]> 7.5.1 Spectral properties of the system operator

#### 7.5.1 Spectral properties of the system operator

It follows directly from Deﬁnition 2.3.3 that the system operator (1.11) is self–adjoint, i.e., ${\mathcal{𝒜}}^{\ast }=\mathcal{𝒜}$. Observe that boundary conditions in (1.11) are strictly regular. To be more precise, the case ${1}^{\circ }$ of Corollary 2.9.1 holds. Since $\mathcal{𝒜}$ is self–adjoint, its spectrum consists of real eigenvalues, and the corresponding normalized eigenvectors form an orthonormal basis of $\text{H}$. To ﬁnd this basis we solve the eigenproblem for $\mathcal{𝒜}$,

$\mathcal{𝒜}q=\lambda q,\phantom{\rule{1em}{0ex}}\lambda \in \mathbb{ℝ},\phantom{\rule{1em}{0ex}}q\in D\left(\mathcal{𝒜}\right),\phantom{\rule{1em}{0ex}}q\ne 0$

which by (1.11) takes the form

 $\left\{\begin{array}{cccc}\hfill a{q}^{\prime \prime }\left(\theta \right)-{R}_{a}q\left(\theta \right)& \hfill =\hfill & \lambda q\left(\theta \right),\hfill & \phantom{\rule{1em}{0ex}}0\le \theta \le 1\hfill \\ \hfill {q}^{\prime }\left(1\right)& \hfill =\hfill & {R}_{1}q\left(1\right)\hfill & \hfill \\ \hfill {q}^{\prime }\left(0\right)& \hfill =\hfill & {R}_{0}q\left(0\right)\hfill & \hfill \end{array}\right\}$ (7.20)

Solving (7.20) we get the desired orthonormal basis of eigenvectors ${\left\{{e}_{n}\right\}}_{n\in \mathbb{ℕ}}$,

 ${e}_{n}^{\ast }=\frac{{h}_{n}}{{∥{h}_{n}∥}_{\text{H}}},\phantom{\rule{1em}{0ex}}{h}_{n}\left(\theta \right)=\frac{{R}_{0}}{{\mu }_{n}}sin\left({\mu }_{n}\theta \right)+cos\left({\mu }_{n}\theta \right),\phantom{\rule{2em}{0ex}}0\le \theta \le 1$ (7.21)

where ${\mu }_{n}$ are positive solutions of the equation

 $\left({\mu }^{2}+{R}_{0}{R}_{1}\right)sin\mu =\left({R}_{0}-{R}_{1}\right)\mu cos\mu$ (7.22)

Moreover,

 ${∥{h}_{n}∥}_{\text{H}}^{2}={\int }_{0}^{1}{h}_{n}^{2}\left(\theta \right)d\theta =\frac{1}{2}+\frac{{R}_{0}^{2}}{2{\mu }_{n}^{2}}+\frac{{\mu }_{n}^{2}-{R}_{0}^{2}}{2{\mu }_{n}^{3}}cos{\mu }_{n}sin{\mu }_{n}+\frac{{R}_{0}}{{\mu }_{n}^{2}}{sin}^{2}{\mu }_{n}$ (7.23)

The eigenvalues of $\mathcal{𝒜}$ express by the formula

 ${\lambda }_{n}=-a{\mu }_{n}^{2}-{R}_{a}\le -{R}_{a}<0$ (7.24)

By Theorem 2.3.4, $\mathcal{𝒜}$ generates a linear analytic EXS semigroup ${\left\{{e}^{t\mathcal{𝒜}}\right\}}_{t\ge 0}$ which in the basis ${\left\{{e}_{n}\right\}}_{n\in \mathbb{ℕ}}$ has the following representation

 ${e}^{t\mathcal{𝒜}}q=\sum _{n=1}^{\infty }{e}^{{\lambda }_{n}t}{〈q,{e}_{n}〉}_{\text{H}}{e}_{n}$ (7.25)

Exercise 7.5.1. Conﬁrm (7.24) by showing that the operator $-\mathcal{𝒜}+{R}_{a}I$ is nonnegative. Indeed, for $q\in D\left(\mathcal{𝒜}\right)$ we have

${〈q,\mathcal{𝒜}q〉}_{\text{H}}={\int }_{0}^{1}q\left(\theta \right)\left[a{q}^{\prime \prime }\left(\theta \right)-{R}_{a}q\left(\theta \right)\right]d\theta =$

$=a{q}^{2}\left(1\right){R}_{1}-a{q}^{2}\left(0\right){R}_{0}-{\int }_{0}^{1}a{\left[{q}^{\prime }\left(\theta \right)\right]}^{2}d\theta -{R}_{a}{∥q∥}^{2}\le {R}_{a}{∥q∥}_{\text{H}}^{2}.$