Download Method of external potential in solution of Cauchy mixed problem for

Method of external potential in solution of Cauchy mixed problem
for the heat equation
Kalmenov T.Sh., Tokmagambetov N.E.
Institute of Mathematics and Mathematical Modeling MES RK,
050010, Shevchenko str., 28, Almaty, Kazakhstan
E-mail: [email protected], [email protected]
Abstract . Numerous research works are devoted to study Cauchy mixed problem for
model heat equations because of its theoretical and practical importance. Among them we can
notice monographers [1]-[3] which demonstrate main research methods, such as Fourier
method, integral equations method and method of a priori estimates. But at the same time to
represent the solution of Cauchy mixed problem in integral form by given and known
functions have not been achieved up to now. This paper completes this omission for the one
dimensional heat equation.
Introduction. Partial differential equations of parabolic type are widely
represented in the study of heat conductivity and diffusion process. Numerous
research works are devoted to study Cauchy mixed problem for model heat
equations because of its theoretical and practical importance. Among them we
can notice monographers [1]-[3] which demonstrate main research methods,
such as Fourier method, integral equations method and method of a priori
estimates. But at the same time to represent the solution of Cauchy mixed
problem in integral form by given and known functions have not been achieved
up to now. This paper completes this omission for the one dimensional heat
equation.
Exterior potential method as a special continuation of a solution for all halfspace is widely used under the solution Cauchy mixed problem. Our idea is
based on a representation possibility of general solution only in the form of
volume potential excluding surface integrals. Thus the system of integral
equations obtained by this method allows us to construct the solution in
quadrature.
Material and methods.
Consider the following problem in a plane domain   {0  x  1,0  t  T }
Cauchy mixed problem.
equation in 
To find a regular solution of the following
u ( x, t )  (
 2

)u ( x, t )  f ( x, t ),
t x 2
(1)
with the initial condition
u ( x,0)  0
(2)
and boundary conditions
u (0, t )  u (1, t )  0.
(3)
Our goal is to construct a classical solution of the problem (1)-(3) in a
quadrature. We will seek a solution in form of sum of three volume potentials
u ( x, t )  u f ( x, t )  u g 0 ( x, t )  u g1 ( x, t )
(4)
where
t
1 exp( 
u f ( x, t )  ( 1  f )( x, t )   d 
0
t
u g 0 ( x, t )  ( 1  g 0 )( x, t )   d
0
t
u g1 ( x, t )  ( 1  g1 )( x, t )   d
0
0
(x   )2
)
4(t   )
2  (t   )
0 exp( 


(x   )2
)
4(t   )
2  (t   )
(x   )2
)
  exp( 
4(t   )

1
f ( , )d ,
2  (t   )
g 0 ( , )d ,
g1 ( , )d .
2
x
Here  1 ( x, t )   (t )(2 t ) e x p ( ) is a fundamental solution of the heat
4t
1
,
t

0

equation (1) and  (t )  
is Heaviside theta-function.
0, t  0
1
It should be noticed that the heat potential u f ( x, t ) satisfies the following
boundary condition

t
 (x   )
 ( x   ) 1 
u ( x, t )
u ( , ) 
 
exp  
2
 4(t   ) 
0  4  (t   )
 1
 ( x   ) 1  u ( , ) 
 d  0, ( x, t )  ,


exp  
2  (t   )
 4(t   )    0 
1
where  is a boundary of the domain  . Note that in works [5-6] are
investigated differential operators with non-local boundary conditions as above.
It is easy to verify that the first term in representation (4) u f ( x, t ) is a
solution of non-homogeneous equation (1) and the second and third ones
u g ( x, t ), u g ( x, t ) are solutions of homogeneous equation. Consequently,
representation (4) gives a solution of non-homogeneous equation (1) satisfying
initial condition (2) for arbitrary g 0 ( x, t ) , g1 ( x, t ) .
0
1
Our aim is to choose unknown functions g 0 ( x, t ) and g1 ( x, t ) such that the
solution will satisfy to boundary condition (3).
We will seek functions g 0 ( x, t ) , g1 ( x, t ) in the forms
g 0 ( x, t )  x ( x) g 0 (t ), g1 ( x, t )  ( x  1) ( x  1) g1 (t ).
(5)
Let  (x ) be periodical function with period 2 defining on the interval (0,2 )
x x2
by the formula  ( x)   . This function can be represented by the formula
2 4
k  
 1
1
 ( x)  
exp( ikx) out of the interval [0,2  ].

6 2 k   k 2
We introduce the following notations
2

x
x2 d 2
G0 (t )   ( 0 ( x) 
) 2 exp( 
)dx,
4 dx
4( ) 2 t

2

x
(x   )2 d 2
G1 (t )   ( 0 ( x   ) 
) 2 exp( 
)dx.
2
4

dx
4
(

)
t

F0 (t )  u f ( x, t )
x 0
F1 (t )  u f ( x, t )
t
1
0
0
exp( 
  d 
x 1

2
4(t   )
)
2  (t   )
t
1
0
0
  d 
exp( 
(6)
f ( , )d ,
(1   ) 2
)
4(t   )
2  (t   )
f ( , )d .
Results and discussion. The main result of this paper is the following
(7)
,

Theorem. Let f ( x, t )  C 2 (),0    1, then the solution of the problem
(1)-(3) is represented by the formula (4), where g 0 ( x, t ) and g1 ( x, t ) are given by
formulae (5)
t
g 0 (t ) 
d2
[G0 (t   ) F0 ( )  G1 (t   ) F1 ( )]d ,
dt 2 0
t
g1 (t )  
d2
[G0 (t   ) F1 ( )  G1 (t   ) F0 ( )]d .
dt 2 0
To prove the Theorem, main role plays the following
Lemma. Let unknown functions be given by the formulae (5).
Then g 0 (t ) and g1 (t ) satisfy to system of equations
g 0 ( )
t
2
 (t   )
0
t
2
0
t
g1 ( )
 (t   )
exp( 
d  
2  (t   )
0
t
1
)
4 (t   )
exp( 
d  
1
)
4 (t   )
2  (t   )
0
g1 ( )d 
d
F0 (t ),
dt
g 0 ( )d  
d
F1 (t ),
dt
(8)
where F0 (t ) , F1 (t ) defined by formula (7).
Proof. By substituting the function (4) in the boundary condition (3) and
taking account of (7), we get a system of interval equations with respect to
unknown functions g 0 (t ) and g1 (t )
t
g
0
( )d
0
exp( 


t
0 exp( 
0


4 (t   )
)
2  (t   )
0
 g 0 ( )d
2
(  1) 2
)
4 (t   )
2  (t   )
t

0
1
d   g1 ( )d
t
d   g1 ( )d
0

exp( 
1
4 (t   )
)
2  (t   )
  exp( 

2
(  1) 2
)
4 (t   )
2  (t   )
(  1)d   F0 (t ),
(9)
(  1)d   F1 (t ).
(10)
Integrating by part it is easy to verify correctness of the following equations
0 exp( 


2
4 (t   )
2  (t   )
)
(t   )

2
d  
exp(

)
d



;
4 (t   )
 (t   )  

2
0
 exp( 

2
4 (t   )
)
2  (t   )
1
(t   )
(  1)d 


1

 exp( 

2
)d .
1
2 ( t  )
After putting  : 1   in integrals (9), (10) our system can be transformed
to
t

0
(t   )

t
g 0 ( )d   [
(t   )
0

exp( 
1
4 (t   )
)

1
 exp( 

)d ]g1 ( )d  F0 (t ),
2
1
2 ( t  )
t
[
0
(t   )

exp( 
1
4 (t   )
)
1

t
1
0
2
 exp(  )d ]g 0 ( )d  

(t   )

g1 ( )d  F1 (t ).
2 ( t  )
By differentiating these relations we obtain the system (8). Lemma is proved.
Now we will construct the solution of the system (8) by using Laplace
transformation properties
exp( 
F(
2
)
~
d
4t )   exp( p ) , F ( f  g )  ~
f ( p) g~ ( p), F ( f )  pf ( p),
dt
t
p
system (8) can be transformed in the next form:
g~0 ( p)
p

g~0 ( p)
p

g~1 ( p)
p
~
exp(  p )  pF0 ( p),
exp(  p ) 
g~1 ( p)
p
~
  pF1 ( p),
where g~0 ( p)  F ( g 0 )( p), g~1 ( p)  F ( g1 )( p) are images of Laplace transform. After
solving this system with respect to g~0 ( p) and g~1 ( p) we find
~
~
p 2 F0 ( p)  F1 ( p) exp(  p )
~
g 0 ( p) 
(
),
p
1  exp( 2 p )
~
~
p 2 F1 ( p)  F0 ( p) exp(  p )
~
g1 ( p )  
(
).
p
1  exp( 2 p )
Since Laplace inverse transformation of function
table form, so by using following expansions
1
(11)
1
p 1  exp( 2 p )
has no
1

1

p 1  exp( 2 p )
k 0
1
p
exp( 2 p k ),
exp(  p )
1
p 1  exp( 2 p )


k 0
1
exp( (2k  1) p ),
p
and table values of Laplace inverse transformation
F 1 (
exp( 2k p )
p
1
)
exp( 

( 2k ) 2
(2k  1) 2
)
exp(

)
4t , F 1 ( exp( (2k  1) p ) )  1
4t
.
t
p

t
Then from (11) we obtain
t
g 0 (t ) 
d2
[G0 (t   ) F0 ( )  G1 (t   ) F1 ( )]d ,
dt 2 0
t
g1 (t )  
d2
[G0 (t   ) F1 ( )  G1 (t   ) F0 ( )]d ,
dt 2 0
where

G0 (t )  F 1 (
exp( 2k p )
k 0

G1 (t )  F 1 (
k 0
p

)   ( F 1 (
exp( 2k p )
p
k 0
exp( (2k  1) p )
p
1
)



exp( 
k 0
)) 
1



exp( 
k 0
(2k  1) 2
)
4t
.
t
( 2k ) 2
)
4t ,
t
(12)
(13)
To complete the proof of the theorem we represent functions G0 (t ) and
G1 (t ) by integrals of a known functions.
As long as  ( x  2k ), (k  0,1,2,...) is a tempered distribution ([4], p.
112), then we have
( ( x  2k ), exp( 
x2
( 2k ) 2
))

exp(

).
4t
4 2 t
Therefore, the following relations hold:


x2
(2k ) 2
exp( 
),
 ( ( x  2k ), exp(  4 2t ))  k
4t
k  
 

 ( ( x  (2k  1) ), exp( 
k  
x
2

x2
(2k  1) 2
))

exp(

).

4t
4 2 t
k  
We define function  0 ( x)  
(14)
(15)
x2
on the interval [0,2 ] and then continue
4
it periodically on all axis (, ) . It is shown in ([4], p. 113)

  ( x  2k ) 
k  
where
d2
x2
(

(
x
)

),
0
4
dx 2
d2
is regarded in the general function sense.
dx 2
From here and relation (14) we conclude that


( 2k ) 2
1 d2
x2
x2
1
x2 d 2
x2
exp( 
)  ( 2 ( 0 ( x) 
), exp(  2 ))   ( 0 ( x) 
)
exp(  2 )dx.

4t
2 dx
4
2 
4 dx 2
4 t
4 t
k 0
Analogically, the next relation

 exp( 
k 0

(2k  1) 2
1
(x   )2 d 2
x2
)   ( 0 ( x   ) 
) 2 exp(  2 )dx.
4t
2 
4
dx
4 t
follows from (15).
We obtain formula (6) after substituting the last relations in (12) and (13).
From (13) we have G1 (t )  C  [0,1] , and from (12) (G0 (t ) 
Since f ( x, t )  C
1 ,1

2
(),0    1 , then F0 (t ), F1 (t )  C
difficult to establish g 0 (t ), g1 (t )  C
1
1
1
t
)  C  [0,1] .

2
(0, T ) . Therefore it is not

2
[0, T ] .
Consequently, representation of the solution u ( x, t )  C
(4) is established. Theorem is fully proved.
2  ,1

2
() in the form
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1988.
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equations”, publish. “Nauka”, 1973.
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publish. “Nauka”, 1980.
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