Numerical Solution Tasks with Mixed Operator in Unlimited Area Mikhail Galanin1,2[0000-0003-0264-6270] and Dmitry Sorokin1,2[0000-0002-6273-3985] 1 Keldysh Institute of Applied Mathematics (KIAM), Moscow 2 Bauman Moscow State Technical University (BMSTU), Moscow sorokin.dmitr@yandex.ru Abstract. Methods for solving problems of elliptic equations, based on the in- tegrated Green formula, were analyzed. New methods for solving a problem with a mixed-type operator in an unbounded domain are proposed. On the basis of the proposed methods, programs for solving problems with a mixed type op- erator have been created. The results of computational experiments, showing the correctness of the application of methods, are presented. Moreover, the con- vergence of the developed iterative methods has been studied; protection of the order of the original difference scheme using the developed methods has been shown. Keywords: Unlimited Area, Mixed Operator, Electrodynamics Accelerator, Railgun, Electromagnetic Field, Maxwell’s Equations. 1 Introduction In the study of physical phenomena, it is often necessary to perform modeling in an unlimited area, for example, in the model-roving of the electrostatic field of charges, solving the problem of thermal conductivity, etc. In the case where the phenomenon can be described by means of the simplest linear elliptic operators, the problem is solved and described, for example, in [1, 2]. The setting of non-reflective boundary conditions for wave equations is discussed in detail in [3]. However, for a wider class of problems for which the problem operator may have a mixed type but outside some finite domain, the fundamental solution of the operator is known and easily calculat- ed, the above methods are not applicable. In particular, problems with operators of mixed type arise when modulating the electromagnetic field in electrodynamic accelerators of the rail type. Process of cur- rent flow in conductors in this case can be described by parabolic equation and elec- tromagnetic field in dielectric – elliptical [4, 5]. Copyright © 2020 for this paper by its authors. Use permitted under Creative Commons License Attribution 4.0 International (CC BY 4.0). 371 2 Methods of Solving the Problem in an Unlimited Area In order to solve the problem, three methods of solving are proposed: the method of expansion of the area, the method of setting integral boundary conditions, three-stage method [6]. When solving problems using grid numerical methods, accuracy is determined by grid spacing and accuracy of setting boundary conditions. If you set the same bounda- ry conditions as infinity in the original task when you solve a problem in the target area on the boundary, the error defined by the boundary condition setting will gener- ally decrease as the calculation area increases in size. Therefore, you can use the scope extension method to solve tasks in an unlimited scope, that is, you can find the size of the calculation scope so that the effect of the artificial boundary is less than the application of the numerical method. The method of expansion of the design area is easy to implement, but has a high computational complexity [6]. Alternative methods are methods based on the use of Green formulas. The first is the method of defining the integral boundary condition [6]. It is based on the fact that outside some finite region there is an operator with a known funda- mental solution and a trivial right part, then the solution on the boundary of the calcu- lation region is determined by means of the basic integral Green formula. Thus, solv- ing the problem using the method of defining integral boundary conditions requires solving a system of equations with filled lines corresponding to the boundary. Three-stage method of solution is built in [6]. When using the three-step method, it is necessary to solve two systems, but with sparse lines related to boundary condi- tions. The results of calculations show that when solving problems reduced to solving linear equations with strongly filled matrices, the method is ineffective, but for prob- lems with sparse matrix of the system, the three-step method allows to obtain the result significantly faster than the method of setting the integral boundary condition. 3 Modification of Three-Stage Method. Iterative Method of Problem Solution in Unlimited Area Consider in more detail the three-step algorithm [6] on the example of solving the problem with an operator of mixed type: ut  a u  f , r  DS , 2  (1)  u  0, r  \DS , 2 where DS is the boundary of the region DS (Fig. 1). After applying the finite difference method to approximate the production time in- stead of the system (1), it is necessary to solve the problem:  1 1 u uˆ  2 uˆ   2 fˆ  2 , r  DS ,  a  a a  (2) uˆ  0, r  \DS . 2 372 Fig. 1. Structure of area. In subsequent records of solved tasks, temporary derivatives use approxima- tion (2). At the first stage of the three-stage method it is proposed to solve the system with time-consuming boundary conditions: vt  a 2 vˆ  f , r  DS ;  vˆ  0, r  D \DS ,  vˆ   u  , v(t )  u.  i 1 At the second stage it is assumed that the solution v̂ in the area DS is slightly dif- ferent from the solution û of the initial problem (3), and the difference is related to the error in the boundary condition setting. Based on these considerations and Green's basic integral formula, the objective is: ut  a 2 uˆ  f , r  DS ;  uˆ  0, r  D \DS ,    vˆi  P  uˆ   vˆi     P n  vˆi n  dS ,    u (ti 1 )  u. At the third stage, according to the main integrated Green formula, the value û on the boundary  by the value û in the area D :  uˆ ФPL  uˆ ( P)    ФPL  uˆ  dS . S  n n  The main advantage of the three-stage algorithm is that without the use of an itera- tive process it allows to obtain results of high accuracy [6, 7]. However, the presence of the second stage deliberately creates an error that is not eliminated by iteratively repeating steps 1–3 of the algorithm and can grow with a reduction in the time inte- gration step  . In order to overcome this disadvantage of the method, it is proposed to delete the second step of the method, and to perform the first and third steps several times on each time layer. 373 The first stage of the new iterative algorithm utk  a 2 uˆ k  f , r  DS ;  k uˆ  0, r  D \DS ,  k k 1 uˆ   gˆ  ,  u (ti 1 )  u , k  1..N . On the first iteration, you are prompted to select a lagging boundary condition as the boundary condition ĝ 0  g  .  The second stage of the algorithm  uˆ k ФPL  gˆ k ( P)    ФPL  uˆ k  dS . S  n n  4 Application of Iterative Algorithm to Solving Problems in Unlimited Area The built algorithm is applicable to solving the following problem: ut  a u  f , r  R0 , 2  (3) u  0, r  R0 ,  2 m  22 f (r ,  , t )  a 2   1  r 2 n2  k 2  r k  2 yk  C1r n2  e  n a t cos .  k 2  Initial condition ( r  R0 ):  m2  u (r ,  , 0)     yk r k   C1r  2c J1 (r n )  cos .  k 2  R n 0 2 ( J 0 ( R  0 n )  J 2 ( R  0 n ))  Where m and n — parameters; J j — j order Bessel function ( j  0,1, 2 );  1 Cmk  2 k n n  ,  n — root of equation J1 (n )  0; yk  ;   2; R0 R0k  2  k 2  1 1 m2 m2  1 C1      k  1 yk R0k 1  ; C2    yk R0k 1   C1 R02 ; c  C2 1   . 1   k 2 k 2   The analytical solution is as follows:  m  2  22    yk r   C1r  2c k J1 (r n )  e  n a t cos  , r  R0 ;  k  2 n R02 ( J 0 ( R0 n )  J 2 ( R0 n ))  u (r ,  , t )    2  n a t C cos  , r  R0 . 2 2  r e 374 Let's set m  1, n  2, a  0.1, R0  1.0. Solve the problem with a new iterative method and a three-step algorithm. Table 1 shows the calculation results. We see that all the methods considered lead to the correct result. Table 1. Task (3) results. R0  1.0, R  2.0,   0.001, h  0.2, T  0.008 . uh  u C uh  u C uh  u C uh  u C uC uC uC uC h analytical p p application of p application of p N  definition of application of iterative iterative a three-stage boundary algorithm algorithm algorithm conditions (1 iter) (2 iter) h 3.15114 103 3.90629 103 3.20485 103 3.25578 103 8  (t = 0.211 s.) (t = 0.489 s.) (t = 0.267 s.) (t = 0.555 s.) h/2 4.79565  104 5.05743 104 4.50521 104 4.57497 104 32 2.7 3.0 2.8 2.8  /4 ( t = 1.153 s.) (t = 2.322 s.) (t = 1.367 s.) (t = 2.660 s.) 5 5 5 5 h/4 6.82737 10 6.8335 10 (t 8.60114 10 7.39768 10 128 2.8 2.9 2.4 2.6  /16 ( t = 8.141 s.) = 15.62 s.) (t = 9.013 s.) (t = 17.71 s.) h/8 1.75524 105 ( 1.75524 105 2.65021 105 2.32554 105 512 2.0 2.0 1.7 1.7  /64 t = 114.5 s.) (t = 233.4 s.) (t = 133.1 s.) (t = 260.5 s.)  Here p  log 2  uh  u C / u C   h  u u / u  h C C h 2  — value characterizing the order of the method,    h — value calculated on a grid with step h . Table 2. The convergence of iterative methods in the solution of task (3) in case of selection at each step as the initial approximation of the zero vector   1.5625  10 , h  0.025, 5 T0  0.000125. Error in analytical method of setting boundary conditions is equal to 8.45  106 uh  u C uh  u C uC Iteration number uC (iterative process based (two-step iterative method) on a three-step method) 1 3.82473 101 1.39064 101 2 1.25412 101 1.65792 102 4 1.34993 102 2.42915  104 8 1.80879 104 1.07342 105 16 2.53095  105 1.07263 105 32 2.53009 105 — 64 2.53275  105 — 375 Note that the iterative process could also be built on the basis of a three-stage algo- rithm [6]. However, according to the results in Table 2, this will not help to get the answer with the required accuracy in less time, since one iteration of the three-step method contains almost twice as many deductive operations as the number of opera- tions of the new iterative method. By studying Tables 1 and 2, it can be observed that the iterative methods con- structed have a high convergence rate as long as the distance from the approximation to the true solution is large. In the vicinity of the solution, iterative projects do not converge, which is confirmed by the results of numerical calculations (Table 3). Table 3. Convergence of two-step iterative method. 0.001 0.2 R0  1.0, R  2.0,   , h , T  0.008 64 8 uh  u C uh  u C uh  u C uh  u C uh  u C uC uC uC uC uC analytical application of application of application of application of definition of iterative algo- iterative algo- iterative algo- iterative algo- boundary rithm rithm rithm rithm conditions (1 iter) (2 iter) (3 iter) (4 iter) Error 1.75524 105 2.65021 105 2.32554 105 2.45878  105 4.61259 105 Time (s.) 114.472 133.145 260.535 359.556 486.5 One way to improve convergence rate in the vicinity of the solution (Table 4) is to change the second step of the iterative method, for example: 1  uˆ k ФPL   gˆ k ( P)   gˆ k 1 ( P)    ФPL  uˆ k  dS  . 2  S  n n   It is also worth noting that the newly built method is less demanding to the size of the auxiliary sub area D\DS , which allows reducing its time-measures, and thus reduc- ing the computational complexity of the task. Table 4. Convergence of modified two-step iterative method. 0.001 0.2 R0  1.0, R  2.0,   , h , T  0.008 64 8 uh  u C uh  u C uh  u C uh  u C uh  u C uC uC uC uC uC analytical application of application of application of application of definition of iterative algo- iterative algo- iterative algo- iterative algo- boundary rithm rithm rithm rithm conditions (1 iter) (2 iter) (4 iter) (8 iter) Error 1.75524 105 2.65021 105 2.51139 105 2.4427 105 2.42558  105 Time (s.) 114.472 133.145 236.651 468.127 925.113 376 5 Conclusion On the basis of the method of setting integral boundary conditions and the three-stage method, computational algorithms are built and software implemented to solve a number of problems in an unlimited area. The results of computational experiments confirm the correctness of the methods. 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