**4. Generalized Semi-Hyers–Ulam–Rassias Stability of a Delay Differential Equation of Order One**

We continue to study generalized semi-Hyers–Ulam–Rassias stability of the Equation (1). Let *ϕ* ∈ O. We consider the inequality

$$|\mathbf{x}'(t) + \mathbf{x}(t-1) - f(t)| \le \varphi(t), \quad t \in (0, \infty). \tag{9}$$

**Definition 4.** *The Equation* (1) *is called generalized semi-Hyers–Ulam–Rassias stable if there exists a function k* : (0, ∞) → (0, ∞) *such that for each solution x of the inequality* (9)*, there exists a solution x*<sup>0</sup> *of the Equation* (1) *with*

$$|\mathbf{x}(t) - \mathbf{x}\_0(t)| \le k(t), \forall t \in (0, \infty). \tag{10}$$

**Remark 3.** *A function <sup>x</sup>* : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> *is a solution of* (9) *if, and only if, there exists a function <sup>p</sup>* : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> *such that*

*(1)* |*p*(*t*)| ≤ *ϕ*(*t*), ∀*t* ∈ (0, ∞), *(2) x* (*t*) + *x*(*t* − 1) − *f*(*t*) = *p*(*t*), ∀*t* ∈ (0, ∞).

**Theorem 7.** *If a function <sup>x</sup>* : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> *satisfies the inequality* (9)*, where <sup>f</sup>* , *<sup>ϕ</sup>* ∈ O*, then there exists a solution x*<sup>0</sup> : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> *of* (1) *such that*

$$|\mathbf{x}(t) - \mathbf{x}\_0(t)| \le \int\_0^t \varphi(\tau) \left| \mathcal{L}^{-1} \left( \frac{1}{s + \varepsilon^{-s}} \right) (t - \tau) \right| d\tau,\tag{11}$$

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

**Proof.** Let *<sup>p</sup>* : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup>,

$$p(t) = \mathbf{x}'(t) + \mathbf{x}(t-1) - f(t), \quad t \in (0, \infty). \tag{12}$$

As in Theorem 3, for *x* that is a solution of (9) and Laplace transform of *x*, *x* exists, we have

$$
\mathcal{L}(x) = \frac{\mathcal{L}(p)}{s + e^{-s}} + \frac{\mathcal{L}(f)}{s + e^{-s}}.
$$

and

$$\varkappa\_0(t) = \mathcal{L}^{-1}\left(\frac{\mathcal{L}(f)}{s + \mathfrak{e}^{-s}}\right)(t), \quad \forall t \in (0, \infty), s$$

is a solution of (1).

We have

$$
\mathcal{L}(\mathfrak{x}) - \mathcal{L}(\mathfrak{x}\_0) = \frac{\mathcal{L}(p)}{s + \mathfrak{c}^{-s}}.
$$

hence

$$\begin{split} \left| \left| \mathbf{x}(t) - \mathbf{x}\_{0}(t) \right| \right| &= \left| \mathcal{L}^{-1} \left( \frac{\mathcal{L}(p)}{s + e^{-s}} \right) \right| = \left| \mathcal{L}^{-1} (\mathcal{L}(p)) \ast \mathcal{L}^{-1} \left( \frac{1}{s + e^{-s}} \right) \right| \\ &= \left| p \ast \mathcal{L}^{-1} \left( \frac{1}{s + e^{-s}} \right) \right| = \left| \int\_{0}^{t} p(\tau) \cdot \mathcal{L}^{-1} \left( \frac{1}{s + e^{-s}} \right) (t - \tau) d\tau \right| \\ &\leq \int\_{0}^{t} |p(\tau)| \cdot \left| \mathcal{L}^{-1} \left( \frac{1}{s + e^{-s}} \right) (t - \tau) \right| d\tau \leq \int\_{0}^{t} \rho(\tau) \left| \mathcal{L}^{-1} \left( \frac{1}{s + e^{-s}} \right) (t - \tau) \right| d\tau. \end{split}$$

**Theorem 8.** *Let ϕ* : (0, ∞) → (0, ∞), *ϕ*(*t*) = *t <sup>n</sup>*. *If a function <sup>x</sup>* : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> *satisfies the inequality* (9)*, where f* ∈ O*, then there exists a solution x*<sup>0</sup> : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> *of* (1) *such that*

$$|\mathbf{x}(t) - \mathbf{x}\_0(t)| \le \frac{t^{n+1}}{n+1} + \frac{(t-1)^{n+2}}{(n+1)(n+2)} + \frac{(t-2)^{n+3}}{(n+1)(n+2)(n+3)} + \dots + \frac{(t-[t])^{n+[t]+1}}{(n+1)(n+2)\dots(n+[t]+1)}.$$

**Proof.** From Theorem 7, we have that if *<sup>x</sup>* : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> satisfies the inequality (9), then there exists a solution *<sup>x</sup>*<sup>0</sup> : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> of (1) such that

$$|\varkappa(t) - \varkappa\_0(t)| \le \int\_0^t \pi^n \left| \mathcal{L}^{-1} \left( \frac{1}{s + e^{-s}} \right) (t - \tau) \right| d\tau$$

is satisfied. We have

$$\begin{aligned} &\int\_0^t \tau^\mu \left| \mathcal{L}^{-1} \left( \frac{1}{s + e^{-s}} \right) (t - \tau) \right| d\tau = \int\_0^t \tau^\mu \sum\_{n=0}^{[t-\tau]} \frac{(t - \tau - n)^n}{n!} \\ &= \int\_0^t \tau^\mu \frac{\left(t - \tau - 0\right)^0}{0!} d\tau + \int\_0^{t-1} \tau^\mu \frac{\left(t - \tau - 1\right)^1}{1!} d\tau + \dots + \int\_0^{t - [t]} \tau^n \frac{\left(t - \tau - [t]\right)^{[t]}}{[t]!} d\tau. \end{aligned}$$

We have

$$\int\_0^t \pi^n \frac{(t-\pi-0)^0}{0!} d\tau = \frac{\pi^{n+1}}{n+1} \Big|\_0^t = \frac{t^{n+1}}{n+1}.$$

Integrating by parts, we have

$$\int\_0^{t-1} \pi^n \frac{(t-\tau-1)^1}{1!} d\tau = \int\_0^{t-1} \left(\frac{\tau^{n+1}}{n+1}\right)' (t-\tau-1) d\tau = \underbrace{\sum\_{n+1}^{n+1} (t-\tau-1) \Big|\_0^{t-1}}\_{0} + \int\_0^{t-1} \frac{\tau^{n+1}}{n+1} d\tau$$

$$= \frac{\tau^{n+2}}{(n+1)(n+2)} \Big|\_0^{t-1} = \frac{(t-1)^{n+2}}{(n+1)(n+2)},$$

$$\int\_0^{t-2} \pi^n \frac{(t-\tau-2)^2}{2!} d\tau = \int\_0^{t-2} \frac{\left(\frac{\tau^{n+1}}{n+1}\right)' \left(\frac{t-\tau-2}{2!}\right)^2}{2!} d\tau = \underbrace{\sum\_{n+1}^{\tau^{n+1}} \frac{(t-\tau-2)^2}{2!}}\_{0} \Big|\_0^{t-2} + \int\_0^{t-2} \frac{\tau^{n+1}}{n+1} \frac{2(t-\tau-2)}{2!} d\tau$$

$$= \int\_0^{t-2} \left(\frac{\tau^{n+2}}{(n+1)(n+2)}\right)' \frac{(t-\tau-2)}{1!} d\tau = \frac{\tau^{n+2}}{(n+1)(n+2)} \frac{(t-\tau-2)}{1!} \Big|\_0^{t-2} + \int\_0^{t-2} \frac{\tau^{n+2}}{(n+1)(n+2)} d\tau$$

$$= \frac{(t-2)^{n+3}}{(n+1)(n+2)(n+3)},$$

$$\begin{split} &\int\_{0}^{t-[t]} \tau^{n} \frac{\left(t-\tau-[t]\right)^{[t]}}{[t]!} d\tau = \int\_{0}^{t-[t]} \left(\frac{\tau^{n+1}}{n+1}\right)' \frac{\left(t-\tau-[t]\right)^{[t]}}{[t]!} d\tau \\ &= \underbrace{\frac{\tau^{n+1}}{n+1} \frac{\left(t-\tau-[t]\right)^{[t]}}{[t]!} \Big|\_{0}^{t-[t]}}\_{0} + \int\_{0}^{t-[t]} \frac{\frac{\tau^{n+1}}{n+1} \frac{[t]\left(t-\tau-[t]\right)^{[t]}}{[t]!} d\tau} \\ &= \int\_{0}^{t-[t]} \left(\frac{\tau^{n+2}}{(n+1)(n+2)}\right)' \frac{\left(t-\tau-[t]\right)^{[t]}}{([t]-1)!} d\tau = \dots = \frac{(t-[t])^{n+[t]+1}}{(n+1)(n+2)\cdots(n+[t]+1)}. \end{split}$$
