Stochastic traveling wave solution to stochastic generalized KPP … · 2017-08-29 · Nonlinear Diﬀer. Equ. Appl. 22 (2015), 143–173 c 2014 Springer Basel 1021-9722/15/010143-31

Nonlinear Differ. Equ. Appl. 22 (2015), 143–173c© 2014 Springer Basel1021-9722/15/010143-31published online June 26, 2014DOI 10.1007/s00030-014-0279-9

Nonlinear Differential Equationsand Applications NoDEA

Stochastic traveling wave solutionto stochastic generalized KPP equation

Zhehao Huang and Zhengrong Liu

Abstract. In this paper, we consider a stochastic generalized KPP equa-tion driven by a white noise term. Denote u the solution to the equationwith Heaviside initial condition u0(x) = χ(−∞,0](x). Choosing a suit-able marker of wavefront R(t), we prove that u(t, ·+R(t)) is a stationaryprocess and limt→∞ R(t)/t exists almost surely, which verify the existenceof stochastic traveling wave solution to the equation.

Mathematics Subject Classification (2010). Primary 99Z99;Secondary 00A00.

Keywords. Wavefront, Stationary process, Stochastic traveling wave solu-tion, Stochastic generalized KPP equation.

1. Introduction

The Kolmogorov–Petrovsky–Piscunov (KPP) equation [13,20]

∂

∂tu =

D

2uxx +Ru(K − u), x ∈ R, t ≥ 0, (1.1)

where R > 0 is the reproduction rate, K > 0 the carrying capacity, and D > 0the diffusion coefficient, provides a deterministic model for the density u ofa population living in an environment with a limited carrying capacity. Thenormalization of (1.1),

∂

∂tu = uxx + u(1 − u), (1.2)

is perhaps the simplest equation which possesses the traveling wave solutions.In fact, (1.2) has a solution u(t, x) = φc(x − ct) for any c ≥ 2, where thefunction φc(ξ) converges exponentially to 1 as ξ → −∞ and to 0 as ξ → ∞.In the classical paper [20], Kolmogorov et al. proved that if u is the solution

Supported by the National Natural Science Foundation (No.11171115).

144 Z. Huang and Z. Liu NoDEA

to (1.2) with Heaviside initial condition, u(0, x) = χ(−∞,0](x), then there is afunction m(t), t > 0 with the property

limt→∞

m(t)t

= 2 (1.3)

and

limt→∞ sup

ξ∈R|u(ξ +m(t), t) − φ2(ξ)| = 0. (1.4)

Many authors [2,4,24,37,45] have paid attention to (1.1) and extended theresults of [13,20] to more general equations, especially the delayed versions[3,15,36,46].

A system in reality is usually of uncertainty due to some external noise,which is random. The random effects are considered not only as compensationsfor the detects in some deterministic models, but also rather essential phenom-ena [1,5,6,10,18,26–28,47]. Therefore, one should make the model (1.1) morerealistic by considering environmental noise. As [10,12,34,35], the carryingcapacity is supposed to be stochastic and is given as the form depending ontime, K(t) = c0 + k(t)ζt, where c0 > 0 and ζt is the white noise. SubstitutingK into (1.1) gives a stochastic partial differential equation

du(t, x) =(D

2uxx(t, x) +Ru(t, x)(c0 − u(t, x))

)dt+ k(t)u(t, x)dWt, (1.5)

where Wt is the Brownian motion whose distributional time derivative is ζt.There are other ways introducing environmental noise in (1.5). In the spatiallyhomogeneous case, D = 0, different versions were discussed and compared in[14,21,23,32,33,41]. The case when the carrying capacity K > 0 is constantand R = r0 +kζt has been analysed in [25]. Some works discussed whether theIto or Stratonovich interpretation of the equation is most appropriate.

In [39], the authors considered the KPP equation driven by a particularspace-time white noise term⎧⎨

⎩∂

∂tu = uxx + θu− u2 +

√uζt,x,

u(0, x) = 1 ∧ (−x ∨ 0).(1.6)

The form of the noise term in (1.6) arises from particle branching in a par-ticle approximation. The same term which appears in the stochastic partialdifferential equation describes the density of one dimensional super Brown-ian motion [22]. It has been shown that (1.6) arises as the high densitylimit of particle systems which undergo branching random walks and havean extra death mechanism due to overcrowding. The authors proved thatthere exist stochastic wavelike solutions which travel with a linear speed. Infact, they looked for solutions u to (1.6) which satisfy that the wavefrontR0(u(t)) := sup{x ∈ R : u(t, x) > 0} ∈ (−∞,∞) for all t ≥ 0, u(t, ·+R0(u(t)))is a stationary process in time and limt→∞R0(u(t))/t exists almost surely. Thework related to the compact support property of solutions to the equation [31].Such a solution is called a stochastic traveling wave solution.

Vol. 22 (2015) Stochastic traveling wave solution 145

In [7,29,30], the authors considered another version of stochastic KPPequation ⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

∂

∂tu = uxx + θu− u2 + ε

√u(1 − u)ζt,x,

0 ≤ u(0, x) ≤ 1,

u(0, x) = 1, x < −n,u(0, x) = 0, x > n,

(1.7)

where n > 0 is a constant, ζt,x is also a two-parameter white noise. Note that(1.7) has two different trivial equilibrium solutions, u = 0 and u = 1, whichis different from (1.6). In [29], the authors showed that for ε small enough,limt→∞R0(u(t))/t exists and lies in (0,∞). The limit depends only on x. Thelaw of u(t, · + R0(u(t))) tends to a stationary limit as t → ∞. The authorsalso analyzed the region which is the smallest closed interval containing thepoints x at which 0 < u(t, x) < 1. They showed that the length of this regiontends to a stationary distribution which implies that the wavefront does notdegenerate. In [7], the authors were interested in the behavior of wave speedc(ε) as a function of parameter ε. They proved that for large ε, c(ε) satisfies theinequality lim infε→∞ ε2c(ε) ≥ 2. For small ε ≤ 1/10, it satisfies the inequality2 − N ln ln(1/ε)/(ln ε)2 ≤ c(ε) ≤ 2, where N is a universal constant. In [30],the authors proved that the speed of wavefronts for small ε is 2 − π2| ln ε2|−2

with an error of order (ln | ln ε|)| ln ε|−3.In [38], from a point of random dynamical system, the authors considered

the reaction-diffusion equation in random media

∂

∂tu = uxx + F (θt,xω, u), (1.8)

including time and/or space recurrent, almost periodic, quasi-periodic, peri-odic ones as special cases, where ω ∈ Ω. (Ω,F ,P) is a probability space.(Ω,F ,P, {θt,x}t,x∈R) is a metric dynamical system. F : Ω × R → R is mea-surable. For each ω ∈ Ω, F (θt,xω, u) is Holder continuous in t, x and Lipschitzcontinuous in u. The authors extended the classical notion of traveling wavesolution to traveling wave solution in general random media and adopted apoint of view that random traveling wave solutions are limits of certain wave-like solutions. A general theory of existence of random traveling wave solutionwas established. It shows that the existence of a wave-like solution implies theexistence of a critical traveling wave solution, which is the random travelingwave solution with minimal propagating speed.

In [10,12,34,35], the authors considered a general form of KPP equationperturbed by environmental noise, with Heaviside initial condition,⎧⎨

⎩du(t, x) =

(D

2uxx(t, x) + u(t, x)c(u(t, x))

)dt+ k(t)u(t, x)dWt,

u(0, x) = χ(−∞,l](x),(1.9)

where c : R+ → R. l may be a constant or a random variable. Under theassumption that c ∈ C1(R+) is strictly decreasing, c0 = c(0) > 0, there


is θ0 > 0 such that c(θ) ≤ 0 for θ ≥ θ0, and k is not identically zero,the traveling wave solution to (1.9) was studied in [12] by using Hamilton-Jacobi theory and some computer simulations of behavior of solutions wereproduced when l and k are deterministic. The noise is called a strong noiseif lim inft→∞(1/2t)

∫ t0k2(s)ds > c0. The noise is weak if

∫ ∞0k2(s)ds < ∞.

When the noise is neither strong nor weak, the noise is called a moderatelystrong noise. The authors showed that the asymptotic behavior of the solutiondepends on the strength of the noise. If the noise is strong, the solution tendsto zero. If the noise is moderately strong, the solution may tend to a travelingwave solution or may be destroyed. The solution tends to the same travelingwave solution as the solution of the deterministic equation if the noise is weak.In [34], the authors considered the similar problems but with either l or k beingrandom. Suppose the limit k∞ := limt→∞(1/2t)

∫ t0k2(s)ds exists. The wave-

front for large time is known as x =√D(2c0 − 2k∞)t. There are constants c1,

c2, c3 > 0 such that when x > (√D(2c0 − 2k∞) + h)t,

u(t, x) < exp(−c1t) a.s. (1.10)

and when x < (√D(2c0 − 2k∞) − h)t,

exp(−c3√

2t ln ln t) ≤ u(t, x) ≤ exp(c2√

2t ln ln t) a.s. (1.11)

for any h > 0 and sufficiently large t [10,12,34]. That is, ahead of the front,the solution to (1.9) is exponentially small almost surly. But behind the front,the solution to (1.9) is oscillatory. In [35], the authors obtained the ergodicityand the pathwise property of the solution to (1.9). They let c(u) = c4(1 − u)and k be a constant. Then the front becomes x =

√D(2c4 − k2)t. The authors

showed the ergodicity by

limt→∞

1t

∫ t

0

infx≤(

√D(2c4−k2)−h)t

u(s, x)ds

= limt→∞

1t

∫ t

0

supxu(s, x)ds

= 1 − k2

2c4a.s. (1.12)

and

limt→∞

1t

∫ t

0

supx≥(

√D(2c4−k2)+h)t

u(s, x)ds = 0 a.s. (1.13)

for any h > 0. For the pathwise property of the solution, they showed that

limt→∞ |u(t, x, ω) − Z(θtω)| = 0 a.s. (1.14)

for any h > 0 and x < (√D(2c4 − k2)−h)t, where θt is the canonical Brownian

shift on a probability space (Ω,F ,P) and

Z(ω) =(c4

∫ ∞

0

exp(c4s− 1

2k2s+ kWs

)ds

)−1

. (1.15)


The authors in [10,12,34,35] focused on the profile of the traveling wave solu-tion for large time.

Basing on the results in [10,12,34,35], in this paper, we consider stochas-tic traveling wave solution to (1.9) under the sense as [39], that is, choosing asuitable wavefront R(t) and proving that u(t, · +R(t)) is a stationary processand limt→∞R(t)/t exists almost surely. We pay more attention to the law oftraveling wave solution rather than the profile of it, which is different from[10,12,34,35]. Although the solution to (1.9) does not satisfy the propertyof compact support. But the property of exponentially decay with respect tospacial variable is sufficient to obtain the result, which we will show in thefollowing. In Sect. 2, some preliminaries are given which are foundations tocontinue the following works. In Sect. 3, under the assumption that the solu-tion to (1.9) starting at u0(x) = χ(−∞,0](x) survives forever, defining a suitablemarker of wavefront R(t), we prove that u(t, · + R(t)) is a stationary processso that the existence of stochastic traveling wave solution to (1.9) will be ver-ified. In Sect. 4, we consider the asymptotic property of the wavefront R(t).We show that limt→∞R(t)/t exists almost surely and coincides with the frontobtained in [34], which states that the result we obtain is uniform with theresult in [34].

We mention that traveling wave solutions to some important water waveequations perturbed by noise were also investigated. In the earlier paper [42],the authors studied the stochastic KdV equation

ut + 6uux + uxxx = ζt. (1.16)

They obtained an explicit random soliton solution

u(t, x) = 2η2sech2

(η

(x− 4η2t− x0 − 6

∫ t

0

Wsds

))+Wt (1.17)

and analyzed the asymptotic behavior of expectation of this soliton solution(1.17). Later another paper [16] considered a stochastic, damped KdV equation

ut + 6uux + uxxx = εζt,xR[u] − εγu. (1.18)

The authors applied singular perturbation theory to the study of soliton of(1.18), which is influenced by space and time dependent external noise. Sincethen, some authors made contributions to the study of stochastic versionsof KdV equation, especially the Wick type of the equation [9,43,44]. Somegeneral form of explicit soliton solutions were obtained. They were the naturalextensions of classical soliton solutions.

2. Preliminaries

Throughout the paper, we let Ω := S ′ be the space of tempered distributions,F := B be the σ-algebra on S ′, and (Ω,F ,P) be the white noise probabilityspace as it is defined in [17]. Denote by E the expectation with respect to P.Let W = {Wt : 0 ≤ t < ∞} be the Brownian motion given by the coordinateprocess, ζt be the distributional time derivative of Wt and Ft := σ{Ws : 0 ≤s ≤ t} be the σ-algebra flow on Ω.


Definition 2.1. A stochastic process u : [0,∞)×R×S ′ → R is called a strongsolution of (1.9) if

(a) u(ω) ∈ C0,2((0,∞) × R

)a.s.,

(b) u(t, x), uxx(t, x) ∈ L2(dP) for all (t, x) ∈ (0,∞) × R,

(c) u satisfies (1.9) a.s. in the sense that for all 0 < t < ∞ and x ∈ R,

u(t, x) =u0(x) +∫ t

0

(D

2uxx(s, x) + u(s, x)c(u(s, x))

)ds

+∫ t

0

k(s)u(s, x)dWs a.s.

Since u0 may have discontinuities, u will satisfy the initial condition inthe sense that for almost all x ∈ R, limt↓0 u(t, x, ω) = u0(x) a.s. Assumingk ∈ L2

loc(R+), the existence and uniqueness of strong solution to (1.9) with

piecewise continuous and nonnegative initial condition have been obtained in[34]. The solution enjoys the following properties:

Proposition 2.2. Let u denote the strong solution of (1.9), then(a) u(t, x) ≥ 0 for all (t, x) ∈ R+ × R a.s.,(b) If x → u0(x) is decreasing a.s., then x → u(t, x, ω) is decreasing for

each t ∈ R+ and a.e. ω ∈ S ′.

Define

Et(k) := exp(∫ t

0

k(s)dWs − 12

∫ t

0

k2(s)ds), 0 ≤ t ≤ ∞. (2.1)

Let u be the solution to (1.9) with initial condition u0(x) and v(t, x) :=u(t, x)E−1

t (k). Under the facts(u(t, x) � E∞(k)

)(ω) = u(t, x, ω − k)E∞(k) a.s. (2.2)

andd

dtEt(k) = k(t)ζt � Et(k), 0 < t < ∞, (2.3)

where � is the Wick product, a straightforward calculation shows that∂

∂tv =

D

2vxx + vc(vEt(k)), v(0, x, ω) = u0(x, ω − k) a.s. (2.4)

for (t, x) ∈ R+ × R. Applying the Feynman-Kac formula to (2.4) gives

v(t, x, ω) = E

[u0

(x+

√DBt, ω − k

)

× exp(∫ t

0

c(v(t− s, x+

√DBs

)Et−s(k, ω))ds

)]a.s. (2.5)

for (t, x) ∈ R+ × R. B = {Bt : t ≥ 0} is a Brownian motion defined onan auxiliary probability space, (Ω,F ,P), and E denotes the expectation with


respect to P. Substituting v with u in (2.5) gives a stochastic Feynman-Kacformula for solution to (1.9),

u(t, x, ω) = E

[u0

(x+

√DBt, ω − k

)

× exp(∫ t

0

c(u(t− s, x+

√DBs

))ds

)

× exp(∫ t

0

k(s)dWs − 12

∫ t

0

k2(s)ds)]

a.s. (2.6)

for (t, x) ∈ R+ × R.In the remainder of this section, some definitions of spaces are given on

which we consider the traveling wave solution to (1.9) and the Kolmogorovtightness criterion is stated.

For continuous functions f : R → R, let ‖ f ‖λ= supx∈R{|f(x)e−λ|x||}.Define

C+λ = {f ≥ 0 : f ∈ C(R) and |f(x)e−λ|x|| → 0 as x → ±∞} (2.7)

and

C+tem = ∩λ>0C

+λ . (2.8)

Then C+tem is the space of continuous functions with slower than exponential

growth. The space C+λ has the topology given by the norm ‖ · ‖λ and C+

tem isgiven by norm family {‖ · ‖λ : λ > 0}. Let

C+dec = {f ≥ 0 : ‖f‖λ < ∞ for some λ < 0} (2.9)

be the space of functions with exponential decay.Let (C((0,∞), C+

tem),U ,Ut, U(t)) be continuous path space, which is thecanonical right continuous filtration with the coordinate variables. Next westate the Arzela Ascoli theorem and Kolmogorov tightness criterion for thespaces C+

λ and C+tem.

A set Σ ⊆ C+λ is relatively compact if and only if

(a) {f : f ∈ Σ} are equicontinuous on compacts,(b) limR→∞ supf∈Σ sup|x|≥R |f(x)e−λ|x|| = 0.

A set Σ ⊆ C+tem is (relatively) compact if and only if it is (relatively) compact

in C+λ for all λ > 0. For C < ∞, γ > 0, δ > 0, μ < λ, define

Σ(C, δ, γ, μ)={f : |f(x) − f(x′)|≤C|x−x′|γeμ|x| for all |x−x′| ≤ δ}. (2.10)

Then using the above conditions one can show that Σ(C, δ, γ, μ) ∩ {f :∫Rf(x)e−|x|dx ≤ c} is compact in C+

λ where c is a constant. If {Xn(·)}n∈N

is a C+λ valued process with {∫

RXn(x)e−|x|dx}n∈N tight and with C0 < ∞,

p > 0, γ > 1, μ < λ such that for all n ≥ 1,

E[|Xn(x) −Xn(x′)|p] ≤ C0|x− x′|γeμp|x| (2.11)

for all |x − x′| ≤ 1, x, x′ ∈ R, then {Xn}n∈N is tight. Indeed, if (2.11) holdsand γ < (γ − 1)/p, μ < μ < λ, then there exist deterministic constantsC = C(γ, μ) < ∞, ρ = ρ(γ, γ, p) > 0 and random δ(ω) such that X(ω) ∈


Σ(C, δ, γ, μ) and E[δ−ρ] ≤ C(C0, μ, μ, γ, γ, p) < ∞. Similarly, if {Xn(·, ·)}n∈N

is a C([t1, t2], C+λ ) valued process, {∫

RXn(0, x)e−|x|dx}n∈N is tight and there

are C0 < ∞, p > 0, γ > 2, μ < λ such that for all n ≥ 1,

E[|Xn(t, x) −Xn(t′, x′)|p] ≤ C0(|x− x′|γ + |t− t′|γ)eμp|x| (2.12)

for all |x−x′| ≤ 1, |t− t′| ≤ 1, t, t′ ∈ [t1, t2], then {Xn}n∈N is tight. The C+tem

(respectively C((0,∞), C+tem)) valued process {Xn}n∈N is tight if and only if

it is tight as C+λ (respectively C((0,∞), C+

λ )) valued process for each λ > 0.

3. Stochastic traveling wave solution

In this section, we define a marker of wavefrontR(t) and prove that u(t, ·+R(t))is a stationary process, where u is the solution to (1.9) with initial conditionu0(x) = χ(−∞,0](x). We denoteQf the law of unique solution to (1.9) satisfyingu(0) = f . A probability measure ν on C+

tem may define a probability measureon C((0,∞), C+

tem) by

Qν(A) =∫C+

tem

Qf (A)ν(df) (3.1)

for A ⊆ C((0,∞), C+tem). As in [39], we need to define a marker of wavefront.

Since the solution to (1.9) possesses the noncompact support property [31].So R0(u(t)) is no longer a suitable front for traveling wave solution to (1.9).We have to choose another marker of wavefront. Since the solution to (1.9)with Heaviside initial condition is exponentially small almost surly as x → ∞[34,35], from the stochastic Feynman-Kac formula (2.6), we may choose

R(f) = ln(∫

R

exf(x)dx)

(3.2)

as a suitable marker of wavefront in the following consideration of travelingwave solution to (1.9), which is also a auxiliary front in [39]. Note that R(f(·+c)) = R(f) − c. On C((0,∞), C+

tem), define

R(t) = R(U(t)) (3.3)

and

V (t) =

⎧⎪⎨⎪⎩

0, R(t) = −∞,

U(t, · +R(t)), −∞ < R(t) < +∞,

c0, R(t) = ∞.

(3.4)

Thus V (t) is the wave U(t) shifted so that its wavefront lies at the origin.Now we summarise the method of proof. Taking the Heaviside initial

condition u0(x) = χ(−∞,0](x), define νT the law of T−1∫ T0V (s)ds under law

Qu0 . It has been shown in [12,34] that if the strength of the noise is not strong,the solution to (1.9) may survive forever. We suppose that the solution survivesforever. We shall show that the sequence νT , T = 1, 2, . . . is tight and that anylimit point is nontrivial. We may check that for any limit point ν, Qν is the lawof a traveling wave solution to (1.9). Two ingredients that go into the proof of


tightness are the Kolmogorov tightness criterion for the unshifted waves andthe control on the movement of the wavefront to ensure the shifting of thewave does not destroy the tightness.

First, the next three lemmas give bounds of solution on certain momentneeded to check the Kolmogorov tightness criterion.

Lemma 3.1. Suppose that the solution u to (1.9) with initial condition f ≥ 0survives forever. Then for p ∈ N, t > 0, there are constants C1(k, p, t) andC2(k, c) such that

Qf (Up(t, x)) ≤ C1(k, p, t)( ∫

R

G(t, x− y)f(y)dy + C2(k, c))p

(3.5)

for x ∈ R, where G(t, x) is the Green’s function of heat equation.

Proof. Denote w solves the following equation

∂

∂tw =

D

2wxx + wc(w) − 1

2k2w, w(0, x) = f(x). (3.6)

As in [12], we shall show that

exp(

inf0≤σ≤t∫ t

σ

k(s)dWs

)w(t, x)

≤ u(t, x) ≤ exp(

sup0≤σ≤t

∫ t

σ

k(s)dWs

)w(t, x) a.s. (3.7)

for (t, x) ∈ [0,∞) × R. To obtain a contradiction, we suppose that there is(t′, x′) ∈ (0,∞) × R such that

u(t′, x′) > exp(

sup0≤σ≤t′

∫ t′

σ

k(s)dWs

)w(t′, x′). (3.8)

Then u(t′, x′) > w(t′, x′). For convenience, we denote Xt′,x′s = (t′ − s, x′ +√

DBs) for s ≥ 0. Define a stopping time

τ := inf{s > 0 : u

(Xt′,x′s

)= w

(Xt′,x′s

)}(3.9)

for each ω ∈ Ω. Using (2.6) and by the strong Markov property, we have

u(t′, x′) = E

[u(Xt′,x′

τ ) exp(∫ τ

0

c(u(Xt′,x′s ))ds

)]

× exp(∫ t′

t′−τk(s)dWs − 1

2

∫ t′

t′−τk2(s)ds

)

≤ E

[w(Xt′,x′

τ ) exp(∫ τ

0

c(w(Xt′,x′s ))ds

)]

× exp(∫ t′

t′−τk(s)dWs − 1

2

∫ t′

t′−τk2(s)ds

)

≤ E

[w(Xt′,x′

τ ) exp(∫ τ

0

(c(w(Xt′,x′s )) − 1

2k2(t′ − s))ds

)]


× exp(

sup0≤σ≤t′

∫ t′

t′−σk(s)dWs

)

= exp(

sup0≤σ≤t′

∫ t′

t′−σk(s)dWs

)w(t′, x′) a.s., (3.10)

which contradicts (3.8) and proves the upper bound in (3.7). The lower boundis shown similarly.

Since the property of function c, there exist constants a = a(c) andb = b(c), such that c(x) ≤ a− bx for x ∈ [0,∞). Denote w the solution to thefollowing equation

∂

∂tw =

D

2wxx + w

(a− 1

2k2(t) − bw

), w(0, x) = f(x). (3.11)

Then by comparison method of parabolic PDE, we have w ≥ w for (t, x) ∈[0,∞)×R. Fixing t > 0, for any ε > 0, multiplying G(t−s+ε, x−y) in (3.11)and integrating, we get

∂

∂s

∫R

w(s, y)G(t− s+ ε, x− y)dy

=(a− 1

2k2(s)

) ∫R


−b∫R

w2(s, y)G(t− s+ ε, x− y)dy

≤(a− 1

2k2(s)

) ∫R


−b( ∫

R

w(s, y)G(t− s+ ε, x− y)dy)2

. (3.12)

Denote h(s) =∫Rw(s, y)G(t− s+ ε, x− y)dy. Thus (3.12) becomes

⎧⎪⎪⎨⎪⎪⎩

d

dsh(s) ≤

(a− 1

2infx∈R

k2(s))h(s) − bh2(s),

h(0) =∫R

f(y)G(t+ ε, x− y)dy.(3.13)

From the theory of ODE, we get

h(s) ≤∫R

f(y)G(t+ ε, x− y)dy +a

b− 1

2binfs∈R

k2(s), (3.14)

which implies∫R

w(t, y)G(ε, x− y)dy≤∫R

f(y)G(t+ε, x− y)dy +a

b− 1

2binft∈R

k2(t). (3.15)

Letting ε → 0 in (3.15) gives

w(t, x) ≤∫R

f(y)G(t, x− y)dy +a

b− 1

2binft∈R

k2(t). (3.16)


Then together with (3.7) and (3.16), we get

up(t, x) ≤ exp(p sup

0≤σ≤t

∫ t

σ

k(s)dWs

)

×(∫

R

f(y)G(t, x− y)dy +a

b− 1

2binft∈R

k2(t))p

a.s. (3.17)

Taking the expectation in (3.17) gives

Qf (Up(t, x)) ≤ C1(k, p, t)( ∫

R

G(t, x− y)f(y)dy + C2(k, t, c))p, (3.18)

where

C1(k, p, t) = E

[exp

(p sup

0≤σ≤t

∫ t

σ

k(s)dWs

)], (3.19)

C2(k, c) =a

b− 1

2binft∈R

k2(t), (3.20)

which completes the proof of the lemma. �

Lemma 3.2. Suppose that the solution u to (1.9) with initial condition f ≥ 0survives forever and ϕ has two continuous derivatives with ϕ, ϕxx ∈ C+

dec suchthat α := sup{ϕxx(x)/ϕ(x) : x ∈ R} < ∞. Set β :=

∫Rϕ(x)dx. Then for all

p ≥ 2, t > 0, there is a constant C3(k, p, t, c) such that

Qf(( ∫

R

U(t, x)ϕ(x)dx)p)

≤ C3(k, p, t, c). (3.21)

Proof. Similarly with Lemma 3.1, there are constants a = a(c) and b = b(c)such that c(x) ≤ a− bx for x ∈ [0,∞). Denote u the solution to the followingequation

∂

∂tu = uxx + u(a− bu) + kuζt, u(0, x) = f(x). (3.22)

By the comparison method for SPDE in [19], we have u ≥ u a.s. for (t, x) ∈[0,∞) × R. Multiplying ϕ in (3.22) and integrating, we get∫

R

u(t, x)ϕ(x)dx =∫R

f(x)ϕ(x)dx+∫ t

0

∫R

u(s, x)ϕxx(x)dxds

+a∫ t

0

∫R

u(s.x)ϕ(x)dxds− b

∫ t

0

∫R

u2(s, x)ϕ(x)dxds

+∫ t

0

∫R

k(s)u(s, x)ϕ(x)dxdWs. (3.23)

By Ito’s formula, together with (3.23), we get

d

( ∫R

u(t, x)ϕ(x)dx)p

= p

(∫R

u(t, x)ϕ(x)dx)p−1(∫

R

u(t, x)ϕxx(x)dx)dt

−bp(∫

R


R

u2(t, x)ϕ(x)dx)dt


+(ap+

12p(p− 1)k2(t)

)(∫R

u(t, x)ϕ(x)dx)pdt

+pk(t)( ∫

R

u(t, x)ϕ(x)dx)pdWt. (3.24)

Under the assumption in the lemma, we have the following inequalities

αE

[(∫R

u(t, x)ϕ(x)dx)p]

≥ E

[(∫R


R

u(t, x)ϕxx(x)dx)]

(3.25)

and ∫R

u2(t, x)ϕ(x)dx ≥ 1β

(∫R

u(t, x)ϕ(x)dx)2

. (3.26)

(3.26) implies

1β

E

[( ∫R

u(t, x)ϕ(x)dx)p]1+1/p

≤ E

[( ∫R


R

u2(t, x)ϕ(x)dx)]. (3.27)

Denote by g(t) = E[(∫Ru(t, x)ϕ(x)dx)p]. Taking expectation in (3.24),

together with (3.25) and (3.27), gives that g(t) is continuously differentiableand satisfies

d

dtg(t) ≤

(αp+ ap+

12p(p− 1)k2(t)

)g(t) − bp

βg1+1/p(t). (3.28)

Define κ(y) = inf{t ≥ 0 : g(t) ≤ y}. For t > κ((2β(α+a+(p−1)k2(t)/2)/b)p),we get

g(t) < 2β(α+ a− 1

2k2(t)

)p 1bp. (3.29)

For t ≤ κ((2β(α+ a+ (n− 1)k2(t)/2)/b)p), we have

d

dtg(t) ≤ − bp

2βg1+1/p(t), (3.30)

which implies

g(t) ≤(

2βb

)p 1tp. (3.31)

Associating with (3.29) and (3.31) gives

Qf(( ∫

R

U(t, x)ϕ(x)dx)p)

≤ g(t) ≤ C3(k, p, t, c), (3.32)

where

C3(k, p, t, c) =(

2β(α+ a− 1

2k2(t)

)p 1bp

)∨

((2βb

)p 1tp

). (3.33)


Thus we complete the proof of the lemma. �

Lemma 3.3. Suppose that the solution u to (1.9) with initial condition f ≥ 0survives forever. For p ≥ 2, t > 0, there exists C4(k, p, t, c) such that for all|x− x′| ≤ 1, x, x′ ∈ R,

Qf (|U(t, x) − U(t, x′)|p) ≤ C4(k, p, t, c)|x− x′|p/2−1. (3.34)

Proof. By the Green’s function representation of solution to (1.9), letting x′ >x, |x− x′| ≤ 1, we get

E[|u(s, x) − u(s, x′)|p]≤ 3p−1

∣∣∣∣∫R

(G(s, x− y) −G(s, x′ − y))f(y)dy∣∣∣∣p

+3p−1E

[∣∣∣∣∫ s

0

∫R

(G(s−r, x−y)−G(s−r, x′−y))c(u(r, y))u(r, y)dydr∣∣∣∣p]

+3p−1E

[∣∣∣∣∫ s

0

∫R

(G(s−r, x−y)−G(s−r, x′−y))k(r)u(r, y)dydWr

∣∣∣∣p].

(3.35)

Later we will use the bounds

∫ s

0

∫R

|G(s− r, x− y) −G(s− r, x′ − y)|dydr ≤ C(s)|x− x′|1/2 (3.36)

for |x− x′| ≤ 1. Then using Burkholder’s inequality, we bound the third termin (3.35) by

E

[∣∣∣∣∫ s

0

∫R

(G(s− r, x− y) −G(s− r, x′ − y))k(r)u(r, y)dydWr

∣∣∣∣p]

≤ C(p)E[∣∣∣∣

∫ s

0

(∫R

|G(s−r, x−y)−G(s−r, x′−y)|k(r)u(r, y)dy)2

dr

∣∣∣∣p/2]

≤ C(p)E[∣∣∣∣

∫ s

0

∫R

|G(s−r, x−y)−G(s−r, x′−y)|k2(r)u2(r, y)dydr∣∣∣∣p/2]

≤ C(p)( ∫ s

0

∫R

|G(s− r, x− y) −G(s− r, x′ − y)|dydr)p/2−1

×∫ s

0

∫R

|G(s− r, x− y) −G(s− r, x′ − y)|kp(r)E[up(r, y)]dydr


≤ C(p, s)|x− x′|p/2−1

×∫ s

0

∫R

|G(s− r, x− y) −G(s− r, x′ − y)|kp(r)E[up(r, y)]dydr. (3.37)

Then we bound the second term in (3.35) by

E

[∣∣∣∣∫ s

0

∫R

(G(s− r, x− y) −G(s− r, x′ − y))c(u(r, y))u(r, y)dydr∣∣∣∣p]

≤ E

[∣∣∣∣∫ s

0

∫R

(G(s− r, x− y) −G(s− r, x′ − y))c0u(r, y)dydr∣∣∣∣p]

≤ C(p, s, c)|x− x′|p/2−1

×∫ s

0

∫R

|G(s− r, x− y) −G(s− r, x′ − y)|E[up(r, y)]dydr. (3.38)

The first term in (3.35) is bounded by∣∣∣∣∫R

(G(s, x− y) −G(s, x′ − y)

)f(y)dy

∣∣∣∣p

=∣∣∣∣∫R

∫ x′

x

(y − z)2s

√4πs

e−(y−z)2/4sf(y)dzdy∣∣∣∣p

≤ C ′(p, s)( ∫

R

∫ x′

x

1√se−(y−z)2/5sf(y)dzdy

)p

≤ C ′(p, s)|x− x′|p∫R

1√se−(y−x)2/5sfp(y)dy. (3.39)

Fix t, p, t as in the lemma and set s = t/5. Applying the Markov property attime 4s and the above estimates imply that there are constants C5(k, p, t, c)and C6(k, p, t, c) such that

E[|u(t, x) − u(t, x′)|p]≤ C5(k, p, t, c)|x− x′|p/2−1

∫R

1√se−(y−x)2/5s

E[up(4s, y)]dy

+C6(k, p, t, c)|x− x′|p/2−1

×∫ t

4s

∫R

|G(t− r, x− y) −G(t− r, x′ − y)|E[up(r, y)]dydr. (3.40)

From Lemma 3.1, we have

Qf (U(4s+ r, x)p)

≤ C1(k, p, r)Qf(( ∫

R

U(4s, y)G(r, x− y)dy + C2(k, c))p)

= C1(k, p, r)p∑i=0

p!i!(p− i)!

Cp−i2 (k, c)Qf(( ∫

R

U(4s, y)G(r, x− y)dy)i)

(3.41)


for r ∈ [0, s], x ∈ R. Apply Lemma 3.2 with ϕ(y) = e−(1+(x−y)2)1/2for which

α ≤ 3, β ≤ 2 and note that G(r, x− y) ≤ C ′(s)ϕ(y). This gives that

Qf(( ∫

R

U(4s, y)G(r, x− y)dy)i)

≤ C ′3(k, i, 4s, c), (3.42)

where

C ′3(k, i, 4s, c) = C ′(s)

(4(

3 + a− 12k2(4s)

)i 1bi

)∨

((4b

)i 1(4s)i

). (3.43)

Associating with (3.40)–(3.42), we get

Qf (|U(t, x) − U(t, x′)|p) ≤ C4(k, p, t, c)|x− x′|p/2−1, (3.44)

where

C4(k, p, t, c)

= C5(k, p, t, c)∫R

√5√te−y2/tdy

p∑i=0

p!i!(p− i)!

Cp−i2 (k, c)C ′3

(k, i,

4t5, c

)

+2C6(k, p, t, c)

×∫ t

5

0

∫R

G

(t

5− r, y

) p∑i=0

p!i!(p− i)!

Cp−i2 (k, c)C ′3

(k, i,

4t5, c

)dydr, (3.45)

which completes the proof of the lemma. �

Under the assumption that the solution to (1.9) with Heaviside initialcondition survives forever, now we tend to control the wavefront R(t). Thefollowing two lemmas give some estimations on the movement of the wavefrontensuring the shifting of the wave dose not destroy the tightness.

Lemma 3.4. Suppose that the solution u to (1.9) with initial condition u0(x) =χ(−∞,0](x) survives forever. Then there exists a constant C < ∞ such that

Qu0(R(t)

) ≥ −C(1 + t). (3.46)

Proof. Let ψ0(x) = χ(−1,1)(x). Define ψr(x) = ψ0(x + r) where r ≥ 2. Thenu0 ≥ ∑∞

j=1 ψjr. The comparison method for SPDE implies that Qu0(R(t)) ≥Q

∑∞j=1 ψjr(R(t)). Consider the following equations⎧⎪⎪⎪⎨

⎪⎪⎪⎩

∂

∂tu(j) =

D

2u(j)xx + c

(u(j)

)u(j) + k(t)u(j)ζt,

u(j)(0, x) = ψjr(x),j = 1, 2, . . .

(3.47)

Next we may couple the solution u to (1.9) starting at u0 with a sequence ofindependent processes u(j) solving (3.47). Denote u(1,2) the solution to (1.9)with initial condition ψr + ψ2r. It is easy to see that u(1) ≤ u(1,2), u(2) ≤


u(1,2) a.s. by comparison method. By the stochastic Feynman-Kac formula, weget

u(1,2)(t, x) = E

[ψr(x+

√DBt) + ψ2r(x+

√DBt)

× exp(∫ t

0

c(u(1,2)(t− s, x+√DBs))ds

)

× exp(∫ t

0

k(s)dWs − 12

∫ t

0

k2(s)ds)]ds

≤ E

[ψr(x+

√DBt)

× exp(∫ t

0

c(u(1)(t− s, x+√DBs))ds

)

× exp(∫ t

0

k(s)dWs − 12

∫ t

0

k2(s)ds)]ds

+E

[ψ2r(x+

√DBt)

× exp(∫ t

0

c(u(2)(t− s, x+√DBs))ds

)

× exp(∫ t

0

k(s)dWs − 12

∫ t

0

k2(s)ds)]ds

= u(1)(t, x) + u(2)(t, x) a.s. (3.48)

for (t, x) ∈ [0,∞) × R. By (2.6) and Doob’s inequality, we get

E

[ ∫ − 32 r

−∞u(1)(t, x)exdx

]

≤ ec0t∫ − 3

2 r

−∞P

(−r − x− 1√D

≤ Bt ≤ −r − x+ 1√D

)exdx

≤ ec0t∫ − 3

2 r

−∞exp

(x− (x+ r − 1)2

2Dt

)dx

≤√

2Dt2

exp(c0t+

92Dt− 3 − (r − 2 + 6Dt)2

8Dt

). (3.49)

Therefore for any ε > 0, we can choose suitable constants C1, C2 and setr = C1t+ C2, such that

E

[ ∫ − 32 r


]<ε

4. (3.50)


A similar computation gives that

E

[ ∫ ∞

− 32 r

u(2)(t, x)exdx]

≤√

2Dt2

exp(c0t+

92Dt− 3 − (r − 2 + 6Dt)2

8Dt

).

(3.51)

Then the choice of r implies

E

[ ∫ ∞

− 32 r

u(2)(t, x)exdx]<ε

4. (3.52)

Thus, together with (3.50) and (3.52), we have

E

[∣∣∣∣∫R

u(1,2)(t, x)exdx−∫R

u(1)(t, x)exdx−∫R

u(2)(t, x)exdx∣∣∣∣]

≤ E

[∣∣∣∣∫ − 3

2 r

−∞u(1,2)(t, x)exdx−

∫ − 32 r

−∞u(1)(t, x)exdx−

∫ − 32 r


∣∣∣∣]

+E

[∣∣∣∣∫ ∞

− 32 r

u(1,2)(t, x)exdx−∫ ∞

− 32 r

u(1)(t, x)exdx−∫ ∞

− 32 r

u(2)(t, x)exdx∣∣∣∣]

≤ E

[∣∣∣∣∫ − 3

2 r

−∞u(1,2)(t, x)exdx−

∫ − 32 r


∣∣∣∣]

+E

[∣∣∣∣∫ ∞

− 32 r

u(1,2)(t, x)exdx−∫ ∞

− 32 r

u(1)(t, x)exdx∣∣∣∣]

+E

[∣∣∣∣∫ ∞

− 32 r

u(2)(t, x)exdx∣∣∣∣]

+ E

[∣∣∣∣∫ − 3

2 r


∣∣∣∣]

<ε

4+ε

4+ε

4+ε

4= ε, (3.53)

which implies

P

( ∫R

u(1,2)(t, x)exdx >∫R

u(1)(t, x)exdx+∫R

u(2)(t, x)exdx− ε

)> 1 − ε.

(3.54)

We claim that there exists μ > 0 such that

P

( ∫R

u(t, x)exdx >∫R

u(1,2)(t, x)exdx+ μ

)> μ. (3.55)

To see this, we suppose the contrary: for any μ > 0, there is t = t(μ) such thatP(

∫Ru(t, x)exdx >

∫Ru(1,2)(t, x)exdx+ μ) ≤ μ. But we have

1 − μ < P

(0 ≤

∫R

u(t, x)exdx−∫R

u(1,2)(t, x)exdx ≤ μ

)

≤ P

(0 ≤

∫ ∞

−2r

u(t, x)exdx−∫ ∞

−2r

u(1,2)(t, x)exdx ≤ μ

)


≤ P

(0 ≤

∫ ∞

−2r

u(t, x)dx−∫ ∞

−2r

u(1,2)(t, x)dx ≤ μe2r)

≤ 1. (3.56)

That is

P

( ∫ ∞

−2r

u(t, x)dx−∫ ∞

−2r

u(1,2)(t, x)dx → 0)

→ 1 as μ → 0, (3.57)

which implies

P

(m({x ∈ (−∞,−2r] : u(t, x) − u(1,2)(t, x) > ε}) → 0

)→ 1 as μ → 0.

(3.58)

This contradicts the decreasing property of u with respect to x. Let ε = μ.Similarly with the above estimations, for some m ∈ N, denoting u(1,m) thesolution to (1.9) with initial condition

∑mj=1 ψjr, we get

P(Ω1) > 1 − μ (3.59)

and

P(Ω2) > μ, (3.60)

where

Ω1 ={ ∫

R

u(1,m)(t, x)exdx >m∑j=1

∫R

u(j)(t, x)exdx− μ

}(3.61)

and

Ω2 ={∫

R

u(t, x)exdx >∫R

u(1,m)(t, x)exdx+ μ

}. (3.62)

On the other hand, we claim that there exists δ > 0 such that

Qψ0

(∫R

U(t, x)dx ≥ δ

)≥ δ. (3.63)

To see this, we suppose the contrary: for each δ > 0 there exists t = t(δ) suchthat Qψ0(

∫RU(t, x) dx ≥ δ) < δ. Denote w0 the solution to (3.6) with initial

condition ψ0. We set

δ < min{

12, infs∈[0,∞)

{ ∫R

w0(s, x)dx exp(

− 2E

[∣∣∣∣∫ s

0

k(τ)dWτ

∣∣∣∣])}}

. (3.64)

Then under the assumption, we have

Qψ0

(∫R

U(t, x)dx < δ

)≥ 1

2. (3.65)

By (3.7) and Doob’s inequality, we get

Qψ0

( ∫R

U(t, x)dx < δ

)

≤ P

(exp

(inf

0≤σ≤t

∫ t

σ

k(s)dWs

)∫R

w0(t, x)dx < δ

)


= P

(inf

0≤σ≤t

∫ t

σ

k(s)dWs < ln δ − ln(∫

R

w0(t, x)dx))

≤ P

(sup

0≤σ≤t

∣∣∣∣∫ σ

0

k(s)dWs

∣∣∣∣ > − ln δ + ln(∫

R

w0(t, x)dx))

≤(

− ln δ + ln(∫

R

w0(t, x)dx))−1

E

[∣∣∣∣∫ t

0

k(s)dWs

∣∣∣∣]

<12, (3.66)

which contradicts (3.65). The property of symmetry gives Qψ0(∫ ∞0U(t, x)dx ≥

δ/2) ≥ δ/2. Therefore, we have

P

(R(u(j)(t)) ≥ 1

alnδ

2− jr

)= P

(R(u(0)(t)) ≥ 1

alnδ

2

)

≥ P

(∫ ∞

0

u(0)(t, x)dx ≥ δ

2

)

≥ δ

2. (3.67)

Then together with (3.59), (3.60) and (3.67), we get

P

({R(u(t)) ≤ ln

δ

2−mr

}∩ Ω1 ∩ Ω2

)

≤ P

( m∑j=1

∫R

u(j)(t, x)exdx ≤ exp(

lnδ

2−mr

))

≤ P

(R(u(j)(t)) ≤ ln

δ

2− jr, j = 1, 2, . . . ,m

)

≤(

1 − δ

2

)m. (3.68)

That is

P

(R(u(t)) ≤ ln

δ

2−mC2 −mC1t

)≤

(1 − δ

2

)m 1μ− μ2

, (3.69)

which implies the lower bound in the lemma. �

Lemma 3.5. Suppose that the solution u to (1.9) with initial condition u0(x) =χ(−∞,0](x) survives forever. For t > 0, there exists constant C ′ < ∞ such that

QνT (|R(t)| ≥ d) ≤ C ′

d(3.70)

for all d > 0, T ≥ 1.

Proof. Let v be the solution to the following equation starting at f ,

∂

∂tv =

D

2vxx + c0v + k(t)vζt, v(0, x) = f(x) ≥ 0. (3.71)


The Green’s function representation of v gives

v(t, x)=∫R

ec0tG(t, x−y)f(y)dy+∫ t

0

∫R

k(s)ec0tG(t−s, x−y)dydWs. (3.72)

The comparison method for SPDE shows that

Qf(∫

R

U(t, x)exdx)

≤ E

[ ∫R

v(t, x)exdx]

≤ ec0t∫R

f(y)∫R

1√4πt

e−|x−y|2/4texdxdy

= ec0t∫R

1√4πt

e−|x|2/4texdx∫R

f(x)exdx

= ec0t+t∫R

f(x)exdx. (3.73)

Note that under Qu0

∫R

V (t, x)exdx =∫R

U(t, x+R(t))exdx

= e−R(t)

∫R

U(t, x)exdx

= 1. (3.74)


QνT (R(t) ≥ d) =1T

∫ T

0

Qu0(QV (s)(R(t) ≥ d))ds

=1T

∫ T

0

Qu0

(QV (s)

(e−d

∫R

U(t, x)exdx ≥ 1))

ds

≤ e−d 1T

∫ T

0

Qu0

(QV (s)

( ∫R

U(t, x)exdx))

ds

≤ e−d+c0t+t 1T

∫ T

0

∫R

V (s, x)exdxds

≤ e−d+c0t+t, (3.75)

which proves half of the lemma. On the other hand, Jensen’s inequality gives

Qf (R(t)) ≤ 1a

ln(Qf

( ∫R

U(t, x)exdx))

≤ ln(ec0t+t

∫R

f(x)exdx)

≤ c0t+ t+R(f). (3.76)

An estimation shows

1TQu0

(∫ T+t

t

R(s)ds−∫ T

0

R(s)ds)


=1TQu0

(∫ T

0

(R(s+ t) −R(s))ds)

=1T

∫ T

0

∫{R(u(s+t))−R(u(s))>−d}

(R(u(s+ t)) −R(u(s)))Qu0(du)ds

+1T

∫ T

0

∫{R(u(s+t))−R(u(s))≤−d}


≤ 1T

∫ T

0

∫{R(u(s+t))−R(u(s))>0}


− d

T

∫ T

0

Qu0(R(s+ t) −R(s) ≤ −d)ds

≤ 1T

∫ T

0

∫ ∞

0

Qu0(R(s+ t) −R(s) ≥ y)dyds

− d

T

∫ T

0

Qu0(R(s+ t) −R(s) ≤ −d)ds

=∫ ∞

0

QνT (R(t) ≥ y)dy − dQνT (R(t) ≤ −d). (3.77)

Rearranging (3.77) and applying Lemma 3.4, we get

QνT (R(t) ≤ −d) ≤ 1d

∫ ∞

0

QνT (R(t) ≥ y)dy +1dT

∫ T

0

Qu0(R(s))ds

− 1dT

∫ T+t

t

Qu0(R(s))ds

≤ 1d

∫ ∞

0

e−y+c0t+tdy +1dT

∫ T

0

(R(u0) + c0s+ s)ds

+1dT

∫ T+t

t

C(1 + s)ds

≤ C ′

d, (3.78)

which completes the proof of the lemma. �Associating with Lemma 3.3 and Lemma 3.5, we are sufficient to show

the tightness of {νT , T = 1, 2, . . .}.

Lemma 3.6. Suppose that the solution u to (1.9) with initial condition u0(x) =χ(−∞,0](x) survives forever. Then the sequence {νT , T = 1, 2, . . .} is tight.

Proof. Recall the set Σ(C, δ, γ, μ) and the condition of tightness stated in theintroduction. We have

νT (Σ(C, δ, γ, μ))

=1T

∫ T

0

Qu0(U(t, · +R(t)) ∈ Σ(C, δ, γ, μ))dt


≥ 1T

∫ T

1

Qu0(U(t, · +R(t− 1)) ∈ Σ(Ce−μd, δ, γ, μ), |R(t) −R(t− 1)| ≤ d)

≥ 1T

∫ T

1

Qu0(QV (t−1)(U(1) ∈ Σ(Ce−μd, δ, γ, μ)))dt

− 1T

∫ T

1

Qu0(|R(t) −R(t− 1)| ≥ d)dt

≥ 1T

∫ T

1

Qu0(QV (t−1)(U(1) ∈ Σ(Ce−μd, δ, γ, μ)))dt

− 1T

∫ T

1

Qu0(QV (t−1)(|R(1)| ≥ d))dt. (3.79)

The second term in (3.79) is bounded by QνT (|R(1)| ≥ d) ≤ C ′/d from Lemma3.5. Lemma 3.3 and the criterion of tightness show that given d, μ > 0, wecan choose C, γ and δ to make the first term in (3.79) as close to (T − 1)/Tas desired. On the other hand, we have

νT

{f :

∫R

f(x)e−|x|dx ≤∫R

f(x)exdx = 1}

= 1. (3.80)

So given μ > 0, we can choose C, γ and δ so that νT (Σ(C, δ, γ, μ) ∩ {f :∫Rf(x)e−|x|dx ≤ 1}) is as close to one as desired for large T and d, which

completes the proof of the lemma. �

By Lemma 3.6 we may take a subsequence νT (n) converging to ν. Nowwe state the main argument that u(t, · + R(t)) is a stationary process, whichimplies that Qν is the law of a traveling wave solution to (1.9).

Theorem 3.7. Suppose that the solution u to (1.9) with initial condition u0(x) =χ(−∞,0](x) survives forever. Then u(t, · + R(t)) is a stationary process. Qν isthe law of a traveling wave solution to (1.9).

Proof. First, we prove the following properties of ν.

limT→∞

νT

{f : lim

d→∞

∫ ∞

d

f(x)dx = 0}

= 1. (3.81)

ν

{f :

∫R

f(x)exdx = 1}

= 1. (3.82)

U(t) �= 0, ∀t ≥ 0, Qν a.s. (3.83)

Under the law Qu0 , from (2.6), we have

V (t, x) = E

[u0

(x+

√2Bt

)exp

( ∫ t

0

c(V (t− s, x+√

2Bs))ds)

× exp(

− 12

∫ t

0

k2(s)ds+∫ t

0

k(s)dWs

)]

≤ exp(c0t− 1

2

∫ t

0

k2(s)ds+∫ t

0

k(s)dWs

)P

(Bt ≤ − x√

2

)


= exp(c0t− 1

2

∫ t

0

k2(s)ds+∫ t

0

k(s)dWs

)

× 1√2πt

∫ −x/√2

−∞e−|y|2/2tdy a.s. (3.84)

for t > 0. Again from (2.6), together with (3.84) and Doob’s inequality, underthe law Qu0 , we get

U(1, x) ≤ exp(c0 − 1

2

∫ 1

0

k2(s)ds+∫ 1

0

k(s)dWs

)E[V (t− 1, x+

√2B1)]

≤ exp(c0t− 1

2

∫ 1

0

k2(s)ds+∫ 1

0

k(s)dWs

)

× exp(

− 12

∫ t−1

0

k2(s)ds+∫ t−1

0

k(s)dWs

)

× 12π

√t− 1

∫ +∞

−∞

∫ −x/√2−z

−∞e−|y|2/2(t−1)dye−|z|2/2dz

= exp(c0t− 1

2

∫ 1

0

k2(s)ds+∫ 1

0

k(s)dWs

)

× exp(

− 12

∫ t−1

0

k2(s)ds+∫ t−1

0

k(s)dWs

)P

(Bt ≤ − x√

2

)

≤ exp(c0t− 1

2

∫ 1

0

k2(s)ds+∫ 1

0

k(s)dWs

)

× exp(

− 12

∫ t−1

0

k2(s)ds+∫ t−1

0

k(s)dWs − x2

4t

)a.s. (3.85)

Integrating (3.85) with respect to x in [d,∞) and taking expectation gives

limd→∞

Qu0

(QV (t−1)

( ∫ ∞

d

U(1, x)dx))

≤ limd→∞

√t exp

(c0t− d2

4t

)= 0, (3.86)

which implies that

Qu0

(QV (t−1)

(limd→∞

∫ ∞

d

U(1, x)dx = 0))

= 1. (3.87)

Therefore, we get

νT

{f : lim

d→∞

∫ ∞

2d

f(x)dx = 0}

=1T

∫ T

0

Qu0

(limd→∞

∫ ∞

2d

V (t, x)dx = 0)dt

=1T

∫ T

0

Qu0

(∀ε > 0, ∃d0,

∫ ∞

2d

V (t, x)dx < ε for ∀d > d0

)dt

≥ 1T

∫ T

1

Qu0

(∀ε > 0, ∃d0,

∫ ∞

d+R(t−1)

U(t, x)dx < ε,


|R(t) −R(t− 1)| ≤ d, for ∀d > d0

)dt

≥ 1T

∫ T

1

Qu0

(QV (t−1)

(limd→∞

∫ ∞

d

U(1, x)dx = 0))

dt

− limd→∞

QνT (|R(1)| ≥ d). (3.88)

Lemma 3.5 implies that the second term on the right hand side of (3.88)tends to zero as d → ∞. Together with (3.87) we get (3.81). Since νT (n){f :∫Rf(x)exdx = 1} = 1, we have ν{f :

∫Rf(x)exdx ≤ 1} = 1. Then

ν

{f :

∫R

f(x)exdx ≥ 1}

≥ ν

{f : lim

d→∞

∫R

f(x)ed−|x−d|dx ≥ 1}

≥ lim supn→∞

νT (n)

{f : lim

d→∞

∫R

f(x)ed−|x−d|dx = 1}

= lim supn→∞

1T (n)

∫ T (n)

0

Qu0

(limd→∞

∫R

V (t, x)ed−|x−d|dx = 1)dt

= lim supn→∞

1T (n)

∫ T (n)

0

Qu0

(limd→∞

∫ ∞

d

(V (t, x)e2d−x − V (t, x)ex)dx = 0)dt

= lim supn→∞

1T (n)

∫ T (n)

0

Qu0

(limd→∞

∫ ∞

d

V (t, x)dx = 0)dt = 1. (3.89)

To prove (3.83), we have

Qν(∃t > 0, U(t) = 0) ≤ Qν(R(t) < −d)≤ lim inf

n→∞ QνT (n)(R(t) < −d)

≤ C ′

d→ 0 as d → ∞. (3.90)

Next we show that u(t, · + R(t)) is a stationary process and Qν is thelaw of a traveling wave solution to (1.9). Let F : C+

tem → R be bounded andcontinuous. Suppose that fn → f . Denote un the solution to (1.9) with initialcondition fn. Then it may be checked that the Kolmogorov tightness criterionholds and any limit point u has law Qf . This proves that Qfn → Qf . Thecontinuity of f → Qf gives that QνT (n) → Qν . Therefore we get

Qν(F (V (t))) = limn→∞QνT (n)(F (V (t)))

= limn→∞

1T (n)

∫ T (n)

0

Qu0(QV (s)(F (V (t))))ds

= limn→∞

1T (n)

∫ T (n)

0

Qu0(F (V (t+ s)))ds

= ν(F ). (3.91)


We have shown that under Qν , the one dimensional marginal of {V (t) : t ≥ 0}has law ν. It is straightforward to check that {V (t) : t ≥ 0} is Markov. Hence{V (t) : t ≥ 0} is stationary. This verifies that Qν is the law of a traveling wavesolution to (1.9). �

4. Asymptotic property of wavefront

In this section, we consider the asymptotic property of the wavefront R(t). Wesuppose that the limit

k∞ = limt→∞

12t

∫ t

0

k2(s)ds (4.1)

exists and 0 ≤ k∞ ≤ c0. Denote γ =√D(2c0 − 2k∞). The following theorem

shows the asymptotic property of R(t).

Theorem 4.1. Suppose that the solution u to (1.9) with initial condition u0(x) =χ(−∞,0](x) survives forever. Then

limt→∞

R(t)t

= γ a.s. (4.2)

Proof. For any h > 0, choose 0 < ε < h2/2D +√

2Dc0 − 2Dk∞ +D2h/D.Then for a.e. ω ∈ Ω, there is T1, such that

exp(

− k2

2t− εt

)≤ Et(k, ω) ≤ exp

(− k2

2t+ εt

)(4.3)

for t ≥ T1. So together with (2.6), we have

u(t, x) ≤ exp(c0t− 1

2

∫ t

0

k2(s)ds+ εt

)E[χ(−∞,0](x+

√DBt)]

= exp(c0t− 1

2

∫ t

0

k2(s)ds+ εt

)P

(Bt ≤ − x√

D

)

≤ exp(c0t− 1

2

∫ t

0

k2(s)ds+ εt− x2

2Dt

)a.s. (4.4)

for t ≥ T1. Let x ≥ (β + h)t, where β is a constant. Multiplying ex in (4.4)and integrating with respect to x in [(β + h)t,∞), we have

∫ ∞

(β+h)t

u(t, x)exdx ≤∫ ∞

(β+h)t

exp(c0t− 1

2

∫ t

0

k2(s)ds+ εt− x2

2Dt+ x

)dx

=√

2Dt exp(c0t− 1

2

∫ t

0

k2(s)ds+ εt+D

2t

)

×∫ ∞

(β+h)t−Dt√2Dt

e−y2dy


≤√

2Dt2

exp((

c0 + ε− β2

2D− βh

D− h2

2D− β − h

)t

)

× exp(

− 12

∫ t

0

k2(s)ds)a.s. (4.5)

for t ≥ T1. We let β =√

2Dc0 − 2Dk∞ +D2 −D. Then we have

limt→∞

∫ ∞

(β+h)t

u(t, x)exdx = 0 a.s. (4.6)

Similarly with the computation of (4.5), we have∫ (β−h)t

(γ+h)t

u(t, x)exdx ≤∫ (β−h)t

(γ+h)t

exp(c0t− 1

2

∫ t

0

k2(s)ds+ εt− x2

2Dt+ x

)dx

=√

2Dt exp(c0t− 1

2

∫ t

0

k2(s)ds+ εt+D

2t

)

×∫ (β−h)t−Dt√

2Dt

(γ+h)t−Dt√2Dt

e−y2dy

≤√

2Dt2

exp(c0t− 1

2

∫ t

0

k2(s)ds+ εt

)

× exp(

− γ2

2Dt− γh

Dt− h2

2Dt+ γt+ ht

)

−√

2Dt2

exp(c0t− 1

2

∫ t

0

k2(s)ds+ εt

)

× exp(

− β2

2Dt+

βh

Dt− h2

2Dt+ βt− ht

)a.s. (4.7)

for t ≥ T1. From the definition of γ and β, (4.7) may reduce to∫ (β−h)t

(γ+h)t

u(t, x)exdx ≤√

2Dt2

exp(k∞t− 1

2

∫ t

0

k2(s)ds+ εt

)

× exp(γt− γht

D− h2t

2D+ ht

)

−√

2Dt2

exp(k∞t− 1

2

∫ t

0

k2(s)ds+ εt

)

× exp(βht

D− h2t

2D− ht

)a.s. (4.8)

for t ≥ T1. Analogously to the derivation of (4.8), we get∫ (γ+h)t

(γ−h)tu(t, x)exdx ≤

√2Dt2

exp(k∞t− 1

2

∫ t

0

k2(s)ds+ εt

)

× exp(γt+

γht

D− h2t

2D− ht

)


−√

2Dt2

exp(k∞t− 1

2

∫ t

0

k2(s)ds+ εt

)

× exp(γt− γht

D− h2t

2D+ ht

)a.s. (4.9)

and∫ (β+h)t

(β−h)tu(t, x)exdx ≤

√2Dt2

exp(k∞t− 1

2

∫ t

0

k2(s)ds+ εt

)

× exp(βht

D− h2t

2D− ht

)

−√

2Dt2

exp(k∞t− 1

2

∫ t

0

k2(s)ds+ εt

)

× exp(

− βht

D− h2t

2D+ ht

)a.s. (4.10)

for t ≥ T1. By (1.11) we know that there exists a T2 > 0 such that for t ≥ T2,x < (γ − h)t,

exp(−c3√

2t ln ln t) ≤ u(t, x) ≤ exp(c2√

2t ln ln t) a.s., (4.11)

which implies that

∫ (γ−h)t

−∞u(t, x)exdx ≤ exp(c2

√2t ln ln t+ γt− ht) a.s. (4.12)

for t ≥ T2. From (4.6) there is a T3 > 0 such that∫ ∞(β+h)t

u(t, x)exdx ≤ 1 fort ≥ T3. Then associating with (4.8), (4.9), (4.10) and (4.12), we get

∫R

u(t, x)exdx ≤ exp(c2√

2t ln ln t+ γt− ht)(2 +H(t) +G(t)) a.s.,(4.13)

where

H(t) =√

2Dt2

exp(k∞ − 1

2

∫ t

0

k2(s)ds+ εt

)

× exp(βht

D− h2t

2D− c2

√2t ln ln t− γt

)(4.14)

and

G(t) =

√2Dt2

exp(k∞ − 1

2

∫ t

0

k2(s)ds+ εt

)

× exp(

− γht

D− h2t

2D+ 2ht− c2

√2t ln ln t

). (4.15)


By the arbitrariness of h and ε, there is a T4 > 0 such that H(t) ≤ 1 a.s. fort ≥ T4. On the other hand, we have

1t

lnG(t) =12t

ln 2Dt− 1t

ln 2 +1t

(k∞ − 1

2

∫ t

0

k2(s)ds)

+ε− γh

D− h2

2D+ 2h− 1

tc2

√2t ln ln t a.s. (4.16)

By the arbitrariness of h and ε, (4.16) implies

limt→∞

1t

lnG(t) = 0 a.s. (4.17)

So (4.13) gives that

1tR(t) ≤ 1

tc2

√2t ln ln t+ γ − h+

1t

ln 8 +1t

lnG(t) a.s., (4.18)

which implies that

lim supt→∞

1tR(t) ≤ γ a.s. (4.19)

By (4.11), a straightforward estimation shows that

1tR(t) ≥ −1

tc3

√2t ln ln t+ γ − h a.s. (4.20)

Since the arbitrariness of h, (4.20) implies that

lim inft→∞

1tR(t) ≥ γ a.s. (4.21)


limt→∞

1tR(t) = γ a.s., (4.22)

which completes the proof of the theorem. �

Theorem 4.1 implies that the wavefront R(t) coincides with the wavefrontobtained in [12,34] after sufficiently large time. Thus the theorem suggests thatthe choice of the wavefront in Sect. 2 is suitable.

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Zhehao Huang and Zhengrong LiuDepartment of MathematicsSouth China University of TechnologyGuangzhou, 510640GuangdongPeople’s Republic of Chinae-mail: [email protected]

Zhengrong Liue-mail: [email protected]

Received: 12 January 2014.

Accepted: 3 June 2014.

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Stochastic traveling wave solution to stochastic generalized KPP … · 2017-08-29 · Nonlinear Diﬀer. Equ. Appl. 22 (2015), 143–173 c 2014 Springer Basel 1021-9722/15/010143-31