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

We are now studying the generalized semi-Hyers–Ulam–Rassias stability of the Equation (5). Let *ϕ* ∈ O. We consider the inequality

$$\left|\mathbf{x}^{\prime\prime}(t) + \mathbf{x}^{\prime}(t-1) - f(t)\right| \le \varphi(t), \quad t \in (0, \infty). \tag{13}$$

**Definition 5.** *The Equation* (5) *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* (13)*, there exists a solution x*<sup>0</sup> *of the Equation* (5) *with*

$$|x(t) - x\_0(t)| \le k(t), \forall t \in (0, \infty). \tag{14}$$

**Remark 4.** *A function <sup>x</sup>* : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> *is a solution of* (13) *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 9.** *Let <sup>f</sup>* : <sup>R</sup> <sup>→</sup> <sup>R</sup> *such that* L−<sup>1</sup> <sup>L</sup>(*f*) *s*2+*se*−*<sup>s</sup>* (0) <sup>=</sup> 0. *If a function <sup>x</sup>* : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> *satisfies the inequality* (13)*, where <sup>f</sup>* , *<sup>ϕ</sup>* ∈ O*, then there exists a solution <sup>x</sup>*<sup>0</sup> : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> *of* (5) *such that* 

$$|\mathbf{x}(t) - \mathbf{x}\_0(t)| \le \int\_0^t \varrho(\tau) \left| \mathcal{L}^{-1} \left( \frac{1}{s^2 + s\mathcal{e}^{-s}} \right) (t - \tau) \right| d\tau,\tag{15}$$

*that is the Equation* (5) *is generalized semi-Hyers–Ulam–Rassias stable.*

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

$$p(t) = \mathbf{x}^{\prime\prime}(t) + \mathbf{x}^{\prime}(t-1) - f(t), \quad t \in (0, \infty). \tag{16}$$

As in Theorem 6, for *x* that is a solution of (13) and Laplace transform of *x*, *x* , *x* exists, we have

$$
\mathcal{L}(\mathfrak{x}) = \frac{\mathcal{L}(p)}{s^2 + s e^{-s}} + \frac{\mathcal{L}(f)}{s^2 + s e^{-s}},
$$

and

$$x\_0(t) = \mathcal{L}^{-1}\left(\frac{\mathcal{L}(f)}{s^2 + s e^{-s}}\right)(t), \quad \forall t \in (0, \infty), t$$

is a solution of (5).

We have

$$
\mathcal{L}(x) - \mathcal{L}(x\_0) = \frac{\mathcal{L}(p)}{s^2 + s\varepsilon^{-s}},
$$

hence

$$\begin{split} \left| \left| x(t) - x\_0(t) \right| \right| &= \left| \mathcal{L}^{-1} \left( \frac{\mathcal{L}(p)}{s^2 + s e^{-s}} \right) \right| = \left| \mathcal{L}^{-1} (\mathcal{L}(p)) \ast \mathcal{L}^{-1} \left( \frac{1}{s^2 + s e^{-s}} \right) \right| \\ &= \left| p \ast \mathcal{L}^{-1} \left( \frac{1}{s^2 + s e^{-s}} \right) \right| = \left| \int\_0^t p(\tau) \cdot \mathcal{L}^{-1} \left( \frac{1}{s^2 + s e^{-s}} \right) (t - \tau) d\tau \right| \\ &\leq \int\_0^t \left| p(\tau) \right| \cdot \left| \mathcal{L}^{-1} \left( \frac{1}{s^2 + s e^{-s}} \right) (t - \tau) \right| d\tau \leq \int\_0^t \boldsymbol{\varrho}(\tau) \left| \mathcal{L}^{-1} \left( \frac{1}{s^2 + s e^{-s}} \right) (t - \tau) \right| d\tau. \\ \square \end{split}$$

**Theorem 10.** *Let <sup>f</sup>* : <sup>R</sup> <sup>→</sup> <sup>R</sup> *such that <sup>f</sup>* ∈ O *and* L−<sup>1</sup> <sup>L</sup>(*f*) *s*2+*se*−*<sup>s</sup>* (0) = 0. *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* (13)*, then there exists a solution x*<sup>0</sup> : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> *of* (5) *such that*

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

**Proof.** From Theorem 9, we have that if *<sup>x</sup>* : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> satisfies the inequality (13), then there exists a solution *<sup>x</sup>*<sup>0</sup> : (0, <sup>∞</sup>) <sup>→</sup> <sup>R</sup> of (5) such that

$$|x(t) - x\_0(t)| \le \int\_0^t \tau^n \left| \mathcal{L}^{-1} \left( \frac{1}{s^2 + s e^{-s}} \right) (t - \tau) \right| d\tau.$$

is satisfied. We have

$$\begin{aligned} &\int\_0^t \tau^n \left| \mathcal{L}^{-1} \left( \frac{1}{s^2 + s e^{-s}} \right) (t - \tau) \right| d\tau = \int\_0^t \tau^n \sum\_{n=0}^{[t-\tau]} \frac{(t - \tau - n)^{n+1}}{(n+1)!} \\ &= \int\_0^t \tau^n \frac{\left( t - \tau - 0 \right)^1}{1!} d\tau + \int\_0^{t-1} \tau^n \frac{\left( t - \tau - 1 \right)^2}{2!} d\tau + \dots + \int\_0^{t-[t]} \tau^n \frac{\left( t - \tau - [t] \right)^{[t]+1}}{([t]+1)!} d\tau. \end{aligned}$$

Integrating by parts, we have

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

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

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

$$\int\_{0}^{\pi} \pi^{n} \frac{\frac{\pi^{n} \cdot \pi^{n-2} \cdot \pi^{n}}{2!} d\pi}{2!} d\pi = \int\_{0}^{\frac{\pi}{n}} \left( \frac{\frac{\pi}{n+1}}{n+1} \right)^{\frac{\pi}{n}} \frac{\frac{\pi^{n+1} \cdot \pi^{n} \cdot \pi^{n}}{2!} d\pi} d\pi = \underbrace{\left[ \frac{\pi^{n} \cdot \pi^{n+1} \cdot \pi^{n}}{n+1} \right]\_{0}^{\frac{\pi}{n}}}\_{0} + \int\_{0}^{\frac{\pi}{n}} \frac{\frac{\pi^{n+1} \cdot \pi^{n+2} \cdot \pi^{n}}{2!} d\pi}{\frac{\pi^{n+2} \cdot \pi^{n} \cdot \pi^{n}}{2!} d\pi} d\pi$$
 
$$= \int\_{0}^{t-1} \left( \frac{\pi^{n+2}}{(n+1)(n+2)} \right)' \frac{(t-\pi-2)}{1!} d\pi = \frac{\pi^{n+2}}{(n+1)(n+2)} \frac{(t-\pi-1)}{1!} \Big|\_{0}^{t-1} + \int\_{0}^{t-1} \frac{\pi^{n+2}}{(n+1)(n+2)} d\pi$$
  $= \frac{(t-1)^{n+3}}{(n+1)(n+2)(n+3)},$ 

···

$$\begin{split} &\int\_{0}^{t-[t]} \tau^{n} \frac{\left(t-\tau-[t]\right)^{[\ell]+1}}{\left([t]+1\right)!} d\tau = \int\_{0}^{t-[t]} \left(\frac{\tau^{n+1}}{n+1}\right)^{\prime} \frac{\left(t-\tau-[t]\right)^{[\ell]+1}}{\left([t]+1\right)!} d\tau \\ &= \underbrace{\frac{\tau^{n+1}}{n+1} \frac{\left(t-\tau-[t]\right)^{[\ell]+1}}{\left([t]+1\right)!} \Big|\_{0}^{t-[t]}}\_{0} + \int\_{0}^{t-[t]} \frac{\tau^{n+1}}{n+1} \frac{\left([t]+1\right)! \left(t-\tau-[t]\right)^{[\ell]}}{\left([t]+1\right)!} d\tau \\ &= \int\_{0}^{t-[t]} \left(\frac{\tau^{n+2}}{(n+1)(n+2)}\right)^{\prime} \frac{\left(t-\tau-[t]\right)^{[\ell]-1}}{[t]!} d\tau = \dots = \frac{(t-[t])^{n+[\ell]+2}}{(n+1)(n+2)\cdots(n+[\ell]+2)}. \end{split}$$
