Nonreflecting time-dependent boundary conditions on artificial boundaries of varying location and shape

Nonreflecting time-dependent boundary conditions on artificial boundaries of varying location and shape

Applied Numerical Mathematics 33 (2000) 481–492 Nonreflecting time-dependent boundary conditions on artificial boundaries of varying location and sha...

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Applied Numerical Mathematics 33 (2000) 481–492

Nonreflecting time-dependent boundary conditions on artificial boundaries of varying location and shape Victor Ryaben’kii 1,2 Keldysh Institute of Applied Mathematics, Moscow, Russia

Abstract In the framework of finite differences, we propose a methodology for the exact transfer of boundary conditions from infinity or remote physical boundary of the original domain to the artificial boundary of the computational subdomain. We analyze the problem in a most general formulation and allow the artificial boundary to change its shape and/or spatial location in the course of time. Outside the computational subdomain, the governing finitedifference equations are assumed linear and homogeneous.  2000 IMACS. Published by Elsevier Science B.V. All rights reserved. Keywords: Artificial boundary conditions; Time-dependent problems; Difference potentials

1. Introduction There is a wide class of unsteady initial boundary value problems formulated either on the entire Euclidean space or a large domain D (with some boundary conditions), for which the solution u(t, x) is needed to be known only on a bounded subdomain of the original domain. Without much loss of generality, we additionally assume that the governing differential equations outside this computational subdomain, as well as boundary conditions at infinity or distant physical boundary, are linear and homogeneous. Let us denote by Din = Din (t) the computational subdomain on which the solution u(t, x) needs to be calculated. This subdomain can change its shape and also move in the course of time. Examples of the corresponding problems arise in different physical and engineering applications. Suppose one subjects a homogeneous material to the strong time-dependent stress concentrated in some local area. Then, plastic deformations or fractures may arise in or near the stress region, while at some distance the deformations are sufficiently small to be described by the Lame system of linear elasticity theory. 1 E-mail: [email protected] 2 Supported by Director’s Discretionary Fund, NASA Langley Research Center, Hampton, VA 23681-2199, USA.

0168-9274/00/$20.00  2000 IMACS. Published by Elsevier Science B.V. All rights reserved. PII: S 0 1 6 8 - 9 2 7 4 ( 9 9 ) 0 0 1 1 6 - 6

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Typically, it is interesting to calculate the solution only on the subdomain, in which large deformations are concentrated. Similarly, when computing external fluid flows around finite configurations, one, as a rule, is interested in calculating the solution only in the near field. At the same time, in the far field the flow can be described by the gas dynamics equations linearized around the unperturbed background solution. Other examples are given by the problems of wave diffraction or scattering (acoustic or electromagnetic). In all these problems, it would be most beneficial from the standpoint of numerical performance to solve the equations only in a relatively small subdomain of interest provided that the influence of the remaining outer part of the domain along with the boundary conditions at infinity or remote physical boundary is compensated for by the equivalent conditions at the artificial boundary. In this paper, we consider the problem of how to transfer the boundary conditions from infinity to the artificial boundary of the computational subdomain for abstract systems of time-dependent difference equations. These equations (systems) are of a general type and subject only to the requirements of solvability, uniqueness of the solution, and linearity outside the computational subdomain. Finite-difference or finite-element approximations of many unsteady physical problems belong to this class. In the paper, we construct a special difference artificial boundary of the computational subdomain so that one can exactly transfer to it boundary conditions from the remote physical boundary or from infinity; we also describe the actual construction of the artificial boundary conditions themselves. Usually, boundary conditions of this type are called the Non-Reflecting Artificial Boundary Conditions (NRABC). It is interesting to note that in the case of explicit difference schemes the resulting NRABC are also explicit.

2. Formulation of the problem 2.1. Grids Let us consider the space R × [−∞, T ] that consists of points (x, t), x ∈ R, −∞ < t < T . Suppose D ⊂ R is the original domain which can, in particular, coincide with the entire R. Let us take τ > 0 and for every hyperplane t = j τ , j = 0, ±1, ±2, . . . , consider discrete set M j ⊂ D. S S Denote M = M j , j = 0, ±1, ±2, . . . , and for every natural p consider the set M (p) = M j , j 6 p. With each point m ∈ M j we associate some finite set of points Nm . These points belong to levels t = j τ, (j − 1)τ, . . . , (j − k)τ , where k is some fixed natural number and does not depend on j and m. S The set Nm is a stencil of the finite-difference scheme at the node m. Denote N = Nm , m ∈ M. Denote N j the subset of N that belongs to the level t = j τ . Now assume that for every t there is a bounded j domain Din = Din (t) ⊂ D, which can move and change its shape in the course of time t. Denote by Min j j the set of those points from M j that are inside Din (tj ). Let Mex = M j \Min . Denote Min = Nex =

[ [

j

[

Min ,

Nin =

Nm ,

m ∈ Mex ,

Nm ,

m ∈ Min ,

γ = Nin ∩ Nex .

V. Ryaben’kii / Applied Numerical Mathematics 33 (2000) 481–492

483 j

j Denote the intersection of these sets with hyperplanes t = j τ , j = 0, ±1, . . . , by Nin , Nex , and γ j , respectively. Let

M (p) = N (p) =

[

[

(p)

[

(p)

[

Mj ,

Min =

Nj,

Nin =

j

Min ,

(p) Mex =

j

(p) Nex =

Nin ,

[

[

j Mex ,

γ (p) =

j Nex ,

j 6 p.

[

γ j,

2.2. The spaces of functions Let us introduce the linear spaces FM , FMin and FMex of all vector functions fM , fMin and fMex of fixed dimension defined on the grids M, Min and Mex . Also introduce the linear spaces VN , VNin , VNex and Vγ of all vector functions vN , vNin , vNex and vγ defined on the sets N , Nin , Nex and γ , respectively. Denote j j (p) j j (p) the restrictions of these spaces to the sub-grids M j , Min , Mex , N j , Nin , Nex , γ j , M (p) , Min , Mex , N (p) , (p) j j j j j j (p) (p) (p) (p) (p) (p) (p) Nin , Nex and γ (p) by FM , FMin , FMex , VN , VNin , VNex , Vγj , FM , FMin , FMex , VN , VNin , VNex and Vγ(p) , respectively. Suppose also that all these spaces consist of only zero elements for as p < 0. Let us impose for each j separately some additional linear conditions on the functions vN , vNin and j vNex . These conditions Aj vN are split into two groups: j

(α)

Ajex vN = 0,

(β)

Ain vN = 0.

j

j

j

j We assume that (α) connects the components of vN only at the points n ∈ Nex \γ j and β connects j j the components of vN only at the points n ∈ Nin \γ j . Thus, each function vγ ∈ Vγ after having been j complemented by zero on the entire N belongs to VN for every j . j j Denote by Vin the space of vector functions defined on Nin and satisfying conditions (β), and by Vexj j the space of vector functions defined on Nex and satisfying conditions (α). Let us note that the extension j j j by zero of every function from Vin or Vexj to the entire N j yields the function vN ∈ VN .

2.3. The original difference boundary problem Let the finite-difference boundary value problem with the unknowns vN ∈ VN be formulated as follows: The solution has to satisfy relations X

amn vn = fm ,

m ∈ Mex ,

(1)

ϕm (vNm ) = fm ,

m ∈ Min ,

(2)

n∈Nm

X

n∈Nm

on the domain and, additionally, relation (α) and (β) that are interpreted as boundary conditions, we rewrite them here for the convenience of referencing Ajex vN = 0,

j = 0, ±1, . . . ,

(3)

j Ain vN

j = 0, ±1, . . . .

(4)

= 0,

Here, vn is the value of the solution at n ∈ Nex and vNm are the values of vN at points on the stencil Nm , m ∈ Min . We assume that the boundary value problem (1)–(4) has one and only one solution vN ∈ VN for an arbitrary right-hand side fM ∈ FM .

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2.4. Formulation of the problem of nonreflecting boundary conditions (NRABC) Consider problem (1)–(4) under the additional assumption fm = 0

as m ∈ Mex .

(5)

Suppose we need to know the solution uN of (1)–(5) only on Nin ⊂ N ; hereafter, Nin is called the (p−1) computational subdomain. Assume that the solution uNin on Nin and the values vγ(p−1) of the solution on the grid boundary γ (p−1) have already been found and are stored in the computer memory. Let us formulate the problem of constructing the linear operators lγp : Vγp 7→ Vγp

ψγp : Vγ(p−1) 7→ Vγp

and

(6)

such that the solution of the problem j

ϕm (vNm ) = fm ,

m ∈ Min ,

 (p−1)

lγp vγp = ψ p vγ j j vNin ∈ VNin , j

j 6 p,

,

(7) (8)

6 p,

(9)

(p) Nin

coincides on ⊂ N (p) with the solution of the problem (1)–(5). Thus, boundary condition (8) in problem (7)–(9) has to be equivalent to equations (1), m ∈ Mex , with boundary condition (3). Definition 1. Boundary condition (8) that possesses the foregoing properties will be called the NRABC on the grid boundary γ of the computational subdomain Nin . 2.5. The universality of NRABC Here we will show that NRABC of type (8) obtained for the original problem (1)–(5) are, in fact, the NRABC for a class of problems. Therefore, the NRABC for a given problem can be replaced by the NRABC for every other problem from this class. One can make use of the freedom in choosing this other problem; we will refer to it as to the auxiliary problem when constructing the NRABC for the original problem. The auxiliary problem should basically be chosen so that the construction of NRABC for it be easier then for the original problem. Let us consider two problems of type (1)–(4) and formulate the requirements for them to belong to the same class of problems, i.e., have the same NRABC on the boundary γ of the computational subdomain Nin . Let the first problem be X

amn vn = fm ,

m ∈ Mex ,

(10)

m ∈ Min ,

(11)

n∈Nm

ϕm (vNm ) = fm , vN ∈ VN ,

vNex ∈ VNex ,

vNin ∈ VNin ,

(12)

and the second problem be X

amn vn = fm , n∈Nm ϕm0 (vNm0 ) = fm ,

m ∈ Mex ,

(13)

m ∈ Min0 ,

(14)

V. Ryaben’kii / Applied Numerical Mathematics 33 (2000) 481–492

vN 0 ∈ VN 0 ,

vNex ∈ VNex ,

vNin0 ∈ VNin0 .

485

(15)

Regarding these problems we assume that both are uniquely solvable for an arbitrary right-hand side. It is also assumed that subsystem (10) of the first problem coincides with subsystem (13) of the second problem and the boundary γ , γ = Nex ∩ Nin = Nex ∩ Nin0 , is the same for both subsystems. The spaces VNex and VNex0 coincide but the sets Min and Min0 , stencils Nm , m ∈ Min , and Nm0 , m ∈ Min0 , functions ϕm and ϕm0 , and spaces VNin and VNin0 , can be different. Theorem 2. Let



lnj vγj = ψnj vγ(j −1) ,

n ∈ γ j,

j = −1, 0, 1, . . . ,

(16)

be the NRABC for problem (13)–(15) on the boundary γ of the computational subdomain Nin0 . Then, (16) is also the NRABC for problem (10)–(12) on the boundary γ of the computational subdomain Nin . Conversely, let (16) be the NRABC for problem (10)–(12) on the boundary γ of the computational subdomain Nin . Then, (16) also is the NRABC for problem (13)–(15) on the boundary γ of the computational subdomain Nin0 . Proof. According to Definition 1, boundary condition (16) is the NRABC for problem (10)–(12) (or problem (13)–(15)) if those and only those functions vγ that satisfy this condition can be defined on the entire N as a function vN ∈ VN (or on N 0 as vN 0 ∈ VN 0 , respectively) that satisfies the following homogeneous equation: X

amn vn = 0,

m ∈ Mex .

(17)

n∈Nm

But any vγ that can be extended as vN ∈ VN satisfying (17) can also be interpreted as some function vN 0 ∈ VN 0 provided that equality (17) holds; the converse is true as well. We will prove only the first, direct, statement. Suppose vN ∈ VN is the aforementioned extension of a given vγ . Then, vN 0 ∈ VN 0 that satisfies (17) can be obtained as vN 0 |n = vN |n ,

if n ∈ Nex ,

vN 0 |n = 0,

if n ∈ Nin0 \γ .

(18)

Thus formulae (16) provide the NRABC simultaneously for both problems (10)–(12) and (13)– (15). 2 3. Construction of NRABC 3.1. General case In this section, we construct the NRABC for problem (1)–(5) subject to an additional condition. In the class of problems that have the same NRABC in accordance with Theorem 2, let us choose the linear problem, which will, in some respect, be most convenient for computations. We will construct

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V. Ryaben’kii / Applied Numerical Mathematics 33 (2000) 481–492

the NRABC for this linear problem and thus obtain the NRABC for the original problem as well, as guaranteed by Theorem 2. For simplicity let us assume that problem (1)–(5) itself is linear and has the form X

j

amn vn = fm ,

m ∈ Min ,

(19)

amn vn = 0,

j m ∈ Mex ,

(20)

= 0,

j = . . . , −1, 0, 1, . . . ,

(21)

= 0,

j = . . . , −1, 0, 1, . . . .

(22)

n∈Nm

X

n∈Nm j j lex vN j j lin vN

We need to construct the NRABC of type (16) such that the solution of the problem (j = . . . , 0, 1, . . .) X

j

amn vn = fm ,

n∈Nm

m ∈ Min ,



lnj vγj = ψnj vγ(j −1) ,

(23) (24)

j

vNin ∈ VN j

(25)

in

coincide on Nin with the solution of problem (19)–(22) for fm = 0, m ∈ Mex . Assume that the linear expressions lγj vγj and function ψγj (v (j −1) ) of (16) have already been determined for j 6 p − 1. Assume also that the solution of problem (23)–(25) has actually been computed for j 6 p − 1 and the values of vγ(p−1) are stored in the computer memory. Now, we are going to construct the expressions lγp vγp and ψγp (vγ(p−1) ). Consider the following auxiliary problem with respect to the unknown function ωNex , j 6 p: X X j wN

amn wn = 0, amn wn = fm ,

j m ∈ Mex ,

(26)

j ∈ Min ,

(27)

m

j ∈ VN .

(28)

Take some function-parameter zγ and define the right-hand side fm by the formula fm =

X

amn xn ,

j

m ∈ Min , j 6 p,

(29)

n∈Nm

where xn =

  vn , 

0, zn ,

if n ∈ γ j , j 6 p − 1, j if n ∈ Nin \γ j , j 6 p, if n ∈ γ p .

(30)

The solution wnp of problem (26)–(30) on the upper time level j = p depends linearly on vnj , n ∈ γ j , j 6 p − 1, and zn , n = γ p : 

wnp = wnp zγ p , vγ(p−1) .

(31)

V. Ryaben’kii / Applied Numerical Mathematics 33 (2000) 481–492

487 p

p

p Theorem 3. The function zn , n ∈ γ p , can be extended to the entire Nex as some vNex ∈ VNex such that the equation

X

amn vn = 0,

j m ∈ Mex ,

(32)

n∈Nm

is satisfied for j 6 p if and only if zn , n ∈ γ p , satisfies the equality 

wnp zγ p , vγ(p−1) = zn ,

n ∈ γ p.

(33) p

p Proof. Assume that Eq. (33) holds for a given zn , n ∈ γ p . Let wNex be the restriction on Nex ⊂ N p of the p p p solution wN to problem (26)–(30). Then, wNex is the function zn , n ∈ γ , from Theorem 3. p p p Conversely, assume that zn , n ∈ γ p , can be defined on the entire Nex as a function vNex ∈ VNex , p j vNex |n = zn , n ∈ γ p . Moreover, the function vn(p) , n ∈ Nex , j 6 p, satisfies Eq. (32). Let us show that (p) p then Eq. (33) holds. Extend the previously constructed function vNex on the set Nin \γ p by zero. The (p) (p) (p) (p−1) extended function vN belongs to VN . Therefore, by constructing the function vN , the solution vN (p) (p) of problem (26)–(28) for j 6 p − 1 has been extended to time level j = p as a function vN ∈ VN that (p) satisfies system (26)–(28) with the right-hand side fm , m ∈ Min , of type (29), (30). As the solution wnp to problem (26)–(28) is unique, it coincides with the constructed vnp for all n ∈ N p , in particular, for n ∈ γ p . But vnp = zn as n ∈ γ p according to our construction. Thus, wnp = zn , n ∈ γ p , and (33) holds. 2

Let us now transform the left-hand side of (33) and write this equality in a more convenient way. Due to the linearity we have 



wn zγ , vγ(p−1) = wn (zγ , 0) + wn 0, vγ(p−1) .

(34)

p Thus, the function wN (zγ , 0) is the solution of system (26)–(28) on one particular time level Nex

X

b n = 0, amn w

n∈Nm

X

bn = amn w

X

p m ∈ Mex ,

amn tn ,

p

m ∈ Min ,

(35) (36)

n∈Nm

b N ∈ VN , w

where



tn =

zn , 0,

(37)

n ∈ γ p, p n ∈ Nin \γ p .

(38)

b N linearly depends on zγ . Hence, one can write The solution w bn = w

X

p

αnk zk .

(39)

k∈γ p p

The coefficients αnk from (39) can be found by calculating solutions to problem (35)–(38) for the appropriate set of values of zn .

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V. Ryaben’kii / Applied Numerical Mathematics 33 (2000) 481–492

Theorem 4. Suppose lnp vγp has the following form: 

lnp vγp = vnp −

X

p

αnk vk .

(40)

k∈γ p

Further, assume that ψnp (vγ(p−1)) is defined as follows: 



ψnp vγ(p−1) = wnp 0, vγ(p−1) ,

n ∈ γ,

(41)

where wnp (0, vγ(p−1) ) is the value of the solution to problem (26)–(30) for zn = 0. Then, the equality 



lnp vγp = ψnp vγ(p−1) ,

n ∈ γ p,

(42)

with the left-hand side from (40) and right-hand side from (41), respectively, is the NRABC on the level γ p of the boundary γ of subdomain Nin . Proof. Consider Eq. (33) in the form



zn − wnp (zγ p , 0) = wnp 0, vγ(p−1) .

(43)

Using relation (40) it is easy to see that the left-hand side of (43) coincides with the left-hand side of Eq. (42) for zn = vγp |n , n ∈ γ p . The right-hand side of (43) coincides with the right-hand side of (42) because of (41). But according to the properties of Eq. (33) proven in Theorem 3, and since boundary conditions (33) and (42) are equivalent, this means that (42) is the NRABC on the level γ p of the boundary γ . 2 The following extension of Theorem 4 may be useful for constructing the NRABC in some applications. e n , n ∈ N j , j 6 p, is the solution of Eqs. (26)–(29) with the following xn : Theorem 5. Suppose w

xn =

  vn , 

ξn , 0,

if n ∈ γ j , j 6 p − 1, j if n ∈ Nin \γ j , j 6 p − 1, if j = p.

(44) j

j

Here ξn is an arbitrary function such that for the function xn , n ∈ Nin , from (44) one has xNin ∈ VNin , (p−1) j 6 p − 1. Define the function ψnp (vNin ), n ∈ γ p , by the formula (p−1) 

ψnp vNin

e np , =w

n ∈ γ p.

(45)

Then, the following equality 

(p−1) 

lnp vγp = ψnp vNex

,

n ∈ γ,

(46)

is the NRABC on the level γ p of the boundary γ for any choice of ξn in (44). Proof. First, let us note that in the case of ξn ≡ 0, n ∈ Nex \γ , the function (p−1) 

ψnp vNex



≡ ψnp vγ(p−1) , (p−1)

n ∈ Nex ,

i.e., the function ψnp (vNex ) coincides with the one from Theorem 4. In virtue of Theorem 4, Eq. (46) is the NRABC on the level γ p of the boundary γ . 2

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Now we show that the values of ψnp (vNex ) at n ∈ γ p do not depend on the choice ξn , n ∈ Nin \γ , if ξn e np of satisfies the condition formulated above. Indeed, due to the linearity of formulae (44), the solution w b np . The function wnp is the problem (26)–(29), (44) can be represented as a sum of two functions wnp and w p solution of problem (26)–(29) on γ with xn given by the formula   vn ,

xn =



if n ∈ γ j , j 6 p − 1, if n ∈ / γ j , j 6 p − 1, if j = p.

0, 0,

(47)

b np is the solution of problem (26)–(29) with xn given by the formula The function w   0,

if n ∈ γ j , j 6 p − 1, xn = ξn , if n ∈ Ninj \γ j , j 6 p − 1,  0, if j = p. (p−1)

Therefore, the expression for ψnp (vNin (p−1) 

ψnp vNin

b np , = wnp + w

(48)

) from (45) can be rewritten as the sum

n ∈ γ p.

(49)

Note that the function wnp from (49) is the same function that has been used to define the right-hand side = wnp , n ∈ γ p , of the NRABC (42) in Theorem 4. Thus, to prove Theorem 5 it is sufficient b np , vanishes at n ∈ γ p . To do that, we to show that the second term in formula (49), i.e., the function w j j extend the function ξn , n ∈ / Nin , to the entire N by setting ξn ≡ 0, n ∈ Nex . Denote this new function ξn , j p b ξn ∈ VN . Note that the solution wn of problem (26)–(29) with the right-hand side ψnp (vγ(p−1) )

fm =

  0, X 

amn ξn ,

m ∈ Mex , m ∈ Min ,

(50)

n∈Min

can be considered as the solution of the problem X

b n = fm , amn w

m ∈ M,

b N ∈ VN w

(51)

n∈Nm

with the right-hand side fm =

X

amn ξn ,

m ∈ M,

(52)

n∈Nm

where ξn are the values of ξN at n ∈ N . Indeed, formulae (50) and (52) coincide as for m ∈ Mex the stencil Nm has its points only in the domain Nex , in which ξn = 0, and hence formula (52) yields fm = 0, m ∈ Mex , or in other words, coincides with (50). One can write Eq. (51) in the form X

bn = amn w

X

amn ξn ,

m ∈ M,

(53)

n∈Nm

b N ∈ VN . w

(54)

Obviously, the function ξN ∈ VN is a solutions to problem (53), (54). But this problem has only b n = ξn , n ∈ N . On the other hand, ξn ≡ 0, n ∈ Nex . In particular, ξn = 0, one solution, therefore, w b np = 0, n ∈ γ p . n ∈ γ ⊂ Nex . Hence, w

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3.2. The case of explicit schemes Let vn = (vn1 , . . . , vns ) be a vector function. Assume that the difference scheme is explicit and normalized such that for every level j , X

amn vn = [vn0 (1),1, . . . , vn0 (s),s ]T ,

(55)

j n∈NM

where n0 (r) ∈ NM , r = 1, 2, . . . , s, is some grid node on the upper level j of the stencil Nm . Let us note that in the case of a scalar function vn the foregoing property means that one of the coefficients amn , n ∈ Nmj , m ∈ M j , equals to one but all other coefficients equal to zero. b n = (w b n1 , . . . , w b ns ) takes the form (k = 1, 2, . . . , s) One can see from Eq. (35) that formula (39) for w j



b nk = w

zn0 (k),k , 0,

if n = n0 (k), if n = 6 n0 (k).

(56)

Thus, the NRABC (42) on the level p now looks as follows (r = 1, 2, . . . , s): 

vnpr =

ψnpr (vγ(p−1) ), arbitrary,

if n 6= n0 (r), n ∈ γ p , if n = n0 (r), n ∈ γ p .

(57)

Theorem 6. If vγ(p−1) is the restriction to the boundary of the solution to the homogeneous problem X

amn vn = 0,

n∈Nm j j vNex ∈ VNex ,

j m ∈ Mex , j 6 p − 1,

j 6 p − 1,

then the following equality holds (r = 1, 2, . . . , s): 

ψnpr vγ(p−1) = 0,

if n = n0 (r), n ∈ γ p .

(58)

Proof. Let us write (33) in the form 

wnr (zγ , 0) + ψnpr vγ(p−1) = zn,r ,

n ∈ γ p,

r = 1, 2, . . . , s,

where wnr (zγ , 0) = zn,r if n = n0 (r). This implies (58) by virtue of (40).

(59) 2

3.3. Construction of the NRABC using the Green function We always assume that the auxiliary finite-difference problem (26)–(28) has a unique solution for an arbitrary right-hand side. Let fm = (fm1 , . . . , fms )T , s > 1, be the right-hand side at the node m. Obviously, there is a matrix function Gnm (Green’s function) such that Gnm fm yields the solution at the node n. In addition, Gnm vanishes if the time coordinate for the point n is greater than the one for the point m. Assume that the j solution vNex of the auxiliary problem for j 6 p − 1 has already been computed and its values vγ(p−1) on the boundary γ j , j 6 p − 1, are stored in the computer memory. The construction of NRABC on the next level 

lnp vγp = ψ vγ(p−1)



(60)

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491

requires, in particular, the calculation of ψnp (vγ(p−1) ). This function can be written as the right-hand side of Eq. (41) in the following form: 

ψ vγ(p−1) =

 X

p−1 X j =−∞



X

Gnm fm +

Gnm fm0 ,

(61)

p m∈Min

j m∈Min

where fm , m ∈ Min , j 6 p − 1, are given by formula (29) and fm0 , m ∈ Min , are given by (29) with zn ≡ 0. Let us emphasize that fm , m ∈ M j , j 6 p − 1, do not depend on p. We also note that sometimes it is possible to speed up the calculations by formula (61) using some particular features of the problem. For example, in the case of translationally invariant structure of the problem we have Gn,m ≡ Gn−m , i.e., the Green function depends only on one argument. Moreover, the following theorem is useful: j

p

Theorem 7. Assume that n0 ∈ γ p is a node for which (58) holds. Then, formula (61) can be written as follows: 



p

vnp = ψnp vγ(p−1) = ψnp vγ(p−1) − ψn0 vγ(p−1) p−1 X

=

j =−∞

 X



(Gnm − G

n0 m

)fm +



(Gnm − Gn0 m )fm0 .

p

j

m∈Min

m∈Min

Proof. It is sufficient to see that

X





p



vnp = ψnp vγ(p−1) = ψnp vγ(p−1) − ψn0 vγ(p−1) . Then subtract, the right-hand sides in (61) for n and n0 .

2

If n0 is chosen in an appropriate way, then the foregoing formula allows us to reduce the amount of computations by neglecting the terms for those j , for which these terms are sufficiently small. 4. Bibliographical comments The basic constructions and results discussed in this paper can be interpreted as an application and development of the difference potentials method proposed by the author in [1]. The NRABC for explicit schemes and immobile boundary of the computational subdomain are outlined in [2] with no proofs and in a slightly different form. In a number of papers by the author and his colleagues the NRABC based on the theory of difference potentials have been constructed and investigated for some steady problems of mathematical physics. Generally, there are many different approaches to constructing the NRABC available in the literature. An extensive review of the methodologies and results in this area can be found in [3]. References [1] V.S. Ryaben’kii, Difference Potentials Method for Some Problems of Continuous Media Mechanics, Nauka, Moscow, 1987 (in Russian).

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[2] V.S. Ryaben’kii, Exact transfer of difference boundary conditions, Functional Anal. Appl. 24 (3) (1990) 251– 253. [3] S.V. Tsynkov, Numerical solution of problems on unbounded domains. A review, Appl. Numer. Math. 27 (1998) 465–532.