Seria I: PRACE MATEMATYCZNE XLV (1) (2005), 87-105
Karolina Kropielnicka
Numerical method of bicharacteristics for quasilinear hyperbolic functional differential
systems
Abstract. Classical solutions of mixed problems for first order partial functional differential systems in two independent variables are approximated in the paper with solutions of a difference problem of the Euler type. The mesh for the approximate solutions is obtained by a numerical solving of equations of bicharacteristics. The convergence of explicit difference schemes is proved by means of consistency and sta- bility arguments. It is assumed that given functions satisfy nonlinear estimates of the Perron type. Differential systems with deviated variables and differential integral systems can be obtained from a general model by specializing given operators.
2000 Mathematics Subject Classification: 35R10, 65M25.
Key words and phrases: initial boundary value problems, bicharacteristics, interpo- lating operators.
1. Introduction. For any metric spaces X and Y we denote by C(X, Y ) the class of all continuous functions defined on X and taking values in Y . We use vectorial inequalities with the understanding that the same inequalities hold between their corresponding components. Let a, b > 0, d0, d ≥ 0 be given. Write c = b + d.
Let us define the sets:
E = [0, a] × [−b, b], E0= [−d0, 0] × [−c, c], ∂0E = [0, a] × ([−c, c] \ (−b, b)), Ω = E ∪ E0∪ ∂0E,
and
D = [−d0, 0] × [−d, d], Y = E × C(D, Rk).
For a function z : Ω → Rk, z = (z1, . . . , zk), and for a point (t, x) ∈ E we define a function z(t,x): D → Rk as follows
z(t,x)(s, y) = z(t + s, x + y), (s, y) ∈ D.
The function z(t,x) is the restriction of z to the set [t − d0, t] × [x − d, x + d], and this restriction is shifted to the set D. Elements of space C(D, Rk) will be denoted by w, ¯w and so on. Suppose that
f : Y → Rk, f = (f1, . . . , fk),
% : Y → Rk, % = (%1, . . . , %k), and
ϕ : E0× ∂0E → Rk, ϕ = (ϕ1, . . . , ϕk), ψ0: [0, a] → R, ψ1: E → R,
are given functions. We assume that 0 ≤ ψ0(t) ≤ t for t ∈ [0, a] and ψ1(t, x) ∈ [−b, b] for (t, x) ∈ E. Write ψ(t, x) = (ψ0(t), ψ1(t, x)), (t, x) ∈ E. We consider the functional differential system
(1) ∂tzτ(t, x) + %τ(t, x, zψ(t,x))∂xzτ(t, x) = fτ(t, x, zψ(t,x)), τ = 1, . . . , k, with the initial boundary condition
(2) z(t, x) = ϕ(t, x) on E0∪ ∂0E.
A function v : Ω → Rk, v = (v1, . . . , vk) is called a classical solution of the above problem if:
(i) v ∈ C(Ω, Rk) and v is of class C1on E,
(ii) v satisfies equation (1) on E and initial boundary condition (2) holds.
We are interested in a numerical solving of problem (1),(2).
In recent years, a number of papers concerning numerical methods for functional partial differential equations have been published. Difference methods and monotone iterative methods for nonlinear parabolic problems were studied in [6]-[9]. Quasi- linear first order functional differential systems and a general class of difference schemes with suitable interpolating operators were considered in [3]. The mono- graph [4] contains an exposition of the theory of difference methods for hyperbolic functional differential problems. The main question in these investigations is to find a difference functional equation which is stable and satisfies consistency conditions with respect to the original problem on sufficiency regular functions. A compar- ison technique is used in the investigation of the stability of functional difference problems.
In the paper we investigate a new class of numerical methods for (1), (2). We transform the initial boundary value problem into a system of integral functional equations along bicharacteristics, to find this solution. This leads to difference prob- lems of the Euler type. The mesh for approximate solutions is obtained by a nu- merical solving of equations of bicharacteristics. The main problem is, that it is impossible to obtain firstly the whole mesh and secondly the approximate solutions, because these bicharacteristics depend on the numerical solution. This is why we have to compute the mesh and the approximate solution simultaneously.
We use in the paper general ideas concerning numerical methods for first order partial differential equations which were introduced in [2], [4], [5].
Existence results for quasilinear hyperbolic functional differential problems can be found in [1], [4] (Chapter III), [10].
2. Discretization of mixed problems. For p ∈ Rk, p = (p1, . . . , pk), we write kpk = max{|pi| : 1 ≤ i ≤ k}.
The maximum norm in the space C(D, Rk) will be denoted by k·k0. We will denote by F (X, Y ) the class of all functions defined on X and taking values in Y , where X and Y are arbitrary sets. Let N and Z be the sets of natural numbers and integers respectively. We formulate now a difference problem corresponding to (1),(2). We define a mesh on the set E0∪ ∂0E in the following way. Let h = (h0, h1) where h0, h1> 0 stand for steps of the mesh. For h = (h0, h1) and (r, m) ∈ Z2we define nodal points as follows:
t(r)= rh0, x(m)= mh1.
Let us denote by H the set of all h = (h0, h1) such that there are N0∈ Z, N ∈ N with the properties: N0h0= d0, N h1= b.
Suppose that K ∈ N is defined by the relations Kh0≤ a < (K + 1)h0. Write R2h= {(t(r), x(m)) : (r, m) ∈ Z2}, Ih= {t(r): 0 ≤ r ≤ K},
Eh.0= E0∩ R2, ∂0Eh= ∂0E ∩ R2h.
For functions ϕh: Eh.0∪∂0Eh→ Rkand η : Ih→ R we write ϕ(r,m)h = ϕh(t(r), x(m)) and η(r)= η(t(r)). Set
Sh.0= {(t(r), x(m)) : r = 0, −N ≤ m ≤ N }, Sh.+= {(t(r), x(m)) : 0 ≤ r ≤ K, m = N }, Sh.−= {(t(r), x(m)) : 0 ≤ r ≤ K, m = −N },
Sh= Sh.0∪ Sh.+∪ Sh.−.
The numerical method of bicharacteristics consist in replacing problem (1),(2) with a system of difference equations for unknown functions (η1, . . . , ηk) = η of one variable and (z1, . . . , zk) = z depending on two variables. Now we present difference equations for η and z.
Functional differential problems considered in the paper have the following prop- erty. System (1) contains the functional variable z(t,x) which is an element of the space C(D, Rk). Numerical solutions of (1),(2) are functions defined on finite sets.
Therefore we need a family of interpolating operators Th = (Th.1, . . . , Th.k) in a system of difference equations. The operators Th are strictly connected with the unknown function η = (η1, . . . , ηk).
Suppose that (t(j), x(m)) ∈ Sh, 0 ≤ j ≤ K − 1, and ϕh: Eh.0∪ ∂0Eh→ Rk, ϕh= (ϕh.1, . . . , ϕh.k), is a given function. Let us consider the system of difference equa- tions
(3) η(r+1)τ = ητ(r)+ h0%τ(t(r), ητ(r), Th[z]ψ(t(r),η(r)
τ )), 1 ≤ r ≤ k, where j ≤ r ≤ K − 1, with initial conditions
(4) ητ(j)= x(m), 1 ≤ τ ≤ k.
Corresponding difference equations for (z1, . . . , zk) have the form zτ(t(r+1), ητ(r+1)) =
(5) zτ(t(r), ητ(r)) + h0fτ(t(r), η(r)τ , Th[z]
ψ(t(r),η(r)τ )), 1 ≤ τ ≤ k, where j ≤ r ≤ K − 1 and
(6) z(r,m)= ϕ(r,m)h on Eh.0∪ ∂0Eh. Now we formulate assumptions on Th. Let us denote by
gh(·, t(j), x(m)) = (gh.1(·, t(j), x(m)), . . . , gh.k(·, t(j), x(m))) and zh= (zh.1, . . . , zh.k) a solution of problem (3)-(6). Write
Eh.τ = {(t(i), y) ∈ E : 0 ≤ i ≤ K,
there is (t(j), x(m)) ∈ Shsuch that y = gh.τ(t(i), t(j), x(m))}, Ωh.τ = Eh.τ∪ Eh.0∪ ∂0Eh, 1 ≤ τ ≤ k,
and
Ω(r)h.τ = Ωh.τ× ([−d0, t(r)] × R), 1 ≤ τ ≤ k,
where 0 ≤ r ≤ K. We will assume that %, f ∈ C(Y, Rk), ϕ ∈ C(E0∪ ∂0E, Rk) and that
%τ(t, b, w) < 0 and %τ(t, −b, w) > 0 (7) for t ∈ [0, a], w ∈ C(D, Rk), 1 ≤ τ ≤ k.
Note that if condition (7) is satisfied then under natural assumptions on regularity of %, f and ϕ there exists a classical solution of (1),(2) and it is unique.
Assumption H[Th]. Suppose that the operators Th= (Th.1, . . . , Th.k) where Th.τ : F (Ωh.τ, R) → F (Ω, R)
satisfy the conditions:
1) if w ∈ F (Ωh.τ, R) then Th.τ[w] ∈ C(Ω, R), 1 ≤ τ ≤ k,
2) for each τ , 1 ≤ τ ≤ k, Th.τ satisfies the following Volterra condition: if (t(r), y) ∈ Ω
and w, ¯w ∈ F (Ωh.τ, R) are such functions that w
Ω(r)h.τ = ¯w Ω(r)h.τ
then
Th.τ[w](t(r), y) = Th.τ[ ¯w](t(r), y).
Note that problem (3)-(6) has the following property: if Assumptions H0[%, f ] and H[Th] are satisfied then there exist exactly one solution gh= (gh.1, . . . , gh.k), zh= (zh.1, . . . , zh.k) of (3)-(6).
Suppose that v : Ω → Rk, v = (v1, . . . , vk), is of class C1. We will denote by g[v](·, t, x) = (g1[v](·, t, x), . . . , gk[v](·, t, x))
the set of bicharacteristics of system (1) corresponding to v. Then for each τ , 1 ≤ τ ≤ k, the function gτ[v](·, t, x) is the solution of the Cauchy problem
(8) w0(s) = %τ(s, w(s), vψ(s,w(s))), w(t) = x.
Write
E˜h.τ = {(t(i), y) ∈ E : 0 ≤ i ≤ K,
and there is (t, x) ∈ Shsuch that y = gτ[v](t(i), t, x)}, 1 ≤ τ ≤ k, and
Ω˜h.τ= ˜Eh.τ∪ Eh.0∪ ∂0Eh, 1 ≤ τ ≤ k, Ω˜(r)h.τ= ˜Ωh.τ∩ ([−b0, t(r)] × R)), 1 ≤ τ ≤ k, where 0 ≤ r ≤ K.
Numerical procedure (3)-(6) generate two sets of functions: {gh}h∈Hand {zh}h∈H. The functions gh= (gh.1, . . . , gh.k), h ∈ H, are used for the construction of the family of sets (Ωh.1, . . . , Ωh.k) whereas zh= (zh.1, . . . , zh.k) are considered as approximate solutions to problem (1)-(2) and
zh.τ: Ωh.τ → R, 1 ≤ τ ≤ k.
Suppose that v : Ω → Rk, v = (v1, . . . , vk), is of class C1 and consider the set of bicharacteristics g[v](·, t, x) = (g1[v](·, t, x), . . . , gk[v](·, t, x)) where (t, x) ∈ Sh. Then the family of sets ( ˜Ωh.1, . . . , ˜Ωh.k) is defined and we consider the function
˜
vh= (˜vh.1, . . . , ˜vh.k) where ˜vh.τ is the restriction of vτ to the set ˜Ωh.τ. The functions g[v] and ˜vh may be considered as approximate solutions of (3)-(6). Note that for each τ, 1 ≤ τ ≤ k, the domains of the functions zh.τ and ˜vh.τ are not the same. This leads to the following notion of an error of the method (3)-(6). Let us introduce the following notation.
[|v − zh|](r)h.E= max
1≤τ ≤kmax{|vτ(t(i), gτ[v](t(i), t, x)) − zh.τ(t(i), gh.τ(t(i), t, x)| : 0 ≤ i ≤ r, (t, x) ∈ Sh, t ≤ t(i−1)}
[|v − zh|](r)h.∂= max
1≤τ ≤kmax{|vτ(t(i), x(m)) − zh.τ(t(i), x(m))| : (t(i), x(m)) ∈ E0.h∪ ∂0Eh, i ≤ r},
and
[|v − zh|](r)h = max{[|v − zh|](r)h.E, [|v − zh|](r)h.∂}, 0 ≤ r ≤ K.
[|g[v] − gh|](r)h = max
1≤τ ≤kmax{|gτ[v](t(i), t, x) − gh.τ(t(i), t, x) :
0 ≤ i ≤ r, (t, x) ∈ Sh, t ≤ t(i−1)}.
Then the error of the method (3)-(6) is defined by
(9) ε(r)h = [|v − zh|](r)h + [|g[v] − gh|](r)h , 0 ≤ r ≤ K.
We prove that for v satisfying (1),(2) and for sufficiently regular f and % we have lim
h→0ε(r)h = 0 uninformly with respect to r, 0 ≤ r ≤ K.
For a function z ∈ C(Ω, Rk) and for a point t ∈ [0, a] we put
||z||t= max||z(τ, y)|| : (τ, y) ∈ Ω ∩ ([−d0, t] × T ) .
Assumption H[Th, ˜Th]. Suppose that the operators ˜Th= ( ˜Th.1, . . . , ˜Th.k) where T˜h.τ : F ( ˜Ωh.τ, R) → F (Ω, R), 1 ≤ τ ≤ k,
satisfy the conditions:
1) if w ∈ F ( ˜Ωh.τ, R) then ˜Th.τ[w] ∈ C(Ω, R), 1 ≤ τ ≤ k,
2) for each τ , 1 ≤ τ ≤ k, ˜Th.τ satisfies the following Volterra condition: if (t(r), y) ∈ Ω and w, ¯w ∈ F ( ˜Ωh.τ, R) are such functions that
w ˜
Ω(r)h.τ = ¯w ˜
Ω(r)h.τ
then
T˜h.τ[w](t(r), y) = ˜Th.τ[ ¯w](t(r), y).
3) if u : Ω → Rk, u = (u1, . . . , uk), is of class C1then there is ˜β : H → R+such
that
T˜h[˜uh] − u t(r)
≤ ˜β(h), 0 ≤ r ≤ K, 1 ≤ τ ≤ k and
lim
h→0
β(h) = 0,˜
where ˜uh= (˜uh.1, . . . , ˜uh.k), ˜Th[˜uh] = ˜Th.1[˜uh.1], . . . , ˜Th.k[˜uh.k]) and ˜uh.τ is the restriction of uτ to the set ˜Ωh.τ,
4) Assumption H[Th] is satisfied and for each v : Ω → Rk which is of class C1, v = (v1, . . . , vk), and for gh= (gh.1, . . . , gh.k), zh= (zh.1, . . . , zh.k) satisfying (3)-(6) there are ˜C ∈ R+, ˜γ : H → R+such that
T˜h[˜vh] − Th[zh]
t(r)≤ [|˜vh− zh|](r)h + (10) C[|g[v] − g˜ h|](r)h + ˜γ(h), 0 ≤ r ≤ K,
(11) lim
h→0γ(h) = 0,˜
where ˜vh= (˜vh.1, . . . , ˜vh.k) and ˜vh.τ is the restriction of vτ to the set ˜Ωh.τ.
Examples of Thand ˜Thwhich satisfy Assumption H[Th, ˜Th] are given in Section 4.
3. Convergence of the numerical method. The main assumptions on f , % and ψ are the following:
Assumption H[f , %, ψ]. Suppose that the functions f : Y → Rk and % : Y → Rk are continuous and that there is σ : [0, a] × R+→ R+such that
1) σ is continuous and it is nondecreasing with respect to both variables, 2) σ(t, 0) = 0 for t ∈ [0, a] and for each κ ≥ 1 the the maximal solution of the
Cauchy problem
ζ0(t) = κσ(t, κζ(t)), ζ(0) = 0 is ζ(t) = 0 for t ∈ [0, a],
3) the estimates
kf (t, x, w) − f (t, ¯x, ¯w)k ≤ σ(t, ||x − ¯x|| + ||w − ¯w||0), k%(t, x, w) − %(t, ¯x, ¯w)k ≤ σ(t, ||x − ¯x|| + ||w − ¯w||0).
are satisfied on Y .
4) for (t, w) ∈ [0, a] × C(D, Rk) we have
%τ(t, b, w) < 0 and %τ(t, −b, w) > 0, 1 ≤ τ ≤ k
5) the functions ψ0 ∈ C([0, a], R), ψ1 ∈ C(E, R) are such that 0 ≤ ψ0(t) ≤ t, ψ1(t, x) ∈ [−b, b] for (t, x) ∈ E and there is p ∈ R+such that
|ψ1(t, x) − ψ1(t, ¯x)| ≤ p |x − ¯x| on E.
Theorem 3.1 Suppose that Assumption H[f, %], H[Th, ˜Th] are satisfied and 1) the function ϕ : E0∪ ∂0E → Rkis of class C1, the function v : Ω → Rk is the
solution of (1)-(2) and v is of class C1 on Ω,
2) zh = (zh.1, . . . , zh.k) and gh = (gh.1, . . . , gh.k) are the solution of (3) - (6), where zh.τ: Ωh.τ → R and gh.τ : Ih→ R, and there is a function α0: H → R+ such that
ϕ(r,m)− ϕ(r,m)h
≤ α0(h) on E0.h∪ ∂0Ehand lim
h→0α0(h) = 0 Then there is ε0> 0 and a function α : H → R+such that for ||hk < ε0we have (12) [|gh− g[v]|](r)h + [|zh− v|](r)h ≤ α(h), 0 ≤ r ≤ N0and lim
h→0
α(h) = 0, where g[v] = (g1[v], . . . , gk[v]) is the set of bicharacteristic of system (1),(2) corre- sponding to v.
Proof We will write a difference inequality for the function εh defined by (9).
Suppose that (t, x) ∈ Sh. Then we have
vτ(t(r+1), gτ[v](t(r+1), t, x)) = vτ(t(r), gτ[v](t(r), t, x))
(13) +
Z t(r+1) t(r)
fτ(s, gτ[v](s, t, x), vψ(s,gτ[v](s,t,x)))ds, 1 ≤ τ ≤ k, and
gτ[v](t(r+1), t, x) = gτ[v](t(r), t, x)
(14) +
Z t(r+1) t(r)
%τ(s, gτ[v](s, t, x), vψ(s,gτ[v](s,t,x)))ds, 1 ≤ τ ≤ k.
It follows from (3),(5), that
zh.τ(t(r+1), gh.τ(t(r+1), t, x)) = zh.τ(t(r), gh.τ(t(r), t, x)) (15) +h0fτ(t(r), gh.τ(t(r), t, x), Th[zh]ψ(t(r),gh.τ(t(r),t,x))), 1 ≤ τ ≤ k, and
gh.τ(t(r+1), t, x) = gh.τ(t(r), t, x)
(16) +h0%τ(t(r), gh.τ(t(r), t, x), Th[zh]ψ(t(r),gh.τ(t(r),t,x))), 1 ≤ τ ≤ k.
Substracting (13) and (15) we find that
vτ(t(r+1), gτ[v](t(r+1), t, x)) − zh.τ(t(r+1), gh.τ(t(r+1), t, x)) = vτ(t(r), gτ[v](t(r), t, x)) − zh.τ(t(r), gh.τ(t(r), t, x))+
h0
fτ(t(r), gτ[v](t(r), t, x), vψ(t(r),gτ[v](t(r),t,x)))−
(17) fτ(t(r), gh.τ(t(r), t, x), Th[zh]ψ(t(r),gh.τ(t(r),t,x)))
+ Γ(r)h.τ, 1 ≤ τ ≤ k, where
Γ(r)h.τ = Z t(r+1)
t(r)
fτ(s, gτ[v](s, t, x), vψ(s,gτ[v](s,t,x)))−
fτ(t(r), gτ[v](t(r), t, x), vψ(t(r),gτ[v](t(r),t,x)))
ds 1 ≤ τ ≤ k.
Write
Γ(r)h = (Γ(r)h.1, . . . , Γ(r)h.k), 0 ≤ r ≤ K − 1.
It follows that there is γ : H → R+such that
||Γ(r)h || ≤ h0γ(h), for 0 ≤ r ≤ K − 1 and lim
h→0γ(h) = 0.
We conclude from (17) and from Assumption H[f, %] that
vτ(t(r+1), gτ[v](t(r+1), t, x)) − zh.τ(t(r+1), gh.τ(t(r+1), t, x))
≤ h0σ t(r), [|g[v] − gh|](r)h + ||vψ(t(r),gτ[v](t(r),t,x))− Th[zh]ψ(t(r),gh.τ(t(r),t,x))||0
(18) +[|˜vh− zh|](r)h.E+ h0γ(h), 1 ≤ τ ≤ k.
Write ˜vh= (˜vh.1. . . , ˜vh.k) where ˜vh.τ is the restriction of vτ to the set ˜Ωh.τ. We thus get
||v(t(r),gτ[v](t(r),t,x))− Th[zh](t(r),gh.τ(t(r),t,x))||0≤
≤ A(r)(h) + B(r)(h) + C(r)(h) where
A(r)(h) = ||vψ(t(r),gτ[v](t(r),t,x))− vψ(t(r),gh.τ(t(r),t,x))||0, B(r)(h) = ||vψ(t(r),gh.τ(t(r),t,x))− ˜Th[˜vh]ψ(t(r),gh.τ(t(r),t,x))||0, C(r)(h) = || ˜Th[˜vh]ψ(t(r),gh.τ(t(r),t,x))− Th[zh]ψ(t(r),gh.τ(t(r),t,x))||0.
Let ˜c ∈ R+be such a constant that k∂xv(t, x)k ≤ ˜c for (t, x) ∈ Ω. Then we have A(r)(h) ≤ ˜cp [|g[v] − gh|](r)h .
It follows from Assumption H[Th, ˜Th] that there is γ0: H → R+such that B(r)(h) ≤ γ0(h), 0 ≤ r ≤ K − 1, and lim
h→0γ0(h) = 0.
Since
C(r)(h) ≤ ||Th[˜vh] − Th[zh]||t(r), 0 ≤ r ≤ K − 1,
Assumption H[Th, ˜Th] shows that there are ˜C ∈ R+and ˜γ : H → R+such that C(r)(h) ≤ [|˜vh− zh|](r)h + ˜C[|g[v] − gh|](r)h + ˜γ(h)
and
lim
h→0γ(h) = 0.˜ We conclude from (18) that
|vτ(t(r+1), gτ[v](t(r+1), t, x)) − zh.τ(t(r+1), gh.τ(t(r+1), t, x))|
≤ [|˜vh− zh|](r)h + h0σ(t(r), (1 + ˜c + ˜C)[|g[v] − gh|](r)h |+
[|˜vh− zh|](r)h + γ0(h) + ˜γ(h)) + h0γ(h), 1 ≤ τ ≤ k, and consequently
[|˜vh− zh|](r+1)h ≤ [|˜vh− zh|](r)h +
(19) h0σ(t(r), ¯cε(r)h + γ0(h) + ˜γ(h)) + h0γ(h), 0 ≤ r ≤ K − 1,
where ¯c = 1 + ˜cp + ˜C.
In a similar way we prove that there is ¯γ : H → R+such that [|g[v] − gh|](r+1)h ≤ [|g[v] − gh|](r)h +
(20) h0σ(t(r), ¯cε(r)h + γ0(h) + ˜γ(h)) + h0¯γ(h), 0 ≤ r ≤ K − 1, and
lim
h→0γ(h) = 0.¯ Write
β0(h) = γ0(h) + ˜γ(h), β(h) = γ(h) + ¯γ(h).
It follows from (19), (20) that the function εhsatisfies the difference inequality (21) ε(r+1)h ≤ ε(r)h + 2h0σ(t(r), ¯cε(r)h + β0(h)) + h0β(h), 0 ≤ r ≤ K − 1, and the initial estimate ε(0)h ≤ α0(h) is satisfied.
Consider the Cauchy problem
(22) ζ0(τ ) = 2σ(τ, ¯cζ(τ ) + β0(h)) + β(h), ζ(0) = α0(h)
It follows from Assumption H[f ,%] that there is ε0> 0 such that for h0+ h1< ε0 there exists the maximal solution ωh of (22) and ωhis defined on [0, a]. Moreover we have
lim
h→0ωh(τ ) = 0 uninformly on [0, a].
The function ωhsatisfies the recurrent inequality
(23) ωh(r+1)≥ ω(r)h + 2h0σ(t(r), ¯cω(r)h + β0(h)) + β(h), 0 ≤ r ≤ K − 1.
It follows from (21), (23) that
ε(r)h ≤ ωh(r)f or 0 ≤ r ≤ K and consequently
[|˜vh− zh|](r)h + [|g[h] − gh|](r)≤ ωh(a), 0 ≤ r ≤ K.
Then assertion (12) is satisfied with α(h) = ωh(a) and the theorem is proved.
Remark 3.2 If all the assumptions of Theorem 1 are satisfied with σ(t, s) = Ls, (t, s) ∈ [0, a] × R+, then we have the estimates
ε(r)h ≤ α0(h)e2L¯ca+ (2Lβ0(h) + β(h))e2L¯ca− 1
2L¯c for L > 0 and
ε(r)h ≤ α0(h) + (2Lβ0(h) + β(h)) for L = 0 where 0 ≤ r ≤ K.
The above inequalities are consequences of (12).
4. Interpolating operators. We give examples of the operators Thand ˜Th satisfying Assumption H[Th, ˜Th]. Our investigations start with the construction of ( ˜Th.1, . . . , ˜Th.k) = ˜Th.
Suppose that α, β ∈ Sh and α = (α0, α1), β = (β0, β1). The relation α ≈ β means that
i) α0= β0= 0 and α1= x(m), β1= x(m+1) where −N ≤ m ≤ N − 1 ii) α1= β1= b and α0= t(i), β0= t(i+1)where 0 ≤ i ≤ K − 1 iii) α1= β1= −b and β0= t(i), α0= t(i+1) where 0 ≤ i ≤ K − 1
Suppose that Assumption H[f, %,ψ] with σ(t, s) = Ls, (t, s) ∈ [0, a] × R+is satisfied and that v : Ω → Rk is of class C1. Let g[v](·, ¯t, ¯x) = (g1[v](·, ¯t, ¯x), . . . , gk[v](·, ¯t, ¯x)) denote the set of bicharacteristics corresponding to v passing through (¯t, ¯x) ∈ E.
Suppose that w ∈ F ( ˜Ωh.τ, R), 1 ≤ τ ≤ k. We have divided the definition of T˜h.τ[w] : Ω → R into following cases.
I Suppose that (t, x) ∈ E0∪ ∂0E and there is (t(i), x(m)) ∈ Eh.0∪ ∂0Ehsuch that t(i)≤ t ≤ t(i+1), x(m)≤ x ≤ x(m+1). Write Th.τ[w](t, x) = Φ(i,m)(t, x) where
Φ(i,m)(t, x) = t(i+1)− t h0
x(m+1)− x
h1 w(i,m)+t(i+1)− t h0
x − x(m)
h1 w(i,m+1)+
(24) t − t(i) h0
x(m+1)− x
h1 w(i+1,m)+t − t(i) h0
x − x(m)
h1 w(i+1,m+1). II Suppose that (t, x) ∈ ¯E and there is 0 ≤ i ≤ K such that t(i)≤ t ≤ t(i+1).
Assume that there are points α, β ∈ Sh, α = (α0, α1), β = (β0, β1) such that α ≈ β and
L(i,α)h (t) ≤ x ≤ L(i,β)h (t) where
L(i,α)(t) = gτ[v](t(i), α) +t − t(i)
h0 gτ[v](t(i+1), α) − gτ[v](t(i), α) and
L(i,β)(t) = gτ[v](t(i), β) + t − t(i)
h0 gτ[v](t(i+1), β) − gτ[v](t(i), β) Write
A(i,m)(t, x) = (1 −t − t(i)
h0 )(1 − x − gτ[v](t(i), α) gτ[v](t(i), β) − gτ[v](t(i), α)), B(i,m)(t, x) =t − t(i)
h0 (1 − x − gτ[v](t(i+1), α) gτ[v](t(i+1), β) − gτ[v](t(i+1), α)), C(i,m)(t, x) = t − t(i)
h0
x − gτ[v](t(i+1), α) gτ[v](t(i+1), β) − gτ[v](t(i+1), α),
D(i,m)(t, x) = (1 −t − t(i)
h0 ) x − gτ[v](t(i), α) gτ[v](t(i), β) − gτ[v](t(i), α), and
T˜h.τ[w](t, x) = A(i,m)(t, x)w(t(i), gτ[v](t(i), α))+
B(i,m)(t, x)w(t(i+1), gτ[v](t(i+1), α)) + C(i,m)(t, x)w(t(i+1), gτ[v](t(i+1), β))+
(25) D(i,m)(t, x)w(t(i), gτ[v](t(i), β)).
III Suppose that (t, x) ∈ E and t(i) ≤ t ≤ t(i+1) where 0 ≤ i ≤ K − 1. Assume that there is β ∈ Sh.+, β = (β0, β1) such that
L(i,β)h (t) ≤ x ≤ b.
Then ˜Th.τ[w](t, x) is defined by (25) with α = (β0− h0, β1).
If (t, x) ∈ E, t(i)≤ t ≤ t(i+1)and
−b ≤ x ≤ L(i,α)h (x)
where α ∈ Sh.−, α = (α0, α1). Then ˜Th.τ[w](t, x) is defined by (25) with β = (α0− h0, α1).
IV If (t, x) ∈ E and t(k)< t ≤ a then we put ˜Th.τ[w](t, x) = ˜Th.τ[w](t(k), x).
Then we have defined ˜Th.τ[w] : Ω → R where 1 ≤ τ ≤ k.
The interpolating operators (Th.1, . . . , Th.k) = Th are defined in the same way with ghinstead of g[v].
Lemma 4.1 Suppose that Assumption H[f, %] is satisfied and that v : Ω → Rk, v = (v1, . . . , vk), is of class C1. Then there is ˜β : H → R+such that
(26) || ˜Th[˜vh] − v||t(r)≤ ˜β(h), 0 ≤ r ≤ K, and
(27) lim
h→0
β(h) = 0˜
where ˜vh= (˜vh.1, . . . , ˜vh.k) and ˜vh.τ is the restriction of vτ to the set ˜Ωh.τ.
Proof Suppose that (t, x) ∈ Ω and that ˜Th.τ[˜vh.τ](t, x) is defined by (25). It is easily seen that
A(i,m)(t, x) + B(i,m)(t, x) + C(i,m)(t, x) + D(i,m)(t, x) = 1,
(t(i)− t)(A(i,m)(t, x) + D(i,m)(t, x) + (t(i+1)− t)(B(i,m)(t, x) + C(i,m)(t, x)) = 0 and
A(i,m)(t, x)(gτ[v](t(i), α) − x) + B(i,m)(t, x)(gτ[v](t(i+1), α) − x)+
+C(i,m)(t, x)(gτ[v](t(i+1), β) − x) + D(i,m)(t, x)(gτ[v](t(i), β) − x) = 0.
We conclude from the above relations and from the Taylor theorem that there are intermediate points An, Bn, Cn, Dnsuch that
T˜h.τ[˜vh.τ](t, x) − vτ(t, x) =
= A(i,m)(t, x)
(t(i)− t)
∂tvτ(An) − ∂tvτ(t, x)
+
(gτ[v](t(i), α) − x)
∂xvτ(An) − ∂xvt(t, x)
+
B(i,m)(t, x)
(t(i+1)− t)
∂tvτ(Bn) − ∂tvτ(t, x)
+
(gτ[v](t(i+1), α) − x)
∂xvτ(Bn) − ∂xvt(t, x)
+
C(i,m)(t, x)
(t(i+1)− t)
∂tvτ(Cn) − ∂tvτ(t, x)
+
(gτ[v](t(i+1), β) − x)
∂xvτ(Cn) − ∂xvt(t, x)
+
D(i,m)(t, x)
(t(i)− t)
∂tvτ(Dn) − ∂tvτ(t, x)
+
(gτ[v](t(i), β) − x)
∂xvτ(Dn) − ∂xvt(t, x)
. Then there is ˜β : F → R+satisfying (26), (27).
The same conclusion can be drawn if ˜Th.τ[˜vh.τ](t, x) is given by (I) or (III). Assumption ˜H[f ,%,ψ]. Suppose that Assumption H[f ,%,ψ] is satisfied with σ(t, s) = Ls, (t, s) ∈ [0, a] × R+and for each ˜C > 0 there is ˜L > 0 such that for each z ∈ C(Ω, R), ||z(t, x)|| ≤ ˜C on Ω, we have
(28) ||%(t, x, z(t,s)) − %(t, x, z(t,¯s))|| ≤ ˜L|s − ¯s|
where t ∈ [0, a], x, s, ¯s ∈ [−b, b].
Remark 4.2 Suppose that the function ¯% : E × Rk→ Rk is continuous and there is ¯L ∈ R+such that
||¯%(t, x, ξ) − ¯%(t, x, ¯ξ)|| ≤ ¯L||ξ − ¯ξ||
where (t, x) ∈ E, ξ, ¯ξ ∈ Rk. Write
%(t, x, w) = ¯%
t, x,
Z
D
w(ξ, η) dξdη
. Then
%(t, x, z(t,s)) = ¯%
t, x,
Z
D
z(t + ξ, s + η) dξdη
. and condition (28) is satisfied.
Lemma 4.3 Suppose that Assumption ˜H[f, %,ψ] is satisfied and that v : Ω → Rk, v = (v1, . . . , vk), is of class C1. Then for sufficiently small h0 there are ˜C ∈ R+,
˜
γ : H → R+such that conditions (10),(11) are satisfied where ˜v = (˜v1, . . . , ˜vk) and
˜
vh.τ is the restriction of vτ to set ˜Ωh.τ and gh, zhsatisfy (3)-(6).
Proof There is ε0> 0 such that for h0< ε0and for α, β ∈ Sh, α ≈ β, we have gh.τ(t(r), α) < gh.τ(t(r), β), 1 ≤ τ ≤ k
on a common domain of gh.τ(·, α) and gh.τ(·, β).
Suppose that 0 ≤ r ≤ K is fixed. There are t(i) ∈ [0, a], y ∈ [−c, c] and τ , 1 ≤ τ ≤ k, such that
|| ˜Th[˜vh] − Th[zh]||t(r)= (29) T˜h.τ[˜vh.τ](t(i), y) − Th.τ[zh.τ](t(i), y).
It is easy to see that estimate (10) is satisfied if y ∈ [−c, −b] ∪ [b, c].
Assume that y ∈ (−b, b). There are α, β ∈ Sh, α ≈ β such that (30) gh.τ(t(i), α) ≤ y ≤ gh.τ(t(i), β)
Let ¯y ∈ [−b, b] be defined by the relations
(31) gτ[v](t(i), α) ≤ ¯y ≤ gτ[v](t(i), β) and
(32) y − g¯ τ[v](t(i), α)
gτ[v](t(i), β) − gτ[v](t(i), α) = y − gh.τ(t(i), α) gh.τ(t(i), β) − gh.τ(t(i), α) Then we have
T˜h.τ[˜vh.τ](t(i), y) − Th.τ[zh.τ](t(i), y) =
= ˜Th.τ[˜vh.τ](t(i), y) − Th.τ[˜vh.τ](t(i), ¯y)+
(33) Th.τ[˜vh.τ](t(i), ¯y) − Th.τ[zh.τ](t(i), y).
It follows from (25) and (30)-(32) that
Th.τ[˜vh.τ](t(i), ¯y) − Th.τ[zh.τ](t(i), y) =
1 − y − g¯ τ[v](t(i), α) gτ[v](t(i), β) − gτ[v](t(i), α)
vτ(t(i), gτ[v](t(i), α))+
¯
y − gτ[v](t(i), α)
gτ[v](t(i), β) − gτ[v](t(i), α)vτ(t(i), gτ[v](t(i), β))−
1 − y − gh.τ(t(i), α) gh.τ(t(i), β) − gh.τ(t(i), α)
zh.τ(t(i), gh.τ(t(i), α))+
y − gh.τ(t(i), α)
gh.τ(t(i), β) − gh.τ(t(i), α)zh.τ(t(i), gh.τ(t(i), β))+
and consequently
Th.τ[˜vh.τ](t(i), ¯y) − Th.τ[zh.τ](t(i), y) ≤ [|˜vh− zh|](r)h .
It follows from (30),(31) that there is ¯γ : H → R+such that
|y − ¯y| ≤ [|g[v] − gh|](r)h + ¯γ(h) and
lim
h→0γ(h) = 0.¯
Let ¯c ∈ R+be such a constant that ||∂xv(t, x)|| ≤ ¯c for (t, x) ∈ Ω. Then we have
| ˜Th.τ[˜vh.τ](t(i), y) − ˜Th.τ[˜vh.τ](t(i), ¯y)| ≤
¯
c[|g[v] − gh|](r)h + ¯c¯γ(h).
We conclude from (29),(33) that
[| ˜Th[˜vh] − Th[zh]|]t(r)≤ [|˜vh− zh|](r)h + ¯c[|g[v] − gh|](r)h + ¯c¯γ(h).
This is our claim.
Now we give an example of Th and ˜Th for problem (1),(2) with ψ0(t) = t, ψ1(t, x) = x. Then the corresponding difference equations have the form
(34) ητ(r+1)= η(r)τ + h0%τ(t(r), ητ(r), Th[z]
(t(r),ητ(r))), 1 ≤ τ ≤ k with initial condition (4) where (t(j), x(m)) ∈ Shand
(35) zτ(t(r+1), η(r+1)τ ) = zτ(t(r), ητ(r)) + h0fτ(t(r), ητ(r), Th[z](t(r),η(r)
τ )), 1 ≤ τ ≤ k with initial boundary condition (6).
Lemma 4.4 Suppose that
1) Assumption H[f ,%,ψ] is satisfied with σ(t, s) = Ls, (t, s) ∈ [0, a] × R+, where L ∈ R+,
2) v : Ω → Rk is of class C1, ϕh= ϕ and there is ˜c ∈ R+such that
||ϕ(t, x) − ϕ(¯t, ¯x)|| ≤ ˜c(|t − ¯t| + |x − ¯x|) on E0∪ ∂0E
Then for suffitiently small h0there are ˜C ∈ R+, ˜γ : H → R+such that condition (10),(11) are satisfied where ˜v = (˜vh.1, . . . , ˜vh.k) and ˜vh.τ is the restriction of vτ to the set ˜Ωh.τ and gh, zh satisfy (34),(4) and (35),(6) and g[v] = (g1[v], . . . , gk[v]) satisfy (8) with ψ0(t) = t, ψ1(t, x) = x.
Proof We prove now that there is ˜L ∈ R+such that
(36) ||gh(t(r), t(i), x(m)) − gh(t(r), t(i), x(m−1))|| ≤ ˜Lh1
(37) ||gh(t(r), t(i), x(m)) − gh(t(r), t(i−1), x(m))|| ≤ ˜Lh0 on the domain of ghand
(38) ||zh(t(r), gh(t(r), t(i), x(m))) − zh(t(r), gh(t(r), t(i), x(m−1)))|| ≤ ˜Lh1
(39) ||zh(t(r), gh(t(r), t(i), x(m))) − zh(t(r), gh(t(r), t(i−1), x(m)))|| ≤ ˜Lh0 on the domain of zh. Write
K˜(r)= max
h−11 ||gh(t(j), t(i), x(m)) − gh(t(j), t(i), x(m−1))|| : 0 ≤ j ≤ r
where 0 ≤ r ≤ K.
K˜(r)= max
h−11 ||zh(t(j), gh(t(j), t(i), x(m))) − zh(t(j), gh(t(j), t(i), x(m−1)))|| : 0 ≤ j ≤ r
It follows that
K(r+1)≤ K(r)+ h0(K(r)+ ˜K(r))
K˜(r+1)≤ ˜K(r)+ h0(K(r)+ ˜K(r)), 0 ≤ r ≤ K − 1.
Then there is C > 0 such that K(r)+ ˜K(r)≤ C for 0 ≤ r ≤ K and estimates (36), (38) follow. In a similar way we prove (37), (39). It follows from (38), (39) that there is ¯C ∈ R+such that
(40) ||Th[zh](t, x) − Th[zh](t, ¯x)|| ≤ ¯C||x − ¯x||.
we conclude from assumption 1) of the Lemma and (40) that there is ε0> 0 such that for 0 ≤ h0≤ ε0and for α, β ∈ Sh, α ≈ β, we have
gh.τ(t(r), α) < gh.τ(t(r), β), 1 ≤ τ ≤ k
on a common domain of gh.τ(·, α) and gh.τ(·, β). Suppose that 0 ≤ r ≤ K is fixed.
There are t(i)∈ [0, a], y ∈ [−c, c] and τ , 1 ≤ τ ≤ k, such that (29) holds. It is easy to see that conditions (10),(11) are satisfied if the y ∈ [−c, −b] ∪ [b, c]. The proof of (10),(11) in the case when y ∈ [−b, b] is similar to the proof at Lemma 2. Details
are omitted.
5. Numerical examples. Write
E = [0, 0.5] × [−1, 1], E0= {0} × [−1, 1], ∂0E = [0, 0.5] × {−1, 1}.
Consider the differential equation with deviated variables
∂tz(t, x) = x
2∂xz(t, x) + e4t
z
t,x + 1
2
+ z
t,x − 1
2
+
(41) f (t, x)z(t, x) + g(t, x)
and the initial boundary condition
(42) z(t, x) = 1 on E0∪ ∂0E
where f (t, x) = 4x2(1 − t) − 4 and g(t, x) = −et(x+1)2− et(x−1)2. The solution of (41),(42) is known, it is u(t, x) = e4t(x2−1). Write
t(r)= rh0, 0 ≤ r ≤ K, x(m)= mh1, −N ≤ m ≤ N,
where Kh0= 0.5, N h1= 1. Let us denote by (zh, ηh) the solution which is obtained by solving the numerical method oh bicharacteristics for (41),(42). Set
Eh = {(t(i), y) ∈ E : 0 ≤ i ≤ K, there is (t(j), x(m)) ∈ E0∪ ∂0E such that y = ηh(t(i), t(j), x(m))}.
The classical difference method for (41),(42) has the form z(r+1,m)= 1
2
z(r,m+1)+ z(r,m−1)
+ h0 x(m) 4h1
z(r,m+1)− z(r,m−1) +
h0 h
e4t(r)
Th[z](t(r), 0.5(x(m)+ 1)) + Th[z](t(r), 0.5(x(m)− 1)) + f(r,m)z(r,m)+ g(r,m)
i , (43)
and
(44) z(r,m)= 1 on E0.h∪ ∂0Eh,
where This the interpolating operator defined by (24).
Let vhdenote the solution of (43),(44). Write Ih(r)=
n
y : (t(r), y) ∈ Eh o
, 0 ≤ r ≤ K, and
ε(r)h = max n
|u(t(r), y) − zh(t(r), y)| : y ∈ Ih(r) o
,
˜
ε(r)h = maxn
|u(t(r), x(m)) − vh(t(r), x(m))| : −N ≤ m ≤ No
where 0 ≤ r ≤ K. The numbers ε(r)h and ˜ε(r)h are the maximal errors of the methods with fixed t(r).
In the Table we give experimental values of the functions ε(r)h and ˜ε(r)h and we write ”X” if ˜ε(r)h > 1.
Table of errors (εh; ˜εh) h0 = 0.002, h1= 0.0005
t = 0.30 (0.002201; X) t = 0.35 (0.002524; X) t = 0.40 (0.003032; X) t = 0.45 (0.003947; X) t = 0.50 (0.005774; X)
If h0< 2h1 then the classical difference method is convergent. In the Table we give the experimental values of the errors for such steps (h0, h1) that the stability condition for (43),(44) is not satisfied. Then the classical method is not applicable and the numerical method of bicharacteristics is convergent.
