This study aimed to investigate the risk-return relationship, provided volatility feedback was taken into account, in the South African market. Volatility feedback, a stronger measure of volatility, was treated as an important source of asymmetry in the investigation of the risk-return relationship. This study analyzed the JSE ALSI excess returns and realized variance for the sample period from 15 October 2009 to 15 October 2019. This study modelled the novel and robust Bayesian approach in a parametric and nonparametric framework. A parametric model has modelling assumptions, such as normality, and a finite sample space. A nonparametric approach relaxes modelling assumptions and allows for an infinite sample space; thus, taking into account every possible asymmetric risk-return relationship. Given that South Africa is an emerging market, which is subject to higher levels of volatility, the presence of volatility feedback was expected to be more pronounced. However, contrary to expectations, the test results from both the parametric and nonparametric Bayesian model showed that volatility feedback had an insignificant effect in the South African market. The risk-return relationship was then investigated free from empirical distortions that resulted from volatility feedback. The parametric Bayesian model found a positive risk-return relationship, in line with traditional theoretical expectations. However, the nonparametric Bayesian model found no relationship between risk and return, in line with early South African studies. Since the nonparametric Bayesian approach is more robust than the parametric Bayesian approach, this study concluded that there is no risk-return relationship. Therefore, investors can include South Africa in their investment portfolio with higher risk countries in order to spread their risk and derive diversification benefits. In addition, risk averse investors can find a safe environment within the South African market and earn a return in accordance to their risk tolerance.
Citation: Nitesha Dwarika. The risk-return relationship and volatility feedback in South Africa: a comparative analysis of the parametric and nonparametric Bayesian approach[J]. Quantitative Finance and Economics, 2023, 7(1): 119-146. doi: 10.3934/QFE.2023007
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This study aimed to investigate the risk-return relationship, provided volatility feedback was taken into account, in the South African market. Volatility feedback, a stronger measure of volatility, was treated as an important source of asymmetry in the investigation of the risk-return relationship. This study analyzed the JSE ALSI excess returns and realized variance for the sample period from 15 October 2009 to 15 October 2019. This study modelled the novel and robust Bayesian approach in a parametric and nonparametric framework. A parametric model has modelling assumptions, such as normality, and a finite sample space. A nonparametric approach relaxes modelling assumptions and allows for an infinite sample space; thus, taking into account every possible asymmetric risk-return relationship. Given that South Africa is an emerging market, which is subject to higher levels of volatility, the presence of volatility feedback was expected to be more pronounced. However, contrary to expectations, the test results from both the parametric and nonparametric Bayesian model showed that volatility feedback had an insignificant effect in the South African market. The risk-return relationship was then investigated free from empirical distortions that resulted from volatility feedback. The parametric Bayesian model found a positive risk-return relationship, in line with traditional theoretical expectations. However, the nonparametric Bayesian model found no relationship between risk and return, in line with early South African studies. Since the nonparametric Bayesian approach is more robust than the parametric Bayesian approach, this study concluded that there is no risk-return relationship. Therefore, investors can include South Africa in their investment portfolio with higher risk countries in order to spread their risk and derive diversification benefits. In addition, risk averse investors can find a safe environment within the South African market and earn a return in accordance to their risk tolerance.
Let
Oε={x=(x∗,xn+1)|x∗=(x1,…,xn)∈Qand0<xn+1<εg(x∗)}, |
where
γ1≤g(x∗)≤γ2,∀x∗∈¯Q. | (1) |
Denote
{dˆuε−Δˆuεdt=(H(t,x,ˆuε(t))+G(t,x))dt+m∑j=1cjˆuε∘dwj,x∈Oε,t>τ,∂ˆuε∂νε=0,x∈∂Oε, | (2) |
with the initial condition
ˆuε(τ,x)=ˆϕε(x),x∈Oε, | (3) |
where
As
{du0−1gn∑i=1(gu0yi)yidt=(H(t,(y∗,0),u0(t))+G(t,(y∗,0)))dt+m∑j=1cju0∘dwj,y∗=(y1,…,yn)∈Q,t>τ,∂u0∂ν0=0,y∗∈∂Q, | (4) |
with the initial condition
u0(τ,y∗)=ϕ0(y∗),y∗∈Q, | (5) |
where
Random attractors have been investigated in [2,5,10,19,9] in the autonomous stochastic case, and in [3,21,22,23] in the non-autonomous stochastic case. Recently, the limiting dynamical behavior of stochastic partial differential equations on thin domain was studied in [16,20,13,14,11,12,17,4]. However, in [17,13], we only investigated the limiting behavior of random attractors in
Let
We organize the paper as follows. In the next section, we establish the existence of a continuous cocycle in
Here we show that there is a continuous cocycle generated by the reaction-diffusion equation defined on
{dˆuε−Δˆuεdt=(H(t,x,ˆuε(t))+G(t,x))dt+m∑j=1cjˆuε∘dwj,x=(x∗,xn+1)∈Oε,t>τ,∂ˆuε∂νε=0,x∈∂Oε, | (6) |
with the initial condition
ˆuετ(x)=ˆϕε(x),x∈Oε, | (7) |
where
H(t,x,s)s≤−λ1|s|p+φ1(t,x), | (8) |
|H(t,x,s)|≤λ2|s|p−1+φ2(t,x), | (9) |
∂H(t,x,s)∂s≤λ3, | (10) |
|∂H(t,x,s)∂x|≤ψ3(t,x), | (11) |
where
Throughout this paper, we fix a positive number
h(t,x,s)=H(t,x,s)+λs | (12) |
for all
h(t,x,s)s≤−α1|s|p+ψ1(t,x), | (13) |
|h(t,x,s)|≤α2|s|p−1+ψ2(t,x), | (14) |
∂h(t,x,s)∂s≤β, | (15) |
|∂h(t,x,s)∂x|≤ψ3(t,x), | (16) |
where
Substituting (12) into (6) we get for
{dˆuε−(Δˆuε−λˆuε)dt=(h(t,x,ˆuε(t))+G(t,x))dt+m∑j=1cjˆuε∘dwj,x=(x∗,xn+1)∈Oε,∂ˆuε∂νε=0,x∈∂Oε, | (17) |
with the initial condition
ˆuετ(x)=ˆϕε(x),x∈Oε. | (18) |
We now transfer problem (17)-(18) into an initial boundary value problem on the fixed domain
x∗=y∗,xn+1=εg(y∗)yn+1. |
It follows from [18] that the Laplace operator in the original variable
Δxˆu(x)=|J|divy(|J|−1JJ∗∇yu(y))=1gdivy(Pεu(y)), |
where we denote by
Pεu(y)=(guy1−gy1yn+1uyn+1⋮guyn−gynyn+1uyn+1−n∑i=1yn+1gyiuyi+1ε2g(1+n∑i=1(εyn+1gyi)2)uyn+1). |
In the sequel, we abuse the notation a little bit by writing
Fε(t,y∗,yn+1,s)=F(t,y∗,εg(y∗)yn+1,s),F0(t,y∗,s)=F(t,y∗,0,s), |
where
{duε−(1gdivy(Pεuε)−λuε)dt=(hε(t,y,uε(t))+Gε(t,y))dt+m∑j=1cjuε∘dwj,y=(y∗,yn+1)∈O,Pεuε⋅ν=0,y∈∂O, | (19) |
with the initial condition
uετ(y)=ϕε(y)=ˆϕε∘T−1ε(y),y∈O, | (20) |
where
Given
θ1,t(τ)=τ+t,for allτ∈R. | (21) |
Then
Ω={ω∈C(R,R):ω(0)=0}. |
Let
θtω(⋅)=ω(⋅+t)−ω(t),ω∈Ω,t∈R. | (22) |
Then
dz+αzdt=dw(t), | (23) |
for
Lemma 2.1. There exists a
limt→±∞|ω(t)|t=0for allω∈Ω′, |
and, for such
z∗(ω)=−α∫0−∞eαsω(s)ds |
is well defined. Moreover, for
(t,ω)→z∗(θtω)=−α∫0−∞eαsθtω(s)ds=−α∫0−∞eαsω(t+s)ds+ω(t) |
is a stationary solution of (23) with continuous trajectories. In addition, for
limt→±∞|z∗(θtω)|t=0,limt→±∞1t∫t0z∗(θsω)ds=0, | (24) |
limt→±∞1t∫t0|z∗(θsω)|ds=E|z∗|<∞. | (25) |
Denote by
˜Ω=Ω′1×⋯×Ω′mand F=m⊗j=1Fj, |
Then
Denote by
SCj(t)u=ecjtu,foru∈L2(O), |
and
T(ω):=SC1(z∗1(ω))∘⋯∘SCm(z∗m(ω))=em∑j=1cjz∗j(ω)IdL2(O),ω∈Ω′. |
Then for every
T−1(ω):=SCm(−z∗m(ω))∘⋯∘SC1(−z∗1(ω))=e−m∑j=1cjz∗j(ω)IdL2(O). |
It follows that
On the other hand, since
limt→±∞1t∫t0‖T(θτω)‖2dτ=E‖T‖2=m∏j=1E(e2cjz∗j)<∞, |
and
limt→±∞1t∫t0‖T−1(θτω)‖2dτ=E‖T−1‖2=m∏j=1E(e−2cjz∗j)<∞. |
Remark 1. We now consider
Next, we define a continuous cocycle for system (19)-(20) in
{dvεdt−1gdivy(Pεvε)=(−λ+δ(θtω))vε+T−1(θtω)hε(t,y,T(θtω)vε(t))+T−1(θtω)Gε(t,y),y∈O,t>τ,Pεvε⋅ν=0,y∈∂O, | (26) |
with the initial conditions
vετ(y)=ψε(y),y∈O, | (27) |
where
Since (26) is a deterministic equation, by the Galerkin method, one can show that if
Φε(t,τ,ω,ϕε)=uε(t+τ,τ,θ−τω,ϕε)=T(θt+τω)vε(t+τ,τ,θ−τω,ψε),for all(t,τ,ω,ϕε)∈R+×R×Ω×N. | (28) |
By the properties of
Let
(Rεˆϕε)(y)=ˆϕε(T−1εy),∀ˆϕε∈L2(Oε). |
Given
ˆΦε(t,τ,ω,ˆϕε)=R−1εΦε(t,τ,ω,Rεˆϕε). |
The same change of unknown variable
{dv0dt−n∑i=11g(gv0yi)yi=(−λ+δ(θtω))v0+T−1(θtω)h0(t,y∗,T(θtω)v0(t))+T−1(θtω)G0(t,y∗),y∗∈Q,t>τ,∂v0∂ν0=0,y∗∈∂Q, | (29) |
with the initial conditions
v0τ(y∗)=ψ0(y∗),y∗∈Q, | (30) |
where
The same argument as above allows us to prove that problem (4) and (5) generates a continuous cocycle
Now we want to write equation (26)-(27) as an abstract evolutionary equation. We introduce the inner product
(u,v)Hg(O)=∫Oguvdy,for allu,v∈N |
and denote by
For
aε(u,v)=(J∗∇yu,J∗∇yv)Hg(O), | (31) |
where
J∗∇yu=(uy1−gy1gyn+1uyn+1,…,uyn−gyngyn+1uyn+1,1εguyn+1). |
By introducing on
‖u‖H1ε(O)=(∫O(|∇y∗u|2+|u|2+1ε2u2yn+1)dy)12, | (32) |
we see that there exist positive constants
η1∫O(|∇y∗u|2+1ε2u2yn+1)dy≤aε(u,u)≤η2∫O(∇y∗u|2+1ε2u2yn+1)dy | (33) |
and
η1‖u‖2H1ε(O)≤aε(u,u)+‖u‖2L2(O)≤η2‖u‖2H1ε(O). | (34) |
Denote by
D(Aε)={v∈H2(O),Pεv⋅ν=0on∂O} |
as defined by
Aεv=−1gdivPεv,v∈D(Aε). |
Then we have
aε(u,v)=(Aεu,v)Hg(O),∀u∈D(Aε),∀v∈H1(O). | (35) |
Using
{dvεdt+Aεvε=(−λ+δ(θtω))vε+T−1(θtω)hε(t,y,T(θtω)vε(t))+T−1(θtω)Gε(t,y),y∈O,t>τ,vετ=ψε. | (36) |
To reformulate system (29)-(30), we introduce the inner product
(u,v)Hg(Q)=∫Qguvdy∗,for allu,v∈M, |
and denote by
a0(u,v)=∫Qg▽y∗u⋅▽y∗vdy∗. |
Denote by
D(A0)={v∈H2(Q),∂v∂ν0=0on∂Q} |
as defined by
A0v=−1gn∑i=1(gvyi)yiv∈D(A0). |
Then we have
a0(u,v)=(A0u,v)Hg(Q),∀u∈D(A0),∀v∈H1(Q). |
Using
{dv0dt+A0v0=(−λ+δ(θtω))v0+T−1(θtω)h0(t,y∗,T(θtω)v0(t))+T−1(θtω)G0(t,y∗),y∗∈Q,t>τ,v0τ(s)=ψ0(s),s∈[−ρ,0]. | (37) |
Hereafter, we set
limt→−∞ect‖Bi(τ+t,θtω)‖Xi=0, |
where
Di={Bi={Bi(τ,ω):τ∈R,ω∈Ω}:Bi is tempered in Xi}. |
Our main purpose of the paper is to prove that the cocycle
limε→0supuε∈ˆAεinfu0∈A0ε−1‖uε−u0‖2H1(Oε)=0. | (38) |
To prove (38), we only need to show that the cocycle
limε→0distH(Aε(τ,ω),A0(τ,ω))=0, |
which will be established in the last section of the paper.
Furthermore, we suppose that there exists
¯γΔ=λ0−2E(|δ(ω)|)>0. | (39) |
Let us consider the mapping
γ(ω)=λ0−2|δ(ω)|. | (40) |
By the ergodic theory and (39) we have
limt→±∞1t∫t0γ(θlω)dl=Eγ=¯γ>0. | (41) |
The following condition will be needed when deriving uniform estimates of solutions:
∫τ−∞e12¯γs(‖G(s,⋅)‖2L∞(˜O)+‖φ1(s,⋅)‖2L∞(˜O)+‖ψ3(s,⋅)‖2L∞(˜O))ds<∞,∀τ∈R. | (42) |
When constructing tempered pullback attractors, we will assume
limr→−∞eσr∫0−∞e12¯γs(‖G(s+r,⋅)‖2L∞(˜O)+‖φ1(s+r,⋅)‖2L∞(˜O)+‖ψ3(s+r,⋅)‖2L∞(˜O))ds=0,∀σ>0. | (43) |
Since
∫τ−∞e12¯γs(‖G(s,⋅)‖2L∞(˜O)+‖ψ1(s,⋅)‖L∞(˜O)+‖ψ3(s,⋅)‖2L∞(˜O))ds<∞,∀τ∈R | (44) |
and
limr→−∞eσr∫0−∞e12¯γs(‖G(s+r,⋅)‖2L∞(˜O)+‖ψ1(s+r,⋅)‖2L∞(˜O)+‖ψ3(s+r,⋅)‖2L∞(˜O))ds=0, | (45) |
for any
In this section, we recall and generalize some results in [17] and derive some new uniform estimates of solutions of problem (36) or (19)-(20) which are needed for proving the existence of
Lemma 3.1. Assume that (8)-(11), (39) and (42) hold. Then for every
sup−1≤s≤0‖vε(τ+s,τ−t,θ−τω,ψε)‖2H1ε(O)≤R2(τ,ω), | (46) |
where
R2(τ,ω)=r1(ω)R1(τ,ω)+c∫0−∞e¯γr‖T−1(θrω)‖2(‖G(r+τ,⋅)‖2L∞(˜O)+‖ψ3(r+τ,⋅)‖2L∞(˜O))dr, | (47) |
where
R1(τ,ω)=c∫0−∞e∫r0γ(θlω)dl‖T−1(θrω)‖2‖G(r+τ,⋅)‖2L∞(˜O)dr+c∫0−∞e∫r0γ(θlω)dl‖T−1(θrω)‖2‖ψ1(r+τ,⋅)‖2L∞(˜O)dr, | (48) |
and
Proof. The proof is similar as that of Lemma 3.4 in [17], so we only sketch the proof here. Taking the inner product of (36) with
12ddt‖vε‖2Hg(O)≤−aε(vε,vε)+(−λ0+δ(θtω))‖vε‖2Hg(O)+(T−1(θtω)hε(t,y,T(θtω)vε(t)),vε)Hg(O)+(T−1(θtω)Gε(t,y),vε)Hg(O). | (49) |
By (13), we have
ddt‖vε‖2Hg(O)+2aε(vε,vε)+λ02‖vε‖2Hg(O)+2α1γ1‖T−1(θtω)‖2‖uε‖pLp(O)≤(−λ0+2δ(θtω))‖vε‖2Hg(O)+2λ0γ2|˜O|‖T−1(θtω)‖2‖G(t,⋅)‖2L∞(˜O)+2γ2|˜O|‖T−1(θtω)‖2‖ψ1(t,⋅)‖L∞(˜O). | (50) |
Then, we have for any
e∫στγ(θlω)dl‖vε(σ)‖2Hg(O)+2∫στe∫rτγ(θlω)dlaε(vε(r),vε(r))dr+λ02∫στe∫rτγ(θlω)dl‖vε(r)‖2Hg(O)dr+2α1γ1∫στ‖T−1(θrω)‖2e∫rτγ(θlω)dl‖uε(r)‖pLp(O)dr≤‖vε(τ)‖2Hg(O)+2λ0γ2|˜O|∫στe∫rτγ(θlω)dl‖T−1(θrω)‖2‖G(r,⋅)‖2L∞(˜O)dr+2γ2|˜O|∫στe∫rτγ(θlω)dl‖T−1(θrω)‖2‖ψ1(r,⋅)‖2L∞(˜O)dr, | (51) |
where
Thus by the similar arguments as Lemma 3.1 in [17] we get for every
‖vετ(⋅,τ−t,θ−τω,ψ)‖2L2(O)≤c∫0−∞e∫r0γ(θlω)dl‖ψ1(r+τ,⋅)‖2L∞(˜O)dr+c∫0−∞e∫r0γ(θlω)dl‖T−1(θrω)‖2‖G(r+τ,⋅)‖2L∞(˜O)dr+c∫0−∞e∫r0γ(θlω)dl‖T−1(θrω)‖2‖ψ1(r+τ,⋅)‖2L∞(˜O)dr. | (52) |
Moreover, taking the inner product of (36) with
12ddtaε(vε,vε)+‖Aεvε‖2Hg(O)≤(−λ0+δ(θtω))aε(vε,vε)+(T−1(θtω)hε(t,y,T(θtω)vε(t)),Aεvε)Hg(O)+(T−1(θtω)Gε(t,y),Aεvε)Hg(O). | (53) |
By (15)-(16) we have
ddtaε(vε,vε)+‖Aεvε‖2Hg(O)≤(c+2δ(θtω))aε(vε,vε)+c‖T−1(θtω)‖2(‖G(t,⋅)‖2L∞(˜O)+‖ψ3(t,⋅)‖2L∞(˜O)), | (54) |
The left proof is similar of that Lemma 3.4 in [17], so we omit it here.
We are now in a position to establish the uniform estimates for the solution
Lemma 3.2. Assume that (8)-(11), (39) and (42) hold. Then for every
sup−1≤s≤0‖uε(τ+s,τ−t,θ−τω,ϕε)‖2H1ε(O)≤r2(ω)R2(τ,ω), | (55) |
where
Lemma 3.3. Assume that (8)-(11), (39) and (42) hold. Then for every
sup−1≤s≤0‖vε(τ+s,τ−t,θ−τω,ψε)‖pLp(O)+∫ττ−ρ‖vε(r,τ−t,θ−τω,ψε)‖2p−2L2p−2(O)dr≤R3(τ,ω), | (56) |
where
Proof. The proof is similar as that of Lemma 3.6 in [14], so we omit it here.
Lemma 3.4. Assume that (8)-(11), (39) and (42) hold. Then for every
∫0−1eγMp−2s∫{y∈O: vε(s+τ,τ−t,θ−τω,ψε)≥2M}|vε(s+τ,τ−t,θ−τω,ψε)|2p−2dyds≤η, | (57) |
∫0−1eγMp−2s∫{y∈O: vε(s+τ,τ−t,θ−τω,ψε)≤−2M}|vε(s+τ,τ−t,θ−τω,ψε)|2p−2dyds≤η. | (58) |
Proof. Let
1pddt‖(vε−M)+‖pLp(O)+(p−1)∫vε≥M(vε−M)p−2aε(vε,vε)dx≤(δ(θtω)vε,(vε−M)p−1+)+(T−1(θtω)hε(t,y,T(θtω)vε),(vε−M)p−1+)+(T−1(θtω)Gε(t,y),(vε−M)p−1+). | (59) |
For the first term on the right side of (59) we have
|(δ(θtω)vε,(vε−M)p−1+)|≤1p|δ(θrω)|p∫O|vε|pdx+p−1p∫O(vε−M)p+dx. | (60) |
For the second term on the right-hand side of (59), by (8), we obtain, for
hε(t,y,T(θtω)vε) (vε−M)p−1+≤−α1‖T(θtω)‖p−1(vε)p−1(v−M)p−1+ |
+‖T(θtω)‖−1ψ1(t,y∗,εg(y∗)yn+1)(vε)−1(vε−M)p−1+ |
≤−12α1Mp−2‖T(θtω)‖p−1(vε−M)p+−12α1‖T(θtω)‖p−1(vε−M)2p−2+ |
+‖T−1(θtω)‖−1|ψ1(t,y∗,εg(y∗)yn+1)|(vε−M)p−2+ |
which implies
(T−1(θtω)hε(t,y,T(θtω)vε), (vε−M)p−1+) |
≤−12α1Mp−2‖T(θtω)‖p−2∫O(vε−M)p+dx−12α1‖T−1(θtω)‖p−2∫O(vε−M)2p−2+dx |
+‖T(θtω)‖−2∫O|ψ1(t,y∗,εg(y∗)yn+1)|(vε−M)p−2+dx |
≤−12α1Mp−2‖T(θtω)‖p−2∫O(vε−M)p+dx−12α1‖T(θtω)‖p−2∫O(vε−M)2p−2+dx |
+p−2p∫O(vε−M)p+dx+2p‖T(θtω)‖−p∫O|ψ1(t,y∗,εg(y∗)yn+1)|p2dy. | (61) |
The last term in (59) is bounded by
(T−1(θtω)Gε(t,y),(vε−M)p−1+)≤18α1‖T(θtω)‖p−2∫O(vε−M)2p−2+dx+2α1‖T(θtω)‖−p∫vε≥M|Gε(t,y)|2dy. | (62) |
All above estimates yield
ddt‖(vε−M)+‖pLp(O)−(2p−3−12pα1Mp−2‖T(θtω)‖p−2)∫O(vε−M)p+dx+14pα1‖T(θtω)‖p−2∫O(vε−M)2p−2+dx≤|δ(θrω)|p∫O|vε|pdx+2‖T(θtω)‖−p∫O|ψ1(t,y∗,εg(y∗)yn+1)|p2dy+2pα1‖T(θtω)‖−p∫O|Gε(t,y)|2dy. | (63) |
Multiplying (63) by
‖(vε(τ,τ−t,ω,ψε)−M)+‖pLp(O) |
+14pα1∫ττ−1‖T(θζω)‖p−2e−∫ζτ(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr |
×∫O(vε(ζ,τ−t,ω,ψε)−M)2p−2+dxdζ |
≤e−∫τ−1τ(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖(vε(τ−1,τ−t,ω,ψε)−M)+‖pLp(O) |
+∫ττ−1|δ(θζω)|pe−∫ζτ(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖vε(ζ,τ−t,ω,ψε)‖pLp(O)dζ |
+2|O|∫ττ−1‖T(θζω)‖−pe−∫ζτ+ξ(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖ψ1(ζ,⋅)‖p2L∞(˜O)dζ. |
+2p|O|α1∫ττ−1‖T(θζω)‖−pe−∫ζτ+ξ(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖G(ζ,⋅)‖2L∞(˜O)dζ, | (64) |
where
14pα1∫0−1‖T(θζω)‖p−2e−∫ζ0(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr |
×∫O(vε(ζ+τ,τ−t,θ−τω,ψε)−M)2p−2+dxdζ |
≤e−∫−10(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖(vε(τ−1,τ−t,θ−τω,ψε)−M)+‖pLp(O) |
+∫0−1|δ(θζ+ξω)|pe−∫ζ0(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖vε(ζ+τ,τ−t,θ−τω,ψε)‖pLp(O)dζ |
+2|O|∫0−1‖T(θζω)‖−pe−∫ζ0(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖ψ1(ζ+τ,⋅)‖p2L∞(˜O)dζ. |
+2p|O|α1∫0−1‖T(θζω)‖−pe−∫ζ0(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖G(ζ+τ,⋅)‖2L∞(˜O)dζ. | (65) |
Since
c1≤12pα1‖T(θrω)‖p−2≤c2 for all r∈[−ρ−1,0]. | (66) |
By (66) we obtain
ec2Mp−2ζ≤e∫ζ+ξξ12pα1Mp−2‖T(θrω)‖p−2dr≤ec1Mp−2ζ for all ζ∈[−1,0]andξ∈[−ρ,0]. | (67) |
For the left-hand side of (65), by (67) we find that there exists
14pα1∫0−1‖T(θζω)‖p−2e−∫ζ0(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr |
∫O(vε(ζ+τ,τ−t,θ−τω,ψε)−M)2p−2+dxdζ |
≥c3∫0−1ec2Mp−2ζ∫O(vε(ζ+τ,τ−t,θ−τω,ψε)−M)2p−2+dxdζ. | (68) |
For the first term on the right-hand side of (65), by (67) we obtain
e−∫−10(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖(vε(τ−1,τ−t,θ−τω,ψε)−M)+‖pLp(O) |
≤e2p−3e−c1Mp−2‖(vε(τ−1,τ−t,θ−τω,ψε)−M)+‖pLp(O) |
≤e2p−3e−c1Mp−2‖vε(τ−1,τ−t,θ−τω,ψε)‖pLp(O). | (69) |
Similarly, for the second terms on the right-hand side of (65), we have from (67) there exists
∫0−1|δ(θζω)|pe−∫ζ0(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖vε(ζ+τ,τ−t,θ−τω,ψε)‖pLp(O)dζ |
≤c4∫0−1ec1Mp−2ζ‖vε(ζ+τ,τ−t,θ−τω,ψε)‖pLp(O)dζ | (70) |
Since
2|O|∫0−1‖T(θζω)‖−pe−∫ζ0(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖ψ1(ζ+τ,⋅)‖p2L∞(˜O)dζ. |
+2p|O|α1∫0−1‖T(θζω)‖−pe−∫ζ0(2p−3−12pα1Mp−2‖T(θrω)‖p−2)dr‖G(ζ+τ,⋅)‖2L∞(˜O)dζ |
≤c5∫0−1ec1Mp−2ζdζ≤c−11c5M2−p. | (71) |
By (68)-(71) we get from (65) that
c3∫0−1ec2Mp−2ζ∫O(vε(ζ+τ,τ−t,θ−τω,ψε)−M)2p−2+dydζ |
≤e2p−3e−c1Mp−2‖vε(τ−1,τ−t,θ−τω,ψε)‖pLp(O) |
+c4∫0−1ec1Mp−2ζ‖vε(ζ+τ,τ−t,θ−τω,ψε)‖pLp(O)dζ+c−11c5M2−p, |
which together with Lemma 3.2 and Lemma 3.3 implies that there exist
c3∫0−1ec2Mp−2ζ∫O(vε(ζ+τ,τ−t,θ−τω,ψε)−M)2p−2+dxdζ |
≤c6e−c1Mp−2+c6∫0−1ec1Mp−2ζdζ+c−11c5M2−p≤c6e−c1Mp−2+c−11(c5+c6)M2−p. | (72) |
Since
∫0−1ec2Mp−2ζ∫O(vε(ζ+τ,τ−t,θ−τω,ψε)−M)2p−2+dydζ≤η. | (73) |
Note that
∫0−1ec2Mp−2ζ∫{y∈O: vε(ζ+τ,τ−t,θ−τω,ψε)≥2M}|vε(ζ+τ,τ−t,θ−τω,ψε)|2p−2dydζ≤22p−2∫0−1ec2Mp−2ζ∫O(vε(ζ+τ,τ−t,θ−τω,ψε)−M)2p−2+dxdζ≤22p−2η. | (74) |
Similarly, one can verify that there exist
∫0−1ec2Mp−2ζ∫{y∈O: vε(ζ+τ,τ−t,θ−τω,ψε)≤−2M}|vε(ζ+τ,τ−t,θ−τω,ψε)|2p−2dydζ≤22p−2η. | (75) |
Then Lemma 3.4 follows from (3) and (75) immediately.
Note that
0≤λε1≤λε2≤…≤λεn≤⋯→+∞, |
and their associated eigenfunctions
It follows from Corollary 9.7 in [8] that the eigenvalues and the eigenfunctions of
Next, we introduce the spectral projections. We use
Pεn(u)=m∑i=1(u,ϖεi)Yεϖεiforu∈Yε. |
We use
aε(u,u)=(Aεu,u)Hg(O)≤λεn(u,u)Hg(O),∀u∈PεnD(A1/2ε). | (76) |
and
aε(u,u)=(Aεu,u)Hg(O)≥λεm+1(u,u)Hg(O),u∈QεmD(A1/2ε). | (77) |
Let
Lemma 3.5. Assume that (8)-(11), (39) and (42) hold. Then for every
‖uε2(τ,τ−t,θ−τω,ϕε)‖H1(O)≤η. |
Proof. Taking the inner product (36) with
12ddtaε(vε2,vε2)+‖Aεvε2‖2≤(δ(θtω)vε2,Aεvε2)+(QεnT−1(θtω)hε(t,y,T(θtω)vε),Aεvε2)+(QεnT−1(θtω)Gε(t,y),Aεvε2). | (78) |
For the first term on the right-hand side of (78), we have
(δ(θtω)vε2,Aεvε2)≤18‖Aεvε2‖2+2|δ(θtω)|2‖vε2‖2. | (79) |
For the superlinear term, we have from (9) that
(QεnT−1(θtω)hε(t,y,T(θtω)vε),Aεvε2)≤18‖Aεvε2‖2+2‖T−1(θtω)‖2∫O|hε(t,y,T(θtω)vε)|2dy≤18‖Aεvε2‖2+2α2‖T−1(θtω)‖2∫O(|T(θtω)vε|p−1+ψ2(t,y∗,εg(y∗)yn+1))2dy≤18‖Aεvε2‖2+4α2‖T(θtω)‖2p−4‖v‖2p−22p−2+4α2|O|‖T−1(θtω)‖2‖ψ2(t,⋅)‖2L∞(˜O). | (80) |
For the last term on the right-hand side of (78), we have
(QεnT−1(θtω)Gε(t,y),Aεvε2)≤18‖Aεvε2‖2+2|O|‖T−1(θtω)‖2‖G(t,⋅)‖2L∞(˜O) | (81) |
Noting that
ddtaε(vε2,vε2)+λεn+1aε(vε2,vε2)≤4δ2(θtω)‖vε2‖2+8α2‖T(θtω)‖2p−4‖vε‖2p−22p−2+c‖T−1(θtω)‖2(‖ψ2(t,⋅)‖2L∞(˜O)+‖G(t,⋅)‖2L∞(˜O)). | (82) |
Taking
aε(vε2(τ,τ−t,θ−τω,ψε),vε2(τ,τ−t,θ−τω,ψε))≤∫ττ−1eλεn+1(r−τ)aε(vε2(r,τ−t,θ−τω,ψε),vε2(r,τ−t,θ−τω,ψε))dr+4δ2∫ττ−1eλεn+1(r−τ)δ2(θr−τω)aε(vε2(r,τ−t,θ−τω,ψε),vε2(r,τ−t,θ−τω,ψε))dr+8α2∫ττ−1eλεn+1(r−τ)‖T(θr−τω)‖2p−4‖vε(r,τ−t,θ−τω,ψε)‖2p−22p−2dr+c∫ττ−1eλεn+1(r−τ)‖T−1(θr−τω)‖2(‖ψ2(r,⋅)‖2L∞(˜O))dr+c∫ττ−1eλεn+1(r−τ)‖T−1(θr−τω)‖2‖G(r,⋅)‖2L∞(˜O)dr. | (83) |
Since
aε(vε2(τ,τ−t,θ−τω,ψε),vε2(τ,τ−t,θ−τω,ψε))≤c∫0−1eλεn+1r‖vε(r+τ,τ−t,θ−τω,ψε)‖2p−22p−2dr+c∫0−1eλεn+1raε(vε(r+τ,τ−t,θ−τω,ψε),vε(r+τ,τ−t,θ−τω,ψε))dr |
+c∫0−1eλεn+1rdr≤c∫0−1e(λ0n+1−1)r‖vε(r+τ,τ−t,θ−τω,ψε)‖2p−22p−2dr+c∫0−1e(λ0n+1−1)raε(vε(r+τ,τ−t,θ−τω,ψε),vε(r+τ+s,τ−t,θ−τω,ψε))dr+c∫0−1e(λ0n+1−1)r‖vε(r+τ−ρ0(r+τ+s),τ−t,θ−τω,ψε)‖2dr+c∫0−1e(λ0n+1−1)rdr. | (84) |
Given
c∫0−1e(λ0n+1−1)r‖vε(r+τ,τ−t,θ−τω,ψε)‖2p−22p−2dr≤c∫0−1e(λ0n+1−1)r∫{y∈O:|vε|≥2M}|vε(r+τ,τ−t,θ−τω,ψε)|2p−2dydr+c∫0−1e(λ0n+1−1)r∫{y∈O:|vε|<2M}|vε(r+τ,τ−t,θ−τω,ψε)|2p−2dydr≤c∫0−1eγMp−2r∫{y∈O:|vε|≥2M}|vε(r+τ,τ−t,θ−τω,ψε)|2p−2dydr+c∫0−1e(λ0n+1−1)r∫{y∈O:|vε|<2M}|vε(r+τ,τ−t,θ−τω,ψε)|2p−2dydr≤η+c22p−2M2p−2|O|∫0−1e(λ0n+1−1)rdr≤η+c22p−2M2p−2|O|1λ0n+1−1. | (85) |
For the last three terms on the right-hand side of (84), by Lemma 3.1, we find that there exist
c∫0−1e(λ0n+1−1)raε(vε(r+τ,τ−t,θ−τω,ψε),vε(r+τ,τ−t,θ−τω,ψε))dr+c∫0−1e(λ0n+1−1)rdr≤c1∫0−1e(λ0n+1−1)rdr≤c11λ0n+1−1. | (86) |
Since
aε(vε2(τ+s,τ−t,θ−τω,ψε),vε2(τ+s,τ−t,θ−τω,ψε))≤2η, |
which together
In this subsection, we establish the existence of
Lemma 4.1. Suppose (8)-(11), (39) and (43) hold. Then the cocycle
Proof. We first notice that, by Lemma 3.2,
K(τ,ω)={u∈H1(O):‖u‖2H1(O)≤L(τ,ω)}, | (87) |
where
Φε(t,τ−t,θ−tω,D(τ−t,θ−tω))⊆K(τ,ω). |
Thus we find that
Lemma 4.2. Assume that (8)-(11), (39) and (43) hold. Then, the cocycle
Proof. We will show that for every
‖uε2(τ,τ−tn,θ−τω,ϕε)‖H1(O)=‖Qm0uε(τ,τ−tn,θ−τω,ϕε)‖H1(O)<η4. | (88) |
On the other hand, by Lemma 3.2 we find that the sequence
Theorem 4.3. Assume that (8)-(11), (39) and (43) hold. Then, the cocycle
Proof. First, we know from Lemma 4.1 that
Analogous results also hold for the solution of (4)-(5). In particular, we have:
Theorem 4.4. Assume that (8)-(11), (39) and (43) hold. Then, the cocycle
The following estimates are needed when we derive the convergence of pullback attractors. By the similar proof of that of Theorem 5.1 in [14], we get the following lemma.
Lemma 5.1. Assume that (8)-(11) and (39) hold. Then for every
∫tτ‖vε(r,τ,ω,ψε)‖2H1ε(O)dr≤c‖ψε‖2N+c∫τ+Tτ(‖G(r,⋅)‖2L∞(˜O)+‖ψ1(r,⋅)‖2L∞(˜O))dr, |
where
Similarly, one can prove
Lemma 5.2. Assume that (8)-(11) and (39) hold. Then for every
∫tτ‖v0(r,τ,ω,ψ0)‖2H1(Q)dr≤c‖ψ0‖2M+c∫τ+Tτ(‖G(r,⋅)‖2L∞(˜O)+‖ψ1(r,⋅)‖2L∞(˜O))dr, |
where
In the sequel, we further assume the functions
‖Gε(t,⋅)−G0(t,⋅)‖L2(O)≤κ1(t)ε | (89) |
and
‖Hε(t,⋅,s)−H0(t,⋅,s)‖L2(O)≤κ2(t)ε, | (90) |
where
By (12) and (90) we have, for all
‖hε(t,⋅,s)−h0(t,⋅,s)‖L2(O)≤κ2(t)ε. | (91) |
Since
Theorem 5.3. Suppose (8)-(11), (39), and (89)-(90) hold. Given
limn→∞‖Φεn(t,τ,ω,ϕεn)−Φ0(t,τ,ω,ϕ0)‖N=0. |
Proof. Since
‖vεn(t)−v0(t)‖2N≤c‖ϕεn−ϕ0‖2N+cmaxν∈[τ,t]ξ(θνω)∫tτ‖vεn(s)−v0(s)‖2Nds |
+cεnmaxν∈[τ,t]‖T−1(θνω)‖∫tτ(‖vεn(s)‖2H1εn(O)+‖v0(s)‖2H1(Q))ds+cεnmaxν∈[τ,t]‖T−1(θνω)‖∫tτ(κ21(s)+κ22(s))ds+cεn∫tτ(‖vεn(s)‖2H1εn(O)+‖v0(s)‖2H1(Q))ds, | (92) |
where
‖vεn(t)−v0(t)‖2N≤ec(1+maxν∈[τ,τ+T]ξ(θνω))T‖ϕεn−ϕ0‖2N+ϱεnec(1+maxν∈[τ,τ+T]ξ(θνω))T[‖ψ0‖2M+‖ψεn‖2N+∫τ+Tτ(κ21(s)+κ22(s))ds+∫τ+Tτ(‖G(s,⋅)‖2L∞(˜O)+‖ψ1(s,⋅)‖2L∞(˜O))ds]. | (93) |
Notice that, for all
‖uεn(t,τ,ω,ϕε)−u0(t,τ,ω,ϕ0)‖2N≤maxν∈[τ,τ+T]‖T(θνω)‖2‖vεn(t,τ,ω,T−1(θτω)ϕε)−v0(t,τ,ω,T−1(θτω)ϕ0‖2N, |
which together with (93) implies the desired results.
The next result is concerned with uniform compactness of attractors with respect to
Lemma 5.4. Assume that (8)-(11), (39) and (43) hold. If
limn→∞‖uεn−u‖H1(O)=0. |
Proof. Take a sequence
uεn=Φεn(tn,τ−tn,θ−tnω,ϕεn). | (94) |
By Lemma 4.1, we have
‖QεnN1uεn(τ,τ−tn,θ−τω,ϕεn)‖H1(O)≤η. | (95) |
By Lemma 3.2, we have
‖PεnN1uεn(τ,τ−tn,θ−τω,ϕεn)‖H1(O)<M. | (96) |
It follows from (95) and (96) that
limn→∞‖uεn−u‖H1(O)=0. | (97) |
This completes the proof.
Now we are in a position to prove the main result of this paper.
Theorem 5.5. Assume that (8)-(11), (39), (43), and (89)-(90) hold. The attractors
limε→0distH1(O)(Aε(τ,ω),A0(τ,ω))=0. |
Proof. Given
‖u‖2H1ε(O)≤L(τ,ω)for all 0<ε<ε0 and u∈Aε(τ,ω), | (98) |
where
distH1(O)(zn,A0(τ,ω))≥δfor alln∈N. | (99) |
By Lemma 5.4 there exists
limn→∞‖zn−z∗‖H1(O)=0. | (100) |
By the invariance property of the attractor
zn=Φεn(t,τ−t,θ−tω,ytn). | (101) |
By Lemma 5.4 again there exists
limn→∞‖ytn−yt∗‖H1(O)=0. | (102) |
It follows from Theorem 5.3 that for every
limn→∞Φεn(t,τ−t,θ−tω,ytn)=Φ0(t,τ−t,θ−tω,yt∗)inN. | (103) |
By (100), (101), (103) and uniqueness of limits we obtain
z∗=Φ0(t,τ−t,θ−tω,yt∗)inH1(O). | (104) |
Notice that
limsupn→∞‖ytn‖H1(O)≤‖K(τ−t,θ−tω)‖H1(O)≤L(τ−t,θ−tω). | (105) |
By (102) and (105) we get, for every
‖yt∗‖H1(Q)≤L(τ−t,θ−tω). | (106) |
By
distH1(Q)(z∗,A0(τ,ω))=distH1(Q)(Φ0(t,τ−t,θ−tω,yt∗),A0(τ,ω))≤distH1(Q)(Φ0(t,τ−t,θ−tω,K0(τ−t,θ−tω)),A0(τ,ω))→0,ast→∞. | (107) |
This implies that
distH1(O)(zn,A0(τ,ω))≤distH1(O)(zn,z∗)→0, |
a contradiction with (99). This completes the proof.
The authors would like to thank the anonymous referee for the useful suggestions and comments.
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