ODE, SDE and PDE. Shizan Fang Université de Bourgogne, France. February 29, 2008

ODE, SDE and PDE Shizan Fang Université de Bourgogne, France February 29, 28 Introduction A.1. Consider the ordinary differential equation abbreviated as ODE on R d dx t = V X t, X = x..1 When the coefficient V is smooth and with the bounded derivative, then the ODE.1 can be solved by Picard iteration. Let X t x be the solution to.1 with the initial condition x. Then X t : R d R d is a diffeomorphism from R d onto R d and enjoys the property of flows: X t+s = X t X s. Consider now the following transport equation du t + V u t =, u = θ C 1 R d..2 This is a first order Partial Differential Equation. The solution u t t to.2 can be expressed in term of X t, namely u t = θ Xt 1. A.2. When V satisfies the following Osgood condition V x V y C x y log 1 x y, x y δ,.3 the ODE.1 defines a flow of homeomorphisms of R d. If V admits the divergence, then u t = θ Xt 1 for θ CR d satisfies again the transport equation.2, but in the following sense: u t, ϕ L 2 = θ, ϕ L 2 t u s, divϕv L 2 ds, ϕ C c R d, where, L 2 denotes the inner product in L 2 R d with respect to the Lebesgue measure λ d. An immediate consequence of this consideration is that, whenever divv =, the flow X t preserves the Lebesgue measure. A.3. Due to the linearity, the transport equation.2 can be uniquely solved under weak conditions on V ; namely under the condition V W 1,q and divv L,.4 1

using the argument of smoothing. In the opposite direction, the transport equation.2 allows to solve the ODE.1 in order to get a flow of quasi-invariant measurable maps. To this end, we shall follow the method developed by L. Ambrosio, which consists of three steps: i Consider the mass conservation equation µ t t D x V µ t =, µ given..5 ii Construct a coupling measure η on the product space R d W, where W = C[, T ], R d in the sense that π R d η = µ, e t η = µ t,.6 where e t : x, γ γt, π R dx, γ = x, such that γt = x + t V γsds, holds η-a.s..7 iii Show that η is supported by a graph: there exists X : R d W such that Then X t x solves the ODE.1. η = I X µ. B.1. Consider the Itô stochastic differential equation abbreviated as SDE dx t = σx t dw t + V X t.8 defined on a probability space Ω, F, P. When the coefficients are globally Lipschitz, then.8 defines a stochastic flow of homeomorphisms on R d : there is a full measure subset Ω o Ω such that for w Ω o, X t w, : R d R d is a homeomorphism. Under the stronger conditions σ C 2 b and V C1 b, X t t is now a flow of diffeomorphisms of R d. Let P x be the law of w X w, x on C[, T ], R d. For a given probability measure µ, consider P µ = R d P x dµ x. Let e t : C[, T ], R d R d such that e t γ = γt. Then µ t := e t P µ solves the Fokker-Planck equation where L is the formal dual operator of L = 1 2 d i,j=1 µ t t = L µ t, µ t t= = µ,.9 a ij 2 x i x j + d i=1 V i, a = σσ..1 x i B.2. The operator L is related to the weak solutions of SDE.8. A better knowledge, for example elliptic estimates, on partial differential equations abbreviated as PDE makes possible to prove that the weak solution is indeed the strong one in some situation. B.3. In the opposite direction, there are some recent works see [5], [9] on SDE.8 from the Fokker-Planck equation.9 for few regular coefficients. 2

1 Flow of Homeomorphisms under Osgood Conditions 1.1 Classical Case Let V t t [,T ] be a time-dependent vector field on R d. Suppose that with The differential equation V t x V t y Ct x y for x, y R d 1.1.1 T V s x ds < +, dx t T Cs ds < +. 1.1.2 = V X t, X = x 1.1.3 can be solved by Picard iteration: define X 1 t = x + t V sx ds and for n 1 X n+1 t = x + t V s X n s ds. Due to the condition 1.1.2, we see that t X n t is absolutely continuous for n 1. We have X n+1 t X n t t Cs X n s X n 1 s ds. By induction, we prove that there exists a constant C > such that X n+1 t X n t C F tn, where F t = n! t Cs ds. It follows that X n t converges to X t, solution to 1.1.3, uniformly on t [, T ]. Note that t X t is absolutely continuous. Denote now by X t x the solution to 1.1.3 with the initial condition x. Let t [, T ] and Y t x be the solution to the ODE: dy t = V t ty t, Y = x. We check that X t tx and Y t X t x satisfy the same ODE with the initial condition X t x; therefore by uniqueness of solutions, X t t = Y t X t for t t. In the same way, Y t t = X t Y t. Now putting t = t gives that X t Y t = Y t X t = I. Hence X 1 t = Y t. It follows that x X t x is a homeomorphism from R d onto R d. Moreover if V t is supposed to be C 1 with bounded derivative, then x X t x is a diffeomorphism from R d onto R d. To see the derivability of t Xt 1, it is convenient to consider the ODE d Xt, s, x = V txt, s, x, Xs, s, x = x. Then X t x = Xt,, x and Xt 1 x = X, t, x. 3

Now let θ C 1 R d. We denote by θ the differential of θ, that is, for x R d, θ x is a linear map from R d to R; we denote by θ the gradient of θ, that is a vector field on R d such that θx, v = θ x v for v R d. For a vector field V on R d and an application F : R d R m, we denote also by D V F the directional derivative of F along V, that is, D V F x = d t= F X t, where X t is the flow associated to V. Let u t = θxt 1. Then On the other hand, du t = θ Xt 1 dx 1 t. i u t, V t = θ Xt 1 D Vt Xt 1. Now differentiating the equality x = Xt 1 X t x with respect to the time t, we have = dx 1 t X t + Xt 1 X t dx t = dx 1 t X t + D Vt Xt 1 X t. Since X t is bijective, so for each x R d, the above equality gives dx 1 t + D Vt X 1 t =. According to i: du t + V t u t =, u = θ C 1 1.1.4 here we used for the inner product. Conversely, if u t C 1 is a solution of 1.1.4, then d [u tx t ] = du t X t + u tx t dx t = du t X t + V t u t X t =, so that u t X t = θ or u t = θx 1 t : it is the unique solution to 1.1.4. Remark 1.1 The equation 1.1.3 is non linear while the equation 1.1.4 is linear. 1.1.3 1.1.4. 1.2 Osgood Condition In this subsection, we suppose that V : R d R d is a time-independent bounded continuous vector field such that V x V y C x y log 1, x y δ < 1. 1.2.1 x y Then the differential equation dx t admits at least one solution X t t. = V X t, X = x 1.2.2 Proposition 1.2 Under 1.2.1, the differential equation 1.2.2 admits a unique solution. 4

Proof. Let X t t and Y t t be two solutions to 1.2.2 starting from the same point. Set η t = X t Y t and ξ t = η t 2. Let ε > be a parameter and define the function Ψ ε ξ = ξ ds s log 1 ξ δ, s + ε, and Φ ε = e Ψ ε. Then Φ ε = Φ ε ξ log 1 ξ +ε. Define τ = inf{t > : ξ t δ 2 }. Then by 1.2.1, for t τ, η t, V X t V Y t η t C η t log 1 η t = C 2 ξ t log 1 ξ t. Hence d Φ εξ t = Φ εξ t dξ t = Φ εξ t 2 η t, dη t 1 Φ ε ξ t ξ t log 1 ξ t + ε Cξ t log 1 CΦ ε ξ t. ξ t This and ξ = lead to Φ ε ξ t Φ ε ξ e Ct = e Ct, t < τ. ds s log 1 s Letting ε, we get Φ ξ t e Ct. If ξ t > for some given t, we have ξ t impossible. Therefore we must have ξ t =, which means X t = Y t for any t. Choose χ C c R d such that χ, suppχ B1, R d χdx = 1, Ct, which is where Br = {x R d : x r}. For n 1, define χ n x = 2 dn χ2 n x. Then suppχ n B2 n and R d χ n dx = 1. Set V n = V χ n convolution product, then V n is a bounded smooth vector field on R d. Proposition 1.3 There exists ζ > 1 such that sup V n x V x ζ n for n big enough. 1.2.3 x R d Proof. We have V n x V x V x y V y χ n ydy C R d C2 n log 2 n χ n ydy Cζ n B2 n B2 n y log 1 y χ nydy for some ζ > 1. 5

Theorem 1.4 Let X n t, x t be the solution to Then for any T >, dx n = V n X n, X n = x. lim sup sup X n t, x Xt, x =. 1.2.4 n t T x R d Proof. For simplicity, we omit x in X n as well as in X. Set ξ n t = X n t Xt 2 and τ n = inf{t > : ξ n t δ 2 }. Then by 1.2.1 and 1.2.3, for t τ n, dx n t dxt V nx n t V X n t + V X n t V Xt ζ n 1 + C X n t Xt log X n t Xt. Therefore for t τ n, dξ n t = 2 X n t Xt, dx nt Using the lemma below, we get for t τ n T, dxt ξ n t 2δζ n e Ct 2δζ n e CT. 2δζ n + Cξ n t log This last quantity is less than δ 2 for n big enough. It follows that τ n T and Letting n yields the result 1.2.4. sup ξ n t 2δζ n e CT. t T 1 ξ n t. A consequence of 1.2.4 is that x X t x defines a flow of homeomorphisms of R d. In fact, by previous section, the inverse maps Xn 1 as well as Xt 1 satisfy the same type differential equations. In the same way, Xn 1 converges to Xt 1 uniformly with respect to t, x in any compact subset of [, + [ R d. Lemma 1.5 Let ϕ : R +, 1 be a derivable function such that for C >, ϕ t Cϕt log 1 ϕt, 1.2.5 then ϕt ϕ e Ct for t. 6

Proof. log ϕt being negative, we use 1.2.5, ϕ t ϕt log ϕt C. Integrating this inequality between, t, it leads to ϕt ϕ ds log ϕt Ct, or log Ct. s log s log ϕ Therefore log ϕt log ϕ e Ct or ϕt ϕ e Ct. In order to apply Lemma 1.5 in Theorem 1.4, we set and observe that for δ small enough, η n t = 2δ C ζ n + ξ n t Cξ n t log ξ n t + 2δζ n Cη n t log η n t. Assume now the divergence divv L 1 loc Rd exists in the distribution sense: divv ϕ dx = R d ϕ, V dx, R d ϕ Cc R d. 1.2.6 We have d divv n d = V n i = V i y χ n x ydy x i=1 i i=1 R d x i = V y y χ n x ydy = divv yχ n x ydy = divv χ n, R d R d It follows that divv n converges to divv in L 1 loc Rd, as n +. Theorem 1.6 Assume 1.2.1 and divv exists. Let θ CR d. Then u t x = θxt 1 x satisfies the transport equation du t + V u t =, u = θ in the sense that, for any ϕ C Rd, u t, ϕ L 2 = θ, ϕ L 2 t u s, divϕv L 2ds. 1.2.7 Proof. Step 1. Suppose firstly θ C 1. Let X n be given in Theorem 1.4 and set u n t = θxn 1 t. Then by Section 1.1, for any ϕ C Rd, u n t, ϕ L 2 = θ, ϕ L 2 t 7 u n s, divϕv n L 2ds. 1.2.8

Let K be the support of ϕ, then the support of divϕv n = ϕdivv n + ϕ, V n is contained in K. Let R = sup t T sup x K X 1 t, x which is finite. By 1.2.4, for n big enough, Xn 1 t, x BR + 1 for all t T, x K. Therefore for ε >, there exists n such that for n n, sup sup θxn 1 t, x θx 1 t, x < ε. x K t T Letting n in 1.2.8, we get 1.2.7. Step 2. Let θ CR d. Let θ n C 1 R d which converges to θ on any compact set. By Step 1, u n t = θ n Xt 1 satisfies 1.2.7. Now let K = suppϕ. By what was done in the above, we see that θ n Xt 1 converges uniformly to θxt 1 over K. So that letting n in we get the result. u n t, ϕ L 2 = θ n, ϕ L 2 t u n s, divϕv L 2ds, Corollary 1.7 If divv =, then X t preserves the Lebesgue measure. Proof. Take θ C c R d. Let K = suppθ. Then K T = t T X t K is compact. Now for x K T c, then for any t [, T ], Xt 1 x K c, so that θxt 1 x =. Now take ϕ C Rd such that ϕ = 1 on K T. We have for s t T, u s, divϕv L 2 = u s x ϕ, V x dx =, so that θxt 1 dx = u t, ϕ L 2 = θ, ϕ L 2 = θxdx, R d R d which means that Xt 1 leaves the Lebesgue measure invariant, so does X t For the general case, we have see [1] Theorem 1.8 Assume divv L. Then the Lebesgue measure λ d on R d is quasi-invariant under the flow X t : X t λ d = k t λ d ; moreover e t divv k t x e t divv. 1.2.9 Proof. Take a positive function θ C c R d and set u t = θxt 1. As seen in the proof of the above corollary, there exists R > such that u t x = for t [, T ] and x > R. Then for ϕ Cc R d such that ϕx = 1 for x R, we have, for s [, T ], u s, divϕv L 2 = R u d s divv dx. The equation 1.2.7 yields t u t dx = θ dx u s divv dx ds. R d R d R d It is easy to see that t R u d t dx is absolutely continuous and d u t dx divv u t dx. R R d d We deduce that e R t divv θ dx θxt 1 dx e t divv θ dx. d R R d d It follows that X 1 t λ d is absolutely continuous with respect to λ d. Now 1.2.9 follows. 8

2 Diperna-Lions Theory Let 1 p +. We denote by L p R d the space of measurable functions f : R d R such that R fx p dx < +, L p d loc Rd the space of those functions f such that f 1 K L p R d for all compact subsets K R d. Note that L p R d is a Banach space, having L q R d as the dual space, where q [1, + ] is the conjugate number of p: 1/p+1/q = 1 and L p loc Rd is not a Banach space, but a vector space endowed with a complete metric. For a Banach space E, similarly, L p R d, E is the space of the measurable E-valued functions such that R fx p d E dx < +. The space L 1 [, T ], L q loc denotes the space of functions t, x u tx such that T 1/q K u tx q dx is finite for any compact subset K R d. For a sequence u n L 1 [, T ], L q loc, we say that it converges to u L 1 [, T ], L q loc if T lim n + K u n t x u t x q dx 1/q =, for all compact K R d. Definition 2.1 Let V t, L 1 loc Rd, R d be a time-dependent vector field and c t L 1 loc Rd. Let p [1, ] and T > be given. We say that u L [, T ], L p R d is a weak solution to u t t + V t u t + c t u t = in, T R d, 2.1 with the initial function u L p R d, if for any φ C [, T Rd, T u φ T R d t dx u φ, xdx + u divv t φ + c t φdx =. 2.2 R d R d Notice that the dependence of t u t is not necessarily continuous. Remark 2.2 If we take φt, x = αtφx with α = in 2.2, then T T α t uφ dx + αt u divv φ + c φdx =, R d R d or in the distribution sense, d u t φ dx + u t divv φ + c φdx =. 2.3 R d R d Note that divv φ = V, φ + φ divv. Therefore 2.2 makes sense provided we assume c divv L 1 [, T ], L q loc Rd, V L 1 [, T ], L q loc Rd. 2.4 Under the condition 2.4, there exists a constant C φ > and R > such that d ] u t φ dx C φ [ c t divv t L q BR + v t L q BR. R d We shall denote by ξt = R d u t φ dx, mt = c t divv t L q BR + v t L q BR and ξ the distributional derivative of ξ. Then we have T T ξs ds C φ ms ds < +. 9

This means that ξ is in the Sobolev space W 1,1 [, T ]. Now for a function γ L 1 [, T ], we 1 t +η say that t, T is a Lebesgue point of γ if γt = lim η 2η t η γs ds. By Lebesgue derivative theorem, the set of such points is of full measure. Now set L = {t, T ; Lebesgue point of ξ}. Then for t 1 < t 2 in L, we have ξt 2 ξt 1 = t2 t 1 ξs ds. It follows that ξt 2 ξt 1 C φ t2 t 1 ms ds, t 1, t 2 L. Since L is dense in [, T ], we deduce that ξ admits an uniformly continuous extension ξ on [, T ]. Again by above inequality, such an uniform continuous version is absolutely continuous; therefore t ξt is derivable at almost point of, T. Proposition 2.3 Let p [1, ] and u L p R d. Assume 2.4 and c, divv L 1 [, T ], L R d. Then there exists a solution u L [, T ], L p R d to 2.1 with u given. Proof. For simplicity, consider the case c =. For 1 < p <, the function x x p is differentiable. A priori estimate: if 2.1 is satisfied in the classical sense, using x p = p sgnx x p 1, then t u t p + V t u t p =. Integrating this equation over R d, we deduce d u p dx = divv t u t p dx divv t u t p dx, R d R d R d which implies by Gronwall lemma, u t p dx u p t dx e divvs ds. 2.5 R d R d The inequality 2.5 is the so-called a priori estimate on solutions to 2.1. Now we are going to prove the existence of solutions to 2.1 in the space L [, T ], L p. For simplicity, we suppose in the sequel that V L 1 [, T ], L q and consider the case 1 < p < + for other cases, we refer to [2]. Let χ n be a regularizing sequence: suppχ n B2 n, R d χ n xdx = 1 and χ n. Define V n t = V t χ n, u n = u χ n. We have χ n L 1 = χ L 1 = 1, u n L p u L p χ n L 1 = u L p. 1

By harmonic analysis, as n, for t fixed, Vt n V t in L q and u n u in L p. Moreover, Vt n Cb 1Rd, u n C1 b Rd. By Section 1, there exists a unique solution u n C[, T ], Cb 1Rd to u n t t + V t n u n t =, u n t= = u n. Note that divvt n = divv t χ n, we have divvt n divv t. Now applying 2.5 to u n, we see that u n n 1 is bounded in L [, T ], L p. Notice that L [, T ], L p is the dual space of L 1 [, T ], L q. Applying the following result in Functional Analysis, up to a subsequence, u n converge to a function u L [, T ], L p in the sense that T T lim u n tϕtdx = utϕtdx n R d R d for each ϕ L 1 [, T ], L q. Now taking a limit procedure in 2.2, we see that u L [, T ], L p is a solution to 2.2. Theoremfrom Functional Analysis. Let E be a Banach space; then the unit ball in the dual space E is compact for weak topology. More precisely, for any sequence l n E such that l n E R, then there exist l E of l E R and a subsequence n k such that lim l n k x = lx holds for all x E. k + It is often difficult to work with an equation in the distribution sense. The following approximation will play an important role. Theorem 2.4 Let 1 p and q its conjugate number. Suppose furthermore that Then, if we denote by u n = u χ n, u n satisfies V L 1 [, T ], W 1,α loc Rd for some α q. 2.6 where r n in L 1 [, T ], L β loc Rd and β is given by { 1 β = 1 α + 1 p, if α or p < + ; any β < +, if α = β =. u n t + V t u n = r n, 2.7 Proof. Recall that f W 1,α loc Rd if f and its distributional derivative f/ x i are in L α loc Rd. Let w L p loc Rd and B be a vector field in W 1,α loc. Define the term B w χ n by [B w χ n ]x = B w τ x χ n dy = w divτ x χ n B dy R d R d = wy χ n x ydivby + y χ n x y, By dy, R d where τ x fy = fx y. 11

Lemma 2.5 For B W 1,α loc Rd, set r n := B w χ n B w χ n. Then r n in L β loc Rd. Proof. We have r n = w divb χ n + wy By Bx, χ n x y dy. R d By hypothesis, w divb L β loc ; therefore w divb χ n w divb in L β loc. Next we estimate the second term in the right hand of the above expression. Firstly we shall deal with the good case: w and B Cb 1. For n big enough, the integrand is concentrated in the ball Bx, 2 n, therefore I n x := wy By Bx, χ n x y dy R d wx By Bx, τ x χ n dy R d = wx divb τ x χ n dy = wx divb χ n x ydy R d R d wxdivbx as n. Secondly, fix R > and set Q x y = By Bx, χ n x y. Then for x BR, I n x β = Bx,1 β wyq x ydy [λbx, 1] β 1 BR+1 wyq x y β dy, where λ denotes the Lebesgue measure. Let C β = λb, 1 β 1/β. Then, according to Hölder inequality, 1 I n x C β w L p BR+1 Q x y α α dy. We have Q x y By Bx 2 n χ x y 2 n χ By Bx 2 n 1 { y x 2 n }. It follows that there exists a constant C β,α,r > such that I n L β BR C β,α,r w L p BR+1 Now we claim that [ BR x y 2 n [ BR By Bx 2 n x y 2 n By Bx 2 n α ] 1 α dy dx. α ] 1 α dy dx B L α BR+1. 2.8 If B C 1, then By Bx y x 1 Bx + ty x 12

and Therefore x y 2 n = By Bx α dy BR BR BR BR 2 n dx dx dx x y 2 n x y 2 n By Bx 1 x y 2 n 1 B2 n 1 B1 2 n 1 α dy dx Bx + ty x α dy Bx + tz α dz Bx + ty x α dy. Bx α dz = λb1 B α L α BR+1. We get 2.8. Now by density argument, we see that 2.8 holds for w L p loc Rd and B W 1,α loc. Finally I n L β BR C β,α,r w L p BR+1 B L α BR+1. Using this estimate and the first step, we can see that I n w divb in L β loc End of the proof of Theorem 2.4. Since u n = u χ n, u n t = u t χ n and for B W 1,α loc. V t u n = V t u χ n = V t u χ n + r n t. We have u n u t + V t u n = t + V t u χ n + r n t = r n t. Now using the above observation, we get the result. Remark 2.6 The equation 2.7 is understood in the distribution sense. However by the equation, we see that u n W 1,1 loc, T Rd ; therefore for a.e. x R d, t u n t, x is derivable at a.e. t, T in the classical sense. In the sequel, we shall say that u L 1 [, T ], L 1 + L 1 [, T ], L if u = u 1 + u 2 such that u 1 belongs to the first space and u 2 to the second one. Theorem 2.7 Uniqueness Under the hypothesis that and V L 1 [, T ], W 1,q loc, divv L1 [, T ], L R d V 1 + x L1 [, T ], L 1 + L 1 [, T ], L, the uniqueness holds in L [, T ], L p, where p [1, ]. 13

Proof. We prove only the case p > 1. For p = 1, we refer to [2]. Let u L [, T ], L p be a solution to 2.1 such that u t= =. Let u n = u χ n. Using Theorem 2.4, r n = u n t + V u n in L 1 [, T ], L 1 loc Rd. Let β C 1 R with bounded derivative. We have t βu n + V βu n = r n β u n. 2.9 As βu n βu β u n u, βu n converges to βu in L p. Letting n in 2.9 yields βu + V βu = in distribution sense. 2.1 t Let φ C c R d. By the remark at the beginning of this section, we have d βu φ dx = divv t φβu dx. R d R d Take now Φ C Rd, Φ 1, suppφ B2 and Φ 1 on B1. Next, we consider Φ R x = Φ x/r, R 1. Then d βuφ R dx βudivv Φ R dx = βuv Φ R dx. Let M >, put βt = t M p β is Lipschitz continuous but not in C 1 : this problem may be overcome by approximation argument, then d u M p Φ R dx C t u M p Φ R dx + Φ R Next we observe that u M p L [, T ], L 1 L while therefore d where mt = Then 1 R V t, x 1 {R x 2R} R x 2R 4 V t, x 1 1 + x {R x }, u M p Φ R dx C t u M p Φ R dx +mt u M p dx + CM p V 2t,x 1+ x d x R u M p V t, x dx. x R V 1 t, x 1 + x. Let D 1 R = x R u Mp dx and D 2 R = CM p x R u M p Φ R dx C t u M p Φ R dx + mtd 1 R + D 2 R. dx, V 1 t,x 1+ x dx. 14

Gronwall lemma yields u M p Φ R dx t e t s cudu D 1 Rms + D 2 R ds. V Notice that 1 1+ x L1 [, T ], L 1 and lim R + D 1 R = lim R + D 2 R =. Letting R in the above inequality yields u M p dx = which implies that u M = almost everywhere. Letting M + gives the result. In what follows, we shall indicate how to reduce c in 2.1 to the case c =. First, notice that if u t solves u t t + V t u t = c t, 2.11 then d [ ut X t ] = d u tx t + u t X t dx t = d u t + u t V t = c t V t, where X t is the flow associated to V t. It follows that u t X t = c + t c sx s ds or u t = c X 1 t + t c s Xt s 1 ds. 2.12 So the flow X t allows also to solve 2.11 via 2.12. Secondly, let w t be a solution to dw t + V t w t =. 2.13 Consider w t = e ut w t. Then w t t = e ut dw t Therefore, according to 2.11 and 2.13, e ut w t u t t and w t = e ut w t e ut w t u t. or w t solves 2.1. w t t + w t V t = c t w t, To conclude this section, we state the following result Theorem 2.8 Under the hypothesis that V L 1 [, T ], W 1,q loc, c, divv L1 [, T ], L R d and V 1 + x L1 [, T ], L 1 R d, the transport equation 2.1 admits a unique solution u L [, T ], L p if u p [1, + ]. L p, where 15

3 Ambrosio s Representation Formula Denote by M + R d = {positive regular Borel Radon measure on R d }. A measure µ M + R d is locally finite, the suitable topology on it is vague convergence: µ n µ vaguely if for any f C c R d, fdµ n fdµ. Consider now M f + Rd = {µ M + R d : µr d < + }. There are two natural topologies on M f + Rd : i Narrow convergence: fdµ n fdµ for any f C b R d ; ii w -convergence: fdµ n fdµ for any f C R d. It is known that µ n converges narrowly to µ if and only if µ n converges to µ for w topology and µ n R d µr d. In particular, if we denote by PR d the space of probability measures on R d, then the weak convergence in PR d is identical to the narrow convergence, as well as the convergence for w topology. Note that the notion of narrow convergence can be defined on any Polish space E and the following result will be used freely. Prohokov Theorem. Let µ n n 1 M f + E such that sup n 1 µ n E < +. Then the tightness implies the relative compactness with respect to the narrow convergence. Definition 3.1 Let t, x V t x R d a time-dependent Borel vector fields. We say that a family of measures µ t t [,T ] on R d is a solution to the continuity equation if T dµ t + D x V t µ t =, in [, T ] R d, 3.1 K V t dµ t < + for any compact subset K of R d and for any ϕ Cc R d, Note. D x is the formal divergence. d ϕ dµ t = V t ϕ dµ t. 3.2 R d R d Theorem 3.2 Let µ t t,t be a solution to 3.1 such that sup t [,T ] µ t R d < + ; then it admits a version µ t t [,T ] such that t µ t is continuous respect to the w topology. Proof. The left hand side in 3.2 is taken in distribution sense and d ϕ dµ t C ϕ V t dµ t L 1 [, T ]. R d K So t ϕ dµ t is in W 1,1 [, T ]. By discussion in Remark 2.2, it admits a uniformly continuous version: there is a full measure subset L ϕ of, T and a uniformly continuous function ξ ϕ on [, T ] such that ϕ dµ t = ξ ϕ t for t L ϕ. Take a countable subset D C c R d which is dense in C c R d and set L = ϕ D L ϕ. Let φ C c R d and ε >, there is ϕ D such that ϕ φ < ε. For s, t L, φ dµ t φ dµ s ϕ dµ t ϕ dµ s + 2Cε, where C is a 16

constant dominating all µ t R d. Therefore t φ dµ t is uniformly continuous over L. Let ξ φ be the uniform continuous extension over [, T ]. We have for t L ξ φ t = φ dµ t C φ. This inequality extends to [, T ]. By Riesez representation theorem, for each t [, T ], there is a Borel measure µ t such that ξ φ t = φ d µ t. Therefore µ t t [,T ] admits a version of µ t t [,T ], which is continuous with respect to the w topology. Remark 3.3 Suppose that T V t x R d 1 + x dµ tx < +, 3.3 then µ t R d = µ R d for all t [, T ]. In fact, consider a function ϕ C c R d such that ϕ 1 on B1 and ϕ on B2 c. Define ϕ R x = ϕx/r, we have Therefore ϕ R ϕ R d or ϕ R dµ t = ϕ R dµ + ε R, where ε R T ϕ R dµ t 4 ϕ x R 1 {R x 2R} 4 ϕ 1 1 + x { x R}. x R V t x 1 + x dµ tx, V t x 1 + x dµ tx as R +. Then letting R + gives the result. Remark 3.4 Let V t be a smooth vector field such that dx t = V t X t is well defined over [, T ]. For µ a probability measure on R d, set µ t = X t µ. Let ϕ Cc R d. Then d ϕ dµ t = d ϕx t dµ = ϕx t, V t X t dµ = ϕ, V t dµ t. Therefore µ t t satisfies the continuity equation µ t t + D x V t µ t = with µ t= = µ. 17

Remark 3.5 If V L 1 [, T ], L q loc and divv L1 [, T ], L and µ t = ρ t λ d with ρ t L [, T ], L p loc, then 3.1 is reduced to the transport equation ρ t t + V t ρ t + divv t ρ t =. 3.4 In the sequel, we shall always consider the family of probability measures µ t t [,T ]. According to Theorem 3.2, such a solution to the continuity equation 3.1 admits a version µ t t [,T ], which is weakly continuous. Now let W = C[, T ], R d, W x = {γ W : γ = x} and e t : W R d be the evaluation map: e t γ = γt. Given an η PR d W, we define µ η t = e t η for t [, T ]. Let η x dγ be the conditional probability measure given e = x; then ψx, γdηx, γ = ψx, γdη x γ dµ η x. R d W R d W x Theorem 3.6 Let µ t t [,T ] be a solution to 3.1. moment and T Rd V t x α Assume that µ admits the finite first 1 + x dµ tx < + for some α > 1. 3.5 Then there exists η PR d W such that µ η t = µ t for t [, T ] and under η it holds γt = x + Proof. The proof consists of several steps. t V s γsds. 3.6 Step 1 smoothing: Let ρ ε x = 2πε d 2 e x 2 2ε. Recall that µ ρ ε x = R ρ d ε x y dµ y, which is a smooth function. It is easy to check that sup <ε 1 R d x µ ρ ε xdx Similarly V t µ t ρ ε x = R d ρ ε x yv t ydµ t y, which is a smooth vector field. Define µ ε t = µ t ρ ε, V ε t 1. = V tµ t ρ ε µ ε. t i Then µ ε t µ t weakly as ε. We have for ϕ Cc R d, ϕ D x Vt ε µ ε t dx = ϕ, Vt ε µ ε t dx = ϕ, V t µ t ρ ε dx = ϕ D x [V t µ t ρ ε ] dx. It follows that D x Vt ε µ ε t = D x [V t µ t ρ ε ] and d d µε t + D x Vt ε µ ε t = µ t ρ ε + D x V t µ t ρ ε [ ] d = µ t + D x V t µ t ρ ε =, 18

or µ ε t solves the transport equation µ ε t t + V t ε µ ε t + divvt ε µ ε t =. On the other hand, let X ε t, be the solution to dx ε t, = V ε t X ε t,. Consider: ν ε t = X ε t µ ε. By Remark 3.4 and 3.5, the density of νε t solves the same above transport equation with the same initial data as µ ε t. By uniqueness, we get µ ε t = ν ε t for all t. Define η ε = I X ε µ ε, where I X ε : x x, X ε, x R d W. Then R d W ϕγtdη ε = ϕx ε t, xdµ ε = R d ϕ dµ ε t. 3.7 Step 2 tightness: Now we are going to prove that the family {η ε : ε > } is tight. Consider the functional ψ on R d W : ψx, γ = { T x + +, γt α 1+ γt, if γ is absolutely continuous; otherwise. where α > 1. We claim that K M = {x, γ : γ = x, ψx, γ M} is a compact subset of R d W. In fact, for x, γ K M, we have γt = x + t γsds, hence which implies that γt x + t γs ds t γs α 1 x + 1 + γs ds α t 1 1 + γs β β α ds M + M 1 α T 1 β 1 + γ 1 α, γ M + M 1 α T 1 β 1 + γ 1 α. Therefore it exists a constant C = CM, α, T such that 1 + γ C1 + γ 1 α or 1 + γ C α α 1. This means that {γ : γ, γ KM } is uniformly bounded. Now t γu α 1 γt γs 1 + γu du α t 1 1 + γu β β 1 α du C t s s where C > is a constant. By Ascoli theorem, we see that K M is compact. Now T Rd ψx, γdη ε x, γ = x dµ ε + Ẋε t, x α R d 1 + X ε t, x dµε x. 19 s β,

The second term on the right hand side is equal to Note that V ε t x α T Rd v ε t X ε t, x α 1 + X ε t, x so that the above quantity is dominated by T V t y α ρ ε x y dµ t ydx = R d R d 1 + x Therefore = T Rd Vt ε x α 1 + x dµε tx. ρε x y V t y dµ t y α ρε x y V t y α dµ t y ρε x ydµ t y µ ε t x, sup <ε 1 T T ρ ε x y R d R d 1 + x Rd V t y α ψx, γdη ε x, γ < +, dx V t y α dµ t y 1 + y dµ ty < + as ε. which implies that {η ε : ε > } is tight. Now let η be a limit point. By 3.7, we have ϕγtdη = ϕ dµ t. 3.8 R d W Final Step. The condition 3.5 implies that T V t x R d 1+ x dµ t x < +. Fix ε, 1/2; introduce the measure ν on [, T ] R d by T ϕt, x ϕt, xdνt, x = dµ t x. R d 1 + x ε Note that ν is not a probability measure, but finite. It is clear that V L 1 [, T ] R d, ν. Pick a sequence {C n } n 1 of continuous functions with compact support, converging to V in L 1 [, T ] R d, ν. We have for such a C n and t [, T ], γt x t Cn s γsds dη ε R d W 1 + γ t = Xε t, x x Cn s X ε s, xds R d 1 + X ε dµ ε, x x t V ε = s X ε s, x Cs n X ε s, x ds 1 + X ε dµ ε, x x where C n,ε s R d t t Rd Vs ε Cs n 1 + x dµ ε sds Rd Vs ε Cs n,ε dµ ε 1 + x sds + t n,ε Cs Cs n dµ ε R d 1 + x sds is defined in the same way as Vs ε. Since Cs n,ε Cs n uniformly as ε, lim ε t n,ε s C n s C R d 1 + x 2 dµ ε sds =.

Remark that t Rd Vs ε C n,ε and for ε ε, Finally 1 + x R d W s dµ ε sds ρ ε x y R d 1 + x t dx = V s y Cs n y R d R d 1 + x R d ρ ε x 1 + x + y dx 1. 1 + y ε ρ ε x y dµ s ydxds, γt x t Cn s γsds 1 + γ dη ε V C n L 1 ν + r ε, with r ε as ε. Letting ε in the above inequality, we get γt x t Cn s γsds dη V C n 1 + γ L 1 ν. We have R d W R d W t γt x t V sγsds 1 + γ dη R d W V s γs C n s γs 1 + γs dη ds + t V s Cs n R d 1 + x which is less than 2 V C n L 1 ν. Now letting n +, we obtain that γt x t V sγsds dη =. 1 + γ The result follows. R d W dµ s ds, Theorem 3.7 Ambrosio The result in Theorem 3.6 subsists under the condition 3.3: T V t x R d 1 + x dµ tx < +. Proof. The technical hypothesis in the above theorem has been removed in [1]. Theorem 3.8 Let V be a vector field satisfying the conditions: V L 1 [, T ], W 1,α loc, divv L1 [, T ], L, V 1 + x L1 [, T ], L 1. Then there exists a unique flow of measurable maps X t : R d R d such that for a.e. x R d, and Moreover X t x = x + t V s X s xds X t λ d = k t xλ d. e T divv s ds k t x e T divv s ds. 21

Proof. Take µ = γ d λ d the standard Gaussian measure: γ d L 1 L. By Theorem 2.8, there exists a unique u L [, T ], L 1 L which solves the transport equation u t t + V t u t + divv t u t =. Then µ t := u t λ d solves the continuity equation 3.1. It is clear that the condition 3.3 holds. By Theorem 3.6, there is a probability measure η on R d W under which γ t = x + t To complete the proof, we need the following lemma. V s γ s ds. Lemma 3.9 Let η M + R d W be a probability measure such that γt = x + t V s γsds, holds η-a.s. Denote by e t : R d W R d the evaluation map e t x, γ = γt and µ η t = e t η. Suppose that T Then {µ η t } satisfies the continuity equation V t x R d 1 + x dµη t < +. µ t t + D x V t µ t =. Proof. We claim that for η-a.s γ, T V sγs ds < + ; therefore t γt is absolutely continuous. In fact, the above condition can be rewritten in the form: It follows that for η-a.s γ, T T Vt γt E η < +. 1 + γt V tγt 1+ γt < +. For R >, consider the subset A R = {x, γ; γ R}. Then R d W = R> A R. For x, γ R, we have T T V t γt R + 1 V t γt < +. 1 + γt Now γt = x + t V sγs ds yields the absolute continuity of γ. Let ϕ Cc ; then d µη t, ϕ = ϕγt, V t γt dη The proof of the lemma is complete. R d W = ϕ V t, µ η t. End of the proof of Theorem 3.8 Let η x be the conditional probability law of η, given γ = x. Note that for x R d given, η x is concentrated on the subset of those γ such 22

that γt = x + t V sγs ds. We shall prove that η is supported by a graph: there exists a measurable map X : R d W such that η = I, X µ. If η is not supported by a graph, then there exists a C R d with µ C > and for each x C, η x is not a Dirac mass on W x. Then there exists t [, T ] and two disjoint Borel subsets E, E R d such that η x e 1 t E η x e 1 t E > for x C. Note that we can choose C such that η x e 1 t E and η x e 1 t E are bounded below by a positive constant ε > for each x C. Define η x 1 ηx 1 e 1 t E η x 1 = 1 C x η x e 1 t E e 1, η2 t E x = 1 C x η x e 1 t E. Then ηx, 1 ηx 2 are concentrated as η x for x C and we can check that the conditions in lemma 3.9 are satisfied. Therefore µ η1 t, µη2 t satisfy the continuity equation with the same initial measure µ C = µ 1 C. Since ηx 1 and ηx 2 are absolutely continuous with respect to η x, it is easy to see that µ η1 t, µ η1 t admit density kt 1 and kt 2 in L [, T ], L 1 L. But in this class, the uniqueness holds due to Theorem 2.8. Then µ η1 t = µ η2 t, which is impossible. Therefore η is supported by the graph of a measurable map X : R d W ; then X t µ = µ t = u t λ d. Let A R d such that λ d A =, then X t µ A = or 1 A X t dµ = which implies that 1 A X t = a.e.; therefore 1 A X t dx =. In other words, X t λ d admits a density k t with respect to λ d. The estimates for k t are similar to 1.2.9. 4 Stochastic Differential Equations: Strong Solutions Let σ : R d M d,m R be a continuous function, taking values in the space of d, m-matrices and V : R d R d a continuous function. Consider the following stochastic differential equation abbreviated as SDE on R d : dx t ω = σx t ωdb t ω + V X t ω, X ω = x R d, 4.1 where t B t ω is a R m -valued standard Brownian motion. Now the situation is quite different since the Brownian motion is not unique in the sense of paths, and the probability space Ω, F t, P is not unique. How to understand 4.1? Definition 4.1 Strong Solution If for any given Brownian motion B t t defined on a probability space Ω, P, there exists a Ft B t -adapted process X t ω t<τx such that a.s., X t ω = x + t σx s ωdb s + t V X s ωds, t < τ x ω, 4.2 where τ x : Ω [, + ] is the life time, which is a F B -stopping time. 23

Recall that F B t is the σ-field in Ω generated by {B s : s t}; τ x is a F B -stopping time in the sense that {ω : τ x ω t} F B t, for all t. If τ x ω is finite, then lim X tω = +. t τ x ω The uniqueness for strong solutions to 4.1 is understood as the pathwise uniqueness : for two solutions X t ω t<τ 1 x and Y t ω t<τ 2 x to 4.1, such that X = Y, then τ 1 x = τ 2 x and Pω : X t ω = Y t ω for all t < τ 1 xω = 1. The existence of strong solutions is very important in probabilistic modelisations. Definition 4.2 Weak Solution For the coefficients σ, V given, if there exists a filtered probability space Ω, F, F t, P satisfying the usual conditions on which are defined a F t -compatible Brownian motion B t t and a F t -adapted process X t <t<τx such that X t ω = X ω + t σx s ωdb s ω + t V X s ωds, t < τ x ω. 4.3 Here are some explanations on Definition 4.2. i The usual conditions mean that the sub-σ-fields F t are completed by null subsets in F and F t = F t+ = ε> F t+ε. ii B t t is compatible with F t t means Ft B F t and for any s < t, B t B s is independent of F s. It is equivalent to say that Ft B F t and E e i Bt Bs,ξ Fs = e t s ξ 2 /2, s < t. 4.4 iii X t is measurable with respect to F t, not necessarily measurable with respect to F B t. If this latter situation happens, then the weak solutions become strong ones. The suitable notion of uniqueness for weak solutions is the uniqueness in law. For the simplicity of exposition, we consider the case where τ x = +. Then for ω Ω given, t X t ω is in C[, +, R d. The law of the solution is the law of ω X ω on C[, +, R d. The uniqueness in law means that if X is another solution defined on another probability space Ω, F, F t, P. Then lawx = law X. Construction of strong solutions: First case. We suppose that the coefficients satisfy the global Lipschitz conditions: σx σy C x y, V x V y C x y, x, y R d, 4.5 where denotes the Hilbert-Schmi norm for matrices: σ 2 = ij σij 2. In this case, the Picard iteration does work. More precisely, let x R d be given. Define X t, x = x, X n t, x = x + t σx n 1 s, xdb s + 24 t V X n 1 s, xds.

The condition 4.5 implies that the coefficients have linear growths: We have X n t 2 3 x 2 + σx C 1 1 + x, V x C 1 1 + x. 4.6 t Let T >. By Doob s maximal inequality, t E sup t T 2 σx n 1 sdb s + t 2 V X n 1 sds. 4.7 2 T T σx n 1 sdb s 4E σx n 1 s 2 ds 8C1 2 1 + E Xn 1 s 2 ds, t 2 T E V X n 1 sds 2C1T 2 1 + E Xn 1 s 2 ds. sup t T According to 4.7 and by induction, we see that E sup X n t 2 < +. 4.8 t T Now we compute X n+1 t X n t = t σxn s σx n 1 s t db s + V Xn s V X n 1 s ds =: I 1 t + I 2 t. Again by Doob s maximal martingale inequality, t t E sup I 1 s 2 4E σx n s σx n 1 s 2 ds 4C 2 E X n s X n 1 s 2 ds. s t In the same way, t E sup I 2 s 2 T C 2 E X n s X n 1 s 2 ds. s t Therefore E sup X n+1 s X n s 2 2C 2 T + 4 s t By induction, Hence E P By Borel-Cantelli lemma, a.s. sup X n+1 s X n s 2 s T t sup X n+1 t X n t > 2 n t T E sup X n u X n 1 u ds. 2 u s K 2C2 T + 4 n T n. n! K 8C2 T + 4 n. n! sup X n+1 t X n t 2 n as n +. t T 25

It follows that the series X t x, ω := x + + n= Xn+1 t X n t converges uniformly with respect to t [, T ]. Now remark that t t 2 t E σx n sdb s σxsdb s C 2 E X n s X s 2 ds So letting n + in as n +. we get X n+1 t = x + X t x = x + t t σx n sdb s + σx s xdb s + t t V X n sds, V X s xds. A comment on the lifetime τ x. By the above construction, we see that for x R d, τ x = + a.s. More precisely, let A = {x, ω R d Ω : τ x ω = + }. Let ν be a Borel probability measure on R d, then By Fubini theorem, Ω ν PA = 1. [ ] 1 A x, ωdνx dpω = 1. R d It follows that there exists Ω Ω with full probability, such that for each ω Ω, τ x ω = + for ν-a.e. x. By limit procedure, we see that x, ω X t x, ω is measurable with respect to BR d F t ν P. A new definition of strong solution: Let W R d = C[, +, R d and W R d = {w W R d : w = }. Let µ W be the law of the Brownian motion on W R m. We say that the SDE 4.1 has a strong solution if there exists F : R d W R m W R d such that i for any Borel probability measure ν on R d, F is measurable with respect to BR d BW R m ν µ W ; ii for any t, x given, w F x, wt is F W t -measurable, where F W t = σws : s t, and X t w = F X w, B is a strong solution to 4.1 in the sense of Definition 4.1, where X is random variable independent of B. Second case. Now we will give another example, for which the strong solution exists. Suppose that σ and V are bounded and σx σy 2 C x y 2 log 1 x y, V x V y C x y log 1 x y 26 4.9

for any x y δ. In this case, the Euler approximation does work. Let n 1, denote Φ n t = k2 n for t [k2 n, k + 12 n, k. Define X n = x and for t [k2 n, k + 12 n, X n t = X n k2 n + σx n k2 n B t B k2 n + V X n k2 n t k2 n F B t. Using Φ n, X n t can be expressed by X n t = x + t σ X n Φ n s db s + t V X n Φ n s ds. Then we can prove that a.s., X n t converges to X t uniformly with respect to t [, T ]. Therefore, in this case, the strong solution exists. For a proof, we refer to [4]. Theorem 4.3 Under 4.9, the pathwise uniqueness holds. Proof. Let X t and Y t be two solutions to 4.1 starting from the same point. Set η t = X t Y t and ξ t = η t 2. We have In coordinate forms, for i = 1,, d, dη i t = dη t = σx t σy t db t + V X t V Y t. m σ ij X t σ ij Y t db j t + V i X t V i Y t. j=1 Itô stochastic contraction of dη t is given by d dηt i dηt i = σ ij X t σ ij Y t 2 = σx t σy t 2. i,j i=1 Now by Itô formula, We see that dξ t = 2 η t, dη t + dη t dη t = 2 η t, σx t σy t db t + 2 η t, V X t V Y t + σx t σy t 2 = 2 σx t σy t η t, db t + 2 η t, V X t V Y t + σx t σy t 2. Let ε >. Define Ψ ε ξ = t Φ ε dξ t dξ t = 4 σxt σy t η t 2. ds s log 1 s +ε, Φ εξ = e Ψεξ. Then Φ ε = ξ log 1 ξ + ε, Φ ε = Φ ε ξ log 1 ξ + ε + Φ ε log ξ + 1 ξ log 1 ξ + ε 2 = Φ ε log ξ + 2 ξ log 1 ξ + ε 2 if ξ e 2. 27

Define By Itô formula, τ = inf{t > : ξ t e 2 δ 2 }. t τ Φ ε ξ t τ = 1 + = 1 + 2 + t τ t τ Taking the expectation, By Gronwall lemma, Φ εξ s dξ s + 1 2 t τ Φ εξ s dξ s dξ s Φ εξ s σx s σy s η s, db s + 2 Φ εξ s σx s σy s 2 ds + 1 2 t τ t τ Φ εξ s dξ s dξ s. E Φ ε ξ t τ t τ Φ ε Cξ s log 1 ξ 1 + 2E s ξ s log 1 ds 1 + 2C ξ s + ε E Φ ε ξ t τ e 2Ct. Φ εξ s η s, V X s V Y s ds t E Φ ε ξ t τ ds. Letting ε, we get ξ t τ = a.s. If Pτ < T >, then by continuity of the paths, we get ξ τ = on {τ < T }. But ξ τ = δ 2 e 2. The contradiction implies that τ T a.s., and hence ξ t = a.s. 5 Stochastic Flow of Homeomorphisms In contrast to ODE, even under the global Lipschitz condition: σx σy C x y, V x V y C x y, 5.1 for a study of the dependence x X t x, ω, we need the following Kolmogorov modification theorem, one of the fundamental tools in probability theory: Theorem 5.1 Kolmogorov Let {X t : t [, 1] m } be a family of R d -valued random variables. Suppose there exist constants γ 1, C, δ > such that Then X admits a continuous version X t, satisfying E X t X s γ R d C t s m+δ R m. 5.2 [ Xt E sup X s γ ] s t t s α < + for any α In particular, there exists M L γ Ω such that, δ. 5.3 γ X t X s M t s α, t, s [, 1] m. 5.4 28

Remark 5.2 1 X is a version of X means that for each t [, 1] m, X t = X t a.s. 2 X t could take values in a Banach space, but it is crucial that the exponent of t s R m is strictly bigger than m, namely δ >. 3 Let {X n t : t [, 1] m } be a sequence of such family satisfying 5.2, but with C independent of n. Then there exist M n L γ bounded in L γ such that X n t X n s M n t s α R m, t, s [, 1]m. The proof of Theorem 5.1 can be found in any textbook of probability theory. We refer to [6]. Let X t x, ω be the solution to dx t ω = σx t ωdb t + V X t ω, X ω = x. 5.5 Theorem 5.3 Under the global Lipschitz condition 5.1, the solution to 5.5 admits a continuous version X, such that a.s., t, x X t x, ω is continuous and for each x R d, P{ω : t, X t x, ω = X t x, ω} = 1. 5.6 Proof. Let R >, T > and set I = [, T ] [ R, R] d. Consider η t = X t x X t y. We have dη t = σx t x σx t y db t + V X t x V X t y. The Itô stochastic contraction of dη t is given by Let ξ t = η t 2. Then by Itô formula, dη t dη t = σx t x σx t y 2. dξ t = 2 η t, dη t + dη t dη t = 2 η t, σx t x σx t y db t + 2 ηt, V X t x V X t y + σx t x σx t y 2. The Itô stochastic contraction of dξ t is Let p 2. By Itô formula, dξ t dξ t = 4 σx t x σx t y ηt 2. dξ p t = pξ p 1 t = 2pξ p 1 t +pξ p 1 t dξ t + 1 pp 1ξp 2 t dξ t dξ t 2 ηt, σx t x σx t y db t + 2pξ p 1 t σx t x σx t y 2 + 2pp 1ξ p 2 =: I 1 t + I 2 t + I 3 t + I 4 t. t η t, V X t x V X t y σx t x σx t y ηt 2 By the Lipschitz conditions, I 2 t 2pξ p 1 t C η t 2 = 2pCξ p t, I 3 t pξ p 1 t C 2 ξ t = pc 2 ξ p t, I 4 t 2pp 1ξ p 2 t η t 2 C 2 η t 2 = 2pp 1C 2 ξ p t. 29

Therefore for some constant C p >, we have By Gronwall lemma, or Next, there exists a C p > such that X t x 2p C p x 2p + By Burkhölder inequality, we have E sup t T t t Eξ p t x y 2p + C p Eξsds. p Eξ p t x y 2p e C pt for any t T, E X t x X t y 2p x y 2p e CpT. 5.7 t 2p σx s xdb s + t V X s xds 2p 2p t p σx s xdb s C p E σx s x 2 ds C p,t [1 + E sup X t x ]. 2p t T If we denote by u t = E sup s t X s x 2p, then by 5.8, Again by Gronwall lemma, we get or Now for t > s, Again by Bürkhölder inequality: u t C p,t x 2p + t u t C p,t x 2p e tc p,t, u s ds.. 5.8 E sup X t x 2p C p,t x 2p e T C p,t. 5.9 t T X t X s = t s σx r db r + t s V X r dr. t 2p t p E σx r db r C p E σx r dr 2 C p E s s s C p [1 + E sup X t ]t 2p s p. t T [ t ] p 1 + sup X r dr 2 r T Finally we get E X t x X s x 2p C p,t t s p. 3

Combining with 5.7, we have E X t x X s y 2p C p,t t s p + x y 2p. 5.1 Now take p > d + 1, we can apply Kolmogorov s modification theorem: there exists a continuous version X such that for given t, x [, T ] [ R, R] d, a.s. X t x, ω = X t x, ω. Since the two processes are continuous with respect to t, we obtain for given x R d, a.s. X t x, ω = X t x, ω for all t. Remark 5.4 By 5.7, we see that x X t x, ω is 1 ε-hölder continuous: there exists a loss of regularity with respect to the coefficients. The main result of this section is Theorem 5.5 There exists a full measure subset Ω Ω such that for ω Ω and any t, x X t x, ω is a global homeomorphism of R d. Proof. Step 1. Let x y be given, for α <, we estimate E X t x X t y 2α. Take ε, x y 2. Set η t = X t x X t y and ξ t = η t 2. We have ξ = x y 2 > ε. Define the stopping time τ ε = inf{t > : ξ t ε}. Then ξ t τε ε. By Itô formula, ξ α t τ ε t τε = x y 2α + α x y 2α + M t τε + 2α + 1 2 αα 1 t τε =: x y 2α + I 1 + I 2 + I 3. ξs α 1 dξ s + 1 t τε 2 αα 1 t τε ξs α 2 dξ s dξ s ξs α 2 dξ s dξ s ξs α 1 η s, V X s x V X s y ds By Lipschitz condition, I 2 2 α C t τε ξ α s ds. In the same way, there exists a constant C α > such that E ξt τ α t ε x y 2α + C α E ξs τ α ε ds. It follows that E ξt τ α ε x y 2α e tc α. 5.11 Let τ = inf{t > : X t x X t y = }, we see that τ ε τ as ε. 31

Then letting ε in 5.11, we get E ξt τ α x y 2α e tc α. If Pτ < + >, then Pτ T > for some T >. Therefore + = E 1 {τ T }ξt α τ E ξ α T τ x y 2α e T C α < +. The contradiction implies that τ = + a.s., in other words, a.s. X t x X t y for all t. Remark 5.6 Here the a.s. is dependent of the given x y. Step 2. Let δ > be given. Set R δ = {x, y Rd R d : x y δ, x R, y R}. Let X t x be the continuous version. Define For p 2, x, y and x, ỹ R δ, By 5.11, η t x, y = X t x X t y 1. η t x, y η s x, ỹ η t x, yη s x, ỹ X s x X s ỹ X t x X t y Combining with 5.1, we have η t x, yη s x, ỹ X s x X t x + X s ỹ X t y. E η t x, y 4p e C pt δ 4p, E η s x, ỹ 4p e C pt δ 4p. E η t x, y η s x, ỹ p C p,t,δ t s p 2 + x x p + ỹ y p. Taking p 2 > 2d + 1, again by Kolmogorov s modification theorem, η tx, y admits a continuous version η t x, y : t, x, y η t x, y on [, T ] R δ. Since δ >, R > are arbitrary, we have a.s., t, x, y η t x, y is continuous on [, T ], where = {x, y R d R d : x y}. Let D be a dense countable subset of [, T ]. Then there exists a subset Ω Ω of full measure such that for every ω Ω, t, x, y D, η t x, y = η t x, y, or X t x X t y = η t x, y 1. By continuity, it holds for all t, x, y [, T ] ; especially X t x X t y = η t x, y 1. In conclusion, we get a full measure subset Ω independent of x, y, such that for each ω Ω, t, x y, we have X t x X t y. Step 3 Surjectivity: By considering ˆx = η t ˆx = x x 2 and { 1 + X t x 1, if ˆx ;, if ˆx =, 32

the same machinery shows that x X t x is continuous at. Set R d = R d { } which is homeomorphic to S d. Then almost surely for all t, x X t x, ω is continuous from R d onto R d. But X, ω = Id, therefore X t, ω is homotopic to Id. By a result from General Topology, X t R d = R d. In particular, X t R d = R d. Due to Remark 5.4, in order that for a.s ω Ω, x X t x, ω is C 2, we have to suppose that σ C 2+δ b, V C 2+δ b, the space of bounded functions having bounded first and second order derivatives and the second order derivative being δ-hölder continuous; for more detail, we refer to Kunita [8]. In what follows, we assume that the coefficients satisfy these conditions. Let ϕ Cc R d, consider P t ϕx = EϕX t x,. Then t, x P t ϕx is in Cb 2. By Itô formula, where E ϕx t x = ϕx + Lϕx = 1 2 t d 2 ϕ a ij x x + x i x j i,j=1 E LϕX s x ds, d i=1 V i x ϕ x i x with a = σσ. Now let µ be a probability measure on R d and P µ the diffusion distribution on W = C[, T ], R d defined by P µ = R P d x dµ x, where P x is the law of ω X x, ω. Define µ t PR d by ϕ dµ t = ϕγtdp µ = E ϕx t x, ω dµ x, for all ϕ C b R d. R d W R d Therefore for ϕ Cb 2Rd, d ϕ dµ t = E LϕX t x dµ x = Lϕ dµ t. 5.12 R d R d R d Definition 5.7 Let µ t t [,T ] be a family of probability measures on R d. We say that µ t satisfies the following Fokker-Planck equation: µ t t = L µ t, µ t= = µ, 5.13 if for ψ C c, T R d, ψ t dµ t = Lψ dµ t. 5.14 As what was done in Theorem 3.2, µ t admits a version µ t such that t µ t is weakly continuous. Therefore the equation 5.14 holds for ψ C 2 b [, T ] Rd. Theorem 5.8 Suppose that a C 2+δ b and V C 2+δ b. Then the Fokker-Planck equation 5.13 admits a unique solution. 33

Proof. By linearity, it is sufficient to prove that µ t = for all t [, T ] if µ =. Fix ϕ Cc R d, t [, T ]. Then the following backward PDE admits a unique solution f Cb 2: { f t + Lf = in [, t ] R d, 5.15 ft, = ϕ. In fact, ft, x = P t tϕx is the candidate. Let α Cc, T. By 5.14, t αf dµ t = α Lf dµ t, or T T α t ft, xdµ t x = αt f R d R d t + Lf dµ t. It follows that d ft, xdµ t x = f R d R d t + Lf dµ t =. Therefore = f, xdµ x = ft, xdµ t x = ϕxdµ t x. R d R d R d It follows that µ t =. As t is arbitrary, we get the result. 6 Stochastic Differential Equations: Weak Solutions Let σ : R d M d,m R and V : R d R d be bounded Borel functions. Suppose that X t, B t solves dx t ω = σx t ωdb t + V X t ω. 6.1 Then by Itô formula, for any f C 2 c R d, M f t := fx t fx t LfX s ds = t σ X s fx s, db s is a L 2 -martingale, where Lfx = 1 2 d 2 f a ij x x + x i x j i,j=1 d i=1 V i x f x i x, a = σσ. 6.2 In general, for a given operator L such as in 6.2, we say that a continuous adapted process X t t defined on a probability space Ω, F, F t, P is a solution to L-martingale problem if for any f Cc 2 R d, M f t := fx t fx t LfX s ds is a continuous martingale. If furthermore the matrix a admits a decomposition a = σσ with σ : R d M d,m R, then there is a Brownian motion compatible with F t such that X t solves 6.1. This topic is well treated in Stroock-Varadhan s book [11]. In what follows, we will construct a weak solution. 34

Theorem 6.1 Let σ, V be bounded continuous and µ PR d with compact support. Then the SDE 6.1 admits a weak solution X t, B t such that µ = lawx Proof. First there exists a probability space Ω, F, F t, P on which are defined a compatible Brownian motion B t and a random variable ξ F with lawξ = µ. Next for n 1, define X n = ξ and for t [k2 n, k + 12 n, X n t = X n k2 n + σ X n k2 n B t B k2 n + V X n k2 n t k2 n. Using Φ n t = k2 n for t [k2 n, k + 12 n, we express X n t as X n t = ξ + t σ X n Φ n s db s + t V X n Φ n s ds. As σ and V are bounded, for T > fixed, by Bürkhölder inequality, sup E n sup t T X n t 2p < +, sup E X n t X n s 2p C p,t t s p. n By Kolmogorov s modification theorem, there exists M n L 2p bounded in L 2p such that Let X n t X n s M n t s α, α < p 1 2p. 6.3 K R = {w C[, T ], R d : w R, wt ws R t s α }. By Ascoli theorem, K R is compact in C[, T ], R d. Let ν n = lawx n. Then But for R big enough, and according to 6.3, ν n K c R ν n w > R + ν n t s, wt ws > R t s α. ν n w > R = P X n > R = µ x > R =, ν n t s, wt ws > R t s α = P t s, X n t X n s > R t s α PM n R M n 2p L 2p R 2p C p R 2p. Therefore sup n ν n KR c < ε for n big enough. The family {ν n : n 1} is tight. Up to a subsequence, ν n converges to ν weakly. Now let f Cc 2 R d and F be a bounded continuous function from C[, T ], R d to R which is Fs W -measurable. We have t ] E ν [fwt fws Lfw u du F s [ t ] = lim E fx n t fx n s LfX n udu F X n. 6.4 n s 35