
Minimizing informal recycling activities is critical for the sustainable end-of-life treatment of electronics. Recent studies have started to revisit the concept of informality in recycling and reported empirical examples where informal sectors coordinate with formal sectors, jointly contributing to a greener recycling solution. This case study examines the systematic effort to transform and integrate the informal sector into the formal recycling industry for managing e-waste in Guiyu, China. This paper analyzes the policy design, implementation, technology development and market establishment of Guiyu's formal sector that enabled the evolution of the local informal recycling industry. The results show that the salient success factor is to offer advanced and centralized e-waste treatment by constructing a formal recycling sector while maintaining the competitive characteristics of the old informal businesses, including manual dismantling and private e-waste collection networks. Those characteristics ensured increased reuse value and sufficient e-waste sources. Meanwhile, the study found that many challenges and conflicts during this transition are rooted in the often-overlooked societal and historical contexts that profoundly shaped the local recycling industry. Authorities of regions facing challenges regulating informal recycling of e-waste, especially developing countries, could initiate similar systems based on local realities and the collaboration between formal and informal sectors to minimize the environmental and societal consequences of unregulated informal e-waste recycling.
Citation: Congying Wang, Fu Zhao, Carol Handwerker. Transforming and integrating informal sectors into formal e-waste management system: A case study in Guiyu, China[J]. Clean Technologies and Recycling, 2022, 2(4): 225-246. doi: 10.3934/ctr.2022012
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Minimizing informal recycling activities is critical for the sustainable end-of-life treatment of electronics. Recent studies have started to revisit the concept of informality in recycling and reported empirical examples where informal sectors coordinate with formal sectors, jointly contributing to a greener recycling solution. This case study examines the systematic effort to transform and integrate the informal sector into the formal recycling industry for managing e-waste in Guiyu, China. This paper analyzes the policy design, implementation, technology development and market establishment of Guiyu's formal sector that enabled the evolution of the local informal recycling industry. The results show that the salient success factor is to offer advanced and centralized e-waste treatment by constructing a formal recycling sector while maintaining the competitive characteristics of the old informal businesses, including manual dismantling and private e-waste collection networks. Those characteristics ensured increased reuse value and sufficient e-waste sources. Meanwhile, the study found that many challenges and conflicts during this transition are rooted in the often-overlooked societal and historical contexts that profoundly shaped the local recycling industry. Authorities of regions facing challenges regulating informal recycling of e-waste, especially developing countries, could initiate similar systems based on local realities and the collaboration between formal and informal sectors to minimize the environmental and societal consequences of unregulated informal e-waste recycling.
In this paper, we consider the following initial-boundary value problem
{ut−Δut−Δu=|x|σ|u|p−1u,x∈Ω,t>0,u(x,t)=0,x∈∂Ω,t>0,u(x,0)=u0(x),x∈Ω | (1) |
and its corresponding steady-state problem
{−Δu=|x|σ|u|p−1u,x∈Ω,u=0,x∈∂Ω, | (2) |
where
1<p<{∞,n=1,2;n+2n−2,n≥3,σ>{−n,n=1,2;(p+1)(n−2)2−n,n≥3. | (3) |
(1) was called homogeneous (inhomogeneous) pseudo-parabolic equation when
The homogeneous problem, i.e.
Li and Du [12] studied the Cauchy problem of equation in (1) with
(1) If
(2) If
Φα:={ξ(x)∈BC(Rn):ξ(x)≥0,lim inf|x|↑∞|x|αξ(x)>0}, |
and
Φα:={ξ(x)∈BC(Rn):ξ(x)≥0,lim sup|x|↑∞|x|αξ(x)<∞}. |
Here
In view of the above introductions, we find that
(1) for Cauchy problem in
(2) for zero Dirichlet problem in a bounded domain
The difficulty of allowing
σ>(p+1)(n−2)2−n⏟<0 if n≥3 |
for
The main results of this paper can be summarized as follows: Let
(1) (the case
(2) (the case
(3) (arbitrary initial energy level) For any
(4) Moreover, under suitable assumptions, we show the exponential decay of global solutions and lifespan (i.e. the upper bound of blow-up time) of the blowing-up solutions.
The organizations of the remain part of this paper are as follows. In Section 2, we introduce the notations used in this paper and the main results of this paper; in Section 3, we give some preliminaries which will be used in the proofs; in Section 4, we give the proofs of the main results.
Throughout this paper we denote the norm of
‖ϕ‖Lγ={(∫Ω|ϕ(x)|γdx)1γ, if 1≤γ<∞;esssupx∈Ω|ϕ(x)|, if γ=∞. |
We denote the
Lp+1σ(Ω):={ϕ:ϕ is measurable on Ω and ‖u‖Lp+1σ<∞}, | (4) |
where
‖ϕ‖Lp+1σ:=(∫Ω|x|σ|ϕ(x)|p+1dx)1p+1,ϕ∈Lp+1σ(Ω). | (5) |
By standard arguments as the space
We denote the inner product of
(ϕ,φ)H10:=∫Ω(∇ϕ(x)⋅∇φ(x)+ϕ(x)φ(x))dx,ϕ,φ∈H10(Ω). | (6) |
The norm of
‖ϕ‖H10:=√(ϕ,ϕ)H10=√‖∇ϕ‖2L2+‖ϕ‖2L2,ϕ∈H10(Ω). | (7) |
An equivalent norm of
‖∇ϕ‖L2≤‖ϕ‖H10≤√λ1+1λ1‖∇ϕ‖L2,ϕ∈H10(Ω), | (8) |
where
λ1=infϕ∈H10(Ω)‖∇ϕ‖2L2‖ϕ‖2L2. | (9) |
Moreover, by Theorem 3.2, we have
for p and σ satisfying (4), H10(Ω)↪Lp+1σ(Ω) continuously and compactly. | (10) |
Then we let
Cpσ=supu∈H10(Ω)∖{0}‖ϕ‖Lp+1σ‖∇ϕ‖L2. | (11) |
We define two functionals
J(ϕ):=12‖∇ϕ‖2L2−1p+1‖ϕ‖p+1Lp+1σ | (12) |
and
I(ϕ):=‖∇ϕ‖2L2−‖ϕ‖p+1Lp+1σ. | (13) |
By (3) and (10), we know that
We denote the mountain-pass level
d:=infϕ∈NJ(ϕ), | (14) |
where
N:={ϕ∈H10(Ω)∖{0}:I(ϕ)=0}. | (15) |
By Theorem 3.3, we have
d=p−12(p+1)C−2(p+1)p−1pσ, | (16) |
where
For
Jρ={ϕ∈H10(Ω):J(ϕ)<ρ}. | (17) |
Then, we define the set
Nρ={ϕ∈N:‖∇ϕ‖2L2<2(p+1)ρp−1},ρ>d. | (18) |
For
λρ:=infϕ∈Nρ‖ϕ‖H10,Λρ:=supϕ∈Nρ‖ϕ‖H10 | (19) |
and two sets
Sρ:={ϕ∈H10(Ω):‖ϕ‖H10≤λρ,I(ϕ)>0},Sρ:={ϕ∈H10(Ω):‖ϕ‖H10≥Λρ,I(ϕ)<0}. | (20) |
Remark 1. There are two remarks on the above definitions.
(1) By the definitions of
(2) By Theorem 3.4, we have
√2(p+1)dp−1≤λρ≤Λρ≤√2(p+1)(λ1+1)ρλ1(p−1). | (21) |
Then the sets
‖sϕ‖H10≤√2(p+1)dp−1⇔s≤δ1:=√2(p+1)dp−1‖ϕ‖−1H10,I(sϕ)=s2‖∇ϕ‖2L2−sp+1‖ϕ‖p−1Lp+1σ>0⇔s<δ2:=(‖∇ϕ‖2L2‖ϕ‖p+1Lp+1σ)1p−1,‖sϕ‖H10≥√2(p+1)(λ1+1)ρλ1(p−1)⇔s≥δ3:=√2(p+1)(λ1+1)ρλ1(p−1)‖ϕ‖−1H10,I(sϕ)=s2‖∇ϕ‖2L2−sp+1‖ϕ‖p−1Lp+1σ<0⇔s>δ2. |
So,
{sϕ:0<s<min{δ1,δ2}}⊂Sρ,{sϕ:s>max{δ2,δ3}}⊂Sρ. |
In this paper we consider weak solutions to problem (1), local existence of which can be obtained by Galerkin's method (see for example [22,Chapter II,Sections 3 and 4]) and a standard limit process and the details are omitted.
Definition 2.1. Assume
∫Ω(utv+∇ut⋅∇v+∇u⋅∇v−|x|σ|u|p−1uv)dx=0 | (22) |
holds for any
u(⋅,0)=u0(⋅) in H10(Ω). | (23) |
Remark 2. There are some remarks on the above definition.
(1) Since
(2) Denote by
(3) Taking
‖u(⋅,t)‖2H10=‖u0‖2H10−2∫t0I(u(⋅,s))ds,0≤t≤T, | (24) |
where
(4) Taking
J(u(⋅,t))=J(u0)−∫t0‖us(⋅,s)‖2H10ds,0≤t≤T, | (25) |
where
Definition 2.2. Assume (3) holds. A function
∫Ω(∇u⋅∇v−|x|σ|u|p−1uv)dx=0 | (26) |
holds for any
Remark 3. There are some remarks to the above definition.
(1) By (10), we know all the terms in (26) are well-defined.
(2) If we denote by
Φ={ϕ∈H10(Ω):J′(ϕ)=0 in H−1(Ω)}⊂(N∪{0}), | (27) |
where
With the set
Definition 2.3. Assume (3) holds. A function
J(u)=infϕ∈Φ∖{0}J(ϕ). |
With the above preparations, now we can state the main results of this paper. Firstly, we consider the case
(1)
(2)
(3)
Theorem 2.4. Assume (3) holds and
‖∇u(⋅,t)‖L2≤√2(p+1)J(u0)p−1,0≤t<∞, | (28) |
where
V:={ϕ∈H10(Ω):J(ϕ)≤d,I(ϕ)>0}. | (29) |
In, in addition,
‖u(⋅,t)‖H10≤‖u0‖H10exp[−λ1λ1+1(1−(J(u0)d)p−12)t]. | (30) |
Remark 4. Since
J(u0)>p−12(p+1)‖∇u0‖2L2>0. |
So the equality (28) makes sense.
Theorem 2.5. Assume (3) holds and
limt↑Tmax∫t0‖u(⋅,s)‖2H10ds=∞, |
where
W:={ϕ∈H10(Ω):J(ϕ)≤d,I(ϕ)<0} | (31) |
and
Tmax≤4p‖u0‖2H10(p−1)2(p+1)(d−J(u0)). | (32) |
Remark 5. There are two remarks.
(1) If
(2) The sets
f(s)=J(sϕ)=s22‖∇ϕ‖2L2−sp+1p+1‖ϕ‖p+1Lp+1σ,g(s)=I(sϕ)=s2‖∇ϕ‖2L2−sp+1‖ϕ‖p+1Lp+1σ. |
Then (see Fig. 2)
(a)
maxs∈[0,∞)f(s)=f(s∗3)=p−12(p+1)(‖∇ϕ‖L2‖ϕ‖Lp+1σ)2(p+1)p−1≤d⏟By (14) since s∗3ϕ∈N, | (33) |
(b)
maxs∈[0,∞)g(s)=g(s∗1)=p−1p+1(2p+1)2p−1(‖∇ϕ‖L2‖ϕ‖Lp+1σ)2(p+1)p−1, |
(c)
f(s∗2)=g(s∗2)=p−12p(p+12p)2p−1(‖∇ϕ‖L2‖ϕ‖Lp+1σ)2(p+1)p−1, |
where
s∗1:=(2‖∇ϕ‖2L2(p+1)‖ϕ‖p+1Lp+1σ)1p−1<s∗2:=((p+1)‖∇ϕ‖2L22p‖ϕ‖p+1Lp+1σ)1p−1<s∗3:=(‖∇ϕ‖2L2‖ϕ‖p+1Lp+1σ)1p−1<s∗4:=((p+1)‖∇ϕ‖2L22‖ϕ‖p+1Lp+1σ)1p−1. |
So,
Theorem 2.6. Assume (3) holds and
G:={ϕ∈H10(Ω):J(ϕ)=d,I(ϕ)=0}. | (34) |
Remark 6. There are two remarks on the above theorem.
(1) Unlike Remark 5, it is not easy to show
(2) To prove the above Theorem, we only need to show
Theorem 2.7. Assume (3) holds and let
Secondly, we consider the case
Theorem 2.8. Assume (3) holds and the initial value
(i): If
(ii): If
Here
Next, we show the solution of the problem (1) can blow up at arbitrary initial energy level (Theorem 2.10). To this end, we firstly introduce the following theorem.
Theorem 2.9. Assume (3) holds and
Tmax≤8p‖u0‖2H10(p−1)2(λ1(p−1)λ1+1‖u0‖2H10−2(p+1)J(u0)) | (35) |
and
limt↑Tmax∫t0‖u(⋅,s)‖2H10ds=∞, |
where
ˆW:={ϕ∈H10(Ω):J(ϕ)<λ1(p−1)2(λ1+1)(p+1)‖ϕ‖2H10}. | (36) |
and
By using the above theorem, we get the following theorem.
Theorem 2.10. For any
The following lemma can be found in [11].
Lemma 3.1. Suppose that
F″(t)F(t)−(1+γ)(F′(t))2≥0 |
for some constant
T≤F(0)γF′(0)<∞ |
and
Theorem 3.2. Assume
Proof. Since
We divide the proof into three cases. We will use the notation
Case 1.
H10(Ω)↪Lp+1(Ω) continuously and compactly. | (37) |
Then we have, for any
‖u‖p+1Lp+1σ=∫Ω|x|σ|u|p+1dx≤Rσ‖u‖p+1Lp+1≲‖u‖p+1H10, |
which, together with (37), implies
Case 2.
H10(Ω)↪L(p+1)rr−1(Ω) continuously and compactly, | (38) |
for any
‖u‖p+1Lp+1σ=∫Ω|x|σ|u|p+1dx≤(∫B(0,R)|x|σrdx)1r(∫Ω|u|(p+1)rr−1dx)r−1r≤{(2σr+1Rσr+1)1r‖u‖p+1L(p+1)rr−1≲‖u‖p+1H10,n=1;(2πσr+2Rσr+2)1r‖u‖p+1L(p+1)rr−1≲‖u‖p+1H10,n=2, |
which, together with (38), implies
Case 3.
−σn<1r<1−(p+1)(n−2)2n. |
By the second inequality of the above inequalities, we have
(p+1)rr−1=p+11−1r<p+1(p+1)(n−2)2n=2nn−2. |
So,
H10(Ω)↪L(p+1)rr−1(Ω) continuously and compactly. | (39) |
Then by Hölder's inequality, for any
‖u‖p+1Lp+1σ=∫Ω|x|σ|u|p+1dx≤(∫B(0,R)|x|σrdx)1r(∫Ω|u|(p+1)rr−1dx)r−1r≤(ωn−1σr+nRσr+n)1r‖u‖p+1L(p+1)rr−1≲‖u‖p+1H10, |
which, together with (39), implies
Theorem 3.3. Assume
d=p−12(p+1)C2(p+1)p−1pσ, |
where
Proof. Firstly, we show
infϕ∈NJ(ϕ)=minϕ∈H10(Ω)∖{0}J(s∗ϕϕ), | (40) |
where
s∗ϕ:=(‖∇ϕ‖2L2‖ϕ‖p+1Lp+1σ)1p−1. | (41) |
By the definition of
On one hand, since
minϕ∈H10(Ω)∖{0}J(s∗ϕϕ)≤minϕ∈NJ(s∗ϕϕ)=minϕ∈NJ(ϕ). |
On the other hand, since
infϕ∈NJ(ϕ)≤infϕ∈H10(Ω)∖{0}J(s∗ϕϕ). |
Then (40) follows from the above two inequalities.
By (40), the definition of
d=minϕ∈H10(Ω)∖{0}J(s∗ϕϕ)=p−12(p+1)minϕ∈H10(Ω)∖{0}(‖∇ϕ‖L2‖ϕ‖Lp+1σ)2(p+1)p−1=p−12(p+1)C−2(p+1)p−1pσ. |
Theorem 3.4. Assume (3) holds. Let
√2(p+1)dp−1≤λρ≤Λρ≤√2(p+1)(λ1+1)ρλ1(p−1). | (42) |
Proof. Let
λρ≤Λρ. | (43) |
Since
d=infϕ∈NJ(ϕ)=p−12(p+1)infϕ∈N‖∇ϕ‖2L2≤p−12(p+1)infϕ∈Nρ‖ϕ‖2H10=p−12(p+1)λ2ρ, |
which implies
λρ≥√2(p+1)dp−1 |
On the other hand, by (8) and (18), we have
Λρ=supϕ∈Nρ‖ϕ‖H10≤√λ1+1λ1supϕ∈Nρ‖∇ϕ‖L2≤√λ1+1λ1√2(p+1)ρp−1. |
Combining the above two inequalities with (43), we get (42), the proof is complete.
Theorem 3.5. Assume (3) holds and
Proof. We only prove the invariance of
For any
‖∇ϕ‖2L2<‖ϕ‖p+1Lp+1σ≤Cp+1pσ‖∇ϕ‖p+1L2, |
which implies
‖∇ϕ‖L2>C−p+1p−1pσ. | (44) |
Let
I(u(⋅,t))<0,t∈[0,ε]. | (45) |
Then by (24),
J(u(⋅,t))<d for t∈(0,ε]. | (46) |
We argument by contradiction. Since
J(u(⋅,t0))<d | (47) |
(note (25) and (46),
‖∇u(⋅,t0)‖L2≥C−p+1p−1pσ>0, |
which, together with
J(u(⋅,t0))≥d, |
which contradicts (47). So the conclusion holds.
Theorem 3.6. Assume (3) holds and
‖∇u(⋅,t)‖2L2≥2(p+1)p−1d,0≤t<Tmax, | (48) |
where
Proof. Let
By the proof in Theorem 3.3,
d=minϕ∈H10(Ω)∖{0}J(s∗ϕϕ)≤minϕ∈N−J(s∗ϕϕ)≤J(s∗uu(⋅,t))=(s∗u)22‖∇u(⋅,t)‖2L2−(s∗u)p+1p+1‖u(⋅,t)‖p+1Lp+1σ≤((s∗u)22−(s∗u)p+1p+1)‖∇u(⋅,t)‖2L2, |
where we have used
d=minϕ∈H10(Ω)∖{0}J(s∗ϕϕ)≤minϕ∈N−J(s∗ϕϕ)≤J(s∗uu(⋅,t))=(s∗u)22‖∇u(⋅,t)‖2L2−(s∗u)p+1p+1‖u(⋅,t)‖p+1Lp+1σ≤((s∗u)22−(s∗u)p+1p+1)‖∇u(⋅,t)‖2L2, |
Then
d≤max0≤s≤1(s22−sp+1p+1)‖∇u(⋅,t)‖2L2=(s22−sp+1p+1)s=1‖∇u(⋅,t)‖2L2=p−12(p+1)‖∇u(⋅,t)‖2L2, |
and (48) follows from the above inequality.
Theorem 3.7. Assume (3) holds and
Proof. Firstly, we show
12‖∇u0‖2L2−1p+1‖u0‖p+1Lp+1σ=J(u0)<λ1(p−1)2(λ1+1)(p+1)‖u0‖2H10≤p−12(p+1)‖∇u0‖2L2, |
which implies
I(u0)=‖∇u0‖2L2−‖u0‖p+1Lp+1σ<0. |
Secondly, we prove
J(u0)<λ1(p−1)2(λ1+1)(p+1)‖u0‖2H10<λ1(p−1)2(λ1+1)(p+1)‖u(⋅,t0)‖2H10≤p−12(p+1)‖∇u(⋅,t0)‖2L2. | (49) |
On the other hand, by (24), (12), (13) and
J(u0)≥J(u(⋅,t0))=p−12(p+1)‖∇u(⋅,t0)‖2L2, |
which contradicts (49). The proof is complete.
Proof of Theorem 2.4. Let
J(u0)≥J(u(⋅,t))≥p−12(p+1)‖∇u(⋅,t)‖2L2,0≤t<Tmax, |
which implies
‖∇u(⋅,t)‖L2≤√2(p+1)J(u0)p−1,0≤t<∞. | (50) |
Next, we prove
ddt(‖u(⋅,t)‖2H10)=−2I(u(⋅,t))=−2(‖∇u(⋅,t)‖2L2−‖u(⋅,t)‖p+1Lp+1σ)≤−2(1−Cp+1pσ‖∇u(⋅,t)‖p−1L2)‖∇u(⋅,t)‖2L2≤−2(1−Cp+1pσ(√2(p+1)J(u0)p−1)p−1)‖∇u(⋅,t)‖2L2=−2(1−(J(u0)d)p−12)‖∇u(⋅,t)‖2L2≤−2λ1λ1+1(1−(J(u0)d)p−12)‖u(⋅,t)‖2H10, |
which leads to
‖u(⋅,t)‖2H10≤‖u0‖2H10exp[−2λ1λ1+1(1−(J(u0)d)p−12)t]. |
The proof is complete.
Proof of Theorem 2.5. Let
Firstly, we consider the case
ξ(t):=(∫t0‖u(⋅,s)‖2H10ds)12,η(t):=(∫t0‖us(⋅,s)‖2H10ds)12,0≤t<Tmax. | (51) |
For any
F(t):=ξ2(t)+(T∗−t)‖u0‖2H10+β(t+α)2,0≤t≤T∗. | (52) |
Then
F(0)=T∗‖u0‖2H10+βα2>0, | (53) |
F′(t)=‖u(⋅,t)‖2H10−‖u0‖2H10+2β(t+α)=2(12∫t0dds‖u(⋅,s)‖2H10ds+β(t+α)),0≤t≤T∗, | (54) |
and (by (24), (12), (13), (48), (25))
F″(t)=−2I(u(⋅,t))+2β=(p−1)‖∇u(⋅,t)‖2L2−2(p+1)J(u(⋅,t))+2β≥2(p+1)(d−J(u0))+2(p+1)η2(t)+2β,0≤t≤T∗. | (55) |
Since
F′(t)≥2β(t+α). |
Then
F(t)=F(0)+∫t0F′(s)ds≥T∗‖u0‖2H10+βα2+2αβt+βt2,0≤t≤T∗. | (56) |
By (6), Schwartz's inequality and Hölder's inequality, we have
12∫t0dds‖u(⋅,s)‖2H10ds=∫t0(u(⋅,s),us(⋅,s))H10ds≤∫t0‖u(⋅,s)‖H10‖us(⋅,s)‖H10ds≤ξ(t)η(t),0≤t≤T∗, |
which, together with the definition of
(F(t)−(T∗−t)‖u0‖2H10)(η2(t)+β)=(ξ2(t)+β(t+α)2)(η2(t)+β)=ξ2(t)η2(t)+βξ2(t)+β(t+α)2η2(t)+β2(t+α)2≥ξ2(t)η2(t)+2ξ(t)η(t)β(t+α)+β2(t+α)2≥(ξ(t)η(t)+β(t+α))2≥(12∫t0dds‖u(⋅,s)‖2H10ds+β(t+α))2,0≤t≤T∗. |
Then it follows from (54) and the above inequality that
(F′(t))2=4(12t∫0dds‖u(s)‖2H10ds+β(t+α))2≤4F(t)(η2(t)+β),0≤t≤T∗. | (57) |
In view of (55), (56), and (57), we have
F(t)F″(t)−p+12(F′(t))2≥F(t)(2(p+1)(d−J(u0))−2pβ),0≤t≤T∗. |
If we take
0<β≤p+1p(d−J(u0)), | (58) |
then
T∗≤F(0)(p+12−1)F′(0)=T∗‖u0‖2H10+βα2(p−1)αβ. |
Then for
α∈(‖u0‖2H10(p−1)β,∞), | (59) |
we get
T∗≤βα2(p−1)αβ−‖u0‖2H10. |
Minimizing the above inequality for
T∗≤βα2(p−1)αβ−‖u0‖2H10|α=2‖u0‖2H10(p−1)β=4‖u0‖2H10(p−1)2β. |
Minimizing the above inequality for
T∗≤4p‖u0‖2H10(p−1)2(p+1)(d−J(u0)). |
By the arbitrariness of
Tmax≤4p‖u0‖2H10(p−1)2(p+1)(d−J(u0)). |
Secondly, we consider the case
Proof of Theorems 2.6 and 2.7. Since Theorem 2.6 follows from Theorem 2.7 directly, we only need to prove Theorem 2.7.
Firstly, we show
d=infϕ∈NJ(ϕ)=p−12(p+1)infϕ∈N‖∇ϕ‖2L2. |
Then a minimizing sequence
limk↑∞J(ϕk)=p−12(p+1)limk↑∞‖∇ϕk‖2L2=d, | (60) |
which implies
(1)
(2)
Now, in view of
limk↑∞J(ϕk)=p−12(p+1)limk↑∞‖∇ϕk‖2L2=d, | (60) |
We claim
‖∇φ‖2L2=‖φ‖p+1Lp+1σ i.e. I(φ)=0. | (62) |
In fact, if the claim is not true, then by (61),
‖∇φ‖2L2<‖φ‖p+1Lp+1σ. |
By the proof of Theorem 3.3, we know that
J(s∗φφ)≥d, | (63) |
where
J(s∗φφ)≥d, | (63) |
On the other hand, since
J(s∗φφ)=p−12(p+1)(s∗φ)2‖∇φ‖2L2<p−12(p+1)‖∇φ‖2L2≤p−12(p+1)lim infk↑∞‖∇ϕk‖2L2=d, |
which contradicts to (63). So the claim is true, i.e.
limk↑∞‖∇ϕk‖2L2=‖φ‖p+1Lp+1σ, |
which, together with
Second, we prove
limk↑∞‖∇ϕk‖2L2=‖φ‖p+1Lp+1σ, |
Then
A:={τ(s)(φ+sv):s∈(−ε,ε)} |
is a curve on
A:={τ(s)(φ+sv):s∈(−ε,ε)} |
where
ξ:=2∫Ω∇(φ+sv)⋅∇vdx‖φ+sv‖p+1Lp+1σ,η:=(p+1)∫Ω|x|σ|φ+sv|p−1(φ+sv)vdx‖∇(φ+sv)‖2L2. |
Since (62), we get
τ′(0)=1(p−1)‖φ‖p+1Lp+1σ(2∫Ω∇φ∇vdx−(p+1)∫Ω|x|σ|φ|p−1φvdx). | (65) |
Let
ϱ(s):=J(τ(s)(φ+sv))=τ2(s)2‖∇(φ+sv)‖2L2−τp+1(s)p+1‖φ+sv‖p+1Lp+1σ,s∈(−ε,ε). |
Since
0=ϱ′(0)=τ(s)τ′(s)‖∇(φ+sv)‖2L2+τ2(s)∫Ω∇(φ+sv)⋅∇vdx|s=0−τp(s)τ′(s)‖φ+sv‖p+1Lp+1σ−τp+1(s)∫Ω|x|σ|φ+sv|p−1(φ+sv)vdx|s=0=∫Ω∇φ⋅∇vdx−∫Ω|x|σ|φ|p−1φvdx. |
So,
Finally, in view of Definition 2.3 and
d=infϕ∈Φ∖{0}J(ϕ). | (66) |
In fact, by the above proof and (27), we have
d=infϕ∈NJ(ϕ) |
and
Proof of Theorem 2.8. Let
ω(u0)=∩t≥0¯{u(⋅,s):s≥t}H10(Ω) |
the
(i) Assume
v(x,t)={u(x,t), if 0≤t≤t0;0, if t>t0 |
is a global weak solution of problem (1), and the proof is complete.
We claim that
I(u(⋅,t))>0,0≤t<Tmax. | (67) |
Since
I(u(⋅,t))>0,0≤t<t0 | (68) |
and
I(u(⋅,t0))=0, | (69) |
which together with the definition of
‖u(⋅,t0)‖H10≥λρ. | (70) |
On the other hand, it follows from (24), (68) and
‖u(⋅,t)‖H10<‖u0‖H10≤λρ, |
which contradicts (70). So (67) is true. Then by (24) again, we get
‖u(⋅,t)‖H10≤‖u0‖H10,0≤t<Tmax, |
which implies
By (24) and (67),
limt↑∞‖u(⋅,t)‖H10=c. |
Taking
∫∞0I(u(⋅,s))ds≤12(‖u0‖2H10−c)<∞. |
Note that
limn↑∞I(u(⋅,tn))=0. | (71) |
Let
u(⋅,tn)→ω in H10(Ω) as n↑∞. | (72) |
Then by (71), we get
I(ω)=limn↑∞I(u(⋅,tn))=0. | (73) |
As the above, one can easily see
‖ω‖H10<λρ≤λJ(u0),J(ω)<J(u0)⏟⇒ω∈JJ(u0), |
which implies
limt↑∞‖u(⋅,t)‖H10=limn↑∞‖u(⋅,tn)‖H10=‖ω‖H10=0. |
(ⅱ) Assume
I(u(⋅,t))<0,0≤t<Tmax. | (74) |
Since
I(u(⋅,t))<0,0≤t<t0 | (75) |
and
I(u(⋅,t0))=0. | (76) |
Since (75), by (44) and
‖∇u(⋅,t0)‖L2≥C−p+1p−1pσ, |
which, together with the definition of
‖u(⋅,t0)‖H10≤Λρ. | (77) |
On the other hand, it follows from (24), (75) and
‖u(⋅,t)‖H10>‖u0‖H10≥Λρ, |
which contradicts (77). So (74) is true.
Suppose by contradiction that
limt↑∞‖u(⋅,t)‖H10=˜c, |
Taking
−∫∞0I(u(⋅,s))ds≤12(˜c−‖u0‖2H10)<∞. |
Note
limn↑∞I(u(⋅,tn))=0. | (78) |
Let
u(⋅,tn)→ω in H10(Ω) as n↑∞. | (79) |
Since
limt↑∞‖u(⋅,t)‖H10=limn↑∞‖u(⋅,tn)‖H10=‖ω‖H10. |
Then by (78), we get
I(ω)=limn↑∞I(u(⋅,tn))=0. | (80) |
By (24), (25) and (74), one can easily see
‖ω‖H10>‖u0‖H10≥Λρ≥ΛJ(u0),J(ω)<J(u0)⏟⇒ω∈JJ(u0), |
which implies
Proof of Theorem 2.9. Let
‖∇u(⋅,t)‖2L2≥λ1λ1+1‖u(⋅,t)‖2H10≥λ1λ1+1‖u0‖2H10,0≤t<Tmax. | (81) |
The remain proofs are similar to the proof of Theorem 2.9. For any
F″(t)=−2I(u(⋅,t))+2β=(p−1)‖∇u(⋅,t)‖2L2−2(p+1)J(u(⋅,t))+2β≥λ1(p−1)λ1+1‖u0‖2H10−2(p+1)J(u0)+2(p+1)η2(t)+2β,0≤t≤T∗. | (82) |
We also have (56) and (57). Then it follows from (56), (57) and (82) that
F(t)F″(t)−p+12(F′(t))2≥F(t)(λ1(p−1)λ1+1‖u0‖2H10−2(p+1)J(u0)−2pβ),0≤t≤T∗. |
If we take
0<β≤12p(λ1(p−1)λ1+1‖u0‖2H10−2(p+1)J(u0)), | (83) |
then
T∗≤F(0)(p+12−1)F′(0)=T∗‖u0‖2H10+βα2(p−1)αβ. |
Then for
α∈(‖u0‖2H10(p−1)β,∞), | (84) |
we get
T∗≤βα2(p−1)αβ−‖u0‖2H10. |
Minimizing the above inequality for
T∗≤βα2(p−1)αβ−‖u0‖2H10|α=2‖u0‖2H10(p−1)β=4‖u0‖2H10(p−1)2β. |
Minimizing the above inequality for
T∗≤8p‖u0‖2H10(p−1)2(λ1(p−1)λ1+1‖u0‖2H10−2(p+1)J(u0)). |
By the arbitrariness of
Tmax≤8p‖u0‖2H10(p−1)2(λ1(p−1)λ1+1‖u0‖2H10−2(p+1)J(u0)). |
Proof of Theorem 2.10. For any
‖αψ‖2H10>2(λ1+1)(p+1)λ1(p−1)M. | (85) |
For such
J(s∗3ϕ)≥M−J(αψ), | (86) |
where (see Remark 5)
J(s∗3ϕ)≥M−J(αψ), | (86) |
which can be done since
J(s∗3ϕ)=p−12(p+1)(‖∇ϕ‖L2‖ϕ‖Lp+1σ)2(p+1)p−1 |
and
By Remark 5 again,
J({sϕ:0≤s<∞})=(−∞,J(s∗3ϕ)]. | (87) |
By (87) and (86), we can choose
J(u0)=J(v)+J(αψ)=M |
and (note (85))
J(u0)=M<λ1(p−1)2(λ1+1)(p+1)‖αψ‖2H10≤λ1(p−1)2(λ1+1)(p+1)(‖αψ‖2H10+‖v‖2H10)=λ1(p−1)2(λ1+1)(p+1)‖u0‖2H10. |
Let
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