# A Study of Some Properties of Fuzzy Laplace Transform with Their Applications in Solving the Second-Order Fuzzy Linear Partial Differential Equations

Authors: Elhassan Eljaoui; Said Melliani
Journal: Advances in Fuzzy Systems
 (2023)

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## Abstract

In this paper, several results and theorems about the high-order strongly generalized Hukuhara differentiability of function defined via the fuzzy Riemann improper integral (in the sense of Wu) have been established. Then, some properties dealing with the partial derivatives of fuzzy Laplace transform for a fuzzy function of two real variables have been proved. Afterwards, an algorithm of fuzzy Laplace transform for solving second-order fuzzy partial differential equations has been proposed. Finally, two numerical examples, including the heat equation under fuzzy initial conditions, have been studied to justify the efficiency of the algorithm.


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## Body

## 1. Introduction

Partial differential equations (PDEs) are extremely useful for the modeling of a variety of natural, physical, and biological phenomena. They have several engineering applications and intervene in many domains of science. Many researchers have investigated fuzzy differential equations (FDEs) and fuzzy partial differential equations (FPDEs) using the rigorous tool of the fuzzy Laplace transform. For example, Allahviranloo et al., Eljaoui et al., and Salahshour et al. have extended this method to solve different kinds of fuzzy differential problems: FDEs of first or second order, FPDEs, and fuzzy integral differential equations [1–5].In this vein, we have studied the improper fuzzy Riemann integrals by establishing some important results about the continuity and differentiability of a fuzzy improper integral depending on a given crisp parameter in [6]. These results have been then employed to prove some fuzzy Laplace transform properties, which we have used to solve linear FPDEs of first order, under generalized Hukuhara differentiability.For recent works about partial differential equations, their theory, and necessary materials, one can see [7–9] and the references therein.The main purpose of this article is to present a fuzzy Laplace transform method for solving FPDEs of second order. To achieve this goal, we begin by developing some results about high-order Hukuhara differentiability of a function defined by a fuzzy improper integral.The remainder of this work is organized as follows. Section2 is reserved for preliminaries and recalls some important results about the continuity and differentiability of fuzzy improper integrals that we will need in the sequel. In Section 3, the main results about the high-order differentiability of fuzzy improper integral are studied and new properties of fuzzy Laplace transform are proved. Then, in Section 4, the procedure for solving FPDEs of second order by the fuzzy Laplace transform is proposed. Section 5 deals with numerical examples. Section 6 is reserved for the discussion of the obtained results. In the last section, we present conclusion and further research topic.


## 2. Preliminaries

Now we recall some basic results which are useful through the rest of this paper.


### 2.1. Fuzzy Numbers and Functions

A fuzzy number is a functionu:ℝ⟶0,1 verifying the following four assumptions:(1)
u is normal, i.e., ∃t0∈ℝ for which ut0=1(2)
u is fuzzy convex(3)
u is upper semicontinuous(4)
The closure of its supportsuppu=t∈ℝ|ut>0 is compact (see [10])DenoteE as the space of all fuzzy numbers.For0<α≤1, let uα=t∈ℝ|ut≥α be the α-level set of u∈E. Then, uα is a nonempty compact interval of ℝ. For all u,u1,u2∈E,k∈ℝ,0≤α≤1, we have(1)u1+u2α=u1α+u2α,kuα=kuα.Define the Hausdorff distance onE by Du1,u2=sup0≤α≤1maxu¯1α−u¯2α,u¯1α−u¯2α.Then,E,D is a complete metric space (for more details, see [11]).Definition 1 (see [1]).
We define a fuzzy numberu in parametric form as a couple u¯,u¯ of mappings u¯α,u¯α, 0≤α≤1, verifying the following properties:(1)
u¯α is bounded increasing, left continuous in 0,1, and right continuous at 0(2)
u¯α is bounded decreasing, left continuous in 0,1, and right continuous at 0(3)
u¯α≤u¯α, for all 0≤α≤1
The length ofu∈E is level-wise given by lenu,α=u¯α−u¯α≥0.Theorem 1 (see [12]).
Letfx=f¯x,α,f¯x,α be a fuzzy function defined on a,∞. Suppose that for all α∈0,1, the maps f¯x,α,f¯x,α are integrable on a,b, and ∃K¯α,K¯α>0:∫abf¯x,αdx≤K¯α and ∫abf¯x,αdx≤K¯α, for every b≥a. Then, f is fuzzy Riemann integrable on a,∞,∫a∞fxdx∈E and we have(2)∫a∞fxdx=∫a∞f¯x,αdx,∫a∞f¯x,αdx.Foru1,u2∈E, if there exists an element u3 in E such that u1=u2+u3, then u3 is called the Hukuhara difference of u1 and u2, which we denote by u1⊖u2.Definition 2 (see [1]).
A mappingf:a,b⟶E is said to be strongly generalized differentiable at x∈a,b, if there exists f′x∈E such that for all h>0 very small, there exist the H-differences(i)
fx+h⊖fx; fx⊖fx−h and the limitslimh⟶0+fx+h⊖fx/h=limh⟶0+fx⊖fx−h/h=f′x(ii)
fx⊖fx+h; fx−h⊖fx and the limitslimh⟶0+fx⊖fx+h/−h=limh⟶0+fx−h⊖fx/−h=f′x(iii)
fx+h⊖fx; fx−h⊖fx and the limitslimh⟶0+fx+h⊖fx/h=limh⟶0+fx−h⊖fx/−h=f′x(iv)
fx⊖fx+h;fx⊖fx−h and the limitslimh⟶0+fx⊖fx+h/−h=limh⟶0+fx⊖fx−h/h=f′xTheorem 2 (see [10]).
Iff:a,b⟶E is a strongly generalized differentiable function on a,b in the sense of Definition 2, (iii) or (iv), then f′x∈ℝ, for each x∈a,b.So, we can consider only Case (i) or (ii) of Definition2 almost everywhere in a,b.Theorem 3 (see [13]).
Letfx=f¯x,α,f¯x,α be a fuzzy strongly generalized differentiable function on a,b; then, f¯x,α and f¯x,α are differentiable. Moreover,(1)
Iff is (i)-differentiable, then f′x=f¯′x,α,f¯′x,α.(2)
Iff is (ii)-differentiable, then f′x=f¯′x,α,f¯′x,α.Definition 3.
We say that a mappingf:a,b⟶E is strongly generalized differentiable of the n-th order at x0∈a,b if there exists fkx0∈E, for all k∈1,2,…,n, such that for all h>0 very small, there exist the H-differences(i)
fk−1x0+h⊖fk−1x0,fk−1x0⊖fk−1x0−h and the limitslimh⟶0+fk−1x0+h⊖fk−1x0/h=limh⟶0+fk−1x0⊖fk−1x0−h/h=fkx0(ii)
fk−1x0⊖fk−1x0+h,fk−1x0−h⊖fk−1x0 and the limitslimh⟶0+fk−1x0⊖fk−1x0+h/−h=limh⟶0+fk−1x0−h⊖fk−1x0/−h=fkx0(iii)
fk−1x0+h⊖fk−1x0,fk−1x0−h⊖fk−1x0 and the limitslimh⟶0+fk−1x0+h⊖fk−1x0/h=limh⟶0+fk−1x0−h⊖fk−1x0/−h=fkx0(iv)
fk−1x0⊖fk−1x0+h,fk−1x0⊖fk−1x0−h and the limitslimh⟶0+fk−1x0⊖fk−1x0+h/−h=limh⟶0+fk−1x0⊖fk−1x0−h/h=fkx0Definition 4 (see [1]).
Iff:0,∞⟶E is a continuous mapping such that e−sxfx is fuzzy Riemann integrable on 0,∞, then Lfx=∫0∞e−sxfxdx is called the fuzzy Laplace transform of f. Notice that Lfx=ℒf¯x,α,ℒf¯x,α, where ℒkx is the Laplace transform of a crisp function kx.Theorem 4 (see [1, 2]).
Letf:0,∞⟶E be a fuzzy-valued function and f′ and f″ be its derivatives on 0,∞. If f is (i)-differentiable, then Lf′x=sLfx⊖f0, or if f is (ii)-differentiable, then Lf′x=−f0⊖−sLfx.Moreover, iff and f′ are (i)-differentiable, then Lf″x=p2Lfx⊖pf0⊖f′0. If f is (i)-differentiable and f′ is (ii)-differentiable, then(3)Lf″x=−f′0⊖−p2Lfx⊖−pf0.Iff is (ii)-differentiable and f′ is (i)-differentiable, then(4)Lf″x=−pf0⊖−p2Lfx⊖f′0.Iff and f′ are (ii)-differentiable, then Lf′′x=−f′0⊖pf0⊖p2Lfx.


### 2.2. Continuity and Differentiability of Fuzzy Improper Integral

In the sequel,I denotes one of the intervals −∞,b or b,∞ or −∞,∞, where b∈ℝ, J denotes another interval, and A is a nonempty subset of ℝ.Let us recall the properties of continuity and differentiability of a function defined by a fuzzy improper integral that we had established and proved in [6].Theorem 5 (see [6]).
LetFx,t:A×I⟶E satisfying the following conditions:(i)
H1 For all x∈A,t⟼Fx,t is continuous on I(ii)
H2 For each t∈I,x⟼Fx,t is continuous on A⊂ℝ(iii)
H3 For all α∈0,1, there exist a couple of nonnegative, continuous crisp functions φαt and ψαt, which are integrable on I verifying, for all x∈A,t∈I:(5)F¯x,t,α≤φαtandF¯x,t,α≤ψαt.Therefore, the fuzzy mappingϕx=∫IFx,tdt is continuous on A.Theorem 6 (see [6]).
LetFx,t:J×I⟶E verifying the following assumptions:(i)
A1 For all x∈J,t⟼Fx,t is continuous and fuzzy Riemann integrable on I(ii)
A2 For all t∈I,x⟼Fx,t is (i)-differentiable on the interval J(iii)
A3 For all x∈J,t⟼∂F/∂xx,t is continuous on I(iv)
A4 For all t∈I,x⟼∂F/∂xx,t is continuous on J(v)
A5 For all α∈0,1, there exist a couple of continuous crisp functions φαt and ψαt, which are integrable on I verifying, for all x∈J,t∈I:(6)∂F¯∂xx,t,α≤φαt,∂F¯∂xx,t,α≤ψαt.Therefore, the fuzzy mappingϕx=∫IFx,tdt is (i)-differentiable on J and(7)ϕ′x=∫I∂F∂xx,tdt,∀x∈J.Moreover, if we replace the assumptionA2 by the alternative condition(i)
A2′ For all t∈I,x⟼Fx,t is (ii)-differentiable on Jthen the fuzzy function ϕx is (ii)-differentiable on J and equation (7) remains true.Theorem 7 (see [6]).
Letuξ,τ:0,∞×0,∞⟶E be a fuzzy function such that Fξ,τ=e−sτuξ,τ satisfies the assumptions A1−A5 above, for all s≥s0>0.
LetLτuξ,τ or Luξ,τ (for short) denote the fuzzy Laplace transform of uξ,τ with respect to the time variable τ. Then,(8)Lτuξξ,τ=∂∂ξLτuξ,τ.Theorem 8 (see [6]).
Letuξ,τ be a fuzzy-valued function on 0,∞×0,∞ into E. Suppose that the mappings τ⟼Fξ,τ=e−sτuξ,τ and τ⟼Gξ,τ=e−sτuτξ,τ are fuzzy Riemann integrable on 0,∞, for all s≥s0 for some s0>0.(a)
Ifuξ,τ is (i)-differentiable with respect to τ, then(9)Lτuτξ,τ=sLτuξ,τ⊖uξ,0.(b)
Ifux,τ is (ii)-differentiable with respect to τ, then(10)Lτuτξ,τ=−uξ,0⊖−sLτuξ,τ.


## 2.1. Fuzzy Numbers and Functions

A fuzzy number is a functionu:ℝ⟶0,1 verifying the following four assumptions:(1)
u is normal, i.e., ∃t0∈ℝ for which ut0=1(2)
u is fuzzy convex(3)
u is upper semicontinuous(4)
The closure of its supportsuppu=t∈ℝ|ut>0 is compact (see [10])DenoteE as the space of all fuzzy numbers.For0<α≤1, let uα=t∈ℝ|ut≥α be the α-level set of u∈E. Then, uα is a nonempty compact interval of ℝ. For all u,u1,u2∈E,k∈ℝ,0≤α≤1, we have(1)u1+u2α=u1α+u2α,kuα=kuα.Define the Hausdorff distance onE by Du1,u2=sup0≤α≤1maxu¯1α−u¯2α,u¯1α−u¯2α.Then,E,D is a complete metric space (for more details, see [11]).Definition 1 (see [1]).
We define a fuzzy numberu in parametric form as a couple u¯,u¯ of mappings u¯α,u¯α, 0≤α≤1, verifying the following properties:(1)
u¯α is bounded increasing, left continuous in 0,1, and right continuous at 0(2)
u¯α is bounded decreasing, left continuous in 0,1, and right continuous at 0(3)
u¯α≤u¯α, for all 0≤α≤1
The length ofu∈E is level-wise given by lenu,α=u¯α−u¯α≥0.Theorem 1 (see [12]).
Letfx=f¯x,α,f¯x,α be a fuzzy function defined on a,∞. Suppose that for all α∈0,1, the maps f¯x,α,f¯x,α are integrable on a,b, and ∃K¯α,K¯α>0:∫abf¯x,αdx≤K¯α and ∫abf¯x,αdx≤K¯α, for every b≥a. Then, f is fuzzy Riemann integrable on a,∞,∫a∞fxdx∈E and we have(2)∫a∞fxdx=∫a∞f¯x,αdx,∫a∞f¯x,αdx.Foru1,u2∈E, if there exists an element u3 in E such that u1=u2+u3, then u3 is called the Hukuhara difference of u1 and u2, which we denote by u1⊖u2.Definition 2 (see [1]).
A mappingf:a,b⟶E is said to be strongly generalized differentiable at x∈a,b, if there exists f′x∈E such that for all h>0 very small, there exist the H-differences(i)
fx+h⊖fx; fx⊖fx−h and the limitslimh⟶0+fx+h⊖fx/h=limh⟶0+fx⊖fx−h/h=f′x(ii)
fx⊖fx+h; fx−h⊖fx and the limitslimh⟶0+fx⊖fx+h/−h=limh⟶0+fx−h⊖fx/−h=f′x(iii)
fx+h⊖fx; fx−h⊖fx and the limitslimh⟶0+fx+h⊖fx/h=limh⟶0+fx−h⊖fx/−h=f′x(iv)
fx⊖fx+h;fx⊖fx−h and the limitslimh⟶0+fx⊖fx+h/−h=limh⟶0+fx⊖fx−h/h=f′xTheorem 2 (see [10]).
Iff:a,b⟶E is a strongly generalized differentiable function on a,b in the sense of Definition 2, (iii) or (iv), then f′x∈ℝ, for each x∈a,b.So, we can consider only Case (i) or (ii) of Definition2 almost everywhere in a,b.Theorem 3 (see [13]).
Letfx=f¯x,α,f¯x,α be a fuzzy strongly generalized differentiable function on a,b; then, f¯x,α and f¯x,α are differentiable. Moreover,(1)
Iff is (i)-differentiable, then f′x=f¯′x,α,f¯′x,α.(2)
Iff is (ii)-differentiable, then f′x=f¯′x,α,f¯′x,α.Definition 3.
We say that a mappingf:a,b⟶E is strongly generalized differentiable of the n-th order at x0∈a,b if there exists fkx0∈E, for all k∈1,2,…,n, such that for all h>0 very small, there exist the H-differences(i)
fk−1x0+h⊖fk−1x0,fk−1x0⊖fk−1x0−h and the limitslimh⟶0+fk−1x0+h⊖fk−1x0/h=limh⟶0+fk−1x0⊖fk−1x0−h/h=fkx0(ii)
fk−1x0⊖fk−1x0+h,fk−1x0−h⊖fk−1x0 and the limitslimh⟶0+fk−1x0⊖fk−1x0+h/−h=limh⟶0+fk−1x0−h⊖fk−1x0/−h=fkx0(iii)
fk−1x0+h⊖fk−1x0,fk−1x0−h⊖fk−1x0 and the limitslimh⟶0+fk−1x0+h⊖fk−1x0/h=limh⟶0+fk−1x0−h⊖fk−1x0/−h=fkx0(iv)
fk−1x0⊖fk−1x0+h,fk−1x0⊖fk−1x0−h and the limitslimh⟶0+fk−1x0⊖fk−1x0+h/−h=limh⟶0+fk−1x0⊖fk−1x0−h/h=fkx0Definition 4 (see [1]).
Iff:0,∞⟶E is a continuous mapping such that e−sxfx is fuzzy Riemann integrable on 0,∞, then Lfx=∫0∞e−sxfxdx is called the fuzzy Laplace transform of f. Notice that Lfx=ℒf¯x,α,ℒf¯x,α, where ℒkx is the Laplace transform of a crisp function kx.Theorem 4 (see [1, 2]).
Letf:0,∞⟶E be a fuzzy-valued function and f′ and f″ be its derivatives on 0,∞. If f is (i)-differentiable, then Lf′x=sLfx⊖f0, or if f is (ii)-differentiable, then Lf′x=−f0⊖−sLfx.Moreover, iff and f′ are (i)-differentiable, then Lf″x=p2Lfx⊖pf0⊖f′0. If f is (i)-differentiable and f′ is (ii)-differentiable, then(3)Lf″x=−f′0⊖−p2Lfx⊖−pf0.Iff is (ii)-differentiable and f′ is (i)-differentiable, then(4)Lf″x=−pf0⊖−p2Lfx⊖f′0.Iff and f′ are (ii)-differentiable, then Lf′′x=−f′0⊖pf0⊖p2Lfx.


## 2.2. Continuity and Differentiability of Fuzzy Improper Integral

In the sequel,I denotes one of the intervals −∞,b or b,∞ or −∞,∞, where b∈ℝ, J denotes another interval, and A is a nonempty subset of ℝ.Let us recall the properties of continuity and differentiability of a function defined by a fuzzy improper integral that we had established and proved in [6].Theorem 5 (see [6]).
LetFx,t:A×I⟶E satisfying the following conditions:(i)
H1 For all x∈A,t⟼Fx,t is continuous on I(ii)
H2 For each t∈I,x⟼Fx,t is continuous on A⊂ℝ(iii)
H3 For all α∈0,1, there exist a couple of nonnegative, continuous crisp functions φαt and ψαt, which are integrable on I verifying, for all x∈A,t∈I:(5)F¯x,t,α≤φαtandF¯x,t,α≤ψαt.Therefore, the fuzzy mappingϕx=∫IFx,tdt is continuous on A.Theorem 6 (see [6]).
LetFx,t:J×I⟶E verifying the following assumptions:(i)
A1 For all x∈J,t⟼Fx,t is continuous and fuzzy Riemann integrable on I(ii)
A2 For all t∈I,x⟼Fx,t is (i)-differentiable on the interval J(iii)
A3 For all x∈J,t⟼∂F/∂xx,t is continuous on I(iv)
A4 For all t∈I,x⟼∂F/∂xx,t is continuous on J(v)
A5 For all α∈0,1, there exist a couple of continuous crisp functions φαt and ψαt, which are integrable on I verifying, for all x∈J,t∈I:(6)∂F¯∂xx,t,α≤φαt,∂F¯∂xx,t,α≤ψαt.Therefore, the fuzzy mappingϕx=∫IFx,tdt is (i)-differentiable on J and(7)ϕ′x=∫I∂F∂xx,tdt,∀x∈J.Moreover, if we replace the assumptionA2 by the alternative condition(i)
A2′ For all t∈I,x⟼Fx,t is (ii)-differentiable on Jthen the fuzzy function ϕx is (ii)-differentiable on J and equation (7) remains true.Theorem 7 (see [6]).
Letuξ,τ:0,∞×0,∞⟶E be a fuzzy function such that Fξ,τ=e−sτuξ,τ satisfies the assumptions A1−A5 above, for all s≥s0>0.
LetLτuξ,τ or Luξ,τ (for short) denote the fuzzy Laplace transform of uξ,τ with respect to the time variable τ. Then,(8)Lτuξξ,τ=∂∂ξLτuξ,τ.Theorem 8 (see [6]).
Letuξ,τ be a fuzzy-valued function on 0,∞×0,∞ into E. Suppose that the mappings τ⟼Fξ,τ=e−sτuξ,τ and τ⟼Gξ,τ=e−sτuτξ,τ are fuzzy Riemann integrable on 0,∞, for all s≥s0 for some s0>0.(a)
Ifuξ,τ is (i)-differentiable with respect to τ, then(9)Lτuτξ,τ=sLτuξ,τ⊖uξ,0.(b)
Ifux,τ is (ii)-differentiable with respect to τ, then(10)Lτuτξ,τ=−uξ,0⊖−sLτuξ,τ.


## 3. High-Order Differentiability of Fuzzy Improper Integral

Theorem 9.
We consider a fuzzy-valued functionFx,t:J×I⟶E, verifying the following assumptions:(i)
B1 For all x∈J,t⟼Fx,t is continuous and fuzzy Riemann integrable on I(ii)
B2 For all t∈I,x⟼Fx,t is strongly generalized differentiable of the second order on the interval J(iii)
B3 For all x∈J,t⟼∂F/∂xx,t and t⟼∂2F/∂x2x,t are continuous on I(iv)
B4 For all t∈I,x⟼∂F/∂xx,t and x⟼∂2F/∂x2x,t are continuous on J(v)
B5 For all α∈0,1, there exist four continuous crisp functions φαt,ψαt,ηαt, and θαt, which are integrable on I verifying, for all x∈J,t∈I:(11)∂F¯∂xx,t,α≤φαt,∂F¯∂xx,t,α≤ψαt,∂2F¯∂x2x,t,α≤ηαt,∂2F¯∂x2x,t,α≤θαt.Therefore, the fuzzy mappingϕx=∫IFx,tdt is strongly generalized differentiable of the second order on J and(12)ϕ′′x=∫I∂2F∂x2x,tdt,∀x∈J.Proof.
Using Theorem6 and since F verifies assumptions A1−A5, then the fuzzy mapping ϕx is strongly generalized differentiable on J and(13)ϕ′x=∫I∂F∂xx,tdt,∀x∈J.
Then, from the assumptionsB1−B5, the function ∂F/∂x satisfies the conditions A1−A5. Hence, and by Theorem 6, ϕ′x is strongly differentiable on J and for all x∈J, we have(14)ϕ′′x=ϕ′′x=∫I∂∂x∂F∂xx,tdt=∫I∂2F∂x2x,tdt.Theorem 10.
Letu:0,∞×0,∞⟶E be a fuzzy function. Suppose that the mapping Fx,t=e−stux,t satisfies the assumptions B1−B5 above, for all s≥s0 for some s0>0. Let Ltux,t or Lux,t (for short) denote the fuzzy Laplace transform of ux,t with respect to the time variable t. Then,(15)Lt∂2u∂x2x,t=∂2∂x2Ltux,t.Proof.
For fixeds≥s0, then using Theorem 9, we have(16)Lt∂2u∂x2x,t=∫0∞e−st∂2u∂x2x,tdt=∫0∞∂2F∂x2x,tdt,Lt∂2u∂x2x,t=∂2∂x2∫0∞Fx,tdt=∂2∂x2Ltux,t.Theorem 11.
Letux,t be a fuzzy-valued function on 0,∞×0,∞ into E. Suppose that the mappings t⟼Fx,t=e−stux,t,t⟼Gx,t=e−stutx,t, and t⟼Hx,t=e−stuttx,t are fuzzy Riemann integrable on 0,∞, for all s≥s0 for some s0>0.(a)
Ifux,t and utx,t are (i)-differentiable with respect to t, thenLt∂2ux,t/∂t2=s2Ltux,t⊖sux,0⊖utx,0.(b)
Ifux,t is (i)-differentiable and utx,t is (ii)-differentiable with respect to t, thenLt∂2ux,t/∂t2=−utx,0⊖−s2Ltux,t⊖−sux,0.(c)
Ifux,t is (ii)-differentiable and utx,t is (i)-differentiable with respect to t, thenLt∂2ux,t/∂t2=−sux,0⊖−s2Ltux,t=⊖utx,0.(d)
Ifux,t and utx,t are (ii)-differentiable with respect to t, thenLt∂2u/∂t2x,t=−utx,0⊖sux,0⊖s2Ltux,t.Proof.
This obviously results from Theorem4, by fixing x≥0 and taking the Laplace transforms and derivations with respect to the variable t.Theorem 12.
We consider a fuzzy-valued functionFx,t:J×I⟶E, verifying the following assumptions:(i)
H1 For all x∈J,t⟼Fx,t is continuous and fuzzy Riemann integrable on I(ii)
H2 For all t∈I,x⟼Fx,t is strongly generalized differentiable of the n-th order on J(iii)
H3 For all k∈0,1,…,n and for each x∈J,t⟼∂kF/∂xkx,t is continuous on I(iv)
H4 For all k∈0,1,…,n and for each t∈I,x⟼∂kF/∂xkx,t is continuous on J(v)
H5 For all α∈0,1 and for all k∈0,1,…,n, there exist a couple of continuous and integrable crisp functions φα,kt,ψα,kt on I verifying, for all x∈J,t∈I:(17)∂kF¯∂xkx,t,α≤φα,kt,∂kF¯∂xkx,t,α≤ψα,kt.Therefore, the fuzzy mappingϕx=∫IFx,tdt is strongly generalized differentiable of the n-th order on J and we have(18)ϕnx=∫I∂nF∂xnx,tdt,∀x∈J.Proof.
According to Theorem6, the result holds true for n=1. Also, by induction, assume that the result is true to the n−1-th order. In addition, let a function Fx,t:J×I⟶E, satisfying the conditions H1−H5. Then, ϕx=∫IFx,tdt is strongly generalized differentiable of the n-th order on J and(19)ϕn−1x=∫I∂n−1F∂xn−1x,tdt,∀x∈J.
From the assumptionsH1−H5 and using Theorem 6 and since ∂n−1F/∂xn−1 verifies assumptions A1−A5, then the fuzzy mapping ϕn−1x is strongly generalized differentiable on J, that is, ϕx is strongly generalized differentiable to the n-th order on J and we have(20)ϕnx=ddxϕn−1x=∫I∂∂x∂n−1F∂xn−1x,tdt=∫I∂nF∂xnx,tdt,∀x∈J.


## 4. Fuzzy Laplace Transform for Second-Order Fuzzy Linear Partial Differential Equations

Our aim now is to solve the following second-order linear FPDE using the fuzzy Laplace transform method:(21)uxxx,t+axuxtx,t+bxuttx,t+c1xuxx,t,+c2xutx,t+dxux,t=fx,t,x≥0,t≥0,ux,0=gx,u0,t=ht,andutx,0=kx,where ux,t is a fuzzy strongly differentiable function of second order, with continuous partial derivatives. a,b,c1,c2, and d are real continuous functions, and fx,t,gx,ht, and kx are continuous fuzzy functions. Without loss of generality, assume that the mappings a,b,c1,c2 are all positive.


### 4.1. Resolution of Equation (21) by Fuzzy Laplace Transform Method

By using fuzzy Laplace transform with respect tot, we get(22)∂2Lt∂x2ux,t+axs∂Lt∂xux,t−axg′x+bxs2Ltux,t−bxsgx−bxkx+c1x∂Lt∂xuxx,t+c2xsLtutx,t−c2xgx+dxLtux,t=Ltfx,t.Therefore, we have to distinguish the following 32 cases for solving this last equation.(a)
Ifu is (i)-differentiable with respect to x and t, ux is (i)-differentiable with respect to x and t, and ut is (i)-differentiable with respect to t, then using Theorems 10 and 11, we get(23)∂2ℒt∂x2u¯x,t,α+axs∂ℒt∂xu¯x,t,α−axg¯′x,α+bxs2ℒtu¯x,t,α−bxsg¯x,α+c1x∂ℒt∂xu¯x,t,α+c2xsℒtu¯tx,t,α−c2xg¯x,α+dxℒtu¯x,t,α=ℒtf¯x,t,α,∂2ℒt∂x2u¯x,t,α+axs∂ℒt∂xu¯x,t,α−axg¯′x,α+bxs2ℒtu¯x,t,α−bxsg¯x,α−bxk¯x,α+c1x∂ℒt∂xu¯x,t,α+c2xsℒtu¯tx,t,α−c2xg¯x,α+dxℒtu¯x,t,α=ℒtf¯x,t,α.DenoteUx,s=Ltux,t and Fx,s=Ltfx,t. Then,(24)∂2U¯∂x2x,s,α+axs+c1x∂U¯∂xx,s,α+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α,∂2U¯∂x2x,s,α+axs+c1x∂U¯∂xx,s,α+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α,satisfying the following initial conditions:(25)U¯0,s,α=ℒu¯0,t,α=ℒh¯t,α,U¯0,s,α=ℒu¯0,t,α=ℒh¯t,α.Assume that this leads to(26)ℒu¯x,t,α=H1x,s,α,ℒu¯x,t,α=K1x,s,α,whereH1x,s,α,K1x,s,α is the solution of system (24) under (25).By the inverse Laplace transform, we obtain(27)u¯x,t,α=ℒ−1H1x,s,α,u¯x,t,α=ℒ−1K1x,s,α.(b)
Ifu is (i)-differentiable with respect to x and t, ux is (i)-differentiable with respect to t and (ii)-differentiable with respect to x, and ut is (i)-differentiable with respect to t, then we obtain(28)∂2ℒt∂x2u¯x,t,α+axs∂ℒt∂xℒtu¯x,t,α−axg¯′x,α+bxs2ℒtu¯x,t,α−bxk¯x,α−bxsg¯x,α+c1x∂ℒt∂xu¯x,t,α+c2xsℒtu¯tx,t,α−c2xg¯x,α+dxℒtu¯x,t,α=ℒtf¯x,t,α,∂2ℒt∂x2u¯x,t,α+axs∂ℒt∂xu¯x,t,α−axg¯′x,α+bxs2ℒtu¯x,t,α−bxk¯x,α−bxsg¯x,α+c1x∂ℒt∂xu¯x,t,α+c2xsℒtu¯tx,t,α−c2xg¯x,α+dxℒtu¯x,t,α=ℒtf¯x,t,α.Thus,(29)∂2U¯x,s,α∂x2+axs+c1x∂U¯x,s,α∂x+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α,∂2U¯x,s,α∂x2+axs+c1x∂U¯x,s,α∂x+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α.Assume that this leads to(30)ℒu¯x,t,α=H2x,s,α,ℒu¯x,t,α=K2x,t,α,whereH2x,s,α,K2x,s,α is the solution of system (29) under (25).By the inverse Laplace transform, we get(31)u¯x,s,α=ℒ−1H2x,s,α,u¯x,s,α=ℒ−1K2x,s,α.(c)
Ifu is (i)-differentiable with respect to x and (ii)-differentiable with respect to t, ux is (i)-differentiable with respect to x and t, and ut is (i)-differentiable with respect to t, then(32)∂2U¯x,s,α∂x2+axs∂U¯x,s,α∂x+c1x∂U¯x,s,α∂x+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α,∂2U¯x,s,α∂x2+axs∂U¯x,s,α∂x+c1x∂U¯x,s,α∂x+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α.Assume that this leads to(33)ℒu¯x,t,α=H3x,s,α,ℒu¯x,t,α=aK3x,s,α,whereH3x,s,α,K3x,s,α is the solution of system (32) under (25).By the inverse Laplace transform, we obtain(34)u¯x,t,α=ℒ−1H3x,s,α,u¯x,t,α=ℒ−1K3x,s,α.(d)
For thep-th case from the 29 remaining cases, with p∈4,5,…,32, we get a differential system similar to one of the previous systems (24) and (29).Assume that this leads to(35)ℒu¯x,t,α=Hpx,s,α,ℒu¯x,t,α=Kpx,s,α,where Hpx,s,α,Kpx,s,α is the solution of the latter system under initial condition (25).By the inverse Laplace transform, we have(36)u¯x,t,α=ℒ−1Hpx,s,α,u¯x,t,α=ℒ−1Kpx,s,α.


### 4.2. Algorithm of Fuzzy Laplace Transform

The steps of the proposed algorithm are as follows:(i)
Choose a case from the 32 possible ones according to the differentiability’s type of each from the functionsu,ux,ux, and ut with respect to x and t, respectively.(ii)
Replace these functions by their parametric forms to transform equation (21) into an equivalent classical differential system of two linear equations with unknown u¯x,t,α,u¯x,t,α, for 0≤α≤1.(iii)
Solve this differential system under the given initial and boundary conditions.(iv)
Calculate the length of the mappingsu,ux,ux,ut,uxx,uxt, and utt.(v)
Deduce the domain of definition for the solutionux,t,α using the nonnegativity of calculated lengths.


## 4.1. Resolution of Equation (21) by Fuzzy Laplace Transform Method

By using fuzzy Laplace transform with respect tot, we get(22)∂2Lt∂x2ux,t+axs∂Lt∂xux,t−axg′x+bxs2Ltux,t−bxsgx−bxkx+c1x∂Lt∂xuxx,t+c2xsLtutx,t−c2xgx+dxLtux,t=Ltfx,t.Therefore, we have to distinguish the following 32 cases for solving this last equation.(a)
Ifu is (i)-differentiable with respect to x and t, ux is (i)-differentiable with respect to x and t, and ut is (i)-differentiable with respect to t, then using Theorems 10 and 11, we get(23)∂2ℒt∂x2u¯x,t,α+axs∂ℒt∂xu¯x,t,α−axg¯′x,α+bxs2ℒtu¯x,t,α−bxsg¯x,α+c1x∂ℒt∂xu¯x,t,α+c2xsℒtu¯tx,t,α−c2xg¯x,α+dxℒtu¯x,t,α=ℒtf¯x,t,α,∂2ℒt∂x2u¯x,t,α+axs∂ℒt∂xu¯x,t,α−axg¯′x,α+bxs2ℒtu¯x,t,α−bxsg¯x,α−bxk¯x,α+c1x∂ℒt∂xu¯x,t,α+c2xsℒtu¯tx,t,α−c2xg¯x,α+dxℒtu¯x,t,α=ℒtf¯x,t,α.DenoteUx,s=Ltux,t and Fx,s=Ltfx,t. Then,(24)∂2U¯∂x2x,s,α+axs+c1x∂U¯∂xx,s,α+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α,∂2U¯∂x2x,s,α+axs+c1x∂U¯∂xx,s,α+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α,satisfying the following initial conditions:(25)U¯0,s,α=ℒu¯0,t,α=ℒh¯t,α,U¯0,s,α=ℒu¯0,t,α=ℒh¯t,α.Assume that this leads to(26)ℒu¯x,t,α=H1x,s,α,ℒu¯x,t,α=K1x,s,α,whereH1x,s,α,K1x,s,α is the solution of system (24) under (25).By the inverse Laplace transform, we obtain(27)u¯x,t,α=ℒ−1H1x,s,α,u¯x,t,α=ℒ−1K1x,s,α.(b)
Ifu is (i)-differentiable with respect to x and t, ux is (i)-differentiable with respect to t and (ii)-differentiable with respect to x, and ut is (i)-differentiable with respect to t, then we obtain(28)∂2ℒt∂x2u¯x,t,α+axs∂ℒt∂xℒtu¯x,t,α−axg¯′x,α+bxs2ℒtu¯x,t,α−bxk¯x,α−bxsg¯x,α+c1x∂ℒt∂xu¯x,t,α+c2xsℒtu¯tx,t,α−c2xg¯x,α+dxℒtu¯x,t,α=ℒtf¯x,t,α,∂2ℒt∂x2u¯x,t,α+axs∂ℒt∂xu¯x,t,α−axg¯′x,α+bxs2ℒtu¯x,t,α−bxk¯x,α−bxsg¯x,α+c1x∂ℒt∂xu¯x,t,α+c2xsℒtu¯tx,t,α−c2xg¯x,α+dxℒtu¯x,t,α=ℒtf¯x,t,α.Thus,(29)∂2U¯x,s,α∂x2+axs+c1x∂U¯x,s,α∂x+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α,∂2U¯x,s,α∂x2+axs+c1x∂U¯x,s,α∂x+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α.Assume that this leads to(30)ℒu¯x,t,α=H2x,s,α,ℒu¯x,t,α=K2x,t,α,whereH2x,s,α,K2x,s,α is the solution of system (29) under (25).By the inverse Laplace transform, we get(31)u¯x,s,α=ℒ−1H2x,s,α,u¯x,s,α=ℒ−1K2x,s,α.(c)
Ifu is (i)-differentiable with respect to x and (ii)-differentiable with respect to t, ux is (i)-differentiable with respect to x and t, and ut is (i)-differentiable with respect to t, then(32)∂2U¯x,s,α∂x2+axs∂U¯x,s,α∂x+c1x∂U¯x,s,α∂x+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α,∂2U¯x,s,α∂x2+axs∂U¯x,s,α∂x+c1x∂U¯x,s,α∂x+bxs2+c2xs+dxU¯x,s,α=axg¯′x,α+bxk¯x,α+bxs+c2xg¯x,α+F¯x,s,α.Assume that this leads to(33)ℒu¯x,t,α=H3x,s,α,ℒu¯x,t,α=aK3x,s,α,whereH3x,s,α,K3x,s,α is the solution of system (32) under (25).By the inverse Laplace transform, we obtain(34)u¯x,t,α=ℒ−1H3x,s,α,u¯x,t,α=ℒ−1K3x,s,α.(d)
For thep-th case from the 29 remaining cases, with p∈4,5,…,32, we get a differential system similar to one of the previous systems (24) and (29).Assume that this leads to(35)ℒu¯x,t,α=Hpx,s,α,ℒu¯x,t,α=Kpx,s,α,where Hpx,s,α,Kpx,s,α is the solution of the latter system under initial condition (25).By the inverse Laplace transform, we have(36)u¯x,t,α=ℒ−1Hpx,s,α,u¯x,t,α=ℒ−1Kpx,s,α.


## 4.2. Algorithm of Fuzzy Laplace Transform

The steps of the proposed algorithm are as follows:(i)
Choose a case from the 32 possible ones according to the differentiability’s type of each from the functionsu,ux,ux, and ut with respect to x and t, respectively.(ii)
Replace these functions by their parametric forms to transform equation (21) into an equivalent classical differential system of two linear equations with unknown u¯x,t,α,u¯x,t,α, for 0≤α≤1.(iii)
Solve this differential system under the given initial and boundary conditions.(iv)
Calculate the length of the mappingsu,ux,ux,ut,uxx,uxt, and utt.(v)
Deduce the domain of definition for the solutionux,t,α using the nonnegativity of calculated lengths.


## 5. Numerical Examples

Example 1.
We consider the heat equation with fuzzy initial and boundary conditions:(37)utx,t=auxxx,t,u0,t,α=uπ,t,α=0,0,α∈0,1,ux,0,α=sinx.α,2−α,x≥0,t≥0,where a is a positive real number. Here, we have to distinguish only 8 cases.(a)
Ifu is (i)-differentiable with respect to x and t and ux is (i)-differentiable with respect to x, then(38)sℒtu¯x,t,α−u¯x,0,α=a∂2ℒtu¯x,t,α∂x2,sℒtu¯x,t,α−u¯x,0,α=a∂2ℒtu¯x,t,α∂x2.Using the conditionsu¯x,0,α=αsinx,u¯x,0,α=2−αsinx leads to(39)sℒtu¯x,t,α=a∂2∂x2ℒtu¯x,t,α+αsinx,sℒtu¯x,t,α=a∂2∂x2ℒtu¯x,t,α+2−αsinx.Solving (39), we get(40)ℒtu¯x,t,α=A1se−s/ax+B1ses/ax+αsinxs+a,ℒtu¯x,t,α=C1se−s/ax+D1ses/ax+2−αsinxs+a.Sinceℒtu¯0,t,α=ℒtu¯π,t,α=ℒtu¯π,t,α=ℒtu¯0,t,α=0, thenA1s=B1s=C1s=D1s=0. Thus,(41)ℒtu¯x,t,α=αsinxs+a,ℒtu¯x,t,α=2−αsinxs+a.By the inverse Laplace transform, we deduce(42)u¯x,t,α=αe−atsinx,u¯x,t,α=2−αe−atsinx.The lengths ofu,ux,ut, and uxx are, respectively, given by(43)lenux,t,α=21−αe−atsinx,lenuxx,t,α=21−αe−atcosx,lenutx,t,α=−2a1−αe−atsinx,lenuxxx,t,α=−21−αe−atsinx.Hence, this solution is invalid becauselenux,t,α×lenuxxx,t,α≤0.(b)
Ifu is (i)-differentiable with respect to x and t and ux is (ii)-differentiable with respect to x, then(44)sℒtu¯x,t,α−αsinx=a∂2ℒtu¯x,t,α∂x2,sℒtu¯x,t,α−2−αsinx=a∂2ℒtu¯x,t,α∂x2.Solving (44), we get(45)ℒtu¯x,t,α=A2se−sax+B2sesax+C2scoss/ax+D2ssins/ax+2s−sα−aαs2−a2sinx,ℒtu¯x,t,α=A2se−s/ax+B2ses/ax−C2scoss/ax−D2ssins/ax+sα−2a+aαs2−a2sinx.Sinceℒtu¯0,t,α=ℒtu¯0,t,α=ℒtu¯π,t,α=ℒtu¯π,t,α=0, thenA2s=B2s=C2s=D2s=0. Hence,(46)ℒtu¯x,t,α=sα−2a+aαs2−a2sinx,ℒtu¯x,t,α=2s−sα−aαs2−a2sinx.By the inverse Laplace transform, we obtain(47)u¯x,t,α=e−at+α−1eatsinx,u¯x,t,α=e−at+1−αeatsinx.The lengths ofu,ux,ut, and uxx are, respectively, given by(48)lenux,t,α=21−αeatsinx,lenuxx,t,α=21−αeatcosx,lenutx,t,α=2a1−αeatsinx,lenuxxx,t,α=21−αeatsinx.So, this solution (called solution 1) is valid over2kπ,π/2+2kπ×ℝ+, where k∈ℤ (see Figures 1–4).(c)
Ifu and ux are (i)-differentiable with respect to x and u is (ii)-differentiable with respect to t, then(49)sℒtu¯x,t,α=a∂2ℒtu¯x,t,α∂x2+αsinx,sℒtu¯x,t,α=x22∂2ℒtu¯x,t,α∂x2+2−αsinx.Solving (49), we obtain(50)ℒtu¯x,t,α=A3se−s/ax+B3ses/ax+C3scoss/ax+D3ssins/ax+sα−2a+aαs2−a2sinx,ℒtu¯x,t,α=A3se−s/ax+B3ses/ax−C3scoss/ax−D3ssins/ax+2s−sα−aαs2−a2sinx.Similarly and as in Case 2, we obtainA3s=B3s=C3s=D3s=0. So,(51)ℒtu¯x,t,α=sα−2a+aαs2−a2sinx,ℒtu¯x,t,α=2s−sα+αs2−a2sinx.By the inverse Laplace transform, we deduce(52)u¯x,t,α=e−at+α−1eatsinx,u¯x,t,α=e−at+1−αeatsinx.Hence, this solution is invalid becauselenux,t,α×lenuxxx,t,α≤0.(d)
Ifu is (i)-differentiable and ux is (ii)-differentiable with respect to x and u is (ii)-differentiable with respect to t, then(53)sℒtu¯x,t,α=a∂2ℒtu¯x,t,α∂x2+αsinx,sℒtu¯x,t,α=a∂2ℒtu¯x,t,α∂x2+2−αsinx.Solving (53), we get(54)ℒtu¯x,t,α=A4se−s/ax+B4ses/ax+αsinxs+a,ℒtu¯x,t,α=C4se−s/ax+D4ses/ax+2−αsinxs+a.As in Case 1, we getA4s=B4s=C4s=D4s=0.Also, by the inverse Laplace transform,(55)u¯x,t,α=αe−atsinx,u¯x,t,α=2−αe−atsinx.Thus, this solution (called solution 2) is valid over2kπ,π/2+2kπ, where k∈Z (see Figures 5–8).(e)
Ifu is (ii)-differentiable and ux is (i)-differentiable with respect to x and is u (i)-differentiable with respect to t, then(56)sℒtu¯x,t,α=a∂2∂x2ℒtu¯x,t,α+αsinx,sℒtu¯x,t,α=a∂2∂x2ℒtu¯x,t,α+2−αsinx.Solving (56), we get(57)ℒtu¯x,t,α=A5se−s/ax+B5ses/ax+C5scoss/ax+D5ssins/ax+sα−2a+aαs2−a2sinx,ℒtu¯x,t,α=A5se−s/ax+B5ses/ax−C5scoss/ax−D5ssins/ax+2s−sα−aαs2−a2sinx.As in Case 2, we getA5s=B5s=C5s=D5s=0.By the inverse Laplace transform,(58)u¯x,t,α=e−at+α−1eatsinx,u¯x,t,α=e−at+1−αeatsinx.Thus, this solution (called solution 2) is valid overπ/2+2kπ,π+2kπ, where k∈ℤ (see Figures 5–8).(f)
Ifu and ux are (ii)-differentiable with respect to x and u is (i)-differentiable with respect to t, then(59)sℒtu¯x,t,α=a∂2∂x2ℒtu¯x,t,α+αsinx,sℒtu¯x,t,α=a∂2∂x2ℒtu¯x,t,α+2−αsinx.Solving (59), we get(60)ℒtu¯x,t,α=A6se−s/ax+B6ses/ax+αsinxs+a,ℒtu¯x,t,α=C6se−s/ax+D6ses/ax.+2−αsinxs+aAs in Case 1, we haveA6s=B6s=C6s=D6s=0. Also, by the inverse Laplace transform,(61)u¯x,t,α=αe−atsinx,u¯x,t,α=2−αe−atsinx.Therefore, this solution is invalid becauselenux,t,α×lenuxxx,t,α≤0.(g)
Ifu is (ii)-differentiable and ux is (i)-differentiable with respect to x and u is (ii)-differentiable with respect to t, then(62)sℒtu¯x,t,α=a∂2ℒtu¯x,t,α∂x2+αsinx,sℒtu¯x,t,α=a∂2ℒtu¯x,t,α∂x2+2−αsinx.Solving (62), we get(63)ℒtu¯x,t,α=A7se−s/ax+B7ses/ax+αsinxs+a,ℒtu¯x,t,α=C7se−s/ax+D7ses/ax+2−αsinxs+a.As in Case 1, we getA7s=B7s=C7s=D7s=0.Also, by the inverse Laplace transform,(64)u¯x,t,α=αe−atsinx,u¯x,t,α=2−αe−atsinx.Hence, this solution (called solution 2) is valid overπ/2+2kπ,π+2kπ, where k∈ℤ (see Figures 5–8).(h)
Ifu and ux are (ii)-differentiable with respect to x and u is (ii)-differentiable with respect to t, then(65)sℒtu¯x,t,α=a∂2ℒtu¯x,t,α∂x2+αsinx,sℒtu¯x,t,α=a∂2ℒtu¯x,t,α∂x2+2−αsinx.Solving (65), we get(66)ℒtu¯x,t,α=A8se−s/ax+B8ses/ax+C8scoss/ax+D8ssins/ax+sα−2a+aαs2−a2sinx,ℒtu¯x,t,α=A8se−s/ax+B8ses/ax−C8scoss/ax−D8ssins/ax+2s−sα−aαs2−a2sinx.As in Case 2, we getA8s=B8s=C8s=D8s=0.Also, by the inverse Laplace transform,(67)u¯x,t,α=e−at+α−1eatsinx,u¯x,t,α=e−at+1−αeatsinx.
Therefore, this solution is invalid becauselenux,t,α×lenuxxx,t,α≤0.Figure 1
Lower partu¯x,t,α of solution 1 for equation (37), with α=0.5.Figure 2
Upper partu¯x,t,α of solution 1 for equation (37), with α=0.5.Figure 3
Lower partu¯x,t,α of solution 1 for equation (37), with α=0.7.Figure 4
Upper partu¯x,t,α of solution 1 for equation (37), with α=0.7.Figure 5
Lower partu¯x,t,α of solution 2 for equation (37), with α=0.5.Figure 6
Upper partu¯x,t,α of solution 2 for equation (37), with α=0.5.Figure 7
Lower partu¯x,t,α of solution 2 for equation (37), with α=0.7.Figure 8
Upper partu¯x,t,α of solution 2 for equation (37), with α=0.7.Remark 1.
Notice that in all cases, if we takeα=1, we find the crisp solution ux,t=e−atsinx of the corresponding classical heat equation(68)utx,t=auxxx,t,x≥0,t≥0,u0,t=uπ,t=0,ux,0=sinx.
For the graph of the crisp solution, see Figures9 and 10.Figure 9
Crisp solutionucx,t over π/2,π for equation (37).Figure 10
Crisp solutionucx,t over −π,π for equation (37).Example 2.
We consider the following FPDE:(69)uxtx,t=0˜=α−1,1−α,u0,t,α=t2,α∈0,1,ux,0,α=x2,x≥0,t≥0.
Analogously, we obtain the solution’s expression by distinguishing the following cases:(a)
Ifu (i)-differentiable with respect to t and ut is (i)-differentiable with respect to x, then(70)u¯x,t,α=α−1xt+x2+t2,u¯x,t,α=1−αxt+x2+t2.Therefore, this solution (called solution 3) is valid all overℝ+2 (see Figures 11–14).(b)
Ifu (i)-differentiable with respect to t and ut is (ii)-differentiable with respect to x, then(71)u¯x,t,α=1−αxt+x2+t2,u¯x,t,α=α−1xt+x2+t2.Thus, this solution is invalid becauselenu≤0.(c)
Ifu (ii)-differentiable with respect to t and ut is (i)-differentiable with respect to x, then(72)u¯x,t,α=1−αxt+x2+t2,u¯x,t,α=α−1xt+x2+t2.Also, this solution is invalid becauselenu≤0.(d)
Ifu (ii)-differentiable with respect to t and ut is (ii)-differentiable with respect to x, then(73)u¯x,t,α=α−1xt+x2+t2,u¯x,t,α=1−αxt+x2+t2.
So, this solution is invalid becauseu is not (ii)-differentiable with respect to t.Figure 11
Lower partu¯x,t,α of solution 3 for equation (69), with α=0.5.Figure 12
Upper partu¯x,t,α of solution 3 for equation (69), with α=0.5.Figure 13
Lower partu¯x,t,α of solution 3 for equation (69), with α=0.999999.Figure 14
Upper partu¯x,t,α of solution 3 for equation (69), with α=0.999999.Remark 2.
Notice that in all cases, if we takeα=1, we find the crisp solution ux,t=x2+t2 of the corresponding classical problem(74)uxtx,t=0,u0,t=t2,ux,0=x2,x≥0,t≥0.


## 6. Discussion

The Laplace transform method was used to compute the analytic solutionux,t,α of two linear FPDEs of second order. First, we decomposed each FPDE into a system of two crisp PDEs with the unknown u¯x,t,α,u¯x,t,α, for 0≤α≤1, for which we calculated the Laplace transforms using the properties proved in Section 3. Then, using the inverse Laplace transform, we obtained the lower and upper solution’s parts u¯x,t,α,u¯x,t,α, respectively. Finally, we determined the definition’s domain of these fuzzy solutions by utilizing the positivity of the length for ux,t,α and its partial derivatives uxx,t,α,utx,t,α,uxtx,t,α,uxxx,t,α, and uttx,t,α.In both numerical examples studied, the fuzzy solutionux,t,α of the second-order FPDE can be expressed as follows:(75)ux,t,α=ucx,t+uFx,t,α,where ucx,t is the crisp solution of the corresponding classical second-order PDE, obtained by letting α=1, and uFx,t,α is an undesirable term, which represents the fuzzy pure part of the fuzzy solution ux,t,α.This fuzzy pure componentuFx,t,α results from the modeling choices and steps using fuzzy tools and theory. It also measures the uncertainty and vagueness in the adopted model due to the imprecisions in the initial and boundary conditions or in the fuzzy (respectively, real) second member of the FPDE (respectively, PDE).On the one hand, we get for Example1:(76)u¯x,t,α=e−atsinx+α−1eatsinx,u¯x,t,α=e−atsinx+1−αeatsinx,that is,(77)ux,t,α=e−atsinx+eatsinx.α−1,1−α,for solution 1 of (37), given in its parametric form. Hence, we have(78)ucx,t=e−atsinx,uFx,t,α=eatsinx.α−1,1−α.On the other hand, we obtain(79)u¯x,t,α=e−atsinx+α−1e−atsinx,u¯x,t,α=e−atsinx+1−αe−atsinx,that is, ux,t,α=e−atsinx+e−atsinx.α−1,1−α, for solution 2 of (37), written in its parametric form. So, we have(80)ucx,t=e−atsinx,uFx,t,α=e−atsinx.α−1,1−α.The presence of the pure fuzzy parts in these two solutions of the fuzzy heat equation can be explained by some uncontrollable parameters or omitted inputs in the modeling of the crisp problem. So, the fuzzy approach is better and more efficient than the ordinary classical way.Furthermore, we have for Example2:(81)u¯x,t,α=x2+t2+α−1xt,u¯x,t,α=x2+t2+1−αxt,that is, ux,t,α=x2+t2+xt.α−1,1−α, for solution 3 of (69), given in its parametric form. Then, we have(82)ucx,t=x2+t2,uFx,t,α=α−1xt.In (37), the length of the fuzzy pure part lenuFx,t,α=21−αeatsinx goes to infinity as t⟶∞ for solution 2, while the length lenuFx,t,α=21−αe−atsinx converges to 0 as t⟶∞ for the fuzzy pure part of solution 1. Thus, the first solution is stable, whereas the second solution is unstable. In other words, solution 2 is most uncertain or vague than solution 1 for Example 1.In Example2, the crisp part of the solution 3 is ucx,t=x2+t2 and its fuzzy pure part is uFx,t,α=xt.α−1,1−α, for which the length lenuFx,t,α=21−αxt diverges to the infinity as x⟶∞ (respectively, t⟶∞). Hence, the uncertainty is increasing as the value of x or t increases, and this solution is unstable.In general, asα approaches 1, the fuzziness and uncertainty become smaller and completely disappear, for α=1, yielding the crisp solution of the classic real problem.Moreover, note that the existence, form, and asymptotic behavior of the solution depend on the choice of the kind of each used fuzzy partial derivative. Indeed, solution 2 of Example1 is unstable, and it is obtained if we assume that u is (i)-differentiable with respect to x and it is (ii)-differentiable with respect to t and ux is (ii)-differentiable with respect to x (see the fourth case (d)), while solution 1 is stable and is obtained provided that u is (i)-differentiable with respect to x and t and ux is (ii)-differentiable with respect to x (see the second case (b)). But, in the other cases, there is no valid fuzzy solution.In Example2, we note that solution 3 exists and is unstable, and it is valid only if we assume that u is (i)-differentiable with respect to t and ut is (ii)-differentiable with respect to x.Furthermore, the uniqueness of the solution is lost in Example1 (we have two (solutions 1 and 2)), although this unicity is preserved in Example 2.Consequently, the study of these examples demonstrates that the type of fuzzy partial derivatives used influences the existence, uniqueness, and stability of the fuzzy solution(s) for a second-order FPDE under the strong generalized differentiability assumption.Now, we proceed to the graphic interpretation of Figures5–8. Firstly, in the solution 2 of (37), the lower and upper parts of this solution are proportional to the crisp solution of the corresponding classic equation, for all α∈0,1. Indeed, we have(83)u¯x,t,α=α−1ucx,t,u¯x,t,α=1−αucx,t.So, all the obtained graphs are in fact images of the crisp solution’s graph (see Figures9 and 10) by the dilation of ratio equal to α−1 for the graph of u¯x,t,α and equal to 1−α for the graph of u¯x,t,α.Secondly, for each valid fuzzy solution (1, 2, and 3), in both examples, the maximum value of the upper solutionu¯x,t,α is greater than the maximum value of the lower solution u¯x,t,α. For instance, we get the following results for solution 2, a=1, and α=0.5:(84)maxu¯x,t,α≃0.35;maxu¯x,t,α≃0.9,and for solution 1 and α=0.5, we have(85)maxu¯x,t,α≃11;maxu¯x,t,α=0,while we get for solution 3 and α=0.5:(86)maxu¯x,t,α≃3500;maxu¯x,t,α≃6000.Finally, the graphs of the solutionsu¯x,t,α and u¯x,t,α are almost equal and coincide with the crisp solution’s graph for α close to 1, as it is shown in the graph of solution 3 for α=0.999999 (see Figures 13 and 14).


## 7. Conclusions

Theorems of high differentiability for a fuzzy function defined via a fuzzy improper integral have been investigated and proved, which have been employed to prove some results related to the partial derivatives of the fuzzy Laplace transform. Then, using the Laplace transform method, the solutions for linear FPDEs of second order have been given. For future research, one can apply this method to solve nonlinear FPDEs of first or high order. The influence of the choice of the kind of the fuzzy partial derivatives on the fuzzy solutions and their existence, uniqueness, and asymptotic behavior have been discussed.



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