From the theorem of intermediate derivatives (see [8,9]) if \(u(x)\in W_{2}^{5} (R;H)\), then \(A^{5-j} \frac {d^{j} u (x)}{dx^{j}} \in L_{2} (R;H), j=\overline {1,5}\) and the following inequalities:

$$ \| A^{5-j} \frac{d^{j} u (x)}{dx^{j}} \|_{L_{2} (R;H)} \le c_{j} \| u \|_{W_{2}^{5} (R;H)}, j=1,2,3,4,5 $$

(2.1)

are correct.

Equation 1.1 has the following operator form: *Q**u*(*x*)≡*Q*_{0}*u*(*x*)+*Q*_{1}*u*(*x*)=*f*(*x*), where

\(Q_{0} = \left (-\frac {d^{2}}{dx^{2}} +A^{2}\right) \left (\frac {d}{dx} +A\right)^{3} \) and \(Q_{1} =\sum _{j=1}^{5}A_{s} \frac {d^{5-j}}{dx^{5-j}} \).

The following theorem provides the association between the norms of operators of intermediate derivatives and the solvability conditions of the problems (1.1) and (1.2).

###
**Theorem 1**

The operator *Q*_{0} isomorphically maps the space \(W_{2}^{5} (R;H)\) onto the space *L*_{2}(*R*;*H*), moreover, for *f*(*x*)∈*L*_{2}(*R*;*H*) and Eq. 1.1 has a solution

$$u(x)=\int_{-\infty}^{+\infty}G (x-s)f (s)ds +u_{0} (x),$$

wher

$$G (x-s)=2^{-4} \left\{\begin{array}{l} (E+2A (x-s)+2A^{2} (x-s)^{2} +\\ +\frac{8}{6} A^{3} (x-s)^{3}) e^{-A (x-s)} A^{-4},\qquad\qquad\qquad\qquad\quad if\, \, x>s,\\ {e^{A (x-s)} A^{-4},\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\quad if\, \, x< s,} \end{array}\right.$$

$$\begin{array}{*{20}l} u_{0} (x)&=-2^{-4} \left(\left(E+2Ax+2A^{2} x^{2} +\frac{4}{3} A^{3} x^{3}\right)\int_{0}^{\infty}e^{-A (x+s)} A^{-4} f (s)ds +\right.\\ &+ \left(E-2As+2A^{2} s^{2} -\frac{4}{3} A^{3} s^{3}\right)\int_{-\infty}^{0}e^{-A (x-s)} A^{-4} f (s)ds+\\ &+2Ax \left(E-2As+2A^{2} s^{2}\right)\int_{-\infty}^{0}e^{-A (x-s)} A^{-4} f (s)ds+\\ &+2A^{2} x^{2} (E-2As)\int_{-\infty}^{0}e^{-A (x-s)} A^{-4} f (s)ds+\\ &\left.+\frac{4}{3} A^{3} x^{3} \int_{-\infty}^{0}e^{-A (x-s)} A^{-4} f (s)ds\right). \end{array} $$

###
*Proof*

Let Eq. 1.1 has a solution *u*(*x*)=*u*_{1}(*x*)+*u*_{0}(*x*), wher

$$u_{1} (x)=\int_{-\infty}^{+\infty}G (x-s)f (s)ds,$$

and

$$\begin{array}{*{20}l}u_{0} (x)&=\phi_{0} e^{-Ax}+\phi_{1}Ax e^{-Ax}+\phi_{2} A^{2}x^{2} e^{-Ax}+\phi_{3} A^{3}x^{3} e^{-Ax},\\ \phi_{0}&=-u_{1} (0),\\ \phi_{1}&=\phi_{0}-A^{-1}u_{1}^{\prime} (0),\\ \phi_{2}&=\phi_{1}-\frac{1}{2}\phi_{0}-\frac{1}{2} A^{-2}u_{1}^{\prime\prime} (0),\\ \phi_{3}&=\frac{1}{6}\phi_{0}-\frac{1}{6}A^{-3}u_{1}^{\prime\prime\prime} (0)-\frac{1}{2}\phi_{1}+\phi_{2},\end{array} $$

First, we find Green’s function of Eq. 1.1 where, *A*_{j}=0, *j*=1,2,3,4,5.

The operator *Q*_{0} can be simplified on the following form:

$$Q_{0}= \left(-\frac{d}{dx} +A\right) \left(\frac{d}{dx} +A\right)^{4}, $$

then applying Fourier transform to the equation *Q*_{0}*u*(*x*)=*f*(*x*), we obtain:

$$(-i\xi E+A) (i\xi E+A)^{4} \overset{\sim}{u} (\xi)=\overset{\sim}{f} (\xi).$$

where *E*- identity operator and \(\overset {\sim }{u} (\xi), \overset {\sim }{f} (\xi)\) are Fourier transforms to the functions *u*(*x*), *f*(*x*), respectively.

Thus, the polynomial operator pencil (*i**ξ**E*−*A*)(*i**ξ**E*+*A*)^{4} is invertible, and moreover,

$$ \overset{\sim}{u} (\xi)= \frac{1}{ (-i\xi E+A)(i\xi E+A)^{4}} \overset{\sim}{f} (\xi), $$

(2.2)

hence,

$$u (x)=\frac{1}{2\pi} \int_{-\infty}^{+\infty} (-i\xi E+A)^{-1} (i\xi E+A)^{-4} \overset{\sim}{f} (\xi) e^{i\zeta (t-s)} d\zeta.$$

Using Cauchy’s theorem of residues,

at *x*<*s*

$$Res= \underset{i\xi\to A}{\lim}\ \frac{e^{i\zeta (t-s)}}{(i\xi E+A)^{4}} = 2^{-4}e^{A (x-s)} A^{-4},$$

at *x*>*s*

$$\begin{array}{*{20}l}Res&=\frac{1}{6}\ \underset{i\xi\to -A}{\lim}\ \frac{d^{3}}{d\xi^{3}} \left(\frac{e^{i\zeta (t-s)}}{ -i\xi E+A}\right)=\\ &=\left(E+2A (x-s)+2A^{2} (x-s)^{2} +\frac{8}{6} A^{3} (x-s)^{3}\right) e^{-A (x-s)} A^{-4}.\end{array} $$

Then from inequality (2.1), it is simple to prove that *Q*_{0} which acts from \(W_{2}^{5} (R;H)\) to *L*_{2}(*R*;*H*) is bounded (see [10]).

Now, we show that \(u (x)\in W_{2}^{5} (R;H)\).

Using the Parseval’s equality and (2.2), we obtain:

$$\| u\|_{W_{2}^{5} (R;H)}^{2} = \| \frac{d^{5} u}{dx^{5}} \|_{L_{2} (R;H)}^{2} + \| A^{5} u\|_{L_{2} (R;H)}^{2} =$$

$$= \| {i\zeta}^{5} \overset{\sim}{u (\xi)} \|_{L_{2} (R;H)}^{2} + \| {A}^{5} \overset{\sim}{u (\xi)} \|_{L_{2} (R;H)}^{2} =$$

$$= \| {i\zeta}^{5} (-i\xi E+A)^{-1} (i\xi E+A)^{-4} \overset{\sim}{f} (\xi) \|^{2}_{L_{2} (R;H)}+ $$

$$+ \| {A}^{5} (-i\xi E+A)^{-1} (i\xi E+A)^{-4} \overset{\sim}{f} (\xi) \|^{2}_{L_{2} (R;H)} \le $$

$$\le {\underset{\zeta \in R}{\sup} \| i\zeta^{5} (-i\zeta {\text{E}} +A)^{-1} (i\zeta {\text{E}} +A)^{-4} \|_{H\to H}^{2} \| \overset{\sim}{f} (\zeta) \|_{L_{2} (R;H)}^{2}}+ $$

$$ + {\underset{\zeta \in R}{\sup} \| A^{5} (-i\zeta {\text{E}} +A)^{-1} (i\zeta {\text{E}} +A)^{-4} \|_{H\to H}^{2} \| \overset{\sim}{f} (\zeta) \|_{L_{2} (R;H)}^{2}}. $$

(2.3)

From the spectral decomposition of the operator A (*σ*(A)– the spectrum of operator A) for *ζ**ε* R (see [11]), we have:

$${\| i\zeta^{5} (-i\zeta {\text{E}} +A)^{-1} (i\zeta {\text{E}} +A)^{-4} \|_{H\to H} =_{\sigma \in \sigma (A)}^{\sup} |i\zeta^{5} (-i\zeta +\sigma)^{-1} (i\zeta +\sigma)^{-4} |\le}$$

$$ \le \underset{\sigma \in \sigma (A)}{\sup} \frac{|\zeta |^{5}}{\left(\zeta^{2} +\sigma^{2}\right)^{\frac{5}{2}}} \le 1, $$

(2.4)

$$\| A^{5} (-i\zeta {\text{E}} +A)^{-1} (i\zeta {\text{E}} +A)^{-4} \|_{H\to H} =\underset{\sigma \in \sigma (A)}{\sup} |{\sigma}^{5} (-i\zeta {\text{E}} +\sigma)^{-1} (i\zeta {\text{E}} +\sigma)^{-4} |\le$$

$$ \le \underset{\sigma \in \sigma (A)}{\sup} \frac{{\sigma}^{5}}{{(\zeta^{2} +\sigma^{2})^{^{\frac{5}{2}}}}} \le 1. $$

(2.5)

From (2.4) and (2.5) into (2.3), we obtain:

$$\|{u}\|_{W_{2}^{5} (R;H)}^{2} \le 2 \| \overset{\sim}{f} (\zeta) \|_{L_{2} (R;H)}^{2} =2 \| f (x) \|_{L_{2} (R;H)}^{2}.$$

Hence, \(u (x)\in W_{2}^{5} (R;H).\)

Using the Banach theorem of the inverse operator, then the operator *Q*_{0} is an isomorphism from \(W_{2}^{5} (R;H)\) to *L*_{2}(*R*;*H*) (see [12]). □

Before we formulate exact conditions on regular solution of the problems (1.1) and (1.2), expressed only by its operator coefficients, we must estimate the norms of intermediate derivative operators participating in the perturbed part of the given equation. Theorem 1 leads to the norm \(\phantom {\dot {i}\!} \| Q_{0} u \|_{L_{2} (R;H)} \) is equivalent to the norm \( \| u \|_{W_{2}^{5} (R;H)} \) in the space \(W_{2}^{5} (R;H)\). Therefore by the norm \(\phantom {\dot {i}\!} \| Q_{0} u \|_{L_{2} (R;H)} \), the theorem on intermediate derivatives is valid as well.

###
**Theorem 2**

When the function \(u (x)\in W_{2}^{5} (R;H)\), so it keeps the following inequalities:

$$ \| A^{j} \frac{d^{5-j} u (x)}{dx^{5-j}} \|_{L_{2} (R;H)} \le b_{j} \| Q_{0} u \|_{L_{2} (R;H)}, j=\overline{1,5} $$

(2.6)

true, where \(b_{1} =b_{4} =\frac {16}{25\sqrt {5}}, b_{2} =b_{3} =\frac {6\sqrt {3}}{25\sqrt {5}}, b_{5}=1\) see [13].

###
*Proof*

To establish the validity of inequalities (2.6), we take *Q*_{0}*u*(*x*)=*f*(*x*) and apply the Fourier transformation as follows:

$$\| A^{j} (i\zeta)^{5-j} (-i\zeta {\text{E}} +A)^{-1} (i\zeta {\text{E}} +A)^{-4} \overset{\sim}{f} (\zeta) \|_{L_{2} (R;H)} \le$$

$$\le \underset{\zeta \in R}{\sup}\ \| A^{j} (i\zeta)^{5-j} (-i\zeta {\text{E}} +A)^{-1} (i\zeta {\text{E}} +A)^{-4} \|_{H\to H} \| \overset{\sim}{f} (\zeta) \|_{L_{2} (R;H)},$$

For *ζ*∈*R*, we have:

$$\| A^{j} (i\zeta)^{5-j} (-i\zeta {\text{E}} +A)^{-1} (i\zeta {\text{E}} +A)^{-4} \|_{H\to H} \le$$

$$\le \underset{\sigma \in \sigma (A)}{\sup}\ |\sigma^{j} (i\zeta)^{5-j} (-i\zeta {\text{E}} +\sigma)^{-1} (i\zeta {\text{E}} +\sigma)^{-4} |=$$

$$= \underset{\eta =\frac{\xi^{2}}{\sigma^{2}} \ge 0}{\sup}\ \frac{\eta^{\left({(5-j)_{/2}}\right)}}{ (\eta +1)^{\frac{5}{2}}}=b_{j} \, \,, j=1,2,3,4,5.$$

Using inequalities (2.7), we have:

$$\| A^{j} (i\zeta)^{5-j} (-i\zeta {\text{E}} +A)^{-1} (i\zeta {\text{E}} +A)^{-4} \overset{\sim}{f (\zeta)} \|_{L_{2} (R;H)} \le$$

$$\le b_{j} \| \overset{\sim}{f (\xi)} \|_{L_{2} (R;H)}, j=1,2,3,4,5.$$

□

###
**Lemma 1**

The operator *Q*_{1} continuously acts from \(W_{2}^{5} (R;H)\) to *L*_{2}(*R*;*H*) provided that the operators *A*_{j}*A*^{−j},*j*=1,2,3,4,5 are bounded in *H*.

Considering the results found up to now see [14], for problems (1.1) and (1.2), we get the possibility to establish regular solvability conditions.

###
**Theorem 3**

Let |*κ*|<2*λ*_{0}(*A*=*A*∗≥ *λ*_{o}*E*,*λ*_{o}>0) for any \(u (t)\in W_{2}^{5} (R;H)\), then holds the inequality

*b*(*k*)(*c*_{1}(*k*)∥*A*_{1}*A*^{−1}∥_{H→H}+*c*_{2}(*k*)∥*A*_{2}*A*^{−2}∥_{H→H}+*c*_{3}(*k*)∥*A*_{3}*A*^{−3}∥_{H→H}++*c*_{4}(*k*)∥*A*_{4}*A*^{−4}∥_{H→H})<1 see [15],

where

$$\begin{array}{l} {c_{1} (k)= \left[1+\frac{4\lambda_{0} |\lambda_{0} +k |}{ (2\lambda_{0} +k)^{2}}\right]^{\frac{3}{2}}}, \, c_{2} (k)=\frac{2\lambda_{0}}{2\lambda_{0} +k} \left[1+\frac{4\lambda_{0} |\lambda_{0} +k |}{ (2\lambda_{0} +k)^{2}} \right], \\ c_{3} (k)=\frac{4\lambda_{0}^{2}}{ (2\lambda_{0} +k)^{2}} \left[1+\frac{4\lambda_{0} |\lambda_{0} +k |}{ (2\lambda_{0} +k)^{2}}\right]^{\frac{1}{2}},\, c_{4} (k)=\frac{8\lambda_{0}^{3}}{(2\lambda_{0} +k)^{3}}, \end{array}$$

and

$$b (k)= \Bigg\{\begin{array}{l} {\frac{\lambda_{0}}{2^{^{\frac{1}{2}}} \left(2\lambda_{0}^{2} -k^{2}\right)^{\frac{1}{2}}},\, \, \,\, \, \,\, if \, \, \, 0\le \frac{k^{2}}{4\lambda_{0}^{2}} < \frac{1}{3},\,} \\ {\, \, \, \, \frac{2\lambda_{0} |k |}{4\lambda_{0}^{2} -k^{2}},\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, if\, \, \, \frac{1}{3} \le \frac{k^{2}}{4\lambda_{0}^{2}} < 1.} \end{array}$$

###
**Theorem 4**

Suppose that the operators \(A_{j} A^{-j}, j=\overline {1,4}\), be bounded in *H* and they hold the inequalit

$$\sum_{j=1}^{4}C_{j}(k)b(k) \| A_{j} A^{-j} \|_{H\to H} < 1,$$

where the numbers *C*_{j}(*k*),*j*=1,2,3,4 and *b*(*k*) are determined in Theorem 3 so the problems (1.1) and (1.2) are regularly solvable (see [16,17]).

###
*Proof*

where \(f(x)\in L_{2} (R;H), u(x)\in W_{2}^{5} (R;H)\) and by Theorem (1.1), there exists a bounded inverse operator to *Q*_{0}, which acts from *L*_{2}(*R*;*H*) to \(W_{2}^{5} (R;H)\); then after replacing *Q*_{0}*u*(*x*)=*w*(*x*) in Eq. 1.1, it can be written as \(\left (E+{Q}_{1} Q_{0}^{-1}\right)w(x)=f(x).\)

Now, we prove under the theorem conditions see [18], that

$$\| {Q}_{1} {Q}_{0}^{-1} \|_{L_{2} (R;H)\to L_{2} (R;H)} <1.$$

By Theorem 3, we have:

$$\begin{array}{l} { \| Q_{1} Q_{0}^{-1} w \|_{L_{2} (R;H)} = \| Q_{1} u \|_{L_{2} (R;H)} \le \sum_{j=1}^{4} \| A_{j} \frac{{d}^{4-j} u}{{dx}^{4-j}} \|_{L_{2} (R;H)} \le} \\ {\le \sum_{j=1}^{4} \| A_{j} A^{-j} \|_{H\to H} \| A^{j} \frac{{d}^{4-j} u}{{dx}^{4-j}} \|_{L_{2} (R;H)} \le} \end{array}$$

$$\begin{array}{l} {\le \sum_{j=1}^{4}C_{j}(k)b(k) \| A_{j} {A}^{-j} \|_{H\to H} \| Q_{0} u \|_{L_{2} (R;H)} =} \\ {=\sum_{j=1}^{4}C_{j}(k)b(k) \| A_{j} {A}^{-j} \|_{H\to H} \| v \|_{L_{2} (R;H)}.} \end{array}$$

Consequently,

\( \| Q_{1} Q_{0}^{-1} \|_{L_{2} (R;H)\to L_{2} (R;H)} \le \sum _{j=1}^{4}C_{j}(k) b(k) \| A_{j} {A}^{-j} \|_{H\to H} < 1. \) Thus, the operator \(E+Q_{1} Q_{0}^{-1} \) is invertible in *L*_{2}(*R*;*H*); therefore, *u*(*x*) can be determined by \({u (x)=Q}_{0}^{-1} \left (E+Q_{1} Q_{0}^{-1}\right)^{-1} f (x)\); moreover:

$$\| u \|_{W_{2}^{5} (R;H)}\! \le\! \| {Q}_{0}^{-1} \|_{L_{2} (R;H)\to W_{2}^{5} (R;H)} \times \| \left(\left(E+Q_{1} Q_{0}^{-1}\right)\right)^{-1} \|_{L_{2} (R;H)\to L_{2} (R;H)} \| f \|_{L_{2} (R;H)}$$

$$\le const \| f \|_{L_{2} (R;H)}$$

□