Here we state that the techniques used in this paper to establish the local stability of solutions for problem (6) come from the methods of Kruzkov’s device of doubling the variables presented in Kruzkov’s paper [14].
Proof of Theorem 1.
For an arbitrary T>0, set ζT=[0,T]×R. Let f(t,x)∈C0∞(ζT). We assume that f(t,x)=0 outside the cylinder
(24)Θ={(t,x)}=[δ,T-2δ]×Hr-2δ, 0<2δ≤min(T,r).
We define
(25)h=f(t+τ2,x+y2)θk(t-τ2)θk(x-y2)=f(⋯)λk(*),
where (⋯)=((t+τ)/2,(x+y)/2) and (*)=((t-τ)/2,(x-y)/2). The function θk(ω) is defined in (13). Note that
(26)ht+hτ=ft(⋯)λk(*), hx+hy=fx(⋯)λk(*).
Taking v=v1(t,x) and α=v2(τ,y) and assuming that f(t,x)=0 outside the cylinder Θ, from Lemma 6, we have
(27)⨌ζT×ζT{-sign(v1(t,x)-v2(τ,y))Jv1(t,x)h|v1(t,x)-v2(τ,y)|ht +sign(v1(t,x)-v2(τ,y)) ×(v13(t,x)-v23(τ,y))hx -sign(v1(t,x)-v2(τ,y)) × Jv1(t,x)h}dx dt dy dτ=0.
Similarly, it has
(28)⨌ζT×ζT{-sign(v2(τ,y)-v1(t,x))Jv2(τ,y)|v2(τ,y)-v1(t,x)|gτ +sign(v2(τ,y)-v1(t,x)) ×(v23(τ,y)-v13(t,x))hy -sign(v2(τ,y)-v1(t,x)) × Jv2(τ,y)h}dx dt dy dτ=0,
from which we obtain
(29)0≤⨌ζT×ζT|v1(t,x)-v2(τ,y)|(ht+hτ)dx dt dy dτ +⨌ζT×ζTsign(v1(t,x)-v2(τ,y)) ×(v13(t,x)-v23(τ,y)) ×[hx+hy]dx dt dy dτ +|⨌ζT×ζTsign(v1(t,x)-v2(τ,y)) ×(Jv1(t,x)-Jv2(τ,y))h dx dt dy dτ∭∫ζT×ζTsign(v1(t,x)-v2(τ,y))|=⨌ζT×ζT(I1+I2)dx dt dy dτ +|⨌ζT×ζTI3 dx dt dy dτ|.
We claim that
(30)0≤∬ζT|v1(t,x)-v2(t,x)|ft+sign(v1(t,x)-v2(t,x)) ×(v13(t,x)-v23(t,x))fxdx dt +|∬ζTsign(v1(t,x)-v2(t,x)) ×[Jv1(t,x)-Jv2(t,x)]f dx dt∬ζTsign(v1(t,x)-v2(t,x))|.
We note that the first two terms in the integrand of (29) can be represented in the form
(31)Hk=H(t,x,τ,y,v1(t,x),v2(τ,y))λk(*).
From Lemma 5, we know that Hk satisfies the Lipschitz condition in v1 and v2, respectively. By the choice of h, we have Hk=0 outside the region as follows:
(32){(t,x;τ,y)}={|x-y|2δ≤t+τ2≤T-2δ,|t-τ|2≤k, |x+y|2≤r-2δ,|x-y|2≤k},⨌ζT×ζTHk dx dt dy dτ =⨌ζT×ζT[H(t,x,τ,y,v1(t,x),v2(τ,y)) - H(t,x,t,x,v1(t,x),v2(t,x))] ×λk(*)dx dt dy dτ +⨌ζT×ζTH(t,x,t,x,v1(t,x),v2(t,x)) ×λk(*)dx dt dy dτ =B1(k)+B2.
Considering the estimate |λ(*)|≤c/k2 and the expression of function Hk, we have(33)|B1(k)|≤c[k+1k2⨌|(t-τ)/2|≤k,δ≤(t+τ)/2≤T-δ,|(x-y)/2|≤k,|(x+y)/2|≤r-δ|v2(t,x)-v2(τ,y)|dx dt dy dτ],k+1k2
where the constant c does not depend on k. Using Lemma 4, we obtain B1(k)→0 as k→0. The integral B2 does not depend on k. In fact, substituting t=α1, (t-τ)/2=β, x=η, (x-y)/2=ξ and noting that
(34)∫-kk∫-∞∞λk(β,ξ)dξ dβ=1,
we have
(35)B2=22∬ζTHk(α1,η,α1,η,v1(α1,η),v2(α1,η)) ×{∫-kk∫-∞∞λk(β,ξ)dξ dβ}dη dα1=4∬ζTH(t,x,t,x,v1(t,x),v2(t,x))dx dt.
Hence
(36)limk→0⨌ζT×ζTHk dx dt dy dτ =4∬ζTH(t,x,t,x,v1(t,x),v2(t,x))dx dt.
Since
(37)I3=sign(v1(t,x)-v2(τ,y)) ×(Jv1(t,x)-Jv2(τ,y))fλk(*),(38)⨌ζT×ζTI3 dx dt dy dτ =⨌ζT×ζT[I3(t,x,τ,y)-I3(t,x,t,x)]dx dt dy dτ +⨌ζT×ζTI3(t,x,t,x)dx dt dy dτ=C1(h)+C2,
we obtain(39)|C1(k)|≤c(k+1k2⨌|(t-τ)/2|≤k,δ≤(t+τ)/2≤T-δ,|(x-y)/2|≤k,|(x+y)/2|≤r-δ|Jv2(t,x)-Jv2(τ,y)|dx dt dy dτ).
By Lemmas 3 and 4, we have C1(h)→0 as k→0. Using (34), we have
(40)C2=22∬ζTI3(α1,η,α1,η,v1(α1,η),v2(α1,η)) ×{∫-hh∫-∞∞λh(β,ξ)dξ dβ}dη dα1=4∬ζTI3(t,x,t,x,v1(t,x),v2(t,x))dx dt=4∬ζTsign(v1(t,x)-v2(t,x)) ×(Jv1(t,x)-Jv2(t,x))f(t,x)dx dt.
From (36) and (38)–(40), we prove that inequality (30) holds.
Let
(41)B(t)=∫-∞∞|v1(t,x)-v2(t,x)|dx.
We define
(42)ρk(σ)=∫-∞σθk(ω)dω, (ρk′(σ)=θk(σ)≥0)
and choose the two numbers τ1 and τ2∈(0,T0), τ1<τ2. In (30), we choose
(43)f=[ρk(t-τ1)-ρk(t-τ2)]χ(t,x), k<min(τ1,T0-τ2),
where
(44)χ(t,x)=χɛ(t,x)=1-ρɛ(|x|+R0t-N+ɛ), ɛ>0.
When ɛ is sufficiently small, we note that function χ(t,x)=0 outside the cone ℶ and f(t,x)=0 outside the set Θ. For (t,x)∈Θ, we have
(45)0=χt+R0|χx|≥χt+R0χx.
Applying (30) and (42)–(45) and suitably choosing large R0, we have the inequality
(46)0≤∬ξT0{[θk(t-τ1)-θk(t-τ2)] ×χɛ|v1(t,x)-v2(t,x)|}dx dt +|∬ξT0[ρk(t-τ1)-ρk(t-τ2)][Jv1(t,x)-Jv2(t,x)] ×E(t,x)χ(t,x)dx dt∬ξT0[ρk(t-τ1)-ρk(t-τ2),Jv1(t,x)-Jv2(t,x)]|,
where E(t,x)=sign[v1(t,x)-v2(t,x)].
From (46), we obtain
(47)0≤∬ξT0{[θk(t-τ1)-θk(t-τ2)] ×χɛ|v1(t,x)-v2(t,x)|}dx dt +∫0T0(ρk(t-τ1)-ρk(t-τ2)) ×|∫-∞∞[Jv1(t,x)-Jv2(t,x)]E(t,x)χ(t,x)dx|dt.
Using Lemma 7, we have
(48)0≤∬ξT0{[θk(t-τ1)-θk(t-τ2)] ×χɛ|v1(t,x)-v2(t,x)|}dx dt +c∫0T0(ρk(t-τ1)-ρk(t-τ2)) ×∫-∞∞|v1-v2|dx dt,
where c>0 is a constant as described in (21).
Letting ɛ→0 in (48) and sending N→∞, we have
(49)0≤∬ξT0{[θk(t-τ1)-θk(t-τ2)] ×|v1(t,x)-v2(t,x)|}dx dt +c∫0T0(ρk(t-τ1)-ρk(t-τ2)) ×∫-∞∞|v1-v2|dx dt.
By the properties of the function θk(ω) for k≤min(τ1,T0-τ1), we have
(50)|∫0T0θk(t-τ1)B(t)dt-B(τ1)| =|∫0T0θk(t-τ1)[B(t)-B(τ1)]dt| ≤c1k∫τ1-kτ1+k|B(t)-B(τ1)|dt⟶0 as k⟶0,
where c is independent of h.
Set
(51)P(τ1)=∫0T0ρk(t-τ1)B(t)dt=∫0T0∫-∞t-τ1θk(σ)dσB(t)dt.
Using the similar proof of (50), we get
(52)P′(τ1)=-∫0T0θk(t-τ1)B(t)dt⟶-B(τ1) as k⟶0,
from which we obtain
(53)P(τ1)⟶P(0)-∫0τ1B(σ)dσ as k⟶0.
Similarly, we have
(54)P(τ2)⟶P(0)-∫0τ2B(σ)dσ as k⟶0.
Then, we get
(55)P(τ1)-P(τ2)⟶∫τ1τ2B(σ)dσ as k⟶0.
Letting τ1→0 and τ2→t, from (49), (50), and (55), for any t∈[0,T0], we have
(56)∫-∞∞|v1(t,x)-v2(t,x)|dx≤∫-∞∞|v1(0,x)-v2(0,x)|dx +c∫0t∫-∞∞|v1-v2|dx,
where c depends on ∥v10∥L1(R), ∥v10∥L∞(R), ∥v20∥L1(R), ∥v20∥L∞(R), and T. Using the Gronwall inequality and (56) completes the proof of Theorem 1.