Inspired by the representation designed for floorplanning problems, in this paper, we proposed a new representation, namely the moving block sequence (MBS), for resource investment project scheduling problems (RIPSPs). Since each activity of a project in RIPSPs has fixed duration and resource demand, we consider an activity as a rectangle block whose width is equal to the duration of the activity and height the resource needed by the activity. Four move modes are designed for activities, by using which the activity can move to the appropriate position. Therefore, the new representation of the project of RIPSPs consists of two parts: an activity list and a move mode list. By initializing the move modes randomly for each activity and moving it appropriately, the activity list can be decoded into valid solutions of RIPSPs. Since the decoding method of MBS guarantees that after moved, each activity is scheduled in the left-most and bottom-most position within a coordinate, which means that each activity in the corresponding project is arranged as early as possible when the precedence constraints and resource demands are satisfied. In addition, the multiagent evolutionary algorithm (MAEA) is employed to incorporate with the newly designed MBS representation in solving RIPSPs. With the intrinsic properties of MBS in mind, four behaviors, namely the crossover, mutation, competition, and self-learning operators are designed for agents in MAEA. To test the performance of our algorithm, 450 problem instances are used and the experimental results demonstrate the good performance of the proposed representation.
Citation: Xiaoxiao Yuan, Jing Liu, Xingxing Hao. 2017: A moving block sequence-based evolutionary algorithm for resource investment project scheduling problems, Big Data and Information Analytics, 2(1): 39-58. doi: 10.3934/bdia.2017007
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Inspired by the representation designed for floorplanning problems, in this paper, we proposed a new representation, namely the moving block sequence (MBS), for resource investment project scheduling problems (RIPSPs). Since each activity of a project in RIPSPs has fixed duration and resource demand, we consider an activity as a rectangle block whose width is equal to the duration of the activity and height the resource needed by the activity. Four move modes are designed for activities, by using which the activity can move to the appropriate position. Therefore, the new representation of the project of RIPSPs consists of two parts: an activity list and a move mode list. By initializing the move modes randomly for each activity and moving it appropriately, the activity list can be decoded into valid solutions of RIPSPs. Since the decoding method of MBS guarantees that after moved, each activity is scheduled in the left-most and bottom-most position within a coordinate, which means that each activity in the corresponding project is arranged as early as possible when the precedence constraints and resource demands are satisfied. In addition, the multiagent evolutionary algorithm (MAEA) is employed to incorporate with the newly designed MBS representation in solving RIPSPs. With the intrinsic properties of MBS in mind, four behaviors, namely the crossover, mutation, competition, and self-learning operators are designed for agents in MAEA. To test the performance of our algorithm, 450 problem instances are used and the experimental results demonstrate the good performance of the proposed representation.
Our goal of this paper is to consider the existence of nodal solution and ground state solution for the following fractional Kirchhoff equation:
{−(a+b‖u‖2K)LKu+V(x)u=|u|2∗α−2u+kf(x,u),x∈Ω,u=0,x∈R3∖Ω, | (1) |
where
LKu(x)=12∫R3(u(x+y)+u(x−y)−2u(x))K(y)dy,x∈R3, |
the kernel
(ⅰ)
(ⅱ) there exists
(ⅲ)
We note that when
(−Δ)αu(x)=−C(α)2∫R3(u(x+y)+u(x−y)−2u(x))|y|3+2αdy |
and in this case
∫R3∫R3|u(x)−u(y)|2K(x−y)dxdy=2C(α)∫R3|(−Δ)α/2u(x)|2dx, |
where
When
{(−Δ)αu+V(x)u=|u|2∗α−2u+kf(x,u),x∈Ω,u=0,x∈R3∖Ω. | (2) |
Equation (2) is derived from the fractional Schrödinger equation and the nonlinearity
(a+b∫R3∫R3|u(x)−u(y)|2|x−y|3+2αdxdy)(−Δ)αu=f(x,u), | (3) |
where
utt+(a+b∫R3∫R3|u(x)−u(y)|2|x−y|3+2αdxdy)(−Δ)αu=f(x,u). | (4) |
As a special significant case, the nonlocal aspect of the tension arises from nonlocal measurements of the fractional length of the string. For more mathematical and physical background on Schrödinger-Kirchhoff type problems, we refer the readers to [6] and the references therein.
In the remarkable work of Caffarelli and Silvestre [2], the authors express this nonlocal operator
In past few years, some researchers began to search for nodal solutions of Schrödinger type equation with critical growth nonlinearity and have got some interesting results. For example, Zhang [20] considered the following Schrödinger-Poisson system:
{−Δu+u+k(x)ϕu=a(x)|u|p−2u+u5,x∈R3,−Δϕ=k(x)u2,x∈R3, | (5) |
where
Wang [17] studies the following Kirchhoff-type equation:
{−(a+b∫Ω|∇u|2dx)Δu=|u|4u+λf(x,u), x∈Ω,u=0, x∈∂Ω, | (6) |
where
However, as for fractional Kirchhoff types equation, to the best of our knowledge, few results involved the existence and asymptotic behavior of ground state and nodal solutions in case of critical growth. If
It's worth noting that, the Brouwer degree method used in [18] strictly depends on the nonlinearity
Throughout this paper, we let
E={u∈X:∫R3V(x)u2dx<∞,u=0a.e.inR3∖Ω}, |
where the space X introduced by Servadei and Valdinoci ([12,13]) denotes the linear space of Lebesgue measurable functions
((x,y)→(u(x)−u(y))√K(x−y)∈L2((R3×R3)∖(Ωc×Ωc),dxdy) |
with the following norm
||u||2X=||u||2L2+∫Q|u(x)−u(y)|2K(x−y)dxdy, |
where
⟨u,v⟩=a2∫R3∫R3(u(x)−u(y))(v(x)−v(y))K(x−y)dxdy+∫ΩV(x)uvdx,∀u,v∈E |
and the norm
‖u‖2=a2∫R3∫R3|u(x)−u(y)|2K(x−y)dxdy+∫ΩV(x)u2dx. |
The following result for the space
Lemma 1.1. ([12]) Let
A weak solution
12(a+b‖u‖2K)∫R3∫R3(u(x)−u(y))(v(x)−v(y))K(x−y)dxdy+∫ΩV(x)u(x)v(x)dx−∫Ω|u(x)|2∗α−2u(x)v(x)dx−k∫Ωf(x,u(x))v(x)dx=0, |
for any
⟨u,v⟩+b‖u‖2K(u,v)K−k∫Ωf(x,u)vdx−∫Ω|u|2∗α−2uvdx=0,∀v∈E. |
As for the function
(f1):
(f2): There exists
(f3):
Remark 1. We note that under the conditions (
The main results can be stated as follows.
Theorem 1.2. Suppose that
Remark 2. The ground state nodal solution
Jk(uk)=ck:=infu∈MkJk(u), |
where
u+=max{u(x),0},u−=min{u(x),0}. |
We recall that the nodal set of a continuous function
Theorem 1.3. Suppose that
Jk(uk)>2c∗, |
where
Comparing with the literature, the above two results can be regarded as a supplementary of those in [3,4,14,17].
The remainder of this paper is organized as follows. In Section 2, we give some useful preliminaries. In Section 3, we study the existence of ground state and nodal solutions of (1) and we prove Theorems 1.1-1.2.
We define the energy functional associated with equation (1) as follows:
Jk(u)=12‖u‖2+b4‖u‖4K−k∫ΩF(x,u)dx−12∗α∫Ω|u|2∗αdx,∀u∈E. |
According to our assumptions on
⟨J′k(u),v⟩=⟨u,v⟩+b‖u‖2K(u,v)K−k∫Ωf(x,u)vdx−∫Ω|u|2∗α−2uvdx,∀u,v∈E. |
Note that, since (1) involves pure critical nonlinearity
For fixed
H(s,t)=⟨J′k(su++tu−),su+⟩,G(s,t)=⟨J′k(su++tu−),tu−⟩. |
The argument generally used is to modify the method developed in [11]. The nodal Nehari manifold is defined by
Mk={u∈E,u±≠0and⟨J′k(u),u+⟩=⟨J′k(u),u−⟩=0}, | (7) |
which is a subset of the Nehari manifold
Jk(u)=Jk(u+)+Jk(u−), |
it brings difficulties to construct a nodal solution.
The following result describes the shape of the nodal Nehari manifold
Lemma 2.1. Assume that
Proof. From
|f(x,t)|≤ε|t|+Cε|t|q−1,∀t∈R. | (8) |
By above equality and Sobolev's embedding theorems, we have
H(s,t):=s2‖u+‖2+b‖su++tu−‖2K(su++tu−,su+)K−∫Ω|su+|2∗αdx−k∫Ωf(x,su+)su+dx−ast2∫R3∫R3[u+(x)u−(y)+u−(x)u+(y)]K(x−y)dxdy≥s2‖u+‖2−C1s2∗α‖u+‖2∗α−kεC2s2‖u+‖2−kCεC3sq‖u+‖q. | (9) |
By choosing
G(s,t):=t2‖u−‖2+b‖su++tu−‖2K(su++tu−,tu−)K−∫Ω|tu−|2∗αdx−k∫Ωf(x,tu−)tu−dx−ast2∫R3∫R3[u+(x)u−(y)+u−(x)u+(y)]K(x−y)dxdy>0, | (10) |
for
H(δ1,t)>0,G(s,δ1)>0. | (11) |
For any
H(δ2,t)≤δ22‖u+‖2+b‖δ2u++tu−‖2K(δ2u++tu−,δ2u+)K−δ22∗α∫R3|u+|2∗αdx+δ2tD(u). |
Similarly, we have
G(s,δ2)≤δ22‖u−‖2+b‖su++δ2u−‖2K(su++δ2u−,δ2u−)K−δ22∗α∫R3|u−|2∗αdx+sδ2D(u). |
By choosing
H(δ2,t)<0,G(s,δ2)<0 | (12) |
for all
Following (11) and (12), we can use Miranda's Theorem (see Lemma 2.4 in [17]) to get a positive pair
Lastly, we will prove that
s2u‖u+‖2+s2uD(u)+s4ub(‖u+‖2K+‖u−‖2K+2D(u))(‖u+‖2K+D(u))≥s2u‖u+‖2+sutuD(u)+b(s2u‖u+‖2K+t2u‖u−‖2K+2sutuD(u))(s2u‖u+‖2K+sutuD(u))=s2∗αu∫R3|u+|2∗αdx+k∫R3f(x,suu+)suu+dx. | (13) |
On the other hand,
‖u+‖2+D(u)+b(‖u+‖2K+‖u−‖2K+2D(u))(‖u+‖2K+D(u))≤∫R3|u+|2∗αdx+k∫R3f(x,u+)u+dx. | (14) |
From (13) - (14), we can see that
(1s2u−1)(‖u+‖2+D(u))≥(s2∗α−4u−1)∫R3|u+|2∗αdx+k∫R3[f(x,suu+)(suu+)3−f(x,u+)(u+)3](u+)4dx. |
So we have
Lemma 2.2. There exists
Proof. For any
‖u±‖2+b‖u‖2K(u,u±)K=k∫R3f(x,u±)u±dx+∫R3|u±|2∗αdx. |
Hence, in view of (8), we have
‖u±‖2≤kεC1‖u±‖2+kC2‖u±‖q+C3‖u±‖2∗α. |
By choosing
‖u±‖≥ρ | (15) |
for some
f(x,t)t−4F(x,t)≥0, | (16) |
and
Jk(u)=14‖u‖2+(14−12∗α)∫R3|u|2∗αdx+k4∫R3[f(x,u)u−4F(x,u)]dx≥14‖u‖2. |
So
Let
s2∗αk∫R3|u+|2∗αdx+t2∗αk∫R3|u−|2∗αdx≤‖sku++tku−‖2+b‖sku++tku−‖4K≤2s2k‖u+‖2+2t2k‖u−‖2+4bs4k‖u+‖4K+4bt4k‖u−‖4K, |
which implies
‖sknu++tknu−‖2+b‖sknu++tknu−‖4K=∫R3|sknu++tknu−|2∗αdx+kn∫R3f(sknu++tknu−)(sknu++tknu−)dx. | (17) |
Because
0≤ck≤Jk(sku++tku−)≤s2k‖u+‖2+t2k‖u−‖2+2bs4k‖u+‖4K+2bt4k‖u−‖4K, |
so
u±n→u±inLp(R3)∀p∈(2,2∗α),u±n(x)→u±(x)a.e.x∈R3. |
Denote
S:=infu∈E∖{0}‖u‖2(∫R3|u|2∗αdx)22∗α. |
Sobolev embedding theorem insures that
By
‖u±n‖2−‖u±n−u±‖2=2⟨u±n,u±⟩−‖u±‖2. |
By taking
limn→∞‖u±n‖2=limn→∞‖u±n−u±‖2+‖u±‖2. |
On the other hand, by (8) we have
∫R3F(x,su±n)dx→∫R3F(x,su±)dx. |
Then,
lim infn→∞Jk(su+n+tu−n)≥s22limn→∞(‖u+n−u+‖2+‖u+‖2)+t22limn→∞(‖u−n−u−‖2+‖u−‖2)+bs44[limn→∞‖u+n−u+‖2K+‖u+‖2K]2+bt44[limn→∞‖u−n−u−‖2K+‖u−‖2K]2+stlim infn→∞D(un)+bs2t24a2lim infn→∞D2(un)+bs2t22lim infn→∞(‖u+n‖2K‖u−n‖2K)+bs3t2alim infn→∞(D(un)‖u+n‖2K)+bst32alim infn→∞(D(un)‖u−n‖2K)−s2∗α2∗αlimn→∞(|u+n−u+|2∗α2∗α+|u+|2∗α2∗α)−t2∗α2∗αlimn→∞(|u−n−u−|2∗α2∗α+|u−|2∗α2∗α)−k∫R3F(x,su+)dx−k∫R3F(x,tu−)dx, |
where
lim infn→∞Jk(su+n+tu−n)≥Jk(su++tu−)+s22limn→∞‖u+n−u+‖2+t22limn→∞‖u−n−u−‖2−s2∗α2∗αlimn→∞|u+n−u+|2∗α2∗α−t2∗α2∗αlimn→∞|u−n−u−|2∗α2∗α+bs42limn→∞‖u+n−u+‖2K‖u+‖2K+bs44(limn→∞‖u+n−u+‖2K)2+bt42limn→∞‖u−n−u−‖2K‖u−‖2K+bt44(limn→∞‖u−n−u−‖2K)2=Jk(su++tu−)+s22A1−s2∗α2∗αB1+t22A2−t2∗α2∗αB2+bs42A3‖u+‖2K+bs44A23+bt42A4‖u−‖2K+bt44A24, |
where
A1=limn→∞‖u+n−u+‖2,A2=limn→∞‖u−n−u−‖2,B1=limn→∞|u+n−u+|2∗α2∗α,B2=limn→∞|u−n−u−|2∗α2∗α,A3=limn→∞‖u+n−u+‖2K,A4=limn→∞‖u−n−u−‖2K. |
From the inequality above, we deduce that
Jk(su++tu−)+s22A1−s2∗α2∗αB1+t22A2−t2∗α2∗αB2+bs42A3‖u+‖2K+bs44A23+bt42A4‖u−‖2K+bt44A24≤ck. | (18) |
We next prove
β=α3S32α≤α3(A1(B1)22∗α)32α. |
It happens that,
α3(A1(B1)22∗α)32α=maxs≥0{s22A1−s2∗α2∗αB1}≤maxs≥0{s22A1−s2∗α2∗αB1+bs42A3‖u+‖2K+bs44A23}. |
The inequality (18) and
β≤maxs≥0{s22A1−s2∗α2∗αB1+bs42A3‖u+‖2K+bs44A23}≤ck<β, |
which is a contradiction. Thus
Then, we consider the key point to the proof of Theorem 1.1, that is
Similarly, we only prove
Case 1:
ˉs22A1−ˉs2∗α2∗αB1+bˉs42A3‖u+‖2K+bˉs44A23=maxs≥0{s22A1−s2∗α2∗αB1+bs42A3‖u+‖2K+bs44A23}, |
ˉt22A2−ˉt2∗α2∗αB2+bˉt42A4‖u−‖2K+bˉt44A24=maxt≥0{t22A2−t2∗α2∗αB2+bt42A4‖u−‖2K+bt44A24}. |
Since
φu(su,tu)=max(s,t)∈[0,ˉs]×[0,ˉt]φu(s,t). |
If
By direct computation, we get
s22A1−s2∗α2∗αB1+bs42A3‖u+‖2K+bs44A23>0, | (19) |
t22A2−t2∗α2∗αB2+bt42A4‖u−‖2K+bt44A24>0, | (20) |
for all
β≤ˉs22A1−ˉs2∗α2∗αB1+bˉs42A3‖u+‖2K+bˉs44A23+t22A2−t2∗α2∗αB2+bt42A4‖u−‖2K+bt44A24, |
β≤ˉt22A2−ˉt2∗α2∗αB2+bs42A3‖u+‖2K+bs44A23+s22A1−s2∗α2∗αB1+bˉt42A4‖u−‖2K+bˉt44A24. |
In view of (18), it follows that
ck≥Jk(suu++tuu−)+s2u2A1−s2∗αu2∗αB1+t2u2A2−t2∗αu2∗αB2+bs4u2A3‖u+‖2K+bs4u4A23+bt4u2A4‖u−‖2K+bt4u4A24>Jk(suu++tuu−)≥ck. |
It is impossible. The proof of Case 1 is completed.
Case 2:
φu(su,tu)=max(s,t)∈[0,ˉs]×[0,∞)φu(s,t). |
We need to prove that
β≤ˉs22A1−ˉs2∗α2∗αB1+bˉs42A3‖u+‖2K+bˉs44A23+t22A2−t2∗α2∗αB2+bt42A4‖u−‖2K+bt44A24. |
Thus also from (20) and
ck≥Jk(suu++tuu−)+s2u2A1−s2∗αu2∗αB1+t2u2A2−t2∗αu2∗αB2+bs4u2A3‖u+‖2K+bs4u4A23+bt4u2A4‖u−‖2K+bt4u4A24>Jk(suu++tuu−)≥ck, |
which is a contradiction.
Since
⟨J′k(u),u±⟩≤lim infn→∞‖u±n‖2+blim infn→∞‖un‖2K(un,u±n)K−limn→∞∫R3f(x,u±n)u±ndx−limn→∞∫R3|u±n|2∗α+lim infn→∞L(un)≤limn→∞⟨J′k(un),u±n⟩=0. |
By Lemma 2.1, we know
ck≤Jk(˜u)−14⟨J′k(˜u),˜u⟩=14(‖suu+‖2+‖tuu−‖2)+(14−12∗α)(|suu+|2∗α2∗α+|tuu−|2∗α2∗α)+k4∫R3[f(x,suu+)(suu+)−4F(x,suu+)]dx+k4∫R3[f(x,tuu−)(tuu−)−4F(x,tuu−)]dx≤lim infn→∞[Jk(un)−14⟨J′k(un),un⟩]=ck. |
So, we have completed proof of Lemma 2.2.
In this section, we will prove main results.
Proof. Since
Jk(su+k+tu−k)<ck. | (21) |
If
‖J′k(v)‖≥θ,forall‖v−uk‖≤3δ. |
We know by the result (15), if
¯ck:=max∂QI∘g<ck. | (22) |
Let
To finish the proof of Theorem 1.1, one of the key points is to prove that
max(s,t)∈ˉQJk(η(1,g(s,t)))<ck. | (23) |
The other is to prove that
Jk(η(1,v))≤Jk(η(0,v))=Jk(v),∀v∈E. | (24) |
For
Jk(η(1,g(s,t)))≤Jk(g(s,t))<ck. |
If
Jk(η(1,g(1,1)))≤ck−ε<ck. |
Thus (23) holds. Then, let
Υ(s,t):=(1s⟨J′k(φ(s,t)),(φ(s,t))+⟩,1t⟨J′k(φ(s,t)),(φ(s,t))−⟩). |
The claim holds if there exists
‖g(s,t)−uk‖2=‖(s−1)u+k+(t−1)u−k‖2≥|s−1|2‖u+k‖2>|s−1|2(6δ)2, |
and
Υ(12,t)=(2⟨J′k(12u+k+tu−k),12u+k⟩,1t⟨J′k(12u+k+tu−k),tu−⟩). |
On the other hand, from (9) and
H(t,t)=(t2−t4)(‖u+‖2+D(u))+(t4−t2∗α)∫R3|u+|2∗αdx+kt4∫R3(f(x,u+)−f(x,tu+)t3)u+dx. |
According to
H(12,t)=‖12u+k‖2+t2D(uk)+b(14‖u+k‖2K+t2‖u−k‖2K+taD(uk))(14‖u+k‖2K+t2aD(uk))−(12)2∗α∫R3|u+k|2∗αdx−k∫R3f(x,12u+k)12u+kdx≥H(12,12)>0, |
which implies that
H(12,t)>0,∀t∈[12,32]. | (25) |
Analogously,
H(32,t)=‖32u+k‖2+3t2D(uk)+b(94‖u+k‖2K+t2‖u−k‖2K+3taD(uk))(94‖u+k‖2K+3t2aD(uk))−(32)2∗α∫R3|u+k|2∗αdx−k∫R3f(x,32u+k)32u+kdx≤H(32,32)<0, |
that is,
H(32,t)<0,∀t∈[12,32]. | (26) |
By the same way,
G(s,12)>0,∀s∈[12,32],andG(s,32)>0,∀s∈[12,32]. | (27) |
From (25)-(27), the assumptions of Miranda's Theorem (see Lemma 2.4 in [17]) are satisfied. Thus, there exists
Proof. Recall that
Firstly, we prove that
β≤˜t22A−˜t2∗α2∗αB:=maxt≥0{t22A−t2∗α2∗αB}≤maxt≥0{t22A−t2∗α2∗αB+bt44(C2+2C‖vk‖2K)}≤c∗<β. |
Which is a contradiction.
Then, we prove that
c∗≤Jk(tvvk)<Jk(tvvk)+t2v2A−t2∗αv2∗αB+bt4v4(C2+2C‖vk‖2K)≤c∗. |
From the above arguments we know
c∗≤Jk(˜v)−14⟨J′k(˜v),˜v⟩≤14‖vk‖2+(14−12∗α)|vk|2∗α2∗α+k4∫R3[f(x,vk)vk−4F(x,vk)]dx=lim infn→∞[Jk(vn)−14⟨J′k(vn),vn⟩]=c∗. |
Therefore,
By standard arguments, we can find
su+u+∈Nk,tu−u−∈Nk. |
Thus, the above fact follows that
2c∗≤Jk(su+u+)+Jk(tu−u−)≤Jk(su+u++tu−u−)<Jk(u++u−)=ck. |
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