**4. Examples**

Two examples are studied to prove the applicability of Theorems 2 and 3.

**Example 1.** *Consider the NFOTDSs* (2)*, with λ*<sup>2</sup> = 0.7*, λ*<sup>1</sup> = 0.2*, ς*(*s*) = 0.1,

$$\boldsymbol{\upsilon}(\boldsymbol{\tau}) = \begin{pmatrix} \mathbf{0}.5, \mathbf{0} \end{pmatrix}^T, \quad \boldsymbol{\zeta}(\boldsymbol{\tau}) = \begin{pmatrix} \mathbf{0}.05, \mathbf{0} \end{pmatrix}^T, \text{for } \boldsymbol{\tau} \in \left[ -\mathbf{0}.1, \mathbf{0} \right],$$

$$F(\mathbf{s}, \mathbf{x}(\mathbf{s}), \mathbf{x}(\mathbf{s} - \boldsymbol{\xi}(\mathbf{s})), \boldsymbol{\upsilon}(\mathbf{s})) = \mathbf{0}.01 \Big( \sin \left( \mathbf{x}\_2(\mathbf{s} - \boldsymbol{\xi}(\mathbf{s})) \right), \sin \left( \mathbf{x}\_1(\mathbf{s}) \right) \Big)^T$$

*and*

$$B\_0 = \begin{pmatrix} 0 & 0.4 \\ 0.1 & 0 \end{pmatrix}, B\_1 = \begin{pmatrix} -0.6 & 0 \\ -0.2 & 0 \end{pmatrix}, B\_2 = \begin{pmatrix} 0.3 & 0 \\ 0.4 & 0 \end{pmatrix}, \mathbb{C} = \begin{pmatrix} 0.2 & 0 \\ -0.1 & 0 \end{pmatrix}.$$

*We get b*<sup>0</sup> = 0.41*, b*<sup>1</sup> = 0.64*, b*<sup>2</sup> = 0.51 *and c* = 0.2236*.*

*For η*<sup>1</sup> = *η*<sup>2</sup> = 1*,*  = 1*, γ*<sup>1</sup> = 0.3 *and γ*<sup>2</sup> = 60*. Moreover, if we calculate δ*, *c*˜1 *and c*˜2*, then <sup>G</sup>*˜(*γ*1, ) 59 < *γ*2, *for T* = 0.61*. Based on theorem 2 it is clear that the NFOTDSs is FTS w.r.t* 0.3, 60, 1, 0.61 *.*

**Example 2.** *Consider the NFOTDSs* (2)*, with λ*<sup>2</sup> = *λ*<sup>1</sup> = 0.6*, ς*(*s*) = 0.1,

$$\boldsymbol{\nu}(\boldsymbol{\pi}) = \begin{pmatrix} 0, 0.5, 0 \end{pmatrix}^T, \quad \boldsymbol{\zeta}(\boldsymbol{\pi}) = \begin{pmatrix} 0.04, 0, 0.02 \end{pmatrix}^T, \text{ for } \boldsymbol{\pi} \in [-0.1, 0], \boldsymbol{\zeta}$$

$$F(\mathbf{s}, \mathbf{x}(\mathbf{s}), \mathbf{x}(\mathbf{s} - \boldsymbol{\zeta}(\mathbf{s})), \boldsymbol{\upsilon}(\mathbf{s})) = 0.01 \Big( \sin \left( \mathbf{x}\_2(\mathbf{s} - \boldsymbol{\zeta}(\mathbf{s})) \right), \sin \left( \mathbf{x}\_3(\mathbf{s} - \boldsymbol{\zeta}(\mathbf{s})), \sin \left( \mathbf{x}\_1(\mathbf{s}) \right) \right) \Big)^T \Big)$$

*and*

$$B\_0 = \begin{pmatrix} 0.01 & -0.2 & 0.25 \\ -0.02 & 0.05 & 0.1 \\ 0.2 & -0.01 & 0.15 \end{pmatrix}, B\_1 = \begin{pmatrix} 0.01 & -0.15 & 0.31 \\ 0.25 & 0.12 & -0.14 \\ 0.13 & -0.12 & 0.22 \end{pmatrix},$$

$$B\_2 = \begin{pmatrix} 0.08 & 0.07 & 0.2 \\ 0.08 & -0.07 & -0.06 \\ -0.12 & -0.03 & -0.14 \end{pmatrix}, C = \begin{pmatrix} 0.1 & 0.2 & 0.03 \\ 0.12 & 0.22 & 0.05 \\ -0.17 & 0.05 & -0.21 \end{pmatrix}.$$

*We get b*<sup>0</sup> = 0.37*, b*<sup>1</sup> = 0.47*, b*<sup>2</sup> = 0.30*, and c* = 0.35*. For*  <sup>=</sup> <sup>1</sup>*, <sup>θ</sup>* <sup>=</sup> <sup>1</sup>*, <sup>γ</sup>*<sup>1</sup> <sup>=</sup> 0.4*, <sup>γ</sup>*<sup>2</sup> <sup>=</sup> <sup>100</sup>*, and T* <sup>=</sup> 1.05*, we get <sup>K</sup>*˜(*γ*1, ) 97 < *γ*2. *Theorem 3 implies that the NFOTDSs is FTS w.r.t* 0.4, 100, 1, 1.05 *.*
