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Exploring endophytes for in vitro synthesis of bioactive compounds similar to metabolites produced in vivo by host plants

  • Endophytes represent microorganisms residing within plant tissues without typically causing any adverse effect to the plants for considerable part of their life cycle and are primarily known for their beneficial role to their host-plant. These microorganisms can in vitro synthesize secondary metabolites similar to metabolites produced in vivo by their host plants. If microorganisms are isolated from certain plants, there is undoubtedly a strong possibility of obtaining beneficial endophytes strains producing host-specific secondary metabolites for their potential applications in sustainable agriculture, pharmaceuticals and other industrial sectors. Few products derived from endophytes are being used for cultivating resilient crops and developing non-toxic feeds for livestock. Our better understanding of the complex relationship between endophytes and their host will immensely improve the possibility to explore their unlimited functionalities. Successful production of host-secondary metabolites by endophytes at commercial scale might progressively eliminate our direct dependence on high-valued vulnerable plants, thus paving a viable way for utilizing plant resources in a sustainable way.

    Citation: Hemant Sharma, Arun Kumar Rai, Divakar Dahiya, Rajen Chettri, Poonam Singh Nigam. Exploring endophytes for in vitro synthesis of bioactive compounds similar to metabolites produced in vivo by host plants[J]. AIMS Microbiology, 2021, 7(2): 175-199. doi: 10.3934/microbiol.2021012

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  • Endophytes represent microorganisms residing within plant tissues without typically causing any adverse effect to the plants for considerable part of their life cycle and are primarily known for their beneficial role to their host-plant. These microorganisms can in vitro synthesize secondary metabolites similar to metabolites produced in vivo by their host plants. If microorganisms are isolated from certain plants, there is undoubtedly a strong possibility of obtaining beneficial endophytes strains producing host-specific secondary metabolites for their potential applications in sustainable agriculture, pharmaceuticals and other industrial sectors. Few products derived from endophytes are being used for cultivating resilient crops and developing non-toxic feeds for livestock. Our better understanding of the complex relationship between endophytes and their host will immensely improve the possibility to explore their unlimited functionalities. Successful production of host-secondary metabolites by endophytes at commercial scale might progressively eliminate our direct dependence on high-valued vulnerable plants, thus paving a viable way for utilizing plant resources in a sustainable way.



    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 such that .

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

    (37)

    Then we have, for any ,

    which, together with (37), implies continuously and compactly.

    Case 2. and or . We can choose . Then by Hölder's inequality and

    (38)

    for any , we have

    which, together with (38), implies continuously and compactly.

    Case 3. and . Then there exists a constant such that

    By the second inequality of the above inequalities, we have

    So,

    (39)

    Then by Hölder's inequality, for any , we have

    which, together with (39), implies continuously and compactly. Here denotes the surface area of the unit ball in .

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

    where is the positive constant defined in (11).

    Proof. Firstly, we show

    (40)

    where is the set defined in (15) and

    (41)

    By the definition of in (15) and in (41), one can easily see that if and for any .

    On one hand, since and for , we have

    On the other hand, since , we have

    Then (40) follows from the above two inequalities.

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

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

    (42)

    Proof. Let and be the set defined in (18). By the definitions of and in (19), it is obvious that

    (43)

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

    which implies

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

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

    Theorem 3.5. Assume (3) holds and is a weak solution to (1). Then the sets and , defined in (31) and (29) respectively, are both variant for , i.e., () for when (), where is the maximal existence time of .

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

    For any , since , it follows from the definition of (see (13)) and (11) that

    which implies

    (44)

    Let be the weak solution of problem (1) with . Since and , there exists a constant small enough such that

    (45)

    Then by (24), for , and then by (25) and , we get

    (46)

    We argument by contradiction. Since , if the conclusion is not true, then there exists a such that for , but and

    (47)

    (note (25) and (46), cannot happen). By (44), we have and , then

    which, together with , implies . Then it follows from the definition of in (14) that

    which contradicts (47). So the conclusion holds.

    Theorem 3.6. Assume (3) holds and is a weak solution to (1) with . Then

    (48)

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

    Proof. Let . Then by Theorem 3.5, for .

    By the proof in Theorem 3.3,

    where we have used in the last inequality. Since , we get from (41) that

    Then

    and (48) follows from the above inequality.

    Theorem 3.7. Assume (3) holds and is a weak solution to (1) with . Then for , where is the maximal existence time of and is defined in (36)

    Proof. Firstly, we show . In fact, by the definition of in (12), , and (8), we get

    which implies

    Secondly, we prove for . In fact, if it is not true, in view of , there must exist a such that for but . Then by (24), we get , which, together with and (8), implies

    (49)

    On the other hand, by (24), (12), (13) and , we get

    which contradicts (49). The proof is complete.

    Proof of Theorem 2.4. Let be a weak solution to (1) with and be its maximal existence time. By Theorem 3.5, for , which implies for . Then it follows from (25), (12) and (13) that

    which implies exists globally (i.e. ) and

    (50)

    Next, we prove decays exponentially, if in addition, . By (24), (13), (11), (50), (16) we have

    which leads to

    The proof is complete.

    Proof of Theorem 2.5. Let be a weak solution to (1) with and be its maximal existence time.

    Firstly, we consider the case and . By Theorem 3.5, for . Let

    (51)

    For any , and , we let

    (52)

    Then

    (53)
    (54)

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

    (55)

    Since , it follow from (24) and the first equality of (54) that

    Then

    (56)

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

    which, together with the definition of , implies

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

    (57)

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

    If we take small enough such that

    (58)

    then . Then, it follows from Lemma 3.1 that

    Then for

    (59)

    we get

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

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

    By the arbitrariness of it follows that

    Secondly, we consider the case and . By the proof of Theorem 3.5, there exists a small enough such that and . Then it follows from the above proof that 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 . By the definition of in (14), we get

    Then a minimizing sequence exists such that

    (60)

    which implies is bounded in . Since is reflexive and continuously and compactly (see (10)), there exists such that

    (1) in weakly;

    (2) in strongly.

    Now, in view of is weakly lower continuous in , taking in the equality (sine ), we get

    (60)

    We claim

    (62)

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

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

    (63)

    where

    (63)

    On the other hand, since , we get from the definitions of in (12) and in (13), , is weakly lower continuous in , (60) that

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

    which, together with is uniformly convex and in weakly, implies strongly in . Then by (60), , which, together with (62) and the definition of in (34), implies , i.e., .

    Second, we prove , where is the set defined in (27). For any , we need to show , i.e. satisfies (26). Fix any and , where is a small constant such that for . Let

    Then . So by the definition of in (15), the set

    is a curve on , which passes when . The function is differentiable and

    where

    Since (62), we get and

    (65)

    Let

    Since for , , , it follows from the definition of that achieves its minimum at , then . So,

    So, , i.e. . Moreover, we have since for any .

    Finally, in view of Definition 2.3 and , to complete the proof, we only need to show

    (66)

    In fact, by the above proof and (27), we have . Then, in view of the definition of in (14), i.e.,

    and for any , we get (66). The proof is complete.

    Proof of Theorem 2.8. Let be the solution of problem (1) with initial value satisfying . We denote by the maximal existence of . If is global, i.e. , we denote by

    the -limit set of .

    (i) Assume (see (20)) with . Without loss of generality, we assume for . In fact it there exists a such that , then it is easy to see the function defined as

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

    We claim that

    (67)

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

    (68)

    and

    (69)

    which together with the definition of in (15) and the assumption that for , implies . Moreover, by using (68), 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

    (70)

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

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

    which implies exists globally, i.e. .

    By (24) and (67), is strictly decreasing for , so a constant exists such that

    Taking in (24), we get

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

    (71)

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

    (72)

    Then by (71), we get

    (73)

    As the above, 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 (73), , then it follows from is strictly decreasing and (72) that

    (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.



    Funding



    This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

    Conflict of interest



    All authors declare no conflicts of interest in this review article.

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