Important Notes for a Fuzzy Boundary Value Problem

Firstly, Zadeh introduced the concept of fuzzy numbers and fuzzy arithmetic [22]. The major application of fuzzy arithmetic is fuzzy differential equations. Fuzzy differential equations are suitable models to model dynamic systems in which there exist uncertainties or vaguness. Fuzzy differential equations can be examined by several approach, such as Hukuhara differentiability, generalized differentiability, the concept of differential inclusion etc [1], [3], [4], [5], [6, 7, 8], [9], [11], [13, 14], [15] [17], [19], [20].

In this paper is on a fuzzy Sturm-Liouville problem with the eigenvalue parameter in the boundary condition. Important notes are given for the problem. Integral equations are found of the problem.

Preliminaries

Definition 1. [18] A fuzzy number is a function u : $ℝ \to [0, 1]$ $\mathbb{R}\to \left[ 0,1 \right]$satisfying the properties:u is normal, u is convex fuzzy set, u is upper semi-continuous on ℝ, $c l \{x ε ℝ |u (x) > 0\}$ $cl\left\{ x\varepsilon \mathbb{R}\left| u\left( x \right)>0 \right. \right\}$is compact, where cl denotes the closure of a subset.

Let R_F denote the space of fuzzy numbers.

Definition 2. [15] Let uɛℝ_F. The a-level set of u, denoted, ${[u]}^{α}, 0 < α \leq 1,$ ${{\left[ u \right]}^{\alpha }},0<\alpha \le 1,$is

{[u]}^{α} = \{x ε ℝ |u (x) \geq α\} .

$${{\left[ u \right]}^{\alpha }}=\left\{ x\varepsilon \mathbb{R}\left| u\left( x \right)\ge \alpha \right. \right\}.$$

If a = 0; the support of u is defined

{[u]}^{0} = c l \{x ε ℝ |u (x) > 0\} .

$${{\left[ u \right]}^{0}}=cl\left\{ x\varepsilon \mathbb{R}\left| u\left( x \right)>0 \right. \right\}.$$

The notation, ${[u]}^{α} = [{\underline{u}}_{α}, {\bar{u}}_{α}]$ ${{\left[ u \right]}^{\alpha }}=\left[ {{\underline{u}}_{\alpha }},{{{\bar{u}}}_{\alpha }} \right]$denotes explicitly the a-level set of u. We refer to $\underline{u}$ $\underline{u}$and ū as the lower and upper branches of u,respectively.

The following remark shows when $[{\underline{u}}_{α}, {\bar{u}}_{α}]$ $\left[ {{\underline{u}}_{\alpha }},{{{\bar{u}}}_{\alpha }} \right]$ is a valid a-level set.

Remark 1. [10,15] The sufficient and necessary conditions for $[{\underline{u}}_{α}, {\bar{u}}_{α}]$ $\left[ {{\underline{u}}_{\alpha }},{{{\bar{u}}}_{\alpha }} \right]$to define the parametric form of a fuzzy number as follows:

${\underline{u}}_{α}$ ${{\underline{u}}_{\alpha }}$is bounded monotonic increasing (nondecreasing) left-continuous function on (0;1] and right-continuous for a = 0,

${\bar{u}}_{α}$ ${{\bar{u}}_{\alpha }}$is bounded monotonic decreasing (nonincreasing) left-continuous function on (0;1] and right-continuous fora = 0,

{\underline{u}}_{α} \leq {\bar{u}}_{α}, 0 \leq α \leq 1.

$${{\underline{u}}_{\alpha }}\le {{\bar{u}}_{\alpha }},0\le \alpha \le 1.$$

Definition 3. [15] For u,vεℝ_F and λ ∈R, the sum u+v and the product λ u are defined by ${[u + v]}^{α} = {[u]}^{α} + {[v]}^{α},$ ${{\left[ u+v \right]}^{\alpha }}={{\left[ u \right]}^{\alpha }}+{{\left[ v \right]}^{\alpha }},$ ${[λ u]}^{α} = λ {[u]}^{α}$ ${{\left[ \lambda u \right]}^{\alpha }}=\lambda {{\left[ u \right]}^{\alpha }}$where means the usual addition of two intervals (subsets) of ℝ and λ [u]^α means the usual product between a scalar and a subset of R.

The metric structure is given by the Hausdorff distance

D : ℝ_{F} \times ℝ_{F} \to ℝ_{+} \cup \{0\},

$$D:{{\mathbb{R}}_{F}}\times {{\mathbb{R}}_{F}}\to {{\mathbb{R}}_{+}}\cup \left\{ 0 \right\},$$

D (u, v) = sup_{α \in [0, 1]} max \{|{\underline{u}}_{α} - {\underline{v}}_{α}|, |{\bar{u}}_{α} - {\bar{v}}_{α}|\} .

$$D\left( u,v \right)=\underset{\alpha \in \left[ 0,1 \right]}{\mathop{\text{sup}}}\,\text{ max}\left\{ \left| {{\underline{u}}_{\alpha }}-{{\underline{v}}_{\alpha }} \right|,\left| {{{\bar{u}}}_{\alpha }}-{{{\bar{v}}}_{\alpha }} \right| \right\}.$$

Definition 4. [18] If A is a symmetric triangular numbers with supports $[\underline{a}, \bar{a}],$ $\left[ \underline{a},\bar{a} \right],$the α–level sets of A is ${[A]}^{α} =$ ${{\left[ A \right]}^{\alpha }}=$ $[\underline{a} + (\frac{\bar{a} - \underline{a}}{2}) α, \bar{a} - (\frac{\bar{a} - \underline{a}}{2}) α] .$ $\left[ \underline{a}+\left( \frac{\bar{a}-\underline{a}}{2} \right)\alpha ,\bar{a}-\left( \frac{\bar{a}-\underline{a}}{2} \right)\alpha \right].$

Definition 5. [16] u,v 2 ℝ_F, ${[u]}^{α} = [{\underline{u}}_{α}, {\bar{u}}_{α}], {[v]}^{α} = [{\underline{v}}_{α}, {\bar{v}}_{α}],$ ${{\left[ u \right]}^{\alpha }}=\left[ {{\underline{u}}_{\alpha }},{{{\bar{u}}}_{\alpha }} \right],{{\left[ v \right]}^{\alpha }}=\left[ {{\underline{v}}_{\alpha }},{{{\bar{v}}}_{\alpha }} \right],$the product uv is defined by

{[u v]}^{α} = {[u]}^{α} {[v]}^{α}, \forall α \in [0, 1],

$${{\left[ uv \right]}^{\alpha }}={{\left[ u \right]}^{\alpha }}{{\left[ v \right]}^{\alpha }},\forall \alpha \in \left[ 0,1 \right],$$

where

\begin{array}{l} {[u]}^{α} {[v]}^{α} = [{\underline{u}}_{a}, {\bar{u}}_{α}] [{\underline{v}}_{a}, {\bar{v}}_{α}] = [{\underline{w}}_{a}, {\bar{w}}_{α}], \\ {\underline{w}}_{a} = min \{{\underline{u}}_{a} {\underline{v}}_{a}, {\underline{u}}_{a} {\bar{v}}_{α}, {\bar{u}}_{α} {\underline{v}}_{a}, {\bar{u}}_{α} {\bar{v}}_{α}\}, \\ {\bar{w}}_{a} = max \{{\underline{u}}_{a} {\underline{v}}_{a}, {\underline{u}}_{a} {\bar{v}}_{α}, {\bar{u}}_{α} {\underline{v}}_{a}, {\bar{u}}_{α} {\bar{v}}_{α}\} . \end{array}

$$\begin{align}& {{\left[ u \right]}^{\alpha }}{{\left[ v \right]}^{\alpha }}=\left[ {{\underline{u}}_{a}},{{{\bar{u}}}_{\alpha }} \right]\left[ {{\underline{v}}_{a}},{{{\bar{v}}}_{\alpha }} \right]=\left[ {{\underline{w}}_{a}},{{{\bar{w}}}_{\alpha }} \right], \\ & {{\underline{w}}_{a}}=\text{min}\left\{ {{\underline{u}}_{a}}{{\underline{v}}_{a}},{{\underline{u}}_{a}}{{{\bar{v}}}_{\alpha }},{{{\bar{u}}}_{\alpha }}{{\underline{v}}_{a}},{{{\bar{u}}}_{\alpha }}{{{\bar{v}}}_{\alpha }} \right\}, \\ & {{\overline{w}}_{a}}=\text{max}\left\{ {{\underline{u}}_{a}}{{\underline{v}}_{a}},{{\underline{u}}_{a}}{{{\bar{v}}}_{\alpha }},{{{\bar{u}}}_{\alpha }}{{\underline{v}}_{a}},{{{\bar{u}}}_{\alpha }}{{{\bar{v}}}_{\alpha }} \right\}. \\ \end{align}$$

Definition 6. [15, 21] Let u,v 2 ℝ_F.If there exists w 2 ℝ_F such that u = v+w; then w is called the Hukuhara difference of fuzzy numbers u and v;and it is denoted by w = u⊖v:

Definition 7. [2,15] Let f : [a;b]→R_F and t₀ ∈ [a,b] :We say that f is Hukuhara differential at t₀, if there exists an element f' ⁰ (t₀) 2 ℝ_F such that for all h > 0 sufficiently small, $\exists f (t_{0} + h) ⊖ f (t_{0}), f (t_{0}) ⊖ f (t_{0} - h)$ $\exists f\left( {{t}_{0}}+h \right)\ominus f\left( {{t}_{0}} \right),\,\,f\left( {{t}_{0}} \right)\ominus f\left( {{t}_{0}}-h \right)$and the limits (in the metric D)

\lim_{h \to 0} \frac{f (t_{0} + h) ⊖ f (t_{0})}{h} = \lim_{h \to 0} \frac{f (t_{0}) ⊖ f (t_{0} - h)}{h} = f^{'} (t_{0}) .

$$\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( {{t}_{0}}+h \right)\ominus f\left( {{t}_{0}} \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( {{t}_{0}} \right)\ominus f\left( {{t}_{0}}-h \right)}{h}={f}'\left( {{t}_{0}} \right).$$

Definition 8. [12] If $p^{'} (x) = 0, r (x) = 1$ ${p}'\left( x \right)=0,\,\,r\left( x \right)=1$and $L y = p (x) y ˝ + q (x) y$ $Ly=p\left( x \right)y''+q\left( x \right)y$in the fuzzy differential equation ${(p (x) y^{'})}^{'} + q (x) y + λ r (x) y = 0, p (x), p^{'} (x), q (x), r (x),$ ${{\left( p\left( x \right){y}' \right)}^{\prime }}+q\left( x \right)y+\lambda r\left( x \right)y=0,p\left( x \right),{p}'\left( x \right),q\left( x \right),r\left( x \right),$are continuous functions and positive, the fuzzydifferential equation

(2.1)

L y + λ y = 0

$$Ly+\lambda y=0$$

is called a fuzzy Sturm-Liouville equation.

Definition 9. [12] ${[y (x, λ_{0})]}^{α} = [\underline{y} (x, λ_{0}), \bar{y} (x, λ_{0})] \neq 0,$ ${{\left[ y\left( x,{{\lambda }_{0}} \right) \right]}^{\alpha }}=\left[ \underline{y}\left( x,{{\lambda }_{0}} \right),\bar{y}\left( x,{{\lambda }_{0}} \right) \right]\ne 0,$we say that λ = λ₀ is eigenvalue of (2.1) if the fuzzy differential equation (2.1) has the nontrivial solutions $\underline{y} (x, λ_{0}) \neq 0, \bar{y} (x, λ_{0}) \neq 0.$ $\underline{y}\left( x,{{\lambda }_{0}} \right)\ne 0,\bar{y}\left( x,{{\lambda }_{0}} \right)\ne 0.$

Findings and Main Results

Consider the fuzzy boundary value problem

τ u = u ˝ + q (x) u,

$$\tau u=u+q\left( x \right)u,$$

(3.1)

τ u + λ u = 0, x \in (- 1, 1)

$$\tau u+\lambda u=0,x\in \left( -1,1 \right)$$

(3.2)

u (- 1) + u^{'} (- 1) = 0,

$$u\left( -1 \right)+{u}'\left( -1 \right)=0,$$

(3.3)

λ β u (1) + γ u^{'} (1) = 0,

$$\lambda \beta u\left( 1 \right)+\gamma {u}'\left( 1 \right)=0,$$

where q(x) is continuous function and positive on [-1,1] , λ > 0 and β; γ > 0.

Let u₁ (x;λ) and u₂ (x;λ) be linearly independent solutions of the classical differential equation τu+λu=0. Then, the general solution of the fuzzy differential equation (3.1) is

{[u (x, λ)]}^{α} = [{\underline{u}}_{α} (x, λ), {\bar{u}}_{α} (x, λ)],

$${{\left[ u\left( x,\lambda \right) \right]}^{\alpha }}=\left[ {{\underline{u}}_{\alpha }}\left( x,\lambda \right),{{{\bar{u}}}_{\alpha }}\left( x,\lambda \right) \right],$$

{\underline{u}}_{α} (x, λ) = a_{α} (λ) u_{1} (x + λ) + b_{α} (λ) u_{2} (x + λ),

$${{\underline{u}}_{\alpha }}\left( x,\lambda \right)={{a}_{\alpha }}\left( \lambda \right){{u}_{1}}\left( x+\lambda \right)+{{b}_{\alpha }}\left( \lambda \right){{u}_{2}}\left( x+\lambda \right),$$

Also,

{\bar{u}}_{α} (x, λ) = c_{α} (λ) u_{1} (x, λ) + d_{α} (λ) u_{2} (x, λ) .

$${{\bar{u}}_{\alpha }}\left( x,\lambda \right)={{c}_{\alpha }}\left( \lambda \right){{u}_{1}}\left( x,\lambda \right)+{{d}_{\alpha }}\left( \lambda \right){{u}_{2}}\left( x,\lambda \right).$$

{[φ (x, λ)]}^{α} = [{\underline{φ}}_{α} (x, λ), {\bar{φ}}_{α} (x, λ)]

$${{\left[ \varphi \left( x,\lambda \right) \right]}^{\alpha }}=\left[ {{\underline{\varphi }}_{\alpha }}\left( x,\lambda \right),{{{\bar{\varphi }}}_{\alpha }}\left( x,\lambda \right) \right]$$

be the solution of the equation (3.1) satisfying the conditions

(3.4)

u (- 1) = 1, u ´ (- 1) = - 1

$$u\left( -1 \right)=1,u\acute{\ }\left( -1 \right)=-1$$

and

{[χ (x, λ)]}^{α} = [{\underline{χ}}_{α} (x, λ), {\bar{χ}}_{α} (x, λ)]

$${{\left[ \chi \left( x,\lambda \right) \right]}^{\alpha }}=\left[ {{\underline{\chi }}_{\alpha }}\left( x,\lambda \right),{{{\bar{\chi }}}_{\alpha }}\left( x,\lambda \right) \right]$$

be the solution of the equation (3.1) satisfying the conditions

(3.5)

u (1) = γ, u^{'} (1) = - λ β

$$u\left( 1 \right)=\gamma ,{u}'\left( 1 \right)=-\lambda \beta $$

where

\begin{array}{l} {\underline{φ}}_{α} (x, λ) + {\underline{c}}_{1 α} (λ) u_{1} (x, λ) + {\underline{c}}_{2 α} (λ) u_{2} (x, λ), \\ {\bar{φ}}_{α} (x, λ) + {\bar{c}}_{1 α} (λ) u_{1} (x, λ) + {\bar{c}}_{2 α} (λ) u_{2} (x, λ) \end{array}

$$\begin{align}& {{\underline{\varphi }}_{\alpha }}\left( x,\lambda \right)+{{\underline{c}}_{1\alpha }}\left( \lambda \right){{u}_{1}}\left( x,\lambda \right)+{{\underline{c}}_{2\alpha }}\left( \lambda \right){{u}_{2}}\left( x,\lambda \right), \\ & {{{\bar{\varphi }}}_{\alpha }}\left( x,\lambda \right)+{{{\bar{c}}}_{1\alpha }}\left( \lambda \right){{u}_{1}}\left( x,\lambda \right)+{{{\bar{c}}}_{2\alpha }}\left( \lambda \right){{u}_{2}}\left( x,\lambda \right) \\ \end{align}$$

\begin{array}{l} {\underline{χ}}_{α} (x, λ) = {\underline{c}}_{3 α} (λ) u_{1} (x, λ) + {\underline{c}}_{4 α} (λ) u_{2} (x, λ) \\ {\bar{χ}}_{α} (x, λ) = {\bar{c}}_{3 α} (λ) u_{1} (x, λ) + {\bar{c}}_{4 α} (λ) u_{2} (x, λ) . \end{array}

$$\begin{align}& {{\underline{\chi }}_{\alpha }}\left( x,\lambda \right)={{\underline{c}}_{3\alpha }}\left( \lambda \right){{u}_{1}}\left( x,\lambda \right)+{{\underline{c}}_{4\alpha }}\left( \lambda \right){{u}_{2}}\left( x,\lambda \right) \\ & {{{\bar{\chi }}}_{\alpha }}\left( x,\lambda \right)={{{\bar{c}}}_{3\alpha }}\left( \lambda \right){{u}_{1}}\left( x,\lambda \right)+{{{\bar{c}}}_{4\alpha }}\left( \lambda \right){{u}_{2}}\left( x,\lambda \right). \\ \end{align}$$

From here, yields

(3.6)

W ({\underline{φ}}_{α}, {\underline{χ}}_{α}) (x, λ) = ({\underline{c}}_{1 α} (λ) {\underline{c}}_{4 α} (λ) - {\underline{c}}_{2 α} (λ) {\underline{c}}_{3 α} (λ)) W (u_{1}, u_{2}) (x, λ)

$$W\left( {{\underline{\varphi }}_{\alpha }},{{\underline{\chi }}_{\alpha }} \right)\left( x,\lambda \right)=\left( {{\underline{c}}_{1\alpha }}\left( \lambda \right){{\underline{c}}_{4\alpha }}\left( \lambda \right)-{{\underline{c}}_{2\alpha }}\left( \lambda \right){{\underline{c}}_{3\alpha }}\left( \lambda \right) \right)W\left( {{u}_{1}},{{u}_{2}} \right)\left( x,\lambda \right)$$

(3.7)

W ({\bar{φ}}_{α}, {\bar{χ}}_{α}) (x, λ) = ({\bar{c}}_{1 α} (λ) {\bar{c}}_{4 α} (λ) - {\bar{c}}_{2 α} (λ) {\bar{c}}_{3 α} (λ)) W (u_{1}, u_{2}) (x, λ)

$$W\left( {{{\bar{\varphi }}}_{\alpha }},{{{\bar{\chi }}}_{\alpha }} \right)\left( x,\lambda \right)=\left( {{{\bar{c}}}_{1\alpha }}\left( \lambda \right){{{\bar{c}}}_{4\alpha }}\left( \lambda \right)-{{{\bar{c}}}_{2\alpha }}\left( \lambda \right){{{\bar{c}}}_{3\alpha }}\left( \lambda \right) \right)W\left( {{u}_{1}},{{u}_{2}} \right)\left( x,\lambda \right)$$

Also, since u₁ (x;λ) and u₂ (x;λ) are linearly independent solutions of the classical differential equation $τ u + λ u = 0,$ $\tau u+\lambda u=0,$the solution of the equation is

u (x, λ) = a (λ) u_{1} (x, λ) + b (λ) u_{2} (x, λ) .

$$u\left( x,\lambda \right)=a\left( \lambda \right){{u}_{1}}\left( x,\lambda \right)+b\left( \lambda \right){{u}_{2}}\left( x,\lambda \right).$$

φ(x;λ) be the solution of the classical differential equation τu+lu = 0 satisfying the conditions $u (- 1) =$ $u\left( -1 \right)=$ $1, u^{'} (- 1) = - 1.$ $1,{u}'\left( -1 \right)=-1.$Using boundary conditions, we have

a (λ) u_{1} (- 1, λ) + b (λ) u_{2} (- 1, λ) = 1,

$$a\left( \lambda \right){{u}_{1}}\left( -1,\lambda \right)+b\left( \lambda \right){{u}_{2}}\left( -1,\lambda \right)=1,$$

a (λ) {u^{'}}_{1} (- 1, λ) + b (λ) {u^{'}}_{2} (- 1, λ) = - 1.

$$a\left( \lambda \right){{{u}'}_{1}}\left( -1,\lambda \right)+b\left( \lambda \right){{{u}'}_{2}}\left( -1,\lambda \right)=-1.$$

From this, a(λ) , b(λ) are obtained as

a (λ) = \frac{{u^{'}}_{2} (- 1, λ) + u_{2} (- 1, λ)}{W (u_{1}, u_{2}) (- 1, λ)}, b (λ) = \frac{- u_{1} (- 1, λ) - {u^{'}}_{1} (- 1, λ)}{W (u_{1}, u_{2}) (- 1, λ)}

$$a\left( \lambda \right)=\frac{{{{{u}'}}_{2}}\left( -1,\lambda \right)+{{u}_{2}}\left( -1,\lambda \right)}{W\left( {{u}_{1}},{{u}_{2}} \right)\left( -1,\lambda \right)},b\left( \lambda \right)=\frac{-{{u}_{1}}\left( -1,\lambda \right)-{{{{u}'}}_{1}} \& \left( -1,\lambda \right)}{W\left( {{u}_{1}},{{u}_{2}} \right)\left( -1,\lambda \right)}$$

Then,

\begin{array}{l} φ (x, λ) = \frac{1}{W (u_{1}, u_{2}) (- 1, λ)} \{({u^{'}}_{2} (- 1, λ) + u_{2} (- 1, λ)) u_{1} (x, λ) \\ - (u_{2} (- 1, λ) + {u^{'}}_{1} (- 1, λ)) u_{2} (x, λ)\} . \end{array}

$$\begin{align}& \varphi \left( x,\lambda \right)=\frac{1}{W\left( {{u}_{1}},{{u}_{2}} \right)\left( -1,\lambda \right)}\left\{ \left( {{{{u}'}}_{2}}\left( -1,\lambda \right)+{{u}_{2}}\left( -1,\lambda \right) \right){{u}_{1}}\left( x,\lambda \right) \right. \\ & \left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\left( {{u}_{2}}\left( -1,\lambda \right)+{{{{u}'}}_{1}}\left( -1,\lambda \right) \right){{u}_{2}}\left( x,\lambda \right) \right\}. \\ \end{align}$$

Again,χ (x,λ) be the solution of the classical differential equation τu+lu = 0 satisfying the conditions $u (1) = γ, u (1) = - λ β .$ $u\left( 1 \right)=\gamma ,u\left( 1 \right)=-\lambda \beta .$Similarly, χ(x;λ) is obtained as

\begin{array}{l} χ (x, λ) = \frac{1}{W (u_{1}, u_{2}) (- 1, λ)} \{(λ β u_{2} (1, λ) + γ u_{2}^{´} (1, λ)) u_{1} (x, λ) \\ - (λ β u_{1} (1, λ) + γ u_{1}^{´} (1, λ)) u_{2} (x, λ)\} \end{array}

$$\begin{align}& \chi \left( x,\lambda \right)=\frac{1}{W\left( {{u}_{1}},{{u}_{2}} \right)\left( -1,\lambda \right)}\left\{ \left( \lambda \beta {{u}_{2}}\left( 1,\lambda \right)+\gamma u_{2}^{\acute{\ }}\left( 1,\lambda \right) \right){{u}_{1}}\left( x,\lambda \right) \right. \\ & \left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\left( \lambda \beta {{u}_{1}}\left( 1,\lambda \right)+\gamma u_{1}^{\acute{\ }}\left( 1,\lambda \right) \right){{u}_{2}}\left( x,\lambda \right) \right\} \\ \end{align}$$

Thus,

{[φ (x, λ)]}^{α} = [{\underline{φ}}_{α} (x, λ), {\bar{φ}}_{α} (x, λ)] = [c_{1} (α), c_{2} (α)] φ (x, λ)

$${{\left[ \varphi \left( x,\lambda \right) \right]}^{\alpha }}=\left[ {{\underline{\varphi }}_{\alpha }}\left( x,\lambda \right),{{{\bar{\varphi }}}_{\alpha }}\left( x,\lambda \right) \right]=\left[ {{c}_{1}}\left( \alpha \right),{{c}_{2}}\left( \alpha \right) \right]\varphi \left( x,\lambda \right)$$

is the solution of the equation (3.1) satisfying the conditions (3.4) and

{[χ (x, λ)]}^{α} = [{\underline{χ}}_{α} (x, λ), {\bar{χ}}_{α} (x, λ)] = [c_{1} (α), c_{2} (α)] χ (x, λ)

$${{\left[ \chi \left( x,\lambda \right) \right]}^{\alpha }}=\left[ {{\underline{\chi }}_{\alpha }}\left( x,\lambda \right),{{{\bar{\chi }}}_{\alpha }}\left( x,\lambda \right) \right]=\left[ {{c}_{1}}\left( \alpha \right),{{c}_{2}}\left( \alpha \right) \right]\chi \left( x,\lambda \right)$$

is the solution of the equation (3.1) satisfying the conditions (3.5), where $[c_{1} (α), c_{2} (α)] = {[1]}^{α} .$ $\left[ {{c}_{1}}\left( \alpha \right),{{c}_{2}}\left( \alpha \right) \right]={{\left[ 1 \right]}^{\alpha }}.$We take ${[1]}^{α} = [α, 2 - α] .$ ${{\left[ 1 \right]}^{\alpha }}=\left[ \alpha ,2-\alpha \right].$From here,

(3.8)

{\underline{W}}_{α} (x, λ) = α^{2} (φ (x, λ) χ^{'} (x, λ) - χ (x, λ) φ^{'} (x, λ))

$${{\underline{W}}_{\alpha }}\left( x,\lambda \right)={{\alpha }^{2}}\left( \varphi \left( x,\lambda \right){\chi }'\left( x,\lambda \right)-\chi \left( x,\lambda \right){\varphi }'\left( x,\lambda \right) \right)$$

(3.9)

{\bar{W}}_{α} (x, λ) = {(2 - α)}^{2} (φ (x, λ) χ^{'} (x, λ) - χ (x, λ) φ^{'} (x, λ))

$${{\overline{W}}_{\alpha }}\left( x,\lambda \right)={{\left( 2-\alpha \right)}^{2}}\left( \varphi \left( x,\lambda \right){\chi }'\left( x,\lambda \right)-\chi \left( x,\lambda \right){\varphi }'\left( x,\lambda \right) \right)$$

are obtained. Computing the value $φ (x, λ) χ^{'} (x, λ) - χ (x, λ) φ^{'} (x, λ),$ $\varphi \left( x,\lambda \right){\chi }'\left( x,\lambda \right)-\chi \left( x,\lambda \right){\varphi }'\left( x,\lambda \right),$we have

\begin{array}{l} \frac{W (u_{1}, u_{2}) (x, λ)}{W (u_{1}, u_{2}) (- 1, λ) W (u_{1}, u_{2}) (1, λ)} \{(u_{1} (- 1, λ) + u_{1}^{´} (- 1, λ)) (λ β u_{2} (1, λ) + γ u_{2}^{´} (1, λ)) \\ - (u_{2} (- 1, λ) + u_{2}^{´} (- 1, λ)) (λ β u_{1} (1, λ) + γ u_{1}^{´} (1, λ))\} \end{array}

$$\begin{align}& \frac{W\left( {{u}_{1}},{{u}_{2}} \right)\left( x,\lambda \right)}{W\left( {{u}_{1}},{{u}_{2}} \right)\left( -1,\lambda \right)W\left( {{u}_{1}},{{u}_{2}} \right)\left( 1,\lambda \right)}\left\{ \left( {{u}_{1}}\left( -1,\lambda \right)+u_{1}^{\acute{\ }}\left( -1,\lambda \right) \right)\left( \lambda \beta {{u}_{2}}\left( 1,\lambda \right)+\gamma u_{2}^{\acute{\ }}\left( 1,\lambda \right) \right) \right. \\ & -\left. \left( {{u}_{2}}\left( -1,\lambda \right)+u_{2}^{\acute{\ }}\left( -1,\lambda \right) \right)\left( \lambda \beta {{u}_{1}}\left( 1,\lambda \right)+\gamma u_{1}^{\acute{\ }}\left( 1,\lambda \right) \right) \right\} \\ \end{align}$$

Considering the 3.7), the value is ${\underline{c}}_{1 α} (λ) {\underline{c}}_{4 α} (λ) - {\underline{c}}_{2 α} (λ) {\underline{c}}_{3 α} (λ)$ ${{\underline{c}}_{1\alpha }}\left( \lambda \right){{\underline{c}}_{4\alpha }}\left( \lambda \right)-{{\underline{c}}_{2\alpha }}\left( \lambda \right){{\underline{c}}_{3\alpha }}\left( \lambda \right)$

\begin{array}{l} \frac{α^{2}}{W (u_{1}, u_{2}) (- 1, λ) W (u_{1}, u_{2}) (- 1, λ)} \{(u_{1} (- 1, λ) + u_{1}^{´} (- 1, λ)) (λ β u_{2} (1, λ) + γ u_{2}^{´} (1, λ)) \\ - (u_{2} (- 1, λ) + u_{2}^{´} (- 1, λ)) (λ β u_{1} (1, λ) + γ u_{1}^{´} (1, λ))\} \end{array}

$$\begin{align}& \frac{{{\alpha }^{2}}}{W\left( {{u}_{1}},{{u}_{2}} \right)\left( -1,\lambda \right)W\left( {{u}_{1}},{{u}_{2}} \right)\left( -1,\lambda \right)}\left\{ \left( {{u}_{1}}\left( -1,\lambda \right)+u_{1}^{\acute{\ }}\left( -1,\lambda \right) \right)\left( \lambda \beta {{u}_{2}}\left( 1,\lambda \right)+\gamma u_{2}^{\acute{\ }}\left( 1,\lambda \right) \right) \right. \\ & \left. -\left( {{u}_{2}}\left( -1,\lambda \right)+u_{2}^{\acute{\ }}\left( -1,\lambda \right) \right)\left( \lambda \beta {{u}_{1}}\left( 1,\lambda \right)+\gamma u_{1}^{\acute{\ }}\left( 1,\lambda \right) \right) \right\} \\ \end{align}$$

and the value ${\bar{c}}_{1 α} (λ) {\bar{c}}_{4 α} (λ) - {\bar{c}}_{2 α} (λ) {\bar{c}}_{3 α} (λ)$ ${{\bar{c}}_{1\alpha }}\left( \lambda \right){{\bar{c}}_{4\alpha }}\left( \lambda \right)-{{\bar{c}}_{2\alpha }}\left( \lambda \right){{\bar{c}}_{3\alpha }}\left( \lambda \right)$is

\begin{array}{l} \frac{{(2 - α)}^{2}}{W (u_{1}, u_{2}) (- 1, λ) W (u_{1}, u_{2}) (- 1, λ)} \{(u_{1} (- 1, λ) + u_{1}^{´} (- 1, λ)) (λ β u_{2} (1, λ) + γ u_{2}^{´} (1, λ)) \\ - (u_{2} (- 1, λ) + u_{2}^{´} (- 1, λ)) (λ β u_{1} (1, λ) + γ u_{1}^{´} (1, λ))\} \end{array}

$$\begin{align}& \frac{{{\left( 2-\alpha \right)}^{2}}}{W\left( {{u}_{1}},{{u}_{2}} \right)\left( -1,\lambda \right)W\left( {{u}_{1}},{{u}_{2}} \right)\left( -1,\lambda \right)}\left\{ \left( {{u}_{1}}\left( -1,\lambda \right)+u_{1}^{\acute{\ }}\left( -1,\lambda \right) \right)\left( \lambda \beta {{u}_{2}}\left( 1,\lambda \right)+\gamma u_{2}^{\acute{\ }}\left( 1,\lambda \right) \right) \right. \\ & \left. -\left( {{u}_{2}}\left( -1,\lambda \right)+u_{2}^{\acute{\ }}\left( -1,\lambda \right) \right)\left( \lambda \beta {{u}_{1}}\left( 1,\lambda \right)+\gamma u_{1}^{\acute{\ }}\left( 1,\lambda \right) \right) \right\} \\ \end{align}$$

Consequently,

W ({\underline{φ}}_{α}, {\underline{χ}}_{α}) (x, λ) = \frac{α^{2}}{{(2 - α)}^{2}} W ({\bar{φ}}_{α}, {\bar{χ}}_{α}) (x, λ) .

$$W\left( {{\underline{\varphi }}_{\alpha }},{{\underline{\chi }}_{\alpha }} \right)\left( x,\lambda \right)=\frac{{{\alpha }^{2}}}{{{\left( 2-\alpha \right)}^{2}}}W\left( {{{\bar{\varphi }}}_{\alpha }},{{{\bar{\chi }}}_{\alpha }} \right)\left( x,\lambda \right).$$

Theorem 1. The Wronskian functions $W ({\underline{φ}}_{α}, {\underline{χ}}_{α})$ $W\left( {{\underline{\varphi }}_{\alpha }},{{\underline{\chi }}_{\alpha }} \right)$(x;λ) and W $({\bar{φ}}_{α}, {\bar{χ}}_{α})$ $\left( {{{\bar{\varphi }}}_{\alpha }},{{{\bar{\chi }}}_{\alpha }} \right)$(x;λ) are independent of variable x for x ∈ (–1;1), where functions $\underline{φ_{α}}, \underline{χ_{α}}, {\bar{φ}}_{α}, {\bar{χ}}_{α}$ $\underline{{{\varphi }_{\alpha }}},\underline{{{\chi }_{\alpha }}},{{\bar{\varphi }}_{\alpha }},{{\bar{\chi }}_{\alpha }}$are the solution of the fuzzy boundary value problem (3.1)-(3.3).

Proof. Derivating of equations W $({\underline{φ}}_{α}, {\underline{χ}}_{α})$ $\left( {{\underline{\varphi }}_{\alpha }},{{\underline{\chi }}_{\alpha }} \right)$(x;λ) and W $({\bar{φ}}_{α}, {\bar{χ}}_{α})$ $\left( {{{\bar{\varphi }}}_{\alpha }},{{{\bar{\chi }}}_{\alpha }} \right)$(x;λ) according to variable x and using the functions ${[φ (x, λ)]}^{α}, {[χ (x, λ)]}^{α}$ ${{\left[ \varphi \left( x,\lambda \right) \right]}^{\alpha }},{{\left[ \chi \left( x,\lambda \right) \right]}^{\alpha }}$are the solutions of the equation (3.1)

W ´ ({\underline{φ}}_{α}, {\underline{χ}}_{α}) (x, λ) = 0 a n d W ´ ({\bar{φ}}_{α}, {\bar{χ}}_{α}) (x, λ) = 0.

$$W\acute{\ }\left( {{\underline{\varphi }}_{\alpha }},{{\underline{\chi }}_{\alpha }} \right)\left( x,\lambda \right)=0\ and\,W\acute{\ }\left( {{{\bar{\varphi }}}_{\alpha }},{{{\bar{\chi }}}_{\alpha }} \right)\left( x,\lambda \right)=0.$$

are obtained. The proof is complete.

(3.10)

{\underline{W}}_{α} (λ) = W ({\underline{φ}}_{α}, {\underline{χ}}_{α}) (x, λ) = {\bar{W}}_{α} (λ) = W ({\bar{φ}}_{α}, {\bar{χ}}_{α}) (x, λ) .

$${{\underline{W}}_{\alpha }}\left( \lambda \right)=W\left( {{\underline{\varphi }}_{\alpha }},{{\underline{\chi }}_{\alpha }} \right)\left( x,\lambda \right)={{\overline{W}}_{\alpha }}\left( \lambda \right)=W\left( {{{\bar{\varphi }}}_{\alpha }},{{{\bar{\chi }}}_{\alpha }} \right)\left( x,\lambda \right).$$

Theorem 2. The eigenvalues of the fuzzy boundary value problem (3.1)-(3.3) if and only if are consist of the zeros of functions ${\underline{W}}_{α} (λ)$ ${{\underline{W}}_{\alpha }}\left( \lambda \right)$and ${\bar{W}}_{α} (λ) .$ ${{\overline{W}}_{\alpha }}\left( \lambda \right).$

Proof. Let be λ = λ₀ is the eigenvalue. We show that ${\underline{W}}_{α} (λ_{0}) = 0$ ${{\underline{W}}_{\alpha }}\left( {{\lambda }_{0}} \right)=0$and ${\bar{W}}_{α} (λ_{0}) = 0.$ ${{\overline{W}}_{\alpha }}\left( {{\lambda }_{0}} \right)=0.$We assume that ${\underline{W}}_{α} (λ_{0}) \neq 0 or {\bar{W}}_{α} (λ_{0}) \neq 0 .$ ${{\underline{W}}_{\alpha }}\left( {{\lambda }_{0}} \right)\ne 0\,\text{or}\,{{\overline{W}}_{\alpha }}\left( {{\lambda }_{0}} \right)\ne 0\,.$Let be ${\underline{W}}_{α} (λ_{0}) \neq 0.$ ${{\underline{W}}_{\alpha }}\left( {{\lambda }_{0}} \right)\ne 0.$Then, the functions ${\underline{φ}}_{α} (x, λ_{0})$ ${{\underline{\varphi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right)$and ${\underline{χ}}_{α} (x, λ_{0})$ ${{\underline{\chi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right)$are linearly independent. So, the general solution of the equation (3.1)

(3.11)

{[u (x, λ_{0})]}^{α} = [{\underline{u}}_{α} (x, λ_{0}), {\bar{u}}_{α} (x, λ_{0})],

$${{\left[ u\left( x,{{\lambda }_{0}} \right) \right]}^{\alpha }}=\left[ {{\underline{u}}_{\alpha }}\left( x,{{\lambda }_{0}} \right),{{{\bar{u}}}_{\alpha }}\left( x,{{\lambda }_{0}} \right) \right],$$

(3.12)

{\underline{u}}_{α} (x, λ_{0}) = a_{α} (λ_{0}) {\underline{φ}}_{α} (x, λ_{0}) + b_{α} (λ_{0}) {\underline{χ}}_{α} (x, λ_{0}),

$${{\underline{u}}_{\alpha }}\left( x,{{\lambda }_{0}} \right)={{a}_{\alpha }}\left( {{\lambda }_{0}} \right){{\underline{\varphi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right)+{{b}_{\alpha }}\left( {{\lambda }_{0}} \right){{\underline{\chi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right),$$

(3.13)

{\bar{u}}_{α} (x, λ_{0}) = c_{α} (λ_{0}) {\bar{φ}}_{α} (x, λ_{0}) + d_{α} (λ_{0}) {\bar{χ}}_{α} (x, λ_{0}) .

$${{\bar{u}}_{\alpha }}\left( x,{{\lambda }_{0}} \right)={{c}_{\alpha }}\left( {{\lambda }_{0}} \right){{\bar{\varphi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right)+{{d}_{\alpha }}\left( {{\lambda }_{0}} \right){{\bar{\chi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right).$$

Using the boundary condition (3.2) and using the solution function ${[φ (x, λ_{0})]}^{α} = [{\underline{φ}}_{α} (x, λ_{0}), {\bar{φ}}_{α} (x, λ_{0})]$ ${{\left[ \varphi \left( x,{{\lambda }_{0}} \right) \right]}^{\alpha }}=\left[ {{\underline{\varphi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right),{{{\bar{\varphi }}}_{\alpha }}\left( x,{{\lambda }_{0}} \right) \right]$satisfies the boundary condition (3.2),

b_{α} (λ_{0}) ({\underline{χ}}_{α} (- 1, λ_{0}) + {\underline{χ}}^{'}_{α} (- 1, λ_{0})) = 0,

$${{b}_{\alpha }}\left( {{\lambda }_{0}} \right)\left( {{\underline{\chi }}_{\alpha }}\left( -1,{{\lambda }_{0}} \right)+{{\underline{\chi }}^{\prime }}_{\alpha }\left( -1,{{\lambda }_{0}} \right) \right)=0,$$

d_{α} (λ_{0}) ({\bar{χ}}_{α} (- 1, λ_{0}) + {\bar{χ}}^{'}_{α} (- 1, λ_{0})) = 0

$${{d}_{\alpha }}\left( {{\lambda }_{0}} \right)\left( {{{\bar{\chi }}}_{\alpha }}\left( -1,{{\lambda }_{0}} \right)+{{{\bar{\chi }}}^{\prime }}_{\alpha }\left( -1,{{\lambda }_{0}} \right) \right)=0$$

are obtained. Again, using (3.4), (3.10), we have

b_{α} (λ_{0}) {\underline{W}}_{α} (λ_{0}) = 0, d_{α} (λ_{0}) {\bar{W}}_{α} (λ_{0}) = 0.

$${{b}_{\alpha }}\left( {{\lambda }_{0}} \right){{\underline{W}}_{\alpha }}\left( {{\lambda }_{0}} \right)=0,{{d}_{\alpha }}\left( {{\lambda }_{0}} \right){{\overline{W}}_{\alpha }}\left( {{\lambda }_{0}} \right)=0.$$

From this, since ${\underline{W}}_{α} (λ_{0}) \neq 0, b_{α} (λ_{0}) = 0$ ${{\underline{W}}_{\alpha }}\left( {{\lambda }_{0}} \right)\ne 0,{{b}_{\alpha }}\left( {{\lambda }_{0}} \right)=0$is obtained. Similarly, using the boundary condition (3.3), we obtained $a_{α} (λ_{0}) = 0.$ ${{a}_{\alpha }}\left( {{\lambda }_{0}} \right)=0.$Thus, ${\underline{u}}_{α} (x . λ_{0}) = 0, λ_{0}$ ${{\underline{u}}_{\alpha }}\left( x.{{\lambda }_{0}} \right)=0,{{\lambda }_{0}}$is not an eigenvalue. That is, we have a contradiction. Similarly, if ${\bar{W}}_{α} (λ_{0}) \neq 0, {\bar{u}}_{α} (x, λ_{0}) = 0$ $\,{{\overline{W}}_{\alpha }}\left( {{\lambda }_{0}} \right)\ne 0\,,{{\bar{u}}_{\alpha }}\left( x,{{\lambda }_{0}} \right)=0\,$is obtained. λ₀ is not an eigenvalue.

Let λ₀ be zero of ${\underline{W}}_{α} (λ)$ ${{\underline{\text{W}}}_{\alpha }}\left( \lambda \right)$and ${\bar{W}}_{α} (λ) .$ ${{\overline{W}}_{\alpha }}\left( \lambda \right).$Then,

(3.14)

{\underline{φ}}_{α} (x, λ_{0}) = k_{1} {\underline{χ}}_{α} (x, λ_{0}), {\bar{φ}}_{α} (x, λ_{0}) = k_{2} {\bar{χ}}_{α} (x, λ_{0}) .

$${{\underline{\varphi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right)={{k}_{1}}{{\underline{\chi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right),{{\bar{\varphi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right)={{k}_{2}}{{\bar{\chi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right).$$

That is, the functions ${\underline{φ}}_{α}, {\underline{χ}}_{α}$ ${{\underline{\varphi }}_{\alpha }},{{\underline{\chi }}_{\alpha }}$and ${\bar{φ}}_{α}, {\bar{χ}}_{α}$ ${{\bar{\varphi }}_{\alpha }},{{\bar{\chi }}_{\alpha }}$are linearly dependent. Also, since ${[χ (x, λ)]}^{α}$ ${{\left[ \chi \left( x,\lambda \right) \right]}^{\alpha }}$satisfies the boundary condition (3.3), ${\underline{χ}}_{α} (x, λ_{0})$ ${{\underline{\chi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right)$and ${\bar{χ}}_{α} (x, λ_{0})$ ${{\bar{\chi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right)$satisfy the boundary condition (3.3). In addition, from (3.14) the functions ${\underline{φ}}_{α}$ ${{\underline{\varphi }}_{\alpha }}$x;λ₀) and ${\bar{φ}}_{α} (x, λ_{0})$ ${{\bar{\varphi }}_{\alpha }}\left( x,{{\lambda }_{0}} \right)$satisfy the boundary condition (3.3). So, ${[φ (x, λ_{0})]}^{α}$ ${{\left[ \varphi \left( x,{{\lambda }_{0}} \right) \right]}^{\alpha }}$satisfies the boundary condition (3.3). Hence, [φ(x;λ₀)]a is the solution of the boundary value problem (3.1)-(3.3) for λ = λ₀. Thus, λ = λ₀ is the eigenvalue. The proof is complete.

Lemma 1. Let λ = s². The lower and the upper solutions ${\underline{φ}}_{α} (x, λ), {\bar{φ}}_{α} (x, λ)$ ${{\underline{\varphi }}_{\alpha }}\left( x,\lambda \right),{{\bar{\varphi }}_{\alpha }}\left( x,\lambda \right)$satisfy the following integral equations for k=0 and k=1:

(3.15)

\begin{array}{l} {({\underline{φ}}_{α} (x, λ))}^{(k)} = {(C o s (s (x + 1)))}^{(k)} - \frac{1}{s} {(S i n (s (x + 1)))}^{(k)} \\ + \frac{1}{s} \int_{- 1}^{x} {(S i n (s (x - y)))}^{(k)} q (y) {({\underline{φ}}_{α} (x, λ))}^{(k)} d y, \end{array}

$$\begin{align}& {{\left( {{\underline{\text{ }\!\!\varphi\!\!\text{ }}}_{\alpha }}\left( x,\lambda \right) \right)}^{\left( k \right)}}={{\left( Cos\left( s\left( x+1 \right) \right) \right)}^{\left( k \right)}}-\frac{1}{s}{{\left( Sin\left( s\left( x+1 \right) \right) \right)}^{\left( k \right)}} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\,\,+\frac{1}{s}\int\limits_{-1}^{x}{{{\left( Sin\left( s\left( x-y \right) \right) \right)}^{\left( k \right)}}q\left( y \right)}{{\left( {{\underline{\text{ }\!\!\varphi\!\!\text{ }}}_{\alpha }}\left( x,\lambda \right) \right)}^{\left( k \right)}}dy, \\ \end{align}$$

(3.16)

\begin{aligned} {({\bar{φ}}_{α} (x, λ))}^{(k)} = {(C o s (s (x + 1)))}^{(k)} - \frac{1}{s} {(Si n (s (x + 1)))}^{(k)} + \\ \frac{1}{s} \int_{- 1}^{x} {(Si n (s (x - y)))}^{(k)} q (y) {({\bar{φ}}_{α} (x, λ))}^{(k)} d y . \end{aligned}

$$\begin{align}& {{\left( {{{\bar{\varphi}}}_{\alpha }}\left( x,\lambda \right) \right)}^{\left( k \right)}}={{\left( Cos\left( s\left( x+1 \right) \right) \right)}^{\left( k \right)}}-\frac{1}{s}{{\left( \operatorname{Si}n\left( s\left( x+1 \right) \right) \right)}^{\left( k \right)}}+ \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\,\,\,\,\,\,\,\,\frac{1}{s}\int\limits_{-1}^{x}{{{\left( \operatorname{Si}n\left( s\left( x-y \right) \right) \right)}^{\left( k \right)}}q\left( y \right)}{{\left( {{\bar{\varphi}}_{\alpha }}\left( x,\lambda \right) \right)}^{\left( k \right)}}dy. \\ \end{align}$$

Proof. Since

{[φ (x, λ)]}^{α} = [{\underline{φ}}_{α} (x, λ), {\bar{φ}}_{α} (x, λ)]

$${{\left[ \varphi \left( x,\lambda \right) \right]}^{\alpha }}=\left[ {{\underline{\varphi }}_{\alpha }}\left( x,\lambda \right),{{{\bar{\varphi }}}_{\alpha }}\left( x,\lambda \right) \right]$$

is the solution of the equation (3.1), the equation

{[{\underline{φ}}_{α} (y, λ), {\bar{φ}}_{α} (y, λ)]}^{' ​'} + q (y) [{\underline{φ}}_{α} (y, λ), {\bar{φ}}_{α} (y, λ)] + λ [{\underline{φ}}_{α} (y, λ), {\bar{φ}}_{α} (y, λ)] = 0

$${{\left[ {{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right),{{{\bar{\varphi }}}_{\alpha }}\left( y,\lambda \right) \right]}^{\prime \prime }}+q\left( y \right)\left[ {{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right),{{{\bar{\varphi }}}_{\alpha }}\left( y,\lambda \right) \right]+\lambda \left[ {{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right),{{{\bar{\varphi }}}_{\alpha }}\left( y,\lambda \right) \right]=0$$

is provided. Using the Hukuhara differentiability and fuzzy arithmetic,

[{\underline{φ}}^{' ​'}_{α} (y, λ), {\bar{φ}}^{' ​'}_{α} (y, λ)] + [q (y) {\underline{φ}}_{α} (y, λ), q (y) {\bar{φ}}_{α} (y, λ)] + [λ {\underline{φ}}_{α} (y, λ), λ {\bar{φ}}_{α} (y, λ)] = 0

$$\left[ {{\underline{\varphi }}^{\prime \prime }}_{\alpha }\left( y,\lambda \right),{{{\bar{\varphi }}}^{\prime \prime }}_{\alpha }\left( y,\lambda \right) \right]+\left[ q\left( y \right){{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right),q\left( y \right){{{\bar{\varphi }}}_{\alpha }}\left( y,\lambda \right) \right]+\left[ \lambda {{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right),\lambda {{{\bar{\varphi }}}_{\alpha }}\left( y,\lambda \right) \right]=0$$

is obtained. From here, yields

\begin{array}{l} {\underline{φ}}^{' ​'}_{α} (y, λ) + q (y) {\underline{φ}}_{α} (y, λ) + λ {\underline{φ}}_{α} (y, λ) = 0, \\ {\bar{φ}}^{' ​'}_{α} (y, λ) + q (y) {\bar{φ}}_{α} (y, λ) + λ {\bar{φ}}_{α} (y, λ) = 0. \end{array}

$$\begin{align}& {{\underline{\varphi }}^{\prime \prime }}_{\alpha }\left( y,\lambda \right)+q\left( y \right){{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right)+\lambda {{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right)=0, \\ & {{{\bar{\varphi }}}^{\prime \prime }}_{\alpha }\left( y,\lambda \right)+q\left( y \right){{{\bar{\varphi }}}_{\alpha }}\left( y,\lambda \right)+\lambda {{{\bar{\varphi }}}_{\alpha }}\left( y,\lambda \right)=0. \\ \end{align}$$

Substituing the identity $q (y) {\underline{φ}}_{α} (y, λ) = - λ {\underline{φ}}_{α} (y, λ) - {\underline{φ}}^{' '}_{a} (y, λ)$ $q\left( y \right){{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right)=-\lambda {{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right)-{{\underline{\varphi }}^{\prime \prime }}_{a}\left( y,\lambda \right)$in the right side of (3.15)

\begin{array}{l} \int_{- 1}^{x} Si n (s (x - y)) q (y) {\underline{φ}}_{α} (y, λ) d y = - s^{2} \int_{- 1}^{x} Si n (s (x - y)) {\underline{φ}}_{α} (y, λ) d y \\ - \int_{- 1}^{x} Si n (s (x - y)) \underline{φ^{″}}_{α} (x, λ) d y \end{array}

$$\begin{align} & \int\limits_{-1}^{x}{\text{Si}n\left( s\left( x-y \right) \right)q\left( y \right){{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right)dy=-{{s}^{2}}}\int\limits_{-1}^{x}{\text{Si}n\left( s\left( x-y \right) \right){{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right)dy} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\int\limits_{-1}^{x}{\text{Si}n\left( s\left( x-y \right) \right)\underline{{{\varphi }''}}{{ \& }_{\alpha }}\left( x,\lambda \right)dy} \\ \end{align}$$

On integrating by parts twice and using (3.4)

\begin{array}{l} \int_{- 1}^{x} Si n (s (x - y)) {\underline{φ}}^{' ​'}_{α} (x, λ) d y = Si n (s (x + 1)) - s {\underline{φ}}_{α} (x, λ) + s C o s (s (x + 1)) \\ + s^{2} \int_{- 1}^{x} Si n (s (x - y)) {\underline{φ}}_{α} (y, λ) d y \end{array}

$$\begin{align}& \int\limits_{-1}^{x}{\text{Si}n\left( s\left( x-y \right) \right){{\underline{\varphi }}^{\prime \prime }}_{\alpha }\left( x,\lambda \right)dy}=\text{Si}n\left( s\left( x+1 \right) \right)-s{{\underline{\varphi }}_{\alpha }}\left( x,\lambda \right)+sCos\left( s\left( x+1 \right) \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +{{s}^{2}}\int\limits_{-1}^{x}{\text{Si}n}\left( s\left( x-y \right) \right){{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right)dy \\ \end{align}$$

is obtained. Substituing this back into the previous equality yields

\int_{- 1}^{x} Si n (s (x - y)) q (y) {\underline{φ}}_{α} (y, λ) d y = Si n (s (x + 1)) - s {\underline{φ}}_{α} (x, λ) + s C o s (s (x + 1))

$$\int\limits_{-1}^{x}{\text{Si}n\left( s\left( x-y \right) \right)q\left( y \right){{\underline{\varphi }}_{\alpha }}\left( y,\lambda \right)dy=\text{Si}n\left( s\left( x+1 \right) \right)-s{{\underline{\varphi }}_{\alpha }}\left( x,\lambda \right)+sCos\left( s\left( x+1 \right) \right)}$$

From here, we have

{\underline{φ}}_{α} (x, λ) = C o s (s (x + 1)) - \frac{1}{s} Si n (s (x + 1)) + \int_{- 1}^{x} Si n (s (x - y)) q (y) {\underline{φ}}_{α} (x, λ) d y .

$${{\underline{\varphi }}_{\alpha }}\left( x,\lambda \right)=Cos\left( s\left( x+1 \right) \right)-\frac{1}{s}\text{Si}n\left( s\left( x+1 \right) \right)+\int\limits_{-1}^{x}{\text{Si}n\left( s\left( x-y \right) \right)q\left( y \right){{\underline{\varphi }}_{\alpha }}\left( x,\lambda \right)dy.}$$

Similarly ${\bar{φ}}_{α} (x, λ)$ ${{\bar{\varphi }}_{\alpha }}\left( x,\lambda \right)$is found. Derivating in these equations according to x, the derivative equations are obtained.

Lemma 2. Let λ = s². The lower and the upper solutions ${\underline{χ}}_{α} (x, λ), {\bar{χ}}_{α} (x, λ)$ ${{\underline{\chi }}_{\alpha }}\left( x,\lambda \right),{{\bar{\chi }}_{\alpha }}\left( x,\lambda \right)$satisfy the following integral equations for k=0 and k=1:

(3.17)

\begin{aligned} {({\underline{χ}}_{α} (x, λ))}^{(k)} = - β {(C o s (s (x + 1)))}^{(k)} + \frac{λ α}{s} {(S i n (s (x + 1)))}^{(k)} \\ - \frac{1}{s} \int_{x}^{1} {(S i n (s (x - y)))}^{(k)} q (y) {({\underline{χ}}_{α} (x, λ))}^{(k)} d y \end{aligned}

$$\begin{align}& {{\left( {{\underline{\chi }}_{\alpha }}\left( x,\lambda \right) \right)}^{\left( k \right)}}=-\beta {{\left( Cos\left( s\left( x+1 \right) \right) \right)}^{\left( k \right)}}+\frac{\lambda \alpha }{s}{{\left( Sin\left( s\left( x+1 \right) \right) \right)}^{\left( k \right)}} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\,\,-\frac{1}{s}\int\limits_{x}^{1}{{{\left( Sin\left( s\left( x-y \right) \right) \right)}^{\left( k \right)}}q\left( y \right)}{{\left( {{\underline{\chi }}_{\alpha }}\left( x,\lambda \right) \right)}^{\left( k \right)}}dy \\ \end{align}$$

(3.18)

\begin{aligned} {({\bar{χ}}_{α} (x, λ))}^{(k)} = - β {(C o s (s (x + 1)))}^{(k)} + \frac{λ α}{s} {(S i n (s (x + 1)))}^{(k)} \\ - \frac{1}{s} \int_{x}^{1} {(S i n (s (x - y)))}^{(k)} q (y) {({\bar{χ}}_{α} (x, λ))}^{(k)} d y . \end{aligned}

$$\begin{align}& {{\left( {{{\bar{\chi }}}_{\alpha }}\left( x,\lambda \right) \right)}^{\left( k \right)}}=-\beta {{\left( Cos\left( s\left( x+1 \right) \right) \right)}^{\left( k \right)}}+\frac{\lambda \alpha }{s}{{\left( Sin\left( s\left( x+1 \right) \right) \right)}^{\left( k \right)}} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\,\,-\frac{1}{s}\int\limits_{x}^{1}{{{\left( Sin\left( s\left( x-y \right) \right) \right)}^{\left( k \right)}}q\left( y \right)}{{\left( {{{\bar{\chi }}}_{\alpha }}\left( x,\lambda \right) \right)}^{\left( k \right)}}dy. \\ \end{align}$$

Proof. Substituing the identity $q (y) {\underline{χ}}_{α} (y, λ) = - λ {\underline{χ}}_{α} (y, λ) - {\underline{χ}}^{' '}_{a} (y, λ)$ $q\left( y \right){{\underline{\chi }}_{\alpha }}\left( y,\lambda \right)=-\lambda {{\underline{\chi }}_{\alpha }}\left( y,\lambda \right)-{{\underline{\chi }}^{\prime \prime }}_{a}\left( y,\lambda \right)$in the right side of (3.17), integrating by parts twice and using (3.5) yields (3.17) for k=0. Similarly, the equation (3.18) is found for k=0. Derivating in these equations according to x, the derivative equations are obtained.

eISSN:: 2444-8656
Idioma:: Inglés

Calendario de la edición:: Volume Open
Temas de la revista:: Life Sciences, other, Mathematics, Applied Mathematics, General Mathematics, Physics

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Important Notes for a Fuzzy Boundary Value Problem

Publicado en línea: 12 ago 2019

Páginas: 305 - 314

Recibido: 24 abr 2019

Aceptado: 08 may 2019

DOI: https://doi.org/10.2478/AMNS.2019.2.00027

Palabras claveFuzzy logic, Fuzzy Sturm-Liouville equation, Sturm-Liouville Theory

© 2019 Hülya Gültekin Çitil, published by Sciendo

This work is licensed under the Creative Commons Attribution 4.0 Public License.

Palabras clave
Fuzzy logic, Fuzzy Sturm-Liouville equation, Sturm-Liouville Theory