We witness an increased emphasis on the integration of different areas of manufacturing firms with the intention of avoiding suboptimal solutions. In particular, due to new technologies which make it easy to change the price of a product in real time, the integration of pricing and production planning may be garnering the most interest. We are proposing in this paper a way to model the dynamics of the price. Thus, the price and the inventory level are considered as state variables whereas the supply (production) rate is the control variable. The demand rate is dynamic and state-dependent. Using a model predictive control approach, the optimal supply rate, and thus the optimal price and inventory level, are obtained. Different examples are provided under different scenarios for the supply rate and for the demand rate.
Citation: Messaoud Bounkhel, Lotfi Tadj. Model predictive control based integration of pricing and production planning[J]. AIMS Mathematics, 2024, 9(1): 2282-2307. doi: 10.3934/math.2024113
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We witness an increased emphasis on the integration of different areas of manufacturing firms with the intention of avoiding suboptimal solutions. In particular, due to new technologies which make it easy to change the price of a product in real time, the integration of pricing and production planning may be garnering the most interest. We are proposing in this paper a way to model the dynamics of the price. Thus, the price and the inventory level are considered as state variables whereas the supply (production) rate is the control variable. The demand rate is dynamic and state-dependent. Using a model predictive control approach, the optimal supply rate, and thus the optimal price and inventory level, are obtained. Different examples are provided under different scenarios for the supply rate and for the demand rate.
Let H and K be complex infinite-dimensional separable Hilbert spaces, and let B(H,K) (resp., C(H,K), C+(H,K)) be the set of all bounded (resp., closed, closable) operators from H to K. If K=H, we use B(H), C(H), and C+(H) as usual. The domain of T∈C(H) is denoted by D(T). A closed operator T is said to be left (resp., right) Fredholm if its range R(T) is closed and α(T)<∞ (resp., β(T)<∞), where α(T) and β(T) denote the dimension of the null space N(T) and the quotient space H/R(T), respectively. T is said to be Fredholm if it is both left and right Fredholm. If T is a left or right Fredholm operator, then we define the index of T by ind(T)=α(T)−β(T). T is called Weyl if it is a Fredholm operator of index zero. The left essential spectrum, right essential spectrum, essential spectrum, and Weyl spectrum of T are, respectively, defined by
σle(T)={λ∈C:T−λI is not left Fredholm},σre(T)={λ∈C:T−λI is not right Fredholm},σe(T)={λ∈C:T−λI is not Fredholm},σw(T)={λ∈C:T−λI is not Weyl}. |
Iterates T2,T3,⋯ of T∈C(H) are defined by Tnx=T(Tn−1x) for x∈D(Tn) (n=2,3,⋯), where
D(Tn)={x:x,Tx,⋯,Tn−1x∈D(T)}. |
It is easy to see that
N(Tn)⊂N(Tn+1),R(Tn+1)⊂R(Tn). |
If n≥0, we follow the convention that T0=I (the identity operator on H, with D(I)=H). Then N(T0)={0} and R(T0)=H. It is also well known that if N(Tk)=N(Tk+1), then N(Tn)=N(Tk) for n≥k. In this case, the smallest nonnegative integer k such that N(Tk)=N(Tk+1) is called the ascent of T, and is denoted by asc(T). If no such k exists, we define asc(T)=∞. Similarly, if R(Tk)=R(Tk+1), then R(Tn)=R(Tk) for n≥k. Thus, we can analogously define the descent of T denoted by des(T), and define des(T)=∞ if R(Tn+1) is always a proper subset of R(Tn). We call T left (resp., right) Browder if it is left (resp., right) Fredholm and asc(T)<∞ (resp., des(T)<∞). The left Browder spectrum, right Browder spectrum, and Browder spectrum of T are, respectively, defined by
σlb(T)={λ∈C:T−λI is not left Browder},σrb(T)={λ∈C:T−λI is not right Browder},σb(T)={λ∈C:T−λI is not Browder}. |
We write σ(T),σap(T), and σδ(T) for the spectrum, approximate point spectrum, and defect spectrum of T, respectively.
For the spectrum and Browder spectrum, the perturbations of 2×2 bounded upper triangular operator matrix MC=[AC0B], acting on a Hilbert or Banach space, have been studied by numerous authors (see, e.g., [5,8,9,12,13,18]). However, the characterizations of the unbounded case are still unknown. In this paper, the unbounded operator matrix
TB=[AB0D]:D(A)⊕D(D)⊂H⊕H⟶H⊕H |
is considered, where A∈C(H), D∈C(K), and B∈C+D(K,H)={B∈C+(K,H):D(B)⊃D(D)}. It is obvious that TB is a closed operator matrix. Our main goal is to present some sufficient and necessary conditions for TB to be Browder (resp., invertible) for some closable operator B∈C+D(K,H) applying a space decomposition technique. Further, the sufficient and necessary condition, which is completely described by the diagonal operators A and D, for
σ∗(TB)=σ∗(A)∪σ∗(D) | (1.1) |
is characterized, where σ∗ is the Browder spectrum (resp., spectrum). This is an extension of the results from [3,7,10,12]. In addition, this result is applied to a Hamiltonian operator matrix from elasticity theory.
Definition 1.1. [4] Let T:D(T)⊂H⟶K be a densely defined closed operator. If there is an operator T†:D(T†)⊂K⟶H such that D(T†)=R(T)⊕R(T)⊥,N(T†)=R(T)⊥ and
T†Tx=P¯R(T†)x,x∈D(T),TT†y=P¯R(T)y,y∈D(T†) |
then T† is called the maximal Tseng inverse of T.
For the proof of the main results in the next sections, we need the following lemmas:
Lemma 1.1. [2] Let TB=[AB0D]:D(A)⊕D(D)⊂H⊕H⟶H⊕H be a closed operator matrix such that A,D are closed operators with dense domains and B is a closable operator. Then,
(1) σle(A)∪σre(D)⊂σe(TB)⊂σe(A)∪σe(D).
(2) σw(TB)⊂σw(A)∪σw(D).
(3) σb(TB)⊂σb(A)∪σb(D).
Lemma 1.2. [17] Let TB=[AB0D]:D(A)⊕D(D)⊂H⊕H⟶H⊕H be a closed operator matrix such that A,D are closed operators with dense domains and B is a closable operator. Then,
(1) asc(A)≤asc(TB)≤asc(A)+asc(D),
(2) des(D)≤des(TB)≤des(A)+des(D).
Lemma 1.3. [16] Let T∈C(H). Suppose that either α(T) or β(T) is finite, and that asc(T) is finite. Then, α(T)≤β(T).
Lemma 1.4. [6] Let T∈C(H). Then, the following inequalities hold for any non negative integer k:
(1) α(Tk)≤asc(TB)⋅α(T);
(2) β(Tk)≤des(TB)⋅β(T).
Lemma 1.5. [11] Suppose that the closed operator T is a Fredholm operator and B is T-compact. Then,
(1) T+B is a Fredholm operator,
(2) ind(T+B)=ind(T).
Lemma 1.6. [16] Let T∈C(H). Suppose that asc(T) is finite, and that α(T)=β(T)<∞. Then, des(T)=asc(T).
In this section, some sufficient and necessary conditions for TB to be Browder for some closable operator B with D(B)⊃D(D) are given and the set ⋂B∈C+D(K,H)σb(TB) is estimated.
Theorem 2.1. Let A∈C(H) and D∈C(K) be given operators with dense domains. If β(A)=∞, then TB is left Browder for some B∈C+D(K,H) if and only if A is left Browder.
Proof. The necessity is obvious by Lemmas 1.1–1.3.
Now we verify the sufficiency. Let p=asc(A)<∞, then α(Ap)≤p⋅α(A)<∞ from Lemma 1.4. Thus, we get dim(R(A)+N(Ap))⊥=∞, since β(A)=∞. It follows that there are two infinite dimensional subspaces Δ1 and Δ2 of (R(A)+N(Ap))⊥ such that (R(A)+N(Ap))⊥=Δ1⊕Δ2. Define an operator B:K→H by
B=TB=[0U0]:K→[R(A)+N(Ap)Δ1Δ2] |
where U:K→Δ1 is a unitary operator. Then, we can claim that TB is left Browder.
First we prove that N(TB)=N(A)⊕{0}, which implies α(TB)<∞. It is enough to verify that N(TB)⊂N(A)⊕{0}. Let (xy)∈N(TB), then Ax+By=0 and Dy=0. Thus, Ax=−By∈R(A)∩Δ1={0}, and hence x∈N(A) and By=0. From the definition of B, we get y=0, which means N(TB)⊂N(A)⊕{0}.
Next, we check that R(TB) is closed. We only need to check that the range of [U0D] is closed since R(A) is closed. Suppose that
[U0D]yn→(y10y3)(n→∞) |
where yn∈D(D). Then, Uyn→y1(n→∞) and Dyn→y3(n→∞). Thus, we have that yn→U−1y1(n→∞) from U is a unitary operator. We also obtain U−1y1∈D(D) and DU−1y1=y3 since D is a closed operator. Hence,
(y10y3)=[U0D]U−1y1 |
which implies R(TB) is closed.
Last, we verify that N(TpB)=N(Tp+1B0), which induces asc(TB)<∞. Let (xy)∈N(Tp+1B), then
{Ap+1x+ApBy+Ap−1BDy+⋅⋅⋅+ABDp−1y+BDpy=0,Dp+1y=0. |
Thus, Dpy∈N(D)∩R(Dp) and BDpy∈(R(A)+N(Ap))⊥⊂R(A)⊥. This implies that Ap+1x+ApBy+Ap−1BDy+⋅⋅⋅+ABDp−1y=−BDpy∈R(A)∩(R(A)+R(Ap))⊥⊂R(A)∩R(A)⊥={0}, and then
{Ap+1x+ApBy+Ap−1BDy+⋅⋅⋅+ABDp−1y=0,BDpy=0. |
We can obtain Dpy=0 by the definition of B. According to Ap+1x+ApBy+⋅⋅⋅+ABDp−1y=A(Apx+Ap−1By+⋅⋅⋅+BDp−1y)=0, we have
Apx+Ap−1By+⋅⋅⋅+BDp−1y∈N(A). |
Let x1:=Apx+Ap−1By+⋅⋅⋅+BDp−1y, then
Apx+Ap−1By+⋯+ABDp−2y−x1+BDp−1y=0. |
It follows that Apx+Ap−1By+⋯+ABDp−2y−x1= −BDp−1y∈(R(A)+N(A))∩(R(A)+N(Ap))⊥⊂(R(A)+N(Ap))∩(R(A)+N(Ap))⊥={0}. Then, Apx+Ap−1By+⋯+ABDp−2y−x1=−BDp−1y=0, i.e.,
{Apx+Ap−1By+Ap−1BDy+⋅⋅⋅+ABDp−2y=x1,BDp−1y=0. |
Let x2:=Ap−1x+Ap−2By+Ap−1BDy+⋅⋅⋅+BDp−2y, then x2∈N(A2), as Ax2=x1 and x1∈N(A). This implies Ap−1x+Ap−2By+Ap−3BDy+⋅⋅⋅+BDp−3y−x2=−BDp−2y∈(R(A)+N(A2))∩(R(A)+N(Ap))⊥⊂(R(A)+N(Ap))∩(R(A)+N(Ap))⊥={0}, and then
{Ap−1x+Ap−2By+Ap−3BDy+⋅⋅⋅+BDp−3y=x2,BDp−2y=0. |
Continuing this process, there is xp∈N(Ap) such that
{Ax+By=xp,By=0. |
Thus, x∈N(Ap+1)=N(Ap). This induces that (xy)∈N(TpB), which means that N(TpB)=N(Tp+1B).
Theorem 2.2. Let A∈C(H) and D∈C(K) be given operators with dense domains. If α(D)=∞, then TB is right Browder for some B∈C+D(K,H) if and only if D is right Browder.
Proof. If TB is right Browder for some B∈C+D(K,H), then D is right Fredholm and des(D)<∞ by Lemmas 1.1 and 1.2. According to the assumption α(D)=∞, we can obtain ind(D)≥0, which means D is right Browder.
Now, we verify the reverse implication. Denote q=des(D). By Lemma 1.4, we have β(Dq)≤q⋅β(D)<∞, and then R(Dq) is closed. Thus, K=R(Dq)⊕R(Dq)⊥. This includes N(D)=[N(D)∩R(Dq)]⊕[N(D)∩R(Dq)⊥]. According to the assumption dim(N(D))=∞ and dim(N(D)∩R(Dq)⊥)<∞, we know that dim(N(D)∩R(Dq))=∞. Then, there exist two infinite dimensional subspaces Ω1 and Ω2 such that N(D)∩R(Dq)=Ω1⊕Ω2. Define the operator B0 by
B=[U000]:[Ω1Ω2N(D)∩R(Dq)⊥N(D)⊥]→H |
where U:Ω1→H is a unitary operator. Then, we obtain that R(TB)=[HR(D)] is closed, β(TB)=β(D)<∞, and α(TB)=α(A)+dimΩ2+dim(N(D)∩R(Dq)⊥)=∞. Thus, we also get des(TB)<∞ by des(D)=q<∞. This means TB is right Browder.
Remark 2.1. Let A∈C(H) and D∈C(K) be given operators with dense domains. If TB is Browder for some B∈C+D(K,H), then α(D)<∞ if and only if β(A)<∞. In fact, if TB is Browder for some B∈C+D(K,H), then A is left Fredholm with finite ascent and D is right Fredholm with finite descent, by Lemmas 1.1 and 1.2. It follows that TB admits the following representation:
TB=[A1B11B120B21B2200D1000]:[D(A)N(D)N(D)⊥∩D(A)]⟶[R(A)R(A)⊥R(D)R(D)⊥] |
where D1:D(D)→R(D) is invertible and A1:D(A)→R(A1)(=R(A)) is an operator with closed range. Let A†1:R(A)→H be the maximal Tseng inverse of A1, then A1A†1=IR(A). Set P=[I0−B12D−1100I−B22D−11000I0000I], Q=[I−A†1B1100I000I], then
PTBQ=[A1000B21000D1000]:=TB21. |
Thus, ind(TB)=ind(TB21) since P and Q are both injective. Assume that α(D)<∞ (resp., β(A)<∞), then B21 is a compact operator. By Lemma 1.5, we also obtain ind(TB)=ind(A)+ind(D)=0. Therefore, α(D)<∞ if and only if β(A)<∞.
The next result is given under the hypothesis that β(A) and α(D) are both finite.
Theorem 2.3. Let A∈C(H) and D∈C(K) be given operators with dense domains. Suppose that β(A)<∞ and α(D)<∞. Then, TB is Browder for some B∈C+D(K,H) if and only if A is left Browder, D is right Browder, and α(A)+α(D)=β(A)+β(D).
Proof. Necessity. If TB is Browder for some B∈C+D(K,H), then A is left Fredholm, D is right Fredholm, asc(A)<∞, and des(D)<∞ by Lemmas 1.1 and 1.2. This means A is left Browder and D is right Browder. We can also get α(A)+α(D)=β(A)+β(D) by Remark 2.1.
Sufficiency. There are two cases to consider.
Case Ⅰ: Assume that α(D)≤β(A), then we can define a unitary operator J1:N(D)→M1, where M1⊂R(A)⊥ and dim(M1)=α(D). Set
B=[00J10]:[N(D)N(D)⊥]→[R(A)R(A)⊥]. |
We have α(TB)=α(A) and β(TB)=β(D)+β(A)−α(D). Then, asc(TB)=asc(A)<∞ and α(TB)=β(TB)<∞ by the assumption α(A)+α(D)=β(A)+β(D). Thus, des(TB)<∞ from Lemma 1.6, which means TB is Browder.
Case Ⅱ: Suppose that α(D)>β(A). Define a unitary operator J2:M2→R(A)⊥, where M2⊂N(D) and dim(M2)=β(A). Let
B=[00J20]:[M2M⊥2]→[R(A)R(A)⊥]. |
We get α(TB)=α(A)+α(D)−β(A) and β(TB)=β(D). Then, des(TB)=des(D)<∞ and α(TB)=β(TB)<∞ by the assumption α(A)+α(D)=β(A)+β(D). Hence, α(T∗B)=β(T∗B)<∞ and asc(T∗B)=des(TB)<∞ by [1, Proposition 3.1]. Therefore, we get asc(TB)=asc(T∗∗B)=des(T∗B)<∞ by Lemma 1.6. This means TB is Browder.
According to Theorems 2.1–2.3 and Remark 2.1, we obtain the theorem below, which is the main result of this section.
Theorem 2.4. Let A∈C(H) and D∈C(K) be given operators with dense domains. Then, TB is Browder for some B∈C+D(K,H) if and only if A is left Browder, D is right Browder, and α(A)+α(D)=β(A)+β(D).
Immediately, we get the following corollary, in which the set ⋂B∈C+D(K,H)σb(TB) is estimated:
Corollary 2.1. Let A∈C(H) and D∈C(K) be given operators with dense domains. Then,
⋂B∈C+D(K,H)σb(TB)=σlb(A)∪σrb(D)∪{λ∈C:α(A−λI)+α(D−λI) |
≠β(A−λI)+β(D−λI)}. |
Remark 2.2. Theorem 2.4 and Corollary 2.1 extend the results of bounded case in [5,18].
Theorem 2.5. Let A∈C(H) and D∈C(K) be given operators with dense domains. Then,
σb(TB)=σb(A)∪σb(D) | (2.1) |
for every B∈C+D(K,H), if and only if λ∈σasc(D)∖(σlb(A)∪σrb(D)) implies α(A−λI)+α(D−λI)≠β(A−λI)+β(D−λI), where σasc(D)={λ∈C:asc(D−λI)=∞}.
Proof. Equation (2.1) holds for every B∈C+D(K,H) if and only if
σasc(D)∖(σlb(A)∪σrb(D))⊂σb(TB) |
by [2, Theorem 3.3]. That is,
σasc(D)∖(σlb(A)∪σrb(D))⊂⋂B∈C+D(K,H)σb(TB). |
This induces
σasc(D)∖(σlb(A)∪σrb(D))⊂{λ∈C:α(A−λI)+α(D−λI)≠β(A−λI)+β(D−λI)}. |
Theorem 3.1. Let A∈C(H) and D∈C(K) with dense domains. Then, TB is right invertible for some B∈C+D(K,H) if and only if D is right invertible and α(D)≥β(A).
Proof. Necessity. If TB is right invertible for some B∈C+D(K,H), then D is right invertible and TB has the following representation:
TB=[A1B11B120B21B2200D1]:[D(A)N(D)N(D)⊥∩D(A)]⟶[R(A)R(A)⊥K]. |
Obviously, D1:N(D)⊥∩D(A)→K is a bijection. Set P=[I0−B12D−110I−B22D−1100I], then
PTB=[A1B1100B21000D1]:=ˆTB. |
Thus, R(ˆTB)=H⊕K since P is bijective, and hence α(D)≥β(A).
Sufficiency. There are two cases to consider.
Case Ⅰ: Let α(D)=∞, then we can define a Unitary operator U:N(D)→H. Set
B=[U0]:[N(D)N(D)⊥]→H. |
Clearly, TB is right invertible.
Cases Ⅱ: Assume that α(D)<∞, then we can define the right invertible operator R:N(D)→R(A)⊥. Set
B=[00R0]:[N(D)N(D)⊥]→[R(A)R(A)⊥]. |
Clearly, TB is right invertible.
From the above theorem and Theorems 5.2.1, 5.2.3 of [15], we can obtain the next results immediately.
Theorem 3.2. Let A∈C(H) and D∈C(K) with dense domains. Then, TB is invertible for some B∈C+D(K,H) if and only if A is left invertible, D is right invertible, and α(D)≥β(A).
Corollary 3.1. Let A∈C(H) and D∈C(K) with dense domains. Then,
⋂B∈C+D(K,H)σ(TB)=σap(A)∪σδ(D)∪{λ∈C:α(D−λI)≠β(A−λI)}. |
Remark 3.1. Theorem 3.2 and Corollary 3.1 are also valid for bounded operator matrix B∈B(K,H). These conclusions extend the results in [8,9,12,13].
From Corollary 3.1, we have the following theorem, which extends the results of [3,7,10,12].
Theorem 3.3. Let A∈C(H) and D∈C(K) be given operators with dense domains. Then,
σ(TB)=σ(A)∪σ(D) | (3.1) |
for every B∈C+D(K,H) if and only if
λ∈σp,1(D)∩σr,1(A)⇒α(D−λI)≠β(A−λI) |
where σp,1(D)={λ∈C:N(D−λI)≠{0},N(D−λI)=K} and σr,1(A)={λ∈C:N(A−λI)={0}, R(A−λI)=¯R(A−λI)≠H}.
Proof. Equation (3.1) holds for every B∈C+D(K,H) if and only if
σp,1(D)∩σr,1(A)⊂σ(TB) |
by [7, Corollary 2]. That is,
σp,1(D)∩σr,1(A)⊂⋂B∈C+D(K,H)σ(TB). |
This induces
σp,1(D)∩σr,1(A)⊂{λ∈C:α(D−λI)≠β(A−λI)} |
by Corollary 3.1, σr,1(A)∩σap(A)=∅, and σp,1(D)∩σδ(D)=∅.
Applying Theorems 2.5 and 3.3 for the Hamiltonian operator matrix, we obtain the next result.
Theorem 3.4. Let H=[AB0−A∗]:D(A)⊕D(A∗)⊂H⊕K→H⊕K be a Hamiltonian operator matrix, where A is a densely defined closed operator and B is a self adjoint operator. Then, the following assertions hold:
(1) For every C∈C+A∗(K,H),
σb(H)=−σb(A)∗∪σb(A) | (3.2) |
if and only if
λ∈σasc(A∗)∖(−σlb(A)∪σrb(A∗))⇒α(A−λI)+α(A+¯λI)≠β(A−λI)+β(A+¯λI). |
(2) For every C∈C+A∗(K,H),
σ(H)=−σ(A)∗∪σ(A) | (3.3) |
if and only if
λ∈(−σr,1(A)∗∩σr,1(A))∖{λ∈C:Reλ=0}⇒β(A+¯λI)≠β(A−λI). |
In particular, if σr,1(A) does not include symmetric points about the imaginary axis, then (3.3) holds.
Example 3.1. Consider the plate bending equation in the domain {(x,y):0<x<1,0<y<1}
D(∂2∂x2+∂2∂y2)2ω=0 |
with boundary conditions
ω(x,0)=ω(x,1)=0, |
∂2ω∂x2+∂2ω∂y2=0,y=0,1. |
Set
θ=∂ω∂x,q=D(∂3ω∂x3+∂3ω∂y3),m=−D(∂2ω∂x2+∂2ω∂y2) |
then the equation can be written as the Hamiltonian system (see [19])
∂∂x(ωθqm)=[0100−∂2∂y200−1D000∂2∂y200−10](ωθqm) |
and the corresponding Hamiltonian operator matrix is given by
H=[0100−d2dy200−1D000d2dy200−10]=[AB0−A∗] |
with domain D(A)⊕D(A∗)⊂H⊕H, where H=L2(0,1)⊕L2(0,1), A=AC[0,1], and
A=[01−d2dy20],B=[000−1D], |
D(A)={(ωθ)∈H:ω(0)=ω(1)=0,ω′∈A,ω″∈H}. |
By a simple calculation, we get σr,1(A)=∅ and σ(A)={kπ:k=±1,±2,...}. By Theorem 3.4, we get
σ(H)=−σ(A∗)∪σ(A)={kπ:k=±1,±2,...}. |
On the other hand, we can easily calculate that
σ(H)={kπ:k=±1,±2,...}=−σ(A∗)∪σ(A). |
Certain spectral and Browder spectral properties of closed operator matrix TB=[AB0D] in a Hilbert space are considered in this paper. Specifically, some sufficient and necessary conditions are described under which TB is Browder (resp., invertible) for some closable operator B with D(B)⊃D(D). Additionally, a sufficient and necessary condition is obtained under which the Browder spectrum (resp., spectrum) of TB coincides with the union of the Browder spectrum (resp., spectrum) of its diagonal entries. These results also hold for bounded operator matrix. As an application, the corresponding properties of the Hamiltonian operator matrix from elasticity theory are given.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
We are grateful to the referees for their valuable comments on this paper.
This work is supported by the Basic Scientific Research Funds of Subordinate Universities of Inner Mongolia (Grant No. ZSLJ202213).
No potential conflict of interest was reported by the authors.
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