**4. Examples**

Let *α*1 = 1/3, *α*2 = 1/2, *β*1 = 5/2, (*n* = 3), *β*2 = 13/4, (*m* = 4), *r*1 = 4, 1 = 4/3, *r*2 = 5, 2 = 5/4, *p* = 2, *q* = 1, *γ*0 = 4/3, *γ*1 = 1/4, *γ*2 = 6/5, *δ*0 = 11/5, *δ*1 = 7/6, *<sup>H</sup>*1(*t*) = *t*/3 for all *t* ∈ [0, 1], *<sup>H</sup>*2(*t*) = {1/6, *t* ∈ [0, 2/3); 2/3, *t* ∈ [2/3, <sup>1</sup>]}, *<sup>K</sup>*1(*t*) = {1/4, *t* ∈ [0, 1/3); 9/4, *t* ∈ [1/3, <sup>1</sup>]}.

⎪⎪⎨

We consider the system of fractional differential equations

$$\begin{cases} D\_{0+}^{1/3}(\varphi\_4(D\_{0+}^{5/2}u(t))) + f(t, u(t), v(t)) = 0, & t \in (0, 1), \\ D\_{0+}^{1/2}(\varphi\_5(D\_{0+}^{13/4}v(t))) + \mathfrak{g}(t, u(t), v(t)) = 0, & t \in (0, 1), \end{cases} \tag{33}$$

with the nonlocal boundary conditions

$$\begin{cases} \boldsymbol{u}(0) = \boldsymbol{u}'(0) = 0, \ D\_{0+}^{5/2} \boldsymbol{u}(0) = 0, \ D\_{0+}^{4/3} \boldsymbol{u}(1) = \frac{1}{3} \int\_0^1 D\_{0+}^{1/4} \boldsymbol{u}(t) \, dt + \frac{1}{2} D\_{0+}^{6/5} \boldsymbol{u}\left(\frac{2}{3}\right), \\\ \boldsymbol{v}(0) = \boldsymbol{v}'(0) = \boldsymbol{v}''(0) = 0, \ D\_{0+}^{13/4} \boldsymbol{v}(0) = 0, \ D\_{0+}^{11/5} \boldsymbol{v}(1) = 2 D\_{0+}^{7/6} \boldsymbol{v}\left(\frac{1}{3}\right). \end{cases} \tag{34}$$

We obtain here Δ1 ≈ 0.60331103 > 0 and Δ2 ≈ 1.12479609 > 0. We also find

*g*1(*<sup>t</sup>*,*<sup>s</sup>*) = 1 Γ(5/2) - *t*3/2(1 − *s*)1/6 − (*t* − *s*)3/2, 0 ≤ *s* ≤ *t* ≤ 1, *t*3/2(1 − *s*)1/6, 0 ≤ *t* ≤ *s* ≤ 1, *g*21(*<sup>t</sup>*,*<sup>s</sup>*) = 1 Γ(9/4) - *t*5/4(1 − *s*)1/6 − (*t* − *s*)5/4, 0 ≤ *s* ≤ *t* ≤ 1, *t*5/4(1 − *s*)1/6, 0 ≤ *t* ≤ *s* ≤ 1, *g*22(*<sup>t</sup>*,*<sup>s</sup>*) = 1 Γ(13/10) - *t*3/10(1 − *s*)1/6 − (*t* − *s*)3/10, 0 ≤ *s* ≤ *t* ≤ 1, *t*3/10(1 − *s*)1/6, 0 ≤ *t* ≤ *s* ≤ 1, *g*3(*<sup>t</sup>*,*<sup>s</sup>*) = 1 Γ(13/4) - *t*9/4(1 − *s*)1/20 − (*t* − *s*)9/4, 0 ≤ *s* ≤ *t* ≤ 1, *t*9/4(1 − *s*)1/20, 0 ≤ *t* ≤ *s* ≤ 1, *g*41(*<sup>t</sup>*,*<sup>s</sup>*) = 1 Γ(25/12) - *t*13/12(1 − *s*)1/20 − (*t* − *s*)13/12, 0 ≤ *s* ≤ *t* ≤ 1, *t*13/12(1 − *s*)1/20, 0 ≤ *t* ≤ *s* ≤ 1, G1(*<sup>t</sup>*,*<sup>s</sup>*) = *g*1(*<sup>t</sup>*,*<sup>s</sup>*) + *t*3/2 Δ1 1 3 1 0 *g*21(*<sup>τ</sup>*,*<sup>s</sup>*) *dτ* + 1 2 *g*22 2 3 ,*s*! , *t*,*<sup>s</sup>* ∈ [0, 1], G2(*<sup>t</sup>*,*<sup>s</sup>*) = *g*3(*<sup>t</sup>*,*<sup>s</sup>*) + 2*t*9/4 Δ2 *g*41 1 3 ,*s* , *t*,*<sup>s</sup>* ∈ [0, 1], *h*1(*s*) = 1 Γ(5/2)[(<sup>1</sup> − *s*)1/6 − (1 − *<sup>s</sup>*)3/2], *s* ∈ [0, 1], *h*2(*s*) = 1 Γ(13/4)[(<sup>1</sup> − *s*)1/20 − (1 − *<sup>s</sup>*)9/4], *s* ∈ [0, 1].

In addition we deduce

$$\mathcal{J}\_{1}(s) = \begin{cases} h\_{1}(s) + \frac{1}{\Lambda\_{1}} \left\{ \frac{4}{2\Gamma(9/4)} (1-s)^{1/6} - \frac{4}{2\Gamma(9/4)} (1-s)^{9/4} + \frac{1}{2\Gamma(13/10)} \right\} \\ \quad \times \left[ \left( \frac{2}{3} \right)^{3/10} (1-s)^{1/6} - \left( \frac{2}{3} - s \right)^{3/10} \right], & 0 \le s < \frac{2}{3}, \\\ h\_{1}(s) + \frac{1}{\Lambda\_{1}} \left[ \frac{4}{2\Gamma(9/4)} (1-s)^{1/6} - \frac{4}{2\Gamma(9/4)} (1-s)^{9/4} + \frac{1}{2\Gamma(13/10)} \right. \\ \left. \times \left( \frac{2}{3} \right)^{3/10} (1-s)^{1/6} \right], & 2/3 \le s \le 1, \\\ h\_{2}(s) + \frac{2}{\Lambda\_{2} \Gamma(25/12)} \left[ \left( \frac{1}{3} \right)^{13/12} (1-s)^{1/20} - \left( \frac{1}{3} - s \right)^{13/12} \right], & 0 \le s < \frac{1}{3}, \\\ h\_{2}(s) + \frac{2}{\Lambda\_{2} \Gamma(25/12)} \left( \frac{1}{3} \right)^{13/12} (1-s)^{1/20}, & \frac{1}{3} \le s \le 1. \end{cases}$$

**Example 1.** *We consider the functions*

$$f(t,x,y) = \frac{(x+y)^{3a}}{t^{\eta\_1}(1-t)^{\eta\_2}}, \; g(t,x,y) = \frac{(x+y)^{4b}}{t^{\eta\_3}(1-t)^{\eta\_4}}, \; t \in (0,1), \; x,y \ge 0,\tag{35}$$

*where a* > 1*, b* > 1*, η*1, *η*2 ∈ (0, 1/4)*, η*3, *η*4 ∈ (0, 1/3)*. Here f*(*<sup>t</sup>*, *x*, *y*) = *ζ*1(*t*)*<sup>χ</sup>*1(*<sup>t</sup>*, *x*, *y*)*, g*(*<sup>t</sup>*, *x*, *y*) = *ζ*2(*t*)*<sup>χ</sup>*2(*<sup>t</sup>*, *x*, *y*)*, ζ*1(*t*) = 1 *t η*1 (<sup>1</sup>−*<sup>t</sup>*)*<sup>η</sup>*<sup>2</sup> *, ζ*2(*t*) = 1 *t η*3 (<sup>1</sup>−*<sup>t</sup>*)*<sup>η</sup>*<sup>4</sup> *for all t* ∈ (0, <sup>1</sup>)*, <sup>χ</sup>*1(*<sup>t</sup>*, *x*, *y*)=(*x* + *y*)<sup>3</sup>*a, <sup>χ</sup>*2(*<sup>t</sup>*, *x*, *y*) = (*x* + *y*)<sup>4</sup>*<sup>b</sup> for all t* ∈ [0, 1], *x*, *y* ≥ 0*. By using the Hölder inequality, we obtain*

0 < Λ1 = 10 (1 − *<sup>s</sup>*)*β*<sup>1</sup>−*γ*0−<sup>1</sup>*ϕ*1 (*I<sup>α</sup>*<sup>1</sup> <sup>0</sup>+*ζ*1(*s*)) *ds* = 10 (1 − *s*)1/6 *I*1/3 0+ *<sup>ζ</sup>*1(*s*))1/3*ds* = 1 (Γ(1/3))1/3 10 (1 − *s*)1/6 *s* 0 (*s* − *τ*)−2/3 1 *τη*1 (1 − *τ*)*<sup>η</sup>*<sup>2</sup> *<sup>d</sup>τ*1/3*ds* ≤ 1 (Γ(1/3))1/3 10 (1 − *s*)1/6 ⎡⎣ *s* 0 (*s* − *τ*)−8/9*dτ*3/4 *s* 0 1 *τη*1 (1 − *τ*)*<sup>η</sup>*<sup>2</sup> 4 *<sup>d</sup>τ*1/4⎤⎦1/3*ds* ≤ 1 (Γ(1/3))1/3 10 (1 − *s*)1/6 ((<sup>9</sup>*s*1/9)3/4(*B*(<sup>1</sup> − 4*η*1, 1 − <sup>4</sup>*η*2))1/4)1/3*ds* = 31/2 (Γ(1/3))1/3 (*B*(1 − 4*η*1, 1 − <sup>4</sup>*η*2))1/12*<sup>B</sup>* 3736, 76 < <sup>∞</sup>, 0 < Λ2 = 10 (1 − *<sup>s</sup>*)*β*<sup>2</sup>−*δ*0−<sup>1</sup>*ϕ*2 (*I<sup>α</sup>*2 <sup>0</sup>+*ζ*2(*s*)) *ds* = 10 (1 − *s*)1/20 *I*1/2 0+ *<sup>ζ</sup>*2(*s*)1/4*ds* = 1 (Γ(1/2))1/4 10 (1 − *s*)1/20 *s* 0 (*s* − *τ*)−1/2 1 *τη*3 (1 − *τ*)*<sup>η</sup>*<sup>4</sup> *<sup>d</sup>τ*1/4*ds* ≤ 1 (Γ(1/2))1/4 10 (1 − *s*)1/20 & *s* 0 (*s* − *τ*)−3/4*dτ*2/3 *s* 0 *τ*<sup>−</sup>3*η*<sup>3</sup> (1 − *τ*)−3*η*<sup>4</sup> *<sup>d</sup>τ*1/3'1/4*ds* ≤ 1 (Γ(1/2))1/4 10 (1 − *s*)1/20[(<sup>4</sup>*s*1/4)2/3(*B*(<sup>1</sup> − 3*η*3, 1 − <sup>3</sup>*η*4))1/3]1/4 *ds* = 21/3 (Γ(1/2))1/4 (*B*(1 − 3*η*3, 1 − <sup>3</sup>*η*4))1/12*<sup>B</sup>* 2524, 2120 < ∞.

Hence assumptions (*I*1) and (*I*2) are satisfied.

In addition, in (*<sup>I</sup>*3), for *μ*1 = *μ*2 = 1, we obtain *χ*10 = *χ*20 = 0, and in (*I*4) for [*<sup>a</sup>*1, *<sup>a</sup>*2] ⊂ (0, 1) we have *f i*∞ = ∞ (and *gi*∞ = ∞). Then by Theorem 1, we conclude that problem (33) and (34) with the nonlinearities (35) has at least one positive solution (*u*(*t*), *<sup>v</sup>*(*t*)), *t* ∈ [0, 1].

**Example 2.** *We consider the functions*

$$\begin{cases} f(t, \mathbf{x}, \mathbf{y}) = \frac{c\_0(t+1)}{(t^2+4)\sqrt[5]{t}} [(\mathbf{x}+\mathbf{y})^{\sigma\_1} + (\mathbf{x}+\mathbf{y})^{\sigma\_2}], \ t \in (0, 1], \ \mathbf{x}, \mathbf{y} \ge 0, \\\ g(t, \mathbf{x}, \mathbf{y}) = \frac{d o(2+\sin t)}{(t+3)^4 \sqrt[5]{1-t}} (\mathbf{x}^{\sigma\_3} + \mathbf{y}^{\sigma\_4}), \ t \in [0, 1), \ \mathbf{x}, \mathbf{y} \ge 0, \end{cases} \tag{36}$$

*where c*0 > 0*, d*0 > 0*, σ*1 > 3*, σ*2 ∈ (0, <sup>3</sup>)*, σ*3 > 0*, σ*4 > 0*. Here we have ζ*1(*t*) = 1√5 *t* , *t* ∈ (0, 1]*, <sup>χ</sup>*1(*<sup>t</sup>*, *x*, *y*) = *<sup>c</sup>*0(*<sup>t</sup>*+<sup>1</sup>) *t*2+4 [(*x* + *y*)*<sup>σ</sup>*<sup>1</sup> + (*x* + *y*)*<sup>σ</sup>*<sup>2</sup> ], *t* ∈ [0, 1], *x*, *y* ≥ 0*, ζ*2(*t*) = 1 √4 1−*t* , *t* ∈ [0, <sup>1</sup>)*, <sup>χ</sup>*2(*<sup>t</sup>*, *x*, *y*) = *d*0(<sup>2</sup>+sin *t*) (*t*+<sup>3</sup>)<sup>4</sup> (*xσ*<sup>3</sup> + *y<sup>σ</sup>*<sup>4</sup> ), *t* ∈ [0, 1], *x*, *y* ≥ 0*. By using a computer program, we obtain*

$$\begin{split} \Lambda\_{1} &= \int\_{0}^{1} (1-s)^{\beta\_{1}-\gamma\_{0}-1} \varphi\_{\boldsymbol{\varrho}\_{1}}(I\_{0+}^{a\_{1}}\zeta\_{1}(s)) \, ds = \int\_{0}^{1} (1-s)^{1/6} (I\_{0+}^{1/3}\zeta\_{1}(s))^{1/3} \, ds \\ &= \frac{1}{(\Gamma(1/3))^{1/3}} \int\_{0}^{1} (1-s)^{1/6} \left( \int\_{0}^{s} \tau^{-1/5} (s-\tau)^{-2/3} \, d\tau \right)^{1/3} \, ds \\ &\overset{\mathsf{r}=\mathsf{s}x}{=} \frac{1}{(\Gamma(1/3))^{1/3}} \int\_{0}^{1} (1-s)^{1/6} \left( \int\_{0}^{1} (s\boldsymbol{x})^{-1/5} (s-\boldsymbol{x}\boldsymbol{x})^{-2/3} \boldsymbol{\varsigma} \, d\boldsymbol{x} \right)^{1/3} \, ds \\ &= \frac{1}{(\Gamma(1/3))^{1/3}} \left( \int\_{0}^{1} \boldsymbol{\varsigma}^{2/45} (1-s)^{1/6} \, ds \right) \left( \int\_{0}^{1} \boldsymbol{\varsigma}^{-1/5} (1-\boldsymbol{x})^{-2/3} \, d\boldsymbol{x} \right)^{1/3} \\ &= \frac{1}{(\Gamma(1/3))^{1/3}} B \left( \frac{47}{45}, \frac{7}{6} \right) \left( B \left( \frac{4}{5}, \frac{1}{3} \right) \right)^{1/3} \approx 0.877777, \end{split}$$

$$\begin{split} \Lambda\_{2} &= \int\_{0}^{1} (1-s)^{\beta\_{2}-\delta\_{0}-1} \varrho\_{\ell\_{0}^{2}}(I\_{0+}^{\alpha\_{2}}\zeta\_{2}(s)) \, ds = \int\_{0}^{1} (1-s)^{1/20} (I\_{0+}^{1/2}\zeta\_{2}(s))^{1/4} \, ds \\ &= \frac{1}{(\Gamma(1/2))^{1/4}} \int\_{0}^{1} (1-s)^{1/20} \left( \int\_{0}^{s} (s-\tau)^{-1/2} (1-\tau)^{-1/4} \, d\tau \right)^{1/4} \, ds \\ &\overset{\text{tr}=\text{sr}}{=} \frac{1}{(\Gamma(1/2))^{1/4}} \int\_{0}^{1} (1-s)^{1/20} \left( \int\_{0}^{1} (s-\text{sr})^{-1/2} (1-\text{sr})^{-1/4} \, \text{s} \, \text{d}x \right)^{1/4} \, \text{ds} \\ &= \frac{1}{(\Gamma(1/2))^{1/4}} \int\_{0}^{1} (1-s)^{1/20} \left( s^{1/2} \int\_{0}^{1} (1-\text{s})^{-1/2} (1-\text{s})^{-1/4} \, \text{d}x \right)^{1/4} \, \text{ds} \\ &= \frac{1}{(\Gamma(1/2))^{1/4}} \int\_{0}^{1} (1-s)^{1/20} \left( s^{1/2} \sqrt{\pi}\_{2} F\_{1} \left[ \frac{1}{4}, 1, \frac{3}{2}, s \right] \right)^{1/4} \, \text{ds} \approx 0.901313, \end{split}$$

*where* <sup>2</sup>*F*1[*<sup>a</sup>*, *b*, *c*, *z*] = 1 <sup>Γ</sup>(*b*)Γ(*<sup>c</sup>*−*b*) 10 *sb*−<sup>1</sup>(<sup>1</sup> − *s*)*<sup>c</sup>*−*b*−<sup>1</sup>(<sup>1</sup> − *sz*)−*<sup>a</sup> ds is the regularized hypergeometric function. So* Λ*i* ∈ (0, <sup>∞</sup>), *i* = 1, 2*, and then assumptions* (*I*1) *and* (*I*2) *are satisfied.*

For [*<sup>a</sup>*1, *<sup>a</sup>*2] ⊂ (0, <sup>1</sup>), we find *f i*∞ = <sup>∞</sup>, and if we consider 0 < *ν*1 ≤ 1, 3*<sup>ν</sup>*1 > *σ*2 we obtain *f i*0 = <sup>∞</sup>, and then assumptions (*I*4) and (*I*6) are satisfied. After some computations we deduce

*M*1 = 10 J1(*s*)*ϕ*1 (*I<sup>α</sup>*<sup>1</sup> <sup>0</sup>+*ζ*1(*s*)) *ds* = 10 J1(*s*)*ϕ*4/3 *I*1/3 0+ 1√5 *s ds* = 1 (Γ(1/3))1/3 10 J1(*s*) *s* 0 (*s* − *τ*)−2/3*τ*<sup>−</sup>1/5*dτ*1/3 *ds* = 1 (Γ(1/3))1/3 10 J1(*s*) *s*2/15*<sup>B</sup>* 45, 131/3 *ds* ≈ 0.78160052, *M*2 = 10 J2(*s*)*ϕ*2 (*I<sup>α</sup>*2 <sup>0</sup>+*ζ*2(*s*)) *ds* = 10 <sup>J</sup>2(*s*)*ϕ*5/4(*I*1/2 0+ *ζ*2(*s*)) *ds* = 10 J2(*s*) 1 Γ(1/2) *s* 0 (*s* − *τ*)−1/2(<sup>1</sup> − *τ*)−1/4*dτ*1/4 *ds* = 1 (Γ(1/2))1/4 10 J2(*s*) *s*1/2√*π* 2*F*1 14, 1, 32,*s*!1/4 *ds* ≈ 0.65997289.

In addition, we find *D*0 = max 2*c*05 (2*<sup>σ</sup>*1 + 2*<sup>σ</sup>*2 ), 4*d*0 81 . If *c*0 < min+ 5 <sup>16</sup>*M*31(2*<sup>σ</sup>*1+2*<sup>σ</sup>*2 ), 5 <sup>32</sup>*<sup>M</sup>*42(2*<sup>σ</sup>*1+2*<sup>σ</sup>*2 ), and *d*0 < min + 81 32*M*31 , 81 64*M*42 ,, then the inequalities *D*1/3 0 *M*1 < 12 and *D*1/4 0 *M*2 < 12 are satisfied (that is, assumption (*I*7) is satisfied). For example, if *σ*1 = 4 and *σ*2 = 2, and *c*0 ≤ 0.032 and *d*0 ≤ 5.301, then the above inequalities are satisfied. By Theorem 3, we conclude that problem (33) and (34) with the nonlinearities (36) has at least two positive solutions (*<sup>u</sup>*1(*t*), *<sup>v</sup>*1(*t*)), (*<sup>u</sup>*2(*t*), *<sup>v</sup>*2(*t*)), *t* ∈ [0, 1].
