Semi-Hyers–Ulam–Rassias Stability of the Convection Partial Differential Equation via Laplace Transform

Abstract

In this paper, we study the semi-Hyers–Ulam–Rassias stability and the generalized semi-Hyers–Ulam–Rassias stability of some partial differential equations using Laplace transform. One of them is the convection partial differential equation.

1. Introduction

It is well known that the study of Ulam stability began in 1940, with a problem posed by Ulam concerning the stability of homomorphisms [1]. In 1941, Hyers [2] gave a partial answer in the case of the additive Cauchy equation in Banach spaces.

After that, Obloza [3] and Alsina and Ger [4] began the study of the Hyers–Ulam stability of differential equations. The field continued to develop rapidly. Linear differential equations were studied in [5,6,7], integral equations in [8], delay differential equations in [9], linear difference equations in [10,11], other equations in [12], and systems of differential equations in [13]. A summary of these results can be found in [14].

The Hyers–Ulam stability of linear differential equations was studied using the Laplace transform by H. Rezaei, S. M. Jung, and Th. M. Rassias [15], and by Q. H. Alqifiary and S. M. Jung [16]. This method was also used in [17,18,19].

The study of the stability of partial differential equations began in 2003, with the paper [20] of A. Prastaro and Th.M. Rassias. The Ulam–Hyers stability of partial differential equations was also studied in [21,22,23,24,25,26].

In [27], M. N. Qarawani used the Laplace transform to establish the Hyers–Ulam–Rassias–Gavruta stability of initial-boundary value problem for heat equations on a finite rod:

$$\frac{\partial u}{\partial t}=a^2\frac{{\partial }^{2}u}{\partial x^2},t>0,0<x<l.$$

In [28], D.O. Deborah and A. Moyosola studied nonlinear, nonhomogeneous partial differential equations using the Laplace differential transform method:

$$\frac{d^{2}w\left(x,t\right)}{d{t}^2}+a_n\left(x\right)Rw\left(x,t\right)+b_n\left(x\right)Sw\left(x,t\right)=f\left(x,t\right),t>0,x>0,n\in \mathbb{N},$$

where $a_n\left(x\right),b_n\left(x\right)$ are variable coefficients, $n\in \mathbb{N},R$ is the linear operator, S is the nonlinear operator, and $f\left(x,t\right)$ is the source function.

In [29], E. Bicer used the Sumudu transform to study the equation:

$$y_t-k{y}_{xx}=0,k\mathrm{a}\mathrm{positive}\mathrm{real}\mathrm{constant},\left(x,t\right)\in D,D=\left(x_0,x\right]\times \left(0,\infty \right).$$

In [30], the Poisson partial differential equation

$$u_{xx}\left(x,y\right)+u_{yy}\left(x,y\right)=g\left(x,y\right)$$

is studied via the double Laplace transform method (DLTM).

In the following sections, we will study the semi-Hyers–Ulam–Rassias stability and the generalized semi-Hyers–Ulam–Rassias stability of some partial differential equations using Laplace transform. One of them is the convection partial differential equation:

$$\frac{\partial y}{\partial t}+a\frac{\partial y}{\partial x}=0,a>0,x>0,t>0,y\left(0,t\right)=c,y\left(x,0\right)=0.$$

(1)

A physical interpretation [31] of these equations is a river of solid goo, since we do not want anything to diffuse. The function $y=y(x,t)$ is the concentration of some toxic substance. The variable x denotes the position where $x=0$ is the location of a factory spewing the toxic substance into the river. The toxic substance flows into the river so that at $x=0$, the concentration is always C. We also study the semi-Hyers–Ulam–Rassias stability of the following equation:

$$\frac{\partial y}{\partial t}+\frac{\partial y}{\partial x}-x=0,x>0,t>0,y\left(0,t\right)=0,y\left(x,0\right)=0.$$

(2)

Our results regarding Equation (1) complete those obtained by S.-M. Jung and K.-S. Lee in [22]. In [22], the following equation:

$$a\frac{\partial y\left(x,t\right)}{\partial x}+b\frac{\partial y\left(x,t\right)}{\partial t}+cy\left(x,t\right)+d=0,a,b\in \mathbb{R},b\ne 0,c,d\in \mathbb{C},\mathrm{with}\Re \left(c\right)\ne 0,$$

(3)

where $\Re (c)$ denotes the real part of c, was studied. In our paper, we consider the case $c=0$ in Equation (3). Moreover, we also study the generalized stability. The method used in [22] was the method of changing variables.

2. Preliminaries

We first recall some notions and results regarding the Laplace transform.

Let $f:\left(0,\infty \right)\to \mathbb{R}$ be a piecewise differentiable and of exponential order, that is $\exists M>0$ and ${\alpha }_0\ge 0$ such that

$$\left|f\left(t\right)\right|\le M·e^{{\alpha }_{0}t},\forall t>0.$$

We denote by $\mathcal{L}[f]$ the Laplace transform of the function f, defined by

$$\mathcal{L}[f](s)=F(s)={\int }_0^{\infty }f(t)e^{-st}dt.$$

Let

$$u\left(t\right)=\left\{\begin{array}{c}0,\mathrm{if}t\le 0\hfill \\ 1,\mathrm{if}t>0\hfill \end{array}\right.$$

be the unit step function of Heaviside. We write $f(0)$ instead of the lateral limit $f(0^{+})$. The following properties are used in the paper:

$$\mathcal{L}\left[t^n\right]\left(s\right)=\frac{n!}{s^{n+1}},s>0,n\in \mathbb{N},$$

$${\mathcal{L}}^{-1}\left[\frac{1}{s^n}\right]\left(t\right)=\frac{t^{n-1}}{\left(n-1\right)!}u\left(t\right),$$

$$\mathcal{L}[f^{\prime }](s)=s\mathcal{L}[f](s)-f(0),$$

$$\mathcal{L}\left[f\left(t-a\right)u\left(t-a\right)\right]\left(s\right)=e^{-as}F\left(s\right),a>0,$$

hence,

$${\mathcal{L}}^{-1}[e^{-as}F\left(s\right)]\left(t\right)=f\left(t-a\right)u\left(t-a\right).$$

We now consider the function $y:\left(0,\infty \right)\times \left(0,\infty \right)\to R,y=y\left(x,t\right),$ a piecewise differentiable and of exponential order with respect to t. The Laplace transform of y with respect to t is as follows:

$$\mathcal{L}\left[y\left(x,t\right)\right]={\int }_0^{\infty }y\left(x,t\right)e^{-st}dt,$$

where x is treated as a constant. We also denote the following:

$$\mathcal{L}\left[y\left(x,t\right)\right]=Y\left(x,s\right)=Y\left(x\right)=Y.$$

We treat Y as a function of x, leaving s as a parameter. We then have the following:

$$\mathcal{L}\left[\frac{\partial y}{\partial t}\right]=sY\left(x,s\right)-y\left(x,0\right),$$

$$\mathcal{L}\left[\frac{{\partial }^{2}y}{\partial t^2}\right]=s^{2}Y\left(x,s\right)-sy\left(x,0\right)-\frac{\partial y}{\partial t}(x,0).$$

Since we transform with respect to t, we can move $\frac{\partial }{\partial x}$ to the front of the integral; hence, we have:

$$\mathcal{L}\left[\frac{\partial y}{\partial x}\right]=\frac{dY}{dx}=Y^{\prime }\left(x\right).$$

Similarly,

$$\mathcal{L}\left[\frac{{\partial }^{2}y}{\partial x^2}\right]={\int }_0^{\infty }\frac{{\partial }^{2}y}{\partial x^2}{e}^{-st}dt=\frac{d}{d{x}^2}{\int }_0^{\infty }y\left(x,t\right)e^{-st}dt=\frac{dY}{d{x}^2}=Y^{″}\left(x\right).$$

For the Laplace transform properties and applications, see [31,32].

3. Semi-Hyers–Ulam–Rassias Stability of the Convection Partial Differential Equation

Let $\epsilon >0$. We also consider the following inequality:

$$\left|\frac{\partial y}{\partial t}+a\frac{\partial y}{\partial x}\right|\le \epsilon ,$$

(4)

or the equivalent

$$-\epsilon \le \frac{\partial y}{\partial t}+a\frac{\partial y}{\partial x}\le \epsilon .$$

(5)

Analogous to [33], we give the following definition:

Definition 1.

The Equation (1) is called semi-Hyers–Ulam–Rassias stable if there exists a function $\phi :\left(0,\infty \right)\times \left(0,\infty \right)\to \left(0,\infty \right)$, such that for each solution y of the inequality (4), there exists a solution $y_0$ for the Equation (1) with

$$\left|y\left(x,t\right)-y_0\left(x,t\right)\right|\le \phi (x,t),\forall x>0,t>0.$$

Theorem 1.

If a function $y:\left(0,\infty \right)\times \left(0,\infty \right)\to \mathbb{R}$ satisfies the inequality (4), then there exists a solution $y_0:\left(0,\infty \right)\times \left(0,\infty \right)\to \mathbb{R}$ for (1), such that

$$\left|y\left(x,t\right)-y_0\left(x,t\right)\right|\le \left\{\begin{array}{c}\epsilon t,t<\frac{x}{a}\\ \epsilon \frac{x}{a},t\ge \frac{x}{a}\end{array}\right.,$$

(6)

that is, the Equation (1) is considered semi-Ulam–Hyers–Rassias stable.

Proof. 

We apply the Laplace transform with respect to t in (5); thus, we have the following:

$$-\frac{\epsilon }{s}\le sY\left(x\right)-y\left(x,0\right)+a{Y}^{\prime }\left(x\right)\le \frac{\epsilon }{s}.$$

Since $y\left(x,0\right)=0,$ dividing by a we get the following:

$$-\frac{\epsilon }{as}\le Y^{\prime }\left(x\right)+\frac{s}{a}Y\left(x\right)\le \frac{\epsilon }{as}.$$

We now multiply by $e^{\frac{s}{a}x}$ and we obtain this equation:

$$-\frac{\epsilon }{as}{e}^{\frac{s}{a}x}\le e^{\frac{s}{a}x}{Y}^{\prime }\left(x\right)+\frac{s}{a}{e}^{\frac{s}{a}x}Y\left(x\right)\le \frac{\epsilon }{as}{e}^{\frac{s}{a}x},$$

hence,

$$-\frac{\epsilon }{as}{e}^{\frac{s}{a}x}\le \frac{d}{dx}\left(e^{\frac{s}{a}x}Y\left(x\right)\right)\le \frac{\epsilon }{as}{e}^{\frac{s}{a}x}.$$

Integrating from 0 to x we get the following:

$$-\frac{\epsilon }{as}\frac{e^{\frac{s}{a}x}}{\frac{s}{a}}{|}_0^x<e^{\frac{s}{a}x}Y\left(x\right){|}_0^x\le \frac{\epsilon }{as}{e}^{\frac{s}{a}x}{|}_0^x,$$

that is,

$$-\epsilon \left(\frac{e^{\frac{s}{a}x}}{s^2}-\frac{1}{s^2}\right)\le e^{\frac{s}{a}x}Y\left(x\right)-Y\left(0\right)\le \epsilon \left(\frac{e^{\frac{s}{a}x}}{s^2}-\frac{1}{s^2}\right).$$

But $Y\left(0\right)=\mathcal{L}[y\left(0,t\right)]=\mathcal{L}[c]=\frac{c}{s}$, so we obtain:

$$-\epsilon \left(\frac{e^{\frac{s}{a}x}}{s^2}-\frac{1}{s^2}\right)\le e^{\frac{s}{a}x}Y\left(x\right)-\frac{c}{s}\le \epsilon \left(\frac{e^{\frac{s}{a}x}}{s^2}-\frac{1}{s^2}\right).$$

We now multiply by $e^{-\frac{s}{a}x}$ and we obtain the equation below:

$$-\epsilon \left(\frac{1}{s^2}-\frac{e^{-\frac{s}{a}x}}{s^2}\right)\le Y\left(x\right)-c\frac{e^{-\frac{s}{a}x}}{s}\le \epsilon \left(\frac{1}{s^2}-\frac{e^{-\frac{s}{a}x}}{s^2}\right).$$

We apply the inverse Laplace transform and we obtain the following:

$$-\epsilon \left[t-\left(t-\frac{x}{a}\right)u\left(t-\frac{x}{a}\right)\right]\le y\left(x,t\right)-c·u\left(t-\frac{x}{a}\right)\le \epsilon \left[t-\left(t-\frac{x}{a}\right)u\left(t-\frac{x}{a}\right)\right],$$

that is,

$$\left|y\left(x,t\right)-c·u\left(t-\frac{x}{a}\right)\right|\le \epsilon \left[t-\left(t-\frac{x}{a}\right)u\left(t-\frac{x}{a}\right)\right].$$

We then put

$$y_0\left(x,t\right)=c·u\left(t-\frac{x}{a}\right)=\left\{\begin{array}{c}0,t<\frac{x}{a}\\ c,t\ge \frac{x}{a}\end{array}\right..$$

This is the solution of (1) and the equation below:

$$\left|y\left(x,t\right)-y_0\left(x,t\right)\right|\le \left\{\begin{array}{c}\epsilon t,t<\frac{x}{a}\\ \epsilon \frac{x}{a},t\ge \frac{x}{a}\end{array}\right..$$

□

4. Generalized Semi-Hyers–Ulam–Rassias Stability of the Convection Partial Differential Equation

Let $\varphi :$$\left(0,\infty \right)\times \mathbb{R}\to \left(0,\infty \right),$ and $\mathcal{L}\left[\varphi \left(x,t\right)\right]=\Phi \left(x,s\right)$. We consider the following inequality:

$$\left|\frac{\partial y}{\partial t}+a\frac{\partial y}{\partial x}\right|\le \varphi \left(x,t\right),$$

(7)

or the equivalent

$$-\varphi \left(x,t\right)\le \frac{\partial y}{\partial t}+a\frac{\partial y}{\partial x}\le \varphi \left(x,t\right),\forall x>0,t>0.$$

(8)

Definition 2.

The Equation (1) is called generalized semi-Hyers–Ulam–Rassias stable if there exists a function $\phi :\left(0,\infty \right)\times \left(0,\infty \right)\to \left(0,\infty \right)$, such that for each solution y of the inequality (7), there exists a solution $y_0$ for the Equation (1) with

$$\left|y\left(x,t\right)-y_0\left(x,t\right)\right|\le \phi (x,t),\forall x>0,t>0.$$

Theorem 2.

Assume that

$${\int }_0^x{e}^{\frac{s}{a}x}\Phi \left(x,s\right)dx\le \Phi \left(x,s\right),\forall x>0,s>0.$$

(9)

If a function $y:\left(0,\infty \right)\times \left(0,\infty \right)\to \mathbb{R}$ satisfies the inequality (7), then there exists a solution $y_0:\left(0,\infty \right)\times \left(0,\infty \right)\to \mathbb{R}$ for (1), such that

$$\left|y\left(x,t\right)-y_0\left(x,t\right)\right|\le \frac{1}{a}\varphi \left(x,t-\frac{x}{a}\right),\forall x>0,t>0,$$

that is, the Equation (1) is considered generalized semi-Hyers–Ulam–Rassias stable.

Proof. 

We apply the Laplace transform with respect to t in (8), so we have the following:

$$-\Phi \left(x,s\right)\le sY\left(x\right)-y\left(x,0\right)+a{Y}^{\prime }\left(x\right)\le \Phi \left(x,s\right).$$

Since $y\left(x,0\right)=0,$ dividing by a we get the equation below:

$$-\frac{1}{a}\Phi \left(x,s\right)\le Y^{\prime }\left(x\right)+\frac{s}{a}Y\left(x\right)\le \frac{1}{a}\Phi \left(x,s\right).$$

We now multiply by $e^{\frac{s}{a}x}$ and we obtain the following:

$$-\frac{e^{\frac{s}{a}x}}{a}\Phi \left(x,s\right)\le e^{\frac{s}{a}x}{Y}^{\prime }\left(x\right)+\frac{s}{a}{e}^{\frac{s}{a}x}Y\left(x\right)\le \frac{e^{\frac{s}{a}x}}{a}\Phi \left(x,s\right),$$

hence,

$$-\frac{e^{\frac{s}{a}x}}{a}\Phi \left(x,s\right)\le \frac{d}{dx}\left(e^{\frac{s}{a}x}Y\left(x\right)\right)\le \frac{e^{\frac{s}{a}x}}{a}\Phi \left(x,s\right).$$

Integrating from 0 to x we get the following equation:

$$-\frac{1}{a}{\int }_0^x{e}^{\frac{s}{a}x}\Phi \left(x,s\right)dx\le e^{\frac{s}{a}x}Y\left(x\right){|}_0^x\le {\int }_0^x\frac{1}{a}{e}^{\frac{s}{a}x}\Phi \left(x,s\right)dx.$$

Using (9), we have

$$-\frac{1}{a}\Phi \left(x,s\right)\le e^{\frac{s}{a}x}Y\left(x\right)-Y\left(0\right)\le \frac{1}{a}\Phi \left(x,s\right).$$

But $Y\left(0\right)=L[y\left(0,t\right)]=L[c]=\frac{c}{s}$, so we obtain

$$-\frac{1}{a}\Phi \left(x,s\right)\le e^{\frac{s}{a}x}Y\left(x\right)-\frac{c}{s}\le \frac{1}{a}\Phi \left(x,s\right).$$

We now multiply by $e^{-\frac{s}{a}x}$ and we obtain the following equation:

$$-\frac{1}{a}{e}^{-\frac{s}{a}x}\Phi \left(x,s\right)\le Y\left(x\right)-c\frac{e^{-\frac{s}{a}x}}{s}\le \frac{1}{a}{e}^{-\frac{s}{a}x}\Phi \left(x,s\right).$$

We apply the inverse Laplace transform and we obtain:

$$-\frac{1}{a}\varphi \left(x,t-\frac{x}{a}\right)\le y\left(x,t\right)-c·u\left(t-\frac{x}{a}\right)\le \frac{1}{a}\varphi \left(x,t-\frac{x}{a}\right),$$

that is,

$$\left|y\left(x,t\right)-c·u\left(t-\frac{x}{a}\right)\right|\le \frac{1}{a}\varphi \left(x,t-\frac{x}{a}\right).$$

We then put the following:

$$y_0\left(x,t\right)=c·u\left(t-\frac{x}{a}\right)=\left\{\begin{array}{c}0,t<\frac{x}{a}\\ c,t\ge \frac{x}{a}\end{array}\right..$$

This is the solution of Equation (1) and the equation below:

$$\left|y\left(x,t\right)-c{y}_0\left(x,t\right)\right|\le \frac{1}{a}\varphi \left(x,t-\frac{x}{a}\right).$$

□

5. Semi-Hyers–Ulam–Rassias Stability of Equation (2)

Let $\epsilon >0$. We also consider the following inequality:

$$\left|\frac{\partial y}{\partial t}+\frac{\partial y}{\partial x}-x\right|\le \epsilon ,$$

(10)

or the equivalent

$$-\epsilon \le \frac{\partial y}{\partial t}+\frac{\partial y}{\partial x}-x\le \epsilon .$$

(11)

Definition 3.

The Equation (2) is called semi-Hyers–Ulam–Rassias stable if there exists a function $\phi :\left(0,\infty \right)\times \left(0,\infty \right)\to \left(0,\infty \right)$, such that for each solution y of the inequality (10), there exists a solution $y_0$ for the Equation (2) with the following:

$$\left|y\left(x,t\right)-y_0\left(x,t\right)\right|\le \phi (x,t),\forall x>0,t>0.$$

Theorem 3.

If a function $y:\left(0,\infty \right)\times \left(0,\infty \right)\to \mathbb{R}$ satisfies the inequality (10), then there exists a solution $y_0:\left(0,\infty \right)\times \left(0,\infty \right)\to \mathbb{R}$ for (2), such that

$$\left|y\left(x,t\right)-y_0\left(x,t\right)\right|\le \left\{\begin{array}{c}\epsilon t,t<x\\ \epsilon x,t\ge x\end{array}\right.,$$

that is, the Equation (2) is considered semi-Hyers–Ulam–Rassias stable.

Proof. 

We apply the Laplace transform with respect to t in (11), so we have the equation below:

$$-\frac{\epsilon }{s}\le sY\left(x\right)-y\left(x,0\right)+Y^{\prime }\left(x\right)-x\frac{1}{s}\le \frac{\epsilon }{s}.$$

Since $y\left(x,0\right)=0,$ we get the following:

$$-\frac{\epsilon }{s}\le Y^{\prime }\left(x\right)+sY\left(x\right)-x\frac{1}{s}\le \frac{\epsilon }{s}.$$

We now multiply by $e^{sx}$ and we obtain the following equation:

$$-\frac{\epsilon }{s}{e}^{sx}\le e^{sx}{Y}^{\prime }\left(x\right)+s{e}^{sx}Y\left(x\right)-x\frac{e^{sx}}{s}\le \frac{\epsilon }{s}{e}^{sx}.$$

hence,

$$-\frac{\epsilon }{s}{e}^{sx}\le \frac{d}{dx}\left(e^{sx}Y\left(x\right)\right)-x\frac{e^{sx}}{s}\le \frac{\epsilon }{s}{e}^{sx}.$$

Integrating from 0 to x, we get the following:

$$-\frac{\epsilon }{s}\frac{e^{sx}}{s}{|}_0^x\le e^{sx}Y\left(x\right){|}_0^x-\frac{1}{s}{\int }_0^{x}x{e}^{sx}dx\le \frac{\epsilon }{s}\frac{e^{sx}}{s}{|}_0^x.$$

Integrating by parts, we get the equation below:

$${\int }_0^{x}x{e}^{sx}dx=\frac{\left(xs-1\right)e^{sx}}{s^2}+\frac{1}{s^2},$$

hence,

$$-\epsilon \left(\frac{e^{sx}}{s^2}-\frac{1}{s^2}\right)\le e^{sx}Y\left(x\right)-Y\left(0\right)-\frac{1}{s}\left[\frac{\left(xs-1\right)e^{sx}}{s^2}+\frac{1}{s^2}\right]\le \epsilon \left(\frac{e^{sx}}{s^2}-\frac{1}{s^2}\right).$$

But $Y\left(0\right)=\mathcal{L}[y\left(0,t\right)]=0$, so we obtain the following:

$$-\epsilon \left(\frac{e^{sx}}{s^2}-\frac{1}{s^2}\right)\le e^{sx}Y\left(x\right)-\frac{1}{s}\left[\frac{\left(xs-1\right)e^{sx}}{s^2}+\frac{1}{s^2}\right]\le \epsilon \left(\frac{e^{sx}}{s^2}-\frac{1}{s^2}\right).$$

We now multiply by $e^{-sx}$ and we obtain the following:

$$-\epsilon \left(\frac{1}{s^2}-\frac{e^{-sx}}{s^2}\right)\le Y\left(x\right)-\frac{1}{s}\left[\frac{xs-1}{s^2}+\frac{e^{-sx}}{s^2}\right]\le \epsilon \left(\frac{1}{s^2}-\frac{e^{-sx}}{s^2}\right),$$

hence,

$$-\epsilon \left(\frac{1}{s^2}-\frac{e^{-sx}}{s^2}\right)\le Y\left(x\right)-\frac{x}{s^2}+\frac{1}{s^3}-\frac{e^{-sx}}{s^3}\le \epsilon \left(\frac{1}{s^2}-\frac{e^{-sx}}{s^2}\right).$$

We apply the inverse Laplace transform and we obtain the following equation:

$$-\epsilon \left[t-\left(t-x\right)u\left(t-x\right)\right]\le y\left(x,t\right)-xt+\frac{1}{2}{t}^2-\frac{1}{2}{\left(t-x\right)}^{2}u\left(t-x\right)\le \epsilon \left[t-\left(t-x\right)u\left(t-x\right)\right].$$

We then put the following:

$$y_0\left(x,t\right)=xt-\frac{1}{2}{t}^2+\frac{1}{2}{\left(t-x\right)}^{2}u\left(t-x\right)=\left\{\begin{array}{c}xt-\frac{1}{2}{t}^2,t<x\hfill \\ \frac{1}{2}{x}^2,t\ge x\hfill \end{array}\right..$$

This is the solution of (2) and the equation below:

$$\left|y\left(x,t\right)-y_0\left(x,t\right)\right|\le \left\{\begin{array}{c}\epsilon t,t<x\\ \epsilon x,t\ge x\end{array}\right..$$

□

6. Conclusions

In this paper, we studied the semi-Hyers–Ulam–Rassias stability of Equations (1) and (2) and the generalized semi-Hyers–Ulam–Rassias stability of Equation (1) using the Laplace transform. To the best of our knowledge, the Hyers-Ulam-Rassias stability of Equations (1) and (2) has not been discussed in the literature with the use of the Laplace transform method. Our results complete those of Jung and Lee [22]. In [22], the Equation (3) was studied for $\Re \left(c\right)\ne 0$. We considered the case $c=0$ in Equation (3). We can apply our results to the convection equation in the sense that for every solution y of (4), which is called an approximate solution, there exists an exact solution $y_0$ of (1), such that the relation (6) is satisfied. From a different perspective, the approximate solution can be viewed in relation to the perturbation theory, as any approximate solution of (4) is an exact solution of the perturbed equation $\frac{\partial y}{\partial t}+a\frac{\partial y}{\partial x}=h(x,t),\left|h(x,t)\right|\le \epsilon ,a>0,x>0,t>0,y\left(0,t\right)=c,y\left(x,0\right)=0$.

We intend to study other partial differential equations as well as other integro-differential equations using this method. We have already applied this method to [34], where we investigated the semi-Hyers–Ulam–Rassias stability of a Volterra integro-differential equation of order I with a convolution-type kernel.