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# On the interval differential equation: novel solution methodology

## Abstract

In this study, by a parametric representation of interval numbers, two parametric representations for interval-valued functions are presented. Then, using these representations, the calculus of interval-valued functions and interval differential equations are investigated with two different approaches. In the first approach, the interval differential is transformed into a crisp problem. In the second approach, two solutions are obtained with the characterization of the solutions of two ordinary differential equation systems.

## Introduction

The interval-valued analysis and interval differential equations are specific cases of set-valued analysis and set differential equations, respectively . Interval analysis is introduced as an attempt to handle interval uncertainty, while interval differential equations are natural models for describing dynamic systems under uncertainty, and this approach is useful in many applications areas, such as physics and engineering [4, 5]. As in classical real analysis, the importance of the derivative of an interval-valued function in the study of interval differential equations is well known. On the other hand, the inversions of addition and multiplication are fundamental in interval arithmetic, interval analysis, and the concept of interval differentiability.

In interval arithmetic, standard Minkowski addition and multiplication are not invertible operations, and the idea of finding some inverse operations has been a field that has long been of interest . The Hukuhara difference, which was introduced by Hukuhara in , has been a starting point for this purpose, which exists under restrictive conditions; for more details, see . To overcome this shortcoming, Stefanini and Bede  proposed the generalized Hukuhara difference of two interval numbers, which has a large advantage over the peer concept, namely, that it always exists. In a similar discussion, the concept of Hukuhara derivative is extended to the generalized Hukuhara derivative. The same remark holds if the concept of differentiability for differential equations in the interval setting is regarded. The diameter of the solution of an interval differential equation with a Hukuhara derivative is a nondecreasing function with respect to (w.r.t.) time . The generalized Hukuhara differentiability for an interval-valued function allows us to obtain the solutions to interval differential equations with a decreasing diameter [6, 7]. Recently, several studies [6, 10] have been performed on generalized Hukuhara differentiability.

A key point in our investigation is the parametric representations for interval numbers. By this notation, interval arithmetic operations are defined, and two parametric representations for interval-valued functions are expressed. The results are applied to introduce the concepts of the derivative and integral of interval-valued functions. Additionally, several properties of the new concepts are investigated and compared with other recently presented results. Moreover, the relationship between the new interval derivative and the interval integral is studied, and Newton-Leibniz type formulas are nontrivially extended to the interval case. Finally, an application to solve interval differential equations is shown in two different aspects: in the first approach, the interval problem has a unique solution, which is obtained by solving an ordinary differential equation, and the results in the second approach are equivalent to the obtained results using the gH-differentiability concept. Furthermore, the existence and uniqueness of the solutions are discussed.

This paper is organized in five sections. In Section 2, the parametric representations of interval numbers are defined, and interval arithmetic operations are presented. The interval-valued functions in the parametric form and their properties, derivatives, integrals and the relations between them are studied in Section 3. Section 4 describes an application of the new results to interval differential equations, and the paper ends with conclusions in Section 5.

## Arithmetic of intervals

In this section, some basic definitions and existing arithmetic operations between intervals in the parametric form are presented. Let $$K_{c}$$ be the space of nonempty compact and convex sets of $$\mathbb{R}$$, i.e.,

$$K_{c}=\bigl\{ [a,b]\mid a,b \in\mathbb{R} \mbox{ and } a\leq b\bigr\} .$$

Suppose $$A=[\underline{a},\overline{a}]\in K_{c}$$, where $$\underline{a}$$ and mean the lower and upper bounds of A, respectively. Obviously, two parametric representations for any $$A \in K_{c}$$ can be considered:

1. i.

$$A=\{a(t)\mid a(t)=\underline{a}+t(\overline{a}-\underline{a}), t \in[0,1]\}$$ (increasing representation or IR),

2. ii.

$$A=\{a(t)\mid a(t)=\overline{a}+t(\underline{a}-\overline{a}), t \in[0,1]\}$$ (decreasing representation or DR).

### Definition 2.1

Let $$\{a(t) \mid t \in[0,1]\}$$ and $$\{b(t) \mid t \in[0,1]\}$$ be the IRs (DRs) of $$A=[\underline{a},\overline{a}]$$ and $$B=[\underline{b},\overline{b}]$$, respectively, and λ be a real number. The parametric arithmetics in $$K_{c}$$ can be defined as

1. 1.

$$A\oplus B=\{a(t_{1})+b(t_{2}) \mid t_{1}, t_{2} \in[0,1]\}$$,

2. 2.

$$A\otimes B=\{a(t_{1})b(t_{2}) \mid t_{1}, t_{2} \in[0,1]\}$$,

3. 3.

$$A\oslash B=\{a(t_{1})/b(t_{2}) \mid b(t_{2})\neq0, t_{1}, t_{2} \in[0,1]\}$$,

4. 4.

$$\lambda\odot A=\{\lambda a(t) \mid t \in[0,1]\}$$,

5. 5.

$$A\ominus_{p}B=\{a(t)-b(t) \mid t \in[0,1]\}$$,

6. 6.

$$A=B \Leftrightarrow \{a(t) \mid t\in[0,1]\}=\{b(t) \mid t\in[0,1]\}$$.

### Remark 2.2

Note that if A has IR (DR) and $$\lambda\geq0$$, then $$\lambda\odot A$$ has IR (DR), and if A has IR (DR) and $$\lambda\leq0$$, then $$\lambda\odot A$$ has DR (IR). Moreover, if $$a(t)-b(t)$$ is an increasing (a decreasing) function, then $$A\ominus_{p}B$$ has IR (DR).

### Remark 2.3

By Definition 2.1, two intervals A and B are equal if and only if $$a(t)=b(t)$$, for all $$t \in[0,1]$$.

### Example 2.1

Assume that $$A=[-1,2]$$ and $$B=[1,5]$$. According to IR

$$A=\bigl\{ -1+3t \mid t\in[0,1]\bigr\} , \qquad B=\bigl\{ 1+4t \mid t\in[0,1]\bigr\} ,$$

and thus

1. 1.

$$A\oplus B=\{(-1+3t_{1})+(1+4t_{2}) \mid t_{1}, t_{2} \in[0,1]\}=\{ 3t_{1}+4t_{2} \mid t_{1}, t_{2} \in[0,1]\}$$, because $$3t_{1}+4t_{2}$$ is a continuous function w.r.t. $$t_{1}$$, $$t_{2}$$, $$A\oplus B=[\min_{t_{1},t_{2}} 3t_{1}+4t_{2}, \max_{t_{1},t_{2}} 3t_{1}+4t_{2}]=[0,7]$$,

2. 2.

$$A\otimes B=\{(-1+3t_{1})(1+4t_{2}) \mid t_{1}, t_{2} \in[0,1]\}=[-5,10]$$, because of continuity of the function $$-1-4t_{2}+3t_{1}+12t_{1}t_{2}$$ w.r.t. $$t_{1}$$, $$t_{2}$$,

3. 3.

$$A\oslash B=\{(-1+3t_{1})/(1+4t_{2}) \mid t_{1}, t_{2} \in[0,1]\}=[-1,2]$$,

4. 4.

$$-2 \odot A=\{-2(-1+3t) \mid t\in[0,1]\}=\{2-6t \mid t\in[0,]\}$$, due to the decreasing function $$2-6t$$ and DR, $$-2 \odot A=[-4,2]$$,

5. 5.

$$A\ominus_{p} B=\{(-1+3t)-(1+4t) \mid t\in[0,1]\}=\{-2-t \mid t\in[0,1]\} =[-3,-2]$$, from DR,

6. 6.

$$B\ominus_{p} A=\{(1+4t)-(-1+3t) \mid t\in[0,1]\}=\{2+t \mid t\in[0,1]\} =[2,3]$$, from IR.

If $$\lambda=-1$$, then the interval $$(-1)A$$ gives the additive inverse of A, which is denoted by −A, where $$-A=\{-a(t) \mid t\in[0,1]\}$$. Note that, generally, $$A\oplus(-1)A \neq\{0\}$$. Therefore, another difference, $$A-B=A\oplus(-1)B$$, which is constructed with the following parametric form, can be defined:

$$A-B= \bigl\lbrace a(t_{1})-b(t_{2}) \mid a(t_{1})= \underline{a}+t_{1}(\overline {a}-\underline{a}), b(t_{2})= \underline{b}+ t_{2}(\overline{b}-\underline {b}), t_{1}, t_{2} \in[0,1] \bigr\rbrace .$$

Because $$a(t_{1})-b(t_{2})$$ is a continuous function w.r.t. $$t_{1}$$ and $$t_{2}$$, we have

$$A-B= \Bigl[\min_{t_{1},t_{2} \in[0,1]} \bigl(a(t_{1})-b(t_{2}) \bigr), \max_{t_{1},t_{2} \in[0,1]} \bigl(a(t_{1})-b(t_{2}) \bigr) \Bigr].$$

If $$A=[\underline{a},\overline{a}]$$ and $$B=[\underline{b},\overline {b}]$$, then $$a(t_{1})-b(t_{2})=\underline{a}-\underline{b}+t_{1}(\overline {a}-\underline{a})-t_{2}(\overline{b}-\underline{b})$$, thus

$$\min_{t_{1},t_{2} \in[0,1]} \bigl(a(t_{1})-b(t_{2}) \bigr)= \underline {a}-\underline{b}-(\overline{b}-\underline{b})=\underline{a}- \overline{b}$$

and

$$\max_{t_{1},t_{2} \in[0,1]} \bigl(a(t_{1})-b(t_{2}) \bigr)= \underline {a}-\underline{b}+(\overline{a}-\underline{a})=\overline{a}- \underline{b}.$$

Hence, $$A-B=[\underline{a}-\overline{b},\overline{a}-\underline{b}]$$, which is the same as the Minkowski difference . Note that, in general, $$(A\oplus B)-B \neq A$$, while this equality is valid for the parametric difference.

### Remark 2.4

Obviously, $$A\ominus_{p} B=A-B$$ and $$B \ominus_{p} A=B-A$$, whenever $$A \in \mathbb{R}$$ and $$B \in K_{c}$$.

### Proposition 2.5

The parametric difference $$\ominus_{p}$$ has the following properties:

1. (i)

$$A\ominus_{p}A=\{0\}$$,

2. (ii)

$$A\ominus_{p}B=-(B\ominus_{p}A)$$,

3. (iii)

$$(A\oplus B)\ominus_{p}B=A$$, $$A\ominus_{p}(A\oplus B)=-B$$, $$A\ominus_{p}(A-B)=B$$,

4. (iv)

$$A\ominus_{p}B=B\ominus_{p}A$$ if and only if $$C=-C$$ and $$A\ominus_{p}B=C$$, and particulary $$C=\{0\}$$ if and only if $$A=B$$,

5. (v)

$$A\oplus(B \ominus_{p}A)=B$$ or $$B-(B\ominus_{p}A)=A$$,

6. (vi)

if $$A \ominus_{p} B$$ has IR (DR) and $$A \ominus _{p} B=C$$, then $$A\ominus_{p}C=B$$ ($$A-C=B$$),

7. (vii)

if $$A-B=C$$, then $$A \ominus_{p} C=B$$.

### Proof

Suppose $$A=[\underline{a},\overline{a}]=\{a(t) \mid a(t)=\underline {a}+t(\overline{a}-\underline{a})\}$$ and $$B=[\underline{b},\overline {b}]=\{b(t) \mid b(t)=\underline{b}+t{(\overline{b}-\underline{b})}\}$$. Property (i) is an immediate consequence of Definition 2.1. To prove (ii), one has to show that

$$\bigl\{ a(t)-b(t) \mid t \in[0,1]\bigr\} =\bigl\{ -\bigl(b(t)-a(t)\bigr) \mid t \in[0,1]\bigr\} ,$$

which can be obtained as

\begin{aligned} \bigl\{ a(t)-b(t) \mid t\in[0,1]\bigr\} &=\bigl\{ \underline{a}+t(\overline{a}- \underline {a})-\underline{b}-t(\overline{b}-\underline{b}) \mid t\in[0,1]\bigr\} \\ &=\bigl\{ - \bigl(\underline{b}-\underline{a}-t(\overline{a}-\underline {a})+t( \overline{b}-\underline{b}) \bigr) \mid t\in[0,1]\bigr\} \\ &=\bigl\{ - \bigl(\underline{b}+t(\overline{b}-\underline{b})-\bigl(\underline {a}+t(\overline{a}-\underline{a})\bigr) \bigr) \mid t \in[0,1]\bigr\} \\ &=\bigl\{ -\bigl(b(t)-a(t)\bigr) \mid t \in[0,1]\bigr\} . \end{aligned}

The first part of (iii) is evidently proved by

$$(A\oplus B)\ominus_{p} B=\bigl\{ \underline{a}+\underline{b}+t\bigl(( \overline {a}+\overline{b})-(\underline{a}+\underline{b})\bigr)-\underline {b}-t( \overline{b}-\underline{b}) \mid t \in[0,1]\bigr\} =\bigl\{ a(t) \mid t \in[0,1] \bigr\} =A.$$

The second part of (iii) follows from a similar argument. To prove the third part, it is sufficient to note that $$A-B=[\underline {a}-\overline{b},\overline{a}-\underline{b}]$$. To denote the first part of (vi), suppose $$A\ominus_{p}B=B\ominus_{p}A$$; then by Definition 2.1

$$\bigl\{ a(t)-b(t) \mid t\in[0,1]\bigr\} =\bigl\{ b(t)-a(t) \mid t\in[0,1]\bigr\} .$$

On the other hand, $$A\ominus_{p}B=C$$, and thus it follows that $$\{ c(t) \mid t\in[0,1]\}=\{-c(t) \mid t\in[0,1]\}$$. This gives $$C=-C$$, which proves our assertion, and because of property (ii) the reverse holds true. For the second part of (vi), from $$C=\{0\}$$ and $$A\ominus _{p}B=C$$, it can be concluded that $$\{a(t)-b(t) \mid t\in[0,1]\}=\{0\}$$, and thus, by Remark 2.3, it can be deduced that $$a(t)-b(t)=0$$ and $$a(t)=b(t)$$, for all $$t\in[0,1]$$. Hence,

$$\bigl\{ a(t) \mid t\in[0,1]\bigr\} =\bigl\{ b(t) \mid t\in[0,1]\bigr\} ,$$

which shows $$A=B$$. Conversely, let $$A=B$$; thus by property 1, $$C=A \ominus_{p}B=\{0\}$$. To deduce (v), if $$B\ominus_{p}A$$ has IR then $$A\oplus(B \ominus_{p}A)=B$$, and if it has DR, then $$B-(B\ominus _{p}A)=A$$. Next, let us to prove (vi). First, consider that $$A \ominus _{p} B=C$$ has IR thus $$C=A \ominus_{p} B= \{ c(t) \mid c(t)=a(t)-b(t)=\underline{a}- \underline{b}+t(\overline {a}-\underline{a}-\overline{b}-\underline{b}) \}$$, where $$\overline {a}-\underline{a}-\overline{b}-\underline{b}$$ is positive. Therefore,

\begin{aligned} \begin{aligned} A\ominus_{p}C&=\bigl\{ a(t)-c(t) \mid t \in[0,1]\bigr\} \\ &=\bigl\{ \underline{a}+t(\overline{a}-\underline{a})-\bigl(\underline {a}- \underline{b}+t(\overline{a}-\underline{a}- \overline{b}-\underline{b})\bigr) \mid t\in[0,1]\bigr\} \\ &=\bigl\{ \underline{b}+t(\overline{b}-\underline{b}) \mid t\in[0,1]\bigr\} =B. \end{aligned} \end{aligned}

With a similar argument as above, one can prove the second case. To see (vii), assume that $$A-B=C$$. Thus, $$C=[\underline{a}-\overline {b},\overline{a}-\underline{b}]$$, and by Definition 2.1

\begin{aligned} A\ominus_{p} C&=\bigl\{ a(t)-c(t) \mid t\in[0,1]\bigr\} \\ &=\bigl\{ \underline{a}+t(\overline{a}-\underline{a})-\bigl(\underline{a}- \overline {b}+t(\overline{a}-\underline{b}-\underline{a} +\overline{b})\bigr) \mid t\in[0,1]\bigr\} \\ &=\bigl\{ \overline{b}+t(\underline{b}-\overline{b}) \mid t \in[0,1]\bigr\} =B. \end{aligned}

The last equality follows from DR of B. □

Stefanini  proposed a generalization of the Hukuhara difference for interval numbers, namely gH-difference. In the next preposition, it will be shown that the concepts of gH-difference and parametric difference are equivalent. So the parametric difference is equivalent with the concept of difference defined by Markov  and with π-difference .

### Proposition 2.6

The parametric difference is equivalent to the gH-difference.

### Proof

Let $$A=[\underline{a},\overline{a}]$$, $$B=[\underline{b},\overline{b}]$$, and $$C=[\underline{c},\overline{c}]=A\ominus_{p}B$$. It is sufficient to show that $$\underline{c}=\min \{\underline{a}-\underline{b},\overline {a}-\overline{b}\}$$ and $$\overline{c}=\max \{\underline{a}-\underline {b},\overline{a}-\overline{b}\}$$. There are two cases, which are dependent on whether the function $$a(t)-b(t)=\underline{a}- \underline {b}+t(\overline{a}-\underline{a}-\overline{b}-\underline{b})$$ is increasing or decreasing w.r.t. t.

1. 1.

In the increasing case, we have $$\overline{a}-\overline{b}\geq \underline{a}-\underline{b}$$ and by IR, and thus

\begin{aligned}& \underline{c}=\underline{a}-\underline{b}=\min \{\underline {a}-\underline{b}, \overline{a}-\overline{b}\}, \\& \overline{c}-\underline{c}=\overline{a}-\underline{a} -\overline {b}+ \underline{b}\quad \Rightarrow \quad \overline{c}=\overline{a}-\overline {b}=\max \{\underline{a}-\underline{b},\overline{a}-\overline{b}\}. \end{aligned}
2. 2.

In the decreasing case, we have $$\overline{a}-\overline{b}\leq \underline{a}-\underline{b}$$ and by DR, it can be concluded that

\begin{aligned}& \overline{c}=\underline{a}-\underline{b}=\max \{\underline{a}-\underline {b}, \overline{a}-\overline{b}\}, \\& \underline{c}-\overline{c}=\overline{a}-\underline{a} -\overline {b}+ \underline{b}\quad \Rightarrow\quad \underline{c}=\overline{a}-\overline {b}=\min \{\underline{a}-\underline{b},\overline{a}-\overline{b}\}, \end{aligned}

which is the desired conclusion. □

### Proposition 2.7

For any two intervals $$A, B \in K_{c}$$, the parametric difference $$A\ominus_{p} B=C$$ always exists and is unique.

### Proof

Clearly, from the definition of a parametric difference

$$C=\bigl\{ a(t)-b(t) \mid a(t)-b(t)=\underline{a}-\underline{b}+t\bigl(\overline {a}-\overline{b}-(\underline{a}-\underline{b})\bigr), t \in[0,1]\bigr\} .$$

First, suppose that $$a(t)-b(t)$$ be an increasing function. Then $$\overline{a}-\overline{b}\geq\underline{a}-\underline{b}$$ and by Remark 2.2 the interval number C has IR and $$C=[\underline{a}-\underline{b},\overline{a}-\overline{b}]$$, which is a unique interval in $$K_{c}$$. Secondly, suppose $$a(t)-b(t)$$ be a decreasing function then the assertion can be proved in a similar way. □

For arbitrary intervals $$A,B \in K_{c}$$, the metric $$D: K_{c} \times K_{c} \rightarrow\mathbb{R}_{+} \cup\{0\}$$ is defined as

$$D(A,B)= \max\Bigl\{ \min_{t}\bigl\vert a(t)-b(t)\bigr\vert , \max_{t} \bigl\vert a(t)-b(t)\bigr\vert \Bigr\} ,$$
(1)

where $$A=\{a(t) \mid a(t)=\underline{a}+t(\overline{a}-\underline {a}), t\in[0,1]\}$$ and $$B=\{b(t) \mid b(t)=\underline{b}+t(\overline {b}-\underline{b}), t \in[0,1]\}$$.

### Proposition 2.8

$$(K_{c},D)$$ is a complete metric space that have following properties:

1. 1.

$$D(A\oplus C, B\oplus C)=D(A,B)$$,

2. 2.

$$D(\lambda A,\lambda B)=|\lambda|D(A,B)$$,

3. 3.

$$D(A\oplus B,C\oplus E) \leq D(A,C)+D(B,E)$$,

4. 4.

$$D(A,B)=D(A\ominus_{p}B,\{0\})$$.

### Proof

We prove only the final part; the other parts are trivial. Suppose

$$D\bigl(A\ominus_{p}B,\{0\}\bigr)=\max\Bigl\{ \min_{t} \bigl\vert c(t)\bigr\vert ,\max_{t} \bigl\vert c(t) \bigr\vert \Bigr\} ,$$

where $$C=A\ominus_{p} B$$ and $$c(t)$$ is IR of C. If $$a(t)-b(t)$$ is an increasing function, then $$c(t)=a(t)-b(t)$$ and $$D(A\ominus_{p} B,\{ 0\})=D(A,B)$$. If $$a(t)-b(t)$$ is a decreasing function, then

$$\max_{t} \bigl\vert c(t)\bigr\vert =\min _{t}\bigl\vert a(t)-b(t)\bigr\vert , \qquad \min _{t} \bigl\vert c(t)\bigr\vert =\max_{t} \bigl\vert a(t)-b(t)\bigr\vert ,$$

and $$D(A\ominus_{p} B,\{0\})=D(A,B)$$. □

The Hausdorff distance between two intervals $$A=[\underline{a},\overline {a}]$$, $$B=[\underline{b},\overline{b}]$$ is defined as follows (see ):

$$D_{H}(A,B)=\bigl\{ \vert \underline{a}-\underline{b}\vert ,| \overline{a}-\overline{b}|\bigr\}$$

### Proposition 2.9

The Hausdorff distance is equivalent to the metric D in (1).

### Proof

The proof is clearly depending on whether $$a(t)-b(t)$$ is an increasing or a decreasing function. □

## Parametric representation of interval-valued function

Let $$F:T\subseteq \mathbb{R} \rightarrow K_{c}$$ be an interval-valued function with lower bound $$\underline{f}$$ and upper bound , i.e., $$F(x)=[ \underline {f}(x) , \overline{f}(x) ]$$. Then it is trivial to see that a parametric representation of $$F(x)$$ is

$$\bigl\{ \underline{f}(x)+t \bigl(\overline{f}(x)-\underline{f}(x) \bigr) \mid t\in[0,1] \bigr\} .$$
(2)

An interval-valued function with real independent variable and interval coefficients is special case of $$F:T\subseteq \mathbb{R} \rightarrow K_{c}$$. Using IRs of coefficients, it is possible to write this function as a set of classical functions. As an example, consider $$F(x)=[-2,3]\odot x\oplus[-3,-1]\odot x^{2}\otimes e^{[1,4]\odot x}\oplus \sin([3,6]\odot x)$$. If $$\mathbf{C}_{\nu }^{4}=([-2,3],[-3,-1],[1,4],[3,6])^{T}$$ is a column vector in $$(K_{c})^{4}$$, whose elements are intervals which, respectively, appear in the function F, then by IRs

$$\mathbf{C}_{\nu}^{4}=\bigl\{ (-2+5t_{1},-3+2t_{2},1+3t_{3},3+3t_{4})^{T} \mid \mathbf {t}=(t_{1},t_{2},t_{3},t_{4}) \in[0,1]^{4}\bigr\} .$$

Hence, $$F(x)$$ can be redefined as

\begin{aligned} F_{\mathbf{C}_{\nu}^{4}}(x) =&\bigl\{ f_{\mathbf{c}(\mathbf{t})}(x) \mid f_{\mathbf{c}(\mathbf{t})}(x)=(-2+5t_{1})x+(-3+2t_{2})x^{2}e^{(1+3t_{3})x} \\ &{}+\sin\bigl((3+3t_{4})x\bigr); \mathbf{c}(\mathbf{t}) \in \mathbf{C}_{\nu}^{4}\bigr\} . \end{aligned}

In the light of the concept of this example, if $$\mathbf{C}_{\nu}^{k} \in (K_{c})^{k}$$ denotes the intervals that are presented in an interval-valued function, then another parametric representation for this class of function, i.e., $$F_{\mathbf{C}_{\nu}^{k}}:T \subseteq\mathbb{R} \rightarrow K_{c}$$, can be written as

$$F_{\mathbf{C}_{\nu}^{k}}(x)=\bigl\{ f_{\mathbf{c}(\mathbf{t})}(x) \mid f_{\mathbf{c}(\mathbf{t})}:T \subseteq\mathbb{R} \rightarrow\mathbb{R}; \mathbf{c}( \mathbf{t})\in\mathbf {C}_{\nu}^{k}\bigr\} ,$$
(3)

where $$\mathbf{C}_{\nu}^{k}=\{\mathbf{c}(\mathbf{t}) \mid \mathbf{c}(\mathbf{t})=(c_{1}(t_{1}),c_{2}(t_{2}),\ldots,c_{k}(t_{k}))^{T}; c_{j}(t_{j})=\underline {c}_{j}+t_{j}(\overline{c}_{j}-\underline{c}_{j}), \mathbf {t}=(t_{1},t_{2},\ldots,t_{k})\in[0,1]^{k}\}$$. Clearly, $$f_{\mathbf{c}(\mathbf{t})}(x)$$ is a continuous function in t for every fixed x.

Note that the parametric variables $$t_{j}$$, $$j=1, 2, \ldots, k$$ in (3) are real numbers in the interval $$[0,1]$$; however, the parameter t in (2) is a function of x, i.e., $$t:\mathbb {R}\rightarrow[0,1]$$. For more precise look, consider the interval-valued function $$F_{\mathbf{C}_{\nu}^{2}}(x)=[1,2]\odot x\oplus [2,4]\odot x^{2}$$, $$0< x<2$$. By (2) and (3), two parametric representations can be considered:

\begin{aligned}& F_{\mathbf{C}_{\nu}^{2}}(x) =\bigl\{ x+2x^{2}+t \bigl(x+2x^{2}\bigr) \mid t\in[0,1]\bigr\} , \end{aligned}
(4)
\begin{aligned}& F_{\mathbf{C}_{\nu}^{2}}(x) =\bigl\{ (1+t_{1})x+(2+2t_{2})x^{2} \mid \mathbf {t}=(t_{1},t_{2}) \in[0,1]^{2}\bigr\} , \end{aligned}
(5)

respectively. By putting $$t_{1}=0$$ and $$t_{2}=\frac{1}{2}$$ in (5), the function $$x+3x^{2}$$ is obtained, while this function can be acquired by setting $$t=\frac{x^{2}}{x+2x^{2}}$$ in the representation (4).

Next, by the parametric representations (2) and (3), some concepts such as limit, continuity, derivative, and integral for interval-valued functions will be defined in a parametric form. Using these concepts, interval differential equations will be studied by two approaches. Considering the parametric representation (2), the results in [6, 10] are obtained, as we expected.

### Definition 3.1

Let $$F:T \subseteq\mathbb{R} \rightarrow K_{c}$$ be an interval-valued function and $$x_{0} \in T$$. $$L \in K_{c}$$ is the limit of F at $$x_{0}$$, if for every $$\varepsilon>0$$, there exists $$\delta>0$$ such that $$D(F(x),L)<\varepsilon$$, whenever $$|x-x_{0}|<\delta$$ and it is denoted by $$\lim_{x\rightarrow x_{0}}F(x)=L$$.

### Proposition 3.2

For an interval-valued function $$F:T \subseteq\mathbb{R} \rightarrow K_{c}$$,

$$\lim_{x\rightarrow x_{0}}F(x)=L \quad \Leftrightarrow\quad \lim _{x\rightarrow x_{0}}\bigl(F(x)\ominus_{p} L\bigr)=\{0\}.$$

### Proof

It can be proved from Definition 3.1 and Proposition 2.8. □

From the definition of limit for interval-valued functions, it is clear that $$F:T \subseteq\mathbb{R} \rightarrow K_{c}$$ is continuous at point $$x_{0} \in T$$, if $$\lim_{x\rightarrow x_{0}}F(x)=F(x_{0})$$. Furthermore, F is continuous on T if it is continuous at any point in T.

### Proposition 3.3

Suppose $$F_{\mathbf{C}_{\nu}^{k}}:T \subseteq\mathbb{R} \rightarrow K_{c}$$ be an interval-valued function and

$$F_{\mathbf{C}_{\nu}^{k}}(x)=\bigl\{ f_{\mathbf{c}(\mathbf{t})}(x) \mid f_{\mathbf{c}(\mathbf{t})}:T \subseteq \mathbb{R} \rightarrow\mathbb{R}, \mathbf{c}(\mathbf{t})\in\mathbf {C}_{\nu}^{k}\bigr\} .$$

If $$\lim_{x\rightarrow x_{0}}f_{\mathbf{\mathbf{c}(\mathbf{t})}}(x)$$ exists for every $$\mathbf{c}(\mathbf{t}) \in\mathbf{C}_{\nu}^{k}$$, then $$\lim_{x\rightarrow x_{0}}F_{\mathbf{C}_{\nu}^{k}}(x)$$ exists and

$$\lim_{x\rightarrow x_{0}}F_{\mathbf{C}_{\nu}^{k}}(x)=\Bigl\{ \lim _{x\rightarrow x_{0}}f_{\mathbf{c}(\mathbf{t})}(x) \bigm| f_{\mathbf{c}(\mathbf{t})}:T \subseteq \mathbb{R} \rightarrow\mathbb{R}, \mathbf{c}(\mathbf{t})\in\mathbf {C}_{\nu}^{k}\Bigr\} .$$

Moreover, $$F_{\mathbf{C}_{\nu}^{k}}(x)$$ is continuous at $$x_{0}$$, if for every $$\mathbf{c}(\mathbf{t}) \in\mathbf{C}_{\nu}^{k}$$ the function $$f_{\mathbf{c}(\mathbf{t})}$$ is continuous at  $$x_{0}$$.

### Proof

Assume that $$\lim_{x\rightarrow x_{0}}f_{\mathbf{c}(\mathbf{t})}(x)=a(\mathbf{t})$$ for every $$\mathbf{c}(\mathbf{t})\in\mathbf{C}_{\nu }^{k}$$. Thus, for every $$\mathbf{t} \in[0,1]^{k}$$, it can be concluded that

$$\forall\varepsilon>0, \exists\delta>0,\quad 0< |x-x_{0}|< \delta \quad \Rightarrow \quad \bigl\vert f_{\mathbf{c}(\mathbf{t})}(x)-a(\mathbf{t})\bigr\vert < \varepsilon.$$
(6)

Because of continuity of $$f_{\mathbf{c}(\mathbf{t})}$$ and $$a(\mathbf{t})$$ in t, there exist $$\mathbf{t}', \mathbf{t}'' \in[0,1]^{k}$$ such that

$$\min_{\mathbf{t}}\bigl\vert f_{\mathbf{c}(\mathbf{t})}(x)-a(\mathbf{t})\bigr\vert =\bigl\vert f_{\mathbf {c}(\mathbf{t}')}(x)-a\bigl(\mathbf{t}'\bigr) \bigr\vert , \qquad \max_{\mathbf{t}}\bigl\vert f_{\mathbf{c}(\mathbf{t})}(x)-a(\mathbf{t})\bigr\vert =\bigl\vert f_{\mathbf{c}(\mathbf{t}'')}(x)-a \bigl(\mathbf{t}''\bigr)\bigr\vert .$$

By (6), for all $$\varepsilon>0$$, there exist $$\delta_{1}, \delta _{2}>0$$, such that

\begin{aligned}& 0< |x-x_{0}|< \delta_{1} \quad \Rightarrow\quad \bigl\vert f_{\mathbf{c}(\mathbf{t}')}(x)-a\bigl(\mathbf {t}'\bigr)\bigr\vert < \varepsilon, \\& 0< |x-x_{0}|< \delta_{1} \quad \Rightarrow\quad \bigl\vert f_{\mathbf{c}(\mathbf{t}'' )}(x)-a\bigl(\mathbf{t}''\bigr)\bigr\vert < \varepsilon. \end{aligned}

Next, by choosing $$\overline{\delta}=\min\{\delta_{1},\delta_{2}\}$$,

$$D\bigl(F_{\mathbf{C}_{\nu}^{k}}(x),A\bigr)=\max\Bigl\{ \min_{\mathbf{t}} \bigl\vert f_{\mathbf{c}(\mathbf{t})}(x)-a(\mathbf{t})\bigr\vert , \max _{\mathbf{t}}\bigl\vert f_{\mathbf{c}(\mathbf{t})}(x)-a(\mathbf{t})\bigr\vert \Bigr\} < \varepsilon,$$

whenever $$0<|x-x_{0}|<\overline{\delta}$$ and $$A=[\min_{\mathbf {t}} a(\mathbf{t}), \max_{\mathbf{t}} a(\mathbf{t})]$$. Because ε is arbitrary, this means that $$\lim_{x\rightarrow x_{0}}F_{\mathbf{C}_{\nu}^{k}}(x)=A$$, which is due to Definition 3.1. In the same manner, it can be proved that the interval-valued function $$F_{\mathbf{C}_{\nu}^{k}}$$ is continuous at $$x_{0}$$, when for every $$\mathbf{c}(\mathbf{t})\in\mathbf{C}_{\nu}^{k}$$ the function $$f_{\mathbf{c}(\mathbf{t})}$$ is continuous at $$x_{0}$$. □

### Proposition 3.4

If $$F_{\mathbf{C}_{\nu}^{k}}(x)$$ is continuous at $$x_{0} \in T$$ and $$F_{\mathbf{C}_{\nu}^{k}}(x)=\{f_{\mathbf{c}(\mathbf{t})}(x) \mid f_{\mathbf{c}(\mathbf{t})}:T \subseteq\mathbb{R} \rightarrow\mathbb{R}, \mathbf{c}(\mathbf{t})\in\mathbf {C}_{\nu}^{k}\}$$, then for all $$\mathbf{t}\in[0,1]^{k}$$, $$f_{\mathbf{c}(\mathbf{t})}(x)$$ is continuous at $$x_{0}$$.

### Proof

Let $$F_{\mathbf{C}_{\nu}^{k}}(x)$$ be continuous at $$x_{0}$$. By the definition of continuity and Proposition 3.2

\begin{aligned}& \lim_{x\rightarrow x_{0}}F_{\mathbf{C}_{\nu}^{k}}(x)= F_{\mathbf {C}_{\nu}^{k}}(x_{0}) \\& \quad \Leftrightarrow\quad \lim_{x\rightarrow x_{0}}\bigl(F_{\mathbf{C}_{\nu}^{k}}(x) \ominus_{p} F_{\mathbf{C}_{\nu}^{k}}(x_{0})\bigr)=\{0\} \\& \quad \Leftrightarrow\quad \Bigl\{ \lim_{x \rightarrow x_{0}} \bigl(f_{\mathbf{c}(\mathbf{t})}(x)-f_{\mathbf{c}(\mathbf{t})}(x_{0})\bigr) \bigm| f_{\mathbf{c}(\mathbf{t})}:T \subseteq \mathbb{R} \rightarrow\mathbb{R}, \mathbf{c}( \mathbf{t}) \in\mathbf{C}_{\nu}^{k} \Bigr\} =\{0\}. \end{aligned}

Hence, for every $$\mathbf{c}(\mathbf{t}) \in\mathbf{C}_{\nu}^{k}$$,

$$\lim_{x \rightarrow x_{0}} \bigl(f_{\mathbf{c}(\mathbf{t})}(x)-f_{\mathbf{c}(\mathbf{t})}(x_{0}) \bigr)=0,$$

and finally

$$\lim_{x \rightarrow x_{0}} f_{\mathbf{c}(\mathbf{t})}(x)=f_{\mathbf{c}(\mathbf{t})}(x_{0}),$$

which is the desired conclusion. □

### Remark 3.5

Let $$F(x)=\{ \underline{f}(x)+t (\overline{f}(x)-\underline {f}(x)) \mid t\in[0,1] \}$$ be a continuous function at $$x_{0}$$. Then, similar to Proposition 3.4, one can prove that $$\underline {f}(x)$$ and $$\overline{f}(x)$$ are continuous functions at $$x_{0}$$.

Markov defined the differential calculus for interval-valued function $$F(x)$$ including differentiability [12, 14]. In these forms an interval-valued function is expressed as its boundary function $$\underline{f}(x)$$ and $$\overline {f}(x)$$. So the existence of its derivative depends upon the existence of the derivative of the boundary functions. In the following definition, the concept of p-derivative for an interval-valued function will be defined, which depends upon the existence of the derivative of $$f_{\mathbf{c}(\mathbf{t})}(x)$$ for every value t.

### Definition 3.6

The p-derivative of an interval-valued function $$F:(a,b) \rightarrow K_{c}$$ at $$x_{0} \in(a,b)$$ is denoted by $$F'(x_{0})$$ and is defined as

$$F'(x_{0})=\lim_{h\rightarrow0} \frac{1}{h}\bigl[F(x_{0}+h)\ominus_{p} F_{\mathbf{C}_{\nu}^{k}}(x_{0})\bigr],$$
(7)

where h is such that $$x_{0}+h \in(a,b)$$. In this case, F is called p-differentiable at $$x_{0}$$.

### Proposition 3.7

The interval-valued function $$F_{\mathbf{C}_{\nu}^{k}}:(a,b) \rightarrow K_{c}$$ with the parametric representation $$F_{\mathbf{C}_{\nu }^{k}}(x)=\{f_{\mathbf{c}(\mathbf{t})}(x) \mid f_{\mathbf{c}(\mathbf{t})}:(a,b) \rightarrow \mathbb{R}, \mathbf{c}(\mathbf{t})\in\mathbf{C}_{\nu}^{k}\}$$ is p-differentiable at $$x_{0} \in(a,b)$$ if for every $$\mathbf{t} \in[0,1]^{k}$$, $$f_{\mathbf{c}(\mathbf{t})}$$ is differentiable at $$x_{0}$$ and additionally

$$F'_{\mathbf{C}_{\nu}^{k}}(x_{0})= \bigl\lbrace f'_{\mathbf{c}(\mathbf{t})}(x_{0}) \mid f_{\mathbf{c}(\mathbf{t})}:(a,b) \rightarrow\mathbb{R}; \mathbf{c}(\mathbf{t}) \in\mathbf{C}_{\nu}^{k} \bigr\rbrace .$$

### Proof

By Definition 2.1

$$F_{\mathbf{C}_{\nu}^{k}}(x_{0}+h)\ominus_{p} F_{\mathbf{C}_{\nu}^{k}}(x_{0})= \bigl\lbrace f_{\mathbf{c}(\mathbf{t})}(x_{0}+h)-f_{\mathbf{c}(\mathbf{t})}(x_{0}) \mid f_{\mathbf{c}(\mathbf{t})}:(a,b) \rightarrow\mathbb{R}; \mathbf{c}(\mathbf{t}) \in \mathbf{C}_{\nu}^{k} \bigr\rbrace .$$

Because $$f_{\mathbf{c}(\mathbf{t})}(x)$$ is differentiable at $$x_{0}$$, $$\lim_{h\rightarrow0}\frac{f_{\mathbf{c}(\mathbf{t})}(x_{0}+h)-f_{\mathbf{c}(\mathbf{t})} (x_{0})}{h}$$ exists. Hence, by Proposition 3.3, $$\lim_{h\rightarrow0}\frac{F_{\mathbf{C}_{\nu}^{k}}(x_{0}+h)\ominus_{p} F_{\mathbf{C}_{\nu}^{k}}(x_{0})}{h}$$ exists and using Definition 3.6

\begin{aligned} F'_{\mathbf{C}_{\nu}^{k}}(x_{0})&=\lim_{h\rightarrow 0} \frac{F_{\mathbf{C}_{\nu}^{k}}(x_{0}+h)\ominus_{p} F_{\mathbf{C}_{\nu }^{k}}(x_{0})}{h} \\ &=\biggl\{ \lim_{h\rightarrow0}\frac{f_{\mathbf{c}(\mathbf{t})}(x_{0}+h)-f_{\mathbf{c}(\mathbf{t})} (x_{0})}{h} \Bigm| f_{\mathbf{c}(\mathbf{t})}:(a,b) \rightarrow\mathbb{R}; \mathbf{c}(\mathbf{t}) \in \mathbf{C}_{\nu}^{k}\biggr\} \\ &= \bigl\lbrace f'_{\mathbf{c}(\mathbf{t})}(x_{0}) \mid f_{\mathbf{c}(\mathbf{t})}:(a,b) \rightarrow\mathbb{R}; \mathbf{c}(\mathbf{t}) \in \mathbf{C}_{\nu}^{k} \bigr\rbrace , \end{aligned}

which proves the proposition. □

The converse of Proposition 3.7 is not true, i.e., it is possible that $$F_{\mathbf{C}_{\nu}^{k}}(x)$$ is p-differentiable at given point $$x_{0}$$ whenever for every $$\mathbf{t}\in[ 0,1]^{k}$$, the function $$f_{\mathbf{c}(\mathbf{t})}$$ is not differentiable at $$x_{0}$$. For example, assume that $$F_{\mathbf{C}_{\nu}^{1}}(x)=[-2,2]\odot g(x)$$, where $$g: \mathbb{R} \rightarrow\mathbb{R}$$ is a real-valued function such that

$$g(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} x,& x\geq0, \\ -x,& x< 0. \end{array}\displaystyle \right .$$

Obviously, $$F_{\mathbf{C}_{\nu}^{1}}(x)$$ is p-differentiable at $$x_{0}=0$$ and due to Definition 3.6

\begin{aligned} F'_{\mathbf{C}_{\nu}^{1}}(0)&=\lim_{h\rightarrow0} \frac {F_{\mathbf{C}_{\nu}^{1}}(h)\ominus_{p} F_{\mathbf{C}_{\nu}^{1}}(0)}{h} \\ &= \left\lbrace\lim_{h\rightarrow0}\frac{f_{\mathbf {c}(t)}(h)- f_{\mathbf{c}(t)}(0)}{h} \biggm| f_{\mathbf {c}(t)}(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} (-2+4t)(x),& x\geq0, \\ (-2+4t)(-x),& x< 0; \end{array}\displaystyle \right . t \in[0,1] \right\rbrace \\ &= \left\lbrace f_{\mathbf{c}(t)}(x) \Bigm| f_{\mathbf{c}(t)}(x)= \left \{ \textstyle\begin{array}{l@{\quad}l} -2+4t,& x \geq0, \\ 2-4t,& x< 0; \end{array}\displaystyle \right . t \in[0,1] \right\rbrace=[-2,2]. \end{aligned}

But

$$f_{\mathbf{c}(\mathbf{t})}=\left \{ \textstyle\begin{array}{l@{\quad}l} (-2+4t)(x),& x \geq0, \\ (-2+4t)(-x),& x< 0, \end{array}\displaystyle \right .$$

is not differentiable at 0 for every $$t \in[0,1]$$.

In the next proposition, the relationship between gH-differentiability concept  and p-differentiability is expressed.

### Proposition 3.8

If functions $$\underline{f}$$ and are differentiable at $$x_{0}$$, then the interval-valued function $$F:(a,b) \rightarrow K_{c}$$ with $$F(x)=\{ \underline{f}(x)+t (\overline{f}(x)-\underline {f}(x)) \mid t\in[0,1] \}$$ is p-differentiable at $$x_{0}$$, and $$F'(x)=\{ \underline{f}'(x)+t (\overline{f}'(x)-\underline {f}'(x)) \mid t\in[0,1] \}$$. Moreover, in this case, the concepts of p-differentiability and gH-differentiability are equivalent.

### Proof

It can be deduced that

\begin{aligned} F'(x_{0}) =&\biggl\{ \lim_{h\rightarrow0} \frac{\underline {f}(x_{0}+h)+t(\overline{f}(x_{0}+h)-\underline{f}(x_{0}+h))-\underline {f}(x_{0})-t(\overline{f}(x_{0})-\underline{f}(x_{0}))}{h} \Bigm| t\in[0,1]\biggr\} \\ =&\biggl\{ \lim_{h\rightarrow0}\frac{\underline{f}(x_{0}+h)-\underline {f}(x_{0})}{h}+t\lim _{h\rightarrow0}\frac{\overline {f}(x_{0}+h)-\overline{f}(x_{0})}{h} \\ &{}-t\lim_{h\rightarrow0} \frac {\underline{f}(x_{0}+h)-\underline{f}(x_{0})}{h} \Bigm| t \in[0,1]\biggr\} \\ =&\bigl\{ \underline{f}'(x_{0})+t\bigl( \overline{f}'(x_{0})-\underline{f}'(x_{0}) \bigr) \mid t\in [0,1]\bigr\} . \end{aligned}

Next, if $$\overline{f}'(x_{0})-\underline{f}'(x_{0}) \geq0$$, then by IR

$$F'(x_{0})=\bigl[\underline{f}'(x_{0}), \overline{f}'(x_{0})\bigr],$$

that is consistent with (i)-gH-differentiability. If $$\overline {f}'(x_{0})-\underline{f}'(x_{0}) \leq0$$, then by DR

$$F'(x_{0})=\bigl[\overline{f}'(x_{0}), \underline{f}'(x_{0})\bigr],$$

which is in accordance with (ii)-gH-differentiability. □

### Remark 3.9

Due to the definition of the interval derivative (7) and two parametric representations (2) and (3) for interval-valued functions, in general, it cannot be expected that the derivative of an interval-valued function be equal based on these representations. This outcome occurs because the derivative for representation (3) is calculated regardless of the sign of the independent variable x, while the derivative for representation (2) depends on the sign of x. For example, consider the interval-valued function $$F_{\mathbf{C}_{\nu}^{2}}(x)=[-1,2]\odot x \oplus[2,4] e^{x}$$, $$x \in[-2,5]$$. By the parametric representation (3), $$F_{\mathbf{C}_{\nu}^{2}}(x)$$ can be written as $$F_{\mathbf {C}_{\nu}^{2}}(x)=\{(-1+3t_{1})x+(2+2t_{2})e^{x} \mid t_{1},t_{2} \in[0,1]\}$$, and it is possible to calculate its p-derivative from Proposition 3.7 as

\begin{aligned} F'_{\mathbf{C}_{\nu}^{2}}(x) =&\bigl\{ (-1+3t_{1})+(2+2t_{2})e^{x} \mid t_{1},t_{2} \in [0,1]\bigr\} \\ =&[-1,2]\oplus[2,4] \odot e^{x}=\bigl[-1+2e^{x},2+4e^{x}\bigr]. \end{aligned}
(8)

Next, by the parametric representation (2), it follows that

$$F_{\mathbf{C}_{\nu}^{2}}(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} \{2x+2e^{x}+t(2e^{x}-3x) \mid t \in[0,1]\},&-2 \leq x \leq0, \\ \{-x+2e^{x}+t(2e^{x}+3x) \mid t \in[0,1]\},&0< x \leq5, \end{array}\displaystyle \right .$$

and its derivative is given by

\begin{aligned} F'_{\mathbf{C}_{\nu}^{2}}(x)&=\left \{ \textstyle\begin{array}{l@{\quad}l} \{2+2e^{x}+t(2e^{x}-3) \mid t \in[0,1]\},&-2 \leq x \leq0, \\ \{-1+2e^{x}+t(2e^{x}+3) \mid t \in[0,1]\},&0< x \leq5 \end{array}\displaystyle \right . \\ &=\left \{ \textstyle\begin{array}{l@{\quad}l} {[-1+4e^{x}, 2+2e^{x}]},&-2 \leq x \leq0, \\ {[-1+2e^{x},2+4e^{x}]},&0< x \leq5, \end{array}\displaystyle \right . \end{aligned}

which is different from (8).

According to Proposition 3.8, two cases corresponding to increasing and decreasing representations are distinguished for the definition of p-differentiability.

### Definition 3.10

Suppose that $$F:(a,b) \rightarrow K_{c}$$ is an interval-valued function with parametric representation $$F=\{ \underline {f}(x)+t (\overline{f}(x)-\underline{f}(x)) \mid t\in[0,1] \}$$. Then F is (i)-p-differentiable at $$x_{0}$$ if

$$F'(x_{0})=\bigl\{ \underline{f}'(x_{0})+t \bigl(\overline{f}'(x_{0})-\underline {f}'(x_{0})\bigr) \mid t \in[0,1]\bigr\} ,$$
(9)

and it is (d)-p-differentiable at $$x_{0}$$ if

$$F'(x_{0})=\bigl\{ \overline{f}'(x_{0})+t \bigl(\underline{f}'(x_{0})-\overline {f}'(x_{0})\bigr) \mid t \in[0,1]\bigr\} .$$
(10)

The concept of a switching point can be extended by the above definition.

### Definition 3.11

A point $$x_{0}\in(a,b)$$ is said to be a switching point for the differentiability of interval-valued function F, if in any neighborhood N of $$x_{0}$$ there exist points $$x_{1}< x_{0}< x_{2}$$ such that

• (type I) at $$x_{1}$$ (9) holds, while (10) does not hold and at $$x_{2}$$ (10) holds and (9) does not hold, or

• (type II) at $$x_{1}$$ (10) holds, while (9) does not hold and at $$x_{2}$$ (9) holds and (10) does not hold.

### Remark 3.12

Using Definition 3.11 and Proposition 3.8, it is easy to find switching points. For this purpose, it is sufficient to determine the sign of $$\overline{f}'-\underline{f}'$$ in the parametric representation of $$F'$$. More precisely, if $$\overline{f}'(x)-\underline {f}'(x)$$ is positive for $$x < x_{0}$$ and negative for $$x>x_{0}$$, then $$x_{0}$$ is a switching point of type I; and $$x_{0}$$ is a switching point of type II, if $$\overline{f}'(x)-\underline{f}'(x)$$ is negative for $$x< x_{0}$$ and positive for $$x>x_{0}$$.

### Example 3.1

Let us consider the interval-valued function $$F_{\mathbf{C}_{\nu }^{1}}:(-10,10) \rightarrow K_{c}$$ defined by (see Figure 1(a))

$$F_{\mathbf{C}_{\nu}^{1}}(x)=[2,4]\odot\biggl(\cos(x)-\frac{x^{2}}{32}\biggr).$$

Clearly,

\begin{aligned}& \underline{f}(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} 4\cos(x)-\frac{x^{2}}{8}, & -10\leq x < -1.5004, \\ 2\cos(x)-\frac{x^{2}}{16}, & -1.5004\leq x < 1.5004, \\ 4\cos(x)-\frac{x^{2}}{8}, & 1.5004 \leq x < -10, \end{array}\displaystyle \right . \\& \overline{f}(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} 2\cos(x)-\frac{x^{2}}{16}, & -10\leq x < -1.5004, \\ 4\cos(x)-\frac{x^{2}}{8}, & -1.5004\leq x < 1.5004, \\ 2\cos(x)-\frac{x^{2}}{16}, & 1.5004\leq x < -10. \end{array}\displaystyle \right . \end{aligned}

By Remark 3.12 and determining the sign of

$$\overline{f}'(x)-\underline{f}'(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} 2\sin(x)+\frac{x}{8}, & -10\leq x < -1.5004, \\ -2\sin(x)-\frac{x}{8}, & -1.5004\leq x < 1.5004, \\ 2\sin(x)+\frac{x}{8}, & 1.5004 \leq x < -10, \end{array}\displaystyle \right .$$

which is represented in Figure 1(b), it can be deduced that $$x_{1}=-1.5004$$, $$x_{2}=1.5004$$, $$x_{3}=-5.9052$$, $$x_{4}=5.9052$$ are switching points of type II; and $$x_{5}=0$$, $$x_{6}=-3.3527$$, $$x_{7}=3.3527$$ are switching points of type I. Figure 1

Finally, the concept of integral for an interval-valued function in terms of its parameters is introduced as follows.

### Definition 3.13

For a given interval-valued function $$F_{\mathbf{C}_{\nu}^{k}}:T\subseteq \mathbb{R}\rightarrow K_{c}$$ with $$F_{\mathbf{C}_{\nu}^{k}}(x)=\{ f_{\mathbf{c}(\mathbf{t})}(x) \mid f_{\mathbf{c}(\mathbf{t})}:T\subseteq\mathbb{R} \rightarrow\mathbb{R}; \mathbf{c}(\mathbf{t})\in\mathbf{C}_{\nu}^{k} \}$$ and an interval $$[\alpha,\beta] \subseteq T$$, the definite integral $$\int_{\alpha}^{\beta}F_{\mathbf{C}_{\nu}^{k}}(x) \, dx$$ is defined as

$$\int_{\alpha}^{\beta}F_{\mathbf{C}_{\nu}^{k}}(x)\, dx= \biggl\lbrace \int_{\alpha}^{\beta} f_{\mathbf{c}(\mathbf{t})}(x)\, dx \Bigm| f_{\mathbf{c}(\mathbf{t})}(x) \text{ is integrable w.r.t. } x \text{ for every } \mathbf{c}( \mathbf{t})\in\mathbf {C}_{\nu}^{k} \biggr\rbrace .$$

The mentioned issue in Remark 3.9 holds for the concept of a definite integral for interval-valued functions. See the following example.

### Example 3.2

Consider $$F_{\mathbf{C}_{\nu}^{2}}(x)=[-1,2]\odot(1-x)\oplus[-2,4]\odot (3-x)$$ and the interval of integration $$[0,4]$$. Then, by Definition 3.13,

\begin{aligned}& \int_{0}^{4}[-1,2] \odot(1-x)\oplus[-2,4] \odot(3-x)\, dx \\& \quad = \biggl\lbrace \int_{0}^{4}(-1+3t_{1}) (1-x)+(-2+6t_{2}) (3-x)\, dx \Bigm| t_{1},t_{2} \in[0,1] \biggr\rbrace \\& \quad =\bigl\{ -4-12t_{1}+24t_{2} \mid t_{1},t_{2} \in[0,1]\bigr\} =[-16,20], \end{aligned}

while by considering the parametric representation (2) for $$F_{\mathbf{C}_{\nu}^{2}}(x)$$, it follows that

\begin{aligned}& \int_{0}^{4}[-1,2] \odot(1-x)\oplus[-2,4] \odot(3-x)\, dx \\& \quad =\int_{0}^{1}\bigl((3x-7)+t(21-9x)\bigr)\, dx+\int_{1}^{3}\bigl(-4+t(15-3x)\bigr) \, dx \\& \qquad {} +\int_{3}^{4}\bigl(14-6x+t(-21+9x) \bigr)\, dx=\biggl\{ \frac{-41}{2}+\frac{90}{2}t \Bigm| t \in[0,1]\biggr\} \\& \quad =\biggl[\frac{-41}{2},\frac{49}{2}\biggr]. \end{aligned}

### Proposition 3.14

A continuous interval-valued function $$F(x)$$ is integrable.

### Proof

The proof is an immediate consequence of Proposition 3.4 and Remark 3.5. □

The concept of definite integral satisfies following properties:

### Proposition 3.15

Let $$F(x)$$ and $$G(x)$$ be two integrable interval-valued functions. Then:

1. 1.

$$\int_{\alpha}^{\beta}F(x) \,dx=\int_{\alpha}^{\gamma}F(x) \,dx +\int_{\gamma}^{\beta}F(x) \,dx$$; $$\alpha\leq\gamma\leq\beta$$,

2. 2.

$$\int_{\alpha}^{\beta} (a F(x)+b G(x)) \,dx=a \int_{\alpha}^{\beta} F(x) \,dx+b\int_{\alpha}^{\beta} G(x) \,dx$$; $$a,b\in\mathbb{R}$$.

### Proposition 3.16

Let $$F_{\mathbf{C}_{\nu}^{k}}:T \subseteq\mathbb{R} \rightarrow K_{c}$$ be continuous. Then:

1. 1.

the function $$G_{\mathbf{D}_{\nu}^{n}}(x)=\int_{\alpha}^{x} F_{\mathbf {C}_{\nu}^{k}}(z) \, dz$$ is p-differentiable, and $$G'_{\mathbf{D}_{\nu }^{n}}(x)=F_{\mathbf{C}_{\nu}^{k}}(x)$$,

2. 2.

the function $$H_{\mathbf{E}_{\nu}^{m}}(x)=\int_{x}^{\beta}F_{\mathbf {C}_{\nu}^{k}}(z)\, dz$$ is p-differentiable, and $$H'_{\mathbf{E}_{\nu }^{m}}(x)=-F_{\mathbf{C}_{\nu}^{k}}(x)$$.

### Proof

First, assume that

$$G_{\mathbf{D}_{\nu}^{n}}(x)=\int_{\alpha}^{x} F_{\mathbf{C}_{\nu}^{k}}(z)\, dz=\biggl\{ \int_{a}^{x} f_{\mathbf{c}(\mathbf{t})}(z)\, dz \Bigm| f_{\mathbf{c}(\mathbf{t})}:T \rightarrow \mathbb{R}; \mathbf{c}(\mathbf{t})\in\mathbf{C}_{\nu}^{k}\biggr\} ,$$

because of the continuity of $$F_{\mathbf{C}_{\nu}^{k}}$$ and Proposition 3.4, the function $$\int_{\alpha}^{x} f_{\mathbf{c}(\mathbf{t})}(z) \,dz$$ is differentiable for every $$\mathbf{t} \in[0,1]^{k}$$. Hence, according to Proposition 3.7, $$G_{\mathbf{D}_{\nu}^{n}}(x)$$ is p-differentiable, and

$$G'_{\mathbf{D}_{\nu}^{n}}(x)=\bigl\{ f_{\mathbf{c}(\mathbf{t})}(x) \mid \mathbf{c}( \mathbf{t})\in \mathbf{C}_{\nu}^{k}\bigr\} =F_{\mathbf{C}_{\nu}^{k}}(x).$$

The second part is proved in the same way as the first one. □

In addition, Proposition 3.16 holds for the interval-valued function $$F(x)=\{ \underline{f}(x)+t (\overline{f}(x)-\underline {f}(x)) \mid t\in[0,1] \}$$, by using Remark 3.5 and Proposition 3.8.

### Proposition 3.17

If $$F(x)=\{ \underline{f}(x)+t (\overline{f}(x)-\underline{f}(x)) \mid t\in [0,1] \}$$ is p-differentiable with no switching point in the interval $$[\alpha,\beta]$$, then

$$\int_{\alpha}^{\beta}F'(x)\,dx=F(\beta) \ominus_{p} F(\alpha).$$

### Proof

By Definition 3.10, because there is no switching point, F is (i)-p-differentiable or (d)-p-differentiable in the interval $$[\alpha,\beta ]$$. Let F be (i)-p-differentiable (the proof for the (d)-p-differentiable case being similar), then

\begin{aligned} \int_{\alpha}^{\beta}F'(x)\,dx &= \biggl\{ \int_{\alpha}^{\beta}\bigl(\underline {f}'(x)+t \bigl(\overline{f}'(x)-\underline{f}'(x)\bigr)\bigr) \,dx \Bigm| t\in[0,1] \biggr\} \\ &= \bigl\{ \underline{f}(\beta)-\underline{f}(\alpha)+t\bigl(\overline {f}( \beta)-\overline{f}(\alpha)-\underline{f}(\beta)+\underline {f}(\alpha)\bigr) \mid t\in[0,1] \bigr\} \\ &=F(\beta)\ominus_{p} F(\alpha), \end{aligned}

which completes the proof. □

### Theorem 3.18

Let $$F(x)=\{ \underline{f}(x)+t(\overline{f}(x)-\underline{f}(x)) | t \in[0,1]\}$$ be p-differentiable with n switching point at $$\gamma _{i}$$, $$i=1,2,\ldots, n$$, $$\alpha=\gamma_{0}<\gamma_{1}<\gamma_{2}<\cdots<\gamma _{n}<\gamma_{n+1}=\beta$$ and exactly at these points. Then

$$F(\beta)\ominus_{p}F(\alpha)=\sum_{i=1}^{n} \biggl[ \int_{\gamma _{i-1}}^{\gamma_{i}}F'(x) \,dx \ominus_{p} (-1)\int_{\gamma_{i}}^{\gamma_{i+1}}F'(x) \,dx \biggr].$$

Moreover,

$$\int_{\alpha}^{\beta}F'(x) \,dx=\sum _{i=1}^{n+1} \bigl( F(\gamma_{i}) \ominus_{p} F(\gamma_{i-1}) \bigr),$$

and if $$F(\gamma_{i})$$ is a real number for $$i=1,2,\ldots,n$$ then $$\int_{\alpha}^{\beta}F'(x) \,dx=F(\beta)-F(\alpha)$$.

### Proof

To prove this, consider only one switching point, the case of a finite number of switching points follows similarly. Let F be (i)-p-differentiable on $$[\alpha,\gamma]$$ and (d)-p-differentiable on $$[\gamma,\beta]$$. Then, by Proposition 3.17,

\begin{aligned}& \int_{\alpha}^{\gamma}F'(x) \,dx \ominus_{p}(-1)\int_{\gamma}^{\beta}F'(x) \,dx \\& \quad =\bigl(F(\gamma)\ominus_{p}F(\alpha)\bigr) \ominus_{p}(-1) \bigl(F(\beta)\ominus_{p}F(\gamma)\bigr) \\& \quad =\bigl(F(\gamma)\ominus_{p}F(\alpha)\bigr)\ominus_{p} \bigl(F(\gamma)\ominus_{p}F(\beta)\bigr) \\& \quad =\bigl\{ \bigl( \underline{f}(\gamma)-\underline{f}(\alpha)+t\bigl( \overline {f}(\gamma)-\overline{f}(\alpha)-\underline{f}(\gamma)+\underline {f}( \alpha)\bigr) \bigr) \\& \qquad {} - \bigl( \underline{f}(\gamma)-\underline{f}(\beta)+t\bigl(\overline {}(\gamma)-\overline{f}(\beta)-\underline{f}(\gamma)+\underline{f}(\beta )\bigr) \bigr) \mid t\in[0,1] \bigr\} \\& \quad =\bigl\{ \bigl(\underline{f}(\beta)+ t\bigl(\overline{f}(\beta)- \underline{f}(\beta )\bigr)\bigr)-\bigl(\underline{f}(\alpha)+ t\bigl( \overline{f}(\alpha)-\underline{f}(\alpha )\bigr)\bigr) \mid t\in[0,1] \bigr\} \\& \quad =F(\beta)\ominus_{p}F(\alpha). \end{aligned}

Additionally, by Proposition 3.15 and Proposition 3.17,

\begin{aligned} \int_{\alpha}^{\beta}F'(x) \,dx &=\int _{\alpha}^{\gamma}F'(x) \,dx+\int _{\gamma}^{\beta}F'(x) \,dx \\ &=\bigl(F(\gamma)\ominus_{p}F(\alpha)\bigr)+\bigl(F(\beta) \ominus_{p}F(\gamma)\bigr). \end{aligned}

If the values $$F(\gamma_{i})$$ at all the switching points $$\gamma_{i}$$, $$i=1,2,\ldots,n$$ are real numbers, then by Remark 2.4

\begin{aligned} \int_{\alpha}^{\beta}F'(x) \,dx =&\sum _{i=1}^{n+1} \bigl( F(\gamma_{i}) \ominus _{p}F(\gamma_{i-1}) \bigr) \\ =& \bigl( F(\beta)-F(\gamma_{n}) \bigr) + \bigl( F( \gamma_{n})-F(\gamma _{n-1}) \bigr)+\cdots \\ &{} + \bigl( F(\gamma_{2})-F(\gamma_{1}) \bigr)+ \bigl( F( \gamma _{1})-F(\alpha) \bigr) \\ =&F(\beta)-F(\alpha). \end{aligned}

□

## Interval differential equation

In this section, two approaches are proposed to find the solution of the interval differential equation. In the first approach, by using the parametric representation (3) and its corresponding definitions of the p-derivative and integral, the interval differential equation is converted to a crisp problem. The second follows from the notation of the p-derivative based on the parametric representation (2). It is noteworthy that the solutions obtained by the two approaches are distinctive.

### The first approach

In this approach, the following interval differential equation is considered:

\begin{aligned}& Y_{\mathbf{D}_{\nu}^{n}}'(x)=F_{\mathbf{C}_{\nu}^{k}} \bigl(x,Y_{\mathbf{D}_{\nu }^{n}}(x)\bigr), \\& Y_{\mathbf{D}_{\nu}^{n}}(x_{0})=Y_{0}, \end{aligned}
(11)

where $$Y_{0} \in K_{c}$$, $$F_{\mathbf{C}_{\nu}^{k}}:[a,b]\times K_{c} \rightarrow K_{c}$$ and $$Y_{\mathbf{D}_{\nu }^{n}}:[a,b]\rightarrow K_{c}$$.

Let $$\{ y_{\mathbf{d}(\mathbf{t})}(x) \mid y_{\mathbf{d}(\mathbf{t})}:[a,b] \rightarrow \mathbb{R}; \mathbf{d}(\mathbf{t}) \in\mathbf{D}_{\nu}^{n}\}$$, $$\{f_{\mathbf {c}(\mathbf{t}')}(x,y_{\mathbf{d}(\mathbf{t})}(x)) \mid f_{\mathbf{c}(\mathbf{t}')}:[a,b] \times\mathbb{R}\rightarrow\mathbb{R}; \mathbf{c}(\mathbf{t}') \in \mathbf{C}_{\nu}^{k}\}$$ and $$\{\underline{y}_{0}+t''(\overline {y}_{0}-\underline{y}_{0}) \mid t''\in[0,1]\}$$ be parametric representations of $$Y_{\mathbf{D}_{\nu}^{n}}:[a,b]\rightarrow K_{c}$$, $$F_{\mathbf {C}_{\nu}^{k}}(x,Y_{\mathbf{D}_{\nu}^{n}}(x))$$, and $$Y_{0}$$, respectively. From definition of equality given in Definition 2.1, the differential equation (11) can be considered as

\begin{aligned}& \bigl\lbrace y_{\mathbf{d}(\mathbf{t})}'(x) \mid y_{\mathbf{d}(\mathbf{t})}:[a,b] \rightarrow\mathbb{R}; \mathbf{d}(\mathbf{t}) \in\mathbf{D}_{\nu}^{n} \bigr\rbrace \\& \quad = \bigl\lbrace f_{\mathbf{c}(\mathbf{t}')}\bigl(x,y_{\mathbf{d}(\mathbf{t})}(x)\bigr) \mid f_{\mathbf{c}(\mathbf{t}')}:[a,b] \times\mathbb {R}\rightarrow\mathbb{R}; \mathbf{c}\bigl( \mathbf{t}'\bigr) \in\mathbf{C}_{\nu }^{k} \bigr\rbrace , \\& \bigl\{ y_{\mathbf{d}(\mathbf{t})}(x_{0}) \mid y_{\mathbf{d}(\mathbf{t})}:[a,b] \rightarrow\mathbb {R}; \mathbf{d}(\mathbf{t}) \in\mathbf{D}_{\nu}^{n} \bigr\} = \bigl\{ \underline {y}_{0}+t''( \overline{y}_{0}-\underline{y}_{0}) \mid t''\in[0,1]\bigr\} . \end{aligned}

By Remark 2.3, it follows that

\begin{aligned}& y_{\mathbf{d}(\mathbf{t})}'(x)= f_{\mathbf{c}(\mathbf{t}')} \bigl(x,y_{\mathbf{d}(\mathbf{t})}(x)\bigr), \\& y_{\mathbf{d}(\mathbf{t})}(x_{0})= \underline{y}_{0}+t''( \overline{y}_{0}-\underline {y}_{0}). \end{aligned}
(12)

### Theorem 4.1

Let $$f: [x_{0},x_{0}+p] \times\overline{B}([\underline{y}_{0},\overline {y}_{0}],r) \times\overline{B}([0,1]^{k} ,q) \rightarrow\mathbb{R}$$ be Lipschitz in its second and third variables; i.e.,

\begin{aligned}& \exists L_{1} \quad \textit{s.t. } \bigl\Vert f\bigl(x,y, \mathbf{t}'\bigr)-f\bigl(x,w,\mathbf{t}'\bigr) \bigr\Vert \leq L_{1} \| y-w \|, \\& \exists L_{2} \quad \textit{s.t. } \bigl\Vert f\bigl(x,y, \mathbf{t}'\bigr)-f(x,y,\mathbf{s}) \bigr\Vert \leq L_{2} \bigl\Vert \mathbf{t}'-\mathbf{s} \bigr\Vert , \end{aligned}

respectively. Then the initial valued problem

\begin{aligned}& y'(x) =f\bigl(x,y(x),\mathbf{t}'\bigr), \\& y(x_{0}) =\underline{y}_{0}+t''( \overline{y}_{0}-\underline{y}_{0}) \end{aligned}

has a unique solution. Moreover, if f is continuous in $$\mathbf{t}'$$ then the solution $$y(x,\mathbf{t}',t'')$$ is continuous in $$\mathbf{t}'$$ and $$t''$$.

### Proof

See , Theorem 9.6, for the first part. The second part follows immediately from the continuity of f and $$y(x_{0})$$ w.r.t. $$\mathbf{t}'$$ and $$t''$$, respectively. □

Let $$Y_{\mathbf{D}_{\nu}^{n}}(x)$$ and $$F_{\mathbf{C}_{\nu}^{k}}(x,Y_{\mathbf {D}_{\nu}^{n}}(x))$$ be interval-valued functions with parametric representations $$\{y_{\mathbf{d}(\mathbf{t})}(x) \mid y_{\mathbf{d}(\mathbf{t})}:[a,b] \rightarrow\mathbb{R}; \mathbf{d}(\mathbf{t}) \in\mathbf{D}_{\nu}^{n} \}$$ and $$\{ f_{\mathbf{c}(\mathbf{t}')}(x,y_{\mathbf{d}(\mathbf{t})}(x)) \mid f_{\mathbf {c}(\mathbf{t}')}:[a,b] \times\mathbb{R}\rightarrow\mathbb{R}; \mathbf {c}(\mathbf{t}') \in\mathbf{C}_{\nu}^{k}\}$$, respectively. Suppose that $$f_{\mathbf{c}(\mathbf{t}')}(x,y_{\mathbf{d}(\mathbf{t})}(x))$$ is Lipschitz in y and $$\mathbf{t}'$$, because $$f_{\mathbf{c}(\mathbf{t}')}(x,y_{\mathbf{d}(\mathbf{t})}(x))$$ and $$y_{d(t'')}(x_{0})=\underline{y}_{0}+t''(\overline {y}_{0}-\underline{y}_{0})$$ are continuous functions in $$\mathbf{t}'$$ and $$t''$$, respectively, by Theorem 4.1, problem (12) has a unique solution $$y_{\mathbf{d}(\mathbf{t})}(x)=y(x,\mathbf{t}',t'')$$, which is a continuous function in $$\mathbf{t}'$$ and $$t''$$. Therefore, $$\min_{\mathbf{t}',t''} {y(x,\mathbf{t}',t'')}$$ and $$\max_{\mathbf {t}',t''} {y(x,\mathbf{t}',t'')}$$ exist, and

$$Y_{\mathbf{D}_{\nu}^{n}}(x)=\Bigl[\min_{\mathbf{t}',t''} {y\bigl(x, \mathbf {t}',t''\bigr)},\max _{\mathbf{t}',t''} y\bigl(x,\mathbf{t}',t''\bigr)\Bigr].$$
(13)

### Example 4.1

Consider the following interval differential equation:

$$\left \{ \textstyle\begin{array}{l} Y_{\mathbf{D}_{\nu}^{n}}'(x)=-Y_{\mathbf{D}_{\nu}^{n}}(x)\oplus[1,2]\odot \sin(x), \\ Y_{\mathbf{D}_{\nu}^{n}}(0)=[1,3], \quad x\in[0,6]. \end{array}\displaystyle \right .$$
(14)

Its corresponding ordinary differential equation is

\begin{aligned}& y'_{\mathbf{d}(\mathbf{t})}(x)=-y_{\mathbf{d}(\mathbf{t})}(x)+ \bigl(1+t'\bigr)\sin(x), \\& y_{\mathbf{d}(\mathbf{t})}(0)=1+2t'', \quad t',t'' \in[0,1], x \in[0,6]. \end{aligned}
(15)

The general solution (15) is provided by

$$y_{\mathbf{d}(\mathbf{t})}(x)=\bigl(1+t'\bigr) \biggl(\frac{-\cos(x)+\sin(x)}{2} \biggr)+c e^{-x},$$

and so the particular solution would be

$$y_{\mathbf{d}(\mathbf{t})}(x)=\bigl(1+t'\bigr) \biggl(\frac{-\cos(x)+\sin(x)+e^{-x}}{2} \biggr)+\bigl(1+2t''\bigr)e^{-x}.$$

Hence, the unique solution of interval differential equation (14) is obtained as

$$Y_{\mathbf{D}_{\nu}^{2}}(x)=[1,2]\odot\biggl(\frac {-\cos(x)+\sin(x)+e^{-x}}{2}\biggr)\oplus[1,3]\odot e^{-x}.$$

Moreover, by (13),

$$Y_{\mathbf{D}_{\nu}^{n}}(x)= \left \{ \textstyle\begin{array}{l@{\quad}l} {[\frac{-\cos(x)+\sin(x)+e^{-x}}{2}+e^{-x} , -\cos(x)+\sin(x)+4e^{-x} ]},& x\in[0,3.941], \\ {[-\cos(x)+\sin(x)+4e^{-x}, \frac{-\cos(x)+\sin(x)+e^{-x}}{2}+3e^{-x}]},& x\in[3.941,6 ], \end{array}\displaystyle \right .$$

which is illustrated in Figure 2. Figure 2

### The second approach

Now, consider the interval differential equation

\begin{aligned}& Y'(x)=F\bigl(x,Y(x)\bigr), \\& Y(x_{0})=Y_{0}, \end{aligned}
(16)

where $$Y_{0} \in K_{c}$$, $$F:[a,b]\times K_{c} \rightarrow K_{c}$$, and $$Y:[a,b]\rightarrow K_{c}$$. By the parametric representation (2) and rewriting (16), two following systems are obtained from (9) and (10):

1. I.

$$\{\underline{y}'(x)+t(\overline{y}'(x)-\underline{y}'(x)) \mid t\in [0,1]\}=\{\underline{f}(x,\underline{y},\overline{y})+t(\overline {f}(x,\underline{y},\overline{y})-\underline{f}(x,\underline {y},\overline{y})) \mid t\in[0,1]\}$$, $$\{\underline{y}(x_{0})+t(\overline{y}(x_{0})-\underline{y}(x_{0})) \mid t\in [0,1]\}=\{\underline{y}_{0}+t(\overline{y}_{0}-\underline{y}_{0}) \mid t \in [0,1]\}$$,

2. II.

$$\{\overline{y}'(x)+t(\underline{y}'(x)-\overline{y}'(x)) \mid t\in[0,1]\} =\{\underline{f}(x,\underline{y},\overline{y})+t(\overline {f}(x,\underline{y},\overline{y})-\underline{f}(x,\underline {y},\overline{y})) \mid t\in[0,1]\}$$, $$\{\underline{y}(x_{0})+t(\overline{y}(x_{0})-\underline{y}(x_{0})) \mid t\in [0,1]\}=\{\underline{y}_{0}+t(\overline{y}_{0}-\underline{y}_{0}) \mid t \in [0,1]\}$$.

Hence, for fixed x, two systems can be deduced as follows:

\begin{aligned}& \left \{ \textstyle\begin{array}{l} \underline{y}'(x)=\underline{f}(x,\underline{y}(x),\overline{y}(x)), \qquad \underline{y}(x_{0})=\underline{y}_{0}, \\ \overline{y}'(x)=\overline{f}(x,\underline{y}(x),\overline{y}(x)),\qquad \overline{y}(x_{0})=\overline{y}_{0}, \end{array}\displaystyle \right . \end{aligned}
(17)
\begin{aligned}& \left \{ \textstyle\begin{array}{l} \overline{y}'(x)=\underline{f}(x,\underline{y}(x),\overline{y}(x)), \qquad \underline{y}(x_{0})=\underline{y}_{0}, \\ \underline{y}'(x)=\overline{f}(x,\underline{y}(x),\overline{y}(x)),\qquad \overline{y}(x_{0})=\overline{y}_{0}. \end{array}\displaystyle \right . \end{aligned}
(18)

### Theorem 4.2

Let $$F: [x_{0},x_{0}+p]\times\overline{B}(Y_{0},r) \rightarrow K_{c}$$ be nontrivial and continuous. If F satisfies the Lipschitz condition $$D(F(x,Y),F(x,Z))\leq L D(Y,Z)$$ for all $$(x,Y),(x,Z) \in [x_{0},x_{0}+p]\times\overline{B}(Y_{0},r)$$, then the interval differential equation

\begin{aligned}& Y'(x) =F\bigl(x,Y(x)\bigr), \\& Y(x_{0}) =Y_{0}, \end{aligned}
(19)

is equivalent to the union of the systems (17) and (18) on some interval $$[x_{0},x_{0}+p]$$. Here, the equivalence means that $$Y=[\underline{y},\overline{y}]:[x_{0},x_{0}+p] \rightarrow K_{c}$$ is a solution of (19) if and only if $$(\underline{y},\overline {y}):[x_{0},x_{0}+p]\rightarrow\mathbb{R}^{2}$$ is a solution of one of the problems (17) or (18).

See . □

### Example 4.2

Consider interval differential equation (14), for $$x \in [0,6]$$. The corresponding systems are

\begin{aligned}& \left \{ \textstyle\begin{array}{l} \underline{y}'(x)= \left\{ \textstyle\begin{array}{l@{\quad}l} -\overline{y}(x)+\sin(x),& 0 \leq x \leq\pi, \\ -\overline{y}(x)+2\sin(x),& \pi\leq x\leq6, \end{array}\displaystyle \right . \\ \overline{y}'(x)= \left\{ \textstyle\begin{array}{l@{\quad}l} -\underline{y}(x)+2\sin(x),& 0 \leq x \leq\pi, \\ -\underline{y}(x)+\sin(x),& \pi\leq x\leq6, \end{array}\displaystyle \right . \\ \underline{y}(0)=1, \qquad \overline{y}(0)=3, \end{array}\displaystyle \right . \\& \left \{ \textstyle\begin{array}{l} \overline{y}'(x)= \left\{ \textstyle\begin{array}{l@{\quad}l} -\overline{y}(x)+\sin(x),& 0 \leq x \leq\pi, \\ -\overline{y}(x)+2\sin(x),& \pi\leq x\leq6, \end{array}\displaystyle \right . \\ \underline{y}'(x)= \left\{ \textstyle\begin{array}{l@{\quad}l} -\underline{y}(x)+2\sin(x),& 0 \leq x \leq\pi, \\ -\underline{y}(x)+\sin(x),& \pi\leq x\leq6, \end{array}\displaystyle \right . \\ \underline{y}(0)=1,\qquad \overline{y}(0)=3. \end{array}\displaystyle \right . \end{aligned}

By considering (i)-p-differentiability the following solution is obtained for the interval differential equation (14):

$$Y_{1}(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} {[\frac{11}{4}e^{-x}-\frac{1}{2}\cos(x)-\frac{5}{4}e^{x}+\sin(x), \frac {11}{4}e^{-x}-\cos(x)+\frac{5}{4}e^{x}+\frac{1}{2}\sin(x)]}, & 0 \leq x \leq\pi, \\ {[\frac{1}{2}\sin(x)-\cos(x)+ 0.1188e^{\pi- x}-29.4259e^{x-\pi}}, \\ \quad {\sin(x)-\frac{1}{2}\cos(x)+0.1188e^{\pi-x}+29.4259e^{x-\pi}]},&\pi\leq x\leq6, \end{array}\displaystyle \right .$$

which has no switching point. The second solution is achieved by starting with (d)-p-differentiability, which has a switching point at 1.3607 for $$x \in[0,\pi]$$ and

$$Y_{2}(x)= \left \{ \textstyle\begin{array}{l@{\quad}l} {[2e^{-x}-\cos(x)+\sin(x), \frac{7}{2}e^{-x}-\frac{1}{2}\cos(x)+ \frac {1}{2}\sin(x)]},& x \in[0, 1.3607], \\ {[\frac{11}{4}e^{-x}+\sin(x)-\frac{1}{2}\cos(x)+0.0761e^{x}}, \\ \quad {\frac{11}{4}e^{-x}+\frac{1}{2}\sin(x)-\cos(x)+0.0761e^{x}]}, & x \in[1.3607, \pi]. \end{array}\displaystyle \right .$$

Furthermore,

$$Y_{2}(x)= \left \{ \textstyle\begin{array}{l@{\quad}l} {[\frac{1}{2}\sin(x)-\frac{1}{2}\cos(x)-1.6418e^{\pi -x}}, \\ \quad {\sin(x)-\cos(x)+1.8795e^{\pi-x}]},& x \in[\pi, 4.9225], \\ {[\frac{1}{2}\sin(x)-\cos(x)+2.7500e^{-x}+0.0014e^{x}}, \\ \quad {\sin(x)-\frac {1}{2}\cos(x)+2.7500e^{-x}+0.0014e^{x}]},& x \in[4.9225,6], \end{array}\displaystyle \right .$$

for $$x \in[\pi,6]$$ and has one switching point at 4.9225. The solutions are shown in Figure 3. Figure 3

The two proposed approaches are applied to the following examples.

### Example 4.3

Consider the interval differential equation

$$\left \{ \textstyle\begin{array}{l} Y_{\mathbf{D}_{\nu}^{n}}'(x)=-Y_{\mathbf{D}_{\nu}^{n}}(x)\oplus[1,2]\odot x, \\ Y_{\mathbf{D}_{\nu}^{n}}(0)=[0,1], \quad x\in[0,4]. \end{array}\displaystyle \right .$$
(20)

By the first approach, the corresponding ordinary differential equation is obtained as

\begin{aligned}& y'_{\mathbf{d}(\mathbf{t})}(x) =-y_{\mathbf{d}(\mathbf{t})}(x)+ \bigl(1+t'\bigr)x, \\& y_{\mathbf{d}(\mathbf{t})}(0) =t'', \quad t',t'' \in[0,1], x \in[0,4]. \end{aligned}
(21)

The particular solution would be

$$y_{\mathbf{d}(\mathbf{t})}(x)=\bigl(1+t'\bigr)x-\bigl(1+t' \bigr)+\bigl(1+t'+t''\bigr) e^{-x}=\bigl(1+t'\bigr) \bigl(x-1+e^{-x} \bigr)+t''e^{-x}.$$

Hence, the unique solution of interval equation (20) is concluded as

$$Y_{\mathbf{D}_{\nu}^{2}}(x)=[1,2]\odot\bigl(x-1+e^{-x}\bigr)\oplus[0,1]\odot e^{-x},$$

which is represented in Figure 4(a). Next, let us to use the second approach for the interval differential equation (20). Thus, the systems (17) and (18) are

$$\left \{ \textstyle\begin{array}{l} \underline{y}'(x)=-\overline{y}(x)+x, \qquad \underline{y}(0)=0, \\ \overline{y}'(x)=-\underline{y}(x)+2x, \qquad \overline{y}(0)=1 \end{array}\displaystyle \right .$$
(22)

and

$$\left \{ \textstyle\begin{array}{l} \underline{y}'(x)=-\underline{y}(x)+2x, \qquad \underline{y}(0)=0, \\ \overline{y}'(x)=-\overline{y}(x)+x,\qquad \overline{y}(0)=1, \end{array}\displaystyle \right .$$
(23)

respectively. As stated previously, (20) has exactly two solutions; one of them is

\begin{aligned} Y_{1}(x)&=\bigl\{ \underline{y}+t(\overline{y}-\underline {y})=2x+2e^{-x}-e^{x}-1+t\bigl(2e^{x}-x-1\bigr), \mid t \in[0,1]\bigr\} \\ &=\bigl[2x+2e^{-x}-e^{x}-1, x+e^{x}+2e^{-x}-2 \bigr], \end{aligned}

which starts with (i)-p-differentiability and there is no switching point on its trajectory. The second one starts with (d)-p-differentiability and has a switching point at $$x=1$$ so it has to switch to the case (i)-p-differentiability. Therefore

\begin{aligned} Y_{2}(x)&=\bigl\{ \underline{y}+t(\overline{y}-\underline{y})=2x + 2e^{-x}-2+t(1-x) \mid t\in[0,1]\bigr\} \\ &=\bigl[2x + 2e^{-x}-2, x+2e^{-x}-1\bigr], \end{aligned}

for $$0\leq x \leq1$$ and

\begin{aligned} Y_{2}(x)&=\bigl\{ \underline{y}+t(\overline{y}-\underline{y})=2x -e^{x-1}+2e^{-x}-1+t\bigl(2e^{x-1}-x-1\bigr) \mid t \in[0,1]\bigr\} \\ &=\bigl[2x -e^{x-1}+2e^{-x}-1, x+e^{x-1}+2e^{-x}-2 \bigr], \end{aligned}

when $$1 \leq x \leq4$$. The solutions are presented in Figure 4(b). Figure 4

### Example 4.4

As an application in engineering, assume the following problem. A tank initially contains $$30~\mbox{gal}$$ of brine with $$c~\mbox{lb}$$ of salt. Brine that contains $$k~\mbox{lb}$$ of salt per gallon enters the tank at the rate of $$3~\mbox{gal/min}$$, and the well-mixed brine in the tank flows out at the rate of $$3~\mbox{gal/min}$$ (see Figure 5). The problem is the amount of salt that will be present in the tank at a time in the future; in other words, we would like to determinate a function $$y(x)$$ that gives us the amount of salt in the tank at time $$x \geq0$$. We have

$$\left \{ \textstyle\begin{array}{l} y'(x)+ \frac{1}{10}y(x)=3k, \\ y(0)=c, \quad x\in[0,3]. \end{array}\displaystyle \right .$$

According to uncertainty in measurements, the coefficients c, k are considered as interval numbers $$c=[1,3]$$, $$k=[1,4]$$. Then

$$\left \{ \textstyle\begin{array}{l} Y_{\mathbf{D}_{\nu}^{n}}'(x)=-\frac{1}{10}Y_{\mathbf{D}_{\nu}^{n}}(x)\oplus [3,12], \\ Y_{\mathbf{D}_{\nu}^{n}}(0)=[1,3], \quad x\in[0,3]. \end{array}\displaystyle \right .$$
(24)

Using the first approach and without having to calculate the switching points, we obtain a solution that is responsive to the nature of the problem. More precisely, the corresponding ordinary differential equation is obtained as

\begin{aligned}& y'_{\mathbf{d}(\mathbf{t})}(x) =\frac{-1}{10}y_{\mathbf{d}(\mathbf{t})}(x)+ \bigl(3+9t'\bigr), \\& y_{\mathbf{d}(\mathbf{t})}(0) =1+2t'', \quad t',t'' \in[0,1], x \in[0,3]. \end{aligned}

Hence, the solution of interval equation (24) is

$$Y_{\mathbf{D}_{\nu}^{2}}(x)=[30,120]\odot\bigl(1-e^{\frac{-x}{10}}\bigr) \oplus [1,3]\odot\bigl(e^{\frac{-x}{10}}\bigr)$$

and its upper bound and lower bounds are shown in Figure 6(a). In the second approach, two systems (17), (18) are

\begin{aligned}& \left \{ \textstyle\begin{array}{l} \underline{y}'(x)=-\frac{1}{10}\overline{y}(x)+3, \qquad \underline {y}(0)=1, \\ \overline{y}'(x)=-\frac{1}{10}\underline{y}(x)+12, \qquad \overline{y}(0)=3, \end{array}\displaystyle \right . \\& \left \{ \textstyle\begin{array}{l} \overline{y}'(x)=-\frac{1}{10}\overline{y}(x)+3, \quad \underline{y}(0)=1, \\ \underline{y}'(x)=-\frac{1}{10}\underline{y}(x)+12, \quad \overline{y}(0)=3. \end{array}\displaystyle \right . \end{aligned}

First, by starting with (i)-p-differentiability the following solution for interval differential equation (24) is obtained:

$$Y_{1}(x)=\bigl[120- 46 e^{\frac{x}{10}} - 73 e^{\frac{-x}{10}}, 46 e^{\frac {x}{10}} - 73 e^{\frac{-x}{10}} + 30\bigr],$$

which has no switching point. The second solution is achieved by considering (d)-p-differentiability, that has a switching point at 0.2198 and

$$Y_{2}(x)= \left \{ \textstyle\begin{array}{l@{\quad}l} {[120- 119 e^{\frac{-x}{10}}, 30 - 27 e^{\frac{-x}{10}}]},& x \in [0, 0.2198], \\ {[120-44.0217 e^{\frac{x}{10}}-73 e^{\frac{-x}{10}}, 44.0217 e^{\frac {x}{10}}-73 e^{\frac{-x}{10}}+30]}, & x \in[0.2198, 3]. \end{array}\displaystyle \right .$$

The solutions are shown in Figure 6(b). Figure 5 Figure 6

## Conclusion

In this study, two different new parametric representations for interval numbers were investigated. The representations had the advantage of following flexible and easy to control shapes of the interval numbers, and they were applicable in practice. Additionally, computational procedures to determine the derivatives and integrals of interval-valued functions were presented.

An interesting line of work is the study of interval differential equations with two approaches. In the first approach, a unique solution with a decreasing length of support was obtained. This interpretation presented an advantage that allowed us to characterize the main properties of ordinary differential equations in a natural way. In the second approach, the solution of an interval differential equation might have decreasing length of support, and more than one solution exists. The existence of several solutions can be an advantage when a decision-maker search is performed for solutions that have specific properties, such as periodic, almost periodic, or asymptotically stable.

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## Acknowledgements

The authors would like to express their gratitude to the anonymous referees for their helpful comments and suggestions, which have greatly improved the paper.

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Correspondence to Mohadeseh Ramezanadeh.

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