ON SPDES WITH VARIABLE COEFFICIENTS IN ONE SPACE DIMENSION

Transcription

1 ON SPDES WITH VARIABLE COEFFICIENTS IN ONE SPACE DIMENSION KYEONG-HUN KIM AND N.V. KRYLOV Abstract. Stochastic second-order artial differential equations of evolution tye are considered in one sace dimension. Existence and uniqueness results are given in weighted Sobolev saces of functions given on an interval in the sace variable. 1. Introduction This article is a natural continuation of the series of articles [6], [9], [10], and [11]. We are dealing with an L -theory of Itô stochastic artial differential equations of tye du = (au xx + bu x + cu + f) dt + (σu x + νu + g) dw t (1.1) considered for x I = (0, 1), t 0, where w t is a Wiener rocess and the coefficients a, b, c, σ, ν and the free terms f, g are random functions deending on (t, x). Our aroach is based on Sobolev saces with weights and our goal is to rove existence and uniqueness theorems in Sobolev classes with fractional ositive or negative number of derivatives summable to the ower 2 with weights allowing the derivatives to blow u near the endoints of I. The motivation for such setting is discussed at length in [6] and [9]. For quite a number of other issues related to stochastic artial differential equations we refer the reader to the recent collection of articles [1] containing a very extensive literature. In [6], [9], [10] equations of tye (1.1) are considered either in the whole sace with variable coefficients or in half saces with constant coefficients. Quite often (although not always) in the theory of artial differential equations, once we know how to solve equations with constant coefficients in the whole sace and in half saces, then constructing a solvability theory even for nonlinear equations with variable 1991 Mathematics Subject Classification. 60H15,35R60. Key words and hrases. Stochastic artial differential equation, Sobolev saces with weights. The work of the second author was artially suorted by NSF grant DMS

2 2 KYEONG-HUN KIM AND N.V. KRYLOV coefficients becomes a standard and rather unrewarding task esecially if one is satisfied with quasi-linear equations and somewhat sloy assumtions on smoothness of the coefficients. The case of L -theory of SPDEs turns out to be one more excetion from such usual situation if we want to only imose almost necessary conditions. Actually, in [2] we already saw this even for = 2 and nonnegative integral number of derivatives. A version of L -solvability theory ( 2) for equations with variable coefficients in smooth domains in R d, d 1, is resented in [11]. However, while reading somewhat sketchy roofs in [11] we could not reconstruct the argument based on renormalization of saces. The trouble we had is that the renormalization may change dramatically constants in estimates for equations with constant coefficients. As we lan to show, the solvability results of [11] can be obtained differently at least for one dimensional equations. Just in case, if the arguments of [11] admit a comlete exlanation, we show that the continuity requirement of [11] on a and σ in sace variables can be considerably relaxed. A similar fact should be true for multidimensional equations as well. However, in the resent article we only deal with one sace dimension. It turns out that even in this case, in which at least art of difficulties does not aear, achieving our goal requires quite a bit of work. By the way, our results are new even for urely deterministic equations when there is no stochastic art in (1.1). In that case we are able to rove the solvability of equations with coefficients which are retty wild near the endoints of I, allowing, say a to behave near zero like 2 + sin( lnx α ), α (0, 1) (see Remark 2.2). The article is organized as follows. In Sec. 2 we introduce necessary notation and resent our main results when b = c = ν = 0 for equations on I (Theorem 2.14) and R + = (0, ) (Theorem 2.16). The roof of Theorem 2.16 takes a substantial art of the article and is given in Sec. 4 after we reare some auxiliary tools in Sec. 3. Theorem 2.14 is roved in Sec. 5 and in the final Sec. 6 we treat equation (1.1) with all terms. As far as we understand our Theorem 6.2 does not follow from the results of [11] even if the main coefficients are constant. 2. Main results Let (Ω, F, P) be a comlete robability sace, {F t, t 0} be an increasing filtration of σ-fields F t F, each of which contains all (F, P)-null sets. By P we denote the redictable σ-field generated by

3 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 3 {F t, t 0}. We assume that on Ω we are given indeendent onedimensional Wiener rocesses wt 1, w2 t,..., each of which is a Wiener rocess relative to {F t, t 0}. We are going to consider the following equation containing only higher order terms du(t, x) = (a(t, x)u xx (t, x) + f(t, x)) dt + (σ k (t, x)u x (t, x) + g k (t, x)) dwt k, (2.1) k=1 where the functions a, f, σ, g also deend on ω Ω. An information on equations with lower order terms can be obtained from Sec. 6. The functions entering the equation are suosed to be real valued. Furthermore, we suose that σ = (σ 1, σ 2,...) is l 2 -valued. To state further assumtions on a and σ we need the following saces of functions on R. Let B 0 = B be the set of bounded functions. For ν = k + α, where k = 0, 1, 2,... and α (0, 1], let B ν = C k,α be the usual Hölder sace of bounded continuous functions whose kth derivative is α-hölder continuous. We do not secify the range of functions in B ν, ν 0, and we use the same notation for real-valued functions such as a and l 2 -valued functions such as σ. Fix a function δ 0 (ν) 0 defined on [0, ) such that δ 0 (ν) > 0 if ν {0, 1, 2,...}. For ν 0 define ν+ = ν + δ 0 (ν). (2.2) Also, take and fix a C 0 (R)-function η 0 such that η 0 = 1 near x = 0, η 0 = 0 near x = 1 and η 0 (x) + η 0 (1 x) = 1 for x I = (0, 1). Finally, we take and fix some constants K R + = (0, ), R, [2, ) and introduce the following assumtion, in which we set a 0 = η 0 a, σ 0 = η 0 σ. and, naturally, the notation f(e ) is used for the function taking value f(e x ) at oint x. Assumtion 2.1 (). (i) For each x > 0, the functions a 0 (t, x) and σ0 k (t, x) are redictable functions of (ω, t). (ii) For each t 0 and ω Ω a 0 (t, e ) B + + σ 0 (t, e ) B +1 + K. (2.3)

4 4 KYEONG-HUN KIM AND N.V. KRYLOV (iii) The function a 0 (t, e x ) and the l 2 -valued function σ 0 (t, e x ) are uniformly continuous in x R uniformly with resect to (ω, t). In other words, there exists a function κ(ε), ε > 0, such that κ(ε) 0 as ε 0 and a 0 (t, e x ) a 0 (t, e y ) + σ 0 (t, e x ) σ 0 (t, e y ) κ( x y ). Remark 2.2. Observe that, if 0, then (2.3) imlies that a 0 (t, e x ) is uniformly continuous in x, but if = 1 ( 0), the uniform continuity of σ 0 (t, e x ) does not follow from (2.3). Similar situation occurs if 1 and if = 0. Remark 2.3. In [11] the uniform continuity of a(t, x) and σ(t, x) is assumed instead of (iii). To see that this is stronger than (iii) it suffices to consider the function a = 2 + sin ln x. Actually, this function is excluded by condition (2.10) of Theorem 2.14, but a = 2 + sin( lnx α ) satisfies its regularity assumtions for any α (0, 1) and R. Assumtion 2.4 (). Assumtion 2.1() is satisfied if we relace in it a(t, x) and σ(t, x) with a(t, 1 x) and σ(t, 1 x) resectively. To describe the assumtions on f and g k we need to introduce aroriate Banach saces. For θ R define an oerator Q : f(x) f(e x )e xθ/. The following definition, in which needs only be in (1, ), is taken from [9]. Definition 2.5. We write f H ν if and only if Q f = h H ν, where H ν = H ν (R) = (1 ) ν/2 L. We write L = H 0. For f H ν we define f H ν = Q f H ν. (2.4) This definition is also used for l 2 -valued functions g, in which case, of course, Q g H ν = (1 ) ν/2 Q g L, where (1 ) ν/2 Q g = (1 ) ν/2 Q g l2. Remark 2.6. According to [9], for ν = 0, 1, 2,..., θ 0, and u H ν, we have ν u H N x θ 1+i u (i) (x) dx, ν ν i=0 i=0 where N deends only on ν,, and θ. 0 0 x θ 1+i u (i) (x) dx N u H ν,

5 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 5 As in [9], by M α we denote the oerator of multilying by x α, M = M 1. Remark 2.7. As in the case of saces H, ν the following interolation inequality holds: u H 1 N u 1/2 u 1/2 H 2 L. We will be also using the following facts about the saces H ν. They all are roved in [9]. For D = / x, it turns out that the oerator MD is a bounded oerator from H ν to Hν 1 for any ν. Furthermore, if θ 0 this oerator is onto and has a bounded inverse. The same is true for DM if θ. In articular, if θ 0, we have for any ν R and u MH +2, Mu xx H + u x H +1 We also take from [9] the saces N M 1 u H +2. (2.5) H ν (τ) = L ( (0, τ]], P, H ν ), Uν = M1 2/ L (Ω, F 0, H ν 2/ ), and the saces H ν (τ), where τ is any F t-stoing time. Without going into all details the latter sace is described as the set of all functions u MH ν (τ) such that u(0, ) Uν and for an f M 1 H ν 2 (τ) and an l 2 -valued g H ν 1 (τ) we have du(t) = f(t) dt + g k (t) dw k t t τ. in the sense of distributions, where in the second term on the right, as usual, the summation over all ossible values of k is assumed. The norm in H ν (τ) is introduced by u H ν (τ) = M 1 u H ν (τ) + Mf H ν 2 (τ) + g H ν 1 (τ) + u(0, ) U ν. Actually, in [9] the norm in H ν (τ) is introduced by a different formula, but by Remark 1.7 of [9], if θ, the new norm is equivalent to the old one. Starting from this oint we always assume that 0 < θ <. (2.6) Later on we restrict the range of θ even further. We use the notation H ν,0 (τ) for the subset of Hν (τ) consisting of all functions u(t) satisfying u(0) = 0. Choose some functions η 1, η 3 C0 (R), η 2 C0 ((0, 1)) such that η 1 = 0 near x = 1, η 3 = 0 near 0, and η 1 + η 2 + η 3 1 on I. Definition 2.8. We write f H ν (I) if and only if f 1, f 3 H ν and f 2 H ν where f 1 = fη 1, f 3 (x) = (fη 3 )(1 x), f 2 = fη 2. We define f H ν (I) = f 1 H ν + f 3 H ν + f 2 H ν. (2.7)

6 6 KYEONG-HUN KIM AND N.V. KRYLOV In Remark 3.5 we show that H ν (I) is indeendent of η i and the norms (2.7) constructed from different η i are equivalent. Similarly we introduce the saces H ν (I, τ), Uν (I). Next, define ρ(x) = x(1 x). We write u H ν (I, τ) if u ρhν (I, τ) and du = fdt + g k dwt k, u(0, ) = u 0 for some f ρ 1 H ν 2 (I, τ), g Hν 1 (I, τ), u 0 U ν (I). We define u H ν (I,τ) := ρ 1 u H ν (I,τ) + ρf H ν 2 (I,τ) + g H ν 1 (I,τ) + u 0 U ν (I). To finish with introducing saces we agree to dro τ in the notation of aroriate saces if τ = and write L... (...) in lace of H 0...(...). Remark 2.9. It turns out that ρd : H ν (I, τ) Hν 1 (I, τ) is a bounded linear oerator for any ν R (see [11]). Definition Let A be a set of (a, σ), a > 0, σ l 2. We call A to be of A -tye if (i) the set {a : (a, 0) A} is a nonemty interval; (ii) (a, σ) A imlies (a, λσ) A for any λ [ 1, 1]; (iii) the sets A a := {σ : (a, σ) A} satisfy the uniform interior cone condition, that is there exists a constant α 0 (0, 1] such that, for each a, σ A a, λ (0, 1) = λσ + β A a β l 2, β α 0 (1 λ); (2.8) (iv) for any ν R and any bounded redictable A-valued function (a, σ) = (a(t), σ(t)) = (a(ω, t), σ(ω, t)), f M 1 H ν, and g = (g 1, g 2,...) H ν+1, equation (2.1) with a(t, x) = a(t) and σ(t, x) = σ(t) admits a unique solution u in the class H ν+2,0 ; (v) there exists a finite function N 0 (ν), ν R, such that for the solution from (iv) we have M 1 u H ν+2 N 0 (ν)( Mf H + g ). ν H ν+1 Remark In all known alications sets A of A -tye are such that A a is a ball in l 2 centered at the origin. In that case condition (2.8) is obviously satisfied with α 0 equal to the radius of the ball. Remark So far the widest A -tye sets are found in [10]. Fix some constants δ 1, δ 2 (0, 1] and consider the set B δ1,δ 2 of all coules (a, σ) such that δ 1 a δ 1 1, a 1 2 σ 2 δ 2 a.

7 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 7 It turns out that if and θ satisfy 1 1 < θ/ < 1, (2.9) (1 δ 2 ) + δ 2 then the set B δ1,δ 2 is of A -tye. The case in which δ 2 = 1 is the case of urely deterministic arabolic equations. In this case, (2.9) is equivalent to (2.6) and B δ1,1 is the widest set of A -tye. In this case the results of the resent article hold true for any (1, ) rather than [2, ). However, it is known that for general δ 1, δ 2 (0, 1) and [2, ) the set B δ1,δ 2 is not of A -tye, so that in addition to (2.6) one has to imose stronger restrictions on θ like (2.9). Interestingly enough the shar conditions on θ are known if = n is an integer and they involve the first eigenvalue of the Lalace-Beltrami oerator on a secial art of the unit shere in R n. This will be shown in a subsequent aer. Assumtion There exists a set A of A -tye such that (a(t, x), σ(t, x)) A ω, t, x. Theorem Let the above assumtions be satisfied and suose that there exists a constant δ > 0 such that, for any x, t, and ω, and a(t, x) 1 2 σ(t, x) 2 δ ( κ n := su su a(t, e x n ) a(t, e n ) + σ(t, e x n ) σ(t, e n ) x 1 ω,t + a(t, 1 e x n ) a(t, 1 e n ) + σ(t, 1 e x n ) σ(t, 1 e n ) ) 0 (2.10) as n. Take a constant T (0, ) and let τ be a stoing time, τ T. Then (i) for any f ρ 1 H (I, τ), g = (g1, g 2,...) H +1 (I, τ), and u 0 U +2 (I) equation (2.1) with initial data u 0 admits a unique solution u in the class H +2 (I, τ), (ii) for this solution ρ 1 u H +2 (I,τ) N( u 0 U +2 (I) + ρf H (I,τ) + g H +1 where the constant N deends only on α 0, δ,, θ,, K, T, and the functions δ 0, N 0, κ, and κ n. (I,τ)), (2.11) Remark Observe that (2.10) does not follow from (2.3) no matter how large is. Indeed, the function sin ln x shows this. Also notice that condition (2.10) is much weaker than the uniform continuity condition of [11]. Indeed, for any α (0, 1), the function sin( lnx α )

8 8 KYEONG-HUN KIM AND N.V. KRYLOV satisfies (2.10). However, we still do not know to what extent (2.10) is necessary. By the way, notice that (2.10) holds automatically if a and σ are uniformly continuous on I. This is the situation considered in [11]. We will see that Theorem 2.14 is derived in a kind of standard way from the results of [6] on the equations in the whole sace and our second main result concerning equations on half lines. Theorem Let Assumtions 2.1 be satisfied with a and σ in lace of a 0 and σ 0, resectively, and let Assumtion 2.13 be satisfied. Let τ be a stoing time. We assert that there exists an ε 0 (0, 1), deending only on α 0, N 0,, θ,, δ 0, and K, such that, if κ(1) ε 0 (2.12) then (i) for any f M 1 H (τ), g = (g1, g 2,...) H +1 (τ), and u 0 U +2, equation (2.1) with initial data u 0 admits a unique solution u in the class H +2 (τ), (ii) for this solution M 1 u H +2 (τ) N( u 0 U +2 + Mf H (τ) + g H +1 (τ)), (2.13) where the constant N deends only on α 0, N 0,, θ,, δ 0, and K. Remark The conditions of Theorem 2.16 are invariant under arabolic dilation, that is under transformations like (t, x) (c 2 t, cx), where c > 0 is a constant. Indeed, say a(c 2 t, ce x ) = a(c 2 t, e x+α ), where α = lnc. Also observe that Theorem 2.16 does not have analogs in the theory without weights. There one has to either assume that τ is bounded and allow N to also deend on the bound or assume that the coefficients are uniformly close to some constants. The latter is much stronger than the requirement of having small oscillation on intervals of fixed length. Corollary Let Assumtions 2.1 be satisfied with a and σ in lace of a 0 and σ 0, resectively, and let Assumtion 2.13 be satisfied. Let ( κ 0,n := su su a(t, e x n ) a(t, e n ) + σ(t, e x n ) σ(t, e n ) ) 0 x 1 ω,t as n. Then there exist r > 0 and N (0, ) deending only on α 0,, θ,, K, and the functions δ 0, N 0, κ, κ 0,n, such that, if u H +2 (τ) satisfies (2.1) and u(t, x) = 0 for x > r and all ω and t, then M 1 u H +2 (τ) N( u 0 U +2 + Mf H (τ) + g H +1 (τ)). (2.14)

9 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 9 3. Auxiliary results Here we rove two results about artitions of unity, a result concerning ointwise multiliers, and an embedding lemma. In contrast with the rest of the article here (1, ), θ, R. In articular, there is no restrictions on θ. Lemma 3.1. Let constants C, δ (0, ), a function u H, and n be the smallest integer such that + 2 n. (i) Let ζ k C (R + ), k = 1, 2,..., satisfy ζ k (e x ) C, d i (ζ (dx) i k (e x )) C (3.1) k for all x R and i = 1,..., n. Then ζ k u NC u (3.2) H H, k where the constant N deends only on and. (ii) If in addition to the condition in (i) ζ k (x) δ x > 0, k then u N ζ H k u H, k where the constant N deends only on C,, δ, and. Proof. Owing to the definition (2.4) both statements of the lemma follow almost directly from Theorem 2.1 and Remark 2.1 of [3]. New here are only the way the condition (3.1) is written and the statement about the deendence of N s on the data. In [3] the corresonding constants are stated to deend on the set of constants C i which dominate d i (ζ (dx) i k (e x )) (3.3) k for all i 0 including zero. The fact that we need only dominate these quantities for i n follows by insecting the argument in [3]. The same goes for the fact that for i = 0 instead of estimates on (3.3) it suffices to have the first condition in (3.1). By the way, observe that obviously ( ζ k (e x ) ) 1/ ζ k (e x ). k k Also an imrovement imortant for us is the way C enters into (3.2) which is obtained by relacing ζ k with ζ k /C. The lemma is roved. k

10 10 KYEONG-HUN KIM AND N.V. KRYLOV The following lemma is similar to Lemma 2.2 of [5] and will lay the same role as there. Lemma 3.2. Let n 1 be an integer, ε > 0. Then there exists a nonnegative function ϕ C0 (R + ) such that, for all x R and i = 1,..., n, 1 k= ϕ (e k+x ) 3, k= d i (ϕ(e k+x )) (dx) ε. (3.4) i Furthermore there exists r [1, ), deending only on, n, and ε, such that su ϕ (e r, e r ). Proof. It is convenient to ass to the function η(x) := ϕ(e x ). First take any nonnegative ϕ C0 (R +) such that ϕ (e x ) dx = Next, denote χ(t) = [t]. Then η (i) (k + x) = η (i) (χ(t) + x) dt k where P i (x) = Hence, t η (i) (χ(t)+x) η (i) (t+x) dt = t 1 k= η (i+1) (s + x) dsdt = d i (ϕ(e k+x )) (dx) = i η (i) (s) ds + k η (x) dx = 2. t η (i) (t) dt + P i (x), χ(t) η (i+1) (s+x) ds dt η (i+1) (s) ds. η (i) (k + x) η (i+1) (s) ds. Now for δ > 0 denote η δ (x) = δ 1/ η(δx) and ϕ δ (x) = η δ (ln x). Then we have η δ (x) = ϕ δ (e x ) and η (i) δ (t) dt = δ1/+i 1 η (i) (t) dt. This shows that for all sufficiently small δ > 0 the second condition in (3.4) is satisfied. Similarly, η δ (k + x) 2 = η δ (k + x) η δ (x) dx k k

11 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 11 d dx η δ (x) dx = δ d dx η (x) dx. This shows how to choose small δ > 0 to satisfy the first relation in (3.4) and brings the roof of the lemma to an end. Lemma 3.3. (i) Let g(e ) B +. Then there exists a constant N, deending only on, +, and, such that, if f H, then gf H N g(e ) B + f H. (3.5) (ii) Let n R, < n, g(e ) B n, and a constant r 1. Then there exists a constant N deending only on n,, such that, if f H and f(e x ) = 0 for x r, then gf H N su x r g(e x ) (n )/(2n) g(e ) (n+ )/(2n) B f n H. (3.6) (iii) Let = 1, 2,..., g(e ) B, and a constant r 1. Then there exists a constant N, deending only on,, such that, if f H and f(e x ) = 0 for x r, then gf H N su g(e x ) f H + (g(e )) B 1 f H 1. x r Proof. (i) The left-hand side of (3.5) is g(e )Q f H, and our assertion follows from Corollary (ii) of [12]. (ii) We have gf = gf, where g is any function such that g(e x ) = g(e x ) for x < r. In the light of assertion (i), the left-hand side of (3.6) equals gf H N g(e ) B ( +n)/2 Q f H. By interolation inequalities (see, for instance, Sec. 3.2 of [4]) g(e ) B ( +n)/2 N g(e ) (n )/(2n) B g(e ) (n+ )/(2n) 0 B. n After that it only remains to observe that one can find a g such that su g(e x ) N su g(e x ), x x r g(e ) B n N g(e ) B n with N indeendent of r and g. Indeed, this is a matter of continuing functions given on [ r, r] to the whole real line. Since r 1 one can always reduce the roblem to extending functions from [r 1, r] to functions on [r 1, ) and do the same for negative x. Then each time we are dealing with an interval of unit length and this shows that the constant N can be chosen to be indeendent of r 1. This roves (ii). Assertion (iii) is a trivial consequence of the fact that H = W and Leibniz s formula. The lemma is roved.

12 12 KYEONG-HUN KIM AND N.V. KRYLOV Remark 3.4. We use the lemma for l 2 -valued functions as well as for real-valued ones. Assertion (i) of Lemma 3.3 will be often used when g C0 (R) in which case as easy to see g(e ) B n for any n. Another frequent alication is when we have f, h H and f = ϕh on the intersection of R + with the suort of η 1, where ϕ C0 (R). Then fη 1 H = ϕhη 1 H N hη 1 H. In articular, observe that, for any α R, M α fη 1 = ρ α fη 1 ϕ α, ρ α fη 1 = M α fη 1 ϕ α where ϕ β are C 0 (R) functions equal (1 x) β on the suort of η 1. Hence ρ α fη 1 H N Mα fη 1 H, Mα fη 1 H N ρα fη 1 H. (3.7) Remark 3.5. By Remark 1.6 of [9], if 0 < a < b < and su f (a, b), then f H N f H N f H, where N are indeendent of f. Let us use this fact and Remark 3.4 to show that the norms in the saces H (I) constructed from different η 1, η 2, η 3 (described before Definition 2.8) are equivalent. Let ζ 1, ζ 2, ζ 3 be another set of functions satisfying the same assumtions. Let ϕ C0 (R) be such that ϕ = (ζ 1+ζ 2 +ζ 3 ) 1 on I. Then, for i = 2, 3, the closed suort of η 1 ζ i is in (0, 1) and ϕζ 1 + ϕζ 2 + ϕζ 3 = 1 on I. Hence fη 1 H i fη 1 ϕζ i H N i fη 1 ζ i H N (fζ 1)η 1 H +N i 2 fη 1 ζ i H N( fζ 1 H + fζ 2 H + (fη 1 ζ 3 )(1 ) H ) N( fζ 1 H + fζ 2 H + (fζ 3 )(1 ) H ) = N f H, where is the norm constructed on the basis of ζ 1, ζ 2, ζ 3. Similarly one estimates the remaining terms on the right in (2.7) thus finishing showing that the two norms are equivalent indeed. As a result of this argument, by taking η 2 0, η 1 = η 0, η 3 (x) = η 0 (1 x), we obtain that f H (I) fη 0 H + f(1 )η 0 H. (3.8) We can also take η 2 0, η 1 = 2η 2 0, η 3 (x) = 2η 2 0(1 x). Since η 1 + η 3 (η 0 (x) + η 0 (1 x)) 2 = 1 we get that f H (I) fη 2 0 H + f(1 )η2 0 H. (3.9)

13 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 13 Remark 3.6. It follows from Remark 3.5 that for any f H and ϕ C 0 ((0, 1)), we have fϕ H N f H (I), (3.10) where the constant N is indeendent of f. Indeed, one can always let η 2 = ϕ and find aroriate η 1 and η 3 to satisfy the requirement needed in Definition 2.8. Here is a generalization of (3.10): f (k) ϕ H N f H +k (I), (3.11) where k = 1, 2,... and again the constant N is indeendent of f. One gets (3.11) from (3.10) by induction after noticing that (fϕ) (k) H N fϕ H +k and (Leibniz s formula) f (k) ϕ = (fϕ) (k) +... Remark 3.7. Lemma 3.3 allows us to give sufficient conditions in order for a function g to be a multilier in H (I). Assume that we have a function g satisfying g(e )η 0 (e ) B + + g(1 e )η 0 (e ) B + =: K 1 <. Then by (3.8), (3.9), and Lemma 3.3 (i) gf H (I) N gη 0 (η 0 f) H + N g(1 )η 0(f(1 )η 0 ) H NK 1 η 0 f H + NK 1 f(1 )η 0 H, that is gf H (I) NK 1 f H (I), where the constants N deend only on, +, and. In the roof of Theorem 2.14, we use the following fact, which is stated and used in [11]. We give its roof for comleteness. Lemma 3.8. For any T R +, u H +1 (I, T), and t T, we have E su s t u(s, ) H (I) N u H +1 (I,t), (3.12) where N deends only on, θ,, T. In articular, u H (I,t) N t 0 u H +1 (I,s)ds. Proof. By Theorem 7.1 (iii) and Theorem 7.2 (i) of [6], if v (T) and dv = f dt + g k wt k, then H +1 E su t T + f H 1 v(t, ) H N( v H +1 (T) + (T) g + E v(0, H (T) ) ). (3.13) H +1 2/

14 14 KYEONG-HUN KIM AND N.V. KRYLOV Furthermore, by Theorems 2.11 of [9] and [10], if v H +1 (T), and dv = f dt + g k wt k, then E su v(t, ) N( M 1 v + H t T H +1 (T) Mf H 1 (T) + g + H (T) E M2/ 1 v(0, ) ). (3.14) H +1 2/ Actually, in [9] and [10] there is a general restriction on the values of θ. But this restriction is not used in the roofs of Theorems 2.11 of [9] and [10]. Also it is assumed there that u(0, ) = 0, but the roof can be easily carried through for general u(0, ) M 1 2/ L (Ω, H +1 2/ ) as well. It may be also worth noting that estimate (3.14) for > 2 follows directly from Theorem 4.1 of [8]. Now if u H +1 (I, T) and du = f dt + gk dwt k then by using (3.7) and (3.14) we conclude that uη 1 H +1 (T) and E su η 1 u(t, ) N( ρ 1 uη H 1 + ρfη t T H 1 +1 (T) H 1 (T) + gη 1 + H (T) E ρ2/ 1 η 1 u(0, ) ) N u H +1 2/ H +1 (I,T). Similarly, we treat uη 3 and to estimate uη 2 use (3.13). After that we remember the definition of H (I) and finish roving (3.12) for t = T, which is certainly enough. The lemma is roved. 4. Proof of Theorem 2.16 As usual (cf., for instance, [9]) we may assume that u 0 = 0. Furthermore, the existence result for τ = is obviously stronger than for finite τ. The issue of uniqueness is more delicate. For those for which we use ariori estimates techniques, the uniqueness for finite τ follows from these estimates, the derivation of which can be reeated for finite τ without changes. For some values of we rely on the results obtained for other values of, and then uniqueness is also inherited. In this connection it is also worth noting that a general aroach to the roblem of uniqueness is resented in Sec. 5 and is based on Corollary 5.3. Thus, we let u 0 = 0 and τ =. Under this standing assumtion the roof of the theorem is slit into several cases. By N with or without indices we denote various constants under control, that is deending only on the same data as in the statement of the theorem. The cases are dealt with differently. First of all, observe that, by assumtion any oint (a, σ) A can be connected by a straight line (a, λσ) to a oint of tye (a, 0) A and such oints form an interval, so that one can take any oint (a 0, 0) A and have λ(a 0, 0) + (1 λ)(a, 0) A for

15 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 15 any λ [0, 1] and (a, 0) A. Furthermore, for any λ [0, 1], the functions (a, λσ) and λ(a 0, 0) + (1 λ)(a, 0) satisfy Assumtions 2.1 and 2.13 with the same function κ. Therefore as usual, the method of continuity shows that to rove the theorem it suffices to rove the existence of an ε 0 such that condition (2.12) would imly the ariori estimate (2.13) given that the solution already exists. The roof of (2.13) is somewhat different in six different cases, and we start with the one covering most values of. Only in this case and the case = 0 we use Lemma 3.2 to full extent. We will see from the roof that if we assumed that δ 0 (ν) > 0 for all ν 0 (see (2.2)), then the argument in Case 1 would be valid for all and the roof of the theorem would be finished. This is what we meant in the Introduction by mentioning somewhat sloy assumtions on smoothness of the coefficients. The matter of fact is that if is an integer, the equation makes erfect sense for u H +2 already under our original assumtions. Case 1: {0, 1, 2,...}. Here we aly the same method as in the roof of Theorem 1.2 of [5]. Take the least integer m + 4. Also take an ε (0, 1) to be secified later and take a function ϕ from Lemma 3.2 corresonding to m, ε,. Then by Lemma 3.1, with ϕ n (x) = ϕ(e n x), we have M 1 u H +2 N Next, observe that (2.1) imlies where n= M 1 uϕ n. (4.1) H +2 d(uϕ n ) = (a n (uϕ n ) xx + M 1 f n ) dt + (σ k n (uϕ n) x + g k n ) dwk t, f n = (a a n )ϕ n Mu xx 2a n Mϕ nx u x a n M 1 um 2 ϕ nxx + Mfϕ n, g k n = (σk σ k n )ϕ nu x σ k n M 1 umϕ nx + ϕ n g k, and (a n (t, x), σ n (t, x)) = (a(t, e n ), σ(t, e n )). By Assumtion 2.13, for each n, we obtain M 1 uϕ n H +2 N 0 ( f n + g H n ). (4.2) H +1 Furthermore, for x su ϕ n (e ) we have n+x r = r(m, ε, ) and we can safely assume that r 1 (just add 1 if you accidentally found an r(m, ε, ) smaller than 1). Also notice that, for x su ϕ n (e ), obviously a(t, e x ) a(t, e n ) (r + 1)κ, κ := κ(1).

16 16 KYEONG-HUN KIM AND N.V. KRYLOV Hence by (2.3) and Lemma 3.3 (remember is not an integer) (a a n )ϕ n Mu xx (t, ) H Nκδ (r + 1) δ ϕ n Mu xx (t, ) H, (4.3) where δ > 0 is a constant deending only on K,, +, and. Similarly, uon remembering that + 1 is not an integer, we estimate (σ k σn k)ϕ nu x and get M 1 uϕ n H +2 Nκ δ (r + 1) δ( ϕ n Mu xx H + ϕ n u x H +1 ) +N ( Mϕ nx u x + M 2 ϕ H nxx M 1 u + ϕ H n Mf H + Mϕ nx M 1 u + ϕ H +1 n g ) H. +1 Coming back to (4.1) and using again Lemma 3.1 we conclude M 1 u H +2 Nκ δ (r + 1) δ ( Mu xx + u H x ) H +1 +NJ (ϕ)( u x + M 1 u ) + N( Mf + g ), (4.4) H H +1 H H +1 where J(ϕ) is defined by J(ϕ) = su x R ( Notice that c n ( n n Therefore, ( (Mϕ nx )(e x ) ) 1/ + su (M 2 ϕ nxx )(e x ) ) 1/ n x R n + max su ( d i 1 i m 1 (dx) x R i (Mϕnx )(e x ) ) + max 1 i m 2 su x R n n d i (dx) i ( (M 2 ϕ nxx )(e x ) ). c n ), (Mϕ nx )(e x ) = e n+x ϕ (e n+x ) = (ϕ(e n+x )), (M 2 ϕ nxx )(e x ) = e 2(n+x) ϕ (e n+x ) = (ϕ(e n+x )) (ϕ(e n+x )). J(ϕ) 6 max 1 i m su x R n d i (dx) i ( ϕ(e n+x ) ) 6ε, where the last inequality we have by the construction of ϕ. Furthermore, (2.5) and (4.4) yield M 1 u H +2 N 1 (κ δ (r + 1) δ + ε ) M 1 u H +2 +N 2 ( Mf H + g H +1 ). (4.5) We emhasize that the constants N i deend only on N 0,, θ,, and K and, finally, we secify our choice of ε. We take an ε so that N 1 ε 1/4.

17 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 17 Then obviously (2.13) holds in our situation with N = 2N 2 whenever κ(1) ε 0, with ε 0 satisfying N 1 ε δ 0 (r(m, ε, ) + 1) δ 1/4. This finishes the roof of the theorem in the articular case under consideration. Now comes the case which is most interesting in alications to filtering theory. Case 2: = 0. First of all we rove an ariori estimate containing the unknown function on the right. Lemma 4.1. There exist constants N, ε 0 (0, ) deending only on, N 0, K, and θ such that if (2.12) holds, then, for any f M 1 L, g H 1, and solution u H2,0 of (2.1), M 1 u H 2 N( Mf L + g H 1 + M 1 u L ). (4.6) Proof. We start as in Case 1, but this time we do not need anything from ϕ aart from its existence and we take a ϕ corresonding to ε = 1. By r [1, ) we again denote a number such that su ϕ (e r, e r ). As before we arrive at (4.2) which now becomes M 1 uϕ n N H 2 0 ( f n L + g n H). 1 Notice that (4.3) trivially holds again with δ = 1 (since = 0). While estimating the H 1 -norm of (σk σ k n)ϕ n u x we use Lemma 3.3 (iii) to get (again κ := κ(1)) (σ k σn k )ϕ nu x Nκ (r + 1) ϕ H 1 n u x + N ϕ H 1 n u x L. After that in lace of (4.4) we have M 1 u Nκ (r + 1) ( Mu H 2 xx L + u x H) 1 +N( u x L + M 1 u ) + N( Mf H 1 L + g (4.7) H). 1 Next we use Remark 2.7 to conclude u x L N M 1 u H 1 Then (2.5) and (4.7) yield N M 1 u /2 M 1 u /2 H 2 L Nκ (r) M 1 u H 2 + N M 1 u L. M 1 u H 2 N 1 κ (r + 1) M 1 u H 2 +N( Mf L + g H 1 + M 1 u L ), which leads to (4.6) if N 1 κ (r + 1) 1/2. The lemma is roved.

18 18 KYEONG-HUN KIM AND N.V. KRYLOV Now to finish considering the case = 0 we need only estimate M 1 u L. Technically, it is easier to estimate a larger quantity M 1 u Of course, we are free to reduce the ε H. 1 0 which comes from Lemma 4.1. Take an ε (0, 1) to be secified later and take a function ϕ and a constant r 1 from Lemma 3.2 corresonding to 2, ε,. Write (4.1) and (4.2) for = 1. The latter now is M 1 uϕ n N H 1 0 ( f n + g H 1 n L ). We use the fact that f n f H 1 n L and (a a n )ϕ n Mu xx L + (σ σ n )ϕ n u x L κ (r + 1) ( ϕ n Mu xx L + ϕ n u x L ). Then similarly to (4.4) and actually much easier, we obtain M 1 u H 1 Nκ (r + 1) ( Mu xx L + u x L ) +NJ ( ϕ)( u x L + M 1 u L ) + N( Mf L + g L ), where J(ϕ) is introduced by the same formula as after (4.4) in which this time we only need the first two terms on the right. Therefore, uon using again (2.5), we get M 1 u H 1 N(κ (r + 1) + ε ) M 1 u H 2 +N( Mf L + g L ). By substituting this into (4.6), we conclude M 1 u H 2 N(κ (r +1) +ε ) M 1 u +N( Mf H 2 L + g H). 1 This is (4.5) with = 0 and we can finish the roof of the theorem in the case = 0 as after (4.5). Now we investigate imroving the smoothness of solutions. Case 3: = 1, 2,... We again take ϕ from Lemma 3.2 corresonding to any ε > 0, say ε = 1 and again as in Case 2 only modify our comutations in Case 1 after (4.2). Now in lace of (4.3), by Lemma 3.3 we have (a a n )ϕ n Mu xx (t, ) H Nκ (r + 1) ϕ n Mu xx (t, ) H +N a(t, e ) ϕ B + n Mu xx (t, ). H 1

19 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 19 Similarly we estimate the H +1 -norm of (σk σ k n )ϕ nu x and arrive at the following counterart of (4.4): M 1 u H +2 Nκ (r + 1) ( Mu xx + u H x ) H +1 +N( u x H + M 1 u H +1 Using (2.5) yields M 1 u H +2 ) + N( f + g ). H H +1 N 1 κ (r + 1) M 1 u H +2 +N M 1 u + N( f + g ), H +1 H H +1 so that, if κ = κ(1) ε 0 with N 1 ε 0 (r + 1) 1/2, then M 1 u H +2 N M 1 u H +1 + N( f + g ). H H +1 By iterating this estimate with resect to and reducing each time ε 0 if needed, after finitely many stes we conclude M 1 u H +2 N M 1 u H 2 + N( f + g ). H H +1 Now it only remains to use the result of Case 2. Next, we turn to yet another basic case: = 1. Notice that for negative we basically concentrate not on roving ariori estimates, but rather roceed directly, going down ste by ste along the integers = 1, 0, 1,... However, of course, in this way we also get the ariori estimates as a byroduct. Case 4: = 1 and Assumtions 2.1(1) is satisfied with a and σ in lace of a 0 and σ 0. Here we closely follow the roof of Theorem 3.2 of [9] in the case of negative. Notice that Assumtions 2.1(1) is stronger than Assumtions 2.1(0) and therefore we can find ε 0 such that the assertions of the theorem hold for = 0 and = 1 if (2.12) is satisfied. We assume that (2.12) is satisfied with the above ε 0 and will show that its value suits the resent case as well. Let R be the oerator which mas the coules (f, g) (M 1 L ) H 1 into the solutions u H 2,0 of equation (2.1). By the above, this oerator is well defined and bounded. Denote Lu = Mu x and notice that by Remark 2.7, since θ 0, the oerator L is a bounded one-to-one oerator from H ν to Hν 1 with bounded inverse for any ν R. Also since M 1 H ν = Hν + and θ + 0, the oerator L is a bounded one-to-one oerator from M 1 H ν to M 1 H ν 1 with bounded inverse for any ν R. Now take (f, g) (M 1 H 1 ) L. Then by the above reasons, (L 1 f, L 1 g) (M 1 L ) H 1

20 20 KYEONG-HUN KIM AND N.V. KRYLOV and the function ũ = R(L 1 f, L 1 g), (4.8) is well defined and is in H 2. Then, by Remark 2.8 of [9], for we have v H 1,0 and with v = LR(L 1 f, L 1 g) (4.9) dv = (av xx + f + f) dt + (σ k v x + g k + ḡ k ) dw k t, (4.10) f = L(aũ xx ) av xx = (Ma x 2a)ũ xx, ḡ = L(σũ x ) σv x = (Mσ x σ)ũ x. (4.11) Observe that ũ xx M 1 L and ũ x H 1 since ũ H2. Furthermore, (Ma x 2a)ũ xx M 1 L, (Mσ x σ)ũ x H 1, (4.12) since Ma x 2a is bounded and (here is one of the laces where we use that Assumtions 2.1() is satisfied with = 1 rather than = 0) (Mσ x σ)(t, e ) = (σ(t, e )) σ(t, e ) B = B 1. Now owing to (4.12) we can define ū = R( f, ḡ) and u = v ū. Then u H 1,0, u satisfies (2.1), and the estimate (2.13) for 1 in lace of follows from the the formulas defining u and the fact that M 1 v H 1 N M 1 ũ H 2, M 1 ū H 1 N M 1 ū H 2. We ass to roving the uniqueness of solution in the sace H 1. Let u H 1,0 be a solution of (2.1) with f = g = 0. By the uniqueness of solution in the sace H 2, to get that u = 0, it is enough to show that u H 2. Equivalently, it suffices to show that v := L 1 u H 3. (4.13) Observe that, since u H 1, we have v H2 and dv = (av xx + f) dt + (σ k v x + ḡ k ) dw k t, (4.14) where, f = L 1 (au xx ) a(l 1 u) xx, ḡ = L 1 (σu x ) σ(l 1 u) x. However, L f = M 1 (Ma x 2a)(M 1 u u x ) M 1 L, Lḡ = (σ Mσ x )M 1 u H 1.

21 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 21 Thus, f M 1 H 1, ḡ H2. Since Assumtions 2.1() is satisfied with = 0 and = 1, equation (4.14) imlies (4.13) and this finishes the roof in the resent case. Case 5: = 1 with no additional assumtions. In this case we come back to roving an ariori estimate. Here the idea is to aroximate σ with smooth functions. If its modulus of continuity is sufficiently small, the aroximating functions are indeed close to σ and have second order derivatives under control. This allows one to aly the result of Case 4. To imlement this idea we use the interior cone condition from Definition Fix a λ (0, 1) close to 1 and to be secified later. Also take a nonnegative ζ C0 (( 1, 1)) with unit integral and define σ(t, x) from Observe that σ σ κ(1), σ(t, e x ) = ( σ(t, e x )) (n) K σ(t, e y )ζ(x y) dy. Now take f M 1 H 1, g L and let u H 1,0 (2.1). Then du = (au xx + f) dt + (λ σ k u x + g) dw k t, ζ (n) (y) dy =: N(n). be a solution of where g = (σ λ σ)u x + g and, due to σ λ σ = σ σ (λ 1) σ, g L g L + (κ(1) + K(1 λ)) u x L. (4.15) Next, (a(t, x), σ(t, x)) A for all ω, t, x and λ σ = λσ + β with β(t, x) = λ( σ(t, x) σ(t, x)), β(t, x) κ(1). Therefore, if κ(1) α 0 (1 λ), then by Definition 2.10 we have (a(t, x), λ σ(t, x)) A for all ω, t, x. By the result of Case 4 and (4.15) we then infer that there exists an ε 0 such that, if additionally (2.12) holds, then M 1 u H 1 N( Mf H 1 + g L ) N 1 ( Mf H 1 + g L ) + N 2 (κ(1) + 1 λ) M 1 u H 1, (4.16) where the constants ε 0, N 1, N 2 deend only on the quantities listed in the statement of the theorem (and on N(1), N(2)). For convenience we assume that N 2 2. We now claim that if, with the ε 0 found above, we have κ(1) ε 0 (α 0 (4N 2 ) 1 ), (4.17)

22 22 KYEONG-HUN KIM AND N.V. KRYLOV then M 1 u H 1 2N 1 ( Mf H 1 + g L ). (4.18) Indeed, take λ so that 1 λ = (4N 2 ) 1. Then notice that, due to (4.17), κ(1) ε 0 and κ(1) α 0 (4N 2 ) 1 = α 0 (1 λ). It follows that (4.16) holds. Furthermore, N 2 (1 λ) = 1/4 and N 2 κ(1) α 0 /4 1/4. Therefore, (4.18) follows from (4.16). We thus got the desired result in the case = 1. Case 6: = 2, 3,... We roceed as in Case 4 enjoying the fact that enough smoothness on σ is assumed so that no counterart of Case 5 is needed. Notice that Assumtion 2.1() is stronger than Assumtion 2.1( 1) and Assumtion 2.1(0) if 2. Therefore, if the assumtions of the theorem are satisfied for a 2 then according to Cases 2 and 5 there exists an ε 0 (0, 1), deending on the data the way it should, such that if (2.12) holds, then the assertions (i) and (ii) of the theorem hold for 1 and 0 in lace of. We fix the ε 0 which we have thus identified and show that it suits all = 2, 3,... Thus assume that (2.12) holds. Let R be the oerator which mas the coules (f, g) (M 1 H 1 ) L into the solutions u H 1,0 of equation (2.1). We start with = 2 and as in Case 4 denote Lu = Mu x and take (f, g) (M 1 H 2 ) H 1. Then as in Case 4, (L 1 f, L 1 g) (M 1 H 1 ) L and the functions ũ and v are well defined by (4.8) and (4.9), resectively, ũ H 1 and v H0. As before equation (4.10) holds with f and ḡ introduced by the same formulas (4.11). However this time for any t we have (Ma x 2a)(t, e ) = (a(t, e )) 2a(t, e ) B 2 1 = B 1, (Mσ x σ)(t, e ) = (σ(t, e )) σ(t, e ) B = B. In addition, ũ xx M 1 H 1 and ũ x L since ũ H 1. Hence, instead of (4.12) we get (Ma x 2a)ũ xx M 1 H 1, (Mσ x σ)ũ x L. (4.19) Now owing to (4.19) we can define ū = R( f, ḡ) and u = v ū. Then u H 0 satisfies (2.1) and the estimate (2.13) for 2 in lace of follows from the the formulas defining u. Next, we mimic the roof of uniqueness in Case 4. Let u H 0,0 be a solution of (2.1) with f = g = 0. By the uniqueness of solution in

23 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 23 the sace H 1, to get that u = 0, it is enough to show that u H1. Equivalently, it suffices to show that v := L 1 u H 2. (4.20) Observe that, since u H 0, we have that v H1 and v satisfies (4.14) with the same f and ḡ. However, this time L f = M 1 (Ma x 2a)(M 1 u u x ) M 1 H 1, Lḡ = (σ Mσ x )M 1 u L. Thus, f M 1 L and ḡ H 1 and (4.20) follows from (4.14) due to our choice of ε 0. We have roved the theorem for = 2. Going to = 3 and then further down is easier because the smoothness assumtions become definitely stronger, unlike what we had for = 2 when we wanted to use the solvability theory for = 1 and = 0 and σ barely made it for = 0, in which case we needed σ(t, e ) B 1 and this is exactly what is rovided by the assumtions for = 2. We leave the details to the reader and thereby finish roving the theorem. 5. Proof of Theorem 2.14 First we resent the following. Proof of Corollary Take an infinitely differentiable function χ on R such that χ 1, χ 1 on R, χ(x) = x for x 0, and χ(x) = 1 for x 2. Define a m (t, x) and σ m (t, x) from a m (t, e x ) = a(t, e χ(x+m) m ), σ m (t, e x ) = σ(t, e χ(x+m) m ). Then one can easily check that a m (t, e ) B + + σ m (t, e ) B +1 + NK =: K, where the constant N deends only on χ,, +, and Now, we take ε 0 from Theorem 2.16 which corresond to α 0, δ 0, θ,,, N 0, and K. After that fix an n such that κ 0,m ε 0 for all m n 1. Then it turns out that the assertions of Theorem 2.16 hold for a n and σ n in lace of a and σ, resectively. Indeed, since χ 1, we have for x y 1 that a n (t, e x ) a n (t, e y ) + σ n (t, e x ) σ n (t, e y ) = a(t, e z m ) a(t, e m ) + σ(t, e z m ) σ(t, e m ) κ 0,m, where m = n χ(y + n) n 1 and z = χ(x + n) χ(y + n). Now, observe that a n (t, x) = a(t, x) and σ n (t, x) = σ(t, x) for x e n. Take r = e n and take an u H +2 (τ) satisfying (2.1) and such that u = 0 for x r. Then, in (2.1), one can relace a and σ by

24 24 KYEONG-HUN KIM AND N.V. KRYLOV a n and σ n, resectively, and one sees that the corollary indeed follows from Theorem Now we turn our attention to Theorem Everywhere below the assumtions of this theorem are suosed to be satisfied. We use standard methods based on ariori estimates and the method of continuity. The following is a basic ariori estimate. Lemma 5.1. Let τ = T, f ρ 1 H (I, τ), g = (g1, g 2,...) H +1 (I, τ), and u 0 U +2 (I). Assume that we are given a function u H+2 (I, τ) which is a solution of equation (2.1) with initial condition u 0. Then estimate (2.11) holds. Proof. Find r (0, 1) which suits Corollary 2.18 and take η 1, η 3 C 0 (R) and η 2 C 0 (I) satisfying the conditions before Definition 2.8 and such that η 1 (x) = η 3 (1 x) = 0 for x > r. Notice that where d(uη i ) = (a(uη i ) xx + f) dt + (σ k (uη i ) x + ḡ k ) dw k t, f i = 2au x η ix auη ixx + fη i, ḡ k i = σ k uη ix + g k η i. We use Corollary 2.18 and (3.7) to conclude that uη 1 H +2 (T) and, for any t T, M 1 uη 1 H +2 (t) N( M f 1 H (t) + ḡ 1 H +1 (t) + u 0η 1 U +2). Furthermore, η ix, η ixx C0 (I) and owing to Lemma 3.3 (i) and (3.11) we have M f 1 H (t) N u H +1 (I,t) + Mfη 1 H (t). Here the last term is less than ρfη 1 H (t) N ρf H (I,t) owing to (3.7). Similarly we estimate ḡ 1 to find M 1 uη 1 H +2 (t) N( ρf H (I,t) + g H +1 (I,t) + u H +1 (I,t) + u 0 U +2 (I)). In the same way we estimate uη 3 near the right end of I. Here we have to consider an equation for (uη 3 )(t, 1 x) and roerty (ii) of Definition 2.10 becomes imortant. The estimate of uη 2 H +2 (τ) follows from [6]. Combining these estimates and again referring to (3.7) we conclude ρ 1 u H +2 (I,t) N( ρf H (I,t) + g H +1 (I,t)+ u H +1 (I,t)+ u 0 U +2 (I)).

25 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 25 Here the left-hand side can be relaced with u H +2 (I,t). Indeed, the only missing terms of the latter norm are ρau xx H (I,t) and σu x H +1 (I,t) (5.1) and they are taken care of on the basis of Lemma 3.3 (i), Remark 2.9, and the formulas u x = ρ(u/ρ) x u/ρ and ρu xx = ρ(u x ) x. Thus u H +2 (I,t) N( ρf H (I,t) + g H +1 (I,t) + u H +1 (I,t) + u 0 U +2 (I)) for t T. Finally, by using Lemma 3.8 and Gronwall s inequality we get (2.11) and the lemma is roved. This lemma and the method of continuity show that Theorem 2.14 for nonrandom τ is a consequence of the following. Lemma 5.2. Let τ = T, a 1, σ 0. Then for any f ρ 1 H (I, τ), g = (g 1, g 2,...) H +1 (I, τ), and u 0 U +2 (I) equation (2.1) with initial data u 0 admits a solution in the class H +2 (I, τ). Proof. We can aroximate g with functions having only finitely many nonzero entries. The ariori estimate (2.11) shows that the solutions corresonding to the aroximations converge to the solution of the equation with initial g. In the same way, we can aroximate g with even better functions. By remembering that smooth functions with comact suort are dense in H ν and H ν we easily convince ourselves that we may concentrate on g which are bounded on Ω [0, T] I along with each derivative in x and vanish if x is in a neighborhood of the endoints of I. In that case it is well known that v(t, x) := t 0 g k (s, x) dw k s is infinitely differentiable and obviously vanishes near the endoint of I. What is imortant for us is that v H ν (I, T) for any ν. Furthermore, equation (2.1) can be rewritten as dū = (ū xx + f + v xx ) dt, where ū = u v. To rove the lemma it suffices to find a solution ū H +2 (I, T) with initial condition u 0. Thus, we have reduced the general case to the one in which g 0. One can also aroximate u 0 with smooth functions with comact suort. Then considering the difference u u 0 we see that we may assume that u 0 = 0. Finally, the same argument shows that we may assume that f is bounded on Ω [0, T] I along with each derivative in (t, x) and vanish if x is in a neighborhood of the endoints of I.

26 26 KYEONG-HUN KIM AND N.V. KRYLOV Now let T t be the heat semigrou corresonding to the heat equation on I with zero boundary condition. Define u(t, x) = t 0 T t s f(s, x) ds. It is well known that u is the classical solution of the heat equation (which is just (2.1) in our case) with zero boundary and initial data. Since f is zero near the endoints, u is infinitely differentiable in [0, T] I for each ω and its derivatives in (t, x) are estimated by su norms of derivatives of f in (t, x). Since u(t, x) = 0 for x = 0, 1 and, for instance, u(x)/x = 1 0 u (rx) dr, x (0, 1), it follows easily that the derivatives of ρ 1 u are also controlled in the same way. Moreover, classical solutions are certainly solutions in the sense of distributions. We could conclude that u H +2 (I, T) (for any ) if we knew that u is aroriately measurable in ω. However, for each t, r 0, and x I, obviously, T r f(s, x) is a redictable function of (s, ω). Since it is continuous in r, the function I s t T t s f(s, x) is a redictable function of (s, ω). Therefore, u(t, x) is F t -adated for each (t, x). Recalling that it is also smooth in (t, x) and bounded along with its derivatives we conclude that u(t, ) is aroriately measurable as a Banach sace valued function. The lemma is roved. Corollary 5.3. Let τ be a stoing time, τ T, and u H +2 (I, τ). Let du(t) = f(t) dt+g k (t) dwt k. Then there exists a unique ũ H+2 (I, T) such that ũ(t) = u(t) for t τ (a.s.) and, on (0, T), dũ(t) = (ũ xx (t) + f(t)) dt + g k (t)i t τ dw k t, (5.2) where f(t) = (f(t) u xx (t))i t τ. Furthermore, ũ H +2 (I,T) N u H +2 where N is indeendent of u and τ. (I,τ), (5.3) Indeed, ρ f H (I, T) (see the argument about (5.1)), so that, by Lemma 5.2, equation (5.2) with initial condition u(0) has a solution in (I, T), which is unique by Lemma 5.1. To rove that ũ(t) = u(t) for t τ notice that, for t τ, the function v(t) = ũ(t) u(t τ) satisfies the equation H +2 v(t) = t 0 v xx (s) ds, (5.4)

27 ONE-DIMENSIONAL SPDES WITH VARIABLE COEFFICIENTS 27 which is the heat equation with zero initial condition. Fix an (almost any) ω such that, for that ω, the deterministic function v(ω, t, x) is of class H +2,0 (I, τ(ω)) and satisfies (5.4) in the sense of distributions for t (0, τ(ω)). Then v(ω, t, x) = 0 for t τ(ω) by Lemma 5.1, which imlies the assertion of the corollary. Now we can finish the roof of Theorem The assertion about existence and estimate (2.11) becomes stronger if we relace τ with T, which is seen if we relace f and g with fi t τ and gi t τ resectively. In what concerns the uniqueness, it suffices to rove that if f 0, g 0, u 0 = 0, and u H +2 (I, τ) is a solution, then u(t) = 0 for t τ (a.s.). However, for such an u and ũ from Corollary 5.3 we have ũ(0) = 0 and dũ(t) = (a(t)i t τ + I t>τ )ũ xx (t) dt + σ k I t τ ũ x dw k t. By Lemma 5.1 we have ũ = 0, so that u(t) = 0 for t τ. This brings the roof of Theorem 2.14 to an end. 6. Possible extensions One can choose the same way of generalizing Theorem 2.14 as in [11], [6] and consider nonlinear equations. We refer to show just a very articular result on this way, when one encounters equations with singular coefficients. The ossibility to treat this tye of equations was mentioned in [6]. We consider the following linear equation containing lower order terms du(t, x) = (a(t, x)u xx (t, x) + b(t, x)u x (t, x) + c(t, x)u(t, x) + f(t, x)) dt + (σ k (t, x)u x (t, x) + ν k (t, x)u(t, x) + g k (t, x)) dwt k, (6.1) k=1 where a, b, c are real-valued and σ, ν are l 2 -valued functions defined for ω Ω, t 0, and x I. We want to allow certain singularities in lower order terms. Fix a (3, ). Assumtion 6.1. (i) The functions b, c, and ν k are measurable with resect to the roduct of the redictable σ-field in Ω R + and Borel σ-field in I. (ii) There exists a number 3 > 3 such that, for any t, ε > 0, and ω, t+ε [ ρ 3 b + ρ c + ν + ρ ν x ] (s, x) dxds κ(ε). t I As far as we understand Theorem 6.2 does not follow from the results of [11] even if the main coefficients are constant.

28 28 KYEONG-HUN KIM AND N.V. KRYLOV Theorem 6.2. Let = 0 and let all the the assumtions of Theorem 2.14 and Assumtion 6.1 be satisfied. Then for any f ρ 1 L (I, τ), g H 1 (I, τ), and u 0 U 2 (I) equation (6.1) with initial condition u(0, ) = u 0 has a unique solution u H 2 (I, τ). Furthermore, for this solution, we have ρ 1 u H 2 (I,τ) N( u 0 U 2 (I) + ρf L (I,τ) + g H 1 (I,τ)), (6.2) where the constant N deends only on 3, α 0, δ,, θ, K, T, and the functions N 0, κ, and κ n. To rove the theorem we need the following lemma stated under the assumtions of the theorem. Lemma 6.3. For t [0, T] and u H 2 (I, t) define F(u) = bu x + cu and G(u) = νu. Then, for 0 t ε < t T, E t t ε ( ρf(u)(s) L (I) + G(u)(s) H 1 (I) ) ds N1 κ(ε) u H 2 (I,t), (6.3) where N 1 deends only on the same data as in the statement of the theorem. Proof. By Theorem 2.7 of [11] (or by a straightforward combination of Theorem 7.2 of [6] and Theorem 4.1 of [8]), for 2/ < β 1 and t T, E su ρ β 1 u(r) r t H 2 β (I) N u H 2 (I,t). Since > 3 by assumtion, for β close to 2/ we have 2 β 1/ > 1. Therefore, by Proosition 2.2 of [11] it holds that I := E su ρ β 1+θ/ u(r, x) + E su ρ β+θ/ u x (r, x) r t,x I r t,x I NE su ρ β 1 u(r) N u r t H 2 β (I) H 2 (I,t). Furthermore obviously, if s t, E ρf(u)(s) L (I) NE ρ1 β θ/ b(s) L (I) where, for instance, +NE ρ 2 β θ/ c(s) L (I) I su r t,x I su ρ β+θ/ u x (r, x) r t,x I ρ β 1+θ/ u(r, x), ρ 1 β θ/ b(s) L (I) N ρ θ 1 ρ 1 β θ/ b(s) (x) dx I = ρ β 1 b(s) (x) dx.