Theory of Stochastic Processes Vol. 11 (27), no. 3–4, 2005, pp. 29–41 UDC 519.21 VOLODYMYR B. BRAYMAN ON THE EXISTENCE AND UNIQUENESS OF THE SOLUTION OF A DIFFERENTIAL EQUATION WITH INTERACTION GOVERNED BY GENERALIZED FUNCTION IN ABSTRACT WIENER SPACE We consider the following differential equation with interaction governed by a gener- alized function 0 : dx(u, t) dt = a(x(u, t), t ),x(u, 0) = u, t = 0 ◦ x(·,t) −1 . The conditions that guarantee the existence and uniqueness of a solution when mapping a belongs to some Sobolev space are obtained. 1. Introduction Let (X,H,µ) be an abstract Wiener space, i.e. X is a real separable Banach space and µ is a Gaussian measure on X with the Cameron–Martin space H (cf. [1]). Consider the evolution of a material system in X in the case where the behaviour of each particle depends not only on the position of this particle but also on some characteristic of the whole system represented by a generalized function. Examples of such characteristics are mass distributions at some surfaces, their derivatives, ets. Let W k p = W k p (X,H,µ),k ∈ N,p ≥ 1, be the Sobolev space (cf. [1], the precise definition will be given later), and let W −k q =(W k p ) ∗ , where 1 p + 1 q =1, be the space of generalized functions equipped with ∗-weak topology. Denote, by x(u,t), the position of the particle starting from u at time t. Assume that the characteristic of the material system at time t is κ t ∈ W −k q , and the evolution of the system is described by the differential equation with interaction (1) braceleftBigg dx(u,t) dt = a(x(u,t), κ t ) x(u,0) = u, κ t = κ 0 ◦ x(·,t) −1 ,t≥ 0. where a : X × W −k q → H is a measurable transformation. Here, the generalized func- tion κ t = κ 0 ◦ x(·,t) −1 is said to be the image of the generalized function κ 0 under transformation x(·,t) if, for every test function f ∈ W k p ,wehavef ◦ x(·,t) ∈ W k p and 〈f,κ t 〉 = 〈f ◦ x(·,t), κ 0 〉. Note that if κ 0 is a measure on X, then the definition of κ t coincides with the standard definition of the image of a measure. 2000 AMS Mathematics Subject Classification. Primary 60H07,28C20. Key words and phrases. Differential equation with interaction, Sobolev space, generalized function. This research has been partially supported by the Ministry of Education and Science of Ukraine, project N GP/F8/0086. 29 30 VOLODYMYR B. BRAYMAN Definition 1. Measurable mapping x : X ×R → X is said to be a solution of Eq. (1) if 1) for every t ≥ 0, the generalized function κ t = κ 0 ◦ x(·,t) −1 belongs to W −k q . 2) for µ-almost all u, x(u,t)=u + integraldisplay t 0 a(x(u,s), κ s )ds holds for all t ≥ 0; 3) for every t ≥ 0, the measure µ◦x(·,t) −1 is absolutely continuous with respect to µ; Remark. Condition 3) in Definition 1 provides that the solution does not depend on the particular choice of a modification of a. In this article, we obtain some sufficient conditions for the existence and uniqueness of solution of (1). To formulate them, we need to recall some standard constructions and notations from the Malliavin calculus (cf. [1]). For any separable Hilbert space E, we denote, by FC ∞ (X,E), a set of smooth cylindi- cal functions, i.e. functions of the form f(u)= m summationdisplay l=1 ϕ l (〈y 1 ,u〉,...,〈y n ,u〉)e l , where y 1 ,... ,y n ∈ X ∗ ,ϕ 1 ,...,ϕ m ∈ C ∞ b (R n )ande 1 ,... ,e m ∈ E. The derivative ∇ along H is defined, for f ∈FC ∞ (X,E), by ∇f(u)= m summationdisplay l=1 n summationdisplay i=1 ∂ϕ l ∂x i (〈y 1 ,u〉,...,〈y n ,u〉)j ∗ y i ⊗ e l ∈FC ∞ (X,E 1 ), where E 1 = H(H,E) is the space of Hilbert–Schmidt operators from H to E equipped with the Hilbert–Schmidt norm. Define higher order derivatives on FC ∞ (X,E)itera- tively by setting E 0 = E, ∇ 0 =1I FC ∞ (X,E) and, for k ∈ N, E k = H(H,E k−1 ), ∇ k = ∇◦∇ k−1 : FC ∞ (X,E) →FC ∞ (X,E k ). Note that E k can be identified with the space of k-linear Hilbert–Schmidt operators on H with range in E. For any k ∈ N and p ∈ [1, +∞), the operator ∇ k is closable under the norm bardblfbardbl p,k = summationtext k i=0 bardbl∇ i fbardbl L p (X,E i ,µ) . The completion of FC ∞ (X,E) under this norm is a Sobolev space W k p (X,E,µ) ⊂ L p (X,E,µ). The extensions D k : W k p (X,E,µ) → L p (X,E,µ) of derivatives ∇ k to W k p are called stochastic derivatives. By δ : D(δ) ⊂ L q (X,H,µ) → L q (X,R,µ), 1 p + 1 q =1, we denote the divergence operator, i.e. the operator adjoint to D. Denote, by bardbl·bardbl H ,the Hilbert–Schmidt norm in each of E k ,k≥ 1, and, by bardbl·bardbl op , the operator norm in L(H). Now we can formulate the results. Theorem 1. Let a : X × W −k q → H be such that 1) ∃p 0 ≥ 1 ∀κ ∈ W −k q a(·, κ) ∈ W k p 0 (X,H,µ); 2) c 0 =sup u∈X κ∈W −k q bardbla(u, κ)bardbl H < ∞, ∀1 ≤ l ≤ kc l =sup u∈X κ∈W −k q bardblD l a(u, κ)bardbl H < ∞; 3) ∀c>0 θ(c)= sup κ∈W −k q integraltext X exp(c|δa(u, κ)|)µ(du) < ∞; 4) if {κ, κ n ,n≥ 1}⊂W −k q and κ n → κ,n→∞, ∗-weakly in W −k q ,thena(u, κ n ) → a(u, κ),n→∞, in measure µ. Suppose that ∃ε>0 κ 0 ∈ W −k q+ε . ON THE EXISTENCE AND UNIQUENESS OF THE SOLUTION 31 Then Eq. (1) has a solution on [0, +∞). Theorem 2. Let a, κ 0 satisfy the conditions of Theorem 1 and, moreover, ∃L>0 ∃q 1 < q ∀u ∈ X ∀h ∈ H ∀κ 1 , κ 2 ∈ W −k q bardbla(u, κ 1 ) − a(u + h, κ 2 )bardbl H ≤ L(bardblhbardbl H + bardblκ 1 − κ 2 bardbl q 1 ,−k−1 ), ∀1 ≤ l ≤ k bardblD l a(u, κ 1 ) − D l a(u + h, κ 2 )bardbl H ≤ L(bardblhbardbl H + bardblκ 1 − κ 2 bardbl q 1 ,−k−1 ), where bardblκbardbl q,−k =sup f∈W k p bardblfbardbl p,k ≤1 |〈f,κ〉|. Then Eq. (1) has a unique solution on [0, +∞). Remark. If the transformation a in (1) depends only on the first argument, i.e. a(u,µ)= a 0 (u), then Eq. (1) turns to be an ordinary differential equation (1’) braceleftbigg dx(u,t) dt = a 0 (x(u,t)), x(u,0) = u. It is well known that Eq. (1 prime ) has a unique solution if the transformation a is Lipschitz- ian. The sufficient conditions for the existence and uniqueness of solution of (1 prime )were studied in [2–4] in the case where the transformation a belongs to some Sobolev space, instead of being Lipschitzian. In particular, it was proved in [3] that if a 0 ∈ W 1 p (X,H,µ 0 ) and exp(|δa 0 |) ∈ L c (X,H,µ 0 ), exp(bardblDa 0 bardbl op ) ∈ L c (X,H,µ 0 )forsomec>0, then Eq. (1 prime ) has a unique solution. 2. The space of generalized functions W −k q− We shall prove Theorem 1 in a slightly different form involving other spaces of general- ized functions. Note that if tildewidep>p, 1 p + 1 q =1and 1 p + 1 q =1,thenW k p ⊂ W k p ,W −k q ⊂ W −k q . Denote W k p+ = uniontext p>p W k p ,W −k q− = intersectiontext qp, the sequence κ n ,n≥ 1convergestoκ ∗-weakly in W −k p , i.e., for every tildewidep>pand for every test function f ∈ W k p ,wehave〈f,κ n 〉→〈f,κ〉,n→∞. We now can formulate the result in terms of the spaces W −k q− . Theorem 1 prime . Let a : X × W −k q− → H be such that 1) ∃p 0 ≥ 1 ∀κ ∈ W −k q− a(·, κ) ∈ W k p 0 (X,H,µ); 2) c 0 =sup u∈X κ∈W −k q− bardbla(u, κ)bardbl H < ∞, ∀1 ≤ l ≤ kc l =sup u∈X κ∈W −k q− bardblD l a(u, κ)bardbl H < ∞; 3) ∀c>0 θ(c)= sup κ∈W −k q− integraltext X exp(c|δa(u, κ)|)µ(du) < ∞; 4) if {κ, κ n ,n ≥ 1}⊂W −k q− and κ n τ → κ,n→∞, in W −k q− then a(u, κ n ) → a(u, κ),n→∞, in measure µ. Then, for every κ 0 ∈ W −k q− , Eq. (1) has a solution on [0, +∞) such that κ t ∈ W −k q− ,t≥ 0. Remark. Condition 1) in Definition 1 is caused by the fact that the transformation a is defined on X × W −k q . Hence, it is reasonable to replace it by κ t ∈ W −k q− ,t≥ 0inthe setup of Theorem 1 prime . 32 VOLODYMYR B. BRAYMAN Theorem 1 follows from Theorem 1 prime immediately since W −k q+ε− ⊂ W −k q and κ 0 ∈ W −k q+ε ⊂ W −k q+ε− imply κ t ∈ W −k q+ε− ⊂ W −k q ,t≥ 0. Theorem 1 prime shows that the solution of Eq. (1) preserves the space W −k q− , i.e. if the initial value κ 0 belongs to W −k q− , then the images κ t ,t≥ 0, remain the elements of the same space. The following example shows that the solution of Eq. (1) does not preserve the space W −k q , hence the condition κ 0 ∈ W −k q+ε in Theorem 1 cannot be replaced by κ 0 ∈ W −k q . Therefore, the spaces W −k q− are more natural when dealing with Eq. (1) than usual spaces W −k q . Example 1. Let X = R, µ(du)= 1 √ 2π e −u 2 /2 du, and let a generalized function κ 0 be defined by 〈f,κ 0 〉 = integraltext R f(u)1I {u≥0} e u 2 /2q− √ u µ(du), f ∈ W 1 p .Since integraldisplay R (1I {u≥0} e u 2 /2q− √ u ) q µ(du)= 1 √ 2π integraldisplay ∞ 0 e u 2 /2− √ u ·e −u 2 /2 du = 1 √ 2π integraldisplay ∞ 0 e −q √ u du < ∞, we get κ 0 ∈ L q (dµ) ⊂ W −1 q (dµ). Let a(u, κ) ≡ 1. Then Eq. (1) turns out to be an ordinary differential equation which has a unique solution x(u,t)=u + t. We have 〈f,κ t 〉 = integraldisplay R f(u + t)1I {u≥0} e u 2 /2q− √ u µ(du)= 1 √ 2π integraldisplay ∞ t f(v)e (v−t) 2 /2q− √ v−t · ·e −(v−t) 2 /2 dv = integraldisplay R f(v)1I {v≥t} e v 2 /2q+(q−1)vt/q−(q−1)t 2 /2− √ v−t µ(dv). Hence, κ t must be a regular generalized function, but integraldisplay R parenleftBig 1I {v≥t} e v 2 /2q+(q−1)vt/q−(q−1)t 2 /2− √ v−t parenrightBig q µ(dv)= = 1 √ 2π integraldisplay ∞ t e (q−1)vt−(q−1)t 2 /2−q √ v−t dv = ∞ and κ t /∈ W −1 q ,t > 0. On the other hand, for every tildewideq 0 ∀t ∈ [0, 1] ∀g ∈ Y esssup u∈X bardblD 2 x g (u,t)bardbl H ≤tildewidec 2 . Similar calculations prove (3) in succession for every l ≤ k. Fix any n ≥ 1. Let us verify that F(g)(t) ∈ W −k q n , i.e. F(g)(t) is a linear continuous functional on W k p n , where p 1 > ... > p n >p n+1 > ... are taken from the definition of the metric λ. We prove that, for every f ∈ W k p n , the function f ◦x g (·,t) belongs to W k p n+1 and (4) bardblf ◦ x g (·,t)bardbl p n+1 ,k ≤tildewidecbardblfbardbl p n ,k ,f∈ W k p n . To simplify notations, we denote here and thereafter, bytildewidec, any constants which depend on a,p n ,andp n+1 and do not depend on f,g,andt. Denote, by L g t , the density of the measure µ ◦ x g (·,t) −1 with respect to µ. Then, for every f ∈ L p n (X,R,µ)wehave,bytheH¨older inequality, integraldisplay X |f ◦x g (u,t)| p n+1 µ(du)= integraldisplay X |f(u)| p n+1 L g t (u)µ(du) ≤ ≤ parenleftbiggintegraldisplay X |f(u)| p n µ(du) parenrightbigg p n+1 p n · parenleftbiggintegraldisplay X (L g t (u)) p n p n −p n+1 µ(du) parenrightbigg p n −p n+1 p n . By [3, Theorem 5.12] for every c>1and0≤ t ≤ 1, we have integraldisplay X (L g t (u)) c µ(du) ≤ ⎛ ⎝ 1+ c − 1 c sup κ∈W −k q− integraldisplay X exp(c|δa(u, κ)|)µ(du) ⎞ ⎠ e 1/c , hence ∀f ∈ L p n (X,R,µ) f ◦ x g (·,t) ∈ L p n+1 (X,R,µ)andbardblf ◦ x g (·,t)bardbl L p n+1 ≤tildewidecbardblfbardbl L p n . For every f ∈ W 1 p n similarly to [4, Corollary 5.6], we have integraldisplay X bardblD(f ◦ x g (u,t))bardbl p n+1 H µ(du)= integraldisplay X bardblDf ◦ x g (u,t)Dx g (u,t)bardbl p n+1 H µ(du) ≤ ≤ integraldisplay X bardblDf ◦ x g (u,t)bardbl p n+1 H ·tildewidec p n+1 op µ(du)=tildewidec p n+1 op integraldisplay X bardblDf(u)bardbl p n+1 H L g t (u)µ(du) ≤ ≤tildewidec parenleftbiggintegraldisplay X bardblDf(u)bardbl p n H µ(du) parenrightbigg p n+1 p n , ON THE EXISTENCE AND UNIQUENESS OF THE SOLUTION 35 hence ∀f ∈ W 1 p n f ◦ x g (·,t) ∈ W 1 p n+1 and bardblf ◦ x g (·,t)bardbl p n+1 ,1 ≤tildewidecbardblfbardbl p n ,1 . Similarly for every f ∈ W k p n and for every 2 ≤ l ≤ k, integraldisplay X bardblD l (f ◦ x g (u,t))bardbl p n+1 H µ(du) ≤tildewidec l summationdisplay i=1 integraldisplay X bardblD i f ◦ x g (u,t)bardbl p n+1 H µ(du) ≤ ≤tildewidec l summationdisplay i=1 parenleftbiggintegraldisplay X bardblD i f(u)bardbl p n H µ(du) parenrightbigg p n+1 p n . Therefore, ∀f ∈ W k p n f ◦ x g (·,t) ∈ W k p n+1 and bardblf ◦ x g (·,t)bardbl p n+1 ,k ≤tildewidecbardblfbardbl p n ,k , and (4) is proved. Hence, for every f ∈ W k p n , we can define 〈f,F(g)(t)〉 = 〈f ◦x g (·,t), κ 0 〉 because of f ◦ x g (·,t) ∈ W k p n+1 and κ 0 ∈ W −k q− ⊂ W −k q n+1 . Moreover, |〈f,F(g)(t)〉| = |〈f ◦ x g (·,t), κ 0 〉| ≤ bardblκ 0 bardbl q n+1 ,−k ·bardblf ◦ x g (·,t)bardbl p n+1 ,k ≤ ≤tildewidecbardblκ 0 bardbl q n+1 ,−k bardblfbardbl p n ,k = R n bardblfbardbl p n ,k . Thus, we have F(g)(t) ∈∩ n≥1 W −k q n = W −k q− . Proposition 1 is proved. Proposition 2. For every g ∈ Y = C([0, 1],W −k q− ), the function F(g) belongs to Y and, moreover, the family of functions {F(g),g∈ Y } is equicontinuous. Proof. By Proposition 1 for every g ∈ Y , the function F(g)maps[0, 1] to W −k q− . Hence, the first assertion of Proposition 2 follows from the second one. We have to check that ∀ε>0 ∃δ>0 ∀g ∈ Y ∀t 1 ,t 2 ∈ [0, 1] |t 1 − t 2 | <δ⇒ λ(F(g)(t 1 ),F(g)(t 2 )) <ε. Since λ(F(g)(t 1 ),F(g)(t 2 )) ≤ ≤ N summationdisplay n=1 1 2 n parenleftBigg 1 ∧ N summationdisplay m=1 1 2 m (1 ∧|〈f nm ,F(g)(t 1 )〉−〈f nm ,F(g)(t 2 )〉|) parenrightBigg + 1 2 N−1 , it is sufficient to prove that ∀n,m ∈ N ∀ε>0 ∃δ = δ nm > 0 ∀g ∈ Y ∀t 1 ,t 2 ∈ [0, 1] (5) |t 1 − t 2 | <δ⇒|〈f nm ,F(g)(t 1 )〉−〈f nm ,F(g)(t 2 )〉| <ε. Fix n,m ∈ N and f = f nm . By definition of the metric λ,wehavef ∈FC ∞ (X,R) ⊂ W k p n−1 .Then ∀t ∈ [0, 1] f ◦ x g (·,t) ∈ W k p n and ∀t 1 ,t 2 ∈ [0, 1] |〈f,F(g)(t 1 )〉−〈f,F(g)(t 2 )〉| = |〈f ◦ x g (·,t 1 ) − f ◦ x g (·,t 2 ), κ 0 〉| ≤ ≤bardblf ◦ x g (·,t 1 ) − f ◦ x g (·,t 2 )bardbl p n ,k ·bardblκ 0 bardbl q n ,−k . Since f has bounded derivatives of any orders, we have integraldisplay X |f ◦ x g (·,t 1 )− f ◦ x g (·,t 2 )| p n µ(du) ≤ ≤tildewidec f integraldisplay X bardblx g (·,t 1 ) − x g (·,t 2 )bardbl p n H µ(du). 36 VOLODYMYR B. BRAYMAN Here and thereafter, we denote, by tildewidec f , any constants which depend on f but do not depend on g ∈ Y and t 1 ,t 2 ∈ [0, 1]. Also we have integraldisplay X bardblD(f ◦ x g (u,t 1 )) − D(f ◦ x g (u,t 2 ))bardbl p n H µ(du) ≤ ≤ 2 p n −1 integraldisplay X bardblDf ◦ x g (u,t 1 ) − D(f ◦ x g (u,t 2 )bardbl p n H · ·bardblDx g (u,t 1 )bardbl p n op µ(du)+2 p n −1 integraldisplay X bardblDf ◦x g (u,t 1 )bardbl p n H ·bardblDx g (u,t 1 )−Dx g (u,t 2 )bardbl p n H µ(du) ≤ ≤tildewidec f parenleftbiggintegraldisplay X bardblx g (u,t 1 ) − x g (u,t 2 )bardbl p n H µ(du)+ integraldisplay X bardblDx g (u,t 1 )− Dx g (u,t 2 )bardbl p n H µ(du) parenrightbigg and similarly ∀l ≤ k integraldisplay X bardblD l (f ◦ x g (u,t 1 )) − D l (f ◦ x g (u,t 2 ))bardbl p n H µ(du) ≤ ≤tildewidec f parenleftBigg integraldisplay X bardblx g (u,t 1 )− x g (u,t 2 )bardbl p n H µ(du)+ l summationdisplay i=1 integraldisplay X bardblD i x g (u,t 1 ) − D i x g (u,t 2 )bardbl p n H µ(du) parenrightBigg . It remains to check that (6) ∀ε>0 ∃δ>0 ∀g ∈ Y ∀t 1 ,t 2 ∈ [0, 1] |t 1 − t 2 | <δ⇒bardblx g (·,t 1 ) − x g (·,t 2 )bardbl p n ,k <ε. Let 0 ≤ t 1 0 ∀f ∈ W k p n ∀g ∈ Y ∀t ∈ [0, 1] |〈f,F(g)(t)〉| ≤ R n bardblfbardbl p n ,k . ON THE EXISTENCE AND UNIQUENESS OF THE SOLUTION 37 Set K(n)={κ ∈ W −k q n vextendsingle vextendsingle bardblκbardbl q n ,−k ≤ R n }. Then, by the Banach–Alaoglu theorem, K(n)isa∗-weak compact in W −k q n . Therefore, K(n)isacompactinW −k q n with the metric λ n . Let K = ∩ n≥1 K(n). Then (7) implies that, for every t ∈ [0, 1], we have {F(g)(t),g ∈ Y }⊂K. Moreover, K is a compact in W −k q− . Really, fix any sequence {κ m ,m≥ 1}⊂K. For every n ≥ 1, there exist a subsequence which converges in W −k q n since K ⊂ K(n)andK(n)isacompactinW −k q n . Then, by applying the diagonal method, we can find a subsequence which converges in each of W −k q n . Hence, this subsequence converges in W −k q− . Thus, K is a compact in W −k q− . The Arzela–Ascoli theorem implies that F(Y ) is relatively compact. Set Y 0 = C([0, 1],K), where K is a compact constructed in the proof of Proposition 3. Then F(Y ) ⊂ Y 0 ⊂ Y ,andY 0 is a closed convex subset of Y. It is evident that F maps Y 0 to Y 0 . Proposition 4. The transformation F is continuous on Y 0 . Proof. Let g n → g 0 ,n→∞, in Y 0 . We have to check that F(g n ) → F(g 0 ),n→∞, in Y 0 . Since F(Y 0 ) is relatively compact, there exist a subsequence {n i ,i≥ 1} such that F(g n i )convergesinY 0 as i →∞. Thus, it is sufficient to verify that if g n → g 0 and F(g n ) → tildewideg in Y 0 ,n→∞, then tildewideg = F(g 0 ). We prove that, for every f from the definition of the metric λ and for every t ∈ [0, 1], 〈f,F(g 0 )(t)〉 = 〈f,tildewideg(t)〉. This will imply tildewideg = F(g 0 ). Fix any f from the definition of the metric λ for t ∈ [0, 1]. Since 〈f,F(g n )(t)〉→〈f,tildewideg(t)〉,n→∞, it is sufficient to check that 〈f,F(g n )(t)〉→ 〈f,F(g 0 )(t)〉,n→∞, or, equivalently 〈f ◦ x g n (·,t) − f ◦ x g 0 (·,t), κ 0 〉→0,n→∞. Fix any m ∈ N. We will verify that (8) ∀κ ∈ W −k q m 〈f ◦ x g n (·,t) − f ◦ x g 0 (·,t), κ〉→0,n→∞. Since f ∈FC ∞ (X,R) ⊂ W k p m−1 ,wegetf ◦ x g n (·,t) ∈ W k p m and bardblf ◦ x g n (·,t)bardbl p m,k ≤ tildewidecbardblfbardbl p m−1,k ,n≥ 0, where tildewidec is a constant which depends on p m−1 ,p m but does not depend on n. It is sufficient to check (8) for κ from a dense subset of W −k q m , for example for regular generalized functions κ of the form (9) 〈f,κ〉 = integraldisplay X f(u)ρ(u)µ(du), where ρ ∈ L q m (X,R,µ). Fix any κ defined by (9). We have |〈f ◦ x g n (·,t) − f ◦ x g 0 (·,t), κ〉| = vextendsingle vextendsingle vextendsingle vextendsingle integraldisplay X (f ◦ x g n (·,t) − f ◦ x g 0 (·,t))ρ(u)µ(du) vextendsingle vextendsingle vextendsingle vextendsingle ≤ ≤tildewidec f integraldisplay X bardblx g n (u,t)− x g 0 (u,t)bardbl H |ρ(u)|µ(du) ≤ (10) ≤tildewidec f parenleftbiggintegraldisplay X bardblx g n (u,t)− x g 0 (u,t)bardbl p m H µ(du) parenrightbigg 1/p m parenleftbiggintegraldisplay X |ρ(u)| q m µ(du) parenrightbigg 1/q m , where tildewidec f is a constant which depends only on f. Similar to the proof of Theorem 5.21 in [3], it can be checked that integraldisplay X bardblx g n (u,t)− x g 0 (u,t)bardbl p m H µ(du) ≤ ≤ c parenleftbiggintegraldisplay X integraldisplay 1 0 bardbla(u,g n (s)) − a(u,g 0 (s))bardbl p m−1 H µ(du)ds parenrightbigg p m p m−1 , where c is a constant which depends only on c 1 and θ(c) defined in conditions 2), 3) of Theorem 1 prime . 38 VOLODYMYR B. BRAYMAN Note that g n (t) ∈ K,n ≥ 1. Hence, the sequence {g n (t),n≥ 1} is bounded in the norms W −k q m ,m≥ 1. Moreover, by the remark after the definition of the metric λ,the convergence g n (t) → g 0 (t),n→∞, in W −k q− with the metric λ implies the convergence g n (t) τ → g 0 (t),n→∞. By condition 4) of Theorem 1 prime , this implies ∀t ∈ [0, 1] a(u,g n (t)) µ −→ a(u,g 0 (t)),n→∞, and, by the Lebesgue dominated convergence theorem, integraldisplay X integraldisplay 1 0 bardbla(u,g n (s)) − a(u,g 0 (s))bardbl p m−1 H µ(du)ds → 0,n→∞. Therefore, (10) proves (8) for any regular generalized function κ ∈ W −k q m . Since regular generalized functions are dense in W −k q m ,(8)isprovedforeveryκ ∈ W −k q m . In particular, (8) holds for κ = κ 0 . Then F(g 0 )(t)=tildewideg(t),t∈ [0, 1], and the continuity of F is proved. 5. The proof of Theorem 2 Assume that Eq. (1) has solutions x(u,t)andy(u,t). Then x(u,t)=u + integraldisplay t 0 a(x(u,s), κ x s )ds, t ≥ 0, y(u,t)=u + integraldisplay t 0 a(y(u,s), κ y s )ds, t ≥ 0, where κ x s = κ 0 ◦x(·,s) −1 ∈ W −k q , κ y s = κ 0 ◦y(·,s) −1 ∈ W −k q ,s≥ 0. We will find t 0 > 0 which depends only on c 0 ,... ,c k ,Land bardblκ 0 bardbl q,−k from the conditions of Theorems 1 and 2 such that x(u,s)=y(u,s),s≤ t 0 for µ-almost all u. This implies the uniqueness of the solution. Set Δ l x(t) = esssup u∈X sup s≤t bardblD l x(u,s)− D l y(u,s)bardbl H , 0 ≤ l ≤ k, Δκ(t)=sup s≤t bardblκ x s − κ y s bardbl q 1,−k−1 , where q 1 p.Thus, Δ 0 κ(t) ≤tildewidecΔκ(t)t 1/p 2 . Let us estimate Δ 1 κ(t). By the chain rule, we have (13) integraldisplay X bardblD(f ◦ x(u,s)) − D(f ◦ y(u,s))bardbl p H µ(du) ≤ ≤ 2 p−1 integraldisplay X bardblDf ◦ x(u,s)−Df ◦ y(u,s)bardbl p H µ(du)· · parenleftbigg esssup u∈X sup s≤t bardblDx(u,s)bardbl op parenrightbigg p +2 p−1 parenleftbiggintegraldisplay X bardblDf ◦ y(u,s)bardbl p H µ(du) parenrightbigg p/p · · parenleftbiggintegraldisplay X bardblDx(u,s) − Dy(u,s)bardbl q H µ(du) parenrightbigg p/q , where 1 p + 1 q = 1 p and p0which depends only on a such that integraldisplay X (L y s (u)) q 1 µ(du) ≤tildewidec, 0 ≤ s ≤ t. Also we have bardblDx(u,t)− Dy(u,t)bardbl H ≤ integraldisplay t 0 bardblDa(x(u,s), κ x s )bardbl H ·bardblDx(u,s) − Dy(u,s)bardbl H ds+ + integraldisplay t 0 bardblDa(x(u,s), κ x s )− Da(y(u,s), κ y s )bardbl H ·bardblDy(u,s)bardbl op ds ≤ ≤tildewidec parenleftbiggintegraldisplay t 0 (bardblDx(u,s)− Dy(u,s)bardbl H + bardblx(u,s)− y(u,s)bardbl H + bardblκ x s − κ y s bardbl q 1 ,−k−1 )ds parenrightbigg for µ-almost all u, where tildewidec depends only on a and L. Similarly to the proof of Proposition 5.2.1 in [3], we get integraldisplay X bardblDf ◦ x(u,s) − Df ◦ y(u,s)bardbl p H µ(du) ≤tildewidec parenleftbiggintegraldisplay X bardblD 2 f(u)bardbl p 1 H parenrightbigg p/p 1 · · parenleftbiggintegraldisplay s 0 bardblDa(u, κ x r )− Da(u, κ y r )bardbl p 2 H µ(du)dr parenrightbigg p/p 2 ≤tildewidecΔκ(t)t p/p 2 ,s≤ t, where tildewidec is a constant which depends only on a. Hence, (13) implies Δ 1 κ(t) ≤tildewidec(Δκ(t)t 1/p 2 +(Δ 1 x(t)+Δ 0 x(t)+Δκ(t))t). Similarly, Δ l κ(t) ≤tildewidec sup s≤t sup bardblfbardbl p 1,k+1 ≤1 ⎛ ⎝ summationdisplay i≤l integraldisplay X bardblD i f ◦ x(u,s) − D i f ◦ y(u,s)bardbl p H µ(du) ⎞ ⎠ 1/p + +tildewidec summationdisplay i≤l Δ i x(t), integraldisplay X bardblD i f ◦ x(u,s) − D i f ◦ y(u,s)bardbl p H µ(du) ≤tildewidect p/p 2 Δκ(t),s≤ t and (14) Δ i x(t) ≤tildewidect ⎛ ⎝ summationdisplay j≤i Δ j x(t)+Δκ(t) ⎞ ⎠ . Hence, (15) Δ l κ(t) ≤tildewidec ⎛ ⎝ Δκ(t)t 1/p 2 + ⎛ ⎝ summationdisplay i≤l Δ i x(t)+Δκ(t))t ⎞ ⎠ ⎞ ⎠ . By (11), (12), (14), and (15), we have Δκ(t)+ summationdisplay i≤k Δ i x(t) ≤ ct 1/p 2 ⎛ ⎝ Δκ(t)+ summationdisplay i≤k Δ i x(t) ⎞ ⎠ ,t<1, where c depends only on a,bardblκ 0 bardbl q,−k ,andL. Thus, for 0