begin{aligned} \sum _{j_{1}}\mu _{j+j_{1}}(v(\beta +\beta _{1}))a(n_{1},\beta _{1},j,\beta ,j+j_{1},\beta +\beta _{1})a(n_{2},-\beta _{1},j+j_{1},\beta +\beta _{1} ,j,\beta )=\nonumber \end{aligned}$$
$$\begin{aligned} (\mid n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle \mid ^{2} )\mid \delta \mid ^{2}+\mu _{j}(v))a(n_{1}+n_{2},0,j,\beta ,j,\beta )- \end{aligned}$$
(4.5.10)
$$ 2i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\mid \delta \mid ^{2}((\mathrm{e}^{i(n_{1}+n_{2})\zeta }\varphi _{j,v(\beta )}^{^{\prime }}(\zeta )),\varphi _{j,v(\beta )}(\zeta )). $$

Here, the last multiplicand can be estimated as follows:

$$ \ \ \mu _{j}(v)(\varphi _{j,v(\beta )}(\zeta ),\mathrm{e}^{i(n_{1}+n_{2})\zeta } \varphi _{j,v(\beta )}(\zeta ))=(\varphi _{j,v(\beta )}(\zeta ),T_{v}(\mathrm{e}^{i(n_{1} +n_{2})\zeta }\varphi _{j,v(\beta )}(\zeta ))) $$
$$ =(n_{1}+n_{2})^{2}\mid \delta \mid ^{2}(\varphi _{j,v(\beta )}(\zeta ),\mathrm{e}^{i(n_{1} +n_{2})\zeta }\varphi _{j,v(\beta )}(\zeta ))+ $$
$$ 2i(n_{1}+n_{2})\mid \delta \mid ^{2}(\varphi _{j,v(\beta )}(\zeta ),\mathrm{e}^{i(n_{1} +n_{2})\zeta }\varphi _{j,v(\beta )}^{^{\prime }}(\zeta ))+\mu _{j}(v)(\varphi _{j,v(\beta )},\mathrm{e}^{i(n_{1}+n_{2})\zeta }\varphi _{j,v(\beta )}), $$
$$ ((\mathrm{e}^{i(n_{1}+n_{2})\zeta }\varphi _{j,v(\beta )}^{^{\prime }}(\zeta )),\varphi _{j,v(\beta )}(\zeta ))=\frac{n_{1}+n_{2}}{2i}((\mathrm{e}^{i(n_{1}+n_{2})\zeta } \varphi _{j,v(\beta )}(\zeta )),\varphi _{j,v(\beta )}(\zeta )). $$

Using this, (4.5.10), and (4.5.7), we get

$$ \sum _{j_{1}}(a(n_{1},\beta _{1},j,\beta ,j+j_{1,}\beta +\beta _{1})a(n_{2} ,-\beta _{1},j+j_{1},\beta +\beta _{1},j,\beta ))\times $$
$$ (\mid \beta _{1}\mid ^{2}+\mu _{j+j_{1}}(v(\beta +\beta _{1}))-\mu _{j} (v(\beta )))=a(n_{1}+n_{2},0,j,\beta ,j,\beta )\times $$
$$ (\mid \beta _{1}\mid ^{2}+\mid n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\mid ^{2}\mid \delta \mid ^{2}-(n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi })\mid \delta \mid ^{2}(n_{1}+n_{2})) $$
$$ =\left( \mid \beta _{1}\mid ^{2}+\mid \delta \mid ^{2}\left( n_{1}-\frac{\langle \beta _{1} ,\delta ^{*}\rangle }{2\pi }\right) \left( -n_{2}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\right) \right) a(n_{1}+n_{2},0,j,\beta ,j,\beta ). $$

Thus

$$ \overline{C_{1}(j,\lambda _{j,\beta })}=C+O(\rho ^{-3a+2\alpha _{1}}), $$

where

$$\begin{aligned} C=\sum _{\beta _{1},n_{1},n_{2}}\frac{c(n_{1},\beta _{1})c(n_{2},-\beta _{1})a(n_{1}+n_{2},0,j,\beta ,j,\beta )}{4\mid \langle \beta +\tau ,\beta _{1} \rangle \mid ^{2}}\times \end{aligned}$$
(4.5.11)
$$ \left( \mid \beta _{1}\mid ^{2}+\left( n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\right) \left( -n_{2}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\right) \mid \delta \mid ^{2}\right) . $$

Now we consider

$$ \int _{F}\left| f_{\delta ,\beta +\tau }(x)\right| ^{2}\left| \varphi _{n,v}(\langle \delta ,x\rangle )\right| ^{2}\text {dx}, $$

where \(f_{\delta ,\beta +\tau }\) is defined in (4.1.12). By (4.5.5)

$$ f_{\delta ,\beta +\tau }(x)=\sum _{(n_{1},\beta _{1})\in \Gamma _{\delta }^{^{\prime } }(\rho ^{\alpha })}\frac{\beta _{1}+(n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi })\delta }{\langle \beta +\tau ,\beta _{1}\rangle }c(n_{1},\beta _{1})\mathrm{e}^{i\langle \beta _{1}+(n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi })\delta ,x\rangle }. $$

Here \(f_{\delta ,\beta +\tau }(x)\) is a vector of \(\mathbb {R}^{d}\). Using \(\langle \beta ,\delta \rangle =0\) for \(\beta \in \Gamma _{\delta },\) we obtain

$$ \left| f_{\delta ,\beta +\tau }(x)\right| ^{2}=\sum _{(n_{1},\beta _{1}),(n_{2},\beta _{2})\in \Gamma _{\delta }^{^{\prime }}(\rho ^{\alpha })} \frac{\langle \beta _{1},\beta _{2}\rangle +(n_{1}-\frac{\langle \beta _{1} ,\delta ^{*}\rangle }{2\pi })(n_{2}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi })\mid \delta \mid ^{2}}{\langle \beta +\tau ,\beta _{1}\rangle \langle \beta +\tau ,\beta _{2}\rangle }\times $$
$$ c(n_{1},\beta _{1})c(-n_{2},-\beta _{2})\mathrm{e}^{i\langle \beta _{1}-\beta _{2} +(n_{1}-n_{2}-(2\pi )^{-1}\langle \beta _{1}-\beta _{2},\delta ^{*} \rangle )\delta ,x\rangle }. $$

Since \(\varphi _{j,v}(\langle \delta ,x\rangle )\) is a function of \(\langle \delta ,x\rangle ,\) we have

$$ \int _{F}\mathrm{e}^{i\langle \beta _{1}-\beta _{2}+(n_{1}-n_{2}-(2\pi )^{-1}\langle \beta _{1}-\beta _{2},\delta ^{*}\rangle )\delta ,x\rangle }\left| \varphi _{j,v}(\langle \delta ,x\rangle )\right| ^{2}\text {dx}\,=0 $$

for \(\beta _{1}\ne \beta _{2}\). Therefore

$$ \int _{F}\left| f_{\delta ,\beta +\tau }(x)\right| ^{2}\left| \varphi _{j,v}(\langle \delta ,x\rangle )\right| ^{2}\text {dx}\,=\sum _{\beta _{1} ,n_{1},n_{2}}\frac{c(n_{1},\beta _{1})c(-n_{2},-\beta _{1})}{\mid \langle \beta +\tau ,\beta _{1}\rangle \mid ^{2}}\times $$
$$ (\mid \beta _{1}\mid ^{2}+\left( n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\right) \left( n_{2}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\right) \mid \delta \mid ^{2}a(n_{1}-n_{2},0,j,\beta ,j,\beta \rangle . $$

Replacing \(n_{2}\) by \(-n_{2},\) we get

$$ \int _{F}\left| f_{\delta ,\beta +\tau }(x)\right| ^{2}\left| \varphi _{n,v}(\langle \delta ,x\rangle )\right| ^{2}\text {dx}\,=4C $$

[see (4.5.11)] and(4.2.34).

4.1.4 Appendix 4. Asymptotic Formulas for \(T_{v}(Q)\)

It is well known that the large eigenvalues of \(T_{0}(Q)\) lie in \(O(\frac{1}{m^{4}})\) neighborhood of

$$ \mid m\delta \mid +\frac{1}{16\pi \mid m\delta \mid ^{3}}\int _{0}^{2\pi }\left| Q(t)\right| ^{2}\mathrm{d}t $$

for the large values of m (see [5], p. 58). This formula yields the invariant (4.1.16). Using the asymptotic formulas for solutions of the Sturm–Liouville equation (see [5], p. 63), one can easily obtain that

$$ \varphi _{n,v}(\zeta )=\mathrm{e}^{i(n+v)\zeta }\left( 1+\frac{Q_{1}(\zeta )}{2i(n+v)\mid \delta \mid ^{2}}+\frac{Q(\zeta )-Q(0)-\frac{1}{2}Q_{1}^{2}(\zeta )}{4(n+v)^{2}\mid \delta \mid ^{4}}\right) +O\left( \frac{1}{n^{3}})\right) , $$

where

$$ Q_{1}(\zeta )=\int _{0}^{\zeta }Q(t)\mathrm{d}t. $$

From this, by direct calculations, we find \(A_{0}(\zeta ),\) \(A_{1}(\zeta ),\) and \(A_{2}(\zeta )\) [see (4.1.6)] and then using these in (4.1.7), we get the invariant (4.1.15).

Now we consider the eigenfunction \(\varphi _{n,v}(\zeta )\) of \(T_{v}(p)\) in the case \(v\ne 0,\ \frac{1}{2}\) and

$$ p(\zeta )=p_{1}\mathrm{e}^{i\zeta }+p_{-1}\mathrm{e}^{-i\zeta }. $$

The eigenvalues and the eigenfunctions of \(T_{v}(0)\) are \(\ (n+v)^{2} \mid \delta \mid ^{2}\) and \(\ \mathrm{e}^{i(n+v)\zeta }\), for \(n\in \mathbb {Z}\). Since the eigenvalues of \(T_{v}(p)\) are simple for \(v\ne 0,\ \frac{1}{2},\) by the well-known perturbation formula

$$ (\varphi _{n,v}(\zeta ),\ \mathrm{e}^{i(n+v)\zeta })\varphi _{n,v}(\zeta )=\mathrm{e}^{i(n+v)\zeta }+ $$
$$\begin{aligned} \sum _{k=1,2,\ldots }\frac{(-1)^{k+1}}{2i\pi }\int \limits _{C}(T_{v}(0)-\lambda )^{-1}p(x))^{k}(T_{v}(0)-\lambda )^{-1}\mathrm{e}^{i(n+v)\zeta }\mathrm{d}\lambda , \end{aligned}$$
(4.5.12)

where C is a contour containing only the eigenvalue \((n+t)^{2}\mid \delta \mid ^{2}\). Using

$$ (T_{v}(0)-\lambda )^{-1}\mathrm{e}^{i(n+v)\zeta }=\frac{\mathrm{e}^{i(n+v)\zeta }}{(n+v)^{2} \mid \delta \mid ^{2}-\lambda }, $$

we see that the kth (\(k=1,2,3,4\)) term \(F_{k}\) of the series (4.5.12) has the form

$$ F_{1}=\frac{1}{2i\pi }\int \limits _{C}\sum _{m=1,-1}\frac{p_{m}\mathrm{e}^{i(n+m+v)\zeta } }{((n+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+m+v)^{2}\mid \delta \mid ^{2} -\lambda )}\mathrm{d}\lambda , $$
$$ F_{2}=\frac{-1}{2i\pi }\int \limits _{C}\sum _{m,l=1,-1}\frac{p_{m}p_{l} \mathrm{e}^{i(n+m+l+v)\zeta }}{((n+v)^{2}\mid \delta \mid ^{2}-\lambda )}\times $$
$$ \frac{1}{((n+m+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+m+l+v)^{2}\mid \delta \mid ^{2}-\lambda )}\mathrm{d}\lambda , $$
$$ F_{3}=\frac{1}{2i\pi }\int \limits _{C}\sum _{m,l,k=1,-1}\frac{p_{m}p_{l} p_{k}\mathrm{e}^{i(n+m+l+k+v)\zeta }}{((n+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+m+v)^{2}\mid \delta \mid ^{2}-\lambda )}\times $$
$$ \frac{1}{((n+m+l+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+m+l+k+v)^{2}\mid \delta \mid ^{2}-\lambda )}\mathrm{d}\lambda , $$
$$ F_{4}=\frac{-1}{2i\pi }\int \limits _{C}\sum _{m,l,k,r=1,-1}\frac{p_{m}p_{l} p_{k}p_{r}\mathrm{e}^{i(n+m+l+k+r+v)\zeta }}{((n+m+l+k+r+v)^{2}\mid \delta \mid ^{2}-\lambda )}\times $$
$$ \frac{1}{((n+m+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+m+l+v)^{2}\mid \delta \mid ^{2}-\lambda )}\times $$
$$ \frac{1}{((n+m+l+k+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+v)^{2}\mid \delta \mid ^{2}-\lambda )}\mathrm{d}\lambda . $$

Since the distance between \((n+v)^{2}\mid \delta \mid ^{2}\) and \((n^{^{\prime } }+v)^{2}\mid \delta \mid ^{2}\) for \(n^{^{\prime }}\ne n\) is greater than \(c_{17}n\), we can choose the contour C such that

$$ \frac{1}{\mid (n^{^{\prime }}+v)^{2}\mid \delta \mid ^{2}-\lambda \mid }<\frac{c_{18}}{n}, \forall \lambda \in C, \forall n^{^{\prime }}\ne n $$

and the length of C is less than \(c_{19}\). Therefore

$$ F_{5}+F_{6}+\cdots =O(n^{-5}). $$

Now we calculate the integrals in \(F_{1}\), \(F_{2}\), \(F_{3}\), and \(F_{4}\) by the Cauchy integral formula and then decompose the obtained expression in power of \(\frac{1}{n}\). Then

$$ F_{1}=\mathrm{e}^{i(n+v)\zeta }\left( (p_{1}\mathrm{e}^{i\zeta }-p_{-1}\mathrm{e}^{-i\zeta })\frac{1}{\mid \delta \mid ^{2}}\left( \frac{-1}{2n}+\frac{v}{2n^{2}}-\frac{4v^{2}+1}{8n^{3}} +O\left( \frac{1}{n^{4}}\right) \right) \right. + $$
$$ (p_{1}\mathrm{e}^{i\zeta }+p_{-1}\mathrm{e}^{-i\zeta })\frac{1}{\mid \delta \mid ^{2}}\left. \left( \frac{v}{4n^{2}}-\frac{v}{2n^{3}}+\frac{12v^{2}+1}{16n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) \right) . $$

Let \(F_{2,1}\) and \(F_{2,2}\) be the sum of the terms in \(F_{2}\) for which \(m+l=\pm 2\) and \(m+l=0\), respectively, i.e.,

$$ F_{2}=F_{2,1}+F_{2,2}, $$

where

$$ F_{2,1}=\mathrm{e}^{i(n+v)\zeta }\left( ((p_{1})^{2}\mathrm{e}^{2i\zeta }+(p_{-1})^{2}\mathrm{e}^{-2i\zeta } )\frac{1}{\mid \delta \mid ^{4}}\left( \frac{-1}{8n^{2}}+\frac{-v}{4n^{3}} -\frac{12v^{2}+7}{32n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) \right. + $$
$$ ((p_{1})^{2}\mathrm{e}^{2i\zeta }-(p_{-1})^{2}\mathrm{e}^{-2i\zeta })\frac{1}{\mid \delta \mid ^{4} }\left. \left( \frac{-3}{16n^{3}}+O\left( \frac{1}{n^{4}}\right) \right) \right) , $$
$$ F_{2,2}=\mathrm{e}^{i(n+v)\zeta }\left| p_{1}\right| ^{2}\left( \frac{c_{20}}{n^{2} }++\frac{c_{21}}{n^{3}}+\frac{c_{22}}{n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) $$

and \(c_{20},c_{21},\) and \(c_{22}\) are the known constants. Similarly,

$$ F_{3}=F_{3,1}+F_{3,2}, $$

where \(F_{3,1}\) and \(F_{3,2}\) are the sum of the terms in \(F_{3}\) for which \(m+l+k=\pm 3\) and \(m+l+k=\pm 1\), respectively. Hence

$$ F_{3,1}=\mathrm{e}^{i(n+v)\zeta }\left( (p_{1}^{3}\mathrm{e}^{3i\zeta }-p_{-1}^{3}\mathrm{e}^{-i\zeta })\frac{1}{\mid \delta \mid ^{6}}\left( \frac{-1}{48n^{3}}+O\left( \frac{1}{n^{4}}\right) \right) \right. + $$
$$ (p_{1}^{3}\mathrm{e}^{3i\zeta }+p_{-1}^{3}\mathrm{e}^{-3i\zeta })\frac{1}{\mid \delta \mid ^{6} }\left. \left( \frac{1}{16n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) \right) , $$
$$ F_{3,2}=\mathrm{e}^{i(n+v)\zeta }\left( (p_{1}\mathrm{e}^{i\zeta }-p_{-1}\mathrm{e}^{-i\zeta })\left| p_{1}\right| ^{2}\left( \frac{c_{23}}{n^{3}}+\frac{c_{24}}{n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) \right. + $$
$$ (p_{1}\mathrm{e}^{i\zeta }+p_{-1}\mathrm{e}^{-i\zeta })\left| p_{1}\right| ^{2} \left. \left( \frac{c_{25}}{n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) \right) . $$

Similarly

$$ F_{4}=F_{4,1}+F_{4,2}+F_{4,3}, $$

where \(F_{4,1}\), \(F_{4,2}\), \(F_{4,3}\) are the sum of the terms in \(F_{4}\) for which \(m+l+k+r=\pm 4\), \(m+l+k+r=\pm 2\), \(m+l+k+r=0\), respectively. Thus

$$ F_{4,1}=\mathrm{e}^{i(n+v)\zeta }(p_{1}^{4}\mathrm{e}^{4i\zeta }+p_{-1}^{4}\mathrm{e}^{-4i\zeta })\frac{1}{\mid \delta \mid ^{8}}\left( \frac{1}{384n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) , $$
$$ F_{4,2}=\mathrm{e}^{i(n+v)\zeta }(p_{1}^{2}\mathrm{e}^{2i\zeta }+p_{-1}^{2}\mathrm{e}^{-2i\zeta })\left| p_{1}\right| ^{2}\left( \frac{c_{26}}{n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) , $$
$$ F_{4,3}=\mathrm{e}^{i(n+v)\zeta }\left| p_{1}\right| ^{4}\left( \frac{c_{27}}{n^{4} }+O\left( \frac{1}{n^{5}}\right) \right) . $$

Since \({p_{-1}^{k}}\mathrm{e}^{-ik\zeta }\) is conjugate of \({p_{1}^{k}}{\mathrm{e}^{ik\zeta }},\) the real and imaginary parts of \({F_{k}\mathrm{e}^{-i(n+v)\zeta }}\) consist of terms with multiplicands

$$ p_{1}^{k}\mathrm{e}^{ik\zeta }+p_{-1}^{k}\mathrm{e}^{-ki\zeta } \quad \& \quad p_{1} ^{k}\mathrm{e}^{ik\zeta }-p_{-1}^{k}\mathrm{e}^{-ik\zeta }, $$

respectively. Taking into account this and using the above estimations, we get

$$\begin{aligned} |(\varphi _{n,v},\mathrm{e}^{i(n+v)\zeta })\varphi _{n,v}|^{2}&=2\left( \sum _{k=1,2,3,4} \mathbf {Re(}F_{k})+\mathbf {Re(}F_{1}F_{2})+\mathbf {Re(}F_{1}F_{3} )\right) \\&\quad +|F_{1}|^{2}+|F_{2}|^{2}+O(n^{-5})= \end{aligned}$$
$$ 1+\frac{1}{2n^{2}}\frac{1}{\mid \delta \mid ^{2}}(p_{1}\mathrm{e}^{i\zeta }+p_{-1} \mathrm{e}^{-i\zeta }+c_{28}|p_{1}|^{2})+\frac{1}{n^{3}}((p_{1}\mathrm{e}^{i\zeta }+p_{-1} \mathrm{e}^{-i\zeta })c_{29} $$
$$ +c_{30}|p_{1}|^{2})+\frac{1}{n^{4}}((p_{1}\mathrm{e}^{i\zeta }+p_{-1}\mathrm{e}^{-i\zeta } )c_{31}+c_{32}|p_{1}|^{2}+c_{33}|p_{1}|^{4} $$
$$ +c_{34}|p_{1}|^{2}(p_{1}\mathrm{e}^{i\zeta }+p_{-1}\mathrm{e}^{-i\zeta })+(c_{35}+c_{36} |p_{1}|^{2})(p_{1}^{2}\mathrm{e}^{2i\zeta }+p_{-1}^{2}\mathrm{e}^{-2i\zeta }))+O\left( \frac{1}{n^{5} }\right) , $$

where \(\mathbf {Re(}F)\) denotes the real part of F. On the other hand

$$ |(\varphi _{n,v}(\zeta ),\mathrm{e}^{i(n+v)\zeta })|^{2}=\left( c_{37}\frac{1}{n^{2}} +c_{38}\frac{1}{n^{3}}+c_{39}\frac{1}{n^{4}}\right) |p_{1}|^{2}+c_{40}\frac{1}{n^{4} }|p_{1}|^{4}+O\left( \frac{1}{n^{5}}\right) . $$

These equalities imply (4.1.18). The invariant (4.1.19) is a consequence of (4.1.18), (4.1.16), and (4.1.7) for \(k=2,4\).   \(\blacksquare \)

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Veliev, O. (2019). Constructive Determination of the Spectral Invariants. In: Multidimensional Periodic Schrödinger Operator. Springer, Cham. https://doi.org/10.1007/978-3-030-24578-8_4

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