**4. Ulam-Hyers Stabilty**

Two types of Ulam stability for (5) are discussed in this section, namely Ulam-Hyers and generalized Ulam-Hyers stability.

**Definition 4.** *Problem* (5) *is said to be Ulam-Hyers stable if there exists <sup>ω</sup>* <sup>∈</sup> <sup>R</sup>+\{0}*, such that for each* <sup>&</sup>gt; <sup>0</sup> *and solution x* ∈ C<sup>1</sup> <sup>1</sup>−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] *of the inequality*

$$|\mathcal{D}\_{0^{+}}^{r,p;\phi}\mathbf{x}(t) - f(t, \mathbf{x}(t), \mathbf{x}(\gamma t), ^{H}\mathcal{D}\_{0^{+}}^{r,p;\phi}\mathbf{x}(\gamma t))| \le \epsilon, \quad t \in \mathcal{J},\tag{42}$$

*there exists a solution z* ∈ C<sup>1</sup> <sup>1</sup>−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] *of equation* (5)*, such that*

$$|\mathbf{x}(t) - z(t)| \le \omega \epsilon\_{\prime} \quad t \in \mathcal{J}.$$

**Definition 5.** *Problem* (5) *is said to be generalized Ulam-Hyers stable if there exist* <sup>Φ</sup> ∈ C(R+, <sup>R</sup>+)*,* <sup>Φ</sup>*f*(0) = <sup>0</sup>*, such that for each solution <sup>x</sup>* ∈ C<sup>1</sup> <sup>1</sup>−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] *of the* (42)*, there exists a solution <sup>z</sup>* ∈ C<sup>1</sup> <sup>1</sup>−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] *of Equation* (5)*, such that*

$$|\mathbf{x}(t) - z(t)| \le \Phi\_f \epsilon, \quad t \in \mathcal{I}.$$

**Remark 1.** *A function <sup>x</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] *is a solution of the inequality* (42)*, if and only if there exist a function <sup>g</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] *such that:*

*(i)* |*g*(*t*)| ≤ , *t* ∈ J .

$$\mathbf{x}(\text{ii})\,\, ^H \mathcal{D}\_{0^+}^{r,p;\phi} \mathbf{x}(t) = f(t, \mathbf{x}(t), \mathbf{x}(\gamma t), ^H \mathcal{D}\_{0^+}^{r,p;\phi} \mathbf{x}(\gamma t)) + \mathbf{g}(t), \quad t \in \mathcal{I}.$$

**Lemma 9.** *Let* <sup>0</sup> <sup>&</sup>lt; *<sup>r</sup>* <sup>&</sup>lt; <sup>1</sup>*,* <sup>0</sup> <sup>≤</sup> *<sup>p</sup>* <sup>≤</sup> <sup>1</sup>*, if a function <sup>x</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] *is a solution of the inequality* (42)*, then x is a solution of the following integral inequality*

$$\left| \left| \mathbf{x}(t) - A\_{\mathbf{x}} - \frac{1}{\Gamma(r)} \int\_{0^+}^{t} \phi'(\mathbf{s}) (\phi(t) - \phi(\mathbf{s}))^{r-1} T\_{\mathbf{x}} d\mathbf{s} \right| \right| \leq \Omega \epsilon. \tag{43}$$

**Proof.** Clearly it follow from Remark 1 that

$$\begin{aligned} \, ^H \mathcal{D}\_{0^+}^{r, p; \phi} \mathbf{x}(t) &= f(t, \mathbf{x}(t), \mathbf{x}(\gamma t) \prime ^H \mathcal{D}\_{0^+}^{r, p; \phi} \mathbf{x}(\gamma t)) + \mathbf{g}(t) \\ &= T\_{\mathbf{x}}(t) + \mathbf{g}(t) . \end{aligned}$$

and

$$\begin{split} \mathbf{x}(t) &= \frac{\delta \Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r)} (\boldsymbol{\phi}(t) - \boldsymbol{\phi}(0))^{q-1} \sum\_{i=1}^{m} b\_{i} \left( \int\_{0^{+}}^{\mathbb{T}\_{i}} \boldsymbol{\phi}'(s) (\boldsymbol{\phi}(\boldsymbol{\xi}\_{i}) - \boldsymbol{\phi}(s))^{\rho+r-1} \boldsymbol{T}\_{\boldsymbol{x}}(s) ds \\ &+ \int\_{0^{+}}^{\mathbb{T}\_{i}} \boldsymbol{\phi}'(s) (\boldsymbol{\phi}(\boldsymbol{\xi}\_{i}) - \boldsymbol{\phi}(s))^{\rho+r-1} \boldsymbol{g}(s) ds \right) + \frac{1}{\Gamma(r)} \int\_{0^{+}}^{t} \boldsymbol{\phi}'(s) (\boldsymbol{\phi}(t) - \boldsymbol{\phi}(s))^{r-1} \boldsymbol{T}\_{\boldsymbol{x}}(s) ds \\ &+ \frac{1}{\Gamma(r)} \int\_{0^{+}}^{t} \boldsymbol{\phi}'(s) (\boldsymbol{\phi}(t) - \boldsymbol{\phi}(s))^{r-1} \boldsymbol{g}(s) ds. \end{split} \tag{44}$$

Hence

$$\begin{split} & \left| x(t) - A\_x - \frac{1}{\Gamma(r)} \int\_{0^+}^t \phi'(s)(\phi(t) - \phi(s))^{r-1} T\_x ds \right| \\ & = \left| \frac{\delta \Gamma(\rho + q)}{\Gamma(q) \Gamma(\rho + r)} (\phi(t) - \phi(0))^{q-1} \sum\_{i=1}^m b\_i \int\_{0^+}^{\zeta\_i} \phi'(s)(\phi(\zeta\_i) - \phi(s))^{\rho + r - 1} \varrho(s) ds \right| \\ & + \frac{1}{\Gamma(r)} \int\_{0^+}^t \phi'(s)(\phi(t) - \phi(s))^{r-1} \varrho(s) ds \Big| \\ & \leq \frac{|\delta| \Gamma(\rho + q)}{\Gamma(q) \Gamma(\rho + r)} (\phi(t) - \phi(0))^{q-1} \sum\_{i=1}^m |b\_i| \int\_{0^+}^{\zeta\_i} \phi'(s)(\phi(\zeta\_i) - \phi(s))^{\rho + r - 1} |\varrho(s)| ds \\ & + \frac{1}{\Gamma(r)} \int\_{0^+}^t \phi'(s)(\phi(t) - \phi(s))^{r-1} |\varrho(s)| ds \\ & \leq \Omega \epsilon. \end{split} \tag{45}$$

**Theorem 5.** *Suppose that the hypotheses* (*A*1)*,* (*A*3) *and* (*A*4) *are satisfied. Then problem* (5) *is both Ulam-Hyers and generalized Ulam-Hyers stable on* J *.*

**Proof.** Let <sup>&</sup>gt; 0 and *<sup>x</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] be a function which satisfies the inequality (42) and let *<sup>z</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] be a unique solution of the following implicit fractional pantograph differential equation

$$\begin{aligned} \mathcal{I}^{H\_{I}^{r},p\phi}z(t) &= f(t,z(t),z(\gamma t)^{H}\_{\cdot}\mathcal{D}^{r,p\phi}\_{0^{+}}z(\gamma t)) \mid \in \mathfrak{e}, \quad t \in \mathcal{J}, \quad 0 < r < 1, 0 \le p \le 1, \\\mathcal{I}^{1-q\phi\theta}\_{0^{+}}z(0^{+}) &= \mathcal{I}^{1-q\phi\theta}\_{0^{+}}z(0^{+}) = \sum\_{i=1}^{m} b\_{i} \mathcal{I}^{p\phi}\_{0^{+}}z(\underline{\mathfrak{z}}\_{i}), \quad \underline{\mathfrak{z}}\_{i} \in (0,T], \quad q = r + p - rp. \end{aligned}$$

Using Lemma 9, we have

$$z(t) = A\_z - \frac{1}{\Gamma(r)} \int\_{0^+}^{t} \phi'(s) (\phi(t) - \phi(s))^{r-1} T\_z(s) ds,$$

where

$$A\_z = \frac{\delta \Gamma(\rho + q)}{\Gamma(q)\Gamma(\rho + r)} (\phi(t) - \phi(a))^{q-1} \sum\_{i=1}^{m} b\_i \int\_{0^+}^{\tilde{\varsigma}\_i} \phi'(s) (\phi(\zeta\_i^{\underline{r}}) - \phi(s))^{\rho + r - 1} T\_z(s) ds.$$

Clearly, if *<sup>z</sup>*(*ξi*) = *<sup>x</sup>*(*ξi*) and <sup>I</sup>1−*q*;*<sup>φ</sup>* <sup>0</sup><sup>+</sup> *<sup>z</sup>*(0+) = <sup>I</sup>1−*q*;*<sup>φ</sup>* <sup>0</sup><sup>+</sup> *<sup>z</sup>*(0+), we get *Az* <sup>=</sup> *Ax* and that

$$\begin{split} &|A\_{2} - A\_{x}| \\ &= \frac{|\boldsymbol{\delta}|\Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r)} (\phi(t) - \phi(0))^{q-1} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\xi\_{i}} \phi'(\boldsymbol{s}) (\phi(\xi\_{i}) - \phi(\boldsymbol{s}))^{\rho+r-1} |T\_{z}(\boldsymbol{s}) - T\_{x}(\boldsymbol{s})| ds \\ &\leq \frac{|\boldsymbol{\delta}|\Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r)} (\phi(t) - \phi(0))^{q-1} \left(\frac{2K}{1-L}\right) \sum\_{i=1}^{m} b\_{i} \mathbb{Z}\_{0^{+}}^{\rho+r\varphi} |\boldsymbol{z}(\boldsymbol{s}) - \boldsymbol{x}(\boldsymbol{s})| (\xi\_{i}) \\ &= 0. \end{split}$$

Now for any *t* ∈ J and Lemma 9, we have

$$\begin{split} |\mathbf{x}(t) - \mathbf{z}(t)| &= \left| \mathbf{x}(t) - A\_{\mathbf{x}} - \frac{1}{\Gamma(r)} \int\_{0^{+}}^{t} \phi'(s)(\phi(t) - \phi(s))^{r-1} T\_{\mathbf{x}}(s) ds \right| \\ &+ \frac{1}{\Gamma(r)} \int\_{0^{+}}^{t} \phi'(s)(\phi(t) - \phi(s))^{r-1} |T\_{\mathbf{x}}(s) - T\_{\mathbf{z}}(s)| ds \\ &\leq \left| \mathbf{x}(t) - A\_{\mathbf{x}} - \frac{1}{\Gamma(r)} \int\_{0^{+}}^{t} \phi'(s)(\phi(t) - \phi(s))^{r-1} T\_{\mathbf{x}}(s) ds \right| \\ &+ \left( \frac{2K}{1 - L} \right) \frac{1}{\Gamma(r)} \int\_{0^{+}}^{t} \phi'(s)(\phi(t) - \phi(s))^{r-1} |\mathbf{x}(s) - \mathbf{z}(s)| ds \\ &\leq \Omega \mathbf{c} + \left( \frac{2K}{1 - L} \right) \frac{\mathcal{B}(r, q)(\phi(T) - \phi(0))^{r}}{\Gamma(r)} |\mathbf{x}(t) - \mathbf{z}(t)| ds. \end{split}$$

Thus,

$$|x(t) - z(t)| \le a\epsilon\_{\prime}$$

where

$$
\omega = \frac{\Omega (1 - L) \Gamma(r)}{(1 - L)\Gamma(r) - 2K(\phi(T) - \phi(0))^r \mathcal{B}(r, q)}.
$$

Therefore, problem (5) is Ulam-Hyers stable. Moreover, if we set Φ*f*() = *ω* such that Φ*f*(0) = 0, then problem (5) is generalized Ulam-Hyers stable.

#### **5. Examples**

**Example 1.** *Consider the implicit fractional pantograph differential equation which involves φ-Hilfer fractional derivative of the following form:*

$$\begin{cases} {}^{H}\mathcal{D}\_{0^{+}}^{\frac{2}{3},\frac{1}{2}t}z(t) = \frac{1}{3(5^{2t}+5)[1+|z(t)|+|z(\frac{1}{2}t)|+|^{H}\mathcal{D}\_{1^{+}}^{\frac{2}{3},\frac{1}{2}t}z(\frac{1}{2}t))]}, & t \in \mathcal{I} = (0,2],\\ {}^{1-\frac{5}{6}t}z(0) = 3\mathcal{Z}\_{0^{+}}^{\frac{1}{2},\frac{1}{2}t}z(\frac{3}{2}), & \frac{2}{3} \le \frac{5}{6} = \frac{2}{3} + (\frac{1}{2}) - (\frac{2}{3})(\frac{1}{2}). \end{cases} \tag{46}$$

*By comparing* (5) *with* (46)*, we have:*

*r* = <sup>2</sup> <sup>3</sup> *, <sup>p</sup>* <sup>=</sup> *<sup>γ</sup>* <sup>=</sup> *<sup>ρ</sup>* <sup>=</sup> <sup>1</sup> <sup>2</sup> *, <sup>q</sup>* <sup>=</sup> <sup>5</sup> <sup>6</sup> *, T* = 2 *and φ*(·) = *t. Also from the initial condition we can easily see that b*<sup>1</sup> = 3 *since m* = 1*, ξ*<sup>1</sup> = <sup>3</sup> <sup>2</sup> ∈ J *and f* : J × <sup>R</sup><sup>3</sup> <sup>→</sup> <sup>R</sup> *is a function defined by*

$$f(t, u, v, w) = \frac{1}{3(5^{2t} + 5)(1 + |u| + |v| + |w|)}, \\ t \in \mathcal{I}\_{\prime} \quad u, v, w \in \mathbb{R}\_{+} .$$

*Obviously, f is continuous and for all u*, *<sup>v</sup>*, *<sup>w</sup>*, *<sup>u</sup>*¯, *<sup>v</sup>*¯, *<sup>w</sup>*¯ <sup>∈</sup> <sup>R</sup><sup>+</sup> *and t* ∈ J *, we have* <sup>|</sup> *<sup>f</sup>*(*t*, *<sup>u</sup>*, *<sup>v</sup>*, *<sup>w</sup>*) <sup>−</sup> *<sup>f</sup>*(*t*, *<sup>u</sup>*¯, *<sup>v</sup>*¯, *<sup>w</sup>*¯)| ≤ <sup>1</sup> <sup>90</sup> (|*u* − *u*¯| + |*v* − *v*¯| + |*w* − *w*¯|). *Thus, it follows that conditions* (*A*1) *and* (*A*3) *are true with K* = *L* = <sup>1</sup> <sup>90</sup> . *Therefore, by simple calculation, we get* |*δ*| ≈ 0.3935 *and*

$$\left(\frac{2K}{1-L}\right)\Omega \approx 0.0642 < 1.1$$

*Since, all the assumptions of Theorem 4 are satisfied. Then problem* (5) *has a unique solution on* J *. However, we can also find out that* Ω ≈ 2.8551 > 0 *and ω* = 2.9321 > 0*. Hence, by Theorem 5, problem* (5) *is both Ulam-Hyers and also generalized Ulam-Hyers stable.*

**Example 2.** *Consider the implicit fractional pantograph differential equation which involves φ-Hilfer fractional derivative of form:*

$$\begin{cases} \begin{aligned} ^{H}\mathcal{D}\_{0^{+}}^{\frac{1}{2},\frac{2}{3};\sqrt{7}}z(t) &= \frac{2+|z(t)|+|z(\frac{3}{2}t)|+\left|\prescript{H}{}{\mathcal{D}}\_{0^{+}}^{\frac{2}{3},\frac{2}{3};\sqrt{7}}z(\frac{3}{2}t)\right|}{95e^{2t}\cos 2t\left(1+|z(t)|+|z(\frac{3}{2}t)|+\left|\prescript{H}{}{\mathcal{D}}\_{0^{+}}^{\frac{1}{2},\frac{2}{3};\sqrt{7}}z(\frac{3}{2}t)\right|\right)}, & t \in \mathcal{I} = (0,1],\\ \mathcal{I}\_{0^{+}}^{1-\frac{7}{3};\sqrt{7}}z(0) &= z(\frac{1}{2}) + 3z(\frac{4}{5}), \quad \qquad \qquad \qquad \qquad \frac{2}{3} < \frac{7}{9} = \frac{1}{3} + (\frac{2}{3}) - (\frac{1}{3})(\frac{2}{3}). \end{aligned} \end{cases} \tag{47}$$

*By comparing Equation* (47) *with Equation* (5)*, we obtain that: r* = <sup>1</sup> <sup>3</sup> *, <sup>p</sup>* <sup>=</sup> <sup>2</sup> <sup>3</sup> *, <sup>q</sup>* <sup>=</sup> <sup>7</sup> <sup>9</sup> *, <sup>ρ</sup>* <sup>=</sup> <sup>0</sup>*, <sup>γ</sup>* <sup>=</sup> <sup>3</sup> <sup>2</sup> *, <sup>T</sup>* <sup>=</sup> <sup>1</sup> *and <sup>φ</sup>*(·) = <sup>√</sup>*t. Also we can easily see that <sup>b</sup>*<sup>1</sup> <sup>=</sup> <sup>1</sup>*, <sup>b</sup>*<sup>2</sup> <sup>=</sup> <sup>3</sup> *since m* = 2*, ξ*<sup>1</sup> = <sup>1</sup> <sup>2</sup> , *<sup>ξ</sup>*<sup>2</sup> <sup>=</sup> <sup>4</sup> <sup>5</sup> ∈ J *and f* : J × <sup>R</sup><sup>3</sup> <sup>→</sup> <sup>R</sup> *is a function defined by*

$$f(t, u, v, w) = \frac{2 + |u| + |v| + |w|}{95e^{2t} \cos 2t \left(1 + |u| + |v| + |w|\right)'}, \quad t \in \mathcal{J}, \quad u, v, w \in \mathbb{R}\_+.$$

*Thus, f is continuous and we can see that, for all u*, *<sup>v</sup>*, *<sup>w</sup>*, *<sup>u</sup>*¯, *<sup>v</sup>*¯, *<sup>w</sup>*¯ <sup>∈</sup> <sup>R</sup><sup>+</sup> *and t* ∈ J *,* <sup>|</sup> *<sup>f</sup>*(*t*, *<sup>u</sup>*, *<sup>v</sup>*, *<sup>w</sup>*) <sup>−</sup> *<sup>f</sup>*(*t*, *<sup>u</sup>*¯, *<sup>v</sup>*¯, *<sup>w</sup>*¯)| ≤ <sup>1</sup> <sup>95</sup> (|*u* − *u*¯| + |*v* − *v*¯| + |*w* − *w*¯|). *So assumptions* (*A*1) *and* (*A*3) *are fulfilled with K* = *L* = <sup>1</sup> <sup>95</sup> . *Furthermore,*

$$|f(t, \mu, v, w)| \le \frac{1}{95c^{2t} \cos 2t} (2 + |\mu| + |v| + |w|), \quad t \in \mathcal{J}.$$

*The above implies that* (*A*2) *is true with k*(*t*) = <sup>2</sup> 95*e*2*<sup>t</sup>* cos 2*t , l*(*t*) = *m*(*t*) = *n*(*t*) = <sup>1</sup> <sup>95</sup>*e*2*<sup>t</sup>* cos 2*<sup>t</sup> and <sup>k</sup>*<sup>∗</sup> <sup>=</sup> <sup>2</sup> 95 *, l* <sup>∗</sup> = *m*<sup>∗</sup> = *n*<sup>∗</sup> = <sup>1</sup> <sup>95</sup> *. Therefore, all the hypotheses of Theorem 4 are satisfied, which means that problem* (5) *has at least one solution on* J *. Moreover, by using the same procedure as in example* 5.2*, we obtain, that* |*δ*| ≈ 1.1025*,* Ω ≈ 3.6662 > 0 *and*

$$\left(\frac{2K}{1-L}\right)\Omega \approx 0.0782 < 1.1$$

*Thus, all the hypotheses of Theorem 4 holds. Hence, problem* (5) *has a unique solution on* J *.*

**Example 3.** *Consider the implicit fractional pantograph differential equation which involves φ-Hilfer fractional derivative of the following form:*

$$\begin{cases} {}^{H}\mathcal{D}\_{0^{+}}^{\frac{1}{2}\frac{1}{3}t^{2}}z(t) = \frac{1}{4^{t+3}[1+|z(t)|+|z(\frac{1}{6}t)|+|^{H}\mathcal{D}\_{0^{+}}^{\frac{1}{2}\frac{1}{3}t^{2}}z(\frac{1}{6}t)]}, & t \in \mathcal{J} = (0,3],\\ {}^{1-qt}z(0) = \sqrt{2}\mathcal{D}\_{0^{+}}^{\frac{1}{2}\frac{1}{3}t}z(2) + \sqrt{5}\mathcal{D}\_{0^{+}}^{\frac{3}{2}t}z(\frac{5}{2}), & q = \frac{1}{2} + (\frac{1}{3}) - (\frac{1}{2})(\frac{1}{3}). \end{cases} \tag{48}$$

*By comparing Equation* (5) *with Equation* (48)*, we get the followings values: r* = <sup>1</sup> <sup>2</sup> *, <sup>p</sup>* <sup>=</sup> <sup>1</sup> <sup>3</sup> *<sup>γ</sup>* <sup>=</sup> <sup>1</sup> <sup>6</sup> *<sup>ρ</sup>* <sup>=</sup> <sup>2</sup> <sup>5</sup> *, <sup>q</sup>* <sup>=</sup> <sup>2</sup> <sup>3</sup> *, T* = 3 *and φ*(·) = *t. Also from the initial condition we can easily see that <sup>b</sup>*<sup>1</sup> <sup>=</sup> <sup>√</sup><sup>2</sup> *<sup>b</sup>*<sup>1</sup> <sup>=</sup> <sup>√</sup><sup>5</sup> *since m* <sup>=</sup> <sup>2</sup>*, <sup>ξ</sup>*<sup>1</sup> <sup>=</sup> <sup>2</sup> *<sup>ξ</sup>*<sup>2</sup> <sup>=</sup> <sup>5</sup> <sup>2</sup> *and f* : J × <sup>R</sup><sup>3</sup> <sup>→</sup> <sup>R</sup> *is a function defined by*

$$f(t, u, v, w) = \frac{1}{4^{t+3}(1 + |u| + |v| + |w|)}, \\ t \in \mathcal{J}, \quad u, v, w \in \mathbb{R}\_+.$$

*Thus, f is continuous and for all u*, *<sup>v</sup>*, *<sup>w</sup>*, *<sup>u</sup>*¯, *<sup>v</sup>*¯, *<sup>w</sup>*¯ <sup>∈</sup> <sup>R</sup><sup>+</sup> *and t* ∈ J *, yields* <sup>|</sup> *<sup>f</sup>*(*t*, *<sup>u</sup>*, *<sup>v</sup>*, *<sup>w</sup>*) <sup>−</sup> *<sup>f</sup>*(*t*, *<sup>u</sup>*¯, *<sup>v</sup>*¯, *<sup>w</sup>*¯)| ≤ <sup>1</sup> <sup>64</sup> (|*u* − *u*¯| + |*v* − *v*¯| + |*w* − *w*¯|). *Hence, it follows that conditions* (*A*1) *and* (*A*3) *are true with K* = *L* = <sup>1</sup> <sup>90</sup> . *Therefore, by substitution these values, we get* |*δ*| ≈ 0.3456*,* Ω ≈ 7.4535 > 0 *and*

$$
\left(\frac{2K}{1-L}\right)\Omega \approx 0.2366 < 1,
$$

*which implies that, all the assumptions of Theorem 4 are satisfied. Thus, problem* (5) *has a unique solution on* J *.*
