Non-linear noise excitation and intermittency under high disorder
NNon-linear noise excitationand intermittency under high disorder ∗ Davar KhoshnevisanUniversity of Utah Kunwoo KimUniversity of UtahLast update: March 3, 2013
Abstract
Consider the semilinear heat equation ∂ t u = ∂ x u + λσ ( u ) ξ on theinterval [0 , with Dirichlet zero boundary condition and a nice non-random initial function, where the forcing ξ is space-time white noiseand λ > denotes the level of the noise. We show that, when thesolution is intermittent [that is, when inf z | σ ( z ) /z | > ], the expected L -energy of the solution grows at least as exp { cλ } and at most as exp { cλ } as λ → ∞ . In the case that the Dirichlet boundary conditionis replaced by a Neumann boundary condition, we prove that the L -energy of the solution is in fact of sharp exponential order exp { cλ } .We show also that, for a large family of one-dimensional randomly-forced wave equations, the energy of the solution grows as exp { cλ } as λ → ∞ . Thus, we observe the surprising result that the stochasticwave equation is, quite typically, significantly less noise-excitable thanits parabolic counterparts. Keywords:
The stochastic heat equation; the stochastic wave equation;intermittency; non-linear noise excitation.
AMS 2000 subject classification:
Primary 60H15, 60H25; Secondary35R60, 60K37, 60J30, 60B15. ∗ Research supported in part by the NSF grant DMS-1006903 a r X i v : . [ m a t h . P R ] M a r Introduction
The principal aim of this paper is to study the non-linear effect of noise instochastic partial differential equations whose solutions are “intermittent.”Our analysis is motivated in part by a series of computer simulations thatwe would like to present first.Consider the following stochastic heat equation on the interval [0 , withhomogeneous Dirichlet boundary conditions: ∂∂t u t ( x ) = 12 ∂ ∂x u t ( x ) + λu t ( x ) ξ for < x < and t > ,u t (0) = u t (1) = 0 for t > ,u ( x ) = sin( πx ) for < x < . (1.1)Here, ξ denotes space-time white noise on (0 , ∞ ) × [0 , , and λ > is anarbitrary parameter that is known as the level of the noise .The stochastic partial differential equation (1.1) and its variations aresometimes referred to as parabolic Anderson models . Those are a family ofnoise-perturbed partial differential equations that arise in a diverse numberof scientific disciplines. For a representative sample see Balázs et al [1],Bertini and Cancrini [2], Carmona and Molchanov [8], Corwin [9], Cranstonand Molchanov [10], Cranston, Mountford, and Shiga [11, 12], Gärtner andKönig [17], den Hollander [15], Kardar [20], Kardar et al [21], Majda [23],Molchanov [24], and Zeldovich et al [28], together with their substantialbibliographies.Since the initial function is the principle eigenfunction of the DirichletLaplacian on [0 , , it is easy to see that the solution is exactly u t ( x ) =sin( πx ) exp {− π t/ } when λ = 0 ; this is the noise-free case, and a numericalsolution is shown is in Figure 1.Figures 2, 3, and 4 contain simulations of the solution to (1.1) for respec-tive noise levels λ = 0 . , λ = 2 , and λ = 5 .One might notice that, as λ increases, the simulated solution rapidlydevelops tall peaks that are distributed over relatively-small “islands.” Thisgeneral phenomenon is called intermittency; see Gibbons and Titi [18] fora modern general discussion of intermittency and its ramifications, as wellas connections to other important topics [in particular, to turbulence]. Ourdiscussion is motivated in part by the discussion in Blümich [3] on the quite-2igure 1: No noise ( λ = 0 . , maximum height = 1 . )Figure 2: Small noise ( λ = 0 . , maximum height ≈ . )3igure 3: Modest noise ( λ = 2 . , maximum height ≈ . )Figure 4: High noise ( λ = 5 . , maximum height ≈ . × )4ifferent topic of NMR spectroscopy.Presently, we study non-linear noise-excitation phenomena for stochasticpartial differential equations that include the parabolic Anderson model (1.1)as a principle example. Namely, let us choose and fix a length scale L > ,and then consider the stochastic partial differential equation, ∂∂t u t ( x ) = ∂ ∂x u t ( x ) + λσ ( u t ( x )) ξ for < x < L and t > ,u t (0) = u t ( L ) = 0 for t > , (1.2)where σ : R → R is a Lipschitz-continuous function, ξ denotes space-timewhite noise as before, and the initial function u : [0 , L ] → R + is a non-random bounded continuous function that is non-negative everywhere on [0 , L ] and strictly positive on a set of positive Lebesgue measure in (0 , L ) .For the sake of simplicity, we assume further that σ (0) = 0 , though some ofour work remains valid when σ (0) (cid:54) = 0 as well.It is well known that the stochastic heat equation (1.2) has an a.s.-uniquecontinuous solution that has the property that sup x ∈ [0 ,L ] sup t ∈ [0 ,T ] E (cid:16) | u t ( x ) | k (cid:17) < ∞ for all T > and k ∈ [2 , ∞ ) . (1.3)We will be interested in the effect of the level λ of the noise on the [expected]energy E t ( λ ) of the solution at time t ; the latter quantity is defined as E t ( λ ) := (cid:114) E (cid:16) (cid:107) u t (cid:107) L [0 ,L ] (cid:17) ( t > . (1.4)Throughout, we use the following notation: (cid:96) σ := inf z ∈ R \{ } (cid:12)(cid:12)(cid:12)(cid:12) σ ( z ) z (cid:12)(cid:12)(cid:12)(cid:12) , L σ := sup z ∈ R \{ } (cid:12)(cid:12)(cid:12)(cid:12) σ ( z ) z (cid:12)(cid:12)(cid:12)(cid:12) . (1.5)Clearly, (cid:54) (cid:96) σ (cid:54) L σ . Moreover, L σ < ∞ because σ is Lipschitz continuous.The following two theorems contain quantitative descriptions of the non-linear noise excitability of (1.2). These are the main findings of this paper.5 heorem 1.1. For all t > , (cid:96) σ t (cid:54) lim inf λ →∞ λ log E t ( λ ) , lim sup λ →∞ λ log E t ( λ ) (cid:54) L σ t. (1.6)In a companion paper [22] we show that exp { cλ } is a typical lowerbound for the energy of a large number of intermittent complex systems.In particular, the energy of the solution for the stochastic heat equation on [0 , L ] with a periodic boundary condition is shown to be of sharp exponentialorder exp { cλ } . The following theorem says that we can get the same kindof result when we replace a Dirichlet with a Neumann boundary condition,provided additionally that the initial profile remains bounded uniformly awayfrom zero. Theorem 1.2.
Suppose that we replace the Dirichlet boundary condition in (1.2) by a Neumann boundary condition; that is, we suppose that u solvesthe following stochastic heat equation: ∂∂t u t ( x ) = ∂ ∂x u t ( x ) + λσ ( u t ( x )) ξ for < x < L and t > ,∂u t ∂x (0) = ∂u t ∂x ( L ) = 0 for t > . (1.7) If, in addition, we assume that inf x ∈ [0 ,L ] u ( x ) > , then for every t > , (cid:96) σ t π e (cid:54) lim inf λ ↑∞ λ log E t ( λ ) (cid:54) lim sup λ ↑∞ λ log E t ( λ ) (cid:54) L σ t . (1.8)Theorems 1.1 and 1.2 together show that under fairly natural regularityconditions [that include (cid:96) σ > ], the energy behaves roughly as exp { cλ } ,which is a fastly-growing function of the level λ of the noise. In Section 4we document the somewhat surprising fact that, by contrast, the stochasticwave equation has typically an energy that grows merely as exp { cλ } . Inother words, the stochastic wave equation is typically substantially less noiseexcitable than the stochastic heat equation.6 Proof of Theorem 1.1
As is customary, we begin by writing the solution to the stochastic heatequation (1.2) in integral form [also known as the mild form ], u t ( x ) = ( P t u )( x ) + λ (cid:90) (0 ,t ) × (0 ,L ) p t − s ( x , y ) σ ( u s ( y )) ξ (d s d y ) , (2.1)where { P t } t (cid:62) denotes the semigroup of the Dirichlet Laplacian on [0 , L ] and { p t } t> denotes the corresponding heat kernel. That is, in particular, P h ≡ h for every h ∈ L ∞ [0 , L ] , and ( P t h )( x ) := (cid:90) L p t ( x , y ) h ( y ) d y for all t > and x ∈ [0 , L ] . (2.2)One can expand the heat kernel p t ( x , y ) —the fundamental solution to theDirichlet Laplacian on [0 , L ] —in terms of the eigenfunctions of the DirichletLaplacian as follows: p t ( x , y ) := ∞ (cid:88) n =1 φ n ( x ) φ n ( y )e − µ n t ; (2.3)where µ n := (cid:16) nπL (cid:17) , φ n ( x ) := (cid:18) L (cid:19) / sin (cid:16) nπxL (cid:17) , (2.4)for all n (cid:62) and (cid:54) x (cid:54) L . According to the maximum principle: (i) p t ( x , y ) > for all t > and x, y ∈ (0 , L ) ; and (ii) (cid:82) L p t ( x , y ) d x < for all y ∈ [0 , L ] and t > .Next we recall, briefly, how to one establishes the existence of a mildsolution to (1.2).Let u (0) t ( x ) := u ( x ) for all x ∈ [0 , L ] and t (cid:62) , and then define itera-tively, for all k (cid:62) , u ( k +1) t ( x ) = ( P t u )( x ) + λ (cid:90) (0 ,t ) × (0 ,L ) p t − s ( x , y ) σ ( u ( k ) s ( y )) ξ (d s d y ) . (2.5)Then the method of Walsh [27, Chapter 3]—see Dalang [13]—shows that7 u ( n ) } t (cid:62) is locally uniformly Cauchy in L (P) in the sense that for all T > , ∞ (cid:88) k =1 sup x ∈ [0 ,L ] sup t ∈ [0 ,T ] (cid:114) E (cid:16) | u ( k +1) t ( x ) − u ( k ) t ( x ) | (cid:17) < ∞ . (2.6)It follows fairly easily from this that u := lim k →∞ u ( k ) t ( x ) exists in L (P) and solves (2.5). Moreover, one can deduce that u is unique among all mildsolutions that satisfy (1.3)—see Dalang [13] for the details. Proof of the first bound in Theorem 1.1.
We follow an idea that is classicalin the context of PDEs (see Kaplan [19]); in the context of SPDEs, a non-trivial adaptation of this idea was used in Bonder and Groisman [4] in orderto compare the solution of an SPDE to the solution to a one-dimensionaldiffusion in order to describe a certain blow-up phenomenon. In a somewhatloose sense, we follow the same general outline, but need to make a numberof modifications along the way.Recall that { φ n } ∞ n =1 denote the eigenfunctions of the Dirichlet Laplacianon [0 , L ] and { µ n } ∞ n =1 are the corresponding eigenvalues. Since p t ( x , y ) = p t ( y , x ) , an appeal to a stochastic Fubini theorem [27, Theorem 2.6, p. 296]implies the following for all n (cid:62) and t > : With probability one, ( u t , φ n )= ( u , P t φ n ) + λ (cid:90) (0 ,t ) × (0 ,L ) ( P t − s φ n )( y ) σ ( u s ( y )) ξ (d s d y )= e − µ n t ( u , φ n ) + λ (cid:90) (0 ,t ) × (0 ,L ) e − µ n ( t − s ) φ n ( y ) σ ( u s ( y )) ξ (d s d y ) . (2.7)Consequently, the Walsh isometry [27, Theorem 2.5, p. 295] shows us that E (cid:2) ( u t , φ n ) (cid:3) = e − µ n t ( u , φ n ) + λ (cid:90) t d s (cid:90) L d y e − µ n ( t − s ) | φ n ( y ) | E (cid:16) | σ ( u s ( y )) | (cid:17) (cid:62) e − µ n t ( u , φ n ) + λ (cid:96) σ (cid:90) t e − µ n ( t − s ) d s (cid:90) L d y | φ n ( y ) | E (cid:16) | u s ( y ) | (cid:17) . (2.8)We may apply the Cauchy–Schwarz inequality, together with an appeal to8he Fubini theorem, in order to see that (cid:90) L | φ n ( y ) | E (cid:16) | u s ( y ) | (cid:17) d y (cid:62) L E (cid:2) ( u s , φ n ) (cid:3) . (2.9)Thus we see that, for every fixed n (cid:62) , the function F ( t ) := e µ n t E (cid:2) ( u t , φ n ) (cid:3) ( t > (2.10)satisfies the recursion F ( t ) (cid:62) ( u , φ n ) + λ (cid:96) σ (cid:90) t F ( s ) d s for all t > . (2.11)Thus the Gronwall’s inequality [or just recursion, in this case] shows that F ( t ) (cid:62) ( u , φ n ) exp (cid:0) λ (cid:96) σ t (cid:1) ( t > . (2.12)Equivalently, E (cid:2) ( u t , φ n ) (cid:3) (cid:62) ( u , φ n ) exp (cid:0)(cid:2) λ (cid:96) σ − µ n (cid:3) t (cid:1) ( t > . (2.13)Since (cid:80) ∞ n =1 ( u t , φ n ) (cid:54) (cid:107) u t (cid:107) L [0 ,L ] , thanks to Bessel’s inequality, it remainsto prove that the following is strictly positive: c t := ∞ (cid:88) n =1 ( u , φ n ) e − µ n t for all t (cid:62) . (2.14)But c t > for all t (cid:62) simply because we can write c t = (cid:107) P t u (cid:107) L [0 ,L ] , andthis expression is strictly positive, thanks to the maximum principle and thefact that u > on a set of positive measure. This concludes the proof ofthe lower bound on the energy in Theorem 1.1.Our derivation of the upper bound of Theorem 1.1 requires a few ele-mentary preparatory lemmas. Lemma 2.1.
For all β, t > : sup x ∈ [0 ,L ] (cid:90) t e − β ( t − s ) d s (cid:90) L d y [ p t − s ( x , y )] (cid:54) √ β ; (2.15)9 nd (cid:90) t e − β ( t − s ) d s (cid:90) L d x (cid:90) L d y [ p t − s ( x , y )] (cid:54) L √ β . (2.16) Proof.
Because | φ n ( x ) | (cid:54) (2 /L ) / for all x ∈ [0 , L ] and n (cid:62) , (cid:90) t e − β ( t − s ) d s (cid:90) L d y [ p t − s ( x , y )] = (cid:90) t ∞ (cid:88) n =1 [ φ n ( x )] e − β + µ n )( t − s ) d s (cid:54) L ∞ (cid:88) n =1 β + ( nπ/L ) . (2.17)The first bound follows because ∞ (cid:88) n =1 β + ( nπ/L ) (cid:54) (cid:90) ∞ d xβ + ( xπ/L ) = L √ β , (2.18)for all β > . In order to obtain the second bound, we integrate (2.17) [d x ] ,using the fact that (cid:107) φ n (cid:107) L [0 ,L ] = 1 .We are now ready to establish the energy upper bound in Theorem 1.1. Proof of the second bound in Theorem 1.1.
We appeal to the nonlinear renewal-theoretic technique of Foondun and Khoshnevisan [16] in order to establishthe following a priori estimate: For every δ ∈ (0 , , E ( | u t ( x ) | m ) (cid:54) δ − m/ (cid:107) u (cid:107) mL ∞ [0 ,L ] exp (cid:32) tm (cid:18) λ L σ − δ (cid:19) (cid:33) , (2.19)valid uniformly for all m ∈ [2 , ∞ ) , x ∈ [0 , L ] , and t (cid:62) . Once this isestablished, then we set m = 2 in (2.19), and then integrate [d x ] , in order todeduce the upper bound of the theorem.Since L σ < ∞ and σ (0) = 0 , it follows that | σ ( z ) | (cid:54) L σ | z | for all z ∈ [0 , L ] . (2.20)Recall that u ( k ) denotes the k th-stage Picard iteration approximation to u .According to the Burkholder–Davis–Gundy inequality (see Burkholder [5],Burkholder et al [6], and Burkholder and Gundy [7]; see also Foondun and10hoshnevisan [16] for an explanation of these particular constants), (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:90) (0 ,t ) × (0 ,L ) Z s ( y ) ξ (d s d y ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) m (cid:54) m (cid:90) t d s (cid:90) L d y (cid:107) Z s ( y ) (cid:107) m , (2.21)for all predictable random fields Z , all t > , and m ∈ [2 , ∞ ) , where (cid:107) Q (cid:107) m := { E( | Q | m ) } /m for all random variables Q . Therefore, we can deduce from(2.5) that (cid:13)(cid:13)(cid:13) u ( k +1) t ( x ) (cid:13)(cid:13)(cid:13) m (cid:54) | ( P t u )( x ) | + λ (cid:115) m (cid:90) t d s (cid:90) L d y [ p t − s ( x , y )] (cid:13)(cid:13)(cid:13) σ ( u ( k ) s ( y )) (cid:13)(cid:13)(cid:13) m (cid:54) (cid:107) u (cid:107) L ∞ [0 ,L ] + λ (cid:115) m L σ (cid:90) t d s (cid:90) L d y [ p t − s ( x , y )] (cid:13)(cid:13)(cid:13) u ( k ) s ( y ) (cid:13)(cid:13)(cid:13) m . (2.22)Define N k ( β ) := sup t (cid:62) sup x ∈ [0 ,L ] (cid:18) e − βt (cid:13)(cid:13)(cid:13) u ( k ) t ( x ) (cid:13)(cid:13)(cid:13) m (cid:19) / ( k (cid:62) , β > . (2.23)The preceding and Lemma 2.1 together show that (cid:90) t d s (cid:90) L d y [ p t − s ( x , y )] (cid:13)(cid:13)(cid:13) u ( k ) s ( y ) (cid:13)(cid:13)(cid:13) m (cid:54) [ N k ( β )] e βt √ β . (2.24)Therefore, we combine our efforts in order to deduce the bound (cid:13)(cid:13)(cid:13) u ( k +1) t ( x ) (cid:13)(cid:13)(cid:13) m (cid:54) (cid:107) u (cid:107) L ∞ [0 ,L ] + √ m λ L σ e βt/ β − / N k ( β ) . (2.25)Since the right-hand side is independent of x ∈ [0 , L ] and depends on t only through the term exp( βt/ , we can divide both sides by exp( βt/ andoptimize over t and x in order to deduce the recursive inequality, N k +1 ( β ) (cid:54) (cid:107) u (cid:107) L ∞ [0 ,L ] + λ L σ √ mβ / N k ( β ) . (2.26)11o far, β was an auxiliary positive parameter. We now fix it at the value β ∗ := 4 m ( λ L σ ) (1 − δ ) , (2.27)where δ ∈ (0 , is arbitrary but fixed. For this particular choice, N k +1 ( β ∗ ) (cid:54) (cid:107) u (cid:107) L ∞ [0 ,L ] + (1 − δ ) N k ( β ∗ ) ( k (cid:62) . (2.28)Because N ( β ∗ ) = (cid:107) u (cid:107) L ∞ [0 ,L ] , the preceding iteration yields sup k (cid:62) N k ( β ∗ ) (cid:54) (cid:107) u (cid:107) L ∞ [0 ,L ] ∞ (cid:88) j =0 (1 − δ ) j = δ − (cid:107) u (cid:107) L ∞ [0 ,L ] . (2.29)This and Fatou’s lemma together imply that sup t (cid:62) sup x ∈ [0 ,L ] (cid:16) e − β ∗ t (cid:107) u t ( x ) (cid:107) m (cid:17) / (cid:54) δ − (cid:107) u (cid:107) L ∞ [0 ,L ] , (2.30)which, in turn, yields (2.19). Let us now consider the following stochastic heat equation on the interval [0 , L ] , subject to a Neumann boundary condition: ∂∂t u t ( x ) = ∂ ∂x u t ( x ) + λσ ( u t ( x )) ξ for < x < L and t > ,∂u t ∂x (0) = ∂u t ∂x ( L ) = 0 for t > , (3.1)where ξ denotes space-time white noise as before, and the assumptions on σ and u ( x ) are also the same as for the Dirichlet problem (1.2). For the sakeof simplicity, we assume further that σ (0) = 0 , though some of our workremains valid when σ (0) (cid:54) = 0 as well. We may assume that (cid:96) σ > . (3.2)12therwise, the more interesting part of Theorem 1.2, that is the lower boundon the energy, has no content. [The proof of the upper bound will not require(3.2).] Also, let us define ε := inf x ∈ [0 ,L ] u ( x ) . (3.3)According to the hypotheses of Theorem 1.2, ε > .As we did for the Dirichlet problem, we begin by writing the solution tothe stochastic heat equation (3.1) in integral form. Namely, we write u t ( x ) = ( P t u )( x ) + λ (cid:90) (0 ,t ) × (0 ,L ) p t − s ( x , y ) σ ( u s ( y )) ξ (d s d y ) , (3.4)where { P t } t (cid:62) denotes the semigroup of the Neumann Laplacian on [0 , L ] and { p t } t> denotes the corresponding heat kernel. That is, P h ≡ h forevery h ∈ L ∞ [0 , L ] , and ( P t h )( x ) := (cid:90) L p t ( x , y ) h ( y ) d y for all t > and x ∈ [0 , L ] . (3.5)It is well known that we can write the heat kernel for { P t } t (cid:62) in the followingform: p t ( x , y ) := ∞ (cid:88) n = −∞ [Γ t ( x − y − nL ) + Γ t ( x + y − nL )] , (3.6)where Γ denotes the fundamental solution to the heat equation in R ; thatis, Γ t ( z ) := 1 √ πt exp (cid:18) − z t (cid:19) ( t > , z ∈ R ) . (3.7)[This requires an application of the method of images.]The existence and uniqueness of a solution of (3.1) is shown in Walsh[27, Chapter 3]. The following establishes the lower bound of Theorem 1.2. Proposition 3.1 (An energy inequality) . For all t > , there exists a finiteand positive constant K —independent of λ —such that E t ( λ ) (cid:62) K exp (cid:18) ( (cid:96) σ λ ) t π e (cid:19) , (3.8) simultaneously for every λ (cid:62) . roof of Proposition 3.1. Owing to (3.6), the Neumann heat kernel is conser-vative; that is, (cid:82) L p t ( x , y ) d y = 1 . Therefore, we apply the Walsh isometryto find that for all fixed ε, t > and x ∈ [0 , L ] , E (cid:16) | u t ( x ) | (cid:17) (cid:62) ε + λ (cid:90) t d s (cid:90) L d y [ p t − s ( x , y )] E (cid:16) | σ ( u s ( y )) | (cid:17) (cid:62) ε + ( λ(cid:96) σ ) (cid:90) t d s (cid:90) L d y [ p t − s ( x , y )] E (cid:16) | u s ( y ) | (cid:17) . The preceding is a recursive, nonlinear renewal-type, inequality for thefunction ( t , x ) (cid:55)→ E( | u t ( x ) | ) . Even though this renewal inequation is thelower bound’s counterpart to (4.11), we cannot solve this inequation usingthe renewal-theoretic ideas of Foondun and Khoshnevisan [16]. The reasonfor this is that the method of [16] works for large values of the time variable t ; whereas here we have t fixed, but λ large. Instead, we appeal to old ideasof localization of multiple integrals [i.e., classical large deviations] that areprobably due to P.-S. Laplace, though we could not find explicit references.Namely, we expand the self-referential inequality (4.17) and observe thatmost of the contribution to the resulting multiple integrals occur in a smallportion of the space of integration.Now we carry out our program by first expanding, to one term, ourrenewal inequality as follows: E (cid:16) | u t ( x ) | (cid:17) (cid:62) ε + ε ( λ(cid:96) σ ) (cid:90) t d s (cid:90) L d y [ p t − s ( x , y )] (3.9) + ( λ(cid:96) σ ) (cid:90) t d s (cid:90) L d y [ p t − s ( x , y )] (cid:90) s d r (cid:90) L d z [ p s − r ( y , z )] E (cid:16) | u r ( z ) | (cid:17) , and proceed. In order to see better what is happening, let us introduce afamily of linear operators P as follows: For all t > , x ∈ [0 , L ] , and allBorel-measurable functions ψ : (0 , ∞ ) × [0 , L ] → R + , ( P ψ )( t , x ) := (cid:90) t d s (cid:90) L d y [ p t − s ( x , y )] ψ ( s , y ) . (3.10)Also define ( t , x ) := 1 for all t > and x ∈ [0 , L ] . Then our recursion can14e written compactly as follows: E (cid:16) | u t ( x ) | (cid:17) (cid:62) ε ∞ (cid:88) k =0 ( λ(cid:96) σ ) k ( P k )( t , x ); (3.11)where P := and P k +1 := P k P for all k (cid:62) . We integrate (3.11) [d x ] inorder to obtain the following key lower bound: | E t ( λ ) | (cid:62) ε ∞ (cid:88) k =0 ( λ(cid:96) σ ) k (cid:90) L ( P k )( t , x ) d x. (3.12)We examine the preceding sum, term by term, using induction.The first term in the sum is trivial; viz., (cid:90) L ( P )( t , x ) d x = L. (3.13)Therefore, let us consider, as a first step, the second term in the sum, whichis straight forward: (cid:90) L ( P )( t , x ) d x = (cid:90) t d s (cid:90) L d x (cid:90) L d y [ p t − s ( x , y )] := Φ( t ) . (3.14)Before we estimate the third term of the infinite sum in (3.12), let us observethat (cid:90) L ( P )( t , x ) d x (3.15) = (cid:90) t d s (cid:90) L d x (cid:90) L d y [ p t − s ( x , y )] (cid:90) s d r (cid:90) L d z [ p s − r ( y , z )] . (cid:90) L ( P )( t , x ) d x (3.16) (cid:62) (cid:90) tt/ d s (cid:90) L d x (cid:90) L d y [ p t − s ( x , y )] (cid:90) s d r (cid:90) L d z [ p s − r ( y , z )] (cid:62) (cid:90) tt/ d s (cid:90) L d x (cid:90) L d y [ p t − s ( x , y )] (cid:90) ss − t/ d r (cid:90) L d z [ p s − r ( y , z )] (cid:62) (cid:90) tt/ d s (cid:90) L d x (cid:90) L d y [ p t − s ( x , y )] (cid:90) t/ d r (cid:90) L d z [ p r ( y , z )] = (cid:90) t/ d s (cid:90) t/ d r (cid:90) L d x (cid:90) L d y [ p s ( x , y )] (cid:90) L d z [ p r ( y , z )] . It is not important that we have a t/ and in the bounds of our integrals onthe second line; t/ e would have worked equally well for us. The key pointis that most of the contribution to our multiple integral comes about where s ≈ t and r ≈ s in the second line.Now, let us recall that p r ( y , z ) can be thought of as the transition func-tion for Brownian motion in (0 , L ) , reflected upon reaching the boundary { , L } [26]. The Markov property of reflected Brownian motion impliesthe semigroup property of { P t } t (cid:62) and in particular that (cid:82) L [ p r ( y , z )] d z = p r ( y , y ) . And by symmetry, (cid:82) L [ p s ( x , y )] d x = p s ( y , y ) , as well. Conse-quently, we obtain (cid:90) L ( P )( t , x ) d x (cid:62) (cid:90) L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) t/ [ p s ( y , y )] d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d y. (3.17)Once we understand this, we can see how to bound the remaining terms inthe infinite sum in (3.12) as well. In fact, induction shows that (cid:90) L ( P k +1 )( t , x ) d x (cid:62) (cid:90) L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) t/ ( k +1)0 [ p s ( y , y )] d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k +1 d y, (3.18)for all k (cid:62) . Once again we stress that our multiple integrals are boundedfrom below by a very small portion of the region of integration.16inally, we apply Jensen’s inequality in order to see that (cid:90) L ( P k +1 )( t , x ) d x (cid:62) L − k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) L d y (cid:90) t/ ( k +1)0 d s [ p s ( y , y )] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k +1 , (3.19)for all k (cid:62) . Recall (3.14) in order to see that (cid:90) L ( P k )( t , x ) d x (cid:62) L (cid:20) Φ( t/k ) L (cid:21) k , (3.20)for all k (cid:62) and t > . This and (3.12) together yield | E t ( λ ) | (cid:62) ε L · ∞ (cid:88) k =1 (cid:20) ( λ(cid:96) σ ) Φ( t/k ) L (cid:21) k . (3.21)Finally, we estimate the function Φ from below.For all τ > , Φ( τ ) = (cid:90) τ d s (cid:90) L d x (cid:90) L d y [ p s ( x , y )] = (cid:90) τ d s (cid:90) L d x p s ( x , x ) . (3.22)Since p s ( x , x ) (cid:62) Γ s (0) = (8 πs ) − / for all s > —see (3.6)—the precedingyields the pointwise bound Φ( τ ) (cid:62) (cid:90) τ L √ πs d s = L (cid:114) τ π ( τ > . (3.23)Thus, we obtain the following: | E t ( λ ) | (cid:62) ε L · ∞ (cid:88) k =1 (cid:20) ( λ(cid:96) σ ) √ t √ πk (cid:21) k (cid:62) ε L · ∞ (cid:88) j =1 (cid:20) ( λ(cid:96) σ ) √ t √ πj (cid:21) j (cid:62) ε L · ∞ (cid:88) j =1 (cid:20) ( λ(cid:96) σ ) t π e (cid:21) j j ! , (3.24)17hanks to the elementary bound, ( j/ e) j (cid:54) j ! , valid for all integers j (cid:62) .The proposition follows from the preceding and the Taylor series expansionof the exponential function.Before we prove the upper bound in Theorem 1.2, let us state a simpleresult regarding the Neumann heat kernel. Lemma 3.2.
For every ε > there exists a positive and finite constant K := K ε,L such that sup t (cid:62) sup (cid:54) x (cid:54) L (cid:90) t e − βs p s ( x , x ) d s (cid:54) ε √ β for all β (cid:62) K. (3.25) Proof.
In accord with (3.6), for every s > and x ∈ [0 , L ] , p s ( x , x ) = ∞ (cid:88) n = −∞ [Γ s (2 nL ) + Γ s (2 x − nL )] (cid:54) s (0) + 2 ∞ (cid:88) n =1 Γ s (2 nL ) + (cid:88) n ∈ Z : | nL − x | (cid:62) Γ s (2 | nL − x | ) (cid:54) s (0) + C ( L )= 3 √ πs + C ( L ) , (3.26)uniformly for all s > and x ∈ [0 , L ] , where C ( L ) is a positive and finiteconstant that depends only on L . The numerical bound of [in front of Γ s (0) ] accounts for the fact that, depending on the value of x , there canbe at most two choices of n ∈ Z such that | nL − x | < . We integrate thepreceding in order to find that (cid:90) ∞ e − βs p s ( x , x ) d s (cid:54) √ β + C ( L ) β , (3.27)which has the desired effect.We need only to prove the upper bound in Theorem 1.2; the correspond-ing lower bound has already been established. Proof of the second bound in Theorem 1.2.
According to the Walsh isome-18ry, for all t > (cid:0) | u t ( x ) | (cid:1) (3.28) = | ( P t u )( x ) | + λ (cid:90) t d s (cid:90) L d y [ p t − s ( x , y )] E (cid:16) | σ ( u s ( y )) | (cid:17) (cid:54) (cid:107) u (cid:107) L ∞ [0 ,L ] + ( λ L σ ) (cid:90) t d s (cid:90) L d y [ p t − s ( x , y )] E (cid:16) | u s ( y ) | (cid:17) . We solve this inequality by applying the method of Foondun and Khosh-nevisan [16]; namely, let us define, for all β > , N ( β ) := sup t (cid:62) sup (cid:54) x (cid:54) L (cid:104) e − βt E (cid:0) | u t ( x ) | (cid:1)(cid:105) . (3.29)The preceding and Lemma 3.2 together imply that for all ε > there exists K := K ε,L such that for all β (cid:62) K , N ( β ) (cid:54) (cid:107) u (cid:107) L ∞ [0 ,L ] + N ( β )( λ L σ ) sup t (cid:62) sup (cid:54) x (cid:54) L (cid:90) t e − βs [ p s ( x , x )] d s (cid:54) (cid:107) u (cid:107) L ∞ [0 ,L ] + (3 + ε )( λ L σ ) √ β N ( β ) . (3.30)We let < δ < and choose β as β ∗ := (3 + ε ) ( λ L σ ) − δ ) , (3.31)in order to deduce the inequality N ( β ∗ ) (cid:54) δ (cid:107) u (cid:107) L ∞ [0 ,L ] , (3.32)valid as long as λ is large enough to ensure that β ∗ (cid:62) K . In this way wefind that E (cid:0) | u t ( x ) | (cid:1) (cid:54) δ (cid:107) u (cid:107) L ∞ [0 ,L ] exp (cid:18) (3 + ε ) ( λ L σ ) t − δ ) (cid:19) . (3.33)We integrate both sides of (3.33) [ d x ] in order to obtain the upper bound,after some relabeling. 19 A stochastic wave equation
In this section we consider the non-linear stochastic wave equation ∂ ∂t w t ( x ) = ∂ ∂x w t ( x ) + λσ ( w t ( x )) ξ for x ∈ R and t > , (4.1)subject to non-random initial function w ( x ) ≡ and non-random non-negative initial velocity v ∈ L ( R ) ∩ L ( R ) such that (cid:107) v (cid:107) L ( R ) > . It iswell known (see Dalang [13] for the general theory, as well as Dalang andMueller [14], for the existence of an L -valued solution) that there exists aunique continuous solution w to (4.1) that satisfies the moment conditions, sup t ∈ [0 ,T ] sup x ∈ R E (cid:16) | w t ( x ) | k (cid:17) + sup t ∈ [0 ,T ] E t ( λ ) < ∞ , (4.2)for every k ∈ [2 , ∞ ) and T > .The main result of this section are the following bounds on the energy E t ( λ ) := { E( (cid:107) u t (cid:107) L ( R ) ) } / of the solution to (4.1). Theorem 4.1.
For every t > , (cid:96) σ t √ (cid:54) lim inf λ →∞ λ log E t ( λ ) (cid:54) lim sup λ →∞ λ log E t ( λ ) (cid:54) L σ t √ . (4.3)Define H ( r ) := r ∧ r r > . (4.4)The preceding theorem requires the following elementary real-variable fact. Lemma 4.2.
Suppose g is a non-negative element of L ( R ) ∩ L ( R ) . Then,there exist positive and finite constants A and A —depending only on (cid:107) g (cid:107) L ( R ) and (cid:107) g (cid:107) L ( R ) —such that A H ( t ) (cid:54) (cid:90) t − t d z (cid:90) t − t d y ( g ∗ ˜ g ) ( y − z ) (cid:54) A H ( t ) , (4.5) for all t > , where ˜ g ( x ) := g ( − x ) for all x ∈ R .Proof. Define h := g ∗ ˜ g for simplicity, and note that: (i) h (cid:62) ; (ii) h ∈ L ( R ) ∩ L ( R ) with (cid:107) h (cid:107) L ( R ) = (cid:107) g (cid:107) L ( R ) and h (0) = (cid:107) g (cid:107) L ( R ) . Further-more, since h ( x ) is maximized at x = 0 , thanks to well-known facts about20ontinuous, positive-definite functions. Now we put these facts together inorder to see that (cid:90) t − t d y (cid:90) t − t d z h ( y − z ) (cid:54) h (0) t ∧ t (cid:107) h (cid:107) L ( R ) (cid:54) (cid:16) (cid:107) g (cid:107) L ( R ) ∧ (cid:107) g (cid:107) L ( R ) (cid:17) H ( t ) . (4.6)This proves the upper bound [with an explicit A ].On the other hand, (cid:82) t − t d y (cid:82) t − t d z h ( y − z ) = (4 + o (1)) t h (0) as t ↓ and (cid:82) t − t d y (cid:82) t − t d z h ( y − z ) = (2 + o (1)) t (cid:107) h (cid:107) L ( R ) as t ↑ ∞ . The lower boundfollows from these observations, since h is non-negative. Proof of Theorem 4.1.
The solution to the stochastic wave equation (4.1)can be written in mild form as follows: w t ( x ) = W t ( x ) + λ (cid:90) (0 ,t ) × R [0 ,t − s ] ( | x − y | ) σ ( w s ( y )) ξ (d s d y ) , (4.7)where W t ( x ) := (cid:90) t − t v ( x − y ) d y. (4.8)Therefore, the Walsh isometry for stochastic integrals assures us that E (cid:0) | w t ( x ) | (cid:1) = | W t ( x ) | + λ (cid:90) t d s (cid:90) ∞−∞ d y [0 ,t − s ] ( | x − y | )E (cid:0) | σ ( w s ( y )) | (cid:1) , (4.9)whence by Fubini’s theorem, | E t ( λ ) | = (cid:107) W t (cid:107) L ( R ) + λ (cid:90) t ( t − s ) d s (cid:90) ∞−∞ d y E (cid:0) | σ ( w s ( y )) | (cid:1) (cid:54) (cid:107) W t (cid:107) L ( R ) + λ L σ (cid:90) t ( t − s ) d s (cid:90) ∞−∞ d y E (cid:0) | w s ( y ) | (cid:1) . (4.10)Since (cid:107) W t (cid:107) L ( R ) = (cid:82) t − t d y (cid:82) t − t d z ( v ∗ ˜ v )( y − z ) , Lemma 4.2 ensures thatthe squared energy | E t ( λ ) | of the solution to the stochastic wave equation21atisfies the renewal inequality, | E t ( λ ) | (cid:54) A t (cid:107) v (cid:107) L ( R ) + λ L σ (cid:90) t ( t − s ) ( E s ( λ )) d s, (4.11)for all t > .Since sup t (cid:62) (cid:104) t e − βt (cid:105) = 4(e β ) , (4.12)the preceding implies that F ( β ) (cid:54) A (e β ) (cid:107) v (cid:107) L [0 ,L ] + λ L σ F ( β ) (cid:90) t ( t − s )e − β ( t − s ) d s (cid:54) A (cid:107) v (cid:107) L [0 ,L ] (e β ) + λ L σ F ( β ) (cid:90) ∞ s e − βs d s (cid:54) A (cid:107) v (cid:107) L [0 ,L ] (e β ) + λ L σ F ( β )2 β . (4.13)Let us choose and fix an arbitrary δ ∈ (0 , . We can apply an a priori method in order to show that E ( β ∗ ) < ∞ , where β ∗ := λ L σ (cid:112) − δ ) . (4.14)See the derivation of (2.19); we omit the details. In this way, we can solve(4.13) in order to deduce the following: F ( β ∗ ) (cid:54) A (cid:107) v (cid:107) L [0 ,L ] δ (e λ L σ ) . (4.15)Equivalently, | E t ( λ ) | (cid:54) A (cid:107) v (cid:107) L [0 ,L ] δ (e λ L σ ) exp (cid:32) λ L σ t (cid:112) − δ ) (cid:33) . (4.16)This readily yields the upper bound of the theorem.We prove the corresponding lower bound by observing the following coun-22erpart of (4.11), which holds for the same reasons as (4.11) does: | E t ( λ ) | (cid:62) A H ( t ) (cid:107) v (cid:107) L [0 ,L ] + λ (cid:96) σ (cid:90) t ( t − s ) | E s ( λ ) | d s. (4.17)Now we proceed as we did in the proof of Proposition 3.1; we expand ourrenewal inequation as an inequality in terms of an infinite sum of multipleintegrals of increasingly-high powers. And then show that the multiple in-tegrals are large, when λ (cid:29) , mainly because of the contribution of theintegrand in a small region of integration. In this way the remainder of theproof is exactly the same as the derivation of Proposition 3.1. However, asit turns out, one has to be quite careful in order to guess the correct regionof integration, as it will be very significantly larger than the one in the proofProposition 3.1.With the preceding in mind, we begin by writing | E t ( λ ) | (cid:62) A H ( t ) (cid:107) v (cid:107) L [0 ,L ] + λ (cid:96) σ (cid:90) t ( t − s ) ( E s ( λ )) d s (cid:62) A H ( t ) (cid:107) v (cid:107) L [0 ,L ] + A λ (cid:96) σ (cid:107) v (cid:107) L [0 ,L ] (cid:90) t ( t − s ) H ( s ) d s + (cid:0) λ (cid:96) σ (cid:1) (cid:90) t d s (cid:90) s d r ( t − s )( s − r ) | E r ( λ ) | , (4.18)etc. In this way, we obtain the following generous lower bound, | E t ( λ ) | (4.19) (cid:62) A (cid:107) v (cid:107) L [0 ,L ] ∞ (cid:88) n =1 (cid:0) λ (cid:96) σ (cid:1) n (cid:90) tt/ d s (cid:90) s s / d s · · · (cid:90) s n − s n − / d s n S , where S := ( t − s ) × ( s − s ) × · · · × ( s n − − s n ) × H ( t ) , over the range ofthe integral in (4.19). Let us emphasize that the n th term involves an n -foldintegral that is integrate on a large part of the original region of integration;this is in sharp contrast with the proof of Proposition 3.1, where the effectiveregion of integration was extremely small for the n th term when n (cid:29) . Oncethe correct region of integration is identified, we can continue the argumentin the proof of Proposition 3.1. Namely, we obtain the following bounds,23fter we appeal to time reversal: | E t ( λ ) | (cid:62) A (cid:107) v (cid:107) L [0 ,L ] H ( t ) · ∞ (cid:88) n =1 (cid:0) λ (cid:96) σ (cid:1) n × (cid:90) t/ d s (cid:90) s / d s · · · (cid:90) s n − / d s n n (cid:89) j =1 s j . (4.20)A change of variables and induction together yield (cid:90) t/ d s (cid:90) s / d s · · · (cid:90) s n − / d s n n (cid:89) j =1 s j = 4 − n (cid:90) t d s (cid:90) s d s · · · (cid:90) s n − d s n n (cid:89) j =1 s j = t n n · (2 n )!! , (4.21)where (2 n )!! := (2 n ) × (2 n − × · · · × denotes the usual double factorialof n . We require only the simple bound (2 n )!! (cid:54) (2 n )! in order to deducethat | E t ( λ ) | (cid:62) A (cid:107) v (cid:107) L [0 ,L ] H ( t ) · ∞ (cid:88) n =1 (cid:18) λ(cid:96) σ t √ (cid:19) n n )! . (4.22)Since ∞ (cid:88) n =1 (cid:18) λ(cid:96) σ t √ (cid:19) n +1 n + 1)! (cid:54) ∞ (cid:88) n =1 (cid:18) λ(cid:96) σ t √ (cid:19) n n )! , (4.23)for λ (cid:62) √ / ( t(cid:96) σ ) . It follows that | E t ( λ ) | (cid:62) A (cid:107) v (cid:107) L [0 ,L ] H ( t ) · ∞ (cid:88) j =2 (cid:18) λ(cid:96) σ t √ (cid:19) j j ! , (4.24)whenever λ (cid:62) √ / ( t(cid:96) σ ) . 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The AlmightyChance , World Scientific, Singapore, 1990.
Davar Khoshnevisan & Kunwoo Kim
Department of Mathematics, University of Utah, Salt Lake City, UT 84112-0090
Emails & URLs : [email protected] ∼ davar/ [email protected] ∼ kkim/kkim/