Stability results of a swelling porous-elastic system with two nonlinear variable exponent damping

1.   Introduction

Our aim for this work was to investigate the stability analysis for a swelling soil through the application of theory of the porous media. Precisely, we consider the following nonlinear swelling soil system:

where $\gamma, \beta \geq 0$ and the components $z$ and $u$ indicate the displacements of the fluid and the elastic solid material, respectively. The densities of each component are represented by the positive constant coefficients $\rho_u$ and $\rho_z$. The coefficients $a_2 \neq 0$, $a_1 > 0$, and $a_3 > 0$ are positive constants that meet some particular requirements. The variables $p(\cdot)$ and $q(\cdot)$ are exponent functions that satisfy additional requirements that will be stated later.

This problem was proposed for the first time by Iecsan ^[1] and simplified by Quintanilla ^[2], as follows:

where the functions $(P_1, G_1, F_1)$, in that order, stand for the partial tension, internal body forces, and external forces, respectively, that are operating on the displacement. For $(P_2, G_2, F_2)$, but in the case of acting on the elastic solid, the definition is analogous. The constitutive equations for partial tensions are also provided by

where the matrix $A$ has the positive definite property in the sense of $a_1 a_3 \geq a_2^2$. For more information about swelling soils, we refer the reader to ^[3,4,5,6,7]. Regarding the stability, Quintanilla ^[2] established an exponential decay for the system (1.2) where

and $\xi > 0$ is the gain feedback. By using the spectral approach, Wang and Guo ^[8] obtained the exponential stability result for the system (1.2) with

where $\gamma(x)$ is an internal viscous damping function with a positive mean. After that, Ramos et al. ^[9] proved that the system (1.2) with

is exponentially stable provided that the wave speeds of the system are equal. Regarding viscoelastic swelling systems, Al-Mahdi and Al-Gharabli ^[10] and Apalara ^[11] obtained general decay results for Systems (1.2) with

for different classes of the relaxation function $g$. Similarly, Youkana et al. ^[12] considered the system (1.2) with

and they came up with a general decay result without imposing the system's wave speed. Apalara et al. ^[13] established a general decay result for the system (1.2) with

without imposing the system's wave speed. The reader is referred to related research for other outcomes in porous elasticity systems, thermo-porous-elastic systems, Timoshenko systems, and other systems ^{[2,8,14,15,16,17,18,19,20,21,22,23,24,25,26,27]}.

Equations with varying exponents of nonlinearity have drawn increasing amounts of attention in recent years. The applications to the mathematical modeling of non-Newtonian fluids are what have sparked such strong interest. These fluids include electro rheological fluid, which can undergo significant changes in response to an external electromagnetic field. A number of factors, including density, temperature, saturation, electric field, and others, affect the variable exponent of nonlinearity. We cite ^[28,29] for further details on the electro-rheological fluids mathematical model. We briefly mention a few of the many references ^{[30,31,32,33,34,35,36]} that discuss the existence, blow-up, and stability of viscoelastic systems with variable exponents. Regarding swelling systems with variable exponents, Al-Mahdi et al. ^[37] proved that the system (1.1) (with $\beta = 0$) is exponentially and polynomially stable based on the range of the variable exponents. In the present work, we study the interaction between the two nonlinear dampings of variable exponent type in the system (1.1). We prove that one damping is enough to have exponential stability and two dampings do not improve the decay rates. In addition to the stability analysis, we present some numerical examples to illustrate the stability theory.

2.   Preliminary and assumptions

In this section, we take into account the following hypotheses:

● (A1): $p, q:[0, 1] \rightarrow [1, \infty)$ is a continuous function such that

where $1 < p_1 \leq p(x) \leq p_2 < \infty.$

where $1 < q_1 \leq q(x) \leq q_2 < \infty.$ Additionally, by satisfying the log-Hölder continuity condition that is, for any $\lambda$ with $0 < \lambda < 1,$ there exists a constant $\delta > 0$ such that,

● (A2): The coefficients denoted by $a_i, \; i = 1, ..., 3$ satisfy that $a_1 a_3 -a_2^2 > 0$.

Throughout the paper, $\Omega = [0, 1]$ and $\overline{c}$ is a positive constant that depends on the coefficients of the system (1.1).

Lemma 2.1. The energy of the problem (1.1) is defined by

and it satisfies the following

Proof. The proof of Eq (2.3) is straightforward by multiplying (1.1) by $z_t$ and $u_t$ respectively, integrating over the interval $(0, 1)$, using integration by parts, and performing some modifications. □

3.   The main results

In this section, we state our decay results in the following theorems:

Theorem 3.1. Assume that $($A1–A2$)$ hold and $1 < p_1, q_1 < 2$. Then, the energy functional (2.2) satisfies the for positive constants denoted by $C_i, \; i = 1, 2, 3$, and for any $t > 0$

where $\overline{p}_1 = \min\{p_1, q_1\}$.

Theorem 3.2. Assume that $($A1–A2$)$ hold and $p_1, q_1\geq2$. Then, the energy functional (2.2) satisfies the for some positive constants $\lambda_i, \sigma_i, \mu_i > 0, \; i = 1, 2, 3$ and for any $t\geq 0$

and

where $\overline{p}_2 = \min\{p_2, q_2\}$.

Theorem 3.3. Assume that $($A1–A2$)$ hold and $p_1\geq 2$ and $1 < q_1 < 2$. Then, the energy functional (2.2) satisfies the for some positive constants denoted by $\vartheta_i > 0, \; i = 1, ..., 6,$ and for any $t\geq 0$

4.   Technical lemmas

In this section, we establish several lemmas needed for the proofs of our main results.

Lemma 4.1. Assume that $($A1–A2$)$ hold. The functional

satisfies the for $p_1, q_1 \geq2$ and any $\varepsilon_1 > 0$

and for $1 < p_1, q_1 < 2$, the functional satisfies

where $\alpha_0 = a_1 a_3-a_2^2 > 0$ and $\overline{c} > 0$ depends on $a_1, a_2, a_3, \rho_u, \rho_z.$

Proof. By considering Eq (1.1) and integrating by parts, we obtain

Using Young's inequality, we get the for any $\varepsilon_1 > 0$

Applying Young's inequality with $\zeta(x) = \frac{p(x)}{p(x)-1}$ and $\zeta^{*}(x) = p(x)$ helps to estimate the last two terms in (4.4) as follows: For a.e $x\in \Omega$ and any $\delta_1 > 0$, we have

where

Hence,

Next, using Eqs (2.2) and (2.3), Poincaré's inequality and the embedding property, we get

where $c_e$ is the embedding constant,

and

Then, Eqs (4.6) and (4.7) yield

Similarly, we can have

By combining all estimates (4.4)–(4.10), and selecting $\delta_1 = \frac{\alpha_0}{4 a_3 c_1}$, it follows that $c_{\delta}(x)$ remains bounded; then, estimate (4.2) is established.

To prove Eq (4.3), we re-estimate the last two terms in Eq (4.4) as follows:

First, we set

Then, we have

We notice that on $\Omega_1$, we have

Therefore, by using Young's and Poincaré's inequalities, then (4.12) leads to

where

Next, we have the following for the case $p(x) \ge 2$

Therefore, we conclude that

Similarly, we can get

Selecting $\eta = \frac{\alpha}{8 a_3}$, that $c_{\delta}(x)$ remains bounded; and, then, combining Eqs (4.11)–(4.17), estimate (4.3) is established. □

Lemma 4.2. Assume that $($A1–A2$)$ hold. The functional

satisfies the for $p_1 \geq 2$ and any $\varepsilon_2, \delta_2 > 0$:

and for $1 < p_1 < 2$, the functional satisfies

where $c_1$ is defined in Eq (4.8).

Proof. Direct computations using Eq (1.1) give

Hence, Young's inequality and the same estimates for the last term in Eq (4.21) yield Eqs (4.19) and (4.20). □

Lemma 4.3. Assume that $($A1–A2$)$ hold. The functional

satisfies the for $p_1, q_1\geq2$ and any $\eta_1 > 0$:

and for $1 < p_1, q_1 < 2$, the functional satisfies

where $\overline{c} > 0$ depends on $a_1, a_2, a_3, \rho_u, \rho_z.$

Proof. By exploiting (1.1), we have

Using Young's inequality, we get

where $\overline{c} > 0$ depends on $a_1, a_2, a_3, \rho_u, \rho_z.$ To estimate the last two terms in (4.25), we apply Young's inequality with $\zeta(x) = \frac{p(x)}{p(x)-1}$ and $\zeta^{*}(x) = p(x)$. So, for a.e $x\in \Omega$ and any $\delta_3 > 0$, we have

where

Hence,

Using Eqs (2.2) and (2.3), Poincaré's inequality and the embedding property, we find that

where $c_e$ is the embedding constant,

and

Then, Eqs (4.27) and (4.28) yield

In the same way, we get

where $c_1, c_3$ have been defined in Eqs (4.8) and (4.29).

Combining all of the above estimates and selecting $\delta_3 = \frac{a_2^2 \rho_u}{4 c_3}$ and $\omega_3 = \frac{1}{ c_1}$ we arrive at Eq (4.23).

To prove Eq (4.24), we re-estimate the last two terms in Eq (4.25) as in the above calculations to obtain

and

Then, by selecting $\eta = \frac{a_2^2 \rho_u}{8 c_3 }$ and $\lambda = \frac{1}{2 c_1}$, estimate (4.24) is established. □

Lemma 4.4. Assume that $($A1–A2$)$ hold. The functional

satisfies the for some $\varepsilon > 0,$ and $q_1 \geq2$

and for $1 < q_1 < 2$,

Proof. Direct computations using Eq (1.1) yield

Estimates (4.35) and (4.36) can be established in a similar manner as for the above estimations. □

Lemma 4.5. Assume that $($A1–A2$)$ hold. If $p_1, q_1 \geq 2$, then

and

Proof. By recalling Eq (2.3), it is easy to establish Eq (4.38). To prove the first estimate in Eq (4.39), we set the following partitions

Using the Hölder and Young inequalities and Eq (2.2), we obtain the following for $\Omega _{1}$,

and for $\Omega _{2}$, we get

Combining Eqs (4.41) and (4.42), the first estimate in Eq (4.39) can be established; also, repeat the same steps to establish the second estimate in Eq (4.39). □

5.   Proofs of the main results

In this section, we prove our decay results in Theorems 3.1, 3.2 and 3.3.

5.1.   Proof of Theorem 3.1

Proof. To prove Theorem 3.1, let

where $\mu, \mu_1, \mu_2, \mu_3, \mu_4$ are positive constants to be properly chosen. By taking the derivative of the functional $\mathcal{L}$ and using all of the above estimates (4.2)–(4.35), we obtain

Choosing $\varepsilon_i = \mu_i, \; i = 1, 2,$ and $\delta_2 = \frac{1}{\mu_2}$, the above estimate becomes

First, we select $\mu_2$ such that

Then, we choose $\mu_3$ large enough such that

Next, we choose $\mu_1$ large enough such that

Now, we choose $\mu_4$ such that

Select $\varepsilon$ such that

After fixing $\mu_i$, where $i = 1, 2, 3, 4$, we select $\mu$ large enough such that

and $\mathcal{L}\sim E$. That is, we can find two positive constants $\alpha _{1}$ and $\alpha_{2}$ such that

On the other hand, Young's inequality and (2.2) allow us to obtain

Hence, estimate (5.2) becomes as follows for any $t\geq 0$ and some positive constant $\alpha_3$,

Then, from Eqs (5.3) and (5.4), we get the following for some positive constant $\alpha_4$,

Thanks to Eq (5.2), we get the following for any $t\geq 0$ and some positive constant $\alpha_5$,

Recalling Eq (2.3) and multiplying the above equation by $E^\alpha(t)$, where $\alpha > 0$, we obtain

● If $\gamma = 0$ and $\beta \neq 0$, then we have

Using Young's inequality with $\zeta = \frac{1}{q_1-1}$ and $\zeta^* = \frac{1}{2-q_1}$, for any $\varepsilon > 0$, we have

Taking $\alpha = \frac{2-q_1}{q_1-1} > 0$, we have

By taking $\varepsilon$ small enough Eq (5.9) becomes:

where $\mathcal{L}_1 = E^\alpha\mathcal{L}+cE\sim E$. Integrating (5.10) over $(0, t)$, we obtain

where $\alpha = \frac{2-q_1}{q_1-1}.$ Then the first estimate in Eq (3.1) is proved.

● If $\gamma \neq 0$ and $\beta = 0$, then we have

The proof of the second estimate in (3.1) is straightforward. obtained in a similar manner as for the above one.

● If $\gamma \neq 0$ and $\beta \neq 0$, then we have

Now, we discuss two cases:

Case 1: If $p_1 > q_1$, then

Since $E$ is non-increasing, then we get

Then, Eq (5.15) becomes

From Eq (5.16), we have

By taking $\varepsilon$ small enough Eq (5.17) becomes:

where $\mathcal{L}_1 = E^\alpha\mathcal{L}+cE\sim E$. Integrating (5.25) over $(0, t)$, we get

where $\alpha = \frac{2-p_1}{p_1-1}$.

Case 2: If $q_1 < p_1$, we will get

where $\alpha = \frac{2-q_1}{q_1-1} > 0$. So, by taking $\overline{p}_1 = \min\{p_1, q_1\}$, the proof of the last estimate in Eq (3.1) is completed.

□

5.2.   Proof of Theorem 3.2

Proof. To prove Theorem 3.2, we reformulate the integrals $\int_0^1 z_t^2 dx$ and $\int_0^1 u_t^2 dx$ in Eq (5.2) and recall that the integrals $\left(\int_0^1 \vert z_t \vert^{p(x)} dx \right)^{p_1-1} and \left(\int_0^1 \vert u_t \vert^{q(x)} dx \right)^{q_1-1}$ are not relevant in this situation; thus, we have

We shall prove the case that $\gamma, \beta \neq 0$ and the other cases will be straightforward by letting either $\gamma = 0$ or $\beta = 0$. Let us select $\mu_2 = 1$ and $\varepsilon \mu_4 = 1$. Then it is easy to select $\mu_3$ and then $\mu_1$; finally, we can select $\mu$ large enough such that estimate (5.21) becomes

and for two positive constants $\beta_{2}$ and $\beta_{3}$,

By recalling Poincaré's inequality and the energy functional defined in Eq (2.2), estimate Eq (5.22) becomes, for a positive constant $\beta_4$,

and thanks to Eq (5.23), we get the following for any $t\geq 0$

Here, we will discuss two cases:

Case Ⅰ: If $p_2 = q_2 = 2$, then by using Lemma 4.5, we have

This gives

where $\mathcal{L}_1 = (\mathcal{L}+c E) \sim E$. Integrating the last estimate over the interval $\left(0, t\right)$ and using the equivalence properties $\mathcal{L}_1, \mathcal{L} \sim E$, the proof of the last estimate in (3.4) is completed.

Case Ⅱ: If $p_2, q_2 > 2$, then by using Lemma 4.5, we have

Multiplying the last equation by $E^{\alpha}$ where $\alpha = \frac{p_2-2}{2} > 0$, we obtain

Applying Young's inequality twice, we obtain the following for $\varepsilon > 0$

We will discuss two cases:

Case A: If $p_2 < q_2$, we will have

Using the non-increasing property of $E$, we get

Taking $\varepsilon$ small enough, the above estimate becomes:

where $\mathcal{L}_2 = E^{\alpha} \mathcal{L}+cE \sim E$.

Integration over $(0, t)$, using $E \sim \mathcal{L}_2$ gives

where $\alpha = \frac{p_2-2}{2}$.

Case B: If $q_2 < p_2$, we will get

where $\alpha = \frac{q_2-2}{2}$. So, by taking $\overline{p}_2 = \min\{p_2, q_2\}$, the proof of the last estimate in Eq (3.3) is completed. □

5.3.   Proof of Theorem 3.3

The proof of this theorem can be obtained by repeating proofs similar to those in Theorem 3.1 and Theorem 3.2.

6.   Numerical Tests

In the numerical part of this paper, we computationally justify our theoretical results form Theorems 3.1, 3.2 and 3.3. We examine the suggested fourteen cases according to our theorems. For the spatial and temporal discretization of the system (1.6), we use a second-order finite difference method in time and space for the space-time domain $[0, L]\times \lbrack 0, T_{e}] = [0, 1]\times \lbrack 0, 10]$. Thereafter, we implement the conservative Lax-Wendroff scheme. Finally, we discuss the computational confirmation of our theoretical results. Moreover, we compare these fourteen tests accordingly. We would also like to mention that, for references for similar construction, we invite the readers to see ^{[36,37,38,39]}. According to the assumptions and conditions of our theorems, we chose to simulate the temporal evolution of the waves for the following tests:

From Theorem 3.1, we examine the following cases

● TEST 1:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1; {\gamma = 0; \beta = 1}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $p(x) = q(x) = 2-\dfrac{1}{1+x}$.

● TEST 2:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1;{ \gamma = 1; \beta = 0}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $p(x) = q(x) = 2-\dfrac{1}{1+x}$.

● TEST 3:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1; {\gamma = 1; \beta = 1}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $p(x) = q(x) = 2-\dfrac{1}{1+x}$.

From Theorem 3.2, we examine the following cases

● TEST 4:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1;{ \gamma = 0; \beta = 1}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $p(x) = 2+\dfrac{1}{1+x}$ and $q(x) = 2$.

● TEST 5:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1; {\gamma = 1; \beta = 0}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $q(x) = 2+\dfrac{1}{1+x}$ and $p(x) = 2$.

● TEST 6:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1; {\gamma = 1; \beta = 1}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $q(x) = p(x) = 2$.

● TEST 7:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1; {\gamma = 0; \beta = 1}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $p(x) = 2$ and $q(x) = 2+\dfrac{1}{1+x}$.

● TEST 8:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1;{ \gamma = 1; \beta = 0}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $q(x) = 2$ and $p(x) = 2+\dfrac{1}{1+x}$.

● TEST 9:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1; {\gamma = 1; \beta = 1}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $q(x) = p(x) = 2+\dfrac{1}{1+x}$.

From Theorem 3.3, we examine the following cases

● TEST 10:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1; {\gamma = 0; \beta = 1}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $p(x) = 2+\dfrac{1}{1+x}$ and $q(x) = 2-\dfrac{1}{1+x}$.

● TEST 11:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1;{ \gamma = 1; \beta = 0}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $q(x) = 2-\dfrac{1}{1+x}$ and $p(x) = 2$.

● TEST 12:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1; {\gamma = 1; \beta = 0}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $q(x) = 2-\dfrac{1}{1+x}$ and $p(x) = 2+\dfrac{1}{1+x}$.

● TEST 13:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1; {\gamma = 1; \beta = 1;} a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $q(x) = 2+\dfrac{1}{1+x}$ and $p(x) = 2$.

● TEST 14:

$({\rm{i}})$ $\rho_u = 1; \rho_z = 1; {\gamma = 1; \beta = 1}; a_1 = 2; a_2 = 0.5$; and $a_3 = 2$.

$({\rm{ii}})$ $q(x) = 2+\dfrac{1.1}{1+x}$ and $p(x) = 2+\dfrac{1}{1+x}$.

To ensure the numerical stability of the implemented numerical scheme, we chose to design our code to satisfy the spatiotemporal Courant-Friedrichs-Lewy (CFL) condition, given as $\Delta{t} < 0.5\Delta x$, where $\Delta t$ represents the time step and $\Delta x$ is the spatial step. The spatial interval $[0, 1]$ has been subdivided into $500$ subintervals, whereas the temporal interval $[0, T_{e}] = [0, 10]$ was deduced from the stability condition above. We ran our code for $20000$ time steps by using the following initial conditions:

In the first block of the numerical Tests 1–3, we examined the polynomial decay of the energy derived from $u$ and $z$. These results were proved in Theorem 3.1. Given the initial and boundary conditions 6.1 and the parameters mentioned above (see TEST 1–3, (ⅰ) and (ⅱ)), in Figures 1–3, we have plotted the energy function and the three cross sections at $x = 0.25, 0.5$ and at $0.75$, where the polynomial decay of the energy is clearly assured.

In the second block of the numerical Tests 4–6, we evaluated the polynomial decay of the energy. These results were proved in Theorem 3.2. Given the same initial and boundary conditions in 6.1 for TEST 4–6, (ⅰ) and (ⅱ), in Figures 4–6, we have plotted the energy function and the three cross sections at $x = 0.25, 0.5$ and $0.75$, where the exponential decay of the energy has been numerically proved.

In the third block of the numerical Tests 7–9, we examine again the polynomial decay of the energy. These results have been proven in the last three cases of Theorem 3.2. Given the same initial and boundary conditions in 6.1 and the parameters mentioned above (see TEST 7–9, (ⅰ) and (ⅱ)), in Figures 7–9, we have plotted the energy function and the three cross sections $x = 0.25, 0.5$, and $0.75$, where the polynomial decay of the energy has been numerically proved.

In the fourth block of the numerical Tests 10–14, we examined other cases leading to the polynomial decay of the energy. These results have been proven in the last three cases of Theorem 3.3. Under the same initial and boundary conditions in 6.1 and the parameters mentioned above (see TEST 10–14, (ⅰ) and (ⅱ)), in Figures 10–14, we have plotted the energy function and the three cross sections at $x = 0.25, 0.5$ and $0.75$, where the polynomial decay of the energy has been numerically proved.

Finally, it should be stressed that our numerical simulations show the energy decay that was proved in Theorems 3.1, 3.2 and 3.3. Obviously, in some cases the polynomial decay could be easily deduced from the exponential decay behavior of the energy. This result can be accepted, since the required and expected result is the polynomial one. We are pretty sure that for other choices of the initial solutions and a rigorous choice of the functional parameters, we could get a clear discrepancy between the energy functions reflecting the polynomial and exponential decays.

7.   Conclusion

In this study, we considered a swelling elastic system with two nonzero dampings of the variable exponent type. We discussed different cases and proved that the system is exponentially and polynomially stable, and that the stability results depend on the values of $p_1, p_2, q_1, q_2$. In addition, we conclude that the decay estimate is not necessarily improved if the system has two dampings.

Use of AI tools declaration

The authors declare that they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors thank King Fahd University of Petroleum and Minerals (KFUPM) and the University of Sharjah (RGs MASEP & BioInformatics FG) for their continuous supports.

Conflict of interest

The authors declare that there is no conflict of interest.