on the occasion of his 70th birthday
ASYMPTOTIC BEHAVIOUR OF RANDOM HAMILTON-JACOBI EQUATIONS (R.H.-J.E.)
MIRCEA NICA
We analyze the exponential stability of the solutions satisfying R.H.-J.E. provided specific restrictions have been imposed on the defining vector field corresponding to the random system of differential equations. Other complementary results re- lated to the global existence of the characteristic system solution are obtained making use of smooth mappings. The analysis of the asymptotic behaviour of R.H.-J.E. points out the relation between the dissipativity condition of the ran- dom Hamiltonian and the boundedness of the solution of R.H.-J.E..
AMS 2000 Subject Classification: 35L40, 35L45.
Key words: random hyperbolic first order equation.
1. INTRODUCTION
Asymptotic behavior of R.H.-J.E. will be focused on revealing the signifi- cance of backward integral equations and dissipativeness property acting in the random hamiltonian
H(t, x, u, p;ω) =hp, g(t, x, u;ω)i+L(t, x, u;ω).
A global solution {uω(t, x) : t ≥ 0, x ∈ B(0,1) ⊆ Rn, ω ∈ Ω} satisfying a R.H.-J.E.
(A)
(∂tu+H(t, x, u, ∂xu;ω) = 0 :t≥0,
x∈B(0,1)⊆Rn, ω∈Ω, u(0, x) =u0(x),
is constructed using the Cauchy method of characteristic systems. Assume that {(xbω(t;λ),ubω(t;λ))∈Rn×[−1,1], t≥0, λ∈Rn, ω ∈Ω}
is the global solution associated with the corresponding random characteristic system
(B)
dbx
dt =g(t,x,b u;b ω),xbω(0;λ) =λ∈Rn, t≥0, dbu
dt =−L(t,x,b bu;ω),buω(0;λ) =u0(λ), t≥0,
MATH. REPORTS11(61),4 (2009), 335–345
where u0 ∈Cb1(Rn) is given in (A) and satisfies sup|u0(x)|=K0 <1.
The global solution of R.H.-J.E. (A) is constructed combiningbuω(t;λ)∈ [−1,1] with a smooth mapping
(C) {λ=ψω(t, x) :t≥0, x∈B(0,1)}
for each fixedω∈Ω, i.e.,uω(t, x) =ubω(t;ψω(t, x)), whereψω(t, x) = (ψω1(t, x), . . . , ψωn(t, x))∈ Rn consists of n fundamental first integrals corresponding to the random system of differential equations
dxb
dt =g(t,x,b ubω(t;λ);ω), t≥0, xbω(0;λ) =λ∈Rn. One may notice that it leads us to solve the integral equation (D) xbω(t;λ) =x∈B(0,1)⊆Rn, t≥0, ω∈Ω, with respect to unknown λ=ψω(t, x) satisfying
(E) ψiω(0, x) =xi, ∂tψωi(t, x) +h∂xψωi(t, x), g(x, uω(t, x);ω)i= 0
for anyt≥0,x∈B(0,1) and eachω∈Ω. This can be accomplished assuming a particular structure for the vector field g(t, x, u;ω) = αω(t, u)bg(x), where αω ∈ C1([−1,1]) for each ω ∈ Ω and gb ∈ Cb1(Rn;Rn) is either a bounded smooth vector field or bg ∈Cbb1(Rn;Rn) agrees with a linear growth condition
|bg(x)| ≤ C2|x|, x ∈ Rn. The main results given in the third section of this paper concern the exponential stability of the solution satisfying R.H.-J.E. (A) when the vector field g(t, x, u;ω) = αω(t, u)bg(x), u∈[−1,1], x ∈Rn, ω ∈Ω, t≥0, the above given conditions.
Some auxiliary results regarding the global existence of the characteristic system solution and construction of the smooth mapping{λ=ψω(t, x) :t≥0, x∈B(0,1)} are given in Section 2.
2. STATEMENT OF THE PROBLEM AND SOME AUXILIARY RESULTS
We are given an increasing sequence{tj}j≥0,t0 = 0, and a piecewise con- stant bounded process y(t, ω) : [0,∞) × Ω → R, y(t, ω) = y(tj, ω), t ∈ [tj, tj+1), j ≥ 0, such that each yj(ω) : (Ω,Fj) → (R,B) is a random variable, where Fj is a σ-algebra of subsets of Ω, and B stands for the Borel σ-algebra in R. Consider the random (H-J) equation
(1)
(∂tu+H(y(t, ω), x, u, ∂xu) = 0, t≥0, x∈B(0,1)⊆Rn, ω∈Ω, u(0, x) =u0(x), u0 ∈Cb1(Rn),
where H(y, x, u, p) = hp, g(x, u, y)i+L(x, u, y), g ∈ Cbb1(Rn+2;Rn) and L ∈ Cbb1(Rn+2).
Denoted by Cbb1 the space of all continuously differentiable functions f(x, u, y) : Rn+2 → Rk, 1 ≤ k ≤ n such that ∂if(x, u, y) = ∂xif(x, u, y) and ∂uf(x, u, y), i∈ {1, . . . , n}, are bounded functions.
Consider buω(t;λ) = u(t,x(t;b λ)) and bxω(t;λ), t ≥ 0, λ ∈ Rn, ω ∈ Ω, as the solution satisfying the corresponding characteristic system associated with (1), i.e.,
(2)
dbx
dt =g(x,b bu, y(t, ω)),xbω(0;λ) =λ∈Rn, t≥0, dbu
dt =−L(bx,u;b y(t, ω)),buω(0;λ) =u0(λ), t≥0.
One may notice that forz= (x,b bu)∈Rn+1 and Z(z, y) =
g(bx,bu, y)
−L(bx,u;b y)
∈ Rn+1 as the lipschitzian field with respect to z ∈ Rn+1, we get a unique global solution
(3) {zω(t, λ) = (bxω(t, λ),buω(t, λ)∈Rn+1 :t≥0, λ∈Rn} satisfying the characteristic system (2).
Assume thatLand u0 fulfil the conditions (4)
(a)L∈Cbb1(Rn+2), L(x,0, y) = 0, ∂uL(x, u, y)≥0, (x, y)∈Rn+1, u∈[−1,1];
(b)u0 ∈Cb1(Rn),sup|u0(x)|=K0 <1.
Lemma1. Assume bg∈Cbb1(Rn+2;Rn), L∈Cbb1(Rn+2) andu0 ∈Cb1(Rn) fulfil conditions (4). Then the unique global solution {(bxω(t, λ), ubω(t, λ)) : t≥0, λ∈Rn} of the characteristic system(2) satisfies
(5) |ubω(t, λ)| ≤ |u0(λ)| ≤K0 <1, t≥0, λ∈Rn, ω∈Ω.
Proof. A direct computation shows that
(6)
dub
dtω(t;λ) =−L(bxω(t;λ),ubω(t;λ), y(t;ω)) =
− Z 1
0
∂uL(xbω(t;λ), θubω(t;λ), y(t;ω))dθ
buω(t;λ)
=−aω(t;λ)buω(t;λ), t≥0, whereaω(t;λ) =R1
0 ∂uL(bxω(t;λ), θubω(t;λ), y(t;ω))dθ≥0 if ubω(t;λ)∈[−1,1].
Hence |buω(t;λ)| ≤1 and
(7) |buω(t;λ)|=|[exp−aω(s;λ)ds]u0(λ)| ≤ |u0(λ)| ≤K0 <1 for any t≥0,λ∈Rn,ω ∈Ω.
Remark1. When we are interested to obtain a solution for which|∂λubω(t;
λ)| ≤ |∂λu0(λ)|,t≥0,λ∈Rn,is bounded we need to assume (seeL(x, u, y)→ L(u, y)) that
(8) (
(a)L∈C1(R2), L(0, y) = 0, ∂uL(u, y)≥0, u∈[−1,1], y∈R;
(b)u0 ∈Cb1(Rn),sup|u0(x)|=K0 <1.
Lemma 2. Consider
H(y, x, u, p) =hp, g(x, u, y)i+L(u, y),
where g∈Cbb1(Rn+2;Rn). Assume thatLandu0 satisfy(8)and let{buω(t, λ)|: t≥0, λ∈Rn, ω ∈Ω} be the global solution fulfilling (2). Then
(9)
(|ubω(t;λ)| ≤ |u0(λ)| ≤K0 <1, t≥0, λ∈Rn, ω ∈Ω,
|∂λubω(t;λ)| ≤ |∂λu0(λ)| ≤K1, t≥0, λ∈Rn, ω ∈Ω, where K1= sup|∂xu0(x)|.
Proof. By hypothesis, Lemma 1 holds, i.e.,
(10) |ubω(t, λ)| ≤ |u0(λ)| ≤K0 <1, t≥0, λ∈Rn, ω∈Ω.
Denote {vω(t;λ) = ∂λbuω(t, λ) : t ≥ 0, λ ∈ Rn, ω ∈ Ω}. From (2) we get a linear system
(11)
dv
dt =−∂uL(buω(t, λ), y(t, ω))v, t≥0, vω(0;λ) =∂λu0(λ), λ∈Rn,
satisfied by v=vω(t;λ),t≥0,λ∈Rn.Using (8(a)), from (11) we obtain (12) |vω(t, λ)| ≤ |∂λu0(λ)| ≤K1, t≥0, λ∈Rn, ω∈Ω,
and the proof is complete.
Remark 2. Notice that under some weak dissipativity condition on L assumed in Lemma 2 we get a solution of the characteristic system (2) for which both components buω(t;λ) and ∂λubω(t;λ),t≥0, λ∈Rn, are bounded.
Unfortunately, the same weak condition is not sufficient for getting the smooth mapping {λ=ψλ(t, x)∈Rn:t≥0, x∈B(0,1), ω ∈Ω}that leads us to the bounded solution
(13) uω(t, x) =buω(t;ψω(t, x)), t≥0, x∈B(0,1), ω∈Ω, of R.(H.-J.)E. (1).
Next, we shall present two types of conditions which allow us to construct the mentioned smooth mapping λ=ψω(t, x).
The first type includes (14)
((a)L∈C1(R2), L(0, y) = 0,0< γ≤ ∂uL(u, y), u∈[−1,1], y∈R, (b)g(x, u, y) =α(u, y)g(x), αb ∈C1(R2),bg∈Cb1(Rn;Rn), α(0, y) = 0, where bg∈Cb1 is a bounded vector field.
Denote
c1= max{|∂uα(u, y)|:u∈[−1,1],|y| ≤b}
and
c2= sup{|bg(x)|:x∈Rn},
where the given bounded process |y(t, ω)| ≤ b is valued in the compact set [−b, b]. Assume that u0 ∈Cb1(Rn) satisfies
(15)
((a)u0 ∈Cb1(Rn),sup|u0(x)|=K0 <1,sup|∂xu0(x)|=K1;
(b) 1γC1C2K1=ρ, whereρ∈[0,1) is fixed and γ >0 is given in (14(a)).
Lemma 3. Assume that
H(y, x, u, p) =hp, g(x, u, y)i+L(u, y)
satisfies(14)and takeu0∈Cb1(Rn)such that(15) is fulfilled. Then there exist a global solution {(xbω(t;λ),ubω(t;λ)) : t ≥ 0, λ ∈ Rn, ω ∈ Ω} of (2) and a smooth mapping {λ=ψω(t, x) :t≥0, x∈B(0,1)} satisfying
xbω(t;ψω(t, x)) =x∈B(0,1)⊆Rn, t≥0, ω ∈Ω, (16)
|ubω(t;λ)| ≤(exp−γt)|u0(λ)| ≤K0 <1, (17)
|∂λubω(t;λ)| ≤(exp−γt)|∂λu0(λ)| ≤K1, (∀)t≥0.
Proof. The conclusions of Lemma 2 hold. Consider the global solution {(xbω(t;λ),ubω(t;λ)) :t≥0, λ∈Rn}of the characteristic system (2) satisfying (18) |ubω(t;λ)| ≤ |u0(λ)| ≤K0 <1, t≥0, λ∈Rn, ω∈Ω,
and
(19) |∂λubω(t;λ)| ≤ |∂λu0(λ)| ≤K1, t≥0, λ∈Rn, ω∈Ω.
This time, the strong dissipativity condition (14(a)) allow us to replace (18) and (19) by the exponential estimates
(20) |buω(t;λ)| ≤(exp−γt)|u0(λ)| ≤K0(exp−γt), t≥0, λ∈Rn, and
(21) |∂λbuω(t;λ)| ≤(exp−γt)|∂λu0(λ)| ≤K1(exp−γt), t≥0, λ∈Rn, where the constant γ >0 is given in (14(a)).
Both conclusions (20) and (21) are directly obtained from the integral representation
buω(t;λ) =
exp− Z t
0
aω(s;λ)ds
u0(λ), vω(t;λ) =∂λbuω(t;λ) =
exp−
Z t 0
bω(s;λ)ds
∂λu0(λ), where
aω(t;λ) = Z 1
0
∂uL(θbuω(t;λ), y(t;ω))dθ≥0 and
bω(t;λ) =∂uL(ubω(t;λ), y(t;ω))≥0, (∀)t≥0, λ∈Rn.
From now on we shall use an adequate integral representation of the solution {(xbω(t, λ) :t≥0},namely,
(22) bxω(t;λ) =G(τb ω(t;λ))[λ] :t≥0, λ∈Rn,
where {G(σ)[λ] :b σ ∈ R, λ ∈ Rn} is the global flow generated by the field gb∈Cbb1(Rn;Rn) and
τω(t;λ) = Z t
0
α(ubω(s;λ), y(s;ω))ds= (23)
= Z t
0
ubω(s;λ) Z 1
0
∂uα(θubω(s;λ), y(s;ω))dθ
ds, t≥0.
Finding λ = ψω(t, x) such that bxω(t, λ) = x ∈ B(0,1) is equivalent to solving the integral equation
(24) λ=G(−τb ω(t;λ))[x] =Vω(t, x;λ) :t≥0, x∈B(0,1), λ∈Rn. Denote ργ = γ1C1C2,whereC1 andC2 are given in (15). We shall prove that the integral operator {Vω(t, x;λ) :t≥0, x∈B(0,1), λ∈Rn}satisfies
(25) Vω(t, x;λ)∈B(x, ργ)⊆Rn, for any t≥0, λ∈Rn, and is a contractive mapping with respect to λ∈Rn,i.e.,
(26) |Vω(t, x;λ00)−Vω(t, x;λ0)≤ρ|λ00−λ0|, t≥0, λ0, λ00 ∈Rn, x∈B(0,1), where ρ∈(0,1) is given in (15). A direct computation yields
(27) |G(−τb )[x]−x|=
Z −τ 0
bg(G(s)[x])dsb
≤C2|τ|, τ ∈R.
In particular, for τ =τω(t;λ) we get
|Vω(t, x;λ)−x|=|G(−τb ω(t;λ))[x]−x| ≤ |C2τω(t;λ)|
(28)
≤C2 Z t
0
Z 1 0
∂uα(θbuω(s;λ))dθ
|ubω(s;λ)|
ds≤ 1
γC1C2K0 ≤ργ for any t ≥ 0, λ ∈ Rn. On the other hand, we can compute ∂λτω(t;λ) from (23), i.e.,
(29) ∂λτω(t;λ) = Z t
0
∂uα(buω(s;λ), y(s;ω))∂λbuω(s;λ)ds and using (21) we get
(30) ∂λτω(t;λ)≤C1K1
Z t 0
(exp−γs)ds≤ 1 γC1K1
for any t≥0,λ∈Rn,ω ∈Ω.
Further, notice that
∂λVω(t, x;λ) = d
dτ n
G(−τb ω(t;λ))[x]
o
∂λτbω(t;λ) = (31)
=−bg(Vω(t, x;λ))∂λτω(t;λ).
Using (30) we get (32) |∂λVω(t, x;λ)| ≤ 1
γC1C2K1=ρ∈(0,1), (∀)t≥0, x∈B(0,1), λ∈Rn. This shows that Vω(t, x;·) :B(x, ργ) → B(x, ργ) is a contraction. Using the contractive mapping fixed point theorem (Banach principle) we get the smooth mapping λ=ψω(t;x) with
(33) ψω(t;x) =Vω(t, x;ψω(t;x)), t≥0, x∈B(0,1), that solves the corresponding integral equation
In addition, the corresponding derivatives can be computed as (34) ∂tψω(t;x) = [In−∂λVω(t, x;ψω(t;x))]−1[∂tV(t, x;λ)] (λ=ψω(t;x)) and
(35) ∂iψω(t;x) = [In−∂λVω(t, x;ψω(t;x))]−1[∂iV(t, x;λ)] (λ=ψω(t;x)) for any i∈ {1, . . . , n}.The proof is complete.
Remark 3. Hypotheses (14) and (15) of Lemma 3 only involve bounded vector fields bg ∈Cb1(Rn;Rn) andg(x, u, y) =α(u, y)bg(x), x ∈Rn,u, y ∈ R.
A relaxation of them allows vector fields (36) g(x, u, y) =
l
X
k=1
αk(u, y)gbk(x), αk∈C1(R2), bgk∈Cbb1(Rn;Rn), 1≤k≤l,
where each αk, bgk and L(u, y) must verify the conditions below. Consider
|y(t, ω)| ≤b,t≥0, ω∈Ω and
(37)
(a)L∈C1(R2), L(0, y) = 0,0< γ≤∂uL(u, y), u∈[−1,1],|y| ≤b;
(b)bgk∈Cbb1(Rn;Rn),bgk(0) = 0 andhAk(x)z, zi ≥0, z∈Rn,
x∈B(0,1), whereAk(x) = [∂xbgk(x)] + [∂xbgk(x)]t, k∈ {1, . . . , l};
(c) [gbi,bgj](x) = 0, x∈Rn, i, j ∈ {1, . . . , l}, where [·,·] stands for the Lie bracket.
In addition, takeu0∈ Cb1(Rn) such that (38)
((a)u0∈Cb1(Rn),sup|u0(x)|=K0 <1,sup|∂xu0(x)|=K1; (b)γ1C1C2K1=ρ∈(0,1), whereγ >0 is given in (37(a)) and
C1 = max{|∂uα(u, y)|:u∈[−1,1], y∈[−b, b]}, C2= max
1≤k≤l[max{|bgk(x)|:x∈B(0,1)}].
Lemma 4. Consider H(y, x, u, p) =hp, g(x, u, y)i+L(u, y), where g and Lfulfil (36)and(37). Takeu0 ∈Cb1(Rn) such that(38)is satisfied. Then there exist a global solution {(xbω(t, λ),ubω(t, λ)) :t≥0, λ∈Rn, ω ∈Ω} of (2)and a smooth mapping {λ=ψω(t;x) :t≥0, x∈B(0,1)} satisfying
(39) xbω(t;ψω(t;x)) =x∈B(0,1)⊂Rn, t≥0, ω ∈Ω, and
(40)
(|buω(t, λ)| ≤(exp−γt)|u0(λ)| ≤K0 <1, t≥0, λ∈Rn,
|∂λubω(t;λ)| ≤(exp−γt)|∂λu0(λ)| ≤K1, t≥0, λ∈Rn.
Proof. (Hint) To prove inequalities (40) we use the same arguments as in Lemma 3, based on condition (37(a)). To construct a smooth mapping {λ=ψω(t;x) :t≥0, x∈B(0,1)} such that the equations
(41) xbω(t;λ) =x∈B(0,1).
are solved for any t ≥0, we notice that the global solution {xbω(t, λ) : t≥0, λ∈Rn}can be represented as
(42) xbω(t;λ) =Gb1(τ1ω(t;λ))◦. . .◦Gbl(τlω(t;λ))[λ], t≥0, λ∈Rn, where {Gbk(tk)[λ] : t ∈ R, λ ∈ Rn} is the global flow generated by bg ∈ Cbb1(Rn;Rn), k∈ {1, . . . , l}, and
(43) τkω(t;λ) = Z t
0
αk(ubω(s;λ), y(s;ω))ds, k∈ {1, . . . , l}, t≥0, λ∈Rn.
Denote τω(t;λ) = (τ1ω(t;λ), . . . , τlω(t;λ)) and consider
(44) G(p)[x] =b Gb1(t1)◦. . .◦Gbl(tl)[x], p= (t1, . . . tl)∈Rl, x∈Rn. Then the equations
(45) x=xbω(t;λ) =G(τb ω(t;λ))[λ], t≥0, x∈B(0,1), λ∈B(0,1), are equivalent to
(46) λ=G(−τb ω(t;λ))[x], t≥0, x, λ∈B(0,1).
Here, the commuting hypothesis (37(c)) was used. RewriteG(−τb ω(t;λ))[x] as (47) Vω(t, x;λ) =G(−τb ω(t;λ))[x] =U(τω(t;λ);x).
By a direct computation (here, (37(b)) is used) we obtain (48) |U(p;x)| ≤ |x|, t≥0, x∈B(0,1), λ∈Rn, and, in particular, for p=τω(t;λ) from (48) we get
(49) |Vω(t, x;λ)|=|U(τω(t;λ);x)| ≤ |x|, t≥0, x∈B(0,1), λ∈Rn. In addition, the matrixMω(t, x;λ) =∂λVω(t, x;λ) satisfies
(50) |Mω(t, x;λ)| ≤ρ∈(0,1), t≥0, x∈B(0,1), λ∈Rn, provided we notice that this time
(51) Mω(t, x;λ) =−
" l X
k=1
bgk(Vω(t, x;λ))∂λτkω(t;λ)
# . A direct consequence of (51) is
|Mω(t, x;λ)| ≤C2
" l X
k=1
|∂λτkω(t;λ)|
# (52)
≤ 1
γC1C2|∂λu0(λ)| ≤ 1
γC1C2K1 ≤ρ∈(0,1).
Now, the argument used in Lemma 3 allows us to define {λ=ψω(t, x) :t≥0, x∈B(0,1)} such that
(53) ψω(t, x) =Vω(t, x;ψω(t, x)), t≥0, x∈B(0,1), and the proof is complete.
3. MAIN RESULTS
We start with
Theorem 1. Consider H(y, x, u, p) = hp, g(x, u, y)i+L(u, y), where g and L fulfil conditions (14). Take u0 ∈ Cb1(Rn) such that (15) is satisfied.
Consider the global solution {(bxω(t, λ), ubω(t, λ)) : t ≥ 0, λ ∈ Rn} of (2) and the smooth mapping {λ = ψω(t;x) : t ≥ 0, x ∈ B(0,1)} satisfying (16) and (17). Then uω(t, x) = ubω(t;ψω(t, x)), t ≥ 0, x ∈ B(0,1) ⊆ Rn, is an exponentially stable solution of the R.H.-J.E. (1)fulfilling
(54) |uω(t, x)| ≤K0(exp−γt), t≥0.
Proof. (Hint) By hypothesis, the conclusions of Lemma 3 hold. Define a smooth mapping as uω(t, x) = buω(t;ψω(t, x)), t ≥ 0, x ∈ B(0,1) for each ω ∈Ω,whereubω(t;λ) and ψω(t, x) fulfil (16) and (17).
By definition, uω(0, x) =ubω(0, x) =u0(x), x ∈B(0,1). Using (34) and (35), we get that {uω(t, x) :t≥0,x∈B(0,1)}satisfies R.H.-J.E. (1).
Theorem 2. Consider H(y, x, u, p) = hp, g(x, u, y)i+L(u, y), where g and L fulfil conditions (36) and (37). Take u0 ∈ Cb1(Rn) such that (38) is verified. Consider the global solution {(bxω(t, λ),ubω(t, λ)) :t≥0, λ∈Rn} of (2) and the smooth mapping{λ=ψω(t;x) :t≥0, x∈B(0,1)} satisfying(39) and (40).
Thenuω(t, x) =ubω(t;ψω(t, x)),t≥0, x∈B(0,1)⊆Rn, is an exponen- tially stable solution of the R.H.-J.E. (1) satisfying
(55) |uω(t, x)| ≤K0(exp−γt), t≥0, x∈B(0,1), ω∈Ω, and
(56) |∂xuω(t, x)| ≤ K1
1−ρ(exp−γt), t≥0, x∈B(0,1), ω∈Ω.
Proof. (Hint) By hypothesis, the conclusions of Lemma 4 hold. Define (57) uω(t, x) =ubω(t;ψω(t, x)), t≥0, x∈B(0,1), ω∈Ω,
where ubω(t;λ) and ψω(t, x) fulfil (39) and (40). In addition, the differential equalities (34) and (35) also hold and by a direct inspection we obtain that {uω(t, x) : t ≥ 0, x ∈ B(0,1)} is a solution of R.H.-J.E. (1) satisfying (55) and (56).
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[1] V. Dr˘agan and A. Halanay, Stability of Linear Systems. Ed. ALL, Bucharest, 1994.
(Romanian)
[2] C. Vˆarsan,Applications of Lie Algebras to Hyperbolic and Stochastic Differential Equa- tions. Kluwer, Dordrecht, 1999.
Received 7 March 2009 Romanian Academy
“Simion Stoilow” Institute of Mathematics P.O. Box 1-764, Bucharest, Romania
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