It is easy to check that the norms

${\parallel u\parallel}_{{W}_{2}^{3}({R}_{+};H)}$ and

${\parallel (-\frac{d}{dt}+A){(\frac{d}{dt}+A)}^{2}u\parallel}_{{L}_{2}({R}_{+};H)}$ are equivalent on

. Then it follows from the theorem on intermediate derivatives [[

3], Ch.1] that the following numbers are finite:

**Lemma 4.1** ${n}_{0,j}=\frac{2}{3\sqrt{3}}$, $j=1,2$.

*Proof* Passing to the limit as

$\beta \to \frac{27}{4}$ in (3.10), we see that, for any function

, the following inequality holds:

$\frac{27}{4}{\parallel {A}^{3-j}\frac{{d}^{j}u}{d{t}^{j}}\parallel}_{{L}_{2}({R}_{+};H)}^{2}\le {\parallel (-\frac{d}{dt}+A){(\frac{d}{dt}+A)}^{2}u\parallel}_{{L}_{2}({R}_{+};H)}^{2},$

*i.e.*,

${n}_{0,j}\le \frac{2}{3\sqrt{3}}$. We need to show that here we have the equality. To do this, it suffices to show that for any

$\epsilon >0$, there exists a function

such that

${\parallel (-\frac{d}{dt}+A){(\frac{d}{dt}+A)}^{2}{u}_{\epsilon}\parallel}_{{L}_{2}({R}_{+};H)}^{2}-(\frac{27}{4}+\epsilon ){\parallel {A}^{3-j}\frac{{d}^{j}{u}_{\epsilon}}{d{t}^{j}}\parallel}_{{L}_{2}({R}_{+};H)}^{2}<0.$

Note that the procedure of constructing such functions ${u}_{\epsilon}(t)$ is thoroughly described in [17] (in addition, the one for fourth-order equations with multiple characteristic is available in [20]). This method is applicable to our case, too. Therefore, we omit the respective part of the proof. So lemma is proved. □

**Remark 4.2** Since
, then ${n}_{j}\ge {n}_{0,j}=\frac{2}{3\sqrt{3}}$, $j=1,2$. Therefore, there arises the question: When do we have ${n}_{j}=\frac{2}{3\sqrt{3}}$, $j=1,2$?

Denote by ${\mu}_{j}$ the root of the equation ${S}_{j}(\beta ;0,1)=0$ in the interval $(0,\frac{27}{4})$ if such exists.

**Theorem 4.3** *The following relation holds*:

${n}_{j}=\{\begin{array}{cc}\frac{2}{3\sqrt{3}}\hfill & \mathit{\text{for}}\phantom{\rule{0.1em}{0ex}}{S}_{j}(\beta ;0,1)\ne 0,\beta \in (0,\frac{27}{4}),\hfill \\ {\mu}_{j}^{-1/2}\hfill & \mathit{\text{otherwise}}.\hfill \end{array}$

*Proof* If

${n}_{j}=\frac{2}{3\sqrt{3}}$, then it follows from equation (

3.11) that for any function

and for all

$\beta \in [0,\frac{27}{4})$, the following inequality holds:

${\parallel {F}_{j}(\frac{d}{dt};\beta ;A)u\parallel}_{{L}_{2}({R}_{+};H)}^{2}+{({S}_{j}(\beta ;0,1)\tilde{\phi},\tilde{\phi})}_{H}\ge {\parallel {P}_{0}u\parallel}_{{L}_{2}({R}_{+};H)}^{2}(1-\beta {n}_{j}^{2})>0.$

(4.1)

Now let us consider the Cauchy problem

${F}_{j}(\frac{d}{dt};\beta ;A)u(t)=0,$

(4.2)

$\frac{{d}^{k}u(0)}{d{t}^{k}}=0,\phantom{\rule{1em}{0ex}}k=0,1,$

(4.3)

$\frac{{d}^{2}u(0)}{d{t}^{2}}={A}^{-1/2}\tilde{\phi},\phantom{\rule{1em}{0ex}}\tilde{\phi}\in H.$

(4.4)

Since by Theorem 3.1, for

$\beta \in [0,\frac{27}{4})$, the polynomial operator pencil

${F}_{j}(\lambda ;\beta ;A)$ is of the form

${F}_{j}(\lambda ;\beta ;A)={\prod}_{s=1}^{3}(\lambda E-{\omega}_{j,s}(\beta )A)$, where

$Re{\omega}_{j,s}(\beta )<0$,

$s=1,2,3$, then Cauchy problem (4.2)-(4.4) has a unique solution

${u}_{\beta}(t)\in {W}_{2}^{3}({R}_{+};H)$, which can be expressed as:

${u}_{\beta}(t)={e}^{{\omega}_{j,1}(\beta )tA}{\eta}_{1}+{e}^{{\omega}_{j,2}(\beta )tA}{\eta}_{2}+{e}^{{\omega}_{j,3}(\beta )tA}{\eta}_{3},$

where

${\eta}_{1},{\eta}_{2},{\eta}_{3}\in {H}_{5/2}$ are uniquely determined by the conditions at zero (4.3), (4.4). Therefore, if we rewrite the inequality (4.1) for function

${u}_{\beta}(t)$, then for

$\beta \in [0,\frac{27}{4})$ we will have

${\parallel {F}_{j}(\frac{d}{dt};\beta ;A){u}_{\beta}\parallel}_{{L}_{2}({R}_{+};H)}^{2}+{({S}_{j}(\beta ;0,1)\tilde{\phi},\tilde{\phi})}_{H}>0.$

This means that

${({S}_{j}(\beta ;0,1)\tilde{\phi},\tilde{\phi})}_{H}>0$, and therefore,

${S}_{j}(\beta ;0,1)>0$ for all

$\beta \in [0,\frac{27}{4})$. Now let

${S}_{j}(\beta ;0,1)>0$ for all

$\beta \in [0,\frac{27}{4})$. Then it follows from (3.11) that for any function

and for all

$\beta \in [0,\frac{27}{4})$,

${\parallel {P}_{0}u\parallel}_{{L}_{2}({R}_{+};H)}^{2}-\beta {\parallel {A}^{3-j}\frac{{d}^{j}u}{d{t}^{j}}\parallel}_{{L}_{2}({R}_{+};H)}^{2}>0.$

Passing here to the limit as $\beta \to \frac{27}{4}$, we obtain that ${n}_{j}\le \frac{2}{3\sqrt{3}}$, and hence, ${n}_{j}=\frac{2}{3\sqrt{3}}$.

Continuing the proof of the theorem, we suppose that

${n}_{j}>\frac{2}{3\sqrt{3}}$. Then

${n}_{j}^{-2}\in (0,\frac{27}{4})$. Note that for

$\beta \in (0,{n}_{j}^{-2})$, we have

${\parallel {P}_{0}u\parallel}_{{L}_{2}({R}_{+};H)}^{2}-\beta {\parallel {A}^{3-j}\frac{{d}^{j}u}{d{t}^{j}}\parallel}_{{L}_{2}({R}_{+};H)}^{2}\ge {\parallel {P}_{0}u\parallel}_{{L}_{2}({R}_{+};H)}^{2}(1-\beta {n}_{j}^{2})>0.$

Therefore, from (3.11) we find that for

$\beta \in (0,{n}_{j}^{-2})$, the following inequality holds:

${\parallel {F}_{j}(\frac{d}{dt};\beta ;A)u\parallel}_{{L}_{2}({R}_{+};H)}^{2}+{({S}_{j}(\beta ;0,1)\tilde{\phi},\tilde{\phi})}_{H}>0.$

Applying this inequality again to the solution

$u(t)$ of Cauchy problem (4.2)-(4.4), we obtain

${S}_{j}(\beta ;0,1)>0$ for all

$\beta \in [0,{n}_{j}^{-2})$. On the other hand, it follows from the definition of

${n}_{j}$ that, for any

$\beta \in ({n}_{j}^{-2},\frac{27}{4})$, there exists a function

such that

${\parallel {P}_{0}{v}_{\beta}\parallel}_{{L}_{2}({R}_{+};H)}^{2}<\beta {\parallel {A}^{3-j}\frac{{d}^{j}{v}_{\beta}}{d{t}^{j}}\parallel}_{{L}_{2}({R}_{+};H)}^{2}.$

Taking into account this inequality in (3.11), we obtain

${\parallel {F}_{j}(\frac{d}{dt};\beta ;A){v}_{\beta}\parallel}_{{L}_{2}({R}_{+};H)}^{2}+{({S}_{j}(\beta ;0,1){\tilde{\phi}}_{\beta},{\tilde{\phi}}_{\beta})}_{H}<0,$

where

${\tilde{\phi}}_{\beta}={A}^{1/2}\frac{{d}^{2}{v}_{\beta}(0)}{d{t}^{2}}$. Therefore, there exists a vector

${\tilde{\phi}}_{\beta}\in H$ such that

${({S}_{j}(\beta ;0,1){\tilde{\phi}}_{\beta},{\tilde{\phi}}_{\beta})}_{H}<0.$

Thus, ${S}_{j}(\beta ;0,1)<0$ for $\beta \in ({n}_{j}^{-2},\frac{27}{4})$. And since ${S}_{j}(\beta ;0,1)$ is a continuous function of the argument *β* in the interval $[0,\frac{27}{4})$, then ${S}_{j}({n}_{j}^{-2};0,1)=0$. It follows from these arguments that the equation ${S}_{j}(\beta ;0,1)=0$ has a root in the interval $(0,\frac{27}{4})$. Now let ${S}_{j}(\beta ;0,1)=0$ have a root in the interval $(0,\frac{27}{4})$. This means that the inequality ${S}_{j}(\beta ;0,1)>0$ cannot be satisfied for any $\beta \in [0,\frac{27}{4})$. Therefore, according to our earlier reasonings in the proof of this theorem, we have ${n}_{j}>\frac{2}{3\sqrt{3}}$. Obviously, for the root ${\mu}_{j}$ of the equation ${S}_{j}(\beta ;0,1)=0$, we have that ${\mu}_{j}\ge {n}_{j}^{-2}$, because the proof of the theorem for $\beta \in [0,{n}_{j}^{-2})$ implies that ${S}_{j}(\beta ;0,1)>0$. And since ${S}_{j}({n}_{j}^{-2};0,1)=0$, we obtain ${n}_{j}^{-2}={\mu}_{j}$. The theorem is proved. □

**Remark 4.4** From Theorem 4.3, it becomes clear that to find the numbers ${n}_{j}$, $j=1,2$, we must solve the equations ${S}_{j}(\beta ;0,1)=0$, $j=1,2$, together with systems (3.3) respectively. In this case, it is necessary to take into account the properties of the numbers ${\alpha}_{1,j}(\beta )$, ${\alpha}_{2,j}(\beta )$, $j=1,2$.

The following theorem holds.

**Theorem 4.5** ${n}_{1}=\frac{2}{3\sqrt{3}}$, ${n}_{2}=\frac{1}{\sqrt{2}{(\sqrt{5}+1)}^{1/2}}$.

*Proof* In view of Remark 4.4, in the case $j=1$, we have ${n}_{1}=\frac{2}{3\sqrt{3}}$ due to the negativity of ${\alpha}_{2,1}(\beta )$, despite ${\alpha}_{1,1}(\beta )=1\Rightarrow {\alpha}_{2,1}(\beta )=-1\Rightarrow \beta =4\in (0,\frac{27}{4})$. In the case $j=2$, we have ${\alpha}_{1,2}(\beta )=1\Rightarrow {\alpha}_{2,2}(\beta )=-\sqrt{5}$ or ${\alpha}_{2,2}(\beta )=\sqrt{5}$. Then $\beta =-2(\sqrt{5}-1)\notin (0,\frac{27}{4})$ or $\beta =2(\sqrt{5}+1)\in (0,\frac{27}{4})$, respectively. Therefore, ${n}_{2}=\frac{1}{\sqrt{2}{(\sqrt{5}+1)}^{1/2}}$. The theorem is proved. □