On a fuzzy initial value problem

Çitil, Hülya Gültekin

doi:10.2478/ijmce-2025-0002

Full Article

1

Introduction

Fuzzy number and fuzzy arithmetic were introduced by Zadeh in 1965 [1]. Fuzzy differential equations are an important topic in many fields. Thus, many researchers have studied fuzzy differential equation using different approach methods [2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. The first approach is the Hukuhara derivative [12, 13]. This approach has a drawback: the solution becomes fuzzier as time goes. Thus, the fuzzy solution behaves quite differently from the crisp solution. So, the generalized Hukuhara derivative was studied [7, 8, 11, 14, 15, 16]. The generalized Hukuhara derivative allows to resolve the above-mentioned shortcoming. The second approach is the extension principle [17]. The third approach is the differential inclusion one [18]. The Fuzzy Laplace transform (FLM) method is useful to solve fuzzy differential equations. In many articles, the solution of fuzzy differential equation has been studied by using FLM approach [19, 20, 21, 22, 23, 24].

The rest of this paper is organized as follows: preliminaries and basic definitions used in this paper are given in Section 2. In Section 3, some applications and main results are reported. In Section 4, the conclusions of this paper are introduced.

2

Preliminaries

In this part, we present some important definitions and theorems which will be used in this paper.

Definition 1

[25] A fuzzy number is a mapping u:ℝ → [0, 1] satisfying the properties $\bar{{x \in ℝ | u (x) > 0}}$ \overline {\left\{{x \in {\rm{\mathbb R}} \left| {u\left( x \right) > 0} \right.} \right\}} is compact, u is normal, u is convex fuzzy set, u is upper semi-continuous on ℝ. Let's denote the set of all fuzzy numbers with ℝ_F.

Definition 2

[11] Let be u ∈ ℝ_F. The α-level set of u is [u]^α = [u_α, ū_α] = {x ∈ ℝ | u (x) ≥ α}, 0 < α ≤ 1.

Definition 3

[11] The parametric form [u_α, ū_α] of u fuzzy number satisfies the following requirements:

The lower part u_α is bounded non-decreasing left-continuous on (0, 1], right-continuous for α = 0.
The upper part ū_α is bounded non-increasing left-continuous on (0, 1], right-continuous for α = 0.
u_α ≤ ū_α, 0 ≤ α ≤ 1.

Definition 4

[25] If A is asymmietric triangular fuzzy number with support [a, ā], the α–level set of A is ${[A]}^{α} = [\underline{a} + (\frac{\bar{a} - \underline{a}}{2}) α, \bar{a} - (\frac{\bar{a} - \underline{a}}{2}) α]$ {\left[A \right]^\alpha} = \left[{\underline a + \left( {{{\overline a - \underline a} \over 2}} \right)\alpha ,\overline a - \left( {{{\overline a - \underline a} \over 2}} \right)\alpha} \right] .

Definition 5

[26] Let u, v ∈ ℝ_F. The generalized Hukuhara difference (gH-difference) between u and v is the set w ∈ ℝ_F which u⊖_gv = w if and only if u = v + w or v = u + (−1) w.

Definition 6

[11] Let f : [a, b] → ℝ_F and t₀ [a, b]. We say that f is (1)-differentiable at t₀, if there is an element f^′ (t₀) ∈ ℝ_F such that for all h > 0 sufficiently small (near to 0), exist f (t₀ + h) ⊖ f (t₀), f (t₀) ⊖ f (t₀ − h) and the limits $lim_{h \to 0} \frac{f (t_{0} + h) ⊖ f (t_{0})}{h} = lim_{h \to 0} \frac{f (t_{0}) ⊖ f (t_{0} - h)}{h} = f^{'} (t_{0}),$ \mathop {\lim}\limits_{h \to 0} {{f\left( {{t_0} + h} \right) \ominus f\left( {{t_0}} \right)} \over h} = \mathop {\lim}\limits_{h \to 0} {{f\left( {{t_0}} \right) \ominus f\left( {{t_0} - h} \right)} \over h} = {f^{'}}\left( {{t_0}} \right), and f is (2)-differentiable if for all h > 0 sufficiently small (near to 0), exist f (t₀) ⊖ f (t₀ + h), f (t₀ − h) ⊖ f (t₀) and the limits $lim_{h \to 0} \frac{f (t_{0}) ⊖ f (t_{0} + h)}{- h} = lim_{h \to 0} \frac{f (t_{0} - h) ⊖ f (t_{0})}{- h} = f^{'} (t_{0}) .$ \mathop {\lim}\limits_{h \to 0} {{f\left( {{t_0}} \right) \ominus f\left( {{t_0} + h} \right)} \over {- h}} = \mathop {\lim}\limits_{h \to 0} {{f\left( {{t_0} - h} \right) \ominus f\left( {{t_0}} \right)} \over {- h}} = {f^{'}}\left( {{t_0}} \right).

Theorem 1

[27] Let f : [a, b] → ℝ_F be fuzzy function, where ${[f (t)]}^{α} = [{\underline{f}}_{α} (t), {\bar{f}}_{α} (t)]$ {\left[{f\left( t \right)} \right]^\alpha} = \left[{{{\underline f}_\alpha}\left( t \right),{{\overline f}_\alpha}\left( t \right)} \right] , for each α ∈ [0, 1].

(i)
If f is (1)-differentiable then f_α and ${\bar{f}}_{α}$ {\overline f_\alpha} are differentiable functions and ${[f^{'} (t)]}^{α} = [{\underline{f}}_{α}^{^{'}} (t), {\bar{f}}_{α}^{^{'}} (t)]$ {\left[{{f^{'}}\left( t \right)} \right]^\alpha} = \left[{\underline f_\alpha^{'}\left( t \right),\overline f_\alpha^{'}\left( t \right)} \right] ,
(ii)
If f is (2)-differentiable then f_α and ${\bar{f}}_{α}$ {\overline f_\alpha} are differentiable functions and ${[f^{'} (t)]}^{α} = [{\bar{f}}_{α}^{^{'}} (t), {\underline{f}}_{α}^{^{'}} (t)]$ {\left[{{f^{'}}\left( t \right)} \right]^\alpha} = \left[{\overline f_\alpha^{'}\left( t \right),\underline f_\alpha^{'}\left( t \right)} \right] .

Definition 7

[28] The fuzzy Laplace transform of fuzzy function f is $\begin{matrix} F (s) = L (f (t)) = \int_{0}^{\infty} e^{- s t} f (t) d t = [lim_{ρ \to \infty} \int_{0}^{ρ} e^{- s t} \underline{f} (t) d t, lim_{ρ \to \infty} \int_{0}^{ρ} e^{- s t} \bar{f} (t) d t], \\ F (s, α) = L ({[f (t)]}^{α}) = [L ({\underline{f}}_{α} (t)), L ({\bar{f}}_{α} (t))] . \\ L ({\underline{f}}_{α} (t)) = \int_{0}^{\infty} e^{- s t} {\underline{f}}_{α} (t) d t = lim_{ρ \to \infty} \int_{0}^{ρ} e^{- s t} {\underline{f}}_{α} (t) d t, \\ L ({\bar{f}}_{α} (t)) = \int_{0}^{\infty} e^{- s t} {\bar{f}}_{α} (t) d t = lim_{ρ \to \infty} \int_{0}^{ρ} e^{- s t} {\bar{f}}_{α} (t) d t . \end{matrix}$ \matrix{{F\left( s \right) = L\left( {f\left( t \right)} \right) = \int_0^\infty {e^{- st}}f\left( t \right)dt = \left[{\mathop {\lim}\limits_{\rho \to \infty} \int_0^\rho {e^{- st}}\underline f \left( t \right)dt\;,\mathop {\lim}\limits_{\rho \to \infty} \int_0^\rho {e^{- st}}\overline f \left( t \right)dt\;} \right],} \cr {F\left( {s,\alpha} \right) = L\left( {{{\left[{f\left( t \right)} \right]}^\alpha}} \right) = \left[{L\left( {{{\underline f}_\alpha}\left( t \right)} \right),L\left( {{{\overline f}_\alpha}\left( t \right)} \right)} \right].} \cr {L\left( {{{\underline f}_\alpha}\left( t \right)} \right) = \int_0^\infty {e^{- st}}{{\underline f}_\alpha}\left( t \right)dt = \mathop {\lim}\limits_{\rho \to \infty} \int_0^\rho {e^{- st}}{{\underline f}_\alpha}\left( t \right)dt\;,} \cr {L\left( {{{\overline f}_\alpha}\left( t \right)} \right) = \int_0^\infty {e^{- st}}{{\overline f}_\alpha}\left( t \right)dt = \mathop {\lim}\limits_{\rho \to \infty} \int_0^\rho {e^{- st}}{{\overline f}_\alpha}\left( t \right)dt.\;} \cr}

Theorem 2

[19] Let f^′ (t) be an integrable fuzzy function and f (t) is primitive of f^′ (t) on (0, ∞].

1.
If f is (1)-differentiable, L (f^′ (t)) = sL ( f (t)) ⊖ f (0).
2.
If f is (2)-differentiable, L (f^′ (t)) = (− f (0)) ⊖ (−sL (f (t))).

Theorem 3

[28] Let f^″ (t) be an integrable fuzzy function and f (t), f^′ (t) are primitive of f^′ (t), f^″ (t) on (0, ∞].

1.
If f and f^′ are (1)-differentiable, L (f^″ (t)) = s²L (f (t)) ⊖ s f (0) ⊖ f^′ (0).
2.
If f and f^′ are (2)-differentiable, L (f^″ (t)) = s²L(f (t)) ⊖ s f (0) − f^′ (0).
3.
If f is (1)-differentiable and f^′ is (2)-differentiable, L (f^″ (t)) = ⊖ (−s²) L(f (t)) − s f (0) − f^′ (0).
4.
If f is (2)-differentiable and f^′ is (1)-differentiable, L (f^″ (t)) = ⊖ (−s²) L(f (t)) − s f (0) ⊖ f^′ (0).

3

Applications

We investigate the solutions of the problem given as $\begin{matrix} u^{″} (t) = {[λ]}^{α} u (t), t > 0 \\ u (0) = {[ρ]}^{α}, u^{'} (0) = {[ν]}^{α} \end{matrix}$ \matrix{{u''(t) = {{\left[\lambda \right]}^\alpha}u(t),\;\;t > 0} \cr {u\left( 0 \right) = {{\left[\rho \right]}^\alpha},\;u'\left( 0 \right) = {{\left[\nu \right]}^\alpha}} \cr} by the fuzzy Laplace transform and the generalized Hukuhara differentiability, where ${[λ]}^{α} = [{\underline{λ}}_{α}, {\bar{λ}}_{α}]$ {\left[\lambda \right]^\alpha} = \left[{{{\underline \lambda}_\alpha},{{\overline \lambda}_\alpha}} \right] , $({\underline{λ}}_{α} > 0, {\bar{λ}}_{α} > 0)$ \left( {{{\underline \lambda}_\alpha} > 0,\;\;{{\overline \lambda}_\alpha} > 0\;} \right) , ${[ρ]}^{α} = [{\underline{ρ}}_{α}, {\bar{ρ}}_{α}]$ {\left[\rho \right]^\alpha} = \left[{{{\underline \rho}_\alpha},{{\overline \rho}_\alpha}} \right] , ${[ν]}^{α} = [{\underline{ν}}_{α}, {\bar{ν}}_{α}]$ {\left[\nu \right]^\alpha} = \;\left[{{{\underline \nu}_\alpha},{{\overline \nu}_\alpha}} \right] are symmetric triangular fuzzy numbers, u (t) is positive fuzzy function, the fuzzy Laplace transform of fuzzy function u (t) is L(u (t)) = U(s) and (i, j)-solution means that u is (i)-differentiable, u^′ is (j)-differentiable, i, j = 1, 2.

1. (1,1)-solution: Let u, u^′ be (1)-differentiable. Then, the equation $s^{2} U (s) ⊖ s u (0) ⊖ u^{'} (0) = {[λ]}^{α} U (s)$ {s^2}U\left( s \right) \ominus su\left( 0 \right) \ominus u'\left( 0 \right) = {\left[\lambda \right]^\alpha}U\left( s \right) is obtained by using FLT. From this, we have the equations $\begin{matrix} s^{2} {\underline{U}}_{α} (s) - s {\underline{u}}_{α} (0) - {\underline{u}}_{α}^{^{'}} (0) = {\underline{λ}}_{α} {\underline{U}}_{α} (s), \\ s^{2} {\bar{U}}_{α} (s) - s {\bar{u}}_{α} (0) - {\bar{u}}_{α}^{^{'}} (0) = {\bar{λ}}_{α} {\bar{U}}_{α} (s) . \end{matrix}$ \matrix{{{s^2}{{\underline U}_\alpha}\left( s \right) - s{{\underline u}_\alpha}\left( 0 \right) - \underline u_\alpha^{'}\left( 0 \right) = {{\underline \lambda}_\alpha}{{\underline U}_\alpha}\left( s \right),} \cr {{s^2}{{\overline U}_\alpha}\left( s \right) - s{{\overline u}_\alpha}\left( 0 \right) - \overline u_\alpha^{'}\left( 0 \right) = {{\overline \lambda}_\alpha}{{\overline U}_\alpha}\left( s \right).} \cr}

Using the initial conditions, we obtain ${\underline{U}}_{α} (s) = \frac{s {\underline{ρ}}_{α}}{s^{2} - {\underline{λ}}_{α}} + \frac{{\underline{ν}}_{α}}{s^{2} - {\underline{λ}}_{α}}, {\bar{U}}_{α} (s) = \frac{s {\bar{ρ}}_{α}}{s^{2} - {\bar{λ}}_{α}} + \frac{{\bar{ν}}_{α}}{s^{2} - {\bar{λ}}_{α}} .$ {\underline U_\alpha}\left( s \right) = {{s{{\underline \rho}_\alpha}} \over {{s^2} - {{\underline \lambda}_\alpha}}} + {{{{\underline \nu}_\alpha}} \over {{s^2} - {{\underline \lambda}_\alpha}}},\;{\overline U_\alpha}\left( s \right) = {{s{{\overline \rho}_\alpha}} \over {{s^2} - {{\overline \lambda}_\alpha}}} + {{{{\overline \nu}_\alpha}} \over {{s^2} - {{\overline \lambda}_\alpha}}}.

Taking the inverse Laplace transform of these equations, (1, 1)-solution is obtained as $\begin{matrix} {\underline{u}}_{α} (t) = {\underline{ρ}}_{α} cosh (\sqrt{{\underline{λ}}_{α}} t) + \frac{{\underline{ν}}_{α}}{\sqrt{{\underline{λ}}_{α}}} sinh (\sqrt{{\underline{λ}}_{α}} t), {\bar{u}}_{α} (t) = {\bar{ρ}}_{α} cosh (\sqrt{{\bar{λ}}_{α}} t) + \frac{{\bar{ν}}_{α}}{\sqrt{{\bar{λ}}_{α}}} sinh (\sqrt{{\bar{λ}}_{α}} t), \\ {[u (t)]}^{α} = [{\underline{u}}_{α} (t), {\bar{u}}_{α} (t)] . \end{matrix}$ \matrix{{{{\underline u}_\alpha}\left( t \right) = {{\underline \rho}_\alpha}\cosh \left( {\sqrt {{{\underline \lambda}_\alpha}} t} \right) + {{{{\underline \nu}_\alpha}} \over {\sqrt {{{\underline \lambda}_\alpha}}}}\sinh \left( {\sqrt {{{\underline \lambda}_\alpha}} t} \right),{{\overline u}_\alpha}\left( t \right) = {{\overline \rho}_\alpha}\cosh \left( {\sqrt {{{\overline \lambda}_\alpha}} t} \right) + {{{{\overline \nu}_\alpha}} \over {\sqrt {{{\overline \lambda}_\alpha}}}}\sinh \left( {\sqrt {{{\overline \lambda}_\alpha}} t} \right),} \cr {{{\left[{u\left( t \right)} \right]}^\alpha} = \left[{{{\underline u}_\alpha}\left( t \right),{{\overline u}_\alpha}\left( t \right)} \right].} \cr}

2. (1,2)-solution: Let u be (1)-differentiable and u^′ be (2)-differentiable. Then, since $- u^{'} (0) ⊖ (- s^{2} U (s)) - s u (0) = {[λ]}^{α} U (s),$ - u'\left( 0 \right) \ominus \left( {- {s^2}U\left( s \right)} \right) - su\left( 0 \right) = {\left[\lambda \right]^\alpha}U\left( s \right), we have the equations $\begin{matrix} - {\bar{u}}_{α}^{^{'}} (0) + s^{2} {\bar{U}}_{α} (s) - s {\bar{u}}_{α} (0) = {\underline{λ}}_{α} {\underline{U}}_{α} (s), \\ - {\underline{u}}_{α}^{^{'}} (0) + s^{2} {\underline{U}}_{α} (s) - s {\underline{u}}_{α} (0) = {\bar{λ}}_{α} {\bar{U}}_{α} (s) . \end{matrix}$ \matrix{{- \overline u_\alpha^{'}\left( 0 \right) + {s^2}{{\overline U}_\alpha}\left( s \right) - s{{\overline u}_\alpha}\left( 0 \right) = {{\underline \lambda}_\alpha}{{\underline U}_\alpha}\left( s \right),} \cr {- \underline u_\alpha^{'}\left( 0 \right) + {s^2}{{\underline U}_\alpha}\left( s \right) - s{{\underline u}_\alpha}\left( 0 \right) = {{\overline \lambda}_\alpha}{{\overline U}_\alpha}\left( s \right).} \cr}

Using the initial conditions and making the necessary operations, we obtain the equations $\begin{matrix} {\underline{U}}_{α} (s) = \frac{{\bar{λ}}_{α} {\bar{ν}}_{α}}{s^{4} - {\underline{λ}}_{α} {\bar{λ}}_{α}} + \frac{s {\bar{λ}}_{α} {\bar{ρ}}_{α}}{s^{4} - {\underline{λ}}_{α} {\bar{λ}}_{α}} + \frac{s^{2} {\underline{ν}}_{α}}{s^{4} - {\underline{λ}}_{α} {\bar{λ}}_{α}} + \frac{s^{3} {\underline{ρ}}_{α}}{s^{4} - {\underline{λ}}_{α} {\bar{λ}}_{α}}, \\ {\bar{U}}_{α} (s) = \frac{{\underline{λ}}_{α} {\underline{ν}}_{α}}{s^{4} - {\underline{λ}}_{α} {\bar{λ}}_{α}} + \frac{s {\underline{λ}}_{α} {\underline{ρ}}_{α}}{s^{4} - {\underline{λ}}_{α} {\bar{λ}}_{α}} + \frac{s^{2} {\bar{ν}}_{α}}{s^{4} - {\underline{λ}}_{α} {\bar{λ}}_{α}} + \frac{s^{3} {\bar{ρ}}_{α}}{s^{4} - {\underline{λ}}_{α} {\bar{λ}}_{α}} . \end{matrix}$ \matrix{{{{\underline U}_\alpha}\left( s \right) = {{{{\overline \lambda}_\alpha}{{\overline \nu}_\alpha}} \over {{s^4} - {{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}}} + {{s{{\overline \lambda}_\alpha}{{\overline \rho}_\alpha}} \over {{s^4} - {{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}}} + {{{s^2}{{\underline \nu}_\alpha}} \over {{s^4} - {{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}}} + {{{s^3}{{\underline \rho}_\alpha}} \over {{s^4} - {{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}}},} \cr {{{\overline U}_\alpha}\left( s \right) = {{{{\underline \lambda}_\alpha}{{\underline \nu}_\alpha}} \over {{s^4} - {{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}}} + {{s{{\underline \lambda}_\alpha}{{\underline \rho}_\alpha}} \over {{s^4} - {{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}}} + {{{s^2}{{\overline \nu}_\alpha}} \over {{s^4} - {{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}}} + {{{s^3}{{\overline \rho}_\alpha}} \over {{s^4} - {{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}}}.} \cr}

Then, (1, 2)-solution is obtained as $\begin{array}{r} {\underline{u}}_{α} (t) = (\frac{{\bar{λ}}_{α} {\bar{ν}}_{α}}{2 \sqrt[4]{{({\underline{λ}}_{α} {\bar{λ}}_{α})}^{3}}} + \frac{{\underline{ν}}_{α}}{2 \sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})}}) (sin (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t) + sinh (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t)) \\ + (\frac{{\bar{λ}}_{α} {\bar{ρ}}_{α}}{2 \sqrt{({\underline{λ}}_{α} {\bar{λ}}_{α})}} + \frac{{\underline{ρ}}_{α}}{2}) (cos (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t) + cosh (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t)), \\ {\bar{u}}_{α} (t) = (\frac{{\underline{λ}}_{α} {\underline{ν}}_{α}}{2 \sqrt[4]{{({\underline{λ}}_{α} {\bar{λ}}_{α})}^{3}}} + \frac{{\bar{ν}}_{α}}{2 \sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})}}) (sin (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t) + sinh (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t)) \\ + (\frac{{\underline{λ}}_{α} {\underline{ρ}}_{α}}{2 \sqrt{({\underline{λ}}_{α} {\bar{λ}}_{α})}} + \frac{{\bar{ρ}}_{α}}{2}) (cos (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t) + cosh (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t)), \\ {[u (t)]}^{α} = [{\underline{u}}_{α} (t), {\bar{u}}_{α} (t)] . \end{array}$ \matrix{\hfill {{{\underline u}_\alpha}\left( t \right) = \left( {{{{{\overline \lambda}_\alpha}{{\overline \nu}_\alpha}} \over {2\root 4 \of {{{\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}^3}}}} + {{{{\underline \nu}_\alpha}} \over {2\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}}}} \right)\left( {\sin \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right) + \sinh \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right)} \right)} \cr \hfill {+ \left( {{{{{\overline \lambda}_\alpha}{{\overline \rho}_\alpha}} \over {2\sqrt {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}}} + {{{{\underline \rho}_\alpha}} \over 2}} \right)\left( {\cos \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right) + \cosh \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right)} \right),} \cr \hfill {{{\overline u}_\alpha}\left( t \right) = \left( {{{{{\underline \lambda}_\alpha}{{\underline \nu}_\alpha}} \over {2\root 4 \of {{{\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}^3}}}} + {{{{\overline \nu}_\alpha}} \over {2\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}}}} \right)\left( {\sin \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right) + \sinh \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right)} \right)} \cr \hfill {+ \left( {{{{{\underline \lambda}_\alpha}{{\underline \rho}_\alpha}} \over {2\sqrt {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}}} + {{{{\overline \rho}_\alpha}} \over 2}} \right)\left( {\cos \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right) + \cosh \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right)} \right),} \cr \hfill {{{\left[{u\left( t \right)} \right]}^\alpha} = \left[{{{\underline u}_\alpha}\left( t \right),{{\overline u}_\alpha}\left( t \right)} \right].} \cr}

3. (2,1)-solution: Let u be (2)-differentiable and u^′ be (1)-differentiable. From the equation $⊖ (- s^{2} U (s)) - s u (0) ⊖ u^{^{'}} (0) = {[λ]}^{α} U (s),$ \ominus \left( {- {s^2}U\left( s \right)} \right) - su\left( 0 \right) \ominus {u^{'}}\left( 0 \right) = {\left[\lambda \right]^\alpha}U\left( s \right), we have the equations $s^{2} {\bar{U}}_{α} (s) - {\underline{λ}}_{α} {\underline{U}}_{α} (s) = s {\bar{ρ}}_{α} + {\underline{ν}}_{α}, s^{2} {\underline{U}}_{α} (s) - {\bar{λ}}_{α} {\bar{U}}_{α} (s) = s {\underline{ρ}}_{α} + {\bar{ν}}_{α} .$ {s^2}{\overline U_\alpha}\left( s \right) - {\underline \lambda_\alpha}{\underline U_\alpha}\left( s \right) = s{\overline \rho_\alpha} + {\underline \nu_\alpha},{s^2}{\underline U_\alpha}\left( s \right) - {\overline \lambda_\alpha}{\overline U_\alpha}\left( s \right) = s{\underline \rho_\alpha} + {\overline \nu_\alpha}.

From this, making the necessary operations, (2, 1)-solution is obtained as $\begin{array}{r} {\underline{u}}_{α} (t) = (\frac{{\bar{λ}}_{α} {\underline{ν}}_{α}}{2 \sqrt[4]{{({\underline{λ}}_{α} {\bar{λ}}_{α})}^{3}}} + \frac{{\bar{ν}}_{α}}{2 \sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})}}) (sin (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t) + sinh (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t)) \\ + (\frac{{\bar{λ}}_{α} {\bar{ρ}}_{α}}{2 \sqrt{({\underline{λ}}_{α} {\bar{λ}}_{α})}} + \frac{{\underline{ρ}}_{α}}{2}) (cos (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t) + cosh (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t)), \\ {\bar{u}}_{α} (t) = (\frac{{\underline{λ}}_{α} {\bar{ν}}_{α}}{2 \sqrt[4]{{({\underline{λ}}_{α} {\bar{λ}}_{α})}^{3}}} + \frac{{\underline{ν}}_{α}}{2 \sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})}}) (sin (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t) + sinh (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t)) \\ + (\frac{{\underline{λ}}_{α} {\underline{ρ}}_{α}}{2 \sqrt{({\underline{λ}}_{α} {\bar{λ}}_{α})}} + \frac{{\bar{ρ}}_{α}}{2}) (cos (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t) + cosh (\sqrt[4]{({\underline{λ}}_{α} {\bar{λ}}_{α})} t)), \\ {[u (t)]}^{α} = [{\underline{u}}_{α} (t), {\bar{u}}_{α} (t)] . \end{array}$ \matrix{\hfill {{{\underline u}_\alpha}\left( t \right) = \left( {{{{{\overline \lambda}_\alpha}{{\underline \nu}_\alpha}} \over {2\root 4 \of {{{\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}^3}}}} + {{{{\overline \nu}_\alpha}} \over {2\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}}}} \right)\left( {\sin \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right) + \sinh \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right)} \right)} \cr \hfill {+ \left( {{{{{\overline \lambda}_\alpha}{{\overline \rho}_\alpha}} \over {2\sqrt {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}}} + {{{{\underline \rho}_\alpha}} \over 2}} \right)\left( {\cos \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right) + \cosh \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right)} \right),} \cr \hfill {{{\overline u}_\alpha}\left( t \right) = \left( {{{{{\underline \lambda}_\alpha}{{\overline \nu}_\alpha}} \over {2\root 4 \of {{{\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}^3}}}} + {{{{\underline \nu}_\alpha}} \over {2\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}}}} \right)\left( {\sin \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right) + \sinh \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right)} \right)} \cr \hfill {+ \left( {{{{{\underline \lambda}_\alpha}{{\underline \rho}_\alpha}} \over {2\sqrt {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)}}} + {{{{\overline \rho}_\alpha}} \over 2}} \right)\left( {\cos \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right) + \cosh \left( {\root 4 \of {\left( {{{\underline \lambda}_\alpha}{{\overline \lambda}_\alpha}} \right)} t} \right)} \right),} \cr \hfill {{{\left[{u\left( t \right)} \right]}^\alpha} = \left[{{{\underline u}_\alpha}\left( t \right),{{\overline u}_\alpha}\left( t \right)} \right].} \cr}

4. (2,2)-solution: Let u and u^′ be (2)-differentiable. Since $s^{2} U (s) ⊖ s u (0) - u^{^{^{'}}} (0) = {[λ]}^{α} U (s),$ {s^2}U\left( s \right) \ominus su\left( 0 \right) - {u^{^{'}}}\left( 0 \right) = {\left[\lambda \right]^\alpha}U\left( s \right), (2, 2)-solution is obtained as $\begin{matrix} {\underline{u}}_{α} (t) = {\underline{ρ}}_{α} cosh (\sqrt{{\underline{λ}}_{α}} t) + \frac{{\bar{ν}}_{α}}{\sqrt{{\underline{λ}}_{α}}} sinh (\sqrt{{\underline{λ}}_{α}} t), {\bar{u}}_{α} (t) = {\bar{ρ}}_{α} cosh (\sqrt{{\bar{λ}}_{α}} t) + \frac{{\underline{ν}}_{α}}{\sqrt{{\bar{λ}}_{α}}} sinh (\sqrt{{\bar{λ}}_{α}} t), \\ {[u (t)]}^{α} = [{\underline{u}}_{α} (t), {\bar{u}}_{α} (t)] . \end{matrix}$ \matrix{{{{\underline u}_\alpha}\left( t \right) = {{\underline \rho}_\alpha}\cosh \left( {\sqrt {{{\underline \lambda}_\alpha}} t} \right) + {{{{\overline \nu}_\alpha}} \over {\sqrt {{{\underline \lambda}_\alpha}}}}\sinh \left( {\sqrt {{{\underline \lambda}_\alpha}} t} \right),{{\overline u}_\alpha}\left( t \right) = {{\overline \rho}_\alpha}\cosh \left( {\sqrt {{{\overline \lambda}_\alpha}} t} \right) + {{{{\underline \nu}_\alpha}} \over {\sqrt {{{\overline \lambda}_\alpha}}}}\sinh \left( {\sqrt {{{\overline \lambda}_\alpha}} t} \right),} \cr {{{\left[{u\left( t \right)} \right]}^\alpha} = \left[{{{\underline u}_\alpha}\left( t \right),{{\overline u}_\alpha}\left( t \right)} \right].} \cr}

These solutions must be valid fuzzy functions.

Example 3.1

Consider the solutions of the fuzzy problem $\begin{matrix} u^{^{''}} (t) = {[1]}^{α} u (t), \\ u (0) = {[1]}^{α}, u^{^{'}} (0) = {[2]}^{α}, \end{matrix}$ \matrix{{{u^{''}}\left( t \right) = {{\left[1 \right]}^\alpha}u\left( t \right),} \cr {u\left( 0 \right) = {{\left[1 \right]}^\alpha},\;\;{u^{'}}\left( 0 \right) = {{\left[2 \right]}^\alpha},} \cr} where [1]^α = [α, 2 − α], [2]^α = [1 + α, 3 − α].

(1, 1)-solution is $\begin{matrix} {\underline{u}}_{α} (t) = α cosh (\sqrt{α} t) + \frac{1 + α}{\sqrt{α}} sinh (\sqrt{α} t), \\ {\bar{u}}_{α} (t) = (2 - α) cosh (\sqrt{(2 - α)} t) + \frac{3 - α}{\sqrt{(2 - α)}} sinh (\sqrt{(2 - α)} t), \\ {[u (t)]}^{α} = [{\underline{u}}_{α} (t), {\bar{u}}_{α} (t)], \end{matrix}$ \matrix{{{{\underline u}_\alpha}\left( t \right) = \alpha \cosh \left( {\sqrt \alpha t} \right) + {{1 + \alpha} \over {\sqrt \alpha}}\sinh \left( {\sqrt \alpha t} \right),} \cr {{{\overline u}_\alpha}\left( t \right) = \left( {2 - \alpha} \right)\cosh \left( {\sqrt {\left( {2 - \alpha} \right)} t} \right) + {{3 - \alpha} \over {\sqrt {\left( {2 - \alpha} \right)}}}\sinh \left( {\sqrt {\left( {2 - \alpha} \right)} t} \right),} \cr {{{\left[{u\left( t \right)} \right]}^\alpha} = \left[{{{\underline u}_\alpha}\left( t \right),{{\overline u}_\alpha}\left( t \right)} \right],} \cr}

(1, 2)-solution is $\begin{matrix} {\underline{u}}_{α} (t) = (\frac{(2 - α) (3 - α)}{2 \sqrt[4]{{(α (2 - α))}^{3}}} + \frac{1 + α}{2 \sqrt[4]{(α (2 - α))}}) (sin (\sqrt[4]{(α (2 - α))} t) + sinh (\sqrt[4]{(α (2 - α))} t)) \\ + (\frac{{(2 - α)}^{2}}{2 \sqrt{(α (2 - α))}} + \frac{α}{2}) (cos (\sqrt[4]{(α (2 - α))} t) + cosh (\sqrt[4]{(α (2 - α))} t)), \\ {\bar{u}}_{α} (t) = (\frac{α (1 + α)}{2 \sqrt[4]{{(α (2 - α))}^{3}}} + \frac{3 - α}{2 \sqrt[4]{(α (2 - α))}}) (sin (\sqrt[4]{(α (2 - α))} t) + sinh (\sqrt[4]{(α (2 - α))} t)) \\ + (\frac{α^{2}}{2 \sqrt{(α (2 - α))}} + \frac{2 - α}{2}) (cos (\sqrt[4]{(α (2 - α))} t) + cosh (\sqrt[4]{(α (2 - α))} t)), \\ {[u (t)]}^{α} = [{\underline{u}}_{α} (t), {\bar{u}}_{α} (t)], \end{matrix}$ \matrix{{{{\underline u}_\alpha}\left( t \right) = \left( {{{\left( {2 - \alpha} \right)\left( {3 - \alpha} \right)} \over {2\root 4 \of {{{\left( {\alpha \left( {2 - \alpha} \right)} \right)}^3}}}} + {{1 + \alpha} \over {2\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)}}}} \right)\left( {\sin \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right) + \sinh \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right)} \right)} \cr {+ \left( {{{{{\left( {2 - \alpha} \right)}^2}} \over {2\sqrt {\left( {\alpha \left( {2 - \alpha} \right)} \right)}}} + {\alpha \over 2}} \right)\left( {\cos \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right) + \cosh \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right)} \right),} \cr {{{\overline u}_\alpha}\left( t \right) = \left( {{{\alpha \left( {1 + \alpha} \right)} \over {2\root 4 \of {{{\left( {\alpha \left( {2 - \alpha} \right)} \right)}^3}}}} + {{3 - \alpha} \over {2\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)}}}} \right)\left( {\sin \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right) + \sinh \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right)} \right)} \cr {+ \left( {{{{\alpha^2}} \over {2\sqrt {\left( {\alpha \left( {2 - \alpha} \right)} \right)}}} + {{2 - \alpha} \over 2}} \right)\left( {\cos \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right) + \cosh \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right)} \right),} \cr {{{\left[{u\left( t \right)} \right]}^\alpha} = \left[{{{\underline u}_\alpha}\left( t \right),{{\overline u}_\alpha}\left( t \right)} \right],} \cr}

(2, 1)-solution is $\begin{matrix} {\underline{u}}_{α} (t) = (\frac{(2 - α) (1 + α)}{2 \sqrt[4]{{(α (2 - α))}^{3}}} + \frac{(3 - α)}{2 \sqrt[4]{(α (2 - α))}}) (sin (\sqrt[4]{(α (2 - α))} t) + sinh (\sqrt[4]{(α (2 - α))} t)) \\ + (\frac{{(2 - α)}^{2}}{2 \sqrt{(α (2 - α))}} + \frac{α}{2}) (cos (\sqrt[4]{(α (2 - α))} t) + cosh (\sqrt[4]{(α (2 - α))} t)), \\ {\bar{u}}_{α} (t) = (\frac{α (3 - α)}{2 \sqrt[4]{{(α (2 - α))}^{3}}} + \frac{(1 + α)}{2 \sqrt[4]{(α (2 - α))}}) (sin (\sqrt[4]{(α (2 - α))} t) + sinh (\sqrt[4]{(α (2 - α))} t)) \\ + (\frac{α^{2}}{2 \sqrt{(α (2 - α))}} + \frac{2 - α}{2}) (cos (\sqrt[4]{(α (2 - α))} t) + cosh (\sqrt[4]{(α (2 - α))} t)), \\ {[u (t)]}^{α} = [{\underline{u}}_{α} (t), {\bar{u}}_{α} (t)], \end{matrix}$ \matrix{{{{\underline u}_\alpha}\left( t \right) = \left( {{{\left( {2 - \alpha} \right)\left( {1 + \alpha} \right)} \over {2\root 4 \of {{{\left( {\alpha \left( {2 - \alpha} \right)} \right)}^3}}}} + {{\left( {3 - \alpha} \right)} \over {2\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)}}}} \right)\left( {\sin \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right) + \sinh \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right)} \right)} \cr {+ \left( {{{{{\left( {2 - \alpha} \right)}^2}} \over {2\sqrt {\left( {\alpha \left( {2 - \alpha} \right)} \right)}}} + {\alpha \over 2}} \right)\left( {\cos \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right) + \cosh \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right)} \right),} \cr {{{\overline u}_\alpha}\left( t \right) = \left( {{{\alpha \left( {3 - \alpha} \right)} \over {2\root 4 \of {{{\left( {\alpha \left( {2 - \alpha} \right)} \right)}^3}}}} + {{\left( {1 + \alpha} \right)} \over {2\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)}}}} \right)\left( {\sin \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right) + \sinh \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right)} \right)} \cr {+ \left( {{{{\alpha^2}} \over {2\sqrt {\left( {\alpha \left( {2 - \alpha} \right)} \right)}}} + {{2 - \alpha} \over 2}} \right)\left( {\cos \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right) + \cosh \left( {\root 4 \of {\left( {\alpha \left( {2 - \alpha} \right)} \right)} t} \right)} \right),} \cr {{{\left[{u\left( t \right)} \right]}^\alpha} = \left[{{{\underline u}_\alpha}\left( t \right),{{\overline u}_\alpha}\left( t \right)} \right],} \cr} and (2, 2)-solution is $\begin{matrix} {\underline{u}}_{α} (t) = α cosh (\sqrt{α} t) + \frac{3 - α}{\sqrt{α}} sinh (\sqrt{α} t), \\ {\bar{u}}_{α} (t) = (2 - α) cosh (\sqrt{(2 - α)} t) + \frac{1 + α}{\sqrt{(2 - α)}} sinh (\sqrt{(2 - α)} t), \\ {[u (t)]}^{α} = [{\underline{u}}_{α} (t), {\bar{u}}_{α} (t)] . \end{matrix}$ \matrix{{{{\underline u}_\alpha}\left( t \right) = \alpha \cosh \left( {\sqrt \alpha t} \right) + {{3 - \alpha} \over {\sqrt \alpha}}\sinh \left( {\sqrt \alpha t} \right),} \cr {{{\overline u}_\alpha}\left( t \right) = \left( {2 - \alpha} \right)\cosh \left( {\sqrt {\left( {2 - \alpha} \right)} t} \right) + {{1 + \alpha} \over {\sqrt {\left( {2 - \alpha} \right)}}}\sinh \left( {\sqrt {\left( {2 - \alpha} \right)} t} \right),} \cr {{{\left[{u\left( t \right)} \right]}^\alpha} = \left[{{{\underline u}_\alpha}\left( t \right),{{\overline u}_\alpha}\left( t \right)} \right].} \cr}

From Definition 3, according to Figure (1a) and Figure (2b), (1, 1) and (2, 2) solutions are valid fuzzy functions. But, according to Figure (1b) and Figure (2a), (1, 2) and (2, 1) solutions are not valid fuzzy functions.

In these figures, blue is used to symbolize $\to {\bar{y}}_{α} (t)$ \to {\overline y_\alpha}\left( t \right) , red is for → y_α (t) and green is used to explain $\to {\bar{y}}_{1} (t) = {\underline{y}}_{1} (t)$ \to {\overline y_1}\left( t \right) = {\underline y_1}\left( t \right) .

4

Conclusion

In this work, we studied a fuzzy initial problem with fuzzy coefficient. We used the generalized differentiability and the fuzzy Laplace transform. We solved an application on the problem and drew the figures of the solutions. Finally, it is shown that the solutions are valid fuzzy functions or not.

5

Declarations

5.1

Conflict of interest

The authors hereby declare that there is no conflict of interests regarding the publication of this paper.

5.2

Funding

Not applicable.

5.3

Author's contribution

H.G.Ç.-Methodology, Writing-Original Draft, Validation, Conceptualization, Formal Analysis, Investigation, Writing-Review and Editing. All authors read and approved the final submitted version of this manuscript.

5.4

Acknowledgement

The authors deeply appreciate the anonymous reviewers for their helpful and constructive suggestions, which can further help improve this paper.

5.5

Data availability statement

All data that support the findings of this study are included within the article.

5.6

Using of AI tools

The authors declare that they have not used Artificial Intelligence (AI) tools in the creation of this article.

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