STATIONARY MEASURE-VALUED PROCESSES GENERATED

BY A FLOW OF INTERACTED PARTICLES ^{∗}

A.Yu.Pilipenko

Abstract. Let x(u, t), t ∈ ℝ, u ∈ ℝ^{d} be a stationary stochastic flow in ℝ^{d} that generated by a system of SDEs with interaction. Here x(u, t) is a position of a particle indexed by u ∈ ℝ in a moment of time t ∈ ℝ. Let ν be an absolute continuous measure in ℝ^{d}. Consider a stationary measure-valued process ν_{t}(ω) = ν ∘x(·, t)^{−1}. The conditions which ensure that ν_{t}(ω) is absolute continuous w.r.t. the Lebesgue measure are studied.

Let x(u, t) be a stationary solution of the following system of SDE’s with interaction:

(1.1)  \begin{aligned}dx(u, t) = & − α(x(u, t) − u)dt + ∫_{ℝ^{d}} a(x(u, t), x(v, t))μ(dv)dt+ \\ & + ∑_{k=1}^{m} b_{k}(x(u, t))dw_{k}(t), \end{aligned}

where u ∈ ℝ^{2d}, α > 0, a : ℝ^{d} → ℝ^{d}, b_{k} : ℝ^{d} → ℝ^{d}, μ is a probability measure on ℝ^{d}, ω_{1}(t), …, ω_{m}(t), t ∈ ℝ are independent one-dimensional Wiener processes given on a probability space (Ω, F, P).

The value of x(u, t) is identified with a position of the particle indexed by u ∈ ℝ^{d} in a moment of time t ∈ ℝ. The dynamics of random flow x(u, t) is the following. The flow attracts to identity flow x̄(u, t) ≡ u (see term −α(x(u, t) − u)), there is some interaction between particles described by function a and measure μ, each particle is randomly perturbed.

Let ν be an absolute continuous probability measure on ℝ^{d}. Consider the evolution of ν driven by the flow x(u, t). Put ν_{t}(ω) := ν ∘ (x(·, t, ω))^{−1}, ν_{t} is a random measure-valued process.

We study conditions that ensures the absolute continuity of ν_{t}(ω) for a.a.ω.

The corresponding result in a case that “close” to one-dimensional one was researched in [1]. A case of additive noise (i.e. b_{k} = const) was considered in [2]. Moreover, conditions of [2] imply that if initial measure ν has a smooth density then ν_{t}(ω) also possesses smooth density. In §3 we show that, generally, this is not true if b_{k} ≠ const.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 60K35, 60H99.

Key words and phrases. Stationary measure-valued processes, Stochastic flows, Absolute continuity.

∗ This article was supported (in part) by Ministry of Education and Science of Ukraine, project N 01.07/103.

§2. Main results

Definition 2.1. A process x(u, t) is said to be a solution of the equation (1.1) if 1) x = x(u, t, ω) is measurable in a triple (t, u, ω);

2) for all u ∈ ℝ^{d}, s, t ∈ ℝ, s ≤ t :

\begin{aligned}x(u, t) = x(u, s)+ & ∫_{s}^{t} (−α(x(u, z) − u) + ∫_{ℝ^{d}} a(x(u, z), x(v, z))μ(dv)) dz \\ & + ∑_{k=1}^{n} ∫_{s}^{t} b_{k}(x(u, z))dw_{k}(z); \end{aligned}

3) E ∫_{ℝ^{d}} ∥x(u, t)∥^{2}μ(du) < ∞, t ∈ ℝ.

Remark. Condition 3) is technical. The function a will satisfy linear growth condition. So we demand 3) to ensure the existence of integrals.

Definition 2.2. A process x(u, t), u ∈ ℝ^{d}, t ∈ ℝ is called stationary if for all n ∈ ℕ, δ ∈ ℝ, u_{1}, …, u_{n} ∈ ℝ^{d}, t_{1}, …, t_{n} ∈ ℝ the distribution of random vectors (x(u_{1}, t_{1}), …, x(u_{n}, t_{n})) and (x(u_{1}, t_{1} + δ), …, x(u_{n}, t_{n} + δ)) coincides.

The main result of the article is the following theorem.

Theorem 2.1. Assume that 1) probability measure μ has all positive moments;

2) function a is twice continuously differentiable in the first argument and bounded;

3) functions b_{k}, k = 1, …, m are twice continuously differentiable and first derivative is globally bounded;

4) a function a satisfy Lipschitz condition:

∃L ∀u_{1}, u_{2}, v_{1}, v_{2} ∈ ℝ^{d} : ∥a(u_{1}, v_{1}) − a(u_{2}, v_{2})∥ ≤ L(∥u_{1} − u_{2}∥ + ∥v_{1} − v_{2}∥)

5) for each x ∈ ℝ^{d} the rank of the matrix ∥b_{k}(x), ∇b_{k}(x)∥_{k=\overline{1,m}} is equal to d + d^{2};

6) α > 2 sup_{x,y} ∥∂_{x}a(x, y)∥ + 2 ∑_{k} sup_{x} ∥∇b_{k}(x)∥^{2}.

Then there exists the unique stationary solution of (1.1) and for each finite measure ν ≪ λ^{d}, for each t ∈ ℝ the absolute continuity

ν_{t}(ω) = ν ∘ (x(·, t, ω))^{−1} ≪ λ^{d}

holds true for a.a.ω.

Proof. The existence and uniqueness is proved in [1]. Moreover, there exists a modification of x(u, t) which is continuously differentiable in a parameter u ∈ ℝ^{d} and this derivative y(u, t) := \frac{∂x(u,t)}{∂u} is the unique stationary solution of the equation:

(2.1)  \begin{matrix}dy(u, t) = −α(y(u, t) − 1I)dt + ∫_{ℝ^{d}} a′_{1}(x(u, t), x(v, t))μ(dv)y(u, t)dt+ \\ + ∑_{k=1}^{m} ∇b_{k}(x(u, t))y(u, t)dw_{k}(t), \end{matrix}

where 1I is identity d × d matrix.

Remark. Note that if x_{1} is some modification of x, i.e.

∀u ∈ ℝ^{d} ∀t ∈ ℝ : x_{1}(u, t) = x(u, t) a.e.

then ν ∘(x_{1}(·, t))^{−1} = ν ∘(x(·, t))^{−1} a.e. So, properties of a process ν_{t} does not depend on a modification of x(u, t). Further, only continuous modification of x(u, t) is considered.

To prove the theorem we need the following well-known statement.

Proposition 2.1. Assume that ν ≪ λ^{d} is a finite measure in ℝ^{d}, f : ℝ^{d} → ℝ^{d} is differentiable mapping. If det ∥∇f(x)∥ ≠ 0 for ν-a.a. x ∈ ℝ^{d} then the image-measure ν ∘ f^{−1} is absolute continuous w.r.t. λ^{d}.

So, to prove the theorem it suffices to show that for all u ∈ ℝ^{d}, t ∈ ℝ the matrix y(u, t) is not degenerative a.s. To do this we will verify that the distribution of y(u, t) in ℝ^{d^{2}} is absolute continuous with respect to λ^{d^{2}}.

Proposition 2.2. Suppose that assumptions 1) - 5) of the Theorem 2.1 are satisfied. Let x(u, t), y(u, t) be solutions of (1.1), (2.1) , t ∈ [0, T] with initial conditions:

(2.2)  x(u, 0) = φ(u), y(u, 0) = ψ(u),

where φ, ψ are continuous differentiable and

(2.3)  ∫_{ℝ^{d}}(∥φ(u)∥^{2} + ∥ψ(u)∥^{2})μ(du) < ∞.

Assume that

(2.4)  det ∥ψ(u_{0})∥ ≠ 0.

Then for each t ∈ [0, T] the distribution of the pair (x(u_{0}, t), y(u_{0}, t)) in ℝ^{d} × ℝ^{d^{2}} = ℝ^{d+d^{2}} is absolute continuous w.r.t. λ^{d+d^{2}}.

Corollary. If conditions of the Proposition 2.2 are fulfilled. Then for each t > 0 the distribution of y(u_{0}, t) is absolute continuous w.r.t. λ^{d^{2}}. In particular, det y(u_{0}, t) ≠ 0 a.s.

Proof. Let R > 0 be fixed, τ = τ(R) be a moment of exit of a process x(u_{0}, t) from the ball with a radius R and a center in zero. Due to condition 2,3 and 5 of the Theorem 2.1 and continuity in t of trajectories of (x(u_{0}, t), y(u_{0}, t)) there exists a system of F_{t} = σ(ω_{k}(s), k = 1, …, m, s ∈ [0, t]) adapted processes α_{k}(t), 1 ≤ k ≤ m, t ∈ [0, T] such that

∑_{k=1}^{m} ∇^{j}b_{k}(x(u_{0}, t))α_{k}(t) = − ∫_{ℝ^{d}}[(∂_{1})^{j}a](x(u_{0}, t), x(v, t))μ(dv), j ∈ {0, 1}, t ∈ [0, τ(R)∧T]

and

sup_{ω} max_{k} |α_{k}(t)| < ∞, t ∈ [0, τ(R) ∧ T].

Remark. Generally, the choice of processes α_{k}(t), t = 1, …, m is not unique. By Girsanov theorem the processes

ω̄_{k}(t) = ω_{k}(t) − ∫_{0}^{t∧τ(R)} α_{k}(z)dz, t ∈ [0, T]

are independent Wiener processes w.r.t. the measure

dP̃ = exp \{∑_{k=1}^{m} (∫_{0}^{T ∧τ(R)} α_{k}(s)dw_{k}(s) − \frac{1}{2} ∫_{0}^{T ∧τ(R)} α_{k}^{2}(s)ds)\} dP.

Observe that for all t ∈ [0, τ(R)] :

x(u_{0}, t) = x̄(t), y(u_{0}, t) = ȳ(t),

where (x̄(t), ȳ(t)) is a solution of a system

\{ \begin{matrix}dx̄(t) = −α(x̄(t) − u_{0})dt + ∑_{k=1}^{m} b_{k}(x̄(t))dw̄_{k}(t) \\ dȳ(t) = −α(ȳ(t) − 1I)dt + ∑_{k=1}^{m} ∇b_{k}(x̄(t))ȳ(t)dw̄_{k}(t) \\ x̄(0) = φ(u_{0}) \\ ȳ(0) = ψ(u_{0}) \end{matrix}

It is well-known [3] that if coefficient of SDE are smooth, have linear growth and diffusion matrix is not degenerative at the initial moment of time t_{0} = 0 then the distribution of a solution in each moment t > 0 has a density.

Due to condition 5) of the Theorem 2.1 and (2.4):

d^{2} + d = rank∥b_{k}(u_{0}), ∇b_{k}(u_{0})∥_{k=\overline{1,m}} = rank∥b_{k}(u_{0}), ∇b_{k}(u_{0})ψ(u_{0})∥_{k=\overline{1,m}}.

So the distribution of (x̄(t), ȳ(t)) is absolute continuous. Consider (x(u_{0}, t), y(u_{0}, t)) as a function on a space Ω with a measure 1I_{{t≤τ(R)}}dP. Its distribution is absolute continuous also. Observe that τ(R) → ∞ as R → ∞ a.s. Therefore the distribution of (x(u_{0}, t), y(u_{0}, t)), considered in a space Ω with a measure P, has a density and the Proposition 2.2 is proved.

Let us continue the proof of the Theorem 2.1 Let x(u, t), y(u, t) be stationary solution of (1.1), (2.1). Observe that ∀t_{1}, t_{2}, t_{1} > t_{1}∀u ∈ ℝ^{d} :

P (det y(u, t_{2}) ≠ 0/ det y(u, t_{1}) ≠ 0) = 1

by the Corollary to the Proposition 2.2.

Remark. Markov property for stochastic flows, which generated by SDE with interaction, was shown in [7]

The stationarity of y(u, t) implies that the probability p(u) = P (det y(u, t) = 0) does not depend on t. By complete probability formula

\begin{matrix}∀t_{1} < t_{2} ∀u ∈ ℝ^{d} : \\ 1 − p(u) = P (det y(u, t_{2}) ≠ 0) = \end{matrix}

= P (det y(u, t_{2}) ≠ 0/ det y(u, t_{1}) ≠ 0)P (det y(u, t_{1}) ≠ 0)+
+P (det y(u, t_{2}) ≠ 0/ det y(u, t_{1}) = 0)P (det y(u, t_{1}) = 0) =

(2.5)  = 1 − p(u) + P (det y(u, t_{2}) ≠ 0/ det y(u, t_{1}) = 0)p(u)

Observe that

P (det y(u, t_{2}) ≠ 0/ det y(u, t_{1}) = 0) =

= ∫_{{(ϕ,ψ)| det ψ(u)=0)}} P (det y(u, t) ≠ 0/y(·, 0) = ψ(·), x(·, 0) = ϕ)P_{0}(dϕ, dψ),

where P_{0} is a distribution of x(·, 0), y(·, 0), t = t_{2} − t_{1}.

Remark. Of course only continuous differentiable functions ϕ, ψ are considered, that satisfy (2.3). Note that for a.a. ω functions x(·, t, ω) and y(·, t, ω) satisfy this condition.

Analogously to [1] it can be proved that for each pair (ϕ, ψ) the integrand converges to P (det y(u, t) ≠ 0) as t → ∞.

Due to (2.5), to prove p(u) = 0 and hence a Theorem 2.1, it is sufficient to show that

∀u ∈ ℝ^{d} ∀t : P (det y(u, t) ≠ 0) > 0.

Proposition 2.3. Let x(u, t), y(u, t) be solutions of (1.1), (2.1) with initial conditions (2.2), which satisfy (2.3). Then for each u ∈ ℝ^{d} there exists t_{0} > 0 and ε > 0 such that

(2.7)  ∀t ≥ t_{0} : E∥y(u, t) − 1I∥^{2} < 1 − ε.

Remark. If (2.7) holds then for stationary solution y(u, t) of (2.1) the inequality E∥y(u, t)− 1I∥^{2} ≤ 1 − ε < 1 is satisfied. Therefore ∥y(u, t) − 1I∥ < 1 with positive probability and so y(u, t) is invertible with positive probability.

Proof. Write the differential for ∥y(u, t) − 1I∥^{2}, where ∥ · ∥ is Hilbert-Schmidt norm.

d∥y(u, t)−1I∥^{2} = −2α∥y(u, t) − 1I∥^{2}dt+

+ 2 ⟨∫_{ℝ^{d}} a′_{1}(x(u, t), x(v, t))μ(dv)y(u, t), y(u, t) − 1I⟩ dt+

+ ∑_{k=1}^{m} ∥∇b_{k}(x(u, t)), y(u, t)∥^{2}dt + dM_{t},

where M_{t} is a martingale.

So

E∥y(u, t)−1I∥^{2} ≤ E∥y(u, 0) − 1I∥^{2}+

+ ∫_{0}^{t}(−2α + 2 sup_{x,y}(∥a′_{1}(x, y)∥ + ∑_{k=1}^{m} ∥b_{k}(x)∥^{2}))∥y(u, s) − 1I∥^{2}ds+

+ ∫_{0}^{t} 2 sup_{x} ∑_{k} ∥∇b_{k}(x)∥^{2} + 2 sup_{x,y} ∥a′_{1}(x, y)∥E∥y(u, s) − 1I∥ds.

Set

γ(t) := \sqrt{E∥y(u, t) − 1I∥^{2}}, a := 2 sup_{x,y} ∥a′_{1}(x, y)∥, b := 2 sup_{x} ∑_{k=1}^{m} ∥∇b_{k}(x)∥^{2}.

Then

γ^{2}(t) ≤ γ^{2}(0) + 2 ∫_{0}^{t}((−α + a + b)γ^{2}(s) + aγ(s) + b)ds.

The value of a function f(x) = (−α + a + b)x^{2} + ax + b in the point x = 1 is negative. Moreover f(x) < f(1) if x > 1. So the inequality γ^{2}(t) ≥ 1 implies f(γ(t)) ≤ f(1) < 0.

Thus there exists t_{0} ∈ (0; \frac{γ^{2}(0)}{−2f(1)}) and ε > 0 such that γ(t) < 1 − ε for all t ≥ t_{0}, i.e.

E∥y(u, t) − 1I∥^{2} < 1 − ε.

Proposition 2.3 and Theorem 2.1 are proved.

§3. Example of erratic behavior

It was shown in §2 that if x(u, t) is solution of (1.1) with initial condition x(u, 0) = φ(u), where det ∇φ(u) ≠ 0 for λ^{d}-a.a u ∈ ℝ^{d} and ν ≪ λ^{d}, then under some additional assumptions

∀t > 0 : ν_{t} ≪ λ^{d} P -a.s.

It is natural to expect that if coefficients of the equation (1.1) are C^{∞}, det ∇φ(u) ≠ 0, u ∈ ℝ^{d} and ν has C^{∞}-density then ν_{t} has also C^{∞} density. This supposition is true in one dimensional case [1] (d = 1) or in a case of additive noise [2], i.e. b_{k} = const. The purpose of this section is to present one counterexample in the case b_{k} ≠ const. Consider the analogue of (1.1) without interaction

(3.1)  \begin{aligned}dx(u, t) = & − α(x(u, t) − u)dt + a(x(u, t))dt+ \\ & + ∑_{k=1}^{m} b_{k}(x(u, t)) ∘ dw_{k}(t), t > 0 \end{aligned}

with identity initial condition

(3.2)  x(u, 0) = u, u ∈ ℝ^{d}.

All functions a, b_{k} : ℝ^{d} → ℝ^{d} are supposed to be infinitely differentiable and bounded with all their derivatives. There exists a modification [4] of x(u, t) which is infinitely differentiable in u and ∂_{u}^{(k)}x(u, t) is continuous in (u, t) for each k ≥ 0. Assume that the conditions of the Proposition 2.2 are satisfied. Then

det \frac{∂x(u, t)}{∂u} ≠ 0 λ^{d}-a.a u ∈ ℝ^{d}, a.a.ω.

So if \frac{dν}{dλ^{d}} = p ∈ C(ℝ^{d}) then

\frac{dν_{t}(u)}{dλ^{d}} = ∑_{v:x(v,t)=u} p(v) |det \frac{∂x(v, t)}{∂v}|^{−1} .

Suppose that p is a positive function. If there exists u ∈ ℝ^{d} such that det \frac{∂x(u,t)}{∂u} = 0 then \frac{dν_{t}}{dλ^{d}} cannot be even continuous. So we construct our counterexample to conjecture p ∈ C^{∞} ⟹ \frac{dν_{t}}{dλ^{d}} ∈ C^{∞} if we find some conditions to ensure

(3.3)  ∀t_{0} > 0 : P (∃u ∈ ℝ^{d} : det \frac{∂x(u, t_{0})}{∂u} = 0) > 0.

Remark. A set \{∃u ∈ ℝ^{d} : det \frac{∂x(u,t_{0})}{∂u} = 0\} is measurable because \frac{∂x(u,t_{0})}{∂u} is continuous in u.

Remark. If we have homogeneous in space equation

\begin{aligned}dx(u, t) & = a(x(u, t))dt + ∑_{k=1}^{m} b_{k}(x(u, t)) ∘ dw_{k}(t) \\ x(u, 0) & = u \end{aligned}

then [4,5] x(u, t) generates a flow of diffeomorphisms. Obviously, in this situation p ∈ C^{∞} implies \frac{dν_{t}}{dλ^{d}} ∈ C^{∞} a.s.

Proposition 3.1. Assume that a, b_{k} ∈ C_{b}^{∞}(ℝ^{d}, ℝ^{d}) and

∀x, y ∈ ℝ^{d} x ≠ y, rank∥b_{k}(x), ∇b_{k}(x), b_{k}(y), ∇b_{k}(y)∥_{k=\overline{1,m}} = 2d + 2d^{2}.

Then (3.3) holds true.

Proof. Let us verify that

(3.4)  ∀t > 0 : P (∃u_{1} ≠ u_{2} : x(u_{1}, t) = x(u_{2}, t)) > 0.

Observe that (3.4) implies (3.3). Really, assume that

x(u_{1}, t) = x(u_{2}, t), det \frac{∂x(u_{1}, t)}{∂u} det \frac{∂x(u_{2}, t)}{∂u} > 0.

Let ℝ̂^{d} ≡ ℝ^{d} ∪ {∞} be a one point compactification of ℝ^{d}. Analogously to [4] (see Lemmas 4.5.6, 4.5.8 and a reasoning on a p.161) it can be proved that for a.a. ω a mapping x̂(·, ·, ω) : ℝ̂^{d} × [0, ∞) → ℝ̂^{d} defined as following

x̂(u, t) = x(u, t), u ∈ ℝ^{d}
x̂(∞, t) = ∞, u = ∞

is continuous in (u, t). So a function u → x(u, t) is homotopic to u → x(u, 0) = u, i.e. the identity map. Thus the topological degree of a point x(u_{1}, t) = x(u_{2}, t) is equal to one. So there exists u_{3} ∈ ℝ^{d} such that x(u_{3}, t) = x(u_{1}, t) and

det \frac{∂x(u_{3}, t)}{∂u} det \frac{∂x(u_{1}, t)}{∂u} < 0.

A function det \frac{∂x(u,t)}{∂u} is continuous in u. Hence

∃u : det \frac{∂x(u, t)}{∂u} = 0

and (3.3) is a consequence of (3.4).

Let us prove (3.4).

Let φ = (φ_{1}, …, φ_{m}) ∈ C([0, ∞), ℝ^{m}).

Consider the equations:

(3.5)  \frac{∂x^{φ}(u, t)}{∂t} = −α(x^{φ}(u, t) − u) + a(x^{φ}(u, t)) + ∑_{k=1}^{m} b_{k}(x^{φ}(u, t))φ_{k}(t),

(3.6)  \begin{aligned}\frac{∂y^{φ}(u, t)}{∂t} & = −α(y^{φ}(u, t) − 1I) + ∇a(x^{φ}(u, t))y^{φ}(u, t)+ \\ & + ∑_{k=1}^{n} ∇b_{k}(x^{φ}(u, t))y^{φ}(u, t)φ_{k}(t), \end{aligned}

x^{φ}(u, 0) = u, y^{φ}(u, 0) = 1I.

Let t_{0} > 0 be fixed, u_{1}, u_{2} ∈ ℝ^{d}, u_{1} ≠ u_{2}. Find a piece wise smooth function φ such that

det y^{φ}(u_{1}, t_{0}) = det y^{φ}(u_{2}, t_{0}) = 1I,
x^{φ}(u_{1}, t_{0}) = x^{φ}(u_{2}, t_{0}).

The choice of such function can be done by the following way. At first choose a point x_{0} ∈ ℝ^{d} such that

α(x_{0} − u_{1}) + a(u_{1}) ≠ α(x_{0} − u_{2}) + a(u_{2}).

Then consider backward equation for (3.5), (3.6) with φ ≡ 0 and initial condition:

\begin{aligned}x^{φ}(u_{1}, t_{0}) & = x^{φ}(u_{2}, t_{0}) = x_{0}, \\ y^{φ}(u_{1}, t_{0}) & = y^{φ}(u_{2}, t_{0}) = 1I. \end{aligned}

Assume that (t_{0} − t_{1}) is small enough to guarantee the existence of a smooth curve

ψ(t) = (ψ^{x_{1}}(t), ψ^{y_{1}}(t), ψ^{x_{2}}(t), ψ^{y_{2}}(t)), t ∈ [0, t_{1}]

such that (i) it connects points (u_{1}, 1I, u_{2}, 1I) and (x^{φ}(u_{1}, t_{1}), y^{φ}(u_{1}, t_{1}), x^{φ}(u_{2}, t_{1}), y^{φ}(u_{2}, t_{1})), (ii) for each t ∈ [0, t_{1}] matrixes ψ^{y_{1}}(t) and ψ^{y_{2}}(t) are not degenerative, (iii) ∀t ∈ [0, t_{1}] : x^{φ}(u_{1}, t) ≠ x^{φ}(u_{2}, t), y^{φ}(u_{1}, t) ≠ y^{φ}(u_{2}, t). It is not difficult to show the existence of piece-wise smooth function φ(t), t ∈ [0, t_{1}]

such that

ψ^{x_{j}}(t) = x^{φ}(u_{j}, t), ψ^{y_{j}}(t) = y^{φ}(u_{j}, t), j ∈ {1, 2}

where x^{φ}, y^{φ} are defined by (3.4), (3.5).

By theorem on support of stochastic flows [4,6]: ∀R > 0 ∀δ > 0

(3.7)  P ( sup_{∥u∥≤R}(∥x(u, t_{0}) − x^{φ}(u, t_{0})∥ + ∥y(u, t_{0}) − y^{φ}(u, t_{0})∥) < δ) > 0.

Remark. Only homogeneous in space equations were considered in [4,6], but all reasoning can be extended to our situation.

Let R > max(∥u_{1}∥, ∥u_{2}∥). The equalities

x^{φ}(u_{1}, t_{0}) = x^{φ}(u_{2}, t_{0}), \frac{∂x^{φ}(u_{j}, t_{0})}{∂u} = y^{φ}(u_{j}, t_{0}) = 1I

and (3.7) imply:

∀ε > 0 P (∃ũ_{1}, ũ_{2}, |ũ_{j} − u_{j}| < ε, j = 1, 2 : x(ũ_{1}, t_{0}) = x(ũ_{2}, t_{0}) = x^{φ}(u_{1}, t_{0})) > 0.

The inequality (3.4) and Proposition 3.1 are proved.

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Institute of Mathematics,National Academy of Science of Ukraine, Tereshchenkivska 3, 10601, Kiev, Ukraine

E-mail address: apilip@imath.kiev.ua