# Best Constant in Ulam Stability for the Third Order Linear Differential Operator with Constant Coefficients

^{*}

## Abstract

**:**

## 1. Introduction

**Definition**

**1.**

- (i)
- ${\rho}_{A}\left(x\right)=0$ if and only if $x=0$;
- (ii)
- ${\rho}_{A}\left(\lambda x\right)=|\lambda |{\rho}_{A}\left(x\right)$ for all $x\in A,$ $\lambda \in \mathbb{K},$ $\lambda \ne 0,$

**Definition**

**2.**

## 2. Main Results

**distinct roots**in the characteristic equation, the form of a particular solution to the equation $Dy=f$ is

**$p=q\ne r$**, a particular solution is given by

**$p=q=r$**, the form of the particular solution is

**Theorem**

**1.**

**distinct roots**of the characteristic equation with $Rep\ne 0,$ $Req\ne 0,$ and $Rer\ne 0$, and let $\epsilon >0.$ Thenm for every $y\in {\mathcal{C}}^{3}(\mathbb{R},X)$ satisfying

**Proof. Existence.**

- (i)
- Let $Rep>0,$ $Req>0,$ and $Rer>0$. Define ${y}_{0}$ by the relation$${y}_{0}\left(x\right)=\tilde{{C}_{1}}{e}^{px}+\tilde{{C}_{2}}{e}^{qx}+\tilde{{C}_{3}}{e}^{rx},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R},$$$$\begin{array}{ccc}\hfill \tilde{{C}_{1}}& =& {C}_{1}+\frac{r-q}{V}{\int}_{0}^{\infty}f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{2}}& =& {C}_{2}+\frac{p-r}{V}{\int}_{0}^{\infty}f\left(t\right){e}^{-qt}dt;\hfill \\ \hfill \tilde{{C}_{3}}& =& {C}_{3}+\frac{q-p}{V}{\int}_{0}^{\infty}f\left(t\right){e}^{-rt}dt,\hfill \end{array}$$The integrals in the definition of the constants $\tilde{{C}_{k}},$ $1\le k\le 3$ are convergent since $Rep>0,$ $Req>0$, $Rer>0$, and $\parallel f\left(t\right)\parallel \le \epsilon ,$ $\forall t\in \mathbb{R}.$ Then,$$\begin{array}{ccc}\hfill y\left(x\right)-{y}_{0}\left(x\right)& =& \frac{-1}{V}{\int}_{x}^{\infty}((r-q){e}^{p(x-t)}+(p-r){e}^{q(x-t)}+(q-p){e}^{r(x-t)})f\left(t\right)dt.\hfill \end{array}$$Now, letting $t-x=u$ in the above integral, we obtain$$\begin{array}{c}\hfill y\left(x\right)-{y}_{0}\left(x\right)=\frac{-1}{V}{\int}_{0}^{\infty}((r-q){e}^{-pu}+(p-r){e}^{-qu}+(q-p){e}^{-ru})f(x+u)du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.\end{array}$$Consequently,$$\begin{array}{ccc}\hfill \parallel y\left(x\right)-{y}_{0}\left(x\right)\parallel & \le & \frac{\epsilon}{|V|}{\int}_{0}^{\infty}|(r-q){e}^{-pu}+(p-r){e}^{-qu}+(q-p){e}^{-ru}|du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R};\hfill \end{array}$$$$\parallel y-{y}_{0}{\parallel}_{\infty}\le K\epsilon .$$
- (ii)
- Let $Rep>0,$ $Req>0,$ and $Rer<0$. Define ${y}_{0}$ by the relation$${y}_{0}\left(x\right)=\tilde{{C}_{1}}{e}^{px}+\tilde{{C}_{2}}{e}^{qx}+\tilde{{C}_{3}}{e}^{rx},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R},$$$$\begin{array}{ccc}\hfill \tilde{{C}_{1}}& =& {C}_{1}+\frac{r-q}{V}{\int}_{0}^{\infty}f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{2}}& =& {C}_{2}+\frac{p-r}{V}{\int}_{0}^{\infty}f\left(t\right){e}^{-qt}dt;\hfill \\ \hfill \tilde{{C}_{3}}& =& {C}_{3}-\frac{q-p}{V}{\int}_{-\infty}^{0}f\left(t\right){e}^{-rt}dt,\hfill \end{array}$$Then,$$\begin{array}{ccc}\hfill y\left(x\right)-{y}_{0}\left(x\right)& =& \frac{-1}{V}{\int}_{x}^{\infty}((r-q){e}^{p(x-t)}+(p-r){e}^{q(x-t)})f\left(t\right)dt+\frac{q-p}{V}{\int}_{-\infty}^{x}{e}^{r(x-t)}f\left(t\right)dt.\hfill \end{array}$$Letting $t-x=u$ and, respectively, $t-x=-u$ in the above integrals, it follows that$$\begin{array}{c}\hfill y\left(x\right)-{y}_{0}\left(x\right)=\frac{-1}{V}{\int}_{0}^{\infty}((r-q){e}^{-pu}+(p-r){e}^{-qu})f(x+u)du+\frac{q-p}{V}{\int}_{0}^{\infty}{e}^{ru}f(x-u)du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.\end{array}$$Consequently,$$\begin{array}{ccc}\hfill \parallel y\left(x\right)-{y}_{0}\left(x\right)\parallel & \le & \frac{\epsilon}{|V|}\left({\int}_{0}^{\infty}|(r-q){e}^{-pu}+(p-r){e}^{-qu}|du+|q-p|{\int}_{0}^{\infty}|{e}^{ru}|du\right),\phantom{\rule{1.em}{0ex}}x\in \mathbb{R};\hfill \end{array}$$$$\parallel y-{y}_{0}{\parallel}_{\infty}\le K\epsilon .$$
- (iii)
- Let $Rep>0,$ $Req<0,$ and $Rer<0$. Define ${y}_{0}$ by the relation$${y}_{0}\left(x\right)=\tilde{{C}_{1}}{e}^{px}+\tilde{{C}_{2}}{e}^{qx}+\tilde{{C}_{3}}{e}^{rx},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R},$$$$\begin{array}{ccc}\hfill \tilde{{C}_{1}}& =& {C}_{1}+\frac{r-q}{V}{\int}_{0}^{\infty}f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{2}}& =& {C}_{2}-\frac{p-r}{V}{\int}_{-\infty}^{0}f\left(t\right){e}^{-qt}dt;\hfill \\ \hfill \tilde{{C}_{3}}& =& {C}_{3}-\frac{q-p}{V}{\int}_{-\infty}^{0}f\left(t\right){e}^{-rt}dt,\hfill \end{array}$$Then,$$\begin{array}{ccc}\hfill y\left(x\right)-{y}_{0}\left(x\right)& =& -\frac{r-q}{V}{\int}_{x}^{\infty}{e}^{p(x-t)}f\left(t\right)dt+\frac{1}{V}{\int}_{-\infty}^{x}\left((p-r){e}^{q(x-t)}\right)+(q-p){e}^{r(x-t)})f\left(t\right)dt.\hfill \end{array}$$Letting $t-x=u$ and, respectively, $t-x=-u$ in the above integrals, it follows that$$\begin{array}{c}\hfill y\left(x\right)-{y}_{0}\left(x\right)=-\frac{r-q}{V}{\int}_{0}^{\infty}{e}^{-pu}f(x+u)du+\frac{1}{V}{\int}_{0}^{\infty}\left((p-r){e}^{-qu}+(q-p){e}^{ru}\right)f(x-u)du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.\end{array}$$Consequently,$$\begin{array}{ccc}\hfill \parallel y\left(x\right)-{y}_{0}\left(x\right)\parallel & \le & \frac{\epsilon}{|V|}\left(|r-q|{\int}_{0}^{\infty}|{e}^{-pu}|du+{\int}_{0}^{\infty}|(p-r){e}^{qu}+(q-p){e}^{ru}|du\right),\phantom{\rule{1.em}{0ex}}x\in \mathbb{R};\hfill \end{array}$$$$\parallel y-{y}_{0}{\parallel}_{\infty}\le K\epsilon .$$
- (iv)
- Let $Rep<0,$ and $Req<0,$ $Rer<0$. Define ${y}_{0}$ by the relation$$\begin{array}{ccc}\hfill \tilde{{C}_{1}}& =& {C}_{1}-\frac{r-q}{V}{\int}_{-\infty}^{0}f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{2}}& =& {C}_{2}-\frac{p-r}{V}{\int}_{-\infty}^{0}f\left(t\right){e}^{-qt}dt;\hfill \\ \hfill \tilde{{C}_{3}}& =& {C}_{3}-\frac{q-p}{V}{\int}_{-\infty}^{0}f\left(t\right){e}^{-rt}dt,\hfill \end{array}$$Then,$$\begin{array}{ccc}\hfill y\left(x\right)-{y}_{0}\left(x\right)& =& \frac{1}{V}{\int}_{-\infty}^{x}((r-q){e}^{p(x-t)}+(p-r){e}^{q(x-t)}+(q-p){e}^{r(x-t)})f\left(t\right)dt.\hfill \end{array}$$Now, letting $t-x=-u$ in the above integral, we obtain$$\begin{array}{c}\hfill y\left(x\right)-{y}_{0}\left(x\right)=\frac{1}{V}{\int}_{0}^{\infty}((r-q){e}^{pu}+(p-r){e}^{qu}+(q-p){e}^{ru})f(x-u)du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.\end{array}$$Consequently,$$\begin{array}{ccc}\hfill \parallel y\left(x\right)-{y}_{0}\left(x\right)\parallel & \le & \frac{\epsilon}{|V|}{\int}_{0}^{\infty}|(r-q){e}^{pu}+(p-r){e}^{qu}+(q-p){e}^{ru}|du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R};\hfill \end{array}$$$$\parallel y-{y}_{0}{\parallel}_{\infty}\le K\epsilon .$$

**Uniqueness.**Suppose that, for some $y\in {\mathcal{C}}^{3}(\mathbb{R},X)$ satisfying Equation (10), there exist ${y}_{1},{y}_{2}\in kerD$ such that

**Theorem**

**2.**

**$p=q\ne r$**with $Rep\ne 0$ and $Rer\ne 0$ and let $\epsilon >0.$ Then, for every $y\in {\mathcal{C}}^{3}(\mathbb{R},X)$ satisfying

**Proof. Existence.**

- (i)
- Let $Rep>0$ and $Rer>0.$ Define ${y}_{0}$ by the relation$${y}_{0}\left(x\right)=\tilde{{C}_{1}}{e}^{px}+\tilde{{C}_{2}}x{e}^{px}+\tilde{{C}_{3}}{e}^{rx},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R},$$$$\begin{array}{ccc}\hfill \tilde{{C}_{1}}& =& {C}_{1}+\frac{1}{{(r-p)}^{2}}{\int}_{0}^{\infty}((r-p)t-1)f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{2}}& =& {C}_{2}-\frac{1}{{(r-p)}^{2}}{\int}_{0}^{\infty}(r-p)f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{3}}& =& {C}_{3}+\frac{1}{{(r-p)}^{2}}{\int}_{0}^{\infty}f\left(t\right){e}^{-rt}dt.\hfill \end{array}$$The proof of the convergence of the improper integrals is analogous to that given in Theorem 1.Then,$$\begin{array}{ccc}\hfill y\left(x\right)-{y}_{0}\left(x\right)& =& \frac{-{e}^{px}}{{(r-p)}^{2}}{\int}_{x}^{\infty}((x-t)(p-r)-1)f\left(t\right){e}^{-pt}dt+\frac{{e}^{rx}}{{(r-p)}^{2}}{\int}_{x}^{\infty}f\left(t\right){e}^{-rt}dt.\hfill \end{array}$$Now, letting $t-x=u$ in the above integral, we obtain$$\begin{array}{c}\hfill y\left(x\right)-{y}_{0}\left(x\right)=\frac{-1}{{(r-p)}^{2}}{\int}_{0}^{\infty}\left(((r-p)u-1){e}^{-pu}+{e}^{-ru}\right)f(x+u)du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.\end{array}$$Consequently,$$\begin{array}{ccc}\hfill \parallel y\left(x\right)-{y}_{0}\left(x\right)\parallel & \le & \frac{1}{{|r-p|}^{2}}{\int}_{0}^{\infty}\left(\left|((r-p)u-1)f(x+u){e}^{-pu}\right|+\left|{e}^{-ru}f(x-u)\right|\right)du\hfill \\ & \le & \frac{\epsilon}{{|r-p|}^{2}}{\int}_{0}^{\infty}\left(|((r-p)u-1){e}^{-pu}|+|{e}^{-ru}|\right)du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R};\hfill \end{array}$$$$\parallel y-{y}_{0}{\parallel}_{\infty}\le K\epsilon .$$
- (ii)
- Let $Rep>0$ and $Rer<0.$ The proof follows analogously, defining ${y}_{0}$ by the relation$${y}_{0}\left(x\right)=\tilde{{C}_{1}}{e}^{px}+\tilde{{C}_{2}}x{e}^{px}+\tilde{{C}_{3}}{e}^{rx},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R},$$$$\begin{array}{ccc}\hfill \tilde{{C}_{1}}& =& {C}_{1}+\frac{1}{{(r-p)}^{2}}{\int}_{0}^{\infty}((r-p)t-1)f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{2}}& =& {C}_{2}-\frac{1}{{(r-p)}^{2}}{\int}_{0}^{\infty}(r-p)f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{3}}& =& {C}_{3}-\frac{1}{{(r-p)}^{2}}{\int}_{-\infty}^{0}f\left(t\right){e}^{-rt}dt.\hfill \end{array}$$Then,$$\begin{array}{ccc}\hfill y\left(x\right)-{y}_{0}\left(x\right)& =& \frac{-{e}^{px}}{{(r-p)}^{2}}{\int}_{x}^{\infty}((x-t)(p-r)-1)f\left(t\right){e}^{-pt}dt+\frac{{e}^{rx}}{{(r-p)}^{2}}{\int}_{-\infty}^{x}f\left(t\right){e}^{-rt}dt.\hfill \end{array}$$Letting $t-x=u$ and, respectively, $t-x=-u$ in the above integrals, it follows that$$y\left(x\right)-{y}_{0}\left(x\right)=\frac{1}{{(r-p)}^{2}}\left(-{\int}_{0}^{\infty}((r-p)u-1)f(x+u){e}^{-pu}du+{\int}_{0}^{\infty}f(x-u){e}^{ru}du\right),\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.$$Consequently,$$\begin{array}{ccc}\hfill \parallel y\left(x\right)-{y}_{0}\left(x\right)\parallel & \le & \frac{1}{{|r-p|}^{2}}{\int}_{0}^{\infty}|((r-p)u-1)f(x+u){e}^{-pu}|du+{\int}_{0}^{\infty}|f(x-u){e}^{ru}|du\hfill \\ & \le & \frac{\epsilon}{{|r-p|}^{2}}{\int}_{0}^{\infty}\left(|((r-p)u-1){e}^{-pu}|+|{e}^{ru}|\right)du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R};\hfill \end{array}$$$$\parallel y-{y}_{0}{\parallel}_{\infty}\le K\epsilon .$$
- (iii)
- Let $Rep<0$ and $Rer<0.$ Then, we define ${y}_{0}$ by the relation$${y}_{0}\left(x\right)=\tilde{{C}_{1}}{e}^{px}+\tilde{{C}_{2}}x{e}^{px}+\tilde{{C}_{3}}{e}^{rx},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R},$$$$\begin{array}{ccc}\hfill \tilde{{C}_{1}}& =& {C}_{1}-\frac{1}{{(r-p)}^{2}}{\int}_{-\infty}^{0}((r-p)t-1)f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{2}}& =& {C}_{2}+\frac{1}{{(r-p)}^{2}}{\int}_{-\infty}^{0}(r-p)f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{3}}& =& {C}_{3}-\frac{1}{{(r-p)}^{2}}{\int}_{-\infty}^{0}f\left(t\right){e}^{-rt}dt.\hfill \end{array}$$Therefore,$$\begin{array}{ccc}\hfill y\left(x\right)-{y}_{0}\left(x\right)& =& \frac{{e}^{px}}{{(r-p)}^{2}}{\int}_{-\infty}^{x}((x-t)(p-r)-1)f\left(t\right){e}^{-pt}dt+\frac{{e}^{rx}}{{(r-p)}^{2}}{\int}_{-\infty}^{x}f\left(t\right){e}^{-rt}dt.\hfill \end{array}$$Letting $x-t=u$ in the above integrals, it follows that$$\begin{array}{c}\hfill y\left(x\right)-{y}_{0}\left(x\right)=\frac{1}{{(r-p)}^{2}}\left({\int}_{0}^{\infty}((p-r)u-1)f(x-u){e}^{pu}du+{\int}_{0}^{\infty}f(x-u){e}^{ru}du\right),\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.\end{array}$$Consequently,$$\begin{array}{ccc}\hfill \parallel y\left(x\right)-{y}_{0}\left(x\right)\parallel & \le & \frac{1}{{|r-p|}^{2}}{\int}_{0}^{\infty}|((p-r)u-1){e}^{pu}+{e}^{ru}|f(x-u)du\hfill \\ & \le & \frac{\epsilon}{{|r-p|}^{2}}{\int}_{0}^{\infty}\left|((p-r)u-1){e}^{pu}+{e}^{ru}\right|du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R};\hfill \end{array}$$$$\parallel y-{y}_{0}{\parallel}_{\infty}\le K\epsilon .$$

**Uniqueness.**Suppose that, for some $y\in {\mathcal{C}}^{3}(\mathbb{R},X)$ satisfying Equation (13), there exist ${y}_{1},{y}_{2}\in kerD$ such that

**Theorem**

**3.**

**$p=q=r$**with $Rep\ne 0$ and let $\epsilon >0.$ Then, for every $y\in {\mathcal{C}}^{3}(\mathbb{R},X)$ satisfying

**Proof. Existence.**

- (i)
- Let $Rep>0$. Define ${y}_{0}$ by the relation$${y}_{0}\left(x\right)=\tilde{{C}_{1}}{e}^{px}+\tilde{{C}_{2}}x{e}^{px}+\tilde{{C}_{3}}{x}^{2}{e}^{px},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R},$$$$\begin{array}{ccc}\hfill \tilde{{C}_{1}}& =& {C}_{1}+\frac{1}{2}{\int}_{0}^{\infty}{t}^{2}f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{2}}& =& {C}_{2}-{\int}_{0}^{\infty}tf\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{3}}& =& {C}_{3}+\frac{1}{2}{\int}_{0}^{\infty}f\left(t\right){e}^{-pt}dt.\hfill \end{array}$$Then,$$\begin{array}{ccc}\hfill y\left(x\right)-{y}_{0}\left(x\right)& =& \frac{-{e}^{px}}{2}{\int}_{x}^{\infty}{(t-x)}^{2}f\left(t\right){e}^{-pt}dt.\hfill \end{array}$$Now, letting $t-x=u$ in the above integral, we obtain$$y\left(x\right)-{y}_{0}\left(x\right)=\frac{-1}{2}{\int}_{0}^{\infty}{u}^{2}f(x+u){e}^{-pu}du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.$$Consequently,$$\begin{array}{ccc}\hfill \parallel y\left(x\right)-{y}_{0}\left(x\right)\parallel & \le & \frac{\epsilon}{2}{\int}_{0}^{\infty}|{u}^{2}{e}^{-pu}|du=\frac{\epsilon}{2}{\int}_{0}^{\infty}{u}^{2}{e}^{-uRep}du=\frac{\epsilon}{{|Rep|}^{3}},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.\hfill \end{array}$$
- (ii)
- Let $Rep<0$. Define ${y}_{0}$ by the relation$$\begin{array}{ccc}\hfill \tilde{{C}_{1}}& =& {C}_{1}+\frac{1}{2}{\int}_{-\infty}^{0}{t}^{2}f\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{2}}& =& {C}_{2}-{\int}_{-\infty}^{0}tf\left(t\right){e}^{-pt}dt;\hfill \\ \hfill \tilde{{C}_{3}}& =& {C}_{3}+\frac{1}{2}{\int}_{-\infty}^{0}f\left(t\right){e}^{-pt}dt.\hfill \end{array}$$Then,$$\begin{array}{ccc}\hfill y\left(x\right)-{y}_{0}\left(x\right)& =& \frac{{e}^{px}}{2}{\int}_{x}^{\infty}{(t-x)}^{2}f\left(t\right){e}^{-pt}dt.\hfill \end{array}$$Now, letting $x-t=u$ in the above integral, we obtain$$\begin{array}{c}\hfill y\left(x\right)-{y}_{0}\left(x\right)=\frac{1}{2}{\int}_{0}^{\infty}{u}^{2}f(x-u){e}^{pu}du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.\end{array}$$Consequently,$$\begin{array}{ccc}\hfill \parallel y\left(x\right)-{y}_{0}\left(x\right)\parallel & \le & \frac{\epsilon}{2}{\int}_{0}^{\infty}{u}^{2}|{e}^{pu}|du=\frac{\epsilon}{{|Rep|}^{3}},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.\hfill \end{array}$$

**Uniqueness.**Suppose that, for some $y\in {\mathcal{C}}^{3}(\mathbb{R},X)$ satisfying Equation (17), there exist ${y}_{1},{y}_{2}\in kerD$ such that

**Theorem**

**4.**

**distinct roots**of the characteristic equation with $Rep\ne 0,$ $Req\ne 0,$ and $Rer\ne 0.$ Then, the best Ulam constant of D is given by

**Proof.**

- (i)
- First, let $Rep>0,$ $Req>0,$ and $Rer>0.$ Then,$${K}_{D}=\frac{1}{|V|}{\int}_{0}^{\infty}|(r-q){e}^{-pu}+(p-r){e}^{-qu}+(q-p){e}^{-ru}|du.$$Let $h\left(x\right)=(r-q){e}^{-px}+(p-r){e}^{-qx}+(q-p){e}^{-rx},$ $x\in \mathbb{R}.$ Take $s\in X,\parallel s\parallel =1,$ and $\theta >0$ as arbitrarily chosen and consider $f:\mathbb{R}\to X$ defined by$$f\left(x\right)=\frac{\overline{h\left(x\right)}}{|h\left(x\right)|+\theta {e}^{-x}}s,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R},$$$$\tilde{y}\left(x\right)=\frac{-1}{V}{\int}_{0}^{\infty}((r-q){e}^{-pu}+(p-r){e}^{-qu}+(q-p){e}^{-ru})f(x+u)du,\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.$$Since f is bounded and $Rep>0,$ $Req>0$, and $Rer>0$, it follows that $\tilde{y}\left(x\right)$ is bounded on $\mathbb{R}$. Furthermore, $\parallel D\left(\tilde{y}\right){\parallel}_{\infty}\le 1$ and the Ulam stability of D for $\epsilon =1$ with the constant K leads to the existence of ${y}_{0}\in kerD,$ which is given by Equation (3) such that$${\parallel \tilde{y}-{y}_{0}\parallel}_{\infty}\le K.$$If $({C}_{1},{C}_{2},{C}_{3})\ne (0,0,0)$, we get, in view of the boundedness of $\tilde{y}$$$\underset{x\to \infty}{lim}\parallel \tilde{y}\left(x\right)-{y}_{0}\left(x\right)\parallel =+\infty ,$$$$\begin{array}{c}\hfill \parallel \tilde{y}\left(x\right)\parallel \le K,\phantom{\rule{1.em}{0ex}}\mathrm{for}\mathrm{all}x\in \mathbb{R}.\end{array}$$Taking $x=0$ in Equation (25) we get$\parallel \tilde{y}\left(0\right)\parallel \le K$; i.e.,$$\begin{array}{c}\hfill \frac{1}{|V|}\left|{\int}_{0}^{\infty}\left(r-q){e}^{-pu}+(p-r){e}^{-qu}+(q-p){e}^{-ru}\right)f\left(u\right)du\right|\le K,\end{array}$$$$\frac{1}{|V|}\left|{\int}_{0}^{\infty}h\left(u\right)f\left(u\right)du\right|=\frac{1}{|V|}{\int}_{0}^{\infty}\frac{{|h\left(u\right)|}^{2}}{|h\left(u\right)|+\theta {e}^{-u}}du\le K,\phantom{\rule{1.em}{0ex}}\forall \theta >0.$$We show next that$$\underset{\theta \to 0}{lim}{\int}_{0}^{\infty}\frac{{|h\left(u\right)|}^{2}}{|h\left(u\right)|+\theta {e}^{-u}}du={\int}_{0}^{\infty}|h\left(u\right)|du.$$Indeed,$$\begin{array}{ccc}\hfill \left|{\int}_{0}^{\infty}\frac{{|h\left(u\right)|}^{2}}{|h\left(u\right)|+\theta {e}^{-u}}du-{\int}_{0}^{\infty}|h\left(u\right)|du\right|& \le & {\int}_{0}^{\infty}\left|\frac{{|h\left(u\right)|}^{2}}{|h\left(u\right)|+\theta {e}^{-u}}-|h\left(u\right)|\right|du\hfill \\ & =& \theta {\int}_{0}^{\infty}\frac{|h\left(u\right)|{e}^{-u}}{|h\left(u\right)|+\theta {e}^{-u}}du\hfill \\ & \le & \theta {\int}_{0}^{\infty}{e}^{-u}du=\theta ,\phantom{\rule{1.em}{0ex}}\theta >0,\hfill \end{array}$$Consequently, letting $\theta \to 0$ in Equation (26), we get ${K}_{D}\le K,$ which contradicts the supposition $K<{K}_{D}.$
- (ii)
- The case where $Rep<0,$ $Req<0,$ and $Rer<0$ follows analogously for$$f\left(x\right)=\frac{\overline{h(-x)}}{|h(-x)|+\theta {e}^{x}}s,$$$$h\left(x\right)=(r-q){e}^{px}+(p-r){e}^{qx}+(q-p){e}^{rx},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.$$
- (iii)
- Consider $Rep>0,$ $Req>0$, and $Rer<0.$ Let$${h}_{1}\left(x\right)=(r-q){e}^{-px}+(p-r){e}^{-qx},\phantom{\rule{1.em}{0ex}}{h}_{2}\left(x\right)=(q-p){e}^{rx},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R}.$$Take an arbitrary $\theta >0,$ $s\in X,$ $\parallel s\parallel =1$ and define the function $f:X\to \mathbb{R}$ by$$\begin{array}{c}\hfill f\left(x\right)=\left\{\begin{array}{cc}\frac{\overline{{h}_{1}\left(x\right)}}{|{h}_{1}\left(x\right)|+\theta {e}^{-x}}s,\hfill & \mathrm{if}x\in [0,+\infty )\hfill \\ \frac{\overline{-{h}_{2}(-x)}}{|{h}_{2}(-x)|+\theta {e}^{x}}s,\hfill & \mathrm{if}x\in (-\infty ,0)\hfill \end{array}\right.\end{array}$$It can be seen that f is continuous and ${\parallel f\parallel}_{\infty}\le 1.$Let $\tilde{y}$ be the solution to $D\left(y\right)=f$ given by$$\tilde{y}\left(x\right)=\frac{-1}{V}{\int}_{0}^{\infty}((r-q){e}^{-pu}+(p-r){e}^{-qu})f(x+u)du+\frac{1}{V}{\int}_{0}^{\infty}(q-p){e}^{ru}f(x-u)du,$$Since f is bounded, taking account of the signs of the roots $p,q,$ and r, it follows that $\tilde{y}\left(x\right)$ is bounded.On the other hand, all the elements of $kerD$ are unbounded, except $y=0.$ We conclude that the relation$$\parallel \tilde{y}{-y\parallel}_{\infty}\le K$$$$\parallel \tilde{y}\left(x\right)\parallel \le K,\phantom{\rule{1.em}{0ex}}\forall x\in \mathbb{R}.$$For $x=0$, it follows that $\parallel \tilde{y\left(0\right)}\parallel \le K$, which is equivalent to$$\begin{array}{c}\hfill {\displaystyle \frac{1}{|V|}{\int}_{0}^{\infty}\left(\frac{|{h}_{1}{\left(u\right)|}^{2}}{|{h}_{1}\left(u\right)|+\theta {e}^{-u}}+\frac{|{h}_{2}{\left(u\right)|}^{2}}{|{h}_{2}\left(u\right)|+\theta {e}^{-u}}\right)du\le K.}\end{array}$$We prove that$$\underset{\theta \to 0}{lim}{\int}_{0}^{\infty}\left(\frac{|{h}_{1}{\left(u\right)|}^{2}}{|{h}_{1}\left(u\right)|+\theta {e}^{-u}}+\frac{|{h}_{2}{\left(u\right)|}^{2}}{|{h}_{2}\left(u\right)|+\theta {e}^{-u}}\right)du={\int}_{0}^{\infty}\left(|{h}_{1}\left(u\right)|+|{h}_{2}\left(u\right)|\right)du.$$Indeed,$$\begin{array}{c}\hfill \left|{\displaystyle {\int}_{0}^{\infty}\left(\frac{|{h}_{1}{\left(u\right)|}^{2}}{|{h}_{1}\left(u\right)|+\theta {e}^{-u}}+\frac{|{h}_{2}{\left(u\right)|}^{2}}{|{h}_{2}\left(u\right)|+\theta {e}^{-u}}\right)du-{\int}_{0}^{\infty}\left(|{h}_{1}\left(u\right)|+|{h}_{2}\left(u\right)|\right)du}\right|\\ \hfill {\displaystyle \le {\int}_{0}^{\infty}\left|\frac{|{h}_{1}{\left(u\right)|}^{2}}{|{h}_{1}\left(u\right)|+\theta {e}^{-u}}-|{h}_{1}\left(u\right)|+\frac{|{h}_{2}{\left(u\right)|}^{2}}{|{h}_{2}\left(u\right)|+\theta {e}^{-u}}-|{h}_{2}\left(u\right)|\right|du}\\ \hfill {\displaystyle =\theta {\int}_{0}^{\infty}\left(\frac{|{h}_{1}\left(u\right)|}{|{h}_{1}\left(u\right)|+\theta {e}^{-u}}+\frac{|{h}_{2}\left(u\right)|}{|{h}_{2}\left(u\right)|+\theta {e}^{-u}}\right){e}^{-u}du}\\ \hfill {\displaystyle \le 2\theta {\int}_{0}^{\infty}{e}^{-u}du=2\theta .}\end{array}$$Now, letting $\theta \to 0$ in Equation (29), it follows that ${K}_{D}<K,$ which is a contradiction.
- (iv)
- The case where $Rep>0,$ $Req<0,$ and $Rer<0$ follows analogously for$$\begin{array}{c}\hfill f\left(x\right)=\left\{\begin{array}{cc}\frac{\overline{{h}_{1}\left(x\right)}}{|{h}_{1}\left(x\right)|+\theta {e}^{-x}}s,\hfill & ifx\in [0,+\infty )\hfill \\ \frac{\overline{-{h}_{2}(-x)}}{|{h}_{2}(-x)|+\theta {e}^{x}}s,\hfill & ifx\in (-\infty ,0)\hfill \end{array}\right.\end{array}$$$${h}_{1}\left(x\right)=(r-q){e}^{-px},\phantom{\rule{1.em}{0ex}}{h}_{2}\left(x\right)=(p-r){e}^{qu}+(q-p){e}^{rx},\phantom{\rule{1.em}{0ex}}x\in \mathbb{R},$$

**Theorem**

**5.**

**double root**and r a

**s**imple root of the characteristic equation with $Rep\ne 0$ and $Rer\ne 0.$ Then, the best Ulam constant of D is given by

**Proof.**

**Theorem**

**6.**

**triple root**of the characteristic equation with $Rep\ne 0.$ Then, the best Ulam constant of D is given by

**Proof.**

## 3. Conclusions

## Author Contributions

## Funding

## Data Availability Statement

## Conflicts of Interest

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Baias, A.R.; Popa, D.
Best Constant in Ulam Stability for the Third Order Linear Differential Operator with Constant Coefficients. *Axioms* **2023**, *12*, 922.
https://doi.org/10.3390/axioms12100922

**AMA Style**

Baias AR, Popa D.
Best Constant in Ulam Stability for the Third Order Linear Differential Operator with Constant Coefficients. *Axioms*. 2023; 12(10):922.
https://doi.org/10.3390/axioms12100922

**Chicago/Turabian Style**

Baias, Alina Ramona, and Dorian Popa.
2023. "Best Constant in Ulam Stability for the Third Order Linear Differential Operator with Constant Coefficients" *Axioms* 12, no. 10: 922.
https://doi.org/10.3390/axioms12100922