Decay rate of the solutions to the Bresse-Cattaneo system with distributed delay

1.   Introduction and preliminaries

In this paper, we are particularly interested in examining the pace at which the following problem solution degrades

where

under the initial

where the parameters $a, l, m, k_{0}, \gamma_{1}$ and $\beta$ are considered to be positive constants, the function $\theta$ stands for temperature gradient, the functions $\varphi, \psi$ and $\omega$ stand for the vertical displacements of the girder, the tilt angle of the linear filament substance and the longitudinal displacements, respectively, the integral embodies the dispersed delay terms with $\tau_{1}, \tau_{2} > 0$ being a time delay, and $\gamma_{2}$ is a $L^{\infty}$ function.

There are several consequences from the coupling of the Cattaneo law of heat conduction with various systems, which has been discussed by several writers. For instance, see (Bresse-Cattaneo) in ^[6,14], the Bresse concept and the Fourier law of heat conduction (Bresse-Fourier) have both been addressed in ^[13], Timoshenko system with historical data in ^[1,7,9] and Moore-Gibson-Thompson problem in ^[4]. The following papers are recommended to the reader for further information ^[2,3,5,8,16].

In the absence of distributed delay term. The researchers briefly looked into the decay rate of system (1.1) in ^[14], and they presented the results as follows:

● For $\alpha = 0$

● For $\alpha\neq 0$

where

The Bresse-Cattaneo system (1.1) optimality decay rates were subsequently displayed by the authors in ^[6]. Alternatively, they enhanced the approximations (1.3) and (1.4) obtained by incorporating new, extremely sensitive Lyapunov functionals.

● For $\alpha = 0$

● For $\alpha\neq 0$

and they demonstrated that the estimations (1.6) and (1.7) depending on the parameter $\delta$ and under the following supposition

are accurate. After a thorough examination of the concept of dispersed postponement, the following questions seem intuitive: What sort of phrase has systemic suppressive activities? How should one determine the complexities that will enable them to "predict" devaluation? Is the concept of amortization always useful? Could the inclusion of the dispersed delay term have somehow made this type of issue more difficult to solve? This work is an attempt to comprehend the Bresse-Cattaneo framework and the dispersed delay term. The distributed delay term that is shown below, notably in Fourier space, does not apply to the Bresse-Cattaneo with friction attenuation solutions if they are relatively simple.

We aim to demonstrate the decay properties of the solution using the energy method in the Fourier space for the problems (1.1) and (1.2) relying on all recent publications, particularly ^[6,14]. This is one of the first studies we are aware of that looks at the Bresse-Cattaneo system with the dispersed delay factor in the Fourier space.

The sections of this manuscript are as follows: Here, we apply our presumptions and preliminary findings to the major decay conclusion. We build the Lyapunov component and determine the estimation for the Fourier image in the subsequent portion by employing the energy approach in Fourier space.

First, as in ^[11], we introduce the new variable

then we get

Therefore, our problem is expressed as follows

where

using the initial data

Regarding the significance of the delay, we only presumptively determine that

(H1) $\gamma_{2}:[\tau_{1}, \tau_{2}]\rightarrow \mathbb{R}$ is a limited function considering

To support our primary finding, we require the Hausdorff-Young inequality in the following Lemma

Lemma 1.1. ^[10] There is a constant $C > 0$ such that, for each $k, \varsigma\geq 0, c > 0$, guarantees that the estimation given below is true $\forall t\geq0$:

Also, we recall Plancherel's theorem.

Theorem 1.1. (^[15] Plancherel theorem)

The integral of a function's squared modulus is equal to the integral of the squared modulus of its frequency spectrum. That is, if $f(x)$ is a function on the real line, and ${ {\widehat {f}}(\xi)}$ is its frequency spectrum, then

2.   Energy method and decay estimates

In this section, we obtain a degradation estimation for the Fourier image of the remedy to problems (1.9) and (1.10). This approach enables us to provide the decay rate of the solution in the energy space by utilizing Plancherel's theorem together with some integral estimations, such as Lemma (1.1). By utilizing the energy approach in Fourier space, we create the proper Lyapunov functionals for this problem. Ultimately, we substantiate our major finding.

2.1.   The energy method in the Fourier space

Allow us to now incorporate the control values to construct the Lyapunov functional in the Fourier space and for convenience

The system (1.9) then adopts the following structure

with the initial data

where

Hence, for ($\tau_{q}\neq0$) the problem (2.2) and (2.3) is written as

with $U = (v, u, z, y, \phi, \eta, \vartheta, {q}, \mathcal{Y})^{T}, \quad U_{0} = (v_{0}, u_{0}, z_{0}, y_{0}, \phi_{0}, \eta_{0}, \vartheta_{0}, {q_{0}}, f_{0})$ and

Now, we will state the well-posedness result of system (2.4).

Theorem 2.1. Suppose that (1.11). Let $U_0\in H^{s}(\mathbb{R}), s\in\mathbb{N}$ and $s\geq2$, then problem (2.4) has a unique solution U such that

For a complete proof and more information, see ^[12].

When we execute the Fourier transform to (2.4), the underneath respective problem arises:

where the solution $\widehat{U}(\xi, t) = (\widehat{v}, \widehat{u}, \widehat{z}, \widehat{y}, \widehat{\phi}, \widehat{\eta}, \widehat{\theta}, \widehat{q}, \widehat{\mathcal{Y}})^{T}(\xi, t)$ is given by

with

Hence, in order to prove the asymptotic behavior of the solution, it suffices to get a function $\rho(\xi)$ such that

where $C$ and $c$ positive constants. Thus, the behavior of the solution depends on a critical way on the behavior of the function $\rho(\xi)$.

To arrive at this result, we start with $(2.5)_{1}$ where it is rewritten as:

Lemma 2.1. Assume that (1.11) is accurate. Let $\widehat{U}(\xi, t)$ be the solution of (2.5). Then the energy functional $\widehat{E}(\xi, t)$ is thus given by

satisfies

where $C_{1} = \bigg(\gamma_{1}-\int_{\tau_{1}}^{\tau_{2}}\vert \gamma_{2}(s)\vert ds\bigg) > 0$.

Proof. Firstly, multiplying $(2.6)_{1, 2, 3, 4, 5, 6}$ by $\overline{\widehat{v}}, \overline{\widehat{u}}, \overline{\widehat{z}}, \overline{\widehat{y}}, \overline{\widehat{\phi}}$ and $\overline{\widehat{\eta}}$ respectively, and multiplying $(2.6)_{7, 8}$ by ${\widehat{\theta}}, \tau_{q}\overline{\widehat{q}}$, adding these equalities and taking the real part, we get

Secondly, multiplying the Eq $(2.6)_{9}$ by $\overline{\widehat{\mathcal{Y}}}\vert \gamma_{2}(s)\vert$, and integrating the findings with $(0, 1)\times (\tau_{1}, \tau_{2})$

as well as utilizing Young's inequality, we have

by substituting (2.10) and (2.11) into (2.9), we find

then, by (1.11), $\exists C_{1} = (\gamma_{1}-\int_{\tau_{1}}^{\tau_{2}}\vert \gamma_{2}(s)\vert ds) > 0$ so that

Hence, we obtain (2.7) ($\widehat{E}$ is a non-increasing function). □

We now require the following lemmas in order to accomplish our objectives.

Lemma 2.2. The functional

satisfies, for any $\varepsilon_{1} > 0$

Proof. Differentiating $\mathcal{D}_{1}$ and by using (2.6), we get

With the help of Young's inequality, we estimate the terms in the RHS of (2.15) and obtain for every $\varepsilon_{1}, \delta_{1} > 0$

By adding the aforementioned estimations (2.16) to (2.15) and setting $\delta_{1} = \frac{1}{2}$, we find (2.14). □

Lemma 2.3. The functional

where

and

satisfies, for any $\varepsilon_{2}, \varepsilon_{3}, \varepsilon_{4} > 0$

Proof. Firstly, differentiating $\mathcal{F}_{1}, \mathcal{F}_{2}, \mathcal{F}_{3}$ and $\mathcal{F}_{4}$ and by using (2.6), we get

and

Now, differentiating $\mathcal{D}_{2}$ and by exploiting (2.21)–(2.24), gives

where

and

By applying the Young's inequality to the terms on the RHS of (2.25), we obtain for any $\varepsilon_{2}, \varepsilon_{3}, \varepsilon_{4}, \delta_{2}, \delta_{3}, \delta_{4} > 0$

By letting $\delta_{2} = \frac{m^{2}\delta^{2}}{2}, \delta_{3} = \frac{m^{2}\delta^{2}l^{2}}{4}, \delta_{4} = \frac{m^{2}\delta^{2}l^{2}}{4\gamma_{1}}$, we obtain (2.20). □

Lemma 2.4. Assume that (1.8) holds. The functional

where

satisfies

(1) For $\alpha = 0$. Then,

(2) For $\alpha\neq 0$. Then,

Proof. Firstly, a direct differentiation of $\mathcal{F}_{5}, \mathcal{F}_{6}$ and by using (2.6), we get

and

Now, by differentiating $\mathcal{D}_{3}$ and exploiting (2.32), (2.33), (2.21), (2.22) and (2.24), we find

where

At this point, we distinguish two cases according to the values of $\alpha$:

$\mathbf{Case \ 1}. \ (\alpha = 0)$.

In this case, the last term in the RHS of (2.34) is zero. Then, by applying the Young's inequality we obtain for any $\delta_{i}, i = 5, ..., 9 > 0$

By letting $\delta_{5} = \frac{\tau_{q}m^{2}lk_{0}}{20}, \delta_{6} = \frac{\tau_{q}m^{2}lk_{0}}{4\gamma_{1}}, \delta_{7} = \frac{alk_{0}}{2}, \delta_{8} = \frac{alk_{0}}{4}$ and $\delta_{9} = \frac{\tau_{q}m^{2}lk_{0}}{6}$, we find (2.30).

$\mathbf{Case \ 2}. \ (\alpha \neq 0)$.

The last term of the RHS in this situation is estimated as follows in (2.34)

Similarly to (2.35) and by Young's inequality, we get

By letting $\delta_{5} = \frac{\tau_{q}m^{2}lk_{0}}{20}, \delta_{6} = \frac{\tau_{q}m^{2}lk_{0}}{4\gamma_{1}}, \delta_{7} = \frac{alk_{0}}{2}, \delta_{8} = \frac{alk_{0}}{4}$ and $\delta_{9} = \frac{\tau_{q}m^{2}lk_{0}}{8}$, we find (2.31). Lemma 5 has been successfully proved. □

Next, we have the following lemma.

Lemma 2.5. The functional

where

satisfies, for any $\varepsilon_{5}, \varepsilon_{6}, \varepsilon_{7} > 0.$

(1) For $\alpha = 0$. Then,

(2) For $\alpha\neq 0$. Then,

Proof. Firstly, differentiating $\mathcal{F}_{7}$ and by using (2.6), we get

Now, differentiating $\mathcal{D}_{4}$ and by exploiting (2.42), (2.21) and (2.22), gives

where

At this point, we distinguish two cases:

$\mathbf{Case \ 1}. \ (\alpha = 0)$.

In this instance, we obtain for any $\delta_{10}, \delta_{11}, \delta_{12} > 0$ and $\varepsilon_{5}, \varepsilon_{6} > 0$ by applying the Young's inequality to the elements on the RHS of (2.43).

by letting $\delta_{10} = \frac{a^{2}l^{3}m^{2}}{8}, \delta_{11} = \frac{a^{2}l^{3}m^{2}}{4\gamma_{1}}, \delta_{8} = \frac{m^{2}l}{2}$, we find (2.40).

$\mathbf{Case \ 2}. \ (\alpha = 0)$.

The last term of the RHS in this situation is estimated as follows in (2.43), for any $\varepsilon_{7} > 0$

Substituting the inequality (2.45) into the statement (2.43), we find (2.41). Lemma 2.5 has been successfully proved. □

After that, we have the following lemma.

Lemma 2.6. The functional

satisfies,

where $\zeta_{1} > 0$.

Proof. By differentiating $\mathcal{D}_{5}$, with respect to $t$ and we use $(2.6)_{9}$, we have

Using the fact that $\mathcal{Y}(\xi, 0, s, t) = \omega_{t}(\xi, t) = \eta$, and $e^{-s}\leq e^{-s\rho}\leq 1$, for all $0 < \rho < 1$, we obtain

□

We have $-e^{-s}\leq-e^{-\tau_{2}}$, for all $s\in [\tau_{1}, \tau_{2}]$. Since $-e^{-s}$ is an increasing function. Lastly, by setting $\zeta_{1} = e^{-\tau_{2}}$ and remembering (1.11), we obtain (2.46).

At this stage, we define the Lyapunov functionals

● For $\alpha = 0$:

● For $\alpha\neq 0$:

positive constants with values of $N, M, N_{i}, M_{i}, i = 1, ..., 5$ are to be carefully selected subsequently.

Lemma 2.7. There exist $\mu_{i} > 0, i = 1, ..., 6$ such that the functionals $\mathcal{K}_{1}(\xi, t)$ and $\mathcal{K}_{2}(\xi, t)$ given by (2.47) and (2.48) satisfies

● For $\alpha = 0$:

● For $\alpha\neq0$:

where

Proof. Firstly, by differentiating (2.47) and using (2.8), (2.14), (2.20), (2.30), (2.40) and (2.46), with the fact that $\frac{\xi^{2}}{1+\xi^{2}}\leq\min\{1, \xi^{2}\}$ and $\frac{1}{1+\xi^{2}}\leq1$, we find

By setting

we obtain

Now, we fixed $N_{3}$ and choosing $N_{2}$ large enough such that

then we select $N_{4}$ in a size big enough that

then we select $N_{1}, N_{5}$ in a size big enough that

Hence, we arrive at

where $\alpha_{1} = \frac{m^{2}l}{2}N_{4}, \alpha_{6} = \zeta_{1}N_{5}$.

Secondly, we have

Using Young's inequality, the fact that $\frac{\xi^{2}}{1+\xi^{2}}\leq\min\{1, \xi^{2}\}$ and $\frac{1}{1+\xi^{2}}\leq1$, we find

Hence, we get

Now, we select $N$ in a size big enough that

and using the estimations (2.7), (2.54) and (2.55), there is a positive constant $\kappa > 0$, for all $t > 0$ and for all $\xi \in \mathbb{R}$, we have

and

then

Furthermore, we derive the following for any positive constant $\mu_{3} = \frac{\lambda_{1}}{\mu_{2}} > 0$

where $\rho_{1}(\xi) = \frac{\xi^{2}}{(1+\xi^{2})}$, for some $\lambda_{1}, \mu_{i} > 0, i = 1, 2, 3$. The proof of the first result (2.49) is finished.

We demonstrate the second result similarly to the first proof. So we derive (2.48) and using (2.8), (2.14), (2.20), (2.31), (2.41) and (2.46), with the fact that $\frac{\xi^{2}}{1+\xi^{2}}\leq\min\{1, \xi^{2}\}$ and $\frac{1}{1+\xi^{2}}\leq1$, we get

By setting

We obtain

Now, we fixed $M_{3}$ and choosing $M_{2}$ large enough such that

then we select $M_{4}$ in a size big enough that

Likewise we select $M_{1}, M_{5}$ in a size big enough that

Hence, we arrive at

where $\kappa_{1} = \frac{m^{2}l}{2}M_{4}, \kappa_{6} = \zeta_{1}M_{5}$.

As an alternative, we have

Using Young's inequality, the fact that $\frac{\xi^{2}}{1+\xi^{2}}\leq\min\{1, \xi^{2}\}$ and $\frac{1}{1+\xi^{2}}\leq1$, we find

Hence, we get

Now we select $M$ in a size big enough that

and using the estimations (2.7), (2.62) and (2.63), there is a positive constant $\widehat{\kappa} > 0$, for all $t > 0$ and for all $\xi \in \mathbb{R}$, we have

and

then

Furthermore, we derive the following for any positive constant $\mu_{6} = \frac{\lambda_{2}}{\mu_{5}} > 0$

where $\rho_{2}(\xi) = \frac{\xi^{2}}{(1+\xi^{2})^{2}}$, for some $\lambda_{2}, \mu_{i} > 0, i = 4, 5, 6$. The proof of the second result (2.50) is finished.

□

The pointwise estimations of the functional $\widehat{E}(\xi, t)$ provided by the next proposition.

Proposition 2.1. Suppose (1.8) and (1.11) hold. Then, for every $t\geq0$ and $\xi\in\mathbb{R}$, positive constants $k_{1}, k_{2} > 0$ exists such that the energy functional provided by (2.7) meets the following conditions

where $\rho_{1}(\xi) = \dfrac{\xi^{2}}{(1+\xi^{2})}, \quad \rho_{2}(\xi) = \dfrac{\xi^{2}}{(1+\xi^{2})^{2}}$.

Proof. From $(2.49)_{2}$ and $(2.50)_{2}$, we have by integration over $(0, t)$

Hence, by according of $(2.49)_{1}$, $(2.50)_{1}$ and (2.69), (2.70), we establish (2.68). □

2.2.   Decay estimates

Now, we declare and support the following finding

Theorem 2.2. Suppose that $s$ be a nonnegative integer, and let $U_{0}\in H^{s}(\mathbb{R})\cap L^{1}(\mathbb{R})$. Consequently, for any $t\geq0$, the following decay estimates are satisfied by the solution $U$ to problems (2.2) and (2.3)

● When $\alpha = 0$

● When $\alpha\neq 0$

where $k$ and $\ell$ are positive integers that meet the equation $k+\ell\leq s$ and $C > 0$.

Proof. From (2.7), we have $\vert\widehat{U}(\xi, t)\vert^{2}\equiv\widehat{E}(\xi, t)$.

● By using the Plancherel theorem 1.1 and making use of $(2.68)_{1}$, If $\alpha = 0$, we can determine

We now determine $R_{1}, R_{2}$, the low-frequency component$\vert\xi\vert\leq1$, and the high-frequency component $\vert\xi\vert\geq1$, separately.

Firstly, we have $\rho_{1}(\xi)\geq\frac{1}{2}\xi^{2}$, for $\vert\xi\vert\leq1$. Then

Lemma 1.1 allows us to acquire

Secondly, we have $\rho_{1}(\xi)\geq\frac{1}{2}$, for $\vert\xi\vert\geq1$. Then

Substituting (2.75) and (2.77) into (2.73), we find (2.71).

● If $\alpha\neq 0$, similar to the first estimate, we apply the Plancherel theorem 1.1 and exploiting $(2.68)_{2}$, we find

Now, we determine $R_{3}, R_{4}$, the low-frequency component $\vert\xi\vert\leq1$ and the high-frequency component $\vert\xi\vert\geq1$ separately.

Firstly, we have $\rho_{2}(\xi)\geq\frac{1}{4}\xi^{2}$, for $\vert\xi\vert\leq1$. Then

by using Lemma 1.1, we obtain

Secondly, we have $\rho_{2}(\xi)\geq\frac{1}{4}\xi^{-2}$, for $\vert\xi\vert\geq1$. Then

Expoiting the inequality

we get that

Substituting (2.80) and (2.83) into (2.78), we find (2.72).

□

3.   Conclusions

The investigation of the generalized degradation assessment of Bresse-Cattaneo system integration about the distributed delay term is the goal of this research, which employs the energy technique in Fourier space.

The distinct process that emerges from the distributed delay, which determines the system's development of this feature in Fourier space, is what we are interested in with this present work.

The same strategy will be used in the same systems in the upcoming works, but we will use various types of memory because we anticipate getting comparable outcomes.

Acknowledgments

Researchers would like to thank the Deanship of Scientific Research, Qassim University for funding publication of this project.

Conflict of interest

The authors declare no conflicts of interest.