American Journal of Business Management, Economics and Banking 
ISSN (E): 2832-8078                        Volume 24, | May - 2024                               

 

 
P a g e  | 18  www.americanjournal.org 

 

A MIXING PROBLEM FOR A QUASI LINEAR EQUATION WITH 

PARTICULAR DERIVATIVES WITH SOME LATE ARGUMENT 

A. Kalandarov 

GulDU “Matematika” kafedrasi kata o‘qituvchisi 

e-mail: abdukayumkalandarov1948@gmail.com 

 

M. M. Anorbayev 

GulDU “Matematika” kafedrasi o‘qituvchisi 

e-mail: madaminanorbayev254@gmail.com 

 

I. J. Jangibayev 

GulDU “Matematika” kafedrasi o‘qituvchisi 

GDPI“Matematika” kafedrasi o‘qituvchisi 

e-mail: mr.jiu_newlife1992@bk.ru 

 

A B S T R A C T K E Y W O R D S 

In the work, the solutions of the mixed problem set for the quasi – 

linear partial differential equation with some delayed arguments 

were studied, and algorithm for finding its numerical solutions was 

solved by the finite difference method. 

mixed problem, delayed 

argument, quasi linear, 

numerical solutions, 

algorithm, finite diffence 

method, transporent and 

non revealing schemes, 

driving. 

 

 

Introduction 

In the work, the solutions of the mixed problem set for the quasi – linear partial differential equation 

with some delayed arguments were studied, and algorithm for finding its numerical solutions was 

solved by the finite differince method. 

 

НЕКОТОРЫЕ СМЕШАННАЯ ЗАДАЧА ДЛЯ КВАЗИЛИНЕЙНОГО  УРАВНЕНИЯ С 

ЧАСТНЫМИ ПРОИЗВОДНЫМ С ЗАПАЗДЫВАЮЩИМ  АРГУМЕНТОМ 

 

Abstract 

В данной работе решается смешанная задача для квазилинейного дифференциального 

уравнения с запаздывающим аргументом,  задача решается методом конечных разностей. 

Составляется алгоритм численного решения 



American Journal of Business Management, Economics and Banking 
Volume 24 May - 2024 

 

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In the article, the quasi – linear equation of the hyperbolic type with delayed argument   𝑄 =

{𝜏 ≤ 𝑡 ≤ 𝑇, 0 ≤ 𝑥 ≤ 𝑙, 0 ≤ 𝑦 ≤ 𝑚}    in the area 

𝜕2𝑢

𝜕𝑡2
= 𝑎2(

𝜕2𝑢

𝜕𝑥2
+

𝜕2𝑢

𝜕𝑦2
) + 𝑏2 (

𝜕2𝑢(𝑡 − 𝜏, 𝑥, 𝑦)

𝜕𝑥2
+

𝜕2𝑢(𝑡 − 𝜏, 𝑥, 𝑦)

𝜕𝑦2
) + 

+𝑓(𝑡, 𝑥, 𝑦, 𝑢(𝑡, 𝑥, 𝑦), 𝑢(𝑡 − 𝜏, 𝑥, 𝑦), 𝑢𝑡(𝑡, 𝑥, 𝑦), 𝑢𝑡(𝑡 − 𝜏, 𝑥, 𝑦))                               (1) 

            ( 𝑡, 𝑥)𝜖𝐸 = {0 ≤ 𝑡 ≤ 𝜏, 0 ≤ 𝑥 ≤ 𝑙, 0 ≤ 𝑦 ≤ 𝑚}  start when   

                       
𝑢(𝑡, 𝑥, 𝑦) = 𝜑(𝑡, 𝑥, 𝑦)

𝑢𝑡(𝑡, 𝑥, 𝑦) = 𝜑𝑡
′(𝑡, 𝑥, 𝑦)

}                                                (2) 

 Gwin the initial internal conditions when the tsist  

τ≤ 𝑡 ≤ 𝑇 

               
𝑢(𝑡, 0, 𝑦) = 0

𝑢(𝑡, 𝑙, 𝑦) = 0
}   0 ≤ 𝑦 ≤ 𝑚       

𝑢(𝑡, 𝑥, 0) = 0
𝑢(𝑡, 𝑥, 𝑚) = 0

}  0 ≤ 𝑥 ≤ 𝑙     (3) 

The problem of finding a satisfactory solution of homogeneous boundary conditions by the 

finite this overael this issue of the existence and uniqueness of generalized solutions was discussed 

by the author in we introduce the notation for.   Q  Lets’ mesh the area   𝑡𝑘 = 𝑘𝜏, 𝑥𝑖 = 𝑖𝛥, 𝑦𝑗 = 𝑗ℎ  

that, the grid function  𝑢(𝑡𝑘, 𝑥𝑖 , 𝑦𝑖) = 𝑢𝑖𝑗
𝑘   Lits’ enter the designations. 

𝑖 = 0,1,2, … , 𝑁, 𝑗 = 0,1,2, … , 𝑃, 𝑘 = 0,1,2, … , 𝑀, 

𝑀𝜏 = 𝑇, 𝑁𝛥 = 𝑙 , 𝑃ℎ = 𝑚 

(2)  from the initial conditions  (t,x,y)𝜖𝐸  when( 𝑘 = 0 va 𝑘 = 1 da ) 

𝑢𝑖𝑗
0 = 𝜑(0, 𝑥𝑖 , 𝑦𝑗)          

𝑢𝑖𝑗
1 −𝑢𝑖𝑗

0

𝜏
≈ 𝜑𝑡

′(𝜏, 𝑥𝑖, 𝑦𝑗)          (4) 

𝑢𝑖𝑗
1 ≈ 𝑈𝑖𝑗

0 + 𝜏𝜑𝑡
′(𝜏, 𝑖𝛥, 𝑗ℎ) = 𝜑(0, 𝑖𝛥, 𝑗ℎ) + 𝜏𝜑𝑡

′(𝜏, 𝑖𝛥, 𝑗ℎ)         (5) 

 (3)  from the boundary conditions 

              𝑢0𝑗
𝑘 = 0 ,     𝑢𝑁𝑗

𝑘 = 0 ,    𝑢𝑖0
𝑘 = 0 ,     𝑢𝑖𝑝

𝑘 = 0             (6) 

We will have values. The values of  u(t,x,y)   are given on the sides and base of the sphere  Q. 

Using the above, we find the numerical values of     𝑢(𝑡, 𝑥, 𝑦)  at the internal nodes of the field Q. 

    We may use the follaving non – disclosure schemes without general permussion:    

 
𝑢𝑖𝑗

𝑘+1 − 2𝑢𝑖𝑗
𝑘 + 𝑢𝑖𝑗

𝑘−1

𝜏2
= 𝑎2 (

𝑢𝑖+1,𝑗
𝑘+1 − 2𝑢𝑖𝑗

𝑘+1 + 𝑢𝑖−1,𝑗
𝑘+1

𝛥2
+

𝑢𝑖,𝑗+1
𝑘 − 2𝑢𝑖𝑗

𝑘 + 𝑢𝑖,𝑗−1
𝑘

ℎ2
) + 

+𝑏2(
𝑢𝑖+1,𝑗

𝑘 −2𝑢𝑖𝑗
𝑘 +𝑢𝑖−1,𝑗

𝑘

𝛥2 +
𝑢𝑖,𝑗+1

𝑘 −2𝑢𝑖𝑗
𝑘 +𝑢𝑖,𝑗−1

𝑘

ℎ2 ) + 𝑓𝑖𝑗
𝑘                                               (7) 

or 

𝑢𝑖𝑗
𝑘+1 − 2𝑢𝑖𝑗

𝑘 + 𝑢𝑖𝑗
𝑘−1

𝜏2
= 𝑎2 (

𝑢𝑖+1,𝑗
𝑘 − 2𝑢𝑖𝑗

𝑘 + 𝑢𝑖−1,𝑗
𝑘

𝛥2
+

𝑢𝑖,𝑗+1
𝑘+1 − 2𝑢𝑖𝑗

𝑘+1 + 𝑢𝑖,𝑗−1
𝑘+1

ℎ2
) + 

+𝑏2(
𝑢𝑖+1,𝑗

𝑘 −2𝑢𝑖𝑗
𝑘 +𝑢𝑖−1,𝑗

𝑘

𝛥2
+

𝑢𝑖,𝑗+1
𝑘 −2𝑢𝑖𝑗

𝑘 +𝑢𝑖,𝑗−1
𝑘

ℎ2
) + 𝑓𝑖𝑗

𝑘                                               (71) 

comes out. Here  

  𝑓𝑖𝑗
𝑘 = 𝑓(𝑘𝜏, 𝑖𝛥, 𝑗ℎ, 𝑢(𝑘𝜏, 𝑖𝛥, 𝑗ℎ), 𝑢((𝑘 − 1)𝜏, 𝑖𝛥, 𝑗ℎ), (𝑢(𝑘𝜏, 𝑖𝛥, 𝑗ℎ) − 𝑢((𝑘 − 1)𝜏, 𝑖𝛥, 𝑗ℎ))/

τ, (𝑢((𝑘 − 1)𝜏, 𝑖𝛥, 𝑗ℎ) − 𝑢((𝑘 − 2)𝜏, 𝑖𝛥, 𝑗ℎ))/τ))  

 τ≤ 𝑡 ≤ 2𝜏   if we have, we will select the driving mode for the above undisclosed circuit.    

(7) in scheme  𝑘 = 1                  𝑢𝑖𝑗
2 − 2𝑢𝑖𝑗

1 + 𝑢𝑖𝑗
0 = 



American Journal of Business Management, Economics and Banking 
Volume 24 May - 2024 

 

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=
𝑎2𝜏2

𝛥2 (𝑢𝑖+1,𝑗
2 − 2𝑢𝑖𝑗

2 + 𝑢𝑖−1,𝑗
2 ) +

𝑎2𝜏2

ℎ2 (𝑢𝑖,𝑗+1
1 − 2𝑢𝑖𝑗

1 + 𝑢𝑖,𝑗−1
1 ) +

𝑏2𝜏2

𝛥2 (𝑢𝑖+1,𝑗
1 − 2𝑢𝑖,𝑗

1 + 𝑢𝑖−1,𝑗
1 ) +

𝑏2𝜏2

ℎ2 (𝑢𝑖,𝑗+1
1 − 2𝑢𝑖,𝑗

1 + 𝑢𝑖,𝑗−1
1 ) + 𝜏2𝑓𝑖𝑗

1      (8) 

Then the differential equation looks like this: 

𝑎𝑖𝑗
1 𝑢𝑖−1,𝑗

2 + 𝑏𝑖𝑗
1 𝑢𝑖,𝑗

2 + 𝑐𝑖𝑗
1 𝑢𝑖+1,𝑗

2 = 𝐹𝑖,𝑗
1                (9) 

Here, 𝑎𝑖𝑗,
1  𝑏𝑖𝑗

1 , 𝑐𝑖𝑗
1 , 𝐹𝑖𝑗

1 -Coefficients are fixed numbers resulting from the expression (8) from 

(9) as je 1 

                                 𝑎𝑖1
1 𝑢𝑖−1,1

2 + 𝑏𝑖1
1 𝑢𝑖,1

2 + 𝑐𝑖1
1 𝑢𝑖+1,1

2 = 𝐹𝑖,1
1                        (10) 

We use the driving method to solve this differencial equation: 

at  i=1                              𝑎11
1 𝑢01

2 + 𝑏11
1 𝑢11

2 + 𝑐11
1 𝑢21

2 = 𝐹11
1                     (111) 

from this  𝑢11
2   ,  𝑢21

2   we express it linearly 

                         𝑢11
2 = 𝐿11

1 𝑢21
2 + 𝐾11

1                                                     (121) 

we will have patience, in this  

                       𝐿11
1 = −

𝑐11
1

𝑏11
1     ,    𝐾11

1 =
𝐹11

1

𝑏11
1                                            (131) 

in 𝑖 = 2,  (10)  from            

                           𝑎21
1 𝑢11

2 + 𝑏21
1 𝑢21

2 + 𝑐21
1 𝑢31

2 = 𝐹21
1                              (112) 

(121) if we use 

𝑎21
1 (𝐿11

1 𝑢21
2 + 𝐾11

1 ) + 𝑏21
1 𝑢21

2 + 𝑐21
1 𝑈31

2 = 𝐹21
1  

Now  𝑢21
2   ni  𝑢31

2  letus linearey express u by us 

                  𝑢21
2 = 𝐿21

1 𝑢31
2 + 𝐾21

1                                                      (122) 

Is formed, in which,             

       𝐿21
1 = −

𝑐21
1

𝑎21
1 𝐿11

1 +𝑏21
1     ,      𝐾21

1 =
𝐹21

1 −𝑎21
1 𝐾11

1

𝑎21
1 𝐿11

1 +𝑏21
1                                  (132) 

   etc  when   i=n-1   

   𝑎𝑁−1,1
1 𝑢𝑁−2,1

2 + 𝑏𝑁−1,1
1 𝑢𝑁−1,1

2 + 𝑐𝑁−1,1
1 𝑢𝑁1

2 = 𝐹𝑁−1,1
1                   (11𝑁−1) 

𝑎𝑁−1,1
1 (𝐿𝑁−2,1

1 𝑢𝑁−1,1
2 + 𝐾𝑁−2,1

1 ) + 𝑏𝑁−1,1
1 𝑢𝑁−1,1

2 + 𝑐𝑁−1,1
1 𝑢𝑁1

2 = 𝐹𝑁−1,1
1  

  from this.   

                                        𝑢𝑁−1,1
2 = 𝐿𝑁−1,1

1 𝑢𝑁1
2 + 𝐾𝑁−1,1

1                ( 12𝑁−1) 

    𝐿𝑁−1,1
1 = −

𝑐𝑁−1,1
1

𝑎𝑁−1,1
1 𝐿𝑁−2,1

1 +𝑏𝑁−1,1
1 ,    𝐾𝑁−1,1

1 =
𝐹𝑁−1,1

1 −𝑎𝑁−1,1
1 𝐾𝑁−2,1

1

𝑎𝑁−1,1
1 𝐿𝑁−2,1

1 +𝑏𝑁−1,1
1      (13𝑁−1) 

     Driving coefficient -  𝐿𝑖1
1 , 𝐾𝑖1

1  is found in the correct way from the formula (13i) in ascending 

order. This, when 𝑗 –  1, the process is terminated. Them when it is 2, the above process is continued 

and  𝑈, 𝑆 are found , etc. They are found in the first layer. 

2τ≤ 𝑡 ≤ 3𝜏   when, we apply the above (7) driving method to the undisclosed scheme. When 𝑘 = 2 

all the above processes are repeated and  𝑢𝑖𝑗
3   is found in the second layer, and so on. It is calculated 

   𝑢𝑖𝑗
𝑘 .It can be calculated using the differential scheme (71) above  𝑢𝑖𝑗

𝑘 . 

 

References 

1. М.Isroilov, Calculation method. 2nd floor. Toshkent, “O’zbekiston”, 2008. 

2. A.Kalandarov, Smeshannaya zadacha dlya giperbolicheskix uravneniy s zapazdivayushimsya 

argumentami. Baku. Uchyoniye zapiski AGU, 1975 y.