PATHWISE EXISTENCE OF SOLUTIONS TO THE IMPLICIT EULER METHOD FOR THE STOCHASTIC CAHN-HILLIARD 6 EQUATION 1 0 2 DAISUKEFURIHATA,FREDRIKLINDGREN,ANDSHUJIYOSHIKAWA n a Abstract. We consider the implicit Euler approximation of the stochastic J Cahn-HilliardequationdrivenbyadditiveGaussiannoiseinaspatialdomain 7 with smooth boundary indimensiond≤3. We show pathwise existence and 2 uniqueness of solutions for the method under a restriction on the step size thatisindependentofthesizeoftheinitialvalueandoftheincrementsofthe ] Wiener process. Thisresultalsorelaxes theimposedassumptiononthetime A stepforthedeterministicCahn-Hilliardequationassumedinearlierexistence N proofs. . h t a m 1. Introduction [ Let D ⊂ Rd, d ≤ 3, be a bounded spatial domain with smooth boundary ∂D 1 and consider the stochastic Cahn-Hilliard equation written in abstract form, v (1) dX +A(AX +f(X))dt=dW(t), t∈(0,T], X(0)=X0, 9 6 where A is the realisation of the Laplace operator −∆ with homogenous Dirichlet 5 boundaryconditionsinH =L2(D)withinnerproducth·,·iandinducednormk·k. 7 The non-linearity f is given by f(s) = s3−β2s and W is an H-valued Q-Wiener 0 process. . 1 Existenceofsolutionsto(1)isstudiedin[1]andspatialsemi-discretisationwith 0 a finite element method in [7] and [9]. Here, we are interested in existence and 6 uniqueness of the implicit Euler approximationof (1) given by 1 v: (2) Xj+kA2Xj+kAf(Xj)=Xj−1+∆Wj, j ∈IN, X0 =X0, Xi whereIN ={1,...,N},N ∈N,k =T/N and∆Wj =W(tj)−W(tj−1)fortj =jk, r j ∈IN ∪{0}. That is, we study a temporal semi-discretisation. a In the deterministic case, when W = 0, existence is usually proved [3, 11] by the reformulation of (2) as a fixed point problem in a ball {kA1/2xk ≤ M}. If k ≤ k the constructed mapping in the formulation becomes a contraction and 0 existence and uniqueness follows. However, the constant M grows and k shrinks 0 as kA1/2X k grows. 0 (Furihata and Lindgren) Cybermedia Center, Osaka University 1-32 Machikaneyama, Toyonaka,Osaka 560-0043,Japan (Yoshikawa)GraduateSchoolofScienceandEngineering,EhimeUniversity3Bunkyo- cho,Matsuyama,Ehime 790-8577,Japan E-mail addresses: furihata@cmc.osaka-u.ac.jp, fredrik.lindgren1979@gmail.com, yoshikawa@ehime-u.ac.jp. 2010 Mathematics Subject Classification. 60H35,35R60,65J15. Keywordsandphrases. Stochasticpartialdifferentialequation;Cahn-Hilliardequation;Euler method;numericalapproximation;existenceproof. 1 2 PATHWISE EXISTENCE In the present setting this dependence can not be allowed. At every time step, the right hand side of (2) plays the role of the initial value and, being a Gaussian random variable, ∆Wj may be arbitrary large with positive probability. If we would rely on earlier existence results we would be forced to utilise an adaptive time stepping scheme and facing the risk of needing arbitrary small time steps. Instead, we shall prove that the equation (3) u+kA2u+kAf(u)=y has a solution in H2 ∩ H1 as soon as y ∈ Dom(A−1), the domain of A−1. At 0 each time step, u corresponds to Xj and y to Xj−1 + ∆Wj in (2), so for this assumption to hold it is sufficient that X ,∆Wj ∈ Dom(A−1), j ∈ I , a.s. This 0 N holds if, e.g., EkA−1X k2 < ∞ and kA−1Q1/2k < ∞, where k·k denotes the 0 HS HS Hilbert-SchmidtnorminH andQisthecovarianceoperatorofW. Moreprecisely, our main results are the following. Theorem 1.1. Assume y ∈ Dom(A−1) and k < 4/β4, then (3) has a unique solution x∈H2∩H1. 0 Corollary 1.2. If, a.s., X , ∆Wj ∈Dom(A−1), j ∈I and k <4/β4, then there 0 N is an a.s. unique solution to (2) with Xj ∈H2∩H1 for j ∈I . 0 N We shall prove the existence part of Theorem 1.1 by applying Schaefer’s fixed point theorem to the mapping z =T (x) given by y (4) z+kA2z+kAzx2 =y+kβ2Ax. Clearly, a fixed point, z =x, of (4) is a solution of (3). The outline is as follows. In Section 2 we give some necessary definitions and state Schaefer’s fixed point theorem and other required results. In Section 3 we givethe mappingT arigorousmeaning andshowthat itfulfils the assumptions of y Schaefer’s fixed point theorem. 2. Preliminaries We shall use the abbreviation L = L (D) for the standard function spaces on p p D and Hr = Hr(D) refers to the usual Sobolev spaces with all partial derivatives of order ≤ r being square integrable. The space H1 is the completion of C∞(D) 0 b in H1. It hold that the operator A with Dom(A) = H1∩H2 has strictly positive 0 eigenvaluesλ <λ ≤...divergingtoinfinitysoanyrealpowerAs maybedefined 1 2 and As is positive definite and self-adjoint with Dom(As/2)=:H˙s. If s ≥s then 1 2 H˙s1 ⊂H˙s2 and (5) kAs2/2xk≤λ(s1−s2)/2kAs1/2xk. 1 In particular, H1 = H˙1 and H−1 := (H1)∗ = H˙−1. The space H1 is a Hilbert 0 0 0 space with the inner product h·,·i := hA1/2·,A1/2·i. More generally, we have the 1 family of inner products h·,·i :=hAs/2·,As/2·i and induced norms k·k =h·,·i1/2 s s s on H˙s. We shall use h·,·i also for the duality pairing of H˙s and H˙−s. We will frequently utilise the embeddings H1 ⊂L ⊂L with 0 6 3 (6) ckuk ≤kuk ≤Ckuk , L3 L6 1 and the resulting inequality (7) kuk ≤Ckuk . −1 L6/5 PATHWISE EXISTENCE 3 Thefirstinequalityin(6)holdsinarbitraryspatialdimensiondwhilethelatterand (7) hold for d ≤ 3. See [7, Lemma 2.5] for a proof of a finite dimensional version, the proof in our case is almost identical. We also have (8) |hu2v,wi|≤kuk2 kvk kwk , L3 L6 L6 (9) ku2vk≤kuk2 kvk and L6 L6 (10) kuvwk ≤Ckuk kvk kwk , (d≤3) −1 L6 L3 L3 where(8)and(9)holdsforarbitrarydasbeingconsequencesofHo¨lder’sinequality. The third, (10), is a consequence of (7) and Ho¨lder’s inequality. The following theorem can be found in [4, Theorem 4, Section 9.2]. Theorem 2.1 (Schaefer’s fixed point theorem). Assume that X is a real Banach space and that T: X → X is a continuous, compact mapping. If the set F = ∪ {u∈H1 :u=λTu} is bounded, then T has a fixed point. λ∈[0,1] 0 The results does not rely on any probabilistic argumentsin addition to the ones used in the introduction. We refer the reader to [2] (the Hilbert-Schmidt norm is defined in Appendix C) and [10]. 3. Proof of the main theorem To make sure that T : H1 →H1 is well-defined for every y ∈H˙−2, in fact even y 0 0 for every y ∈H˙−3, we let z =T (x) be such that for every v ∈H1, y 0 (11) hz,vi +k hz,vi +hxz,xvi =hkβ2x+A−1y,vi. −1 1 (cid:0) (cid:1) This is of the form B (z,v)= L (v) where B is an inner product and L is a x y,x x y,x bounded linear functional on H1 if x∈L and y ∈H˙−3. That B has the claimed 0 3 x domain follows from (8) and that H1 ⊂ L ⊂ L . From (8) and (6), we get that 0 6 3 kkuk2 ≤ B (u,u) ≤ λ−2+k+Ckxk2 kuk2. Thus, (H1,B ) is a Hilbert space. 1 x (cid:0) 1 L3(cid:1) 1 0 x Lemma 3.1 is then immediate from Riesz representation theorem. Lemma 3.1. If x ∈ L and y ∈ H˙−3/2 then (11) has a unique solution z ∈ H1. 3 0 In particular, T is well defined as a mapping on H1. y 0 We now let z be this solution and consider the system of equations (12) kAw =−z+y (13) Au=w−x2z+β2x. By standard elliptic theory, (12) has a unique weak solution w ∈ H as soon as y ∈ H˙−2. We then get a unique weak solution u ∈ H1 to (13) if also x ∈ L . We 0 3 leave to the reader to check that u=z. From (13) and [4, Theorem 4, Section 6.3] (14) kzkH2 ≤Ckw−x2z+β2xk≤C(cid:0)kwk+kx2zk+kxk(cid:1). Taking v = z in (11) , using the positivity of the third term in the left hand side, the self-adjointness of A1/2 and Ho¨lder’s and Cauchy’s inequalities we compute kzk2 +kkzk2+kkzxk2 =kβ2hx,zi+hA−1y,zi −1 1 =kβ2hA−1/2x,A1/2zi+hA−3/2y,A1/2zi≤ǫkkzk2+C kkxk2 +k−1kyk2 1 (cid:0) −1 −3(cid:1) whith C =C . Clearly, there is an ǫ>0 such that ǫ (15) kzk2 +kkzk2≤C kkxk2 +k−1kA−3/2yk2 . −1 1 (cid:16) −1 (cid:17) 4 PATHWISE EXISTENCE It follows from (12), the properties of A and (15) that kkwk≤kA−1zk+kA−1yk≤λ−1/2kzk +kA−1yk 1 −1 (16) ≤C k1/2kxk +k−1/2kA−3/2yk+kA−1yk . (cid:16) −1 (cid:17) From (9), (6) and (15) we also get (17) kx2zk≤kxk2 kzk ≤Ckxk2kzk ≤Ckxk2 kxk +k−1kA−3/2yk . L6 L6 1 1 1(cid:16) −1 (cid:17) Insert (16) and (17) into (14), use (5) and Young’s inequality, to find that kzkH2 ≤C(cid:16)k−1/2kxk−1+k−3/2kA−3/2yk+k−1kA−1yk+ (18) +kxk2 kxk +k−1kA−3/2yk +kxk ≤C kxk3+kA−1yk3 . 1(cid:0) −1 (cid:1) (cid:17) k(cid:0) 1 (cid:1) Compactness of T then follows from Kondrachov-Rellich’s compactness theorem y [4, Theorem 1, Section 5.7]. We have the following lemma. Lemma 3.2. If y ∈H˙−2, then T is a compact mapping from H1 to H1. y 0 0 We now want to verify that T is continuous. y Lemma 3.3. The mapping T is continuous on H1 if y ∈H˙−3. y 0 Proof. Take x and x in H˙1 and let z =T (x ) and z =T (x ). Consider these 1 2 1 y 1 2 y 2 equationsoftheform(11)andsubtractthelatterfromtheformer,usingv =z −z . 1 2 We then arrive at kz −z k2 +kkz −z k2+khz x2−z x2,z −z i 1 2 −1 1 2 1 1 1 2 2 1 2 (19) 1 =kβ2hx −x ,z −z i≤Ckx −x k2+ kz −z k2 1 2 1 2 1 2 1 2 1 2 −1 after also invoking Ho¨lder’s and Cauchy’s inequalities. Note that z x2 −z x2 = 1 1 2 2 x2(z −z )+(x −x )z (x +x ) and thus 1 1 2 1 2 2 1 2 (20) hz x2−z x2,z −z i=kx (z −z )k2+h(x −x )z (x +x ),z −z i. 1 1 2 2 1 2 1 1 2 1 2 2 1 2 1 2 Further, using Ho¨lder’s and Cauchy’s inequalities and (10) we get (21) |h(x −x )z (x +x ),z −z i|=|hA−1/2(x −x )z (x +x ),A1/2(z −z )i| 1 2 2 1 2 1 2 1 2 2 1 2 1 2 1 1 ≤ k(x −x )z (x +x )k2 + kz −z k2 2 1 2 2 1 2 −1 2 1 2 1 1 ≤Ckx −x k (kz k2 +kx k2 +kx k2 )+ kz −z k2. 1 2 L6 2 L6 1 L6 2 L6 2 1 2 1 Inserting (20) into (19), rearrangingand applying (21) we find that (22) 1 kz −z k2 +kkz −z k2+kkx (z −z )k2 2 1 2 −1 1 2 1 1 1 2 k ≤C kkx −x k (kz k2 +kx k2 +kx k2 )+kx −x k2 + kz −z k2. (cid:0) 1 2 L6 2 L6 1 L6 2 L6 1 2 1(cid:1) 2 1 2 1 Subtracting kkz −z k2 from both sides and multiplying by 2 we conclude that 2 1 2 1 kkz −z k2 ≤C kx −x k kz k2 +kx k2 +kx k2 +kx −x k2 1 2 1 (cid:16) 1 2 L6(cid:0) 2 L6 1 L6 2 L6(cid:1) 1 2 1(cid:17) REFERENCES 5 after also dropping redundant terms in the left hand side. As x , x are in H1 by 1 2 0 assumption and z is in H1 by (18), it follows from (6) that T is continuous on 2 0 y H1. (cid:3) 0 Lemma 3.4. Assume that 4kβ4 < 1 and that y ∈ Dom(A−3/2). If ζ ∈ [0,1] and u=ζT (u), then for some M >0 it must hold that kuk ≤MkA−3/2yk. y 1 Proof. Itistrivialforζ =0soassume0<ζ ≤1andwrite u =T(u)andsubstitute ζ z for u and x for u in (11) and take v =u. Then, ζ 1 kuk2 +kkuk2+kkuk4 =hy,ui +kβ2kuk2. ζ (cid:0) −1 1 L4(cid:1) −1 After multiplication with ζ and similar arguments as above we get C (ζk)2β4 kuk2 +kkuk2 ≤ kyk2 +ǫζ2kuk2+kuk2 + kuk2 :=G(ζ). −1 1 ǫ −3 1 −1 4 1 Itholdsthatsup G(ζ)=G(1)sowithζ =1weseethatundertheassumption 0<ζ≤1 on k we may pick 0<ǫ

Pathwise existence of solutions to the Implicit Euler method for the stochastic Cahn-Hilliard Equation

Similar Pathwise existence of solutions to the Implicit Euler method for the stochastic Cahn-Hilliard Equation