ACADEMIC RESEARCH IN MODERN SCIENCE
International scientific-online conference
153
SOLUTION OF THE THIRD INITIAL-BOUNDARY VALUE PROBLEM
FOR A HYPERBOLIC TYPE EQUATION USING THE FOURIER
METHOD
Daniyarova Gulayim Kuwatbayevna
Mathematic teacher of number 27th secondary school
https://doi.org/10.5281/zenodo.16926860
Let the equation
2
tt
xx
u
a u
(1)
be given in halfway
0
x
l
,
0
t
with the following boundary conditions
(0, )
0
x
u
t
,
( , )
( , )
0
x
u l t
hu l t
,
0
h
(2)
and initial conditions
0
( )
t
u
x
,
( )
t t
u
x
(3)
to solve the given equation, we use the method of separating the solution
from variables. First, the non-trivial solution of equation (1) satisfying
conditions (2) is sought in the form
( , )
( )
( )
u x t
X x T t
(4)
using Fourier's method of separation of variables. Substituting (4) into
equation (1), we obtain the equation
2
( )
( )
( ) ( )
X x T t
a X
x T t
When we divide this equation by
( )
( )
X x T t
, we obtain a simple differential
equation of the form
2
2
( )
( )
0
T t
a
T t
(5)
2
( )
( )
0
X
x
X x
(6)
with respect to the functions
( )
T t
and
( )
X x
, where
is an unknown
constant parameter. Also, the solutions of equations
( )
X x
and
( )
T t
have the
form
( )
cos
sin
X x
A
x
B
x
( )
cos
sin
T t
C
a t
D
a t
.
Substituting (4) into (2), considering that
( )
0
T t
, it follows that the
function
( )
X x
must satisfy the conditions
(0)
0,
( )
( )
0,
X
X l
hX l
From it, solution
( )
cos
X x
x
is obtained. where
k
k
ctg
lh
and
k
k
l
,
it6 means that we have the eigenfunction
( )
cos
k
k
X x
x
ACADEMIC RESEARCH IN MODERN SCIENCE
International scientific-online conference
154
Where,
k
k
l
,
1,2,3,...
k
, then
( )
cos
sin
k
k
k
k
k
T t
C
a t
D
a t
.
Now it follows from the found
( )
k
X x
and
( )
k
T t
according to (4) that
( , )
( )
( )
cos
sin
cos
k
k
k
k
k
k
k
k
u x t
X
x T t
C
a t
D
a t
x
(7)
Since function (1) is a non-trivial solution of equation (2) satisfying
condition (2), then the infinite sum of solutions (7) is also a solution, that is
0
( , )
cos
sin
cos
k
k
k
k
k
k
u x t
C
a t
D
a t
x
(8)
We differentiate (8) with respect to
t
:
0
( , )
sin
cos
cos
t
k
k
k
k
k
k
k
u x t
a
C
a t
D
a t
x
(9)
Assuming
0
t
in (8) and (9), based on the initial conditions (3), we obtain
the following equalities
0
( ,0)
cos
( )
k
k
k
u x
C
x
x
,
0
( ,0)
cos
( )
t
k
k
k
k
u x
a D
x
x
Now let's find the unknowns
,
k
k
C
D
from these equations, then
0
2
( )
( )
l
k
k
C
x X
x dx
l
,
0
2
( )
( )
l
k
k
k
D
x X
x dx
a
,
k
.
Substituting these into (4), we obtain the solution of the third mixed
problem (1), (2), (3).
Literature:
1.
Salohitdinov M.S. Matematik fizika tenglamalari. – T.: O’zbekiston, 2002
2.
Бицадзе А.В., Калинисенко Д.Ф. Сборник задач по уравнениям
математической физики. – М.: Наука, 1978.
3.
Владимиров В.С., Михайлов В.П. и др. Сборник задач по уравнениям
математической физики – М.: Наука, 1974Ладыженская О.А. Краевые
задачи математической физики. Наука, 1973 год.
4.
Петровский И.Г. Лекции об уравнениях с частными производными.
Физматгиз, 1961 год.
5.
Владимиров В.С. Уравнения математической физики. Наука, 1971.
6.
Ш.Г.Касимов, Т.Н.Аликулов, Ш.К.Отаев, Ғ.С.Хаитбоев, М.М.Бабаев
“Математик физиканинг замонавий усуллари” 1-2 том
