Rough semi-uniform spaces and its image proximities

Surabhi Tiwari; Pankaj Kumar Singh; Surabhi Tiwari; Pankaj Kumar Singh

doi:10.3934/era.2020060

Electronic Research Archive

2020, Volume 28, Issue 2: 1095-1106. doi: 10.3934/era.2020060

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Special Issues

Rough semi-uniform spaces and its image proximities

Surabhi Tiwari ^,,
Pankaj Kumar Singh

Department of Mathematics, Motilal Nehru National Institute of Technology Allahabad, Prayagraj-211004, India

Received: 01 February 2020 Revised: 01 May 2020
Primary: 54A05, 54D35, 54D80, 54E15; Secondary: 54C60, 54E05, 54E17

In this paper, we introduce the concept of rough semi-uniform spaces as a supercategory of rough pseudometric spaces and approximation spaces. A completion of approximation spaces has been constructed using rough semi-uniform spaces. Applications of rough semi-uniform spaces in the construction of proximities of digital images is also discussed.

Keywords:

Citation: Surabhi Tiwari, Pankaj Kumar Singh. Rough semi-uniform spaces and its image proximities[J]. Electronic Research Archive, 2020, 28(2): 1095-1106. doi: 10.3934/era.2020060

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Abstract

1. Introduction and statement of the problem

Consider the pseudo-parabolic equation in the domain $\Omega = \{(x, t): 0 < x < l, \ \ t > 0 \}$ :

$\begin{equation} \frac{\partial u}{\partial t} = \frac{\partial^{2} }{\partial t \partial x}\bigg(k(x)\frac{\partial u}{\partial x}\bigg) + \frac{\partial}{\partial x}\bigg(k(x)\frac{\partial u}{\partial x}\bigg), \quad (x, t) \in\Omega, \end{equation}$

(1.1)

with boundary conditions

$\begin{equation} u(0, t) = \ \mu(t), \quad u(l, t) = 0, \quad t > 0, \end{equation}$

(1.2)

and initial condition

$\begin{equation} u(x, 0) = 0, \quad 0\leq x\leq l. \end{equation}$

(1.3)

Assume that the function $k(x)\in C^{2}\big([0, l]\big)$ satisfies the conditions

$k(x) > 0, \quad k'(x)\leq 0, \quad 0\leq x \leq l.$

The condition (1.2) means that there is a magnitude of output given by a measurable real-valued function $\mu(t)$ (See ^[1,2,3] for more information).

Definition 1. If function $\mu(t)\in W_{2}^{1}(\mathbb{R_{+}})$ satisfies the conditions $\mu(0) = 0$ and $|\mu(t)| \leq 1$ , we say that this function is an admissible control.

Problem B. For the given function $\theta(t)$ Problem B consists looking for the admissible control $\mu(t)$ such that the solution $u(x, t)$ of the initial-boundary problem (1.1)-(1.3) exists and for all $t\geq0$ satisfies the equation

$\begin{equation} \int\limits_0^l u(x, t)\, dx = \theta(t). \end{equation}$

(1.4)

One of the models is the theory of incompressible simple fluids with decaying memory, which can be described by equation (1) (see ^[1]). In ^[2], stability, uniqueness, and availability of solutions of some classical problems for the considered equation were studied (see also ^[4,5]). Point control problems for parabolic and pseudo-parabolic equations were considered. Some problems with distributed parameters impulse control problems for systems were studied in ^[3,6]. More recent results concerned with this problem were established in ^{[7,8,9,10,11,12,13,14,15]}. Detailed information on the problems of optimal control for distributed parameter systems is given in ^[16] and in the monographs ^{[17,18,19,20]}. General numerical optimization and optimal boundary control have been studied in a great number of publications such as ^[21]. The practical approaches to optimal control of the heat conduction equation are described in publications like ^[22].

Control problems for parabolic type equations are considered in works ^[13,14] and ^[15]. In this work, such control problems are considered for the pseudo-parabolic equation.

Consider the following eigenvalue problem

$\begin{equation} \frac{d}{d x}\bigg(k(x)\frac{d v_{k}(x)}{d x}\bigg) = -\lambda_{k}\, v_{k}(x), \quad 0 < x < l, \end{equation}$

(1.5)

with boundary condition

$\begin{equation} v_{k}(0) = v_{k}(l) = 0, \quad 0\leq x\leq l. \end{equation}$

(1.6)

It is well-know that this problem is self-adjoint in $L_{2}(\Omega)$ and there exists a sequence of eigenvalues $\{\lambda_{k}\}$ so that $0 < \lambda_{1}\leq\lambda_{2}\leq...\leq\lambda_{k}\rightarrow \infty, \ \ k\rightarrow \infty$ . The corresponding eigenfuction $v_{k}$ form a complete orthonormal system $\{v_{k}\}_{k \epsilon N}$ in $L_{2}(\Omega)$ and these function belong to $C(\bar{\Omega})$ , where $\bar{\Omega} = \Omega\cup\partial\Omega$ (see, ^[23,24]).

2. Main integral equation

Definition 2. By the solution of the problem (1.1)–(1.3) we understand the function $u(x, t)$ represented in the form

$\begin{equation} u(x, t) = \frac{l-x}{l}\mu(t) -v(x, t), \end{equation}$

(2.1)

where the function $v(x, t) \in C_{x, t}^{2, 1}(\Omega)\cap C(\bar{\Omega})$ , $v_{x} \in C(\bar{\Omega})$ is the solution to the problem:

$v_{t} = \frac{\partial^{2}}{\partial t\partial x}\bigg(k(x)\frac{\partial v}{\partial x}\bigg) + \frac{\partial}{\partial x}\bigg(k(x)\frac{\partial v}{\partial x}\bigg)+$

$+\frac{k'(x)}{l}\mu(t)+\frac{k'(x)}{l}\mu'(t)+\frac{l-x}{l}\mu'(t),$

with boundary conditions

$v(0, t) = 0, \quad v(l, t) = 0,$

and initial condition

$v(x, 0) = 0.$

Set

$\begin{equation} \beta_k = \big(\lambda_k a_k -b_k\big)\, \gamma_k, \end{equation}$

(2.2)

where

$\begin{equation} a_{k} = \int\limits_0^l \frac{l-x}{l}v_{k}(x)dx, \quad b_{k} = \int\limits_0^l \frac{k'(x)}{l}\, v_{k}(x)dx, \end{equation}$

(2.3)

and

$\begin{equation} \gamma_k = \int\limits_0^l v_{k}(x)dx. \end{equation}$

(2.4)

Consequently, we have

$\begin{equation} v(x, t) = \sum\limits_{k = 1}^\infty \frac{v_{k}(x)}{1+\lambda_{k}} \int\limits_0^t e^{-\mu_{k}\, (t-s)}\big(\mu'(s)\, a_{k}+\mu'(s)\, b_{k}+\mu(s)\, b_{k}\big)\, ds, \end{equation}$

(2.5)

where $a_k$ , $b_k$ defined by (2.3) and $\mu_{k} = \frac{\lambda_{k}}{1+\lambda_{k}}$ .

From (2.1) and (2.5) we get the solution of the problem (1.1)–(1.3) (see, ^[23,25]):

$u(x, t) = \frac{l-x}{l}\mu(t) -\sum\limits_{k = 1}^\infty \frac{v_{k}(x)}{1+\lambda_{k}} \int\limits_0^t e^{-\mu_{k}\, (t-s)}\big(\mu'(s)\, a_{k}+\mu'(s)\, b_{k}+\mu(s)\, b_{k}\big)\, ds.$

According to condition (1.4) and the solution of the problem (1.1)-(1.3), we may write

$\begin{array}{c} \theta = \int\limits_0^l u(x, t) dx = \mu(t) \int\limits_0^l \frac{l-x}{l}dx \\ -\sum\limits_{k = 1}^\infty \frac{1}{1+\lambda_{k}}\, \bigg( \int\limits_0^t e^{-\mu_{k}\, (t-s)}\big(\mu'(s)\, a_{k}+\mu'(s)\, b_{k}+\mu(s)\, b_{k}\big)\, ds\bigg)\int\limits_0^l v_{k}(x)dx \\ = \mu(t) \int\limits_0^l \frac{l-x}{l}dx-\sum\limits_{k = 1}^\infty \frac{b_{k}\, \gamma_{k}}{1+\lambda_{k}} \int\limits_0^t e^{-\mu_{k}(t-s)}\mu(s)ds \\ -\sum\limits_{k = 1}^\infty \frac{(a_{k}+b_{k})\, \gamma_{k}}{1+\lambda_{k}} \int\limits_0^t e^{-\mu_{k}(t-s)}\, \mu'(s)ds = \mu(t) \int\limits_0^l \frac{l-x}{l}dx \\ -\sum\limits_{k = 1}^\infty \frac{b_{k}\, \gamma_{k}}{1+\lambda_{k}} \int\limits_0^t e^{-\mu_{k}(t-s)}\mu(s)ds-\mu(t)\, \sum\limits_{k = 1}^\infty \frac{(a_{k}+b_{k})\, \gamma_{k}}{1+\lambda_{k}} \\ +\sum\limits_{k = 1}^\infty \frac{(a_{k}+b_{k})\, \lambda_{k}\, \gamma_{k}}{(1+\lambda_{k})^{2}} \int\limits_0^t e^{-\mu_{k}(t-s)}\, \mu(s)ds. \end{array}$

(2.6)

where $\gamma_{k}$ defined by (2.4).

Note that

$\begin{equation} \int\limits_0^l \frac{l-x}{l}dx = \int\limits_0^l \bigg( \sum\limits_{k = 1}^\infty a_k\, v_{k}(x)\bigg)dx = \sum\limits_{k = 1}^\infty a_{k}\, \gamma_{k}. \end{equation}$

(2.7)

Thus, from (2.6) and (2.7) we get

$\begin{equation} \theta(t) = \mu(t)\, \sum\limits_{k = 1}^\infty \frac{\beta_{k}}{1+\lambda_{k}}+ \sum\limits_{k = 1}^\infty \frac{\beta_{k}}{(1+\lambda_{k})^{2}} \int\limits_0^t e^{-\mu_{k}(t-s)}\, \mu(s)ds, \quad t > 0, \end{equation}$

(2.8)

where $\beta_{k}$ defined by (2.2).

Set

$\begin{equation} B(t) = \sum\limits_{k = 1}^\infty \frac{\beta_{k}}{(1+\lambda_{k})^{2}}\, e^{-\mu_{k}\, t}, \quad t > 0, \end{equation}$

(2.9)

and

$\delta = \sum\limits_{k = 1}^\infty \frac{\beta_{k}}{1+\lambda_{k}}.$

According to (2.8) and (2.9), we have the following integral equation

$\begin{equation} \delta\, \mu(t)+\int\limits_0^t B(t-s)\, \mu(s)ds = \theta(t), \quad t > 0. \end{equation}$

(2.10)

Proposition 1. For the cofficients $\{\beta_{k}\}_{k = 1}^{\infty}$ the estimate

$0\leq \beta_k \leq C, \quad k = 1, 2, ...$

is valid.

Proof. Step 1. Now we use (1.5) and (2.3). Then consider the following equality

$\lambda_{k}\, a_{k} = \int\limits_0^l \frac{l-x}{l}\lambda_{k}\, v_{k}(x)dx = - \int\limits_0^l \frac{l-x}{l}\frac{d}{d x}\bigg(k(x)\, \frac{d v_{k}(x)}{d x}\bigg)dx$

$= -\bigg( \frac{l-x}{l}k(x)v_{k}'(x)\bigg|_{x = 0}^{x = l}+ \frac{1}{l} \int\limits_0^l k(x)\, v_{k}'(x)dx\bigg) = k(0)v_{k}'(0)-\frac{1}{l}\int\limits_0^l k(x)\, v_{k}'(x)dx$

$= k(0)v_{k}'(0)-\frac{1}{l}\bigg(k(l)v_{k}(l)-k(0)v_{k}(0)\bigg)+\int\limits_0^l \frac{k'(x)}{l}\, v_{k}(x)dx$

$= k(0)v_{k}'(0)+b_{k}.$

Then we have

$\begin{equation} \lambda_{k}\, a_{k}-b_{k} = k(0)v_{k}'(0). \end{equation}$

(2.11)

Step 2. Now we integrate the Eq. (1.5) from $0$ to $x$

$k(x)v_k'(x)-k(0)v_k'(0) = -\lambda_k\int\limits_0^xv_k(\tau)d\tau,$

and according to $k(x) > 0, \ x\in[0, l]$ , we can write

$\begin{equation} v_k'(x)-\frac{1}{k(x)}k(0)v_k'(0) = -\frac{\lambda_k}{k(x)}\int\limits_0^x v_k(\tau)d\tau. \end{equation}$

(2.12)

Thus, we integrate the Eq. (2.12) from $0$ to $l$ . Then we have

$\begin{equation} v_k(l)-v_k(0)-k(0)v_k'(0)\int\limits_0^l\frac{dx}{k(x)} = -\lambda_k\int\limits_0^l\frac{1}{k(x)}\Big(\int\limits_0^x v_k(\tau)d\tau\Big)dx. \end{equation}$

(2.13)

From (1.6) and (2.13) we get

$k(0)v_k'(0)\int\limits_0^l\frac{dx}{k(x)} = \lambda_k\int\limits_0^l\frac{1}{k(x)}\Big(\int\limits_0^x v_k(\tau)d\tau\Big)dx.$

Then

$\begin{equation} k(0)v_k'(0) = \lambda_k\int\limits_0^l G(\tau)v_k(\tau)d\tau, \end{equation}$

(2.14)

where

$G(\tau) = \int\limits_\tau^l\frac{dx}{k(x)}\Big(\int\limits_0^l\frac{dx}{k(x)}\Big)^{-1}.$

According to $G(\tau) > 0$ and from (2.14) we have (see, ^[24])

$\begin{equation} v_k'(0)\int\limits_0^l v_k(\tau)d\tau \geq 0. \end{equation}$

(2.15)

Consequently, from (2.11) and (2.15) we get the following estimate

$\beta_{k} = (\lambda_{k}\, b_{k}-a_{k})\, \gamma_{k} = k(0)v_{k}'(0)\cdot \int\limits_0^l v_{k}(x)dx \geq0.$

Step 3. It is clear that if $k(x)\in C^1([0, l])$ , we may write the estimate (see, ^[24,26])

$\max\limits_{0\leq x\leq l}|v_k'(x)|\leq C\lambda_k^{1/2}.$

Therefore,

$\begin{equation} |v_k'(0)|\leq C\lambda_k^{1/2}, \quad |v_k'(l)|\leq C\lambda_k^{1/2}, \end{equation}$

(2.16)

Then from Eq. (1.5), we can write

$\begin{equation} k(l)v_k'(l)-k(0)v_k'(0) = -\lambda_k\int\limits_0^lv_k(x)dx = -\lambda_k\, \gamma_{k}. \end{equation}$

(2.17)

According to (2.16) and (2.17) we have the estimate

$|\gamma_k|\leq \Big|\frac{1}{\lambda_k}\Big(k(l)v_k'(l)-k(0)v_k'(0)\Big)\Big|\leq C\lambda_k^{-1/2}.$

Then

$\beta_k \leq k(0)\, \big| v_k'(0)\gamma_k\big| \leq C.$

Proposition 2. A function $B(t)$ is continuous on the half-line $t\geq 0$ .

Proof. Indeed, according to Proposition 1 and (2.9), we can write

$0 < B(t)\leq \text{const} \sum\limits_{k = 1}^\infty \frac{1}{(1+\lambda_{k})^{2}}.$

3. Main result

Denote by $W(M)$ the set of function $\theta\in W_{2}^{2}(-\infty, +\infty)$ , $\theta(t) = 0$ for $t\leq 0$ which satisfies the condition

$\|\theta\|_{W_{2}^{2}(R_{+})}\leq M.$

Theorem 1. There exists $M > 0$ such that for any function $\theta\in W(M)$ the solution $\mu(t)$ of the equation (2.10) exists, and satisfies condition

$|\mu(t)|\ \leq\ 1.$

We write integral equation (2.10)

$\delta\, \mu(t)+\int\limits_0^t B(t-s)\mu(s)ds = \theta(t), \quad t > 0.$

By definition of the Laplace transform we have

$\widetilde{\mu}(p) = \int\limits_0^\infty e^{-pt}\, \mu(t)\, dt.$

Applying the Laplace transform to the second kind Volterra integral equation (2.10) and taking into account the properties of the transform convolution we get

$\widetilde{\theta}(p) = \delta\, \widetilde{\mu}(p)+\widetilde{B}(p)\, \widetilde{\mu}(p).$

Consequently, we obtain

$\widetilde{\mu}(p) = \frac{\widetilde{\theta}(p)}{\delta+\widetilde{B}(p)}, \quad \text{where} \ \ p = a+i\xi, \quad a > 0,$

and

$\begin{equation} \mu(t) = \frac{1}{2\pi i}\int\limits_{a-i\xi}^{a+i\xi}\frac{\widetilde{\theta}(p)}{\delta+\widetilde{B}(p)}\, e^{pt}dp = \frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}\frac{\widetilde{\theta}(a+i\xi)}{\delta+\widetilde{B}(a+i\xi)}\, e^{(a+i\xi)t}d\xi. \end{equation}$

(3.1)

Then we can write

$\widetilde{B}(p) = \int\limits_0^\infty B(t)e^{-pt}\, dt = \sum\limits_{k = 1}^\infty \frac{\beta_{k}}{(1+\lambda_{k})^{2}}\int\limits_0^\infty e^{-(p+\mu_{k})t}\, dt = \sum\limits_{k = 1}^\infty \frac{\rho_{k}}{p+\mu_{k}},$

where $\rho_{k} = \frac{\beta_{k}}{(1+\lambda_{k})^{2}}\geq 0$ and

$\widetilde{B}(a+i\xi) = \sum\limits_{k = 1}^\infty \frac{\rho_{k}}{a+\mu_{k}+i\xi} = \sum\limits_{k = 1}^\infty \frac{\rho_{k}\, (a+\mu_{k})}{(a+\mu_{k})^{2}+\xi^{2}}-i\xi\, \sum\limits_{k = 1}^\infty \frac{\rho_{k}}{(a+\mu_{k})^{2}+\xi^{2}}.$

It is clear that

$(a+\mu_{k})^{2}+\xi^{2}\leq [(a+\mu_{k})^{2}+1](1+\xi^{2}),$

and we have the inequality

$\begin{equation} \frac{1}{(a+\mu_{k})^{2}+\xi^{2}} \geq \frac{1}{1+\xi^{2}}\, \frac{1}{(a+\mu_{k})^{2}+1}. \end{equation}$

(3.2)

Consequently, according to (3.2) we can obtain the estimates

$\begin{array}{c} | \text{Re}(\delta+\widetilde{B}(a+i\xi))| = \delta+\sum\limits_{k = 1}^\infty \frac{\rho_{k}\, (a+\mu_{k})}{(a+\mu_{k})^{2}+\xi^{2}} \\ \geq \frac{1}{1+\xi^{2}}\, \sum\limits_{k = 1}^\infty \frac{\rho_{k}\, (a+\mu_{k})}{(a+\mu_{k})^{2}+1} = \frac{C_{1a}}{1+\xi^{2}}, \end{array}$

(3.3)

and

$\begin{array}{c} | \text{Im}(\delta+\widetilde{B}(a+i\xi))| = |\xi|\sum\limits_{k = 1}^\infty \frac{\rho_{k}}{(a+\mu_{k})^{2}+\xi^{2}} \\ \geq \frac{|\xi|}{1+\xi^{2}}\, \sum\limits_{k = 1}^\infty \frac{\rho_{k}}{(a+\mu_{k})^{2}+1} = \frac{C_{2a}\, |\xi|}{1+\xi^{2}}, \end{array}$

(3.4)

where $C_{1a}$ , $C_{2a}$ as follows

$C_{1a} = \sum\limits_{k = 1}^\infty \frac{\rho_{k}\, (a+\mu_{k})}{(a+\mu_{k})^{2}+1}, \ \ C_{2a} = \sum\limits_{k = 1}^\infty \frac{\rho_{k}}{(a+\mu_{k})^{2}+1}.$

From (3.3) and (3.4), we have the estimate

$|\delta+\widetilde{B}(a+i\xi)|^{2} = | \text{Re}(\delta+\widetilde{B}(a+i\xi))|^{2}+| \text{Im}(\delta+\widetilde{B}(a+i\xi))|^{2}\geq \frac{ \text{min}(C_{1a}^{2}, C_{2a}^{2})}{1+\xi^{2}},$

and

$\begin{equation} |\delta+\widetilde{B}(a+i\xi)|\geq \frac{C_{a}}{\sqrt{1+\xi^{2}}}, \quad \text{where} \quad C_{a} = \text{min}(C_{1a}, C_{2a}). \end{equation}$

(3.5)

Then, by passing to the limit at $a\rightarrow 0$ from (3.1), we can obtain the equality

$\begin{equation} \mu(t) = \frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}\frac{\widetilde{\theta}(i\xi)}{\delta+\widetilde{B}(i\xi)}\, e^{i\xi t}d\xi. \end{equation}$

(3.6)

Lemma 1. Let $\theta(t)\in W(M)$ . Then for the image of the function $\theta(t)$ the following inequality

$\int\limits_{-\infty}^{+\infty} |\widetilde{\theta}(i\xi)|\sqrt{1+\xi^{2}}d\xi \leq C\, \| \theta\|_{W_{2}^{2}(R_{+})},$

is valid.

Proof. We use integration by parts in the integral representing the image of the given function $\theta(t)$

$\widetilde{\theta}(a+i\xi) = \int\limits_0^\infty e^{-(a+i\xi)t}\theta(t)\, dt = - \theta(t)\, \frac{e^{-(a+i\xi)t}}{a+i\xi}\bigg|_{t = 0}^{t = \infty}+\frac{1}{a+i\xi}\int\limits_0^\infty e^{-(a+i\xi)t}\, \theta'(t)\, dt.$

Then using the obtained inequality and multiplying by the corresponding coefficient we get

$(a+i\xi)\, \widetilde{\theta}(a+i\xi) = \int\limits_0^\infty e^{-(a+i\xi)t}\, \theta'(t)\, dt,$

and for $a\rightarrow 0$ we have

$i\xi\, \widetilde{\theta}(i\xi) = \int\limits_0^\infty e^{-i\xi t}\, \theta'(t)\, dt.$

Also, we can write the following equality

$(i\xi)^{2}\, \widetilde{\theta}(i\xi) = \int\limits_0^\infty e^{-i\xi t}\, \theta''(t)\, dt.$

Then we have

$\begin{equation} \int\limits_{-\infty}^{+\infty} |\widetilde{\theta}(i\xi)|^{2}(1+\xi^{2})^{2}d\xi\leq C_{1}\, \| \theta\|_{W_{2}^{2}(R_{+})}^{2}. \end{equation}$

(3.7)

Consequently, according to (3.7) we get the following estimate

$\int\limits_{-\infty}^{+\infty} |\widetilde{\theta}(i\xi)|\sqrt{1+\xi^{2}}d\xi = \int\limits_{-\infty}^{+\infty} \frac{|\widetilde{\theta}(i\xi)|(1+\xi^{2})}{\sqrt{1+\xi^{2}}}$

$\leq \bigg(\int\limits_{-\infty}^{+\infty} |\widetilde{\theta}(i\xi)|^{2}(1+\xi^{2})^{2}d\xi\bigg)^{1/2}\bigg(\int\limits_{-\infty}^{+\infty} \frac{1}{1+\xi^{2}}d\xi\bigg)^{1/2} \leq C\, \| \theta\|_{W_{2}^{2}(R_{+})}.$

Proof of the Theorem 1. We prove that $\mu\in W_2^1(\mathbb{R}_+)$ . Indeed, according to (3.5) and (3.6), we obtain

$\int\limits_{-\infty}^{+\infty}|\widetilde{\mu}(\xi)|^2 (1+|\xi|^2)\, d\xi\ = \ \int\limits_{-\infty}^{+\infty} \left|\frac{\widetilde{\theta}(i\xi)}{\delta+\widetilde{B}(i\xi)}\right|^2 (1+|\xi|^2)\, d\xi\$

$\leq C \int\limits_{-\infty}^{+\infty} |\widetilde{\theta}(i\xi)|^2 (1+|\xi|^2)^2\, d\xi\ = \ C\|\theta\|_{W_2^2(\mathbb{R})}^2.$

Further,

$|\mu(t) - \mu(s)|\ = \ \left|\int\limits_s^t \mu'(\tau)\, d\tau\right|\ \leq \ \|\mu'\|_{L_2} \sqrt{t-s}.$

Hence, $\mu\in \text{Lip}\, \alpha$ , where $\alpha = 1/2$ . Then from (3.5), (3.6) and (3.7), we have

$|\mu(t)|\leq \frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}\frac{|\widetilde{\theta}(i\xi)|}{|\delta+\widetilde{B}(i\xi)|}d\xi\leq \frac{1}{2\pi C_{0}}\int\limits_{-\infty}^{+\infty} |\widetilde{\theta}(i\xi)|\sqrt{1+\xi^{2}}d\xi$

$\leq \frac{C}{2\pi C_{0}}\| \theta\|_{W_{2}^{2}(R_{+})}\leq \frac{C\, M}{2\pi C_{0}} = 1,$

as $M$ we took

$M = \frac{2\pi C_{0}}{C}.$

4. Conclusion

An auxiliary boundary value problem for the pseudo-parabolic equation was considered. The restriction for the admissible control is given in the integral form. By the separation variables method, the desired problem was reduced to Volterra's integral equation. The last equation was solved by the Laplace transform method. Theorem on the existence of an admissible control is proved. Later, it is also interesting to consider this problem in the $n-$ dimensional domain. We assume that the methods used in the present problem can also be used in the $n-$ dimensional domain.

Use of AI tools declaration

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

Conflict of interest

The author declare there is no conflict of interest.

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