APPLICATION OF DIGITAL CELLULAR RADIO FOR MOBILE LOCATION ESTIMATION


IIUM Engineering Journal, Vol. 19, No. 1, 2018 Yu.E.Fayziev 

 168 

ON THE CONTROL OF HEAT CONDUCTION 

FAYZIEV  YUSUF ERGASHEVICH  

National University of Uzbekistan,  

700174, Students Town, Tashkent, Uzbekistan. 

Corresponding author: fayziev.yusuf@mail.ru  

(Received: 9th Feb 2017; Accepted: 15th March 2018; Published on-line: 1st June 2018) 

 https://doi.org/10.31436/iiumej.v19i1.796 

ABSTRACT: Mathematical models of thermo control processes in a rectangular plate 

are considered. In the model under consideration, the temperature inside a plate is 

controlled by heat exchange through one boundary while the other three are insulated. 

The control parameter is a function that satisfies certain integral equations. Sufficient 

conditions for achieving the given projection of the temperature at a fixed point on the 

plate and given average temperature are studied.  

ABSTRAK: Model matematik bagi proses kawalan suhu dalam bekas segi empat tepat 

telah dipilih. Melalui model ini, suhu bekas dikawal dengan menukar haba melalui salah 

satu sisi bekas, manakala tiga sisi lain telah ditebat. Parameter kawalan ini ialah fungsi, 

di mana ia sesuai dengan persamaan sesetengah integral. Keadaan sesuai bagi mencapai 

suhu tetap bekas seperti cadangan dan suhu purata yang diberikan turut dikaji. 

KEYWORDS: non-homogeneous heat equation; control of heat conduction; boundary 

conditions; initial conditions   

1. INTRODUCTION  

Many physical processes and engineering problems are described by heat/diffusion 

equations and the study of the properties of the solutions is important. Methods of the 

solutions of various boundary value problems and problems with the initial conditions can 

be found in [1]. For the first time, a detailed explanation of the problems of control of the 

system with distributed parameters, described by partial differential equations were given 

in [2]. 

In recent years, interest in the study of a system with distributed parameters has 

increased significantly. In works [3-5] by Il’in and Moiseev, the questions of boundary 

control by the various systems described by a wave equation are studied. Notable works 

[6, 7], studied problems related to the process of control associated with the equations of 

parabolic type, particularly the heat transfer process. 

Equation (1), in the homogeneous case, describes the problem of heat control studied 

by Alimov in [8-12]. The problem for the non-homogeneous equation (1), in the case of 

heat transfer, is a rod studied in [13]. 

2.   FORMULATION OF PROBLEMS  

Consider the non-homogeneous heat equation 



IIUM Engineering Journal, Vol. 19, No. 1, 2018 Yu.E.Fayziev 

 169 

2
( ) ( , , )t xx yyu a u u f x y t   ,   1 20 , 0x l y l    ,  0t        

(1) 

with the boundary conditions 

  

1

2

(0, , ) ( ), ( , , ) 0,

( , 0, ) 0, ( , , ) 0,

(0) 0, | ( ) |

u y t t u l y t

u x t u x l t

t M



 

 


 
  

  0t                           

(2) 

and initial conditions  

  ( , , 0) ( , )u x y x y ,         
1 2

0 , 0x l y l    ,                                       (3) 

where  φ(x,y) – a given function and M – a positive number.  

It is known that the boundary condition (2) means µ(t) heat entry from the x = 0 side 

of the rectangle and zero temperature kept in other sides. In the present paper we study 

following problems: 

Problem 1: Let  B > 0 be given number. It is required to find such a temperature µ(t) 

that problems (1) – (3) has a solution satisfying following integral relation (Fig. 1):   

                       
1 2

0 0

( , , )

l l

u x y t dydx B
 (4) 

                           

                                             Fig. 1: Total heat. 

Problem 2: Let B > 0 be a given number. Find a temperature µ(t) such that the 
problem (1) – (3) has a solution that at time t at inside point ( , )

o o
A x y  satisfies the 

condition (Fig. 2): 

                        ( , , )
o o

u x y t B                                          (5) 

 

1 2

0 0

( , , )

l l

u x y t dydx B  

l

1 

l

2 

0 x 

y 

h
e
a
t
 



IIUM Engineering Journal, Vol. 19, No. 1, 2018 Yu.E.Fayziev 

 170 

                         

Fig. 2: The temperature at point ( , )
o o

x y . 

Problem 3: Now we consider equation (1) with the initial condition (3) and the 
boundary conditions as follows 

                             
1

2

(0, , ) 0, ( , , ) ( ),

( , 0, ) 0, ( , , ) 0,

(0) 0, | ( ) |

u y t u l y t t

u x t u x l t

t M



 

 


 
  

                   t ≥ 0        
    (6) 

Let B > 0 be a given number. Find a temperature µ(t) such that problems (1), (3), (6) 

have a solution that satisfies (4). 

Note, that boundary condition (6) means that heat flow µ(t) is coming from the 

boundary x = l1  other boundaries kept with zero temperature.  

3.   THE SOLUTION OF PROBLEM 1 

In this section we solve problem 1. 

Theorem 1: Let B > 0 be given number. At the time t  a solution of the problem (1) –
(3) satisfying the condition (4) if for the function µ(t) following equality is true: 

0

( , ) ( ) ( )

t

K t d g t      (7) 

where   

2

2

( ) ( )21 2
4 2 2

1 1

1 ( 1) 1 ( 1)4
( , ) ( ) nm

n m

a t

nm

n m

l l
K t a e

n m

 
 



 
 

 

     
  

   (8) 

 

( , , )
o o

u x y t B  

l

1 

l

2 

0 x 

y 

h
e
a
t
 

xo 

 yo 



IIUM Engineering Journal, Vol. 19, No. 1, 2018 Yu.E.Fayziev 

 171 

2
( )1 2

2
1 1

1 ( 1) 1 ( 1)
ˆ( ) ( ) nm

n m

a t

nm nm

n m

l l
g t B f t c e

nm





 


 

     
      

   
(9) 

1 2
2

( )

1 2 1 20 0 0

4ˆ ( ) ( , , ) sin sin nm
l lt

a

nm

n m
f t f e d d d

l l l l

    
        (10) 

1 2

1 2 1 20 0

4
( , ) sin sin

l l

nm

n m
c d d

l l l l

   
     

 (11) 

2 2

1 2

nm

n m

l l

 


   
    

   

 (12) 

Proof: Solution of the problem (1) – (3) can be represented as following series:  

2
( )

1 1

( , , ) ( , , ) ( ) sin sin ,nm
a t

nm nm n m

n m

u x y t U x y t f t c e x y


 
 



 

       (13) 

where  

1
2

1

2

( ) 0
( , , )

0 0,

l x
t if y l

lU x y t

if y y l




 
 
    

 

1 2
2 2

1

1 2 10 0 0

4
( ) ( , , ) ( ) sin sin nm

l lt
a

nm n m

l
f t f e d d d

l l l

 
           

 
    

 


 

(14) 

       

1

n

n

l


  ,    

2

m

m

l


   and  nmc  is defined by (11). 

According the Theorem a solution should satisfy condition (4) which means 

         
1 2 1 2

1

10 0 0 0

( , , ) ( )

l l l l
l x

B u x y t dydx t dydx
l




     

        
1 2

2
( )

1 10 0

( ) sin sin .nm
l l

a t

nm nm n m

n m

f t c e x ydydx


 
 



 

    
 

In the second term, we change the order of integration and summation after evaluation to 

get  

2
( )1 2 1 2

2
1 1

1 ( 1) 1 ( 1)
( ) ( )

2
nm

n m
a t

nm nm

n m

l l l l
B t f t c e

n m






 


 

   
     . 

 

Then from (14) and (10) obtain  



IIUM Engineering Journal, Vol. 19, No. 1, 2018 Yu.E.Fayziev 

 172 

2
( )1 2 1 2

2
1 1

1 ( 1) 1 ( 1)
ˆ( ) ( )

2
nm

n m

a t

nm nm

n m

l l l l
B t f t c e

nm






 


 

     
      

   

1 2
2

( ) ( )1
2

1 1 10 0 0

4
( ) sin sinnm

l lt
a t

n m

n m

x l
e x ydydxd

l

 
    



 
 

 

 
   

  
    

1 ( 1) 1 ( 1)
n m

nm

     
  

 . 

Due to µ(0) = 0 the last integrals can be evaluated as follows 

2 2 2
( ) ( ) ( )2

0 0

( ) ( ) ( ) ( ) ,nm nm nm
t t

a a t a

nm
e d t e a e d

    
            

2

2

0

sin 1 ( 1)

l

m

m

l
ydy

m



   
  , 

1

1 1

10

sin

l

n

x l l
xdx

l n





  . 

Thus, we have  

2
( )1 2 1 2

2
1 1

1 ( 1) 1 ( 1)
ˆ( ) ( )

2
nm

n m

a t

nm nm

n m

l l l l
B t f t c e

nm






 


 

     
      

 
 

2

2

( ) ( )21 2
4 2 2

1 1 0

1 ( 1) 1 ( 1)4
( ) ( ) ( ) nm

n mt
a t

nm

n m

l l
t a e d

n m

 
    



 
 

 

         
  

 
 

. 

Write the last equation in a form    

2
( )1 2 1 2

2
1 1

1 ( 1) 1 ( 1)
ˆ( ) ( )

2
nm

n m

a t

nm nm

n m

l l l l
B t f t c e

nm






 


 

     
      

 
 

1 2
4 2 2

1 1

4 8
( )

(2 1) (2 1)i j

l l
t

i j




 

 

 
 

  

       
2

2

( ) ( )21 2
4 2 2

1 10

1 ( 1) 1 ( 1)4
( ) ( ) .nm

n mt
a t

nm

n m

l l
a e d

n m

 
   



 
 

 

         
  

 
 


 

After evaluation of the series in third term and taking into account following notations 

2

2

( ) ( )21 2
4 2 2

1 1

1 ( 1) 1 ( 1)4
( , ) ( ) nm

n m

a t

nm

n m

l l
K t a e

n m

 
 



 
 

 

     
  

  , 



IIUM Engineering Journal, Vol. 19, No. 1, 2018 Yu.E.Fayziev 

 173 

       

2
( )1 2

2
1 1

1 ( 1) 1 ( 1)
ˆ( ) ( ) .nm

n m

a t

nm nm

n m

l l
G t f t c e

nm





 


 

     
    

 
 

To obtain 

        

2 2

1 2 1 2
4

0

32
( ) ( ) ( , ) ( ) ( )

2 8 8

t
l l l l

B t t K t d G t
 

     


      . 

Thus we obtain the following integral equation  

        0
( , ) ( ) ( )

t

K t d B G t      . 

If we denote ( ) ( )g t B G t   then  

         0
( , ) ( ) ( ).

t

K t d g t      

Theorem 1 is proven.  

 

4.  THE SOLUTION OF PROBLEM 2 

In this section we study problem 2. 

Theorem 2: Let B > 0 a given number. At time t a solution of the problem (1) – (3) 
satisfies the condition (5) if function µ(t) satisfies following integral equation  

                          0
( , , , ) ( ) ( ),

t

o o
K x y t d g t      (15) 

where 

2
( ) ( )2

2
1 1

1 ( 1)4
( , , , ) ( ) sin sin ,nm

m

a t

o o nm n o m o

n m

K x y t a e x y
nm

 
   



 
 

 

  
    

   

2
( )

1 1

ˆ( ) ( ) sin sinnm
a t

nm nm n o m o

n m

g t B f t c e x y


 
 



 

    
  . 

Proof: It is known that a solution of problems (1) – (3) has the form of (13). Let this 

solution satisfies condition (5). Then obtain the following equality 

1

1

( , , ) ( )o
o o

l x
B u x y t t

l



    

2
( )

1 1

( ) sin sinnm
a t

nm nm n o m o

n m

f t c e x y


 
 



 

     . 

Using equality (14) obtain following 



IIUM Engineering Journal, Vol. 19, No. 1, 2018 Yu.E.Fayziev 

 174 

2
( )1

1 11

ˆ( ) ( ) sin sinnm
a to

nm nm n o m o

n m

l x
B t f t c e x y

l


  

 


 


     
   

1 2
2

( ) ( )1

1 11 2 10 0 0

4
( ) sin sinnm

l lt
a t

n m

n m

l
e d d d

l l l

 
        

 
 

 

 
   

  
   

2
( )

sin sin .nm
a t

n o m o
e x y


 


   

As it was done in the previous problem, by evaluation of the integrals, we obtain following 

equality 

1

1

( ) ( , , )o
o o

l x
B t G x y t

l



    

2
( ) ( )2

2
1 1 0

1 ( 1)4
( ) ( ) ( ) sin sinnm

m t
a t

nm n o m o

n m

t a e d x y
nm

 
      



 
 

 

     
   

 
   

where 

  

2
( )

1 1

ˆ( , , ) ( ) sin sinnm
a t

o o nm nm n o m o

n m

G x y t f t c e x y


 
 



 

   
 

. 

From this we get  

  

1

1

( ) ( , , )o
o o

l x
B t G x y t

l



    

 

1 2
2

1 1

(2 1)
sin 2sin

4
( )

2 1

o o

n k

x n y k

l l
t

n k

 




 

 

  
    

   
  

    

   

 

2
( ) ( )2

2
1 10

1 ( 1)4
( ) ( ) sin sinnm

mt
a t

nm n o m o

n m

a e x y d
nm

 
     



 
 

 

   
     

 
 
 

 . 

By evaluation of the series in the third term and using following notation 

2
( ) ( )2

2
1 1

1 ( 1)4
( , , , ) ( ) sin sinnm

m

a t

o o nm n o m o

n m

K x y t a e x y
nm

 
   



 
 

 

  
    

   

obtain  

1 1

1 1 0

0

( ) ( ) ( , , , ) ( ) ( , , )

( , , , ) ( ) ( , , ).

t

o o
o o o o

t

o o o o

l x l x
B t t K x y t d G x y t

l l

K x y t d G x y t

     

   

 
    

 





 



IIUM Engineering Journal, Vol. 19, No. 1, 2018 Yu.E.Fayziev 

 175 

If we denote ( ) ( , , )
o o

g t B G x y t  , the obtained integral is equation (15). Theorem 2 is 

proven. 

5.   THE SOLUTION OF PROBLEM 3 

In this section we find the solution of problem 3. 

Theorem 3: Let B > 0 be a given number. At a time t, the problem (1), (3), (6) has a 
solution that satisfies (4) if µ(t) a solution of the following integral equation  

                             0
( , ) ( ) ( )

t

K t d g t      (16) 

where  

2

2
1

( ) ( )21 2
4 2 2

1 1

( 1) 1 ( 1) 1 ( 1)4
( , ) ( ) nm

n n m

a t

nm

n m

l l
K t a e

n m

 
 




 

 

 

      
  

  , 

2
( )1 2

2
1 1

1 ( 1) 1 ( 1)
ˆ( ) ( ) nm

n m

a t

nm nm

n m

l l
g t B f t c e

nm





 


 

     
      

  , 

1 2
2

( )

1 2 1 20 0 0

4ˆ ( ) ( , , ) sin sin nm
l lt

a

nm

n m
f t f e d d d

l l l l

    
       , 

1 2

1 2 1 20 0

4
( , ) sin sin

l l

nm

n m
c d d

l l l l

   
       , 

2 2

1 2

nm

n m

l l

 


   
    

   

. 

 

Proof:  We know that a solution of the problem (1), (3), (6) as follows 

2
( )

1 1

( , , ) ( , , ) ( ) sin sinnm
a t

nm nm n m

n m

u x y t U x y t f t c e x y


 
 



 

       (17) 

where  

2

1

2

( ) 0
( , , )

0 0

x
t if y l

lU x y t

if y and y l




 
 
  

, 

1 2
2 2

1 2 10 0 0

4
( ) ( , , ) ( ) sin sin .nm

l lt
a

nm n m
f t f e d d d

l l l

 
           

 
    

 
  (18) 

According to the conditions of theorem 3, a solution (17) must satisfy (4) which means  



IIUM Engineering Journal, Vol. 19, No. 1, 2018 Yu.E.Fayziev 

 176 

1 2 1 2

10 0 0 0

( , , ) ( )

l l l l
x

B u x y t dydx t dydx
l
     

1 2
2

( )

1 10 0

( ) sin sinnm
l l

a t

nm nm n m

n m

f t c e x ydydx


 
 



 

     . 

In the second term, change the order of summation and integration after evaluation of the 

integrals to obtain the following 

2
( )1 2 1 2

2
1 1

1 ( 1) 1 ( 1)
( ) ( )

2
nm

n m
a t

nm nm

n m

l l l l
B t f t c e

n m






 


 

   
     . 

From (18) and (8), obtain  

2
( )1 2 1 2

2
1 1

1 ( 1) 1 ( 1)
ˆ( ) ( )

2
nm

n m

a t

nm nm

n m

l l l l
B t f t c e

nm






 


 

     
      

   

1 2
2

( ) ( )

2
1 1 10 0 0

4
( ) sin sinnm

l lt
a t

n m

n m

x
e x ydydxd

l

 
    



 
 

 

 
   

  
   

1 ( 1) 1 ( 1)
n m

nm

     
  

 . 

Taking into account  

2 2 2
( ) ( ) ( )2

0 0

( ) ( ) ( ) ( )nm nm nm
t t

a a t a

nm
e d t e a e d

    
           , 

2

2

0

sin 1 ( 1)

l

m

m

l
ydy

m



   
 

,      
1

11

10

sin ( 1) .

l

n

n

x l
xdx

l n





 

 

From the last equality, by introducing the notation  

 

2
( )1 2

2
1 1

1 ( 1) 1 ( 1)
ˆ( ) ( ) nm

n m

a t

nm nm

n m

l l
G t f t c e

nm





 


 

     
    

  . 

Obtain the following equation  

 1 2 ( ) ( )
2

l l
B t G t    

2
1

1 2
4 2 2

1 1

( 1) 1 ( 1) 1 ( 1)4
( )

n n m

n m

l l
t

n m





 

 

      
  

   

2

2
1

( ) ( )21 2
4 2 2

1 10

( 1) 1 ( 1) 1 ( 1)4
( ) ( ) .nm

n n mt
a t

nm

n m

l l
a e d

n m

 
   




 

 

 

          
  

 
 

  

By evaluation of the series in the third term and by denoting  

 



IIUM Engineering Journal, Vol. 19, No. 1, 2018 Yu.E.Fayziev 

 177 

          

2

2
1

( ) ( )21 2
4 2 2

1 1

( 1) 1 ( 1) 1 ( 1)4
( , ) ( ) nm

n n m

a t

nm

n m

l l
K t a e

n m

 
 




 

 

 

      
  

   

obtain 

                               0

( , ) ( ) ( )

t

K t d B G t      . 

If we denote g(t) = B–G(t)  then we obtain integral equation (16). The theorem is proven.  

6.   CONCLUSION 

The results of this study have been analyzed and it can be concluded that both a fixed 

average amount of heat and a fixed heat at a given internal point of the rectangular plate 

can be controlled from the one of the boundaries. In this paper, a control function must be 

a solution of the integral equations found in this study.  

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