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A survey of gradient methods for solving nonlinear optimization

  • The paper surveys, classifies and investigates theoretically and numerically main classes of line search methods for unconstrained optimization. Quasi-Newton (QN) and conjugate gradient (CG) methods are considered as representative classes of effective numerical methods for solving large-scale unconstrained optimization problems. In this paper, we investigate, classify and compare main QN and CG methods to present a global overview of scientific advances in this field. Some of the most recent trends in this field are presented. A number of numerical experiments is performed with the aim to give an experimental and natural answer regarding the numerical one another comparison of different QN and CG methods.

    Citation: Predrag S. Stanimirović, Branislav Ivanov, Haifeng Ma, Dijana Mosić. A survey of gradient methods for solving nonlinear optimization[J]. Electronic Research Archive, 2020, 28(4): 1573-1624. doi: 10.3934/era.2020115

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  • The paper surveys, classifies and investigates theoretically and numerically main classes of line search methods for unconstrained optimization. Quasi-Newton (QN) and conjugate gradient (CG) methods are considered as representative classes of effective numerical methods for solving large-scale unconstrained optimization problems. In this paper, we investigate, classify and compare main QN and CG methods to present a global overview of scientific advances in this field. Some of the most recent trends in this field are presented. A number of numerical experiments is performed with the aim to give an experimental and natural answer regarding the numerical one another comparison of different QN and CG methods.



    In this paper, we consider the following initial-boundary value problem

    {utΔutΔu=|x|σ|u|p1u,xΩ,t>0,u(x,t)=0,xΩ,t>0,u(x,0)=u0(x),xΩ (1)

    and its corresponding steady-state problem

    {Δu=|x|σ|u|p1u,xΩ,u=0,xΩ, (2)

    where ΩRn (n1 is an integer) is a bounded domain with boundary Ω and u0H10(Ω); the parameters p and σ satisfy

    1<p<{,n=1,2;n+2n2,n3,σ>{n,n=1,2;(p+1)(n2)2n,n3. (3)

    (1) was called homogeneous (inhomogeneous) pseudo-parabolic equation when σ=0 (σ0). The concept "pseudo-parabolic" was proposed by Showalter and Ting in 1970 in the paper [20], where the linear case was considered. Pseudo-parabolic equations describe a variety of important physical processes, such as the seepage of homogeneous fluids through a fissured rock [1], the unidirectional propagation of nonlinear, dispersive, long waves [2,23], and the aggregation of populations [17].

    The homogeneous problem, i.e. σ=0, was studied in [3,4,5,7,9,10,13,15,16,21,24,25,26,27,28,29]. Especially, for the Cauchy problem (i.e. Ω=Rn and there is no boundary condition), Cao et al. [4] showed the critical Fujita exponent pc (which was firstly introduced by Fujita in [8]) is 1+2/n, i.e. if 1<ppc, then any nontrivial solution blows up in finite time, while global solutions exist if p>pc. In [28], Yang et al. proved that for p>pc, there is a secondary critical exponent αc=2/(p1) such that the solution blows up in finite time for u0 behaving like |x|α at {|x|} if α=(0,αc); and there are global solutions for for u0 behaving like |x|α at {|x|} if α=(αc,n). For the zero Dirichlet boundary problem in a bounded domain Ω, in [13,25,26], the authors studied the properties of global existence and blow-up by potential well method (which was firstly introduced by Sattinger [19] and Payne and Sattinger [18], then developed by Liu and Zhao in [14]), and they showed the global existence, blow-up and asymptotic behavior of solutions with initial energy at subcritical, critical and supercritical energy level. The results of [13,25,26] were extended by Luo [15] and Xu and Zhou [24] by studying the lifespan (i.e. the upper bound of the blow-up time) of the blowing-up solutions. Recently, Xu et al. [27] and Han [9] extended the previous studies by considering the problem with general nonlinearity.

    Li and Du [12] studied the Cauchy problem of equation in (1) with σ>0. They got the critical Fujita exponent (pc) and second critical exponent (αc) by the integral representation and comparison principle. The main results obtained in [12] are as follows:

    (1) If 1<ppc:=1+(2+σ)/n, then every nontrivial solution blows up in finite time.

    (2) If p>pc, the distribution of the initial data has effect on the blow-up phenomena. More precisely, if u0Φα and 0<α<αc:=(2+σ)/(p1) or u0 is large enough, then the solution blows up in finite time; if u0=μϕ(x), ϕΦα with αc<α<n, 0<μ<μ1, then the solution exists globally, where μ1 is some positive constant,

    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 BC(Rn) is the set of bounded continuous functions in Rn.

    In view of the above introductions, we find that

    (1) for Cauchy problem in Rn, only the case σ0 was studied;

    (2) for zero Dirichlet problem in a bounded domain Ω, only the case σ=0 was studied.

    The difficulty of allowing σ to be less than 0 is the term |x|σ become infinity at x=0. In this paper, we consider the problem in a bounded domain Ω with zero Dirichlet boundary condition, i.e. problem (1), and the parameters satisfies (3), which allows σ to be less than 0. To overcome the singularity of |x|σ at x=0, we use potential well method by introducing the |x|σ weighed-Lp+1(Ω) space and assume there is a lower bound of σ, i.e,

    σ>(p+1)(n2)2n<0 if n3

    for n3.

    The main results of this paper can be summarized as follows: Let J and I be the functionals given in (12) and (13), respectively; d be the mountain-pass level given in (14); Sρ and Sρ be the sets defined in (20).

    (1) (the case J(u0)d, see Fig. 1) If u0H10(Ω) such that (I(u0),J(u0)) is in the dark gray region (BR), then the solution blows up in finite time; if u0H10(Ω) such that (I(u0),J(u0)) is in the light gray region (GR), then the solution exists globally; if u0H10(Ω) such that (I(u0),J(u0))=(0,d), then u0 is a ground-state solution and (1) admits a global solution uu0; there is no u0H10(Ω) such that (I(u0),J(u0)) is in the dotted part (ER).

    Figure 1.  The results for J(u0)d.

    (2) (the case J(u0)>d) If u0Sρ for some ρJ(u0)>d, then the solution exists globally and goes to 0 in H10(Ω) as times goes to infinity; if u0Sρ for some ρJ(u0)>d, then the solution blows up in finite time.

    (3) (arbitrary initial energy level) For any MR, there exits a u0H10(Ω)) satisfying J(u0)=M such that the corresponding solution blows up in finite time.

    (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 Lγ(Ω) for 1γ by Lγ. That is, for any ϕLγ(Ω),

    ϕLγ={(Ω|ϕ(x)|γdx)1γ, if 1γ<;esssupxΩ|ϕ(x)|, if γ=.

    We denote the |x|σ-weighted Lp+1(Ω) space by Lp+1σ(Ω), which is defined as

    Lp+1σ(Ω):={ϕ:ϕ is measurable on Ω and uLp+1σ<}, (4)

    where

    ϕLp+1σ:=(Ω|x|σ|ϕ(x)|p+1dx)1p+1,ϕLp+1σ(Ω). (5)

    By standard arguments as the space Lp+1(Ω), one can see Lp+1σ(Ω) is a Banach space with the norm Lp+1σ.

    We denote the inner product of H10(Ω) by (,)H10, i.e.,

    (ϕ,φ)H10:=Ω(ϕ(x)φ(x)+ϕ(x)φ(x))dx,ϕ,φH10(Ω). (6)

    The norm of H10(Ω) is denoted by H10, i.e.,

    ϕH10:=(ϕ,ϕ)H10=ϕ2L2+ϕ2L2,ϕH10(Ω). (7)

    An equivalent norm of H10(Ω) is ()L2, and by Poincaré's inequality, we have

    ϕL2ϕH10λ1+1λ1ϕL2,ϕH10(Ω), (8)

    where λ1 is the first eigenvalue of Δ with zero Dirichlet boundary condition, i.e,

    λ1=infϕH10(Ω)ϕ2L2ϕ2L2. (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 as the optimal constant of the embedding H10(Ω)Lp+1σ(Ω), i.e.,

    Cpσ=supuH10(Ω){0}ϕLp+1σϕL2. (11)

    We define two functionals J and I on H10(Ω) by

    J(ϕ):=12ϕ2L21p+1ϕp+1Lp+1σ (12)

    and

    I(ϕ):=ϕ2L2ϕp+1Lp+1σ. (13)

    By (3) and (10), we know that J and I are well-defined on H10(Ω).

    We denote the mountain-pass level d by

    d:=infϕNJ(ϕ), (14)

    where N is the Nehari manifold, which is defined as

    N:={ϕH10(Ω){0}:I(ϕ)=0}. (15)

    By Theorem 3.3, we have

    d=p12(p+1)C2(p+1)p1pσ, (16)

    where Cpσ is the positive constant given in (11).

    For ρR, we define the sub-level set Jρ of J as

    Jρ={ϕH10(Ω):J(ϕ)<ρ}. (17)

    Then, we define the set Nρ:=NJρ. In view of (15), (12), (17), we get

    Nρ={ϕN:ϕ2L2<2(p+1)ρp1},ρ>d. (18)

    For ρ>d, we define two constants

    λρ:=infϕNρϕH10,Λρ:=supϕNρϕH10 (19)

    and two sets

    Sρ:={ϕH10(Ω):ϕH10λρ,I(ϕ)>0},Sρ:={ϕH10(Ω):ϕH10Λρ,I(ϕ)<0}. (20)

    Remark 1. There are two remarks on the above definitions.

    (1) By the definitions of Nρ, λρ and Λρ, it is easy to see λρ is non-increasing with respect to ρ and Λρ is non-decreasing with respect to ρ.

    (2) By Theorem 3.4, we have

    2(p+1)dp1λρΛρ2(p+1)(λ1+1)ρλ1(p1). (21)

    Then the sets Sρ and Sρ are both nonempty. In fact, for any ϕH10(Ω){0} and s>0,

    sϕH102(p+1)dp1sδ1:=2(p+1)dp1ϕ1H10,I(sϕ)=s2ϕ2L2sp+1ϕp1Lp+1σ>0s<δ2:=(ϕ2L2ϕp+1Lp+1σ)1p1,sϕH102(p+1)(λ1+1)ρλ1(p1)sδ3:=2(p+1)(λ1+1)ρλ1(p1)ϕ1H10,I(sϕ)=s2ϕ2L2sp+1ϕp1Lp+1σ<0s>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 u0H10(Ω) and (3) holds. Let T>0 be a constant. A function u=u(x,t) is called a weak solution of problem (1) on Ω×[0,T] if u(,t)L(0,T;H10(Ω)), ut(,t)L2(0,T;H10(Ω)) and the following equality

    Ω(utv+utv+uv|x|σ|u|p1uv)dx=0 (22)

    holds for any vH10(Ω) and a.e. t[0,T]. Moreover,

    u(,0)=u0() in H10(Ω). (23)

    Remark 2. There are some remarks on the above definition.

    (1) Since u(,t)L(0,T;H10(Ω))L2(0,T;H10(Ω)), ut(,t)L2(0,T;H10(Ω)), we have uH1(0,T;H10(Ω)). According to [6], uC([0,T];H10(Ω)), then (23) makes sense. Moreover, by (10), all terms in (22) make sense for uC([0,T];H10(Ω)) and utL2(0,T;H10(Ω)).

    (2) Denote by Tmax the maximal existence of u, then u(,t)L(0,T;H10(Ω))C([0,T];H10(Ω)), ut(,t)L2(0,T;H10(Ω)) for any T<Tmax.

    (3) Taking v=u in (22), we get

    u(,t)2H10=u02H102t0I(u(,s))ds,0tT, (24)

    where H10 is defined in (7) and I is defined in (13).

    (4) Taking v=ut in (22), we get

    J(u(,t))=J(u0)t0us(,s)2H10ds,0tT, (25)

    where J is defined in (12).

    Definition 2.2. Assume (3) holds. A function uH10(Ω) is called a weak solution of (2) if

    Ω(uv|x|σ|u|p1uv)dx=0 (26)

    holds for any vH10(Ω).

    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 the set of weak solutions to (2), then by the definitions of J in (12) and N in (15), we have

    Φ={ϕH10(Ω):J(ϕ)=0 in H1(Ω)}(N{0}), (27)

    where J(ϕ)=0 in H1(Ω) means J(ϕ),ψ=0 for all ψH10 and , means the dual product between H1(Ω) and H10(Ω).

    With the set Φ defined above, we can defined the ground-state solution to (2).

    Definition 2.3. Assume (3) holds. A function uH10(Ω) is called a ground-state solution of (2) if uΦ{0} and

    J(u)=infϕΦ{0}J(ϕ).

    With the above preparations, now we can state the main results of this paper. Firstly, we consider the case J(u0)d. By the sign of I(u0), we can classify the discussions into three cases:

    (1) J(u0)d, I(u0)>0 (see Theorem 2.4);

    (2) J(u0)d, I(u0)<0 (see Theorem 2.5);

    (3) J(u0)d, I(u0)=0. In this case, by the definition of d in (14), we have u0=0 or J(u0)=d and I(u0)=0. In Theorem 2.6, we will show problem (1) admits a global solution u(,t)u0.

    Theorem 2.4. Assume (3) holds and u=u(x,t) is a weak solution to (1) with u0V, then u exists globally and

    u(,t)L22(p+1)J(u0)p1,0t<, (28)

    where

    V:={ϕH10(Ω):J(ϕ)d,I(ϕ)>0}. (29)

    In, in addition, J(u0)<d, we have the following decay estimate:

    u(,t)H10u0H10exp[λ1λ1+1(1(J(u0)d)p12)t]. (30)

    Remark 4. Since u0V, we have I(u0)>0. Then it follows from the definitions of J in (12) and I in (13) that

    J(u0)>p12(p+1)u02L2>0.

    So the equality (28) makes sense.

    Theorem 2.5. Assume (3) holds and u=u(x,t) is a weak solution to (1) with u0W. Then Tmax< and u blows up in finite time in the sense of

    limtTmaxt0u(,s)2H10ds=,

    where

    W:={ϕH10(Ω):J(ϕ)d,I(ϕ)<0} (31)

    and Tmax is the maximal existence time of u. If, in addition, J(u0)<d, then

    Tmax4pu02H10(p1)2(p+1)(dJ(u0)). (32)

    Remark 5. There are two remarks.

    (1) If J(ϕ)<0, then we can easily get from the definitions of J and I in (12) and (13) respectively that I(ϕ)<0. So we have ϕW if J(ϕ)<0.

    (2) The sets V and W defined in (29) and (31) respectively are both nonempty. In fact for any ϕH10(Ω){0}, we let

    f(s)=J(sϕ)=s22ϕ2L2sp+1p+1ϕp+1Lp+1σ,g(s)=I(sϕ)=s2ϕ2L2sp+1ϕp+1Lp+1σ.

    Then (see Fig. 2)

    Figure 2.  The graphs of f and g.

    (a) f(0)=f(s4)=0, f(s) is strictly increasing for s(0,s3), strictly decreasing for s(s3,), limsf(s)=, and

    maxs[0,)f(s)=f(s3)=p12(p+1)(ϕL2ϕLp+1σ)2(p+1)p1dBy (14) since s3ϕN, (33)

    (b) g(0)=g(s3)=0, g(s) is strictly increasing for s(0,s1), strictly decreasing for s(s1,), limsg(s)=, and

    maxs[0,)g(s)=g(s1)=p1p+1(2p+1)2p1(ϕL2ϕLp+1σ)2(p+1)p1,

    (c) f(s)<g(s) for 0<s<s2, f(s)>g(s) for s>s2, and

    f(s2)=g(s2)=p12p(p+12p)2p1(ϕL2ϕLp+1σ)2(p+1)p1,

    where

    s1:=(2ϕ2L2(p+1)ϕp+1Lp+1σ)1p1<s2:=((p+1)ϕ2L22pϕp+1Lp+1σ)1p1<s3:=(ϕ2L2ϕp+1Lp+1σ)1p1<s4:=((p+1)ϕ2L22ϕp+1Lp+1σ)1p1.

    So, {sϕ:s1<s<s3}V, {sϕ:s3<s<}W.

    Theorem 2.6. Assume (3) holds and u=u(x,t) is a weak solution to (1) with u0G. Then problem (1) admits a global solution u(,t)u0(), where

    G:={ϕH10(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 G. In fact, if we use the arguments as in Remark 5, we only have J(s3ϕ)d and I(s3ϕ)=0 (see Fig. 2 and (33)). In Theorem 2.7, we will use minimizing sequence argument to show G.

    (2) To prove the above Theorem, we only need to show G is the set of the ground-state solution of (2), which is done in Theorem 2.7.

    Theorem 2.7. Assume (3) holds and let G be the set defined in (34), then G and G is the set of the ground-state solution of (2).

    Secondly, we consider the case J(u0)>d, and we have the following theorem.

    Theorem 2.8. Assume (3) holds and the initial value u0H10(Ω) satisfying J(u0)>d.

    (i): If u0Sρ with ρJ(u0), then problem (1) admits a global weak solution u=u(x,t) and u(,t)H100 as t.

    (ii): If u0Sρ with ρJ(u0), then the weak solution u=u(x,t) of problem (1) blows up in finite time.

    Here Sρ and Sρ are the two sets defined in (20).

    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 u=u(x,t) is a weak solution to (1) with u0ˆW. Then

    Tmax8pu02H10(p1)2(λ1(p1)λ1+1u02H102(p+1)J(u0)) (35)

    and u blows up in finite time in the sense of

    limtTmaxt0u(,s)2H10ds=,

    where

    ˆW:={ϕH10(Ω):J(ϕ)<λ1(p1)2(λ1+1)(p+1)ϕ2H10}. (36)

    and Tmax is the maximal existence time of u.

    By using the above theorem, we get the following theorem.

    Theorem 2.10. For any MR, there exists u0H10(Ω) satisfying J(u0)=M such that the corresponding weak solution u=u(x,t) of problem (1) blows up in finite time.

    The following lemma can be found in [11].

    Lemma 3.1. Suppose that 0<T and suppose a nonnegative function F(t)C2[0,T) satisfies

    F(t)F(t)(1+γ)(F(t))20

    for some constant γ>0. If F(0)>0, F(0)>0, then

    TF(0)γF(0)<

    and F(t) as tT.

    Theorem 3.2. Assume p and σ satisfy (3). Then H10(Ω)Lp+1σ(Ω) continuously and compactly.

    Proof. Since ΩRn is a bounded domain, there exists a ball B(0,R):={xRn:|x|=x21+x2n<R}Ω.

    We divide the proof into three cases. We will use the notation ab which means there exits a positive constant C such that aCb.

    Case 1. σ0. By the assumption on p in (3), one can see

    H10(Ω)Lp+1(Ω) continuously and compactly. (37)

    Then we have, for any uH10(Ω),

    up+1Lp+1σ=Ω|x|σ|u|p+1dxRσup+1Lp+1up+1H10,

    which, together with (37), implies H10(Ω)Lp+1σ(Ω) continuously and compactly.

    Case 2. n<σ<0 and n=1 or 2. We can choose r(1,nσ). Then by Hölder's inequality and

    H10(Ω)L(p+1)rr1(Ω) continuously and compactly, (38)

    for any uH10(Ω), we have

    up+1Lp+1σ=Ω|x|σ|u|p+1dx(B(0,R)|x|σrdx)1r(Ω|u|(p+1)rr1dx)r1r{(2σr+1Rσr+1)1rup+1L(p+1)rr1up+1H10,n=1;(2πσr+2Rσr+2)1rup+1L(p+1)rr1up+1H10,n=2,

    which, together with (38), implies H10(Ω)Lp+1σ(Ω) continuously and compactly.

    Case 3. (p+1)(n2)2n<σ<0 and n3. Then there exists a constant r>1 such that

    σn<1r<1(p+1)(n2)2n.

    By the second inequality of the above inequalities, we have

    (p+1)rr1=p+111r<p+1(p+1)(n2)2n=2nn2.

    So,

    H10(Ω)L(p+1)rr1(Ω) continuously and compactly. (39)

    Then by Hölder's inequality, for any uH10(Ω), we have

    up+1Lp+1σ=Ω|x|σ|u|p+1dx(B(0,R)|x|σrdx)1r(Ω|u|(p+1)rr1dx)r1r(ωn1σr+nRσr+n)1rup+1L(p+1)rr1up+1H10,

    which, together with (39), implies H10(Ω)Lp+1σ(Ω) continuously and compactly. Here ωn1 denotes the surface area of the unit ball in Rn.

    Theorem 3.3. Assume p and σ satisfy (3). Let d be the constant defined in (14), then

    d=p12(p+1)C2(p+1)p1pσ,

    where Cpσ is the positive constant defined in (11).

    Proof. Firstly, we show

    infϕNJ(ϕ)=minϕH10(Ω){0}J(sϕϕ), (40)

    where N is the set defined in (15) and

    sϕ:=(ϕ2L2ϕp+1Lp+1σ)1p1. (41)

    By the definition of N in (15) and sϕ in (41), one can easily see that sϕ=1 if ϕN and sϕϕN for any ϕH10(Ω){0}.

    On one hand, since NH10(Ω) and sϕ=1 for ϕN, we have

    minϕH10(Ω){0}J(sϕϕ)minϕNJ(sϕϕ)=minϕNJ(ϕ).

    On the other hand, since {sϕϕ:ϕH10(Ω){0}}N, we have

    infϕNJ(ϕ)infϕH10(Ω){0}J(sϕϕ).

    Then (40) follows from the above two inequalities.

    By (40), the definition of d in (14), the definition of J in (12), and the definition of Cpσ in (11), we have

    d=minϕH10(Ω){0}J(sϕϕ)=p12(p+1)minϕH10(Ω){0}(ϕL2ϕLp+1σ)2(p+1)p1=p12(p+1)C2(p+1)p1pσ.

    Theorem 3.4. Assume (3) holds. Let λρ and Λρ be the two constants defined in (19). Here ρ>d is a constant. Then

    2(p+1)dp1λρΛρ2(p+1)(λ1+1)ρλ1(p1). (42)

    Proof. Let ρ>d and Nρ be the set defined in (18). By the definitions of λρ and Λρ in (19), it is obvious that

    λρΛρ. (43)

    Since NρN, it follows from the definitions of d, J, I in (14), (12), (13), respectively, that

    d=infϕNJ(ϕ)=p12(p+1)infϕNϕ2L2p12(p+1)infϕNρϕ2H10=p12(p+1)λ2ρ,

    which implies

    λρ2(p+1)dp1

    On the other hand, by (8) and (18), we have

    Λρ=supϕNρϕH10λ1+1λ1supϕNρϕL2λ1+1λ12(p+1)ρp1.

    Combining the above two inequalities with (43), we get (42), the proof is complete.

    Theorem 3.5. Assume (3) holds and u=u(x,t) is a weak solution to (1). Then the sets W and V, defined in (31) and (29) respectively, are both variant for u, i.e., u(,t)W (u(,t)V) for 0t<Tmax when u0W (u0V), where Tmax is the maximal existence time of u.

    Proof. We only prove the invariance of W since the proof of the invariance of V is similar.

    For any ϕW, since I(ϕ)<0, it follows from the definition of I (see (13)) and (11) that

    ϕ2L2<ϕp+1Lp+1σCp+1pσϕp+1L2,

    which implies

    ϕL2>Cp+1p1pσ. (44)

    Let u(x,t) be the weak solution of problem (1) with u0W. Since I(u0)<0 and uC([0,Tmax),H10(Ω)), there exists a constant ε>0 small enough such that

    I(u(,t))<0,t[0,ε]. (45)

    Then by (24), ddtu(,t)2H10>0 for t[0,ε], and then by (25) and J(u0)d, we get

    J(u(,t))<d for t(0,ε]. (46)

    We argument by contradiction. Since u(,t)C([0,Tmax),H10(Ω)), if the conclusion is not true, then there exists a t0(0,Tmax) such that u(,t)W for 0t<t0, but I(u(,t0))=0 and

    J(u(,t0))<d (47)

    (note (25) and (46), J(u(,t0))=d cannot happen). By (44), we have u(,t)C([0,Tmax),H10(Ω)) and u(,t0)¯W, then

    u(,t0)L2Cp+1p1pσ>0,

    which, together with I(u(,t0))=0, implies u(,t0)N. Then it follows from the definition of d in (14) that

    J(u(,t0))d,

    which contradicts (47). So the conclusion holds.

    Theorem 3.6. Assume (3) holds and u=u(x,t) is a weak solution to (1) with u0W. Then

    u(,t)2L22(p+1)p1d,0t<Tmax, (48)

    where W is defined in (31) and Tmax is the maximal existence time of u.

    Proof. Let N:={ϕH10(Ω):I(ϕ)<0}. Then by Theorem 3.5, u(,t)N for 0t<Tmax.

    By the proof in Theorem 3.3,

    d=minϕH10(Ω){0}J(sϕϕ)minϕNJ(sϕϕ)J(suu(,t))=(su)22u(,t)2L2(su)p+1p+1u(,t)p+1Lp+1σ((su)22(su)p+1p+1)u(,t)2L2,

    where we have used I(u(,t))<0 in the last inequality. Since I(u(,t))<0, we get from (41) that

    d=minϕH10(Ω){0}J(sϕϕ)minϕNJ(sϕϕ)J(suu(,t))=(su)22u(,t)2L2(su)p+1p+1u(,t)p+1Lp+1σ((su)22(su)p+1p+1)u(,t)2L2,

    Then

    dmax0s1(s22sp+1p+1)u(,t)2L2=(s22sp+1p+1)s=1u(,t)2L2=p12(p+1)u(,t)2L2,

    and (48) follows from the above inequality.

    Theorem 3.7. Assume (3) holds and u=u(x,t) is a weak solution to (1) with u0ˆW. Then I(u(,t))<0 for 0t<Tmax, where Tmax is the maximal existence time of u and ˆW is defined in (36)

    Proof. Firstly, we show I(u0)<0. In fact, by the definition of J in (12), u0ˆW, and (8), we get

    12u02L21p+1u0p+1Lp+1σ=J(u0)<λ1(p1)2(λ1+1)(p+1)u02H10p12(p+1)u02L2,

    which implies

    I(u0)=u02L2u0p+1Lp+1σ<0.

    Secondly, we prove I(u(,t))<0 for 0<t<Tmax. In fact, if it is not true, in view of uC([0,Tmax),H10(Ω)), there must exist a t0(0,Tmax) such that I(u(,t))<0 for t[0,t0) but I(u(,t0))=0. Then by (24), we get u(,t0)2H10>u02H10, which, together with u0ˆW and (8), implies

    J(u0)<λ1(p1)2(λ1+1)(p+1)u02H10<λ1(p1)2(λ1+1)(p+1)u(,t0)2H10p12(p+1)u(,t0)2L2. (49)

    On the other hand, by (24), (12), (13) and I(u(,t0))=0, we get

    J(u0)J(u(,t0))=p12(p+1)u(,t0)2L2,

    which contradicts (49). The proof is complete.

    Proof of Theorem 2.4. Let u=u(x,t) be a weak solution to (1) with u0V and Tmax be its maximal existence time. By Theorem 3.5, u(,t)V for 0t<Tmax, which implies I(u(,t))>0 for 0t<Tmax. Then it follows from (25), (12) and (13) that

    J(u0)J(u(,t))p12(p+1)u(,t)2L2,0t<Tmax,

    which implies u exists globally (i.e. Tmax=) and

    u(,t)L22(p+1)J(u0)p1,0t<. (50)

    Next, we prove u(,t)H10 decays exponentially, if in addition, J(u0)<d. By (24), (13), (11), (50), (16) we have

    ddt(u(,t)2H10)=2I(u(,t))=2(u(,t)2L2u(,t)p+1Lp+1σ)2(1Cp+1pσu(,t)p1L2)u(,t)2L22(1Cp+1pσ(2(p+1)J(u0)p1)p1)u(,t)2L2=2(1(J(u0)d)p12)u(,t)2L22λ1λ1+1(1(J(u0)d)p12)u(,t)2H10,

    which leads to

    u(,t)2H10u02H10exp[2λ1λ1+1(1(J(u0)d)p12)t].

    The proof is complete.

    Proof of Theorem 2.5. Let u=u(x,t) be a weak solution to (1) with u0W and Tmax be its maximal existence time.

    Firstly, we consider the case J(u0)<d and I(u0)<0. By Theorem 3.5, u(,t)W for 0t<Tmax. Let

    ξ(t):=(t0u(,s)2H10ds)12,η(t):=(t0us(,s)2H10ds)12,0t<Tmax. (51)

    For any T(0,Tmax), β>0 and α>0, we let

    F(t):=ξ2(t)+(Tt)u02H10+β(t+α)2,0tT. (52)

    Then

    F(0)=Tu02H10+βα2>0, (53)
    F(t)=u(,t)2H10u02H10+2β(t+α)=2(12t0ddsu(,s)2H10ds+β(t+α)),0tT, (54)

    and (by (24), (12), (13), (48), (25))

    F(t)=2I(u(,t))+2β=(p1)u(,t)2L22(p+1)J(u(,t))+2β2(p+1)(dJ(u0))+2(p+1)η2(t)+2β,0tT. (55)

    Since I(u(,t))<0, it follow from (24) and the first equality of (54) that

    F(t)2β(t+α).

    Then

    F(t)=F(0)+t0F(s)dsTu02H10+βα2+2αβt+βt2,0tT. (56)

    By (6), Schwartz's inequality and Hölder's inequality, we have

    12t0ddsu(,s)2H10ds=t0(u(,s),us(,s))H10dst0u(,s)H10us(,s)H10dsξ(t)η(t),0tT,

    which, together with the definition of F(t), implies

       (F(t)(Tt)u02H10)(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)2ξ2(t)η2(t)+2ξ(t)η(t)β(t+α)+β2(t+α)2(ξ(t)η(t)+β(t+α))2(12t0ddsu(,s)2H10ds+β(t+α))2,0tT.

    Then it follows from (54) and the above inequality that

    (F(t))2=4(12t0ddsu(s)2H10ds+β(t+α))24F(t)(η2(t)+β),0tT. (57)

    In view of (55), (56), and (57), we have

    F(t)F(t)p+12(F(t))2F(t)(2(p+1)(dJ(u0))2pβ),0tT.

    If we take β small enough such that

    0<βp+1p(dJ(u0)), (58)

    then F(t)F(t)p+12(F(t))20. Then, it follows from Lemma 3.1 that

    TF(0)(p+121)F(0)=Tu02H10+βα2(p1)αβ.

    Then for

    α(u02H10(p1)β,), (59)

    we get

    Tβα2(p1)αβu02H10.

    Minimizing the above inequality for α satisfying (59), we get

    Tβα2(p1)αβu02H10|α=2u02H10(p1)β=4u02H10(p1)2β.

    Minimizing the above inequality for β satisfying (58), we get

    T4pu02H10(p1)2(p+1)(dJ(u0)).

    By the arbitrariness of T<Tmax it follows that

    Tmax4pu02H10(p1)2(p+1)(dJ(u0)).

    Secondly, we consider the case J(u0)=d and I(u0)<0. By the proof of Theorem 3.5, there exists a t0>0 small enough such that J(u(,t0))<d and I(u(,t0))<0. Then it follows from the above proof that u will blow up in finite time. The proof is complete.

    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 G. By the definition of d in (14), we get

    d=infϕNJ(ϕ)=p12(p+1)infϕNϕ2L2.

    Then a minimizing sequence {ϕk}k=1N exists such that

    limkJ(ϕk)=p12(p+1)limkϕk2L2=d, (60)

    which implies {ϕk}k=1 is bounded in H10(Ω). Since H10(Ω) is reflexive and H10(Ω)Lp+1σ continuously and compactly (see (10)), there exists φH10(Ω) such that

    (1) ϕkφ in H10(Ω) weakly;

    (2) ϕkφ in Lp+1σ(Ω) strongly.

    Now, in view of ()L2 is weakly lower continuous in H10(Ω), taking lim infk in the equality ϕk2L2=ϕkp+1Lp+1σ (sine ϕkN), we get

    limkJ(ϕk)=p12(p+1)limkϕk2L2=d, (60)

    We claim

    φ2L2=φp+1Lp+1σ i.e. I(φ)=0. (62)

    In fact, if the claim is not true, then by (61),

    φ2L2<φp+1Lp+1σ.

    By the proof of Theorem 3.3, we know that sφφN, which, together with the definition of d in (14), implies

    J(sφφ)d, (63)

    where

    J(sφφ)d, (63)

    On the other hand, since sφφN, we get from the definitions of J in (12) and I in (13), sφ(0,1), ()L2 is weakly lower continuous in H10(Ω), (60) that

    J(sφφ)=p12(p+1)(sφ)2φ2L2<p12(p+1)φ2L2p12(p+1)lim infkϕk2L2=d,

    which contradicts to (63). So the claim is true, i.e.

    limkϕk2L2=φp+1Lp+1σ,

    which, together with H10(Ω) is uniformly convex and ϕkφ in H10(Ω) weakly, implies ϕkφ strongly in H10(Ω). Then by (60), J(φ)=d, which, together with (62) and the definition of G in (34), implies φG, i.e., G.

    Second, we prove GΦ, where Φ is the set defined in (27). For any φG, we need to show φΦ, i.e. φ satisfies (26). Fix any vH10(Ω) and s(ε,ε), where ε>0 is a small constant such that φ+svp+1Lp+1σ>0 for s(ε,ε). Let

    limkϕk2L2=φp+1Lp+1σ,

    Then I(τ(s)(φ+sv))=0. So by the definition of N in (15), the set

    A:={τ(s)(φ+sv):s(ε,ε)}

    is a curve on N, which passes φ when s=0. The function τ(s) is differentiable and

    A:={τ(s)(φ+sv):s(ε,ε)}

    where

    ξ:=2Ω(φ+sv)vdxφ+svp+1Lp+1σ,η:=(p+1)Ω|x|σ|φ+sv|p1(φ+sv)vdx(φ+sv)2L2.

    Since (62), we get τ(0)=1 and

    τ(0)=1(p1)φp+1Lp+1σ(2Ωφvdx(p+1)Ω|x|σ|φ|p1φvdx). (65)

    Let

    ϱ(s):=J(τ(s)(φ+sv))=τ2(s)2(φ+sv)2L2τp+1(s)p+1φ+svp+1Lp+1σ,s(ε,ε).

    Since τ(s)(φ+sv)N for s(ε,ε), τ(s)(φ+sv)|s=0=φ, ϱ(0)=J(φ)=d, it follows from the definition of d that ϱ(s) (s(ε,ε)) achieves its minimum at s=0, then ϱ(0)=0. So,

    0=ϱ(0)=τ(s)τ(s)(φ+sv)2L2+τ2(s)Ω(φ+sv)vdx|s=0τp(s)τ(s)φ+svp+1Lp+1στp+1(s)Ω|x|σ|φ+sv|p1(φ+sv)vdx|s=0=ΩφvdxΩ|x|σ|φ|p1φvdx.

    So, φΦ, i.e. GΦ. Moreover, we have G(Φ{0}) since φ0 for any φG.

    Finally, in view of Definition 2.3 and J(φ)=d (φG), to complete the proof, we only need to show

    d=infϕΦ{0}J(ϕ). (66)

    In fact, by the above proof and (27), we have GΦ{0}N. Then, in view of the definition of d in (14), i.e.,

    d=infϕNJ(ϕ)

    and J(φ)=d for any φG, we get (66). The proof is complete.

    Proof of Theorem 2.8. Let u=u(x,t) be the solution of problem (1) with initial value u0 satisfying J(u0)>d. We denote by Tmax the maximal existence of u. If u is global, i.e. Tmax=, we denote by

    ω(u0)=t0¯{u(,s):st}H10(Ω)

    the ω-limit set of u0.

    (i) Assume u0Sρ={ϕH10(Ω):ϕH10λρ,I(ϕ)>0} (see (20)) with ρJ(u0). Without loss of generality, we assume u(,t)0 for 0t<Tmax. In fact it there exists a t0 such that u(,t0)=0, then it is easy to see the function v defined as

    v(x,t)={u(x,t), if 0tt0;0, if t>t0

    is a global weak solution of problem (1), and the proof is complete.

    We claim that

    I(u(,t))>0,0t<Tmax. (67)

    Since I(u0)>0, if the claim is not true, there exists a t0(0,Tmax) such that

    I(u(,t))>0,0t<t0 (68)

    and

    I(u(,t0))=0, (69)

    which together with the definition of N in (15) and the assumption that u(,t)0 for 0t<Tmax, implies u(,t0)N. Moreover, by using (68), similar to the proof of (46), we have J(u(,t0))<J(u0), i.e. u(,t0)JJ(u0) (see (17)). Then u(,t0)NJ(u0) (since NJ(u0)=NJJ(u0)) and then u(,t0)H10λJ(u0) (see (19)). By monotonicity (see Remark 1) and ρJ(u0), we get

    u(,t0)H10λρ. (70)

    On the other hand, it follows from (24), (68) and u0Sρ that

    u(,t)H10<u0H10λρ,

    which contradicts (70). So (67) is true. Then by (24) again, we get

    u(,t)H10u0H10,0t<Tmax,

    which implies u exists globally, i.e. Tmax=.

    By (24) and (67), u(,t)H10 is strictly decreasing for 0t<, so a constant c[0,u0H10) exists such that

    limtu(,t)H10=c.

    Taking t in (24), we get

    0I(u(,s))ds12(u02H10c)<.

    Note that I(u(,s))>0 for 0s<, so, for any sequence {tn} satisfying tn as n, if the limit limnI(u(,tn)) exists, it must hold

    limnI(u(,tn))=0. (71)

    Let ω be an arbitrary element in ω(u0). Then there exists a sequence {tn} satisfying tn as n such that

    u(,tn)ω in H10(Ω) as n. (72)

    Then by (71), we get

    I(ω)=limnI(u(,tn))=0. (73)

    As the above, one can easily see

    ωH10<λρλJ(u0),J(ω)<J(u0)ωJJ(u0),

    which implies ωNJ(u0). In fact, if ωNJ(u0), by (19), λJ(u0)ωH10, a contradiction. Since NJ(u0)=NJu0 and ωJJ(u0), we get ωN. Therefore, by the definition of N in (15) and (73), ω=0, then it follows from u(,t)H10 is strictly decreasing and (72) that

    limtu(,t)H10=limnu(,tn)H10=ωH10=0.

    (ⅱ) Assume (see (20)) with . We claim that

    (74)

    Since , if the claim is not true, there exists a such that

    (75)

    and

    (76)

    Since (75), by (44) and , we get

    which, together with the definition of in (15), implies . Moreover, by using (75), similar to the proof of (46), we have , i.e. (see (17)). Then (since ) and then (see (19)). By monotonicity (see Remark 1) and , we get

    (77)

    On the other hand, it follows from (24), (75) and that

    which contradicts (77). So (74) is true.

    Suppose by contradiction that does not blow up in finite time, i.e. . By (24) and (74), is strictly increasing for . If the limit exists, i.e. there exists a constant such that

    Taking in (24), we get

    Note for , so, for any sequence satisfying as , if the limit exists, it must hold

    (78)

    Let be an arbitrary element in . Then there exists a sequence satisfying as such that

    (79)

    Since is strictly increasing, exists and

    Then by (78), we get

    (80)

    By (24), (25) and (74), one can easily see

    which implies . In fact, if , by (19), , a contradiction. Since and , we get . Therefore, by the definition of in (15) and (80), . However, this contradicts . So blows up in finite time. The proof is complete.

    Proof of Theorem 2.9. Let be a weak solution of (1) with and be its maximal existence time, where is defined in (36). By Theorem 3.7, we know that for . Then by (8) and (24), we get

    (81)

    The remain proofs are similar to the proof of Theorem 2.9. For any , and , we consider the functional again (see (52)). We also have (53), (54), but there are some differences in (55), in fact, by (81) and (25), we have

    (82)

    We also have (56) and (57). Then it follows from (56), (57) and (82) that

    If we take small enough such that

    (83)

    then . Then, it follows from Lemma 3.1 that

    Then for

    (84)

    we get

    Minimizing the above inequality for satisfying (84), we get

    Minimizing the above inequality for satisfying (58), we get

    By the arbitrariness of it follows that

    Proof of Theorem 2.10. For any , let and be two arbitrary disjoint open domains. Let , extending to by letting in , then . We choose large enough such that

    (85)

    For such and , we take a (which is extended to by letting in i.e. ) such that

    (86)

    where (see Remark 5)

    (86)

    which can be done since

    and can be chosen such that .

    By Remark 5 again,

    (87)

    By (87) and (86), we can choose such that satisfies . Letting , since and are disjoint, we get

    and (note (85))

    Let be the weak solution of problem (1) with initial value given above. Then by Theorem 2.9, blows up in finite time.



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