Before going to viability terms, we set some notations. For \(\varepsilon>0\), by *ε*-neighborhood of a set \(K\subset \mathbb{R}^{n}\) we mean the following:

$$K^{\varepsilon}:=\bigl\{ x\in\mathbb{R}^{n}:\operatorname{dist}(x,K)< \varepsilon\bigr\} . $$

Let us define the distance between two sets \(A\subset\mathbb{R}^{n}\) and \(B\subset\mathbb{R}^{n}\) as \(\triangle(A,B):=\sup\{\operatorname{dist}(q,B):q\in A\}\). Note that \(\triangle(A,B)\) is not the usual symmetric distance between two sets. Indeed, if \(A\subset B\) then \(\triangle(A,B)=0\) while \(\triangle(B,A)\neq0\).

By definitions of *m*, \(\widetilde{m}\) and Proposition 7 the following proposition is obvious.

### Proposition 8

*Let*
*f*
*be bounded on*
\([0,T]\). *Let us take*
\(\varepsilon>0\)
*and*
\(t_{0}\in (0,T]\)
*such that*

$$\bigl\| \widetilde{m}(t_{0},t)-m(t_{0},t)\bigr\| < \varepsilon. $$

*Then*

$$\triangle\bigl(\operatorname{Graph}\bigl(m(t_{0},\cdot)\bigr),[t_{0},T] \times K\bigr)=0\quad\Leftrightarrow\quad \triangle \bigl(\operatorname{Graph}\bigl(\widetilde{m}(t_{0}, \cdot)\bigr),[t_{0},T]\times K\bigr)< \varepsilon, $$

*which we can write equivalently*

$$m(t_{0},t)\in K\quad\Leftrightarrow\quad\widetilde{m}(t_{0},t)\in K^{\varepsilon}\quad \textit{for all } t\in[t_{0},T]. $$

Similarly as for the ordinary differential equations (see [16]), one can define the viability of a subset with respect to the fractional differential equation (3).

Let us denote by *I* an open interval in ℝ.

### Definition 9

Let \(K\subset\mathbb{R}^{n}\) be a nonempty locally closed set and \(f: I\times K\to\mathbb{R}^{n}\). The subset *K* is *fractionally memo-viable with respect to*
*f* if for any \((t_{0},x_{0})\in I\times K\) equation (3) has at least one solution \(x:[t_{0},T]\to \mathbb{R}^{n}\) satisfying \(m(t_{0},t)\in K\) for \(t\in[t_{0},T]\), where \(t_{0}>0\).

The idea of viability of fractional differential equations can be expressed using the concept of tangent cone. There are many notions of tangency of a vector to a set (see, for example, [16], Section 2.3). We will follow the concept of contingent vectors (see [9]).

Let us recall that for \(K \subset\mathbb{R}^{n}\) and \(x_{0}\in K\) one can define the vector tangent to the set *K* as follows.

### Definition 10

The vector \(\eta\in\mathbb{R}^{n}\) is *contingent* to the set *K* at the point \(x_{0}\) if

$$ \liminf_{h\downarrow0} \frac{1}{h} \operatorname{dist} (x_{0}+ {h}\eta; K )=0 . $$

(17)

The set of all vectors that are contingent to the set *K* at point \(x_{0}\) is a closed cone, see [16], Proposition 2.3.1. This cone, denoted by \(\mathcal{T}_{K} (x_{0})\), is called *contingent cone (Bouligand cone)* to the set *K* at \(x_{0}\in K\). From [16], Proposition 2.3.2, we know that \(\eta\in\mathcal{T}_{K} (x_{0})\) if and only if for every \(\varepsilon>0\) there exist \(h\in(0, \varepsilon)\) and \(p_{h} \in B(0, \varepsilon)\) such that \(x_{0} +{h}(\eta+p_{h})\in K\), where \(B(0, \varepsilon)\) denotes the closed ball in \(\mathbb{R}^{n}\) centered at 0 and of radius \(\varepsilon>0\).

### Theorem 11

*Let*
\(K\subset\mathbb{R}^{n}\)
*be nonempty and*
\(f: I\times K \to\mathbb {R}^{n}\). *If the subset*
*K*
*is fractionally memo*-*viable with respect to*
*f*, *then*
\(g(t_{0},x_{0})\in\mathcal{T}_{K}(m_{0})\), *where*
\(x_{0}=x(t_{0})\)
*and*
\(m_{0}=m(t_{0},t_{0})= (I^{1-q}_{0+}x )(t_{0})\).

### Proof

Let \((t_{0}, m_{0})\in I\times K\) and *K* be fractionally viable of order *q* with respect to *f*. Then there is \(T\in I\), \(T>t_{0}\), and a function \(x: [t_{0},T]\rightarrow K\) satisfying \((I^{1-q}_{0+}x )(t_{0})=m_{0}\) and \(D^{q}_{0+}x(t)=f(t,x(t))\) for every \(t\in[t_{0},T]\). Moreover we have

$$\begin{aligned} &\lim_{h\downarrow0}\frac{1}{h}\bigl\Vert \bigl(I^{1-q}_{0+}x \bigr) (t_{0})+hg(t_{0},x_{0})-I^{1-q}_{0+}x(t_{0}+h) \bigr\Vert \\ &\quad=\lim_{h\downarrow0}\biggl\Vert g (t_{0},x_{0} )- \frac{ (I^{1-q}_{0+}x )(t_{0}+h)- (I^{1-q}_{0+}x )(t_{0})}{h}\biggr\Vert \\ &\quad= \bigl\Vert g (t_{0},x_{0} )-D^{q}_{0+}x(t_{0}) \bigr\Vert =0 . \end{aligned}$$

The above calculation shows that, for every \((t_{0},m_{0})\in I\times K\), \(g(t_{0},x_{0})\in\mathcal{T}_{K}(m_{0})\) and the proof is complete. □

Note that it is difficult to check the condition \(g(t_{0},x_{0})\in\mathcal {T}_{K}(m_{0})\) in Theorem 11. The above theorem is still true for the initial inner value problem with the right-hand side of \(\widetilde{g}\) given by (15). Let

$$ \widetilde{f}\bigl(t,x(t)\bigr)= \left \{ \begin{array}{@{}l@{\quad}l} f(t_{0},x_{0}), & t\in[0,t_{0}), \\ f(t,x(t)), & t \geq t_{0}. \end{array} \right . $$

(18)

Then the following is true.

### Corollary 12

*Let*
\(K\subset K^{\varepsilon}\subset\mathbb{R}^{n}\)
*and*
\(f: I\times K\to \mathbb{R}^{n}\). *If the subset*
*K*
*is fractionally memo*-*viable with respect to*
*f*, *then*
\(g(t_{0},x_{0})\in\mathcal{T}_{K^{\varepsilon}}(\widetilde{m}_{0})\).

From Theorem 11 we get the following weaker result.

### Corollary 13

*Let*
\(K\subset\mathbb{R}^{n}\)
*be nonempty and*
\(f: I\times K\to\mathbb {R}^{n}\). *If the subset*
*K*
*is fractionally memo*-*viable with respect to*
*f*, *then*
\(f(t_{0},x_{0})\in\mathcal{T}_{K}(m_{0})\)
*or*
\(x(0^{+})\in\mathcal {T}_{K}(m_{0})\), *where*
\(x_{0}=x(t_{0})\)
*and*
\(m_{0}=m(t_{0},t_{0})= (I^{1-q}_{0+}x )(t_{0})\).

### Example 14

Let us consider a one-dimensional problem with the set \(K=\mathbb {R}_{+}\cup\{0\}\) and \(f(t,x(t))=c=\operatorname{const}\). Then \(g(t,x(t))=c+\frac{t^{-q}}{\Gamma(1-q)}x(0^{+})\) for \(0< t<T\) and

$$\widetilde{g}\bigl(t,x(t)\bigr)= \begin{cases} c+\frac{t_{0}^{-q}}{\Gamma(1-q)}x(0^{+}) & \mbox{for }0< t<t_{0},\\ c+\frac{t^{-q}}{\Gamma(1-q)}x(0^{+}) & \mbox{for }t_{0}\leqslant t<T. \end{cases} $$

Moreover, for \(x_{0}=x(t_{0})\), we have

$$x(t)=\frac{c}{\Gamma(1+q)}t^{q} \biggl(1-\frac{t_{0}}{t} \biggr)+x_{0} \biggl(\frac {t_{0}}{t} \biggr)^{1-q}+x \bigl(0^{+}\bigr) \biggl[1- \biggl(\frac{t_{0}}{t} \biggr)^{1-q} \biggr] $$

and \(m_{0}=m (t_{0},t_{0} )=ct_{0} (1-\frac{1}{q} )+t_{0}^{1-q}\Gamma(q) [x_{0}+x(0^{+}) (\frac{1}{\Gamma(q)\Gamma(2-q)}-1 ) ] \). Let us take \(m_{0}\geqslant0\), then

$$\mathcal{T}_{K}(m_{0})= \begin{cases} \mathbb{R} & \mbox{for }m_{0}>0,\\ {[}0,+\infty ) & \mbox{for }m_{0}=0. \end{cases} $$

We will show that if \(g(t_{0},x_{0})\notin\mathcal{T}_{K}(m_{0})\), then *K* is not memo-viable. Since for \(m_{0}>0\) we have \(\mathcal{T}_{K}(m_{0})=\mathbb {R}\), it is nothing to show then.

Let \(m_{0}=0\), then \(g(t_{0},x_{0})<0\), *i.e.*, \(c+\frac {t_{0}^{-q}}{\Gamma(1-q)}x(0^{+})<0\). Let us notice that the term \(\frac {t_{0}^{-q}}{\Gamma(1-q)}\) is positive.

Let \(x(0^{+})=0\), then obviously \(c<0\), and we get

$$x(t)=c\frac{t^{q}}{\Gamma(1+q)} \biggl(1-q\frac{t_{0}}{t} \biggr)< 0 \quad\mbox{if only } t_{0}\leq t. $$

Integrating the last term, on the basis of monotonicity of integral one gets

$$m(t_{0},t)= \bigl(I_{0^{+}}^{1-q}x \bigr) (t)=c(t-t_{0})< 0 . $$

Now let \(x(0^{+})>0\) and since \(g(t_{0},x_{0})<0\), which means that \(c<\frac{-t_{0}^{-q}}{\Gamma(1-q)}x(0^{+})\), thus \(c<0\). Therefore one can show the following:

$$m(t_{0},t)=ct+x\bigl(0^{+}\bigr) \biggl(1-q\frac{t_{0}}{t} \biggr)t_{0}^{1-q}< 0 $$

provided \(t>\frac{x(0^{+})}{c} (\frac{1}{\Gamma(q)\Gamma(2-q)}-1 )t_{0}^{1-q}\).

Finally, let us assume \(x(0^{+})<0\), and since \(g(t_{0},x_{0})<0\), we get \(c<\frac{-t_{0}^{-q}}{\Gamma(1-q)}x(0^{+})\). Again one can show the following:

$$m(t_{0},t)=ct+x\bigl(0^{+}\bigr) \biggl(1-q\frac{t_{0}}{t} \biggr)t_{0}^{1-q}< 0 $$

if only \(t_{0}< t< (1-\frac{1}{\Gamma(q)\Gamma(2-q)} )\Gamma (1-q)t_{0} \). Since \(m(t_{0},t)<0\), it follows that *K* is not memo-viable with respect to *f*.

Finally, in order to prove the sufficient condition of viability, we need the following definition and proposition.

### Definition 15

([11])

Let \(K\subset\mathbb{R}^{n}\) be nonempty and \(f: I\times K\to\mathbb {R}^{n}\). The subset *K* is *fractionally viable with respect to*
*f* if for any \((t_{0},x_{0})\in I\times K\) equation (3) has at least one solution \(x:[t_{0},T]\to K\) satisfying \(x(t_{0})=x_{0}\). Such a solution we call *viable with respect to f*.

### Proposition 16

([11])

*Let*
\(K\subset \mathbb {R}^{n}\)
*be a nonempty and locally closed set*, *and let*
\(f:I\times K\rightarrow \mathbb {R}^{n}\)
*be a vector*-*valued continuous function*. *If*
\(f(t_{0},x_{0})\in\mathcal{T}_{K}(x_{0})\)
*for every*
\((t_{0}, x_{0})\in I\times K\), *then*
*K*
*is fractionally viable with respect to*
*f*.

Let us re-scale and shift the set \(K^{\varepsilon}\) in such a way that elements of this set are again from the domain of *f*, namely

$$ \tilde{K}^{\varepsilon}=\frac{t_{0}^{q-1}}{\Gamma(q)}K^{\varepsilon}+ \frac {1-q}{\Gamma(q+1)}t_{0}^{q} f(t_{0},x_{0})+ \frac{1-q}{\Gamma(q+1)\Gamma(1-q)}x\bigl(0^{+}\bigr). $$

(19)

To formulate next theorem, which is a middle step in getting a certain sufficient condition of viability, we use the notion (19).

### Theorem 17

*Let*
\(K\subset K^{\varepsilon}\subset\mathbb{R}^{n}\)
*be nonempty and*
\(f: I\times K \to\mathbb{R}^{n}\). *If the subset*
*K*
*is fractionally memo*-*viable with respect to*
*f*, *then*
\(g(t_{0},x_{0})\in\mathcal{T}_{\tilde {K}^{\varepsilon}}(x_{0})\), *where*
\(x_{0}=x(t_{0})\).

### Proof

Let \((t_{0}, m_{0})\in I\times K\) and *K* be fractionally memo-viable with respect to *f*. Then Corollary 12 gives \(g(t_{0},x_{0})\in\mathcal {T}_{K^{\varepsilon}}(\tilde{m}_{0})\). The latter means that

$$\liminf_{h\downarrow0}\frac{1}{h}\operatorname{dist}\bigl(\tilde {m}_{0}+hg(t_{0},x_{0}),K^{\varepsilon}\bigr)=0 . $$

By formula (16) we can rewrite the above equation in the following way:

$$\min_{y\in K^{\varepsilon}}\frac{1}{h}\biggl\vert \frac{q-1}{q}t_{0} f(t_{0},x_{0})+ \Gamma(q)t_{0}^{1-q}x_{0}+\frac{q-1}{q\Gamma (1-q)}t_{0}^{1-q}x \bigl(0^{+}\bigr)+ hg(t_{0},x_{0})-y\biggr\vert \longrightarrow0 . $$

Let \(\tilde{K}^{\varepsilon}\) be as it is given in (19), *i.e.*,

$$\tilde{K}^{\varepsilon}=\frac{t_{0}^{q-1}}{\Gamma(q)}K^{\varepsilon}+ \frac{1-q}{\Gamma(q+1)}t_{0}^{q} f(t_{0},x_{0})+ \frac{1-q}{\Gamma(q+1)\Gamma (1-q)}x\bigl(0^{+}\bigr) , $$

then for \(\tilde{y}=\frac{q-1}{\Gamma(q+1)}t_{0}^{q} f(t_{0},x_{0})+\frac {q-1}{\Gamma(q+1)\Gamma(1-q)}x(0^{+})- \frac{t_{0}^{q-1}}{\Gamma(q)}y\) and \(\tilde{h}=\frac{\Gamma(q)t_{0}^{1-q}}{h}\), we get \(\tilde{y}\in\tilde {K}^{\varepsilon}\) and \(\tilde{h}\rightarrow0\) when \(h\rightarrow0\), thus

$$\min_{\tilde{y}\in\tilde{K}^{\varepsilon}}\frac{1}{\tilde{h}}\bigl\vert x_{0}+ \tilde{h}g(t_{0},x_{0})-y\bigr\vert \longrightarrow0, $$

while \(\tilde{h}\rightarrow0\). Therefore we get \(g(t_{0},x_{0})\in\mathcal {T}_{\tilde{K}^{\varepsilon}}(x_{0})\). □

The next corollary gives, in fact, a sufficient condition of viability for an equation involving the Caputo derivative that is, however, slightly different from the one that we formulated in [11].

### Corollary 18

*Let*
\(K\subset K^{\varepsilon}\subset\mathbb{R}^{n}\)
*be nonempty and*
\(f: I\times\mathbb{R}^{n} \to\mathbb{R}^{n}\). *Let*
\(\tilde{K}^{\varepsilon}=\frac {t_{0}^{q-1}}{\Gamma(q)}K^{\varepsilon}+\frac{1-q}{\Gamma(q+1)}t_{0}^{q} f(t_{0},x_{0})+\frac{1-q}{\Gamma(q+1)\Gamma(1-q)}x(0^{+})\). *If*
\(g(t_{0},x_{0})\in \mathcal{T}_{\tilde{K}^{\varepsilon}}(x_{0})\), *where*
\(x_{0}=x(t_{0})\), *then*
\(\tilde{K}^{\varepsilon}\)
*is viable with respect to*
*g*.

### Proof

The thesis is a simple consequence of Proposition 16 and Theorem 17. □