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ВЕСТНИК ТОМСКОГО ГОСУДАРСТВЕННОГО УНИВЕРСИТЕТА
2024 Математика и механика № 88
Tomsk State University Journal of Mathematics and Mechanics
МАТЕМАТИКА
М ATHEMATICS
Original article
UDC 519.63 MSC: 65M32
doi: 10.17223/19988621/88/1
Numerical method for restoring the initial condition for the wave equation
Khanlar M. Gamzaev
Azerbaijan State Oil and Industry University, Western Caspian University, Baku, Azerbaijan, [email protected]
Abstract. The inverse problem of restoring the initial condition for the time derivative for the one-dimensional wave equation is considered. As an additional condition, the solution of the wave equation at a finite time is given. First, the discretization of the derivative with respect to the spatial variable is carried out and the initial problem is reduced to a differential-difference problem with respect to functions depending on the time variable. To solve the resulting differential-difference problem, a special representation is proposed, with the help of which the problem splits into two independent differential-difference problems. As a result, an explicit formula is obtained for determining the approximate value of the desired function for each discrete value of a spatial variable. The finite difference method is used for the numerical solution of the obtained differential-difference problems. The presented results of numerical experiments conducted for model problems demonstrate the effectiveness of the proposed computational algorithm. Keywords: wave equation, inverse problem, recovery of the initial condition, differential-difference problem
For citation: Gamzaev, K.M. (2024) Numerical method for restoring the initial condition for the wave equation. Vestnik Tomskogo gosudarstvennogo universiteta. Matematika i mekhanika - Tomsk State University Journal of Mathematics and Mechanics. 88. pp. 5-13. doi: 10.17223/19988621/88/1
Научная статья
Численный метод восстановления начального условия для волнового уравнения
Ханлар Мехвали оглу Гамзаев
Азербайджанский государственный университет нефти и промышленности; Западно-Каспийский университет, Баку, Азербайджан, [email protected]
Аннотация. Рассматривается обратная задача восстановления начального условия для производной по времени для одномерного волнового уравнения. В качестве
© K.M. Gamzaev, 2024
дополнительного условия задается решение волнового уравнения в конечный момент времени. Сначала проводится дискретизация производной по пространственной переменной, и исходная задача сводится к дифференциально-разностной задаче относительно функций, зависящих от временной переменной. Для решения полученной дифференциально-разностной задачи предлагается специальное представление, с помощью которого задача распадается на две независимые дифференциально-разностные задачи. В результате получена явная формула для определения приближенного значения искомой функции при каждом дискретном значении пространственной переменной. Для численного решения полученных дифференциально-разностных задач используется метод конечных разностей. Представленные результаты численных экспериментов, проведенных для модельных задач, демонстрируют эффективность предложенного вычислительного алгоритма. Ключевые слова: волновое уравнение, обратная задача, восстановление начального условия, дифференциально-разностная задача
Для цитирования: Гамзаев Х.М. Численный метод восстановления начального условия для волнового уравнения // Вестник Томского государственного университета. Математика и механика. 2024. № 88. С. 5-13. doi: 10.17223/19988621/88/1
Introduction
It is known that inverse problems for wave equations occur in mathematical modeling of many physical processes in geophysics, seismics, electrodynamics, thermophysics, medicine, and many other fields of science and technology [1-5]. In these inverse problems, in addition to solving the wave equation, it is necessary to determine either the right-hand sides, or coefficients, or initial conditions. It should be noted that a large number of publications have been devoted to the study of the correctness, existence, and unambiguous solvability of coefficient inverse problems and inverse problems for determining the right parts of wave equations [6-12]. At the same time, much less work has been devoted to the inverse problem of restoring the initial conditions for wave equations. In a number of papers [13-16], Dirichlet-type problems for the wave equation are presented as an inverse problem of restoring the initial condition and gradient iterative methods are proposed for the numerical solution of such problems.
In this paper, a non-iterative computational algorithm is proposed for the numerical solution of the inverse problem of restoring the initial condition for the time derivative for a one-dimensional wave equation.
1. Problem statement and solution method
Let a one-dimensional wave equation
^ 4 (k(x) + f (x, t), 0 < x < 1, 0 < t , r , (1)
ot ox ox
be considered with the initial conditions
u(x,0) = 9(x), (2)
^ = v(x) , (3)
ot
and boundary conditions
u(0, t) = q(t), (4)
u(1,t) = p(t). (5)
It is known that the direct problem for equation (1) consists in determining a function u(x,t) from equation (1) with a given coefficient k(x), the right side f (x,t) and conditions (2)-(5).
Suppose that in addition to the function u(x, t), the function v(x) is also unknown and the restoration of this function is required. In this case, as an additional condition, the solution of equation (1) is given at a finite time
u(x, T) = y(x) , (6)
where y(x) is the given function.
Thus, the task is to determine the functions u(x, t) and v(x) satisfying equation (1) and conditions (2)-(6). The problem belongs to the class of inverse problems associated with the restoration of initial conditions for partial differential equations.
First, we transform the assigned task to a semi-discrete task. To this end, we introduce a uniform difference grid in the domain [0 < x < 1] of a variable x
fflj ={x = iAx, i = 0,1,2,...,n}
with a step Ax =1.
n
d du(x t)
The differential expression — (k(x)--—) in equation (7) for x = xi,
dx dx
i = 1,2,..., n -1 is approximated by the "central" difference
l k(x) x=x -
dx dx ' '
1
Ax
k(x | Ax. u(xM, t) -u(x,.,t) Ax u(xi,t) -u(xi-,t)
( x 2 Ax ( x' 2 ) Ax
Denoting ui (t) « u (xi, t), k,.±1/2 = k ^x,. j, equation (1) and conditions (2)-(5)
are written as the following system of ordinary differential equations
d 2u (t ) k,+xn 2k, kt
= 2-ui+l(t)_ ut (t) + -i_f. u._ l(t) + f (t) , 0 < t < T , i = 1 n-1, (7) dt Ax Ax Ax
u. (0) = 9, i = 0,n , (8)
du,. (0) . —
—i-= v., / = 0, n , (9)
dt
u0(t) = q(t), (10)
u„ (t) = p(t) , (11)
u. (T) , i = 0n , (12)
where k = (kMn + k_m) / 2, vt - v(x. ), 9 = 9x,. ), = y(x. ), f (t) = f (x,. , t). In the resulting differential-difference problem, the approximate values of the desired functions v(x) in the nodes of the difference grid , i.e. vt and the functions ut (t),
i = 1, 2,..., n-1, act as unknown. For the decomposition of the differential-difference problem (7)-(12) into mutually independent subtasks, each of which can be solved independently, its solution for each fixed value i = 0,1, 2,..., n, is represented as [17, 18]
u.(t) = w(t) + V 0 (t), i = 0, 1, 2, ..., n, (13)
where wi (t), 0. (t) are unknown functions. Substituting the representation u i (t) into equation (7), we obtain
d2w (t) d20.. (t) k +1/2 k +1/2 2k. 2k
+ V —= "ff Wi+! (t) + v.+, -ff- 0.+, (t) - -f w. (t) - v. —2- 0. (t) + dt2 dt Ax Ax Ax Ax
+-rf w.-1(t) + v.0.-1(t) + ft (t). Ax Ax
Replacing v and v with v , the latter relation is represented as
d2 w,. (t) k
dt2
2k
ТГ w.+¿t ) + ~r w, (t )--¡±-/2 At Ax Ax
2 w+1V) ' . 2 () л 2 W-l(t) - f(t)
+v.
d20,. (t) k
dt2
,+1/2 n
Г 0,+1
Ax
2k k (t)+^ 0, (t) - 0,- i(t) Ax Ax
= 0 .
Substitution of representation (13) into (8)—(11) yields
W (0) + V,. e,. (0) = 9,.,
dw (0) d e (0)
—+ V,. —— = V,. , dt dt
W(t) + V0e0(t) = q(t) ,
w (t )+v e„ (t) = p(t ).
From the obtained relations, it is possible to obtain differential-difference problems for determining auxiliary functions wt(t), et(t), i = 0,1, 2,..., n
d2wt (t) k
2k,.
dt2
d 20, (t ) k, +
dt2
"ff w,+i (t) + —2- w, (t) - -ff w,-i (t ) - f (t ) = 0 , , = 1, n -1, Ax Ax Ax
w (0) =ф,., dw, (0)_Q dt ' w0(t) = <?(t),
wn (t) = ^(t).
(t)+0, (t) - ^ 0,-1 (t)=0, ,=1m-1,
Ax Ax
Ax
2 ,+1
0, (0) = 0 d 0 (0)
= 1,
dt
00 (t) = 0 , 0n (t ) = 0.
(14)
(15)
(16)
(17)
(18)
(19)
(20) (21)
(22) (23)
And substituting representation (13) into (12), we have
w, (T) + v, 0, (T) = y,.
From here we get a formula for determining the value of the desired function v(x) for each fixed value x = xt
y. - w. (T) -
v. =y-i-(-), i = 1, n -1. (24)
1 0 (T)
Thus, the computational algorithm for the numerical solution of the differential-difference problem (7)-(12), by definition ut(t), vt, i = 1, 2,...,n -1, consists of the following:
— the solutions of two independent differential-difference problems (14)—(18) and (19)—(23) with respect to auxiliary functions wt(t), 0.(t), i = 0,1, 2,..., n, are determined on the segment [0, T ];
— according to formula (24), approximate values of the desired function v(x) are determined for x = xi, i.e. vi, i = 1, 2,...,n -1;
— the formula (13) determines the values of the functions ut(t), i = 0,1,2,...,n , on the segment [0, T ].
It should be noted that the approximate values of the desired function v(x) at the boundary points x0 = 0 and xn = 1 cannot be determined by formula (24) due to the fulfillment of conditions (22) and (23). Therefore, the values of the desired function v(x) at the boundary points can be determined by interpolation.
It should be noted that the applicability of the proposed computational algorithm is associated with the fulfillment of the condition
0 (T) * 0, i = 1, n -1. For an equation with a constant coefficient, it is possible to find out in advance the fulfillment of this condition. To do this, it is enough to write a differential approximation of the differential-difference problem (19)—(23) for the case k (x) = k0 = const
d20(x,t) , d20(x, t) „ -^-x-1 = k--r-1, 0 < x < 1, 0 < t < 0.1,
dt2 dx
0x0) = 0 , ^^ = a(x) s 1, dt
0(0, t) = 0 , 0(1, t) = 0 . The exact solution of this problem is determined by the explicit formula
2
0( x, t ) = £-
J a(Ç)sin nr^d
sin nr^k"t sin nrx .
r=1 nr^k ^
It follows that when T ^ 11 the condition 0 (T) ^ 0, i = 1, n -1 is satisfied. However, for an equation with variable coefficients, due to the complexity of constructing an analytical solution, the condition 0, (T) ^ 0, i = 1, n -1 can be fulfilled using a numerical experiment.
-f w/+ + -Lw,+1 --^fw{+ - fi+ = 0 , (25)
For the numerical solution of problems (14)—(18) and (19)—(23), the finite difference method can be used. We introduce a uniform difference grid with a step At on the segment [0, T ] in the variable t
ro, ={t, = jAt, j = 0,1,2,...,m, At = T/m} . The discrete analogs of problems (14)-(18) and (19)—(23) on the grid rat are represented as
W/+' - 2W + W/-' ki+1/2 ^ j+1 , 2ki u,j+1 k, ---w. , +--w.--
At2 Ax2 Ax2 . Ax:
w0 = 9., (26)
1 0 w. - w.
w-!- = 0 , (27)
At
w0+1 = qj+1, (28)
w+1 = j, (29)
0j+1 - 2 0j +0j 1 -k^ 0j++; + 0j+1 - ^ 0j+;=0, (30)
At2 Ax2 .+1 Ax2 . Ax2
00 = 0, (31)
01 - 00
a = 1, (32)
At
0j+1 = 0 , (33)
0n+1 = 0, (34)
where wj * w. (tj), 0j * 0. (tj), fj+1 = f (tj+l).
The obtained difference problems (25)—(29) and (30)—(34) for each fixed value j = 1, 2,...,m -1 are systems of linear algebraic equations with a tridiagonal matrix, the solutions of which can be found by the Thomas method [17].
2. Numerical examples
To find out the effectiveness of the proposed computational algorithm, numerical experiments were carried out for model problems. Calculations were carried out on a space-time difference grid with steps Ax = 0.05, At = 0.0001.
Example 1.
52u(x,t) _ 1 52u(x,t)
dt1 8л2 dx
- + e (1.5 + 3cos2rar), 0 < x < 1, 0 < t < 0.,
u( x, 0) = 2(3 + 2cos2rn-), = v(x),
u(0, t) = 10e05t, u(1, t) = 10e05t, u(x, 0.1) = 2e0 05 (3 + 2 cos 2ror). This problem has an exact solution
u(x,t) = 2e0 5t (3 + 2cos 2nx) , v(x) = 3 + 2 cos 2nx.
Gamzaev K.M. Numerical method for restoring the initial condition for the wave equation Example 2.
d2u(x= 0.025 d2u(x;t) + e05t(10x - 10x2 + 2), 0 < x < 1, 0 < t < 0.1,
dt2 dx
2 du( x,0)
u(x, 0) = 40x -40xz, —= v(x), dt
u (0, t) = 0 , u (1, t) = 0, u(x,0.1) = e005(40x - 40x2). The exact solution to this problem has the form
u(x, t) = e0 5t (40x - 40x2) , v(x) = 20x - 20x2.
Example 3.
d2u(x,t) d 02x du(x,.r -0.2x, ^ n^^m
--— = — (xe -) + sin3t(1 - 45e ) , 0 < x < 1, 0 < t < 0.1,
dt dx dx
u(x,0) = 0 , ^uM = v(x), dt
u(0,t) = 5sln3t, u(1,t) = 5e~02sln3t, u(x,0.1) = 5e~0'2x sin0.3. The exact solution of the problem has the form
u(x, t) = 5e~0-2x sin3t, v(x) = 15e-°1:i. The results of numerical experiments to determine the approximate values of the desired function v(x) at x = x, , i = 1,2,...,n -1, for the examples given are presented in the table. The data in the 2nd and 3rd columns refer to the first example; the data in the 4th and 5th columns, to the second example; and data in the 6th and 7th columns, to the third example.
Numerical results on the determination of the function v( x)
xi v(x) = 3 + 2cqs2kx v(x) = 20x - 20x1 v(x) = \5e-"lx
Exact Calculated Exact Calculated Exact Calculated
0.05 4.902 4.899 0.950 0.952 14.851 14.844
0.10 4.618 4.616 1.800 1.799 14.703 14.699
0.15 4.176 4.174 2.550 2.548 14.557 14.552
0.20 3.618 3.617 3.200 3.198 14.412 14.407
0.25 3.000 3.000 3.750 3.748 14.268 14.264
0.30 2.382 2.383 4.200 4.198 14.126 14.122
0.35 1.824 1.826 4.550 4.548 13.986 13.982
0.40 1.382 1.384 4.800 4.798 13.847 13.842
0.45 1.098 1.100 4.950 4.948 13.709 13.705
0.50 1.000 1.003 5.000 4.998 13.573 13.568
0.55 1.098 1.100 4.950 4.948 13.438 13.433
0.60 1.382 1.384 4.800 4.798 13.304 13.299
0.65 1.824 1.826 4.550 4.548 13.171 13.167
0.70 2.382 2.383 4.200 4.198 13.040 13.036
0.75 3.000 3.000 3.750 3.748 12.911 12.906
0.80 3.618 3.617 3.200 3.198 12.782 12.778
0.85 4.176 4.174 2.550 2.548 12.655 12.655
0.90 4.618 4.616 1.800 1.799 12.529 12.535
0.95 4.902 4.899 0.950 0.952 12.404 12.411
The results of numerical experiments indicate that the values of the desired functions u(x, t) and v(x) are determined with a sufficiently high accuracy. At the same time, the maximum relative error in determining the desired function v(x) in the first example does not exceed 0.08%; in the second example, 0.3%; and in the third example,
0.06.. Analysis of results of the numerical experiments shows that to increase the accuracy of solutions, it is sufficient to use small steps of the difference grid.
Conclusion
The problem of determining the initial condition for the time derivative for a one-dimensional wave equation, according to an additionally specified condition at a finite time, is considered. The proposed computational algorithm, based on the discretization of the problem by a spatial variable and the use of a special representation to solve the resulting differential-difference problem, allows us to find by an explicit formula the approximate value of the desired function for each discrete value of the spatial variable. The proposed computational algorithm can also be used to restore the initial condition in time for the one-dimensional wave equation.
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Information about the author:
Gamzaev Khanlar M. (Doctor of Technical Sciences, Professor, Department of General and Applied Mathematics, Azerbaijan State Oil and Industry University, Baku, Azerbaijan; Research Associate, Research and Innovation Center, Western Caspian University, Baku, Azerbaijan). ORCID: 0000-0002-1228-7892. E-mail: [email protected]
Сведения об авторе:
Гамзаев Ханлар Мехвали оглу - доктор технических наук, профессор кафедры «Общая и прикладная математика» Азербайджанского государственного университета нефти и промышленности, Баку, Азербайджан; научный сотрудник Научно-координационного центра Западно-Каспийского университета, Баку, Азербайджан. ORCID: 0000-0002-1228-7892. E-mail: [email protected]
The article was submitted 04.06.2023; accepted for publication 10.04.2024 Статья поступила в редакцию 04.06.2023; принята к публикации 10.04.2024
