Int. J. Anal. Appl. (2022), 20:61 

 

 

Received: Sep. 13, 2022. 

2010 Mathematics Subject Classification. 34A07. 

Key words and phrases. fuzzy Adomian decomposition method; fuzzy autonomous differential equation; fuzzy series 

solution. 

 

https://doi.org/10.28924/2291-8639-20-2022-61 © 2022 the author(s) 

ISSN: 2291-8639  

1 

 

 

Semi Analytical Solution for Fuzzy Autonomous Differential Equations 

Mazin H. Suhhiem1,*, Raad I. Khwayyit2 

1University of Sumer, Iraq 

2Ministry of Education, Iraq 

*Corresponding author: mazin.suhhiem@yahoo.com 

ABSTRACT. In this work, we have used fuzzy Adomian decomposition method to find the fuzzy semi 

analytical solution of the fuzzy autonomous differential equations with fuzzy initial conditions. This 

method allows for the solution of the fuzzy initial value problems to be calculated in the form of an infinite 

fuzzy series in which the fuzzy components can be easily calculated. The fuzzy series solutions that we 

have obtained are accurate solutions and very close to the fuzzy exact analytical solutions. Some 

numerical results have been given to illustrate the efficiency of the used  method.  

 

1. Introduction 

          The topic of fuzzy semi analytical methods (fuzzy series method) for solving fuzzy initial 

value problems (FIVPs) has been rapidly growing in recent years. Several fuzzy semi analytical 

methods have been proposed to obtain the fuzzy series solution of the linear and non-linear FIVB. 

Fuzzy Adomian decomposition method is one of the fuzzy semi analytical methods used to obtain 

the fuzzy series solution of the FIVBs. 

Researchers and scientists are continuing to develop this method for solving various types of the 

FIVBs because it represents an efficient and effective technique (For more details, see [1, 2, 3, 4, 

6, 7, 8, 10]). 

https://doi.org/10.28924/2291-8639-20-2022-61


2 Int. J. Anal. Appl. (2022), 20:61 

 

 

In this work, we will need many basic concepts in the fuzzy theory. These concepts can be found in 

detail in [5, 8, 11]. 

 

2. Fuzzy Autonomous Differential Equations 

          A fuzzy ordinary differential equation is called autonomous if it is independent of its 

independent crisp variable x. This implies that the nth order fuzzy autonomous differential equation 

is of the form [11]: 

𝑢(𝑛)(x) = f ( 𝑢 (x) , 𝑢′(𝑥) , 𝑢′′(𝑥) , … . , 𝑢(𝑛−1)(𝑥)) , 𝑥 ∈ [𝑥0 , ℎ]                     (2.1)                                                                

with the fuzzy initial conditions: 

 u(𝑥0) = 𝑢0   

 𝑢′(𝑥0) = 𝑢0
′     

 𝑢′′(𝑥0) = 𝑢0
′′   

.

.

.
    

 𝑢(𝑛−1)(𝑥0) = 𝑢0
(𝑛−1)

 

where: 𝑢  is a fuzzy function of the crisp variable  𝑥, f (𝑢 (x) , 𝑢′(𝑥) , 𝑢′′(𝑥) , … . , 𝑢(𝑛−1)(𝑥)) is a 

fuzzy function of  the crisp variable  𝑥  and the fuzzy variable  𝑢, 𝑢(𝑛)(x) is the fuzzy derivative of  

𝑢 (x)  ,  𝑢′(𝑥)  ,   𝑢′′(𝑥) , … ., 𝑢(𝑛−1)(𝑥), and  u(𝑥0)   ,  𝑢
′(𝑥0)  ,  𝑢

′′(𝑥0) , … ,  𝑢
(𝑛−1)(𝑥0) are fuzzy 

numbers. 

The general idea of solving the fuzzy differential equation is based on transforming this equation 

into a system of non-fuzzy (crisp) differential equations.  

Thus, problem (2.1) can be written as: 

𝑢(𝑛)(x) = 𝑓 ( 𝑢 , 𝑢′, 𝑢′′ , … , 𝑢(𝑛−1))                                                                                  (2.2) 

          =  H( 𝑢 , 𝑢′ ,  𝑢′′ , … ,  𝑢(𝑛−1) , 𝑢  , 𝑢
′
 , 𝑢

′′
 , … , 𝑢

(𝑛−1)
 ) 

With the initial conditions:  

 u(𝑥0) =  𝑢 0  

 𝑢′(𝑥0) = 𝑢 0
′   

 𝑢′′(𝑥0) =  𝑢 0
′′   



3 Int. J. Anal. Appl. (2022), 20:61 

 

 

.

.

.
    

𝑢(𝑛−1)(𝑥0) =  𝑢 0
(𝑛−1)

      

𝑢
(𝑛)

(x) =  𝑓 ( 𝑢 , 𝑢′, 𝑢′′ , … , 𝑢(𝑛−1))                                                                  (2.3) 

= G( 𝑢 , 𝑢′ ,  𝑢′′ , … ,  𝑢(𝑛−1) , 𝑢  , 𝑢
′
 , 𝑢

′′
 , … , 𝑢

(𝑛−1)
 ) 

With the initial conditions:                                                                       

𝑢 (𝑥0) = 𝑢0   

𝑢
′
(𝑥0) = 𝑢0

′
   

 𝑢
′′

(𝑥0) = 𝑢0
′′
 

.

.

.
    

 𝑢
(𝑛−1)

(𝑥0) = 𝑢0
(𝑛−1)

                                                                                                                  

Where 

H( 𝑢 , 𝑢′ ,  𝑢′′ , … ,  𝑢(𝑛−1) , 𝑢  , 𝑢
′
 , 𝑢

′′
 , … , 𝑢

(𝑛−1)
 )                                               (2.4) 

=Min{ 𝑓 (𝑥 , 𝑧) ∶ 𝑧 ∈  [𝑢 , 𝑢′ ,  𝑢′′ , … ,  𝑢(𝑛−1) , 𝑢  , 𝑢
′
 , 𝑢

′′
 , … , 𝑢

(𝑛−1)
] }, 

G( 𝑥 , 𝑥 ′ ,  𝑥′′ , … ,  𝑥 (𝑛−1) , 𝑥  , 𝑥
′
 , 𝑥

′′
 , … , 𝑥

(𝑛−1)
 )                                                (2.5) 

=Max{ 𝑓 (𝑥 , 𝑧) ∶ 𝑧 ∈  [𝑢 , 𝑢′ ,  𝑢′′ , … ,  𝑢(𝑛−1) , 𝑢  , 𝑢
′
 , 𝑢

′′
 , … , 𝑢

(𝑛−1)
] } ,   

The parametric form of system (2.4 - 2.5) is given by:  

𝑢(𝑛)(x , r) 

=H( 𝑢(𝑥 , r), 𝑢′(𝑥 , r)  , … ,  𝑢(𝑛−1)(𝑥 , r) , 𝑢 (𝑥 , r) , 𝑢
′
(𝑥 , r)  , … , 𝑢

(𝑛−1)
(𝑥 , r) )              (2.6) 

With the initial conditions:                                                                

u(𝑥0 , r)=𝑢 0(r), 

𝑢′(𝑥0 ,  r) = 𝑢 0
′ (r)   

𝑢′′(𝑥0 , r) =  𝑢 0
′′ (r)  

.

.

.
    

 𝑢(𝑛−1)(𝑥0 , r) =  𝑢 0
(𝑛−1)

(r)                                                                                                                                                                                                                         



4 Int. J. Anal. Appl. (2022), 20:61 

 

 

𝑢
(𝑛)

(x , r)  

=G( 𝑢(𝑥 , r) ,𝑢′(𝑥 , r)  , … ,  𝑢(𝑛−1)(𝑥 , r) , 𝑢 (𝑥 , r) , 𝑢
′
(𝑥 , r)  , … , 𝑢

(𝑛−1)
(𝑥 , r) )              (2.7) 

With the initial conditions:                                                               

𝑢 (𝑥0 , r)= 𝑢0(r)   

 𝑢
′
(𝑥0 , r) = 𝑢0

′
(r)  

 𝑢
′′

(𝑥0 , r) = 𝑢0
′′

(𝑟)  
.
.
.
    

 𝑢
(𝑛−1)

(𝑥0 ,r) = 𝑢0
(𝑛−1)

(r)                                                                                                              

Both equation (2.6) and equation (2.7) have only one solution on the interval [𝑥0 , ℎ]. Therefore, 

equation (2.1) has a unique fuzzy solution on [𝑥0 , ℎ], where 𝑟 ∈ [0 ,1] (For more details, see [11]). 

In order to illustrate the above, we give the following example: 

If we consider the second order fuzzy autonomous differential equation 

𝑢´´ (𝑥) = 4 𝑢´(𝑥) − 4 𝑢(𝑥),   x ∈ [0 , 1]                                                            (2.8)                                             

With the fuzzy initial conditions: 

 𝑢(0) = [2 + r  , 4 − r ], 

 𝑢ˊ(0) = [5 + r  , 7 − r]  and  𝑟 ∈ [0 ,1]. 

To convert problem (2.8) into a system of the second order crisp ordinary differential equations, 

we apply the following steps: 

[ 𝑢´´(𝑥) ]r = [4 𝑢´(𝑥)  −  4 𝑢(𝑥) ]r, for all 𝑟 ∈ [0 ,1]                                                    (2.9)                                                                                      

With the fuzzy initial conditions: 

 [𝑢(0)]r  = [ 2 + r  , 4 − r ],  

 [𝑢ˊ(0) ]r  = [ 5 + r  , 7 − r ]    

Then we get 

[𝑢´´(𝑥)]r = 4[𝑢´(𝑥)]r − 4[𝑢(𝑥)]r, for all 𝑟 ∈ [0 ,1]                                                     (2.10)                                                                       

With the fuzzy initial conditions: 

 [𝑢(0)]r  = [ 2 + r  , 4 − r ],  

 [𝑢ˊ(0) ]r  = [ 5 + r  , 7 − r ] 



5 Int. J. Anal. Appl. (2022), 20:61 

 

 

Then we have 

[ [𝑢´´(𝑥)]r
𝐿  , [𝑢´´(𝑥)]r

𝑈  ] = [ 4[𝑢´(𝑥)]r
𝐿 −  4[𝑢(𝑥)]r

𝐿   , 4[𝑢´(𝑥)]r
𝑈 −  4[𝑢(𝑥)]r

𝑈  ]               (2.11)                                                                                                                                                     

With the fuzzy initial conditions: 

[ [𝑢(0)]r
𝐿  , [𝑢(0)]r

𝑈  ] = [ 2 + r  , 4 − r]    

[ [𝑢ˊ(0)]r
𝐿  , [𝑢ˊ(0)]r

𝑈  ] = [ 5 + r  , 7 − r ]    

Then we get the following system of second order crisp ordinary differential equations: 

[𝑢´´(𝑥)]r
𝐿  =  4[𝑢´(𝑥)]r

𝐿 −  4[𝑢(𝑥)]r
𝐿  ;                                                                     (2.12)                                                                                               

 With the initial conditions: 

 [𝑢(0)]r
𝐿 =  2 + r    

 [𝑢ˊ(0)]r
𝐿 = 5 + r 

[𝑢´´(𝑥)]r
𝑈  =  4[𝑢´(𝑥)]α

𝑈 −  4[𝑢(𝑥)]r
𝑈;                                                                    (2.13)                                                                              

With the initial conditions: 

[𝑢(0)]r
𝑈 =  4 − r    

[𝑢ˊ(0)]r
𝑈 = 7 − r  

Which gives the unique crisp solutions 

[𝑢(𝑥)]r
𝐿 = (2 + 𝑟) 𝑒2𝑥 + (1 − 𝑟 ) 𝑥 𝑒2𝑥                                                          (2.14)                                                                      

[𝑢(𝑥)]r
𝑈 =  (4 −  𝑟) 𝑒2𝑥 + (r − 1 ) 𝑥 𝑒2𝑥                                                        (2.15)                                                             

Then the unique fuzzy solution of problem (8) is  

[𝑢(𝑥)]r = [ (2 + 𝑟) 𝑒
2𝑥 + (1 − 𝑟 ) 𝑥 𝑒2𝑥   , (4 −  𝑟) 𝑒2𝑥 + (r − 1 ) 𝑥 𝑒2𝑥  ]               (2.16)                                       

                                                                                                                                                                                                                  

3. Fuzzy Adomian Decomposition Method 

          To understand the fuzzy Adomian decomposition method, we consider the nth order fuzzy 

differential equation [2,3,7]: 

[𝐹(𝑢(𝑥))]𝑟 = [𝑔(𝑥) ]𝑟                                                              (3.1) 

Where F represents a general nonlinear fuzzy ordinary (or fuzzy partial) differential operator 

including both linear and nonlinear terms, x denotes the independent crisp variable, 𝑢(𝑥) and 𝑔(𝑥) 

are unknown fuzzy functions. 

From section (2), We can conclude that: 



6 Int. J. Anal. Appl. (2022), 20:61 

 

 

[𝐹(𝑢(𝑥))]𝑟 = [ 𝐹(𝑢(𝑥))]𝑟
𝐿 , [𝐹(𝑢(𝑥))]𝑟 

𝑈                                                              (3.2) 

[𝑔(𝑥)]𝑟 = [ 𝑔(𝑥)]𝑟
𝐿  , [𝑔(𝑥)]𝑟 

𝑈 ]                                                                          (3.3) 

where 

[𝐹(𝑢(𝑥))]𝑟
𝐿 = [𝑔(𝑥)]𝑟

𝐿                                                                                   (3.4) 

[𝐹(𝑢(𝑥))]𝑟
𝑈 = 𝑔(𝑥)]𝑟

𝑈                                                                                   (3.5) 

The fuzzy linear terms of (3.1) are decomposed into [𝐿]𝑟 + [𝑅]𝑟, where: [𝐿]𝑟 is invertible and is 

taken as the highest order fuzzy derivative. 

This implies that: 

𝐿(∗) =
𝑑𝑚

𝑑𝑥𝑚
(∗) , 𝑚 = 1,2,3, …                                                                         (3.6) 

𝐿−1(∗) = ∫ ∫
……………

𝑚−𝑡𝑖𝑚𝑒𝑠

𝑥 

0
 

𝑥

0
∫ (∗)𝑑𝑥𝑑𝑥 

𝑥

0
…………… 

𝑚−𝑡𝑖𝑚𝑒𝑠
 𝑑𝑥                                                   (3.7) 

and [𝑅]𝑟 is the remainder of the fuzzy linear operator. 

Thus the equation (3.1) may be written as: 

[𝐿(𝑢)]𝑟 + [𝑅(𝑢)]𝑟 + [𝑁(𝑢)]𝑟 = [𝑔]𝑟                                              (3.8) 

Where [𝑁(𝑢)]𝑟  represents the fuzzy nonlinear terms. 

By the concepts of section (2), We get: 

[𝐿(𝑢)]𝑟
𝐿 + [𝑅(𝑢)]𝑟

𝐿 + [𝑁(𝑢)]𝑟
𝐿 = [𝑔]𝑟

𝐿                                                                 (3.9) 

[𝐿(𝑢)]𝑟
𝑈 + [𝑅(𝑢)]𝑟

𝑈 + [𝑁(𝑢)]𝑟
𝑈 = [𝑔]𝑟

𝑈                                                              (3.10) 

The fuzzy Adomian decomposition method represents the fuzzy solution [𝑢(𝑥)]𝑟 of problem (3.1) 

as a fuzzy series of the form: 

[𝑢(𝑥)]𝑟 = [ 𝑢(𝑥)]𝑟
𝐿  , [𝑢(𝑥)]𝑟 

𝑈 ]                                                                         (3.11) 

where 

[𝑢(𝑥)]𝑟
𝐿 = ∑ [𝑢𝑛 (𝑥)]𝑟

𝐿∞
𝑛=0 = [𝑢0(𝑥)]𝑟

𝐿 + [𝑢1(𝑥)]𝑟
𝐿 + [𝑢2(𝑥)]𝑟

𝐿 + [𝑢3(𝑥)]𝑟
𝐿 + ⋯               (3.12) 

[𝑢(𝑥)]𝑟
𝑈 = ∑ [𝑢𝑛 (𝑥)]𝑟

𝑈∞
𝑛=0 = [𝑢0(𝑥)]𝑟

𝑈 + [𝑢1(𝑥)]𝑟
𝑈 + [𝑢2(𝑥)]𝑟

𝑈 + [𝑢3(𝑥)]𝑟
𝑈 + ⋯             (3.13) 

Such that: 

●) [𝑢0(𝑥)]𝑟 = [ 𝑢0(𝑥)]𝑟
𝐿  , [𝑢0(𝑥)]𝑟 

𝑈 ]                                                                 (3.14) 

where 

[𝑢0(𝑥)]𝑟
𝐿 = [𝜃0]𝑟

𝐿 + 𝐿−1([𝑔(𝑥)]𝑟
𝐿 )                                                                    (3.15) 

[𝑢0(𝑥)]𝑟
𝑈 = [𝜃0]𝑟

𝑈 + 𝐿−1([𝑔(𝑥)]𝑟
𝑈 )                                                                   (3.16) 



7 Int. J. Anal. Appl. (2022), 20:61 

 

 

●) [𝑢1(𝑥)]𝑟 = [ 𝑢1(𝑥)]𝑟
𝐿  , [𝑢1(𝑥)]𝑟 

𝑈 ]                                                                  (3.17) 

where 

 [𝑢1(𝑥)]𝑟
𝐿 = −𝐿−1([𝑅(𝑢0)]𝑟

𝐿 ) −𝐿−1([𝐴0]𝑟
𝐿 )                                                         (3.18) 

[𝑢1(𝑥)]𝑟
𝑈 = −𝐿−1([𝑅(𝑢0)]𝑟

𝑈 ) −𝐿−1([𝐴0]𝑟
𝑈 )                                                         (3.19) 

●) [𝑢2(𝑥)]𝑟 = [ 𝑢2(𝑥)]𝑟
𝐿  , [𝑢2(𝑥)]𝑟 

𝑈 ]                                                                 (3.20) 

where 

[𝑢2(𝑥)]𝑟
𝐿 = −𝐿−1([𝑅(𝑢1)]𝑟

𝐿 ) −𝐿−1([𝐴1]𝑟
𝐿 )                                                          (3.21) 

[𝑢2(𝑥)]𝑟
𝑈 = −𝐿−1([𝑅(𝑢1)]𝑟

𝑈 ) −𝐿−1([𝐴1]𝑟
𝑈 )                                                         (3.22) 

.

.

.
      

●) [𝑢𝑛+1(𝑥)]𝑟 = [ 𝑢𝑛+1(𝑥)]𝑟
𝐿  , [𝑢𝑛+1(𝑥)]𝑟 

𝑈 ] , 𝑛 ≥ 0                                               (3.23) 

where 

[𝑢𝑛+1(𝑥)]𝑟
𝐿 = −𝐿−1([𝑅(𝑢𝑛)]𝑟

𝐿 ) −𝐿−1([𝐴𝑛]𝑟
𝐿 )                                                      (3.24) 

[𝑢𝑛+1(𝑥)]𝑟
𝑈 = −𝐿−1([𝑅(𝑢𝑛)]𝑟

𝑈 ) −𝐿−1([𝐴𝑛]𝑟
𝑈 )                                                     (3.25) 

Note that: 

[𝜃0]𝑟 = [ [𝜃0]𝑟
𝐿  , [𝜃0]𝑟 

𝑈 ]                                                                                 (3.26) 

 and it can be calculated as follows:  

●) If  𝐿 =
𝑑

𝑑𝑥
  , then we have: 

[𝜃0]𝑟
𝐿 = [𝑢(0)]𝑟

𝐿                                                                                         (3.27) 

[𝜃0]𝑟
𝑈 = [𝑢(0)]𝑟

𝑈                                                                                        (3.28) 

●) If  𝐿 =
𝑑2

𝑑𝑥2 
 , then we have: 

 [𝜃0]𝑟
𝐿 = [𝑢(0)]𝑟

𝐿 + 𝑥[𝑢′(0)]𝑟
𝐿                                                                        (3.29) 

[𝜃0]𝑟
𝑈 = [𝑢(0)]𝑟

𝑈 + 𝑥[𝑢′(0)]𝑟
𝑈                                                                        (3.30) 

●) If  𝐿 =
𝑑3

𝑑𝑥3
 , then we have: 

 [𝜃0]𝑟
𝐿 = [𝑢(0)]𝑟

𝐿 + 𝑥[𝑢′(0)]𝑟
𝐿 +

𝑥2

2!
 [𝑢′′(0)]𝑟

𝐿                                                       (3.31) 

 [𝜃0]𝑟
𝑈 = [𝑢(0)]𝑟

𝑈 + 𝑥[𝑢′(0)]𝑟
𝑈 +

𝑥2

2!
[𝑢′′(0)]𝑟

𝑈                                                      (3.32) 



8 Int. J. Anal. Appl. (2022), 20:61 

 

 

.

.

.
  

 ●) If  𝐿 =
𝑑𝑛+1

𝑑𝑥𝑛+1
  , then we have: 

 [𝜃0]𝑟
𝐿 = [𝑢(0)]𝑟

𝐿 + 𝑥[𝑢′(0)]𝑟
𝐿 +

𝑥2

2!
 [𝑢′′(0)]𝑟

𝐿 + ⋯ +
𝑥𝑛

𝑛!
 [𝑢(𝑛)(0)]𝑟

𝐿                             (3.33) 

 [𝜃0]𝑟
𝑈 = [𝑢(0)]𝑟

𝑈 + 𝑥[𝑢′(0)]𝑟
𝑈 +

𝑥2

2!
[𝑢′′(0)]𝑟

𝑈 + ⋯ +
𝑥𝑛

𝑛!
 [𝑢(𝑛)(0)]𝑟

𝑈                           (3.34) 

Also, Note that: 

[𝐴0]𝑟 , [𝐴1]𝑟  , [𝐴2]𝑟  , … , [𝐴𝑛 ]𝑟   are the fuzzy Adomian polynomials, 

which can be found as follows:      

●) [𝐴0]𝑟 = [ 𝐴0]𝑟
𝐿  , [𝐴0]𝑟 

𝑈 ]                                                                                   (3.35) 

where 

[𝐴0]𝑟
𝐿 = [𝑁(𝑢0)]𝑟

𝐿                                                                                       (3.36) 

[𝐴0]𝑟
𝑈 = [𝑁(𝑢0)]𝑟

𝑈                                                                                      (3.37) 

●) [𝐴1]𝑟 = [ 𝐴1]𝑟
𝐿  , [𝐴1]𝑟 

𝑈 ]                                                                             (3.38) 

where 

[𝐴1]𝑟
𝐿 = [𝑢1]𝑟

𝐿  [𝑁′(𝑢0)]𝑟
𝐿                                                                               (3.39) 

[𝐴1]𝑟
𝑈 = [𝑢1]𝑟

𝑈  [𝑁′(𝑢0)]𝑟
𝑈                                                                              (3.40) 

●) [𝐴2]𝑟 = [ 𝐴2]𝑟
𝐿  , [𝐴2]𝑟 

𝑈 ]                                                                             (3.41) 

where 

[𝐴2]𝑟
𝐿 = [𝑢2]𝑟

𝐿  [𝑁′(𝑢0)]𝑟
𝐿 +

1

2!
 ([𝑢1]𝑟

𝐿 )2[𝑁′′(𝑢0)]𝑟
𝐿                                                 (3.42) 

[𝐴2]𝑟
𝑈 = [𝑢2]𝑟

𝑈  [𝑁′(𝑢0)]𝑟
𝑈 +

1

2!
 ([𝑢1]𝑟

𝑈 )2[𝑁′′(𝑢0)]𝑟
𝑈                                               (3.43) 

.

.

.
      

●)[𝐴𝑛 ]𝑟 = [ 𝐴𝑛 ]𝑟
𝐿  , [𝐴𝑛]𝑟 

𝑈 ], 𝑛 = 0,1,2, …                                                             (3.44) 

where 

[𝐴𝑛 ]𝑟
𝐿 =

1

𝑛!
 

𝑑𝑛

𝑑𝛽𝑛
  ( [𝑁( ∑ 𝛽𝑛𝑢𝑛 

∞
𝑛=0 )]𝑟

𝐿  )|𝛽=0                                                         (3.45) 

[𝐴𝑛 ]𝑟
𝑈 =

1

𝑛!
 

𝑑𝑛

𝑑𝛽𝑛
  ( [𝑁( ∑ 𝛽𝑛𝑢𝑛  

∞
𝑛=0 )]𝑟

𝑈  )|𝛽=0                                                        (3.46) 



9 Int. J. Anal. Appl. (2022), 20:61 

 

 

4. Applied Examples 

          In this section, we will solve three fuzzy problems to illustrate the accuracy of the fuzzy 

Adomian decomposition method. 

Example 1: Consider the first order fuzzy autonomous differential equation 

𝑢′(𝑥) = 𝑢(𝑥) + 𝑢2(𝑥) , 𝑥 ∈ [0 , 0.1]  ,                                                                                            

With the fuzzy initial condition: 

[𝑢(0)]𝑟 = [0.96 + 0.04𝑟 , 1.01 − 0.01𝑟] ,    𝑟 ∈ [0,1]. 

Solution:  

We define: 

𝐿(𝑢) =
𝑑

𝑑𝑥
(𝑢)     

 𝑅(𝑢) = [ 𝑅(𝑢)]𝑟
𝐿 = [ 𝑅(𝑢)]𝑟

𝑈 = −𝑢    

 𝑁(𝑢) = [ 𝑁(𝑢)]𝑟
𝐿 = [ 𝑁(𝑢)]𝑟

𝑈 = −𝑢2    

  𝑔(𝑥) = [ 𝑔(𝑥)]𝑟
𝐿 = [ 𝑔(𝑥)]𝑟

𝑈 = 0  

From section (3), We can find: 

[𝜃0]𝑟
𝐿 = 0.96 + 0.04𝑟   

[𝑢0]𝑟
𝐿 = (0.96 + 0.04𝑟)    

[𝐴0]𝑟
𝐿 = −(0.96 + 0.04𝑟  )2    

[𝑢1]𝑟
𝐿 = ((0.96 + 0.04𝑟) + (0.96 + 0.04𝑟  )2)𝑥   

[𝐴1]𝑟
𝐿 = (−2(0.96 + 0.04𝑟  )2 − 2(0.96 + 0.04𝑟  )3)𝑥  

[𝑢2]𝑟
𝐿 = (

1

2
(0.96 + 0.04𝑟) +

3

2
(0.96 + 0.04𝑟  )2 + (0.96 + 0.04𝑟  )3) 𝑥2  

[𝐴2]𝑟
𝐿 = (−2(0.96 + 0.04𝑟  )2 − 5(0.96 + 0.04𝑟  )3 − 3(0.96 + 0.04𝑟  )4)𝑥2  

[𝑢3]𝑟
𝐿 = (

1

6
(0.96 + 0.04𝑟) +

7

6
(0.96 + 0.04𝑟  )2 + 2(0.96 + 0.04𝑟  )3 + (0.96 + 0.04𝑟  )4) 𝑥3  

.

.

.
      

Also, we find: 

[𝜃0]𝑟
𝑈 = 1.01 − 0.01𝑟  

[𝑢0]𝑟
𝑈 = (1.01 − 0.01𝑟)    

[𝐴0]𝑟
𝑈 = −(1.01 − 0.01𝑟 )2    



10 Int. J. Anal. Appl. (2022), 20:61 

 

 

[𝑢1]𝑟
𝑈 = ((1.01 − 0.01𝑟) + (1.01 − 0.01𝑟 )2)𝑥   

[𝐴1]𝑟
𝑈 = (−2(1.01 − 0.01𝑟  )2 − 2(1.01 − 0.01𝑟 )3)𝑥  

[𝑢2]𝑟
𝑈 = (

1

2
(1.01 − 0.01𝑟) +

3

2
(1.01 − 0.01𝑟 )2 + (1.01 − 0.01𝑟 )3) 𝑥2  

[𝐴2]𝑟
𝑈 = (−2(1.01 − 0.01𝑟 )2 − 5(1.01 − 0.01𝑟 )3 − 3(1.01 − 0.01𝑟 )4)𝑥2  

[𝑢3]𝑟
𝑈 = (

1

6
(1.01 − 0.01𝑟) +

7

6
(1.01 − 0.01𝑟  )2 + 2(1.01 − 0.01𝑟  )3 + (1.01 − 0.01𝑟  )4) 𝑥3  

.

.

.
      

Therefore, the fuzzy semi analytical solution is: 

[𝑢(𝑥)]𝑟 = [ 𝑢(𝑥)]𝑟
𝐿  , [𝑢(𝑥)]𝑟 

𝑈 ]  

Where  

[𝑢(𝑥)]𝑟
𝐿 = (0.96 + 0.04𝑟) + ((0.96 + 0.04𝑟) + (0.96 + 0.04𝑟  )2)𝑥 +(

1

2
(0.96 + 0.04𝑟) +

3

2
(0.96 + 0.04𝑟  )2 + (0.96 + 0.04𝑟  )3) 𝑥2 + (

1

6
(0.96 + 0.04𝑟) +

7

6
(0.96 + 0.04𝑟  )2 +

2(0.96 + 0.04𝑟  )3 + (0.96 + 0.04𝑟  )4) 𝑥3 + ⋯      

[𝑢(𝑥)]𝑟
𝑈 = (1.01 − 0.01𝑟) + ((1.01 − 0.01𝑟) + (1.01 − 0.01𝑟  )2)𝑥 +(

1

2
(1.01 − 0.01𝑟) +

3

2
(1.01 − 0.01𝑟  )2 + (1.01 − 0.01𝑟 )3) 𝑥2 + (

1

6
(1.01 − 0.01𝑟) +

7

6
(1.01 − 0.01𝑟  )2 +

2(1.01 − 0.01𝑟  )3 + (1.01 − 0.01𝑟  )4) 𝑥3 + ⋯      

Example 2: Consider the second order fuzzy autonomous differential equation 

𝑢′′(𝑥) + 𝑢(𝑥) = 5 , 𝑥 ∈ [0 , 1], 

With the fuzzy initial conditions: 

[𝑢(0)]𝑟 = [𝑟 , 2 − 𝑟]  

[𝑢′(0)]𝑟 = [1 + 𝑟 , 3 − 𝑟] ,      𝑟 ∈ [0,1].  

Solution:  

We define: 

𝐿(𝑢) =
𝑑2

𝑑𝑥2
(𝑢) 

 𝑅(𝑢) = [ 𝑅(𝑢)]𝑟
𝐿 = [ 𝑅(𝑢)]𝑟

𝑈 = 𝑢    



11 Int. J. Anal. Appl. (2022), 20:61 

 

 

 𝑁(𝑢) = [ 𝑁(𝑢)]𝑟
𝐿 = [ 𝑁(𝑢)]𝑟

𝑈 = 0    

  𝑔(𝑥) = [ 𝑔(𝑥)]𝑟
𝐿 = [ 𝑔(𝑥)]𝑟

𝑈 = 5  

From section (3), We can find: 

[𝜃0]𝑟
𝐿 = 𝑟 + (1 + 𝑟)𝑥   

[𝑢0]𝑟
𝐿 = 𝑟 + (1 + 𝑟)𝑥 +

5

2
𝑥2 

[𝐴0]𝑟
𝐿 = 0  

[𝑢1]𝑟
𝐿 = −

𝑟

2
𝑥2 −

(𝑟+1)

6
𝑥3 −

5

24
𝑥4   

[𝐴1]𝑟
𝐿 = 0  

[𝑢2]𝑟
𝐿 =

𝑟

24
𝑥4 +

(𝑟+1)

120
𝑥5 +

5

720
𝑥6   

[𝐴2]𝑟
𝐿 = 0  

[𝑢3]𝑟
𝐿 = −

𝑟

720
𝑥6 −

(𝑟+1)

5040
𝑥7 −

5

40320
𝑥8   

.

.

.
      

Also, we find: 

[𝜃0]𝑟
𝑈 = (2 − 𝑟) + (3 − 𝑟)𝑥   

[𝑢0]𝑟
𝑈 = (2 − 𝑟) + (3 − 𝑟)𝑥  +

5

2
𝑥2  

[𝐴0]𝑟
𝑈 = 0  

[𝑢1]𝑟
𝑈 = −

(2−𝑟)

2
𝑥2 −

(3−𝑟)

6
𝑥3 −

5

24
𝑥4   

[𝐴1]𝑟
𝑈 = 0  

[𝑢2]𝑟
𝑈 =

(2−𝑟)

24
𝑥4 +

(3−𝑟)

120
𝑥5 +

5

720
𝑥6   

[𝐴2]𝑟
𝑈 = 0  

[𝑢3]𝑟
𝑈 = −

(2−𝑟)

720
𝑥6 −

(3−𝑟)

5040
𝑥7 −

5

40320
𝑥8   

.

.

.
    

Therefore, the fuzzy semi analytical solution is: 

[𝑢(𝑥)]𝑟 = [ 𝑢(𝑥)]𝑟
𝐿  , [𝑢(𝑥)]𝑟 

𝑈 ]  

Where  



12 Int. J. Anal. Appl. (2022), 20:61 

 

 

[𝑢(𝑥)]𝑟
𝐿 = 𝑟 + (𝑟 + 1)𝑥 + (

5−𝑟

2
) 𝑥2 − (

𝑟+1

6
) 𝑥3 + (

𝑟−5

24
) 𝑥4 + (

𝑟+1

120
) 𝑥5 + (

5−𝑟

720
) 𝑥6 −

(
𝑟+1

5040
) 𝑥7 − (

5

40320
) 𝑥8 + ⋯   

= 5 +  (𝑟 − 5)𝑐𝑜𝑠𝑥 + (𝑟 + 1)𝑠𝑖𝑛𝑥   

[𝑢(𝑥)]𝑟
𝑈 = (2 − 𝑟) + (3 − 𝑟)𝑥 + (

3+𝑟

2
) 𝑥2 − (

3−𝑟

6
) 𝑥3 − (

3+𝑟

24
) 𝑥4 + (

3−𝑟

120
) 𝑥5 + (

3+𝑟

720
) 𝑥6 −

(
3−𝑟

5040
) 𝑥7 − (

5

40320
) 𝑥8 + ⋯   

= 5 − (3 + 𝑟)𝑐𝑜𝑠𝑥 + (3 − 𝑟)𝑠𝑖𝑛𝑥    

Which is the fuzzy exact analytical solution.  

Example 3: Consider the second order fuzzy autonomous differential equation 

𝑢′′(𝑥) + [2𝑟 + 1, −2𝑟 + 5]𝑢(𝑥) + √𝑢(𝑥)  = 0 , 𝑥 ∈ [0 , 1]  ,                                                                                            

With the fuzzy initial conditions: 

[𝑢(0)]𝑟 = [0 , 0]  

[𝑢′(0)]𝑟 = [𝑟 + 4 , −𝑟 + 6] ,      𝑟 ∈ [0,1].  

Solution:  

We define: 

𝐿(𝑢) =
𝑑2

𝑑𝑥2
(𝑢)  

[𝑅(𝑢)]𝑟
𝐿 = (2𝑟 + 1)𝑢 

 [𝑅(𝑢)]𝑟
𝑈 = (−2𝑟 + 5)𝑢 

𝑁(𝑢) = [ 𝑁(𝑢)]𝑟
𝐿 = [ 𝑁(𝑢)]𝑟

𝑈 = √𝑢     

  𝑔(𝑥) = [ 𝑔(𝑥)]𝑟
𝐿 = [ 𝑔(𝑥)]𝑟

𝑈 = 0  

From section (3), We can find: 

[𝜃0]𝑟
𝐿 = (𝑟 + 4)𝑥       

[𝑢0]𝑟
𝐿 = (𝑟 + 4)𝑥   

[𝐴0]𝑟
𝐿 = ((𝑟 + 4)𝑥)

1

2  

[𝑢1]𝑟
𝐿 = −

(2𝑟+1)

6(𝑟+4)2
((𝑟 + 4)𝑥)3 −

4

15(𝑟+4)2
((𝑟 + 4)𝑥)

5

2  

[𝐴1]𝑟
𝐿 = −

(2𝑟+1)

12(𝑟+4)2
((𝑟 + 4)𝑥)

5

2  −
2

15(𝑟+4)2
((𝑟 + 4)𝑥)2  



13 Int. J. Anal. Appl. (2022), 20:61 

 

 

[𝑢2]𝑟
𝐿 =

1

90(𝑟+4)4
((𝑟 + 4)𝑥)

4
+

(2𝑟+1)2

120(𝑟+4)4
((𝑟 + 4)𝑥)

5
 + 

(2𝑟+1)

90(𝑟+4)4
((𝑟 + 4)𝑥)

9

2 

[𝐴2]𝑟
𝐿 =

(2𝑟+1)

60(𝑟+4)4
((𝑟 + 4)𝑥)

4
−

1

300(𝑟+4)4
((𝑟 + 4)𝑥)

7

2 + 
(2𝑟+1)2

1440(𝑟+4)4
((𝑟 + 4)𝑥)

9

2 

[𝑢3]𝑟
𝐿 = −

(2𝑟+1)

1080(𝑟+4)6
((𝑟 + 4)𝑥)

6
−

(2𝑟+1)3

5040(𝑟+4)6
((𝑟 + 4)𝑥)

7
+ 

1

7425(𝑟+4)6
((𝑟 + 4)𝑥)

11

2   − 

17(2𝑟+1)2

51480(𝑟+4)6
((𝑟 + 4)𝑥)

13

2  

.

.

.
      

Also, we find: 

[𝜃0]𝑟
𝑈 = (6 − 𝑟)𝑥       

[𝑢0]𝑟
𝑈 = (6 − 𝑟)𝑥   

[𝐴0]𝑟
𝑈 = ((6 − 𝑟)𝑥)

1

2  

[𝑢1]𝑟
𝑈 = −

(5−2𝑟)

6(6−𝑟)2
((6 − 𝑟)𝑥)3 −

4

15(6−𝑟)2
((6 − 𝑟)𝑥)

5

2  

[𝐴1]𝑟
𝑈 = −

(5−2𝑟)

12(6−𝑟)2
((𝑟 + 4)𝑥)

5

2  −
2

15(6−𝑟)2
((6 − 𝑟)𝑥)2  

[𝑢2]𝑟
𝑈 =

1

90(6−𝑟)4
((6 − 𝑟)𝑥)

4
+

(5−2𝑟)2

120(6−𝑟)4
((6 − 𝑟)𝑥)

5
 + 

(5−2𝑟)

90(6−𝑟)4
((6 − 𝑟)𝑥)

9

2 

[𝐴2]𝑟
𝑈 =

(2𝑟+1)

60(6−𝑟)4
((6 − 𝑟)𝑥)

4
−

1

300(6−𝑟)4
((6 − 𝑟)𝑥)

7

2 + 
(2𝑟+1)2

1440(6−𝑟)4
((6 − 𝑟)𝑥)

9

2 

[𝑢3]𝑟
𝑈 = −

(5−2𝑟)

1080(6−𝑟)6
((𝑟 + 4)𝑥)

6
−

(5−2𝑟)3

5040(6−𝑟)6
((6 − 𝑟)𝑥)

7
+ 

1

7425(6−𝑟)6
((6 − 𝑟)𝑥)

11

2   − 

17(5−2𝑟)2

51480(6−𝑟)6
((6 − 𝑟)𝑥)

13

2  

.

.

.
    

Therefore, the fuzzy semi analytical solution is: 

[𝑢(𝑥)]𝑟 = [ 𝑢(𝑥)]𝑟
𝐿  , [𝑢(𝑥)]𝑟 

𝑈 ]  

Where [𝑢(𝑥)]𝑟
𝐿 = (𝑟 + 4) [ 𝑥 − (2𝑟 + 1)  

𝑥3

3!
+ (2𝑟 + 1)2  

𝑥5

5!
− (2𝑟 + 1)3  

𝑥7

7!
 ] + (𝑟 +

4)
1

2  [ −  
4

15
 𝑥

5

2 +
1

90
 (2𝑟 + 1) 𝑥

9

2 −
17

51480 
 (2𝑟 + 1)2 𝑥

13

2  ] +  [  
1

90
 𝑥4 −

1

1080
 (2𝑟 + 1) 𝑥6 +

 
1

7425 
 (𝑟 + 4)

−
1

2 𝑥
11

2  ] + ⋯   



14 Int. J. Anal. Appl. (2022), 20:61 

 

 

[𝑢(𝑥)]𝑟
𝑈 = (6 − 𝑟) [ 𝑥 − (5 − 2𝑟)  

𝑥3

3!
+ (5 − 2𝑟)2  

𝑥5

5!
− (5 − 2𝑟)3  

𝑥7

7!
 ] + (6 − 𝑟)

1

2  [ −  
4

15
 𝑥

5

2 +

1

90
 (5 − 2𝑟) 𝑥

9

2 −
17

51480 
 (5 − 2𝑟)2 𝑥

13

2  ] +  [  
1

90
 𝑥4 −

1

1080
 (5 − 2𝑟) 𝑥6 +  

1

7425 
 (6 −

𝑟)
−

1

2 𝑥
11

2  ] + ⋯   

 

5. Discussion 

          To show the accuracy of the results, we will give a numerical comparison between the 

exact analytical solution and the semi analytical solution.  

If we go back to example 1: 

𝑢′(𝑥) = 𝑢(𝑥) + 𝑢2(𝑥) , 𝑥 ∈ [0 , 0.1]  ,                                                                                                                                                                 

The fuzzy exact analytical solution for this problem is: 

[𝑢(𝑥)]𝑟 = [ 𝑢(𝑥)]𝑟
𝐿  , [𝑢(𝑥)]𝑟 

𝑈 ]  

where  

[𝑢(𝑥)]𝑟
𝐿 =

(1.96+0.04𝑟)

(1.96+0.04𝑟)−(0.96+0.04𝑟)𝑒 𝑥
− 1       

[𝑢(𝑥)]𝑟
𝑈 =

(2.01−0.01𝑟)

(2.01−0.01𝑟)−(1.01−0.01𝑟)𝑒 𝑥
− 1  

While the fuzzy semi analytical solution that we got (at  𝑟 = 0.5 ) is:  

[𝑢(𝑥)]𝑟 = [ 𝑢(𝑥)]𝑟
𝐿  , [𝑢(𝑥)]𝑟 

𝑈 ]  

where  

[𝑢(𝑥)]𝑟
𝐿 = 0.98 + 1.9404𝑥 + 2.871792 𝑥2 + 4.08855216𝑥3 + ⋯      

[𝑢(𝑥)]𝑟
𝑈 = 1.005 + 2.015025𝑥 + 3.032612625 𝑥2 + 4.396163251𝑥3 + ⋯      

We test the accuracy by computing the absolute errors: 

 [𝑒𝑟𝑟𝑜𝑟]𝑟
𝐿  = | [𝑢𝑒𝑥𝑎𝑐𝑡(x)]𝑟

𝐿 −  [𝑢𝑠𝑒𝑟𝑖𝑒𝑠(x)]𝑟
𝐿 |  

 [𝑒𝑟𝑟𝑜𝑟]𝑟
𝑈   = | [𝑢𝑒𝑥𝑎𝑐𝑡(x)]𝑟

𝑈 −  [𝑢𝑠𝑒𝑟𝑖𝑒𝑠(x)]𝑟
𝑈 | 

 

 

 

 

 



15 Int. J. Anal. Appl. (2022), 20:61 

 

 

Table 1: Numerical results for example 1. 

𝑥 [𝑢𝑠𝑒𝑟𝑖𝑒𝑠(x)]𝑟
𝐿 [𝑒𝑟𝑟𝑜𝑟]𝑟

𝐿 [𝑢𝑠𝑒𝑟𝑖𝑒𝑠(x)]𝑟
𝑈 [𝑒𝑟𝑟𝑜𝑟]𝑟

𝑈 

 

0 

 

0.980000000000000 

 

0 

 

1.005000000000000 

 

0 

 

0.01 

 

0.999695267752160 

 

5.90 e-8 

 

1.025457907425751 

 

6.46 e-8 

 

0.02 

 

1.019989425217280 

 

9.57 e-7 

 

1.046548714356008 

 

1.05 e-6 

 

0.03 

 

1.040907003708320 

 

4.92 e-6 

 

1.068298797770277 

 

5.39 e-6 

 

0.04 

 

1.062472534538240 

 

1.58 e-5 

 

1.090734534648064 

 

1.73 e-5 

 

0.05 

 

1.084710549020000 

 

3.91 e-5 

 

1.113882301968875 

 

4.29 e-5 

 

0.06 

 

1.107645578466560 

 

8.23 e-5 

 

1.137768476712216 

 

9.03 e-5 

 

0.07 

 

1.131302154190880 

 

1.55 e-4 

 

1.162419435857593 

 

1.70 e-4 

 

0.08 

 

1.155704807505920 

 

2.69 e-4 

 

1.187861556384512 

 

2.95 e-4 

 

0.09 

 

1.180878069724640 

 

4.37 e-4 

 

1.214121215272479 

 

4.80 e-4 

 

0.10 

 

1.206846472160000 

 

6.78 e-4 

 

1.241224789501000 

 

7.44 e-4 

 

 

 

 

 



16 Int. J. Anal. Appl. (2022), 20:61 

 

 

Table 2: Numerical results for example 1. 

𝑥 [𝑢𝑠𝑒𝑟𝑖𝑒𝑠(x)]𝑟
𝐿 [𝑒𝑟𝑟𝑜𝑟]𝑟

𝐿 [𝑢𝑠𝑒𝑟𝑖𝑒𝑠(x)]𝑟
𝑈 [𝑒𝑟𝑟𝑜𝑟]𝑟

𝑈 

 

0.001 

 

0.981943275880552 

 

5.82 e-12 

 

1.007018062008788 

 

6.37 e-12 

 

0.003 

 

0.985847156518908 

 

4.73 e-10 

 

1.011072487210033 

 

5.18 e-10 

 

0.005 

 

0.989774305869020 

 

3.66 e-9 

 

1.015151489836031 

 

4.01 e-9 

 

0.007 

 

0.993724920181391 

 

1.41 e-8 

 

1.019255280902620 

 

1.54 e-8 

 

0.009 

 

0.997699195706525 

 

3.86 e-8 

 

1.023384071425635 

 

4.23 e-8 

 

0.011 

 

1.001697328694925 

 

8.64 e-8 

 

1.027538072420912 

 

9.47 e-8 

 

0.013 

 

1.005719515397095 

 

1.69 e-7 

 

1.031717494904287 

 

1.85 e-7 

 

0.015 

 

1.009765952063540 

 

3.01 e-7 

 

1.035922549891597 

 

3.29 e-7 

 

0.017 

 

1.013836834944762 

 

4.97 e-7 

 

1.040153448398677 

 

5.45 e-7 

 

0.019 

 

1.017932360291266 

 

7.78 e-7 

 

1.044410401441363 

 

8.53 e-7 

 

0.021 

 

1.022052724353554 

 

1.17 e-6 

 

1.048693620035492 

 

1.28 e-6 

 

 

 

 

 



17 Int. J. Anal. Appl. (2022), 20:61 

 

 

6. Conclusion 

          In this work, we have studied the fuzzy semi analytical solutions of the fuzzy autonomous 

differential equations. We have used the fuzzy Adomian decomposition method to find these 

solutions. Based on the numerical results we obtained, the fuzzy Adomian decomposition method is 

a highly efficient method in solving and gives accurate results, and in some cases, this method gives 

us the exact analytical solution. The accuracy of this method varies from one fuzzy differential 

equation to another, and this depends on the type of fuzzy differential equation, whether it is of 

the first order or the highest order, and also depends on the nature of the fuzzy differential 

equation, whether it is linear or non-linear. 

Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the 

publication of this paper. 

 

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