A Note on the Spectral Analysis of Some Fourth-Order Differential Equations with a Semigroup Approach

Abstract

In this paper, we analyze the solubility of a class of abstract fourth-order in time linear evolution equations, using the roots of the characteristic polynomial that is associated with the equations.

1 Introduction

Let X be a separable Hilbert space and $A:D(A)\subset X\rightarrow X$ be an unbounded linear, closed, densely defined, self-adjoint and for some $a>0$, satisfies $\langle Au,u\rangle \geqslant a \langle u,u\rangle$ or all $u\in D(A)$. We also assume that A has compact resolvent on X. It is well-known that with those hypotheses, we can define the fractional power $A^{\alpha }$ of order $0<\alpha <1$ according to [2] and [11], as a closed linear operator, see e.g. [10, 11, 13, 18]. Denote by $X^{\alpha }=D(A^{\alpha })$ for $0\leqslant \alpha \leqslant 1$ (taking $A^0:=I$ on $X^0:=X$ when $\alpha =0$). Recall that $X^{\alpha }$ is dense in X for all $0\leqslant \alpha \leqslant 1$, for details see [2, Theorem 4.6.5]. The fractional power space $X^\alpha$ endowed with the norm $\Vert \cdot \Vert _{X^\alpha }:=\Vert A^{\alpha } \cdot \Vert _X$ is a Banach space. With this notation, we have $X^{-\alpha }=(X^\alpha )'$ for all $\alpha >0$, see [2, 26, 27] for the characterization of the negative scale.

In this paper we consider the fourth-order linear equation in time

$$\begin{aligned} u_{tttt}+A^{\frac{2}{5}}u+aA^{\frac{1}{5}}u_{ttt}+bA^{\frac{4}{5}}u_{tt}+cAu_t=0, \end{aligned}$$

(1.1)

where $a,b,c\geqslant 0$.

We discuss here the well-posedness of Eq. (1.1) in terms of the scalar parameters a, b and c considering the theory of strongly continuous bounded linear operators on suitable phase spaces.

In recent years higher-order evolution equations in time have attracted the attention of many researchers. This can be explained by the sensitivity of the conditions of well-posedness and regularity of solutions of these equations in infinite-dimensional Banach spaces, see e.g. [3,4,5, 8, 9, 22], see also the Moore-Gibson-Thompson (MGT) equations, see e.g. [1, 6, 7, 12, 15,16,17, 19,20,21, 24, 25] and reference therein.

We also observe that it is possible to find in the literature some works dedicated to fourth-order equations in time motivated by MGT equations, see e.g. [1, 7, 16, 19,20,21]; in these papers are made formulations for fourth-order equations of the MGT type, and results of blow-up of solutions, well-posedness, stability and regularity of solutions are obtained.

Our best knowledge indicates that no information is known for models of the type (1.1) concerning the well-posedness and regularity of solutions. This paper makes this point. Furthermore, abstract differential equations of order greater than two are generally ill-posed in the sense of the theory of semigroups of bounded linear operators. However, we identify cases where (1.1) is well-posed. Namely, cases in which this equation is associated with a strongly continuous semigroup and instances in which it is associated with an analytic semigroup.

We will divide our problem into two cases: $c>0$ and $c=0$. First of all, note that our linear problem (1.1) can be rewritten as a first-order abstract system of the form

$$\begin{aligned} U_t+S_{a,b,c}U=0,\quad t>0, \end{aligned}$$

on the phase space

$$\begin{aligned} Y=X^{1}\times X^{\frac{4}{5}}\times X^{\frac{3}{5}}\times X, \end{aligned}$$

endowed with the usual inner product, where $U=\left[ {\begin{matrix} u\\ u_t\\ u_{tt}\\ u_{ttt} \end{matrix}}\right]$, and the matrix operator $S_{a,b,c}$ can be seen as a unbounded linear operator defined by

$$\begin{aligned} D(S_{a,b,c})=X^1 \times X^1 \times X^{\frac{4}{5}} \times X^{\frac{1}{5}} \end{aligned}$$

and

$$\begin{aligned} S_{a,b,c}=\begin{bmatrix} 0 &{}\quad -I &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad -I&{}\quad 0 \\ 0&{}\quad 0&{}\quad 0&{}\quad -I\\ A^{\frac{2}{5}}&{}\quad cA &{}\quad bA^{\frac{4}{5}}&{}\quad aA^{\frac{1}{5}} \end{bmatrix} \end{aligned}$$

Namely, the unbounded linear operator $S_{a,b,c}:D(S_{a,b,c})\subset Y\rightarrow Y$ is closed and densely defined.

It is easily seen that $\lambda \in \rho (-S_{a,b,c})$ if and only if $\lambda ^4I+a\lambda ^3 A^{\frac{1}{5}}+b\lambda ^2 A^{\frac{4}{5}}+c\lambda A+A^{\frac{2}{5}}$ is bijective. Consequently, if $\lambda \in \mathbb {C}$ is such that

$$\begin{aligned} \lambda ^4+a\mu ^{\frac{1}{5}}\lambda ^3+b\mu ^{\frac{4}{5}}\lambda ^2+c\mu \lambda +\mu ^{\frac{2}{5}}=0 \end{aligned}$$

for some $\mu \in \sigma (A)$, then $\lambda \in \sigma (-S_{a,b,c})$. Since $\sigma (A)$ is countably infinite, we will consider the characteristic polynomials”

$$\begin{aligned} p_n(\lambda )=\lambda ^4+a\mu _n^{\frac{1}{5}}\lambda ^3+b\mu _n^{\frac{4}{5}}\lambda ^2+c\mu _n\lambda +\mu _n^{\frac{2}{5}}. \end{aligned}$$

(1.2)

with $\mu _n\in \sigma (A)$. Recall that $\mu _n>0$ and $\mu _n\rightarrow \infty$ as $n\rightarrow \infty$.

Note that the discriminant of (1.2) is given by

$$\begin{aligned} {\begin{matrix} \Delta _n&{}=(-4b^3c^2)\mu _n^{\frac{22}{5}}+(-27c^4+18abc^3+a^2b^2c^2)\mu _n^4+(16b^4-4a^3c^3 )\mu _n^{\frac{18}{5}}+(144a^2b)\mu _n^{\frac{17}{5}}\\ &{}+(144bc^2-80ab^2c-4a^2b^3)\mu _n^{\frac{16}{5}}+(18a^3bc-6a^2c^2)\mu _n^{\frac{14}{5}}-(128b^2)\mu _n^{\frac{12}{5}}-(192ac)\mu _n^2\\ &{}-27a^4\mu _n^{\frac{8}{5}}+256\mu _n^{\frac{6}{5}}, \end{matrix}} \end{aligned}$$

see [14, Example on the pages 192-193].

We can summarize the behavior of $\Delta _n$ when $n>>0$ as follows (Table 1).

See Tables 1, 2, 3 below

Table 1 Signal of the discriminant $\Delta _n$

This is a key result in our analysis since a geometric localization of the points of the spectral set of $-S_{a,b,c}$ can provide us with information about the generation of strongly continuous semigroups of bounded linear operators associated with the problem (1.1), see e.g. [23, Chapter 1, Corollary 3.8].

This work is organized as follows. In Sect. 2 we consider the case $c>0$ and we treat three subcases $c=ab$, $c>ab$, and $c<ab$. Finally, in Sect. 3 we consider the case $c=0$ and we treat three subcases $a=b=c=0$, $a\ne 0$ and $b=c=0$, $b\ne 0$ and $a=c=0$, and $a,b\ne 0$ and $c=0$.

The tables below summarize our results, where $\checkmark$ means that it is possible to generate the respective strongly continuous semigroups of bounded linear operators on Y, and $\times$ means that it is not possible to generate the respective strongly continuous semigroups of bounded linear operators on Y.

For $c>0$, we have the following results (Table 2):

Table 2 Case $c>0$

For $c=0$, we have the following results (Table 3):

Table 3 Case $c=0$

2 Case $c>0$

Since $\Delta _n<0$ for large n when $c>0$ we guarantee the existence of two complex conjugated roots $z_n$ and $\overline{z_n}$ and two real roots $x_n$ and $y_n$. The following lemma shows the behavior of one of the real roots for the polynomial $p_n$ in (1.2).

Lemma 1.1

If $c>0$, then $y_n\rightarrow 0$ as $n\rightarrow +\infty$.

Proof

Given $\epsilon >0$ we have

$$\begin{aligned} p_n(-\epsilon )=\epsilon ^4-a\mu _n^{\frac{1}{5}}\epsilon ^3+b\mu _n^{\frac{4}{5}}\epsilon ^2-c\mu _n\epsilon +\mu _n^{\frac{2}{5}}. \end{aligned}$$

It is easy to see that $p_n(-\epsilon )\rightarrow -\infty$ as $n\rightarrow +\infty$. Since $p_n(0)>0$ for any n, we guarantee that $y_n\in (-\epsilon ,0)$ for large n; that is, $y_n\rightarrow 0$ as $n\rightarrow +\infty$.

Corollary 1.2

If $(x_n+a\mu _n^{\frac{1}{5}})\rightarrow K$ as $n\rightarrow +\infty$, then $Re(z_n)\rightarrow -\dfrac{K}{2}$ as $n\rightarrow +\infty$.

Proof

It follows directly from Vieta’s formulas

$$\begin{aligned} x_n+y_n+z_n+\overline{z_n}=-a\mu _n^{\frac{1}{5}}. \end{aligned}$$

(2.1)

jointly with Lemma 2.1.

Remark 2.3

It will be useful later to know that for any $\epsilon >0$ we have $p(-a\mu _n^{\frac{1}{5}}-\epsilon )$ given by

$$\begin{aligned}{} & {} (a^2b-ac)\mu _n^{\frac{6}{5}}+(2ab-c)\epsilon \mu _n+(\epsilon ^2b)\mu _n^{\frac{4}{5}}+(a^3\epsilon )\mu _n^{\frac{3}{5}}\\{} & {} \quad +(3a^2\epsilon ^2+1)\mu _n^{\frac{2}{5}}+3a\epsilon ^3\mu _n^{\frac{1}{5}} +\epsilon ^4. \end{aligned}$$

2.1 Subcase $ab=c$

In this subcase the polynomial $p_n$ becomes

$$\begin{aligned} p_n(\lambda )=\lambda ^4+a\mu _n^{\frac{1}{5}}\lambda ^3+b\mu _n^{\frac{4}{5}}\lambda ^2+ab\mu _n+\mu _n^{\frac{2}{5}}. \end{aligned}$$

We will prove that $x_n+a\mu _n^{\frac{1}{5}}\rightarrow 0$ and consequently by Corollary 2.2 we have $Re(z_n)\rightarrow 0$. From this, we will see that it is possible to prove a result of well-posedness for Eq. (1.1).

Proposition 1.4

If $c>0$ and $c=ab$, then $Re(z_n)\rightarrow 0$. This means that, in the best case scenario, the problem (1.1) can only generate a strongly continuous semigroup on Y.

Proof

We will show that $x_n\rightarrow -a\mu _n^{\frac{1}{5}}$ and the other claim follows from the Hille-Yosida Theorem, see e.g. [23, Chapter 1, Theorem 3.1]. Fixing $\epsilon >0$ it is easy to see (Remark 2.3) that $p(-a\mu _n^{\frac{1}{5}}+\epsilon )<0$ for large n. Since $p_n(-a\mu _n^{\frac{1}{5}})>0$ we know that $x_n\in (-a\mu _n^{\frac{1}{5}},-a\mu _n^{\frac{1}{5}}+\epsilon )$ for large n; that is, $x_n\rightarrow -a\mu _n^{\frac{1}{5}}$ as $n\rightarrow +\infty$.

Theorem 1.5

If $c>0$ and $c=ab$, then problem (1.1) is well-posedness on Y; that is, the linear operator associated with this problem is the infinitesimal generator of a strongly continuous semigroup on Y.

Proof

Define

$$\begin{aligned} z=u_t+aA^{\frac{1}{5}}u \end{aligned}$$

(2.2)

and

$$\begin{aligned} v=u_{ttt}+aA^{\frac{1}{5}}u_{tt}+bA^{\frac{4}{5}}u_t+abAu. \end{aligned}$$

(2.3)

Transferring our problem to the phase space

$$\begin{aligned} Z=X\times X\times X\times X \end{aligned}$$

endowed with the usual inner product, we define the vector $W=\left[ {\begin{matrix} Au\\ Az\\ A^{\frac{3}{5}}z_t\\ A^{\frac{3}{5}}v \end{matrix}}\right]$. It is easy to see that (1.1) can be rewritten as a first-order abstract system of the form

$$\begin{aligned} W_t=\Lambda W,\quad t>0, \end{aligned}$$

where $\Lambda$ denotes the unbounded linear operator given by

$$\begin{aligned} \Lambda =\begin{bmatrix} -aA^{\frac{1}{5}} &{}\quad 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad A^{\frac{2}{5}} &{}\quad 0 \\ 0&{}\quad -bA^{\frac{2}{5}} &{}\quad 0 &{}\quad 1\\ -I &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{bmatrix}. \end{aligned}$$

Namely, this linear operator is closed and densely defined. Moreover, it is a maximally dissipative operator (see [23, Chapter 1, Definition 4.1]) on a space in which the norm is equivalent to the norm of space Z. This also implies that $\left[ {\begin{matrix} u\\ u_t\\ u_{tt}\\ u_{ttt} \end{matrix}}\right] \in Y$.

2.2 Subcase $c>ab$

We will show that in this subcase the problem (1.1) is ill-posed, that is, the operator associated with the problem does not generate a strongly continuous semigroup of bounded linear operators.

Proposition 1.6

If $c>0$ and $c>ab$ then $Re(z_n)\rightarrow +\infty$. Consequently, the unbounded linear operator associated with (1.1) does not generate a strongly continuous semigroup on Y.

Proof

It follows from Remark (2.3) that $p_n(-a\mu _n^{\frac{1}{5}}-\mu _n^{\frac{1}{10}})<0$ for large n. Since $p_n(\lambda )\rightarrow +\infty$ as $\lambda \rightarrow -\infty$ we guarantee that $x_n<-a\mu _n^{\frac{1}{5}}-\mu _n^{\frac{1}{10}}$. From (2.1) and Lemma 2.1 we obtain $Re(z_n)\rightarrow +\infty$. This violates a necessary condition for the generation of a strongly continuous semigroup, see e.g. [23, Chapter 1, Corollary 3.8].

2.3 Subcase $c<ab$

Note that in this subcase we have to assume that $a,b,c\ne 0$. We will see that problem (1.1) can be well-posed in the sense of the theory of semigroups of bounded linear operators. However, the semigroup associated with (1.1) cannot be analytic on Y.

Lemma 1.7

If $a,b,c\ne 0$ and $c<ab$ then $x_n\rightarrow \dfrac{-c\mu _n^{\frac{1}{5}}}{b}$ as $n\rightarrow +\infty$.

Proof

Firstly note that

$$\begin{aligned} p_n\left( \dfrac{-c\mu _n^{\frac{1}{5}}}{b}\right) =\dfrac{c^3(c-ab)}{b^4}\mu _n^{\frac{4}{5}}+\mu _n^{\frac{2}{5}} \end{aligned}$$

and therefore $p_n(-cb^{-1}\mu _n^{\frac{1}{5}})<0$ for large n. Fixing $\epsilon >0$ we can easily compute that

$$\begin{aligned} {\begin{matrix} p_n(-cb^{-1}\mu _n^{\frac{1}{5}}-\epsilon )&{}=\epsilon c\mu _n+(c^4b^{-4}-ac^3b^{-3}+\epsilon ^2b)\mu _n^{\frac{4}{5}}+(4c^3b^{-3}\epsilon -3ac^2b^{-2}\epsilon )\mu _n^{\frac{3}{5}}\\ &{}\quad +(1+6\epsilon ^2c^2b^{-2}-3\epsilon ^2acb^{-1})\mu _n^{\frac{2}{5}}+(4\epsilon ^3cb^{-1}-a\epsilon ^2)\mu _n^{\frac{1}{5}}+\epsilon ^4. \end{matrix}} \end{aligned}$$

Hence $p_n(-cb^{-1}\mu _n^{\frac{1}{5}}-\epsilon )>0$ for large n which guarantees that

$$\begin{aligned} x_n\in (-cb^{-1}\mu _n^{\frac{1}{5}}-\epsilon ,-cb^{-1}\mu _n^{\frac{1}{5}}) \end{aligned}$$

for large n; that is, $x_n\rightarrow -cb^{-1}\mu _n^{\frac{1}{5}}$ as $n\rightarrow +\infty$.

Corollary 1.8

If $a,b,c\ne 0$ and $c<ab$ then $Re(z_n)\rightarrow (c-ab)2^{-1}b^{-1}\mu _n^{\frac{1}{5}}$. Therefore the problem (1.1) can be well-posed in the sense of the theory of semigroups of bounded linear operators. Moreover, the solution of (1.1) cannot be analytic on Y in the sense of [11].

Proof

From Vieta’s formulas and Lemma 2.7 it follows immediately that $Re(z_n)\rightarrow (c-ab)2^{-1}b^{-1}\mu _n^{\frac{1}{5}}$ and that

$$\begin{aligned} \left( \dfrac{Im(z_n)}{Re(z_n)}\right) ^2=\dfrac{4b^3\mu _n^{\frac{2}{5}}}{(c-ab)^2}\rightarrow +\infty , \end{aligned}$$

which concludes the proof.

3 Case $c=0$

3.1 Subcase $a=b=c=0$

In this case, the polynomial is given by

$$\begin{aligned} p_n(\lambda )=\lambda ^4+\mu _n^{\frac{2}{5}}=0. \end{aligned}$$

It is easy to see that the roots of $p_n$ are $\mu _n^{\frac{1}{10}}e^{\frac{\pi }{4}i}, \mu _n^{\frac{1}{10}}e^{\frac{3\pi }{4}i}, \mu _n^{\frac{1}{10}}e^{\frac{5\pi }{4}i}$ and $\mu _n^{\frac{1}{10}}e^{\frac{7\pi }{4}i}$. To simplify our view, we can rewrite these roots as

$$\begin{aligned}{} & {} \mu _n^{\frac{1}{10}}\left( \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2}\right) , \mu _n^{\frac{1}{10}}\left( -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2}\right) , \mu _n^{\frac{1}{10}}\left( -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\right) \\{} & {} \quad \text {and}\ \mu _n^{\frac{1}{10}}\left( \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\right) . \end{aligned}$$

Theorem 1.9

If $a=b=c=0$, then problem (1.1) is ill-posed in the sense of the theory of semigroups of bounded linear operators on Y.

Proof

Since $\mu _n\rightarrow +\infty$ our next result follows directly from the Hille-Yosida Theorem, see e.g. [23, Chapter 1, Theorem 3.1].

3.2 Subcase $a\ne 0$ and $b=c=0$

In this case we know that $\Delta _n<0$ for large n and therefore our polynomial

$$\begin{aligned} \lambda ^4+a\mu _n^{\frac{1}{5}}\lambda ^3+\mu _n^{\frac{2}{5}} \end{aligned}$$

have two real roots $x_n$, $y_n$ and two complex roots $z_n$ and $\overline{z}_n$.

Lemma 1.10

If $a\ne 0$ and $b=c=0$ then $x_n\rightarrow -a\mu _n^{\frac{1}{5}}$ as $n\rightarrow +\infty$.

Proof

Given $\epsilon >0$ we have

$$\begin{aligned} p_n(-a\mu _n^{\frac{1}{5}}+\epsilon )=-\epsilon a^3\mu _n^{\frac{3}{5}}+(3a^2\epsilon ^2+1)\mu _n^{\frac{2}{5}}-3a\epsilon ^3\mu _n^{\frac{1}{5}}+\epsilon ^4. \end{aligned}$$

Hence $p_n(-a\mu _n^{\frac{1}{5}}+\epsilon )<0$ for large n. Since $p_n(-a\mu _n^{\frac{1}{5}})>0$ for any n, we conclude that $x_n\in (-a\mu _n^{\frac{1}{5}},-a\mu _n^{\frac{1}{5}}+\epsilon )$ for large n; that is, $x_n\rightarrow -a\mu _n^{\frac{1}{5}}$ as $n\rightarrow +\infty$.

Lemma 1.11

If $a\ne 0$ and $b=c=0$ then $y_n\rightarrow - a^{-\frac{1}{3}}\mu _n^{\frac{1}{15}}$ as $n\rightarrow +\infty$.

Proof

Fix $\epsilon >0$ and note that

$$\begin{aligned} {\begin{matrix} p_n(-a^{-\frac{1}{3}}\mu _n^{\frac{1}{15}}-\epsilon )&{}= -3\epsilon a^{\frac{1}{3}}\mu _n^{\frac{5}{15}}+(a^{-\frac{4}{3}}-3\epsilon ^2a^{\frac{2}{3}})\mu _n^{\frac{4}{15}}+(4\epsilon a^{-1}-a\epsilon ^3)\mu _n^{\frac{3}{15}}\\ &{}\quad +6\epsilon ^2a^{-\frac{2}{3}}\mu _n^{\frac{2}{15}}+4\epsilon ^3a^{-\frac{1}{3}}\mu _n^{\frac{1}{15}}+\epsilon ^4. \end{matrix}} \end{aligned}$$

Thus $p_n(-a^{-\frac{1}{3}}\mu _n^{\frac{1}{15}}-\epsilon )<0$ for large n. Since $p_n(-a^{-\frac{1}{3}}\mu _n^{\frac{1}{15}})=a^{\frac{-4}{3}}\mu _n^{\frac{4}{15}}$, we guarantee that $p_n(-a^{-\frac{1}{3}}\mu _n^{\frac{1}{15}})>0$ for large n and therefore

$$y_n\in (-a^{-\frac{1}{3}}\mu _n^{\frac{1}{15}}-\epsilon ,-a^{-\frac{1}{3}}\mu _n^{\frac{1}{15}})$$

when n is large. This means that $|y_n+a^{-\frac{1}{3}}\mu _n^{\frac{1}{15}}|\rightarrow 0$ as $n\rightarrow +\infty$.

Corollary 1.12

If $a\ne 0$ and $b=c=0$ then $Re(z_n)\rightarrow +\infty$. Consequently, problem (1.1) is ill-posed in the sense of the theory of semigroups of bounded linear operators on Y.

3.3 Subscase $b\ne 0$ and $a=c=0$

In this case we have the polynomial

$$\begin{aligned} \lambda ^4+b\mu _n^{\frac{4}{5}}\lambda ^2+\mu _n^{\frac{2}{5}} \end{aligned}$$

(3.1)

which has $\Delta _n>0$ for large n and $P_n=8b\mu _n^{\frac{4}{5}} >0$; that is, our polynomial has two pairs of non-real complex conjugate roots.

Lemma 1.13

If $b\ne 0$ and $a=c=0$ then all roots are imaginary for large n. Consequently, problem (1.1) can generate a strongly continuous semigroup of bounded linear operators on Y. Moreover, it cannot generate an analytic semigroup on Y in the sense of [11].

Proof

Denoting the four roots by $z_n, \overline{z}_n, w_n, \overline{w}_n$ we know from Vieta’s formulas that

$$\begin{aligned} Re(w_n)=-Re(z_n), \ \ \ |z_n|^2|w_n|^2=\mu _n^{\frac{2}{5}} \end{aligned}$$

(3.2)

and

$$\begin{aligned} Re(w_n)(|z_n|^2-|w_n|^2)=0. \end{aligned}$$

Assume that $Re(w_n)\ne 0$. Then

$$\begin{aligned} |z_n|^2=|w_n|^2=\mu _n^{\frac{1}{5}}. \end{aligned}$$

(3.3)

Writing $w_n$ on its polar form $w_n=r_ne^{i\theta _n}$ and using (3.1) it follows that

$$\begin{aligned} r_n^4e^{i4\theta _n}+b\mu _n^{\frac{4}{5}}r_n^2e^{i2\theta _n}+\mu _n^{\frac{2}{5}}=0 \end{aligned}$$

and therefore

$$\begin{aligned} r_n^2(r_n^2sin(4\theta _n)+b\mu _n^{\frac{4}{5}}sin(2\theta _n))=0. \end{aligned}$$

(3.4)

From (3.1) we have $r_n=|w_n|\ne 0$ and this leads to

$$\begin{aligned} r_n^2sin(4\theta _n)+b\mu _n^{\frac{4}{5}}sin(2\theta _n)=0. \end{aligned}$$

Case $sin(4\theta _n)=0$ we have $sin(2\theta _n)=0$, and $\theta _n=0$ or $\theta _n=\frac{\pi }{2}$. From this, $sin(4\theta _n)\ne 0$; that is,

$$\begin{aligned} r_n^2=\dfrac{- sin(2\theta _n)b\mu _n^{\frac{4}{5}}}{sin(4\theta _n)}. \end{aligned}$$

(3.5)

Putting (3.3) and (3.5) together we obtain

$$\begin{aligned} -sin(2\theta _n)b\mu _n^{\frac{3}{5}}=sin(4\theta _n). \end{aligned}$$

Thus

$$\begin{aligned} sin(2\theta _n)(b\mu _n^{\frac{3}{5}}+2cos(2\theta _n))=0. \end{aligned}$$

Since $b\mu _n^{\frac{3}{5}}+2cos(2\theta _n)\ne 0$ for large n we conclude that $sin(2\theta _n)=0$ and

$$\begin{aligned} \theta _n=0\ \ \text {or}\ \ \theta _n=\dfrac{\pi }{2} \end{aligned}$$

(3.6)

which contradicts our assumption that $w_n\notin \mathbb {R}$ and $Re(w_n)\ne 0$. Therefore

$$Re(w_n)=Re(z_n)=0$$

for large n we conclude the proof.

Theorem 1.14

If $b\ne 0$ and $a=c=0$ then the problem (1.1) generates a strongly continuous semigroup on Y.

Proof

It follows similarly to Theorem 2.5.

3.4 Subcase $a,b\ne 0$ and $c=0$

In this case we have the polynomial

$$\begin{aligned} \lambda ^4+a\mu _n^{\frac{1}{5}}\lambda ^3+b\mu _n^{\frac{4}{5}}\lambda ^2+\mu _n^{\frac{2}{5}} \end{aligned}$$

(3.7)

which has $\Delta _n>0$ for large n. Moreover, $P_n=8b\mu _n^{\frac{4}{5}}-3a^2\mu _n^{\frac{2}{5}}>0$ for large n, which implies that our polynomial has four complex roots $z_n, \overline{z_n}, w_n$ and $\overline{w_n}$.

Lemma 1.15

Two of the four roots of polynomial (3.7); namely $w_n$ and $\overline{w_n}$ satisfy

$$|w_n|=| \overline{w_n}|\rightarrow b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}}\ \ \text {as} \ \ n\rightarrow +\infty .$$

Proof

From Vieta’s formulas, we can easily obtain that

$$\begin{aligned}{} & {} |w_n|^2|z_n|^2=\mu _n^{\frac{2}{5}}, \end{aligned}$$

(3.8)

$$\begin{aligned}{} & {} Re(z_n)|w_n|^2=-Re(w_n)| z_n|^2 \end{aligned}$$

(3.9)

$$\begin{aligned}{} & {} |z_n|^2+|w_n|^2+4Re(z_n)Re(w_n)=b\mu _n^{\frac{4}{5}} \end{aligned}$$

(3.10)

and

$$\begin{aligned} Re(w_n)+Re(z_n)=\dfrac{-a\mu _n^{\frac{1}{5}}}{2}. \end{aligned}$$

(3.11)

Since $w_n,z_n\ne 0$ we can put (3.8) and (3.9) into Eq. (3.10) to obtain that

$$\begin{aligned} \dfrac{\mu _n^{\frac{2}{5}}}{|w_n|^2}+|w_n|^2-\dfrac{4Re(w_n)^2\mu _n^{\frac{2}{5}}}{|w_n|^4}=b\mu _n^{\frac{4}{5}}. \end{aligned}$$

which leads us to

$$\begin{aligned} \mu _n^{\frac{2}{5}}|w_n|^2+|w_n|^6-4Re(w_n)^2\mu _n^{\frac{2}{5}}-b\mu _n^{\frac{4}{5}}|w_n|^4=0. \end{aligned}$$

(3.12)

Now, putting (3.9) and (3.11) together we have

$$\begin{aligned} \dfrac{-a\mu _n^{\frac{1}{5}}|w_n|^2 }{2}=Re(w_n)(|w_n|^2-|z_n|^2). \end{aligned}$$

(3.13)

Since $w_n\ne 0$ we know from (3.13) that $|w_n|^2-|z_n|^2\ne 0$ and consequently

$$\begin{aligned} Re(w_n)=\frac{-a\mu _n^{\frac{1}{5}}|w_n|^2}{2(|w_n|^2-|z_n|^2)}. \end{aligned}$$

Putting this equation together with (3.8) we have (the well-defined) equation

$$\begin{aligned} Re(w_n)=\dfrac{-a\mu _n^{\frac{1}{5}}|w_n|^4}{2(|w_n|^4-\mu _n^{\frac{2}{5}})}. \end{aligned}$$

(3.14)

Finally, putting (3.12) and (3.14) together we obtain

$$\begin{aligned} \mu _n^{\frac{2}{5}}|w_n|^2+|w_n|^6-\dfrac{a^2\mu _n^{\frac{4}{5}}|w_n| ^8}{|w_n| ^8-2\mu _n^{\frac{2}{5}}|w_n| ^4+\mu _n^{\frac{4}{5}}}-b\mu _n^{\frac{4}{5}}|w_n|^4=0. \end{aligned}$$

From this we have

$$\begin{aligned} (\mu _n^{\frac{2}{5}}|w_n|^2+|w_n|^6-b\mu _n^{\frac{4}{5}}|w_n|^4)(|w_n| ^8-2\mu _n^{\frac{2}{5}}|w_n| ^4+\mu _n^{\frac{4}{5}})-a^2\mu _n^{\frac{4}{5}}|w_n| ^8=0 \end{aligned}$$

and consequently

$$\begin{aligned} |w_n|^{14}-b\mu _n^{\frac{4}{5}}|w_n|^{12}-\mu _n^{\frac{2}{5}}\parallel&w_n\parallel ^{10}+(2b\mu _n^{\frac{6}{5}}-a^2\mu _n^{\frac{4}{5}})|w_n|^8\\&-\mu _n^{\frac{4}{5}}|w_n|^6-b\mu _n^{\frac{8}{5}}|w_n|^4+\mu _n^{\frac{6}{5}}|w_n|^2=0. \end{aligned}$$

Dividing both sides by $|w_n|^2$ we finally have

$$\begin{aligned}&|w_n|^{12}-b\mu _n^{\frac{4}{5}}|w_n|^{10}-\mu _n^{\frac{2}{5}}|w_n|^{8}\\&\quad +(2b\mu _n^{\frac{6}{5}}-a^2\mu _n^{\frac{4}{5}})|w_n|^6-\mu _n^{\frac{4}{5}}|w_n|^4-b\mu _n^{\frac{8}{5}}|w_n|^2+\mu _n^{\frac{6}{5}}=0. \end{aligned}$$

Define the polynomial $q_n:\mathbb {R}\rightarrow \mathbb {R}$ as

$$\begin{aligned} q_n(x)=x^{12}-b\mu _n^{\frac{4}{5}}x^{10}-\mu _n^{\frac{2}{5}}x^{8}+(2b\mu _n^{\frac{6}{5}}-a^2\mu _n^{\frac{4}{5}})x^6-\mu _n^{\frac{4}{5}}x^4-b\mu _n^{\frac{8}{5}}x^2+\mu _n^{\frac{6}{5}}. \end{aligned}$$

From

$$\begin{aligned} q_n(b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}})=b^{-6}\mu _n^{-\frac{12}{5}}-2b^{-4}\mu _n^{-\frac{6}{5}}-a^2b^{-3}\mu _n^{-\frac{2}{5}}-2b^{-2} \end{aligned}$$

we obtain that $q_n(b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}})<0$ for large n. It is easy to see that for any $\epsilon >0$ we have $q_n(b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}}-\epsilon )>0$ for large n. Thus, given $\epsilon >0$ we have

$$|w_n|=| \overline{w_n}|\in (b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}}-\epsilon ,b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}})$$

for large n, which completes the proof.

Lemma 1.16

Assume that $w_n$ is a root of polynomial (3.7) such that

$$| w_n| \rightarrow b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}}$$

as $n\rightarrow +\infty$. Then $\dfrac{Re(w_n)}{Im(w_n)}\rightarrow 0$ as $n\rightarrow +\infty$.

Proof

We will denote the polar form of $w_n$ as $r_ne^{i\theta _n}$. Remember that

$$r_n\rightarrow b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}}\ \ \text {as}\ \ n\rightarrow +\infty .$$

We just need to prove that $\theta _n\rightarrow \dfrac{\pi }{2}$ or $\theta _n\rightarrow \dfrac{3\pi }{2}.$

Since $w_n$ is a root of (3.7), we have

$$\begin{aligned} r_n^2(r_n^2cos(4\theta _n)+a\mu _n^{\frac{1}{5}}r_ncos(3\theta _n)+b\mu _n^{\frac{4}{5}}cos(2\theta _n))=-\mu _n^{\frac{2}{5}}. \end{aligned}$$

It is easy to see that a root from (3.7) must be different from zero. Thus

$$\begin{aligned} r_n^2cos(4\theta _n)+a\mu _n^{\frac{1}{5}}r_ncos(3\theta _n)+b\mu _n^{\frac{4}{5}}cos(2\theta _n)=-\mu _n^{\frac{2}{5}}r_n^{-2}. \end{aligned}$$

Since $-\mu _n^{\frac{2}{5}}r_n^{-2}\rightarrow -b\mu _n^{\frac{4}{5}}$ we conclude that

$$\begin{aligned} r_n^2cos(4\theta _n)+a\mu _n^{\frac{1}{5}}r_ncos(3\theta _n)+b\mu _n^{\frac{4}{5}}cos(2\theta _n)\rightarrow -b\mu _n^{\frac{4}{5}}. \end{aligned}$$

From the fact that $r_n\rightarrow b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}}$ it is easy to see that $\theta _n\rightarrow \dfrac{\pi }{2}$ or $\theta _n\rightarrow \dfrac{3\pi }{2}$ as $n\rightarrow +\infty$.

Corollary 1.17

There exist two roots of polynomial (3.7); namely $w_n$ and $\overline{w_n}$ such that $w_n\rightarrow b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}}i$ and $\overline{w_n}\rightarrow -b^{-\frac{1}{2}}\mu _n^{-\frac{1}{5}}i$ as $n\rightarrow +\infty$.

Proof

It follows immediately from Lemmas 3.7 and 3.8.

Corollary 1.18

If $a,b\ne 0$ and $c=0$ then the problem (1.1) can generate a strongly continuous semigroup. Furthermore, the semigroup cannot be of an analytic type on Y in the sense of [11].

Proof

From Corollary 3.9 and Vieta’s formulas, we have that

$$\begin{aligned} Re(z_n)\rightarrow -\dfrac{a}{2}\mu _n^{\frac{1}{5}}\ \text {as}\ n\rightarrow +\infty . \end{aligned}$$

This means that problem (1.1) can generate a strongly continuous semigroup. Once again from Vieta’s formulas we can obtain that

$$\begin{aligned} (Im(z_n))^2\rightarrow b\mu _n^{\frac{4}{5}}-\dfrac{a^2}{4}\mu _n^{\frac{2}{5}}\ \text {as}\ n\rightarrow +\infty , \end{aligned}$$

implying that

$$\begin{aligned} \dfrac{(Im(z_n))^2}{(Re(z_n))^2}\rightarrow +\infty \ \text {as}\ n\rightarrow +\infty . \end{aligned}$$

Therefore, problem (1.1) cannot have an analytic semigroup on Y as a solution operator.