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On a boundary control problem for a pseudo-parabolic equation

  • Previously, boundary control problems for parabolic type equations were considered. A portion of the thin rod boundary has a temperature-controlled heater. Its mode of operation should be found so that the average temperature in some region reaches a certain value. In this article, we consider the boundary control problem for the pseudo-parabolic equation. The value of the solution with the control parameter is given in the boundary of the interval. Control constraints are given such that the average value of the solution in considered domain takes a given value. The auxiliary problem is solved by the method of separation of variables, and the problem under consideration is reduced to the Volterra integral equation. The existence theorem of admissible control is proved by the Laplace transform method.

    Citation: Farrukh Dekhkonov. On a boundary control problem for a pseudo-parabolic equation[J]. Communications in Analysis and Mechanics, 2023, 15(2): 289-299. doi: 10.3934/cam.2023015

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  • Previously, boundary control problems for parabolic type equations were considered. A portion of the thin rod boundary has a temperature-controlled heater. Its mode of operation should be found so that the average temperature in some region reaches a certain value. In this article, we consider the boundary control problem for the pseudo-parabolic equation. The value of the solution with the control parameter is given in the boundary of the interval. Control constraints are given such that the average value of the solution in considered domain takes a given value. The auxiliary problem is solved by the method of separation of variables, and the problem under consideration is reduced to the Volterra integral equation. The existence theorem of admissible control is proved by the Laplace transform method.



    Throughout this article, we only consider simple, undirected, and finite graphs and assume that all graphs are connected. Suppose G is a graph with the vertex set V(G)={v1,v2,,vn} and the edge set E(G)={e1,e2,,em}. The adjacency matrix A(G) is a 01n×n matrix indexed by the vertices of G and defined by aij=1 if and only if vsvtEG. For more notation, one refer to [1].

    The Laplacian matrix of graph G is defined as L(G)=D(G)A(G), and assume that the eigenvalues of L(G) are labeled 0μ1<μ2<<μn.

    (L(G))st={1,s=t;1,st,vsvt;0,otherwise. (1.1)

    The normalized Laplacian matrix is given by

    (L(G))st={1,s=t;1dsdt,st,vsvt;0,otherwise. (1.2)

    The distance, dij=dG(vs,vt), between vertices us and ut of G is the length of the shortest us, ut-path in G. The Wiener index [2,3] is the sum of the distances of any two vertices in the graph G, that is

    W(G)=s<tdst.

    In 1947, the distance-based invariant first appeared in chemistry [2,3] and began to the applied to mathematics 30 years later [5]. Nowadays, the Wiener index is widely used in mathematics [6,7,8] and chemistry [9,10,11].

    In a simple graph G, the degree, dij=dG(vi), of a vertex vi is the number of edges at vi. The Gutman index [12] of the simple graph G is expressed by

    Gut(G)=s<tdsdtdst.

    Klein and Randic initially outlined the concepts associated with the resistance distance of the graph. Assume that each edge is replaced by a unit resistor, and we use rst to denote the resistance distance between two vertices s and t. Similar to the Wiener index, the Kirchhoff index [13,14] of graph G is expressed as the sum of the resistance distances between each two vertices, that is

    Kf(G)=s<trst.

    In 2007, Chen and Zhang [15] defined the multiplicative degree-Kirchhoff index [16,17], that is

    Kf(G)=s<tdsdtrst.

    Phenyl is a conjugated hydrocarbon, and L6,4,4n denote a linear chain, containing n hexagons and 2n1 squares, as shown in Figure 1.

    Figure 1.  Graphs of L6,4,4n and Ln.

    With the rapid changes of the times, organic chemistry has also developed rapidly, which has led to a growing interest in polycyclic aromatic compounds.

    In 1985, the computational method and procedure of the matrix decomposition theorem were proposed by Yang [18]. This led to the solution of some problems in graph networks and allowed the unprecedented development of self-homogeneous linear hydrocarbon chains. For example, in 2021, X.L. Ma [20] got the normalized Laplacian spectrum of linear phenylene, and the linear phenylene containing it has n hexagons and n1 squares. L. Lan [21] explored linear phenylene with n hexagons and n squares. Umar Ali [22] analyzed the strong prism of a graph G, which is the strong product of the complete graph of order 2, and X.Y. Geng [23] obtained the Laplacian spectrum of L6,4,4n, which contains n hexagons and 2n1 squares. J.B. Liu [24] derived the Kirchhoff index and complexity of On, which denoting linear octagonal-quadrilateral networks. C. Liu [25] got the Laplacian spectrum and Kirchhoff index of Ln, and Ln has t hexagons and 3t+1 quadrangles.

    Inspired by these recent works, we try to investigate the Laplacians and the normalized Laplaceians for graph transformations on phenyl dicyclobutadieno derivatives.

    The various sections of this article are as follows: In Section 2, we proposed some concepts and lemmas and used them in subsequent articles. In Section 3 and Section 4, we acquired the Laplacian matrix and the normalized Laplacian matrix, then we made sure the Kirchoff index, the multiplicative degree-Kirchoff index, and the complexity of Ln. In Section 5, we obtained conclusions based on the calculations in this paper.

    In this article, graph Ln and graph L6,4,4n are portrayed in Figure 1. Define the characteristic polynomial of matrix U of order n as PU(x)=det(xIU).

    It is easy to understand that π=(1,1)(2,2)((4n),(4n)) is an automorphism. Let V1={u1,u2,,u4n+1,v1,,v4n},V2={u1,u2,,u4n,v1,,v4n+1}, |V(Ln)|=8n and |E(Ln)|=19n4. Thus, the (normalized) Laplacians matrix can be expressed in the form of a block matrix, that is

    L(Ln)=(LV0V0LV0V1LV0V2LV1V0LV1V1LV1V2LV2V0LV2V1LV2V2),

    where LVsVt and LVsVt are a submatrix consisting of rows corresponding to the vertices in Vs and columns corresponding to the vertices in Vt, s,t=0,1,2. Let

    Q=(It00012I4n12I4n012I4n12I4n),

    then

    QL(LG)Q=(LA(G)00LS(G)),QL(LG)Q=(LA(G)00LS(G)),

    and Q is the transposition of Q.

    LA=LV1V1+LV1V2, LS=LV1V1LV1V2, LA=LV1V1+LV1V2, LS=LV1V1LV1V2

    Lemma 2.1. [20] If G is a graph and suppose that LA(G),LS(G),LA(G), and LS(G) are determined as above, then

    ϑL(Ln)(y)=θLA(G)(y)θLS(G)(y),ϑL(Ln)(y)=θLA(G)(y)θLS(G)(y).

    Lemma 2.2. [26] With the extensive study of the Kirchhoff index, Gutman and Mohar proposed an algorithm based on the relation between the Kirchhoff index and the Laplacian eigenvalues, namely

    Kf(G)=nnt=21ξt,

    and the eigenvalues of L(G) are 0=ξ1<ξ2ξn(n2).

    Lemma 2.3. [14] Suppose that the eigenvalues of L(G) are ξ1ξ2ξn, then its multiplicative degree-Kirchhoff index can be denoted by

    Kf(G)=2mnt=21ξt.

    Lemma 2.4. [1] The number of spanning trees in G can also be called the complexity of G. If G is a graph with |VG|=n and |EG|=m. Let λi(i=2,3,,n) be the eigenvalues of L(G). Then the complexity of G is

    2mτ(G)=ni=1dini=2λi.

    In this section, the main objective is to find out the Kirchhoff index of Ln. Then, combining the definition of the Laplacian matrix and Eq (1.1), we can write these block matrices as follows:

    LV1V1=(3114115115115114115114115113)(4n)×(4n),

    and

    LV1V2=(1110111111111110111110111111)(4n)×(4n).

    Hence,

    LA=(2224224224224224224224224222)(4n)×(4n),

    and

    LS=diag(4,4,6,6,6,4,,6,4,6,4).

    Assume that 0α1<α2α3α4n are the roots of PLA(x)=0, and 0β1β2β3β4n are the roots of PLS(x)=0. By Lemma 2.2, we immediately have

    Kf(L2n)=2(4n)(4ni=21αi+4nj=11βj). (3.1)

    Obviously, 4nj=11βj can be obtained according to LS.

    4nj=11βj=16×(3n2)+14×(n+2)=9n+212. (3.2)

    Next, we focus on computing 4ni=11αi. Let

    PLA(x)=det(xILA)=x(x4n1+a1x4n2++a4n2x+a4n1),a4n10.

    Based on Vieta's theorem of PLA(x), we can exactly get the following equation:

    4ni=21αi=(1)4n2a4n2(1)4n1a4n1. (3.3)

    For the sake of convenience, let Ms be used to express the sth order principal minors of matrix A, and ms=detMs is recorded. We can get m1=2,m2=4,m3=8.

    And

    ms=4ms14ms2,4s4n,

    by further induction, we have

    ms=2s.

    In this way, we can get two theorems.

    Theorem 3.1. (1)4n1a4n1=(4n)24n1.

    Proof. Due to the sum of all the principal minors of order 4n1 of LA is (1)4n1a4n1,

    (1)4n1a4n1=4ns=1detLA[s]=4ns=1det(Ms100U4ns)=4ns=1detMs1detU4ns,

    where

    Ms1=(l11202l22000ls1,s1)(s1)×(s1),
    U4ns=(ls+1,s+10l4n1,4n102l4n,4n)(4ns)×(4ns).

    Let m0=1,detU0=1, because of the symmetry of matrix LA, then detU4ns=detM4ns. Hence

    (1)4n1a4n1=4ns=1detms1detm4ns=(4n)24n1,

    as desired.

    Theorem 3.2. (1)4n2a4n2=(4n1)(4n)(4n+1)24n13.

    Proof. Since the (1)4n2a4n2 is the tatal of all the principal minors of order 4n2 of LA, we have

    (1)4n1a4n1=1s<t4ndetLA[s,t]=1s<t4ndetMs1detNts1detU4nt,

    where

    LA[s,t]=(Mp1000Nts1000U4nt),1s<t4n

    and

    Nts1=|4224224224224224|(ts1)=(ts)2ts1.

    Therefore, we can have

    (1)4n2a4n2=1s<t4ndetMs1detNts1detU4nt=1s<t4n(ts)2ts1detms1m4nt=(4n1)(4n)(4n+1)24n13.

    The proof is over.

    From the results of Theorem 3.1 and Theorem 3.2, we can get

    4ni=21αi=(1)4n2a4n2(1)4n1a4n1=16n2112, (3.4)

    where the eigenvalues of LA are 0α1<α2α3α4n.

    Lemma 3.3. Suppose L6,4,4n be the dicyclobutadieno derivative of phenylenes and the graph Ln be obtained from the transformation of the graph L6,4,4n.

    Kf(Ln)=32n3+18n2+2n3.

    Proof. Substituting Eqs (3.5) and (3.6) into (3.4), the Kirchhoff index of Ln can be expressed

    Kf(Ln)=2(4n)(4ni=21αi+4nj=11βj)=(8n)(9n+212+(4n+1)(4n1)12)=32n3+18n2+2n3.

    The result is as desired.

    The Kirchhoff index of Ln from L1 to L12 is shown in Table 1.

    Table 1.  The Kirchhoff indices of L1,L2,...,L12..
    G Kf(G) G Kf(G) G Kf(G)
    L1 17.3 L5 1486.7 L9 8268.0
    L2 110.7 L6 2524.0 L10 24457.3
    L3 344.0 L7 3957.3 L11 30454.7
    L4 781.3 L8 5850.7 L12 37360.0

     | Show Table
    DownLoad: CSV

    Next, we will further consider the Wiener index of Ln.

    Lemma 3.4. Let L6,4,4n be the dicyclobutadieno derivative of [n]phenylenes, and the graph Ln be obtained from the transformation of the graph L6,4,4n, then

    limnKf(Ln)W(Ln)=14.

    Proof. Consider dst for all vertices. For the sake of convenience, we divide the vertices of the graph into the following five categories:

    Case 1. Vertex 1 of Ln:

    g1(i)=1+2(4n1k=1k).

    Case 2. Vertex 4j3(j=1,2,,n) of Ln, i=4j3:

    g2(i)=1+2(4n1k=1k+4nik=1k).

    Case 3. Vertex 4j2(j=1,2,,n) of Ln, i=4j2:

    g3(i)=1+2(4n1k=1k+4nik=1k).

    Case 4. Vertex 4j1(j=1,2,,n) of Ln, i=4j1:

    g4(i)=1+2(4n1k=1k+4nik=1k).

    Case 5. Vertex 4j(j=1,2,,n) of Ln, i=4j:

    g5(i)=1+2(4n1k=1k+4nik=1k).

    Hence, we have

    W(Ln)=4g1(i)+2i=4j3g2(i)+2i=4j2g3(i)+2i=4j1g4(i)+2i=4jg5(i)2=4(1+24n1k=1k)+2nj=1[1+2(4jkk=1k+4n4j+2k=1k)]2+2nj=1[2+2(4j3k=1k+4n4j+2k=1k)]+2nj=1[1+2(4j2k=1k+4n4j+1k=1k)]2+2n1j=1[1+2(4j1k=1k+4n4jk=1k)]2=128n3+48n25n+33.

    Consider the above results of the Kirchhoff index and the Wiener index. We can get following equation when n tends to infinity:

    limnKf(Ln)W(Ln)=14.

    The result is as desired.

    In this section, we use the eigenvalues of the normalized Laplacian matrix to determine the multiplicative degree-Kirchhoff index of Ln. Besides, we calculate the complexity of Ln. Then

    LV1V1=(11121121120120115151151511201201120151120120112012011151151)(4n)×(4n),

    and

    LV1V2=(131121120120120151515151515151201200120151512012001201201511511513)(4n)×(4n).

    Therefore,

    LA=(2313131151545252545252545151511525451515115154521521523)(4n)×(4n),

    and

    LS=diag(43,1,65,65,65,,65,1,65,43).

    Assume that the roots of PLA(x)=0 are 0ξ1<ξ2ξ3α4n, and 0γ1γ2γ3γ4n are the roots of PLS(x)=0. By Lemma 2.3, we can get

    Kf(L2n)=2(19n4)(4ni=21ξi+4nj=11γj).

    Since Ls is a diagonal matrix. Obviously, its diagonal elements 1,43 and 65 correspond to the eigenvalues of Ls, respectively. Then, it can be clearly obtained as

    4ni=11γi=21n16. (4.1)

    Let

    PLA(x)=det(xILA)=x4n+b1x4n1++b4n1x,b4n10,

    1ξ2,1ξ3,,1ξ4n+1are the roots of the following equation

    b4n1x4n1+b4n2x4n2++b1x+1=0.

    Based on the Vieta' s theorem of PLA(x), we can get

    4ni=21ξi=(1)4n2b4n2(1)4n1b4n1.

    Similarly, we can get another two interesting facts.

    Theorem 4.1. (1)4n1b4n1=259(38n8)(4125)n.

    Proof. Let sp=detFp, then we have s1=23,s2=13,s3=215,s4=475, and

    {s4p=45s4p1425s4p2;s4p+1=45s4p425s4p1;s4p+2=s4p+115s4p;s4p+3=45s4p+215s4p+1.

    After further simplification, the transformation form of the above formula is obtained.

    {s4p=53(4125)p,1pn;s4p+1=23(4125)i,0pn1;s4p+2=13(4125)i,0pn1;s4p+3=115(4125)i,0pn1.

    Similarly, we have t1=23,t2=415,t3=215,t4=475, and

    {t4p=25t4p125t4p2;t4p+1=45t4p425t4p1;t4p+2=45t4p+1425t4p;t4p+3=t4p+215t4p+1.

    Therefore, the transformation form of the above formula is obtained.

    {t4p4=53(4125)p,1pn;t4p3=23(4125)p,0pn1;t4p2=415(4125)p,0pn1;t4p1=215(4125)p,0pn1.

    Since the (1)4n1b4n1 is the total of all the principal minors of order 4n1 of LA, we have

    (1)4n1b4n1=4ni=2detNLA[i]+s4n+t4n=145(38n8)(4125)n.

    The proof of Theorem 4.1 completed.

    Theorem 4.2. (1)4n2b4n2=13240(14520n3+4599n21496n+3)(4125)n.

    Proof. We observe that the sum of all the principal minors of order 4n in LA is (1)4n2b4n2, then

    (1)4n2b4n2=1s<t4ndetLA[s,t]fs1f4nt. (4.2)

    By Eq (4.8), we know that the result of detLA[s,t] will change with the values of s and t. Then we can get the following twenty cases:

    Case 1. i=4s, j=4t, 1s<tn,

    detφ=|451525115154525254525254515151152545|(4t4s1)=10(ts)(4125)ts.

    Case 2. i=4s, j=4t+1, 1stn1,

    detφ=|451515115154525254515151151545252545|(4t4s)=(4t4s+1)(4125)ts.

    Case 3. i=4s, j=4t+2, 1stn1,

    detφ=|451515115154525151151545252545252545|(4t4s+1)=45(2t2s+1)(4125)ts.

    Case 4. i=4s, j=4t+3, 1stn1,

    detφ=|451515115154525154525254525254515151|(4t4s+2)=15(4t4s+3)(4125)ts.

    Case 5. i0, j=4n, 1st,

    detφ=|451525115154525254525254515151152545|(4n4s1)=10(ns)(4125)ns.

    Case 6. i=4s+1, j=4t, 1s<tn,

    detφ=|115154525254525254525254515151151545|(4t4s2)=254(4t4s1)(4125)ts.

    Case 7. i=4s+1, j=4t+1, 1s<tn1,

    detφ=|115154525254525254515151151545252545|(4t4s1)=10(ts)(4125)ts.

    Case 8. i=4s+1, j=4t+2, 1s<tn1,

    detφ=|115154525254525251151545252545252545|(4t4s)=(4t4s+1)(4125)ts.

    Case 9. i=4s+1, j=4t+3, 0stn1,

    detφ=|115154525254525154525254525254515151|(4t4s+1)=(2t2s+1)(4125)ts.

    Case 10. i1, j=4n+1, 0sn,

    detφ=|115154525254525254525254515151151545|(4n4s2)=254(4n4s1)(4125)ns.

    Case 11. i=4s+2, j=4t, 0s<tn,

    detφ=|4525254525254525254525254515151151545|(4t4s3)=25(2t2s1)(4125)ts.

    Case 12. i=4s+2, j=4t+1, 0s<tn1,

    detφ=|4525254525254525151151545252545252545|(4t4s2)=5(4t4s1)(4125)ts.

    Case 13. i=4s+2, j=4t+2, 0s<tn1,

    detφ=|4525254525254525151151545252545252545|(4t4s1)=8(ts)(4125)ts.

    Case 14. i=4s+2, j=4t+3, 0stn1,

    detφ=|4525254525254525154525254525254515151|(4t4s)=(4t4s+1)(4125)ts.

    Case 15. i2, j=4t+3, 0sn1,

    detφ=|4525254525254525254525254515151151545|(4n4s3)=25(2n2s1)(4125)ns.

    Case 16. i=4s+3, j=4t, 0s<tn,

    detφ=|452525451515115254525254515151151545|(4t4s4)=1254(4t4s3)(4125)ts.

    Case 17. i=4s+3, j=4t+1, 0s<tn1,

    detφ=|452525451515115254515151151545151545|(4t4s3)=25(2t2s1)(4125)ts.

    Case 18. i=4s+3, j=4t+2, 0s<tn1,

    detφ=|452525451515115151151545252545252545|(4t4s3)=253(4t4s1)(4125)ts.

    Case 19. i=4s+3, j=4t+3, 0s<tn1,

    detφ=|452525451515115154525254525254515151|(4t4s1)=10(ts)(4125)ts.

    Case 20. i3, j=4t, 0sn1,

    detφ=|45252545151511525452525115151151545|(4n4s4)=1254(4n4s3)(4125)ns.

    Therefore, we can get

    (1)4n2b4n2=E1+E2+E3+E4,

    where

    E1=118(227n3+347n2574n+4)(4125)n1.
    E2=172(908n3+3431n2+523n)(4125)n.
    E3=145(454n3+1375n21079n)(4125)n.
    E4=181(92n3+561n2611n)(4125)n1.

    Hence

    (1)4n2b4n2=E1+E2+E3+E4=13240(14520n3+4599n21496n+4)(4125)n.

    The proof of Theorem 4.2 completed.

    Let 0ξ1ξ2ξ3ξ3n+2 are the eigenvalues of LA. We can get the following exact equation:

    4ni=21ξi=(1)4n2b4n2(1)4n1b4n1=172(14520n3+4599n21496n+838n8).

    Theorem 4.3. Set L6,4,4n be the derivative [n]pheylenes, and the expression of the multiplicative degree-Kirchhoff index is

    Kf(Ln)=(29040n3+8996n23198n+8144).

    Proof. Together with Eq.(4.7) and Theorems 4.1 and 4.2, one can get

    Kf(Ln)=2(19n4)(4ni=21ξi+4ni=11γi)=2(19n4)[172(14520n3+4599n21496n+838n8)+21n16]=(29040n3+8996n23198n+8144).

    The result as desired.

    The multiplicative degree-Kirchhoff indices of Ln from L1 to L12, see Table 2.

    Table 2.  The multiplicative degree-Kirchhoff indices of L1,L2,...,L12..
    G Kf(G) G Kf(G) G Kf(G)
    L1 241.98 L5 26659.15 L9 151875.4
    L2 1818.86 L6 45675.81 L10 207691.9
    L3 5940.68 L7 72077.4 L11 275733.2
    L4 13817.44 L8 107073.9 L12 357209.6

     | Show Table
    DownLoad: CSV

    Then we want to calculate the Gutman index of Ln.

    Theorem 4.4. Suppose that L6,4,4n is the dicyclobutadieno derivative of [n]phenylenes and the graph Ln is obtained from the transformation of the graph L6,4,4n, then

    limnKf(Ln)Gut(Ln)=14.

    Proof. Consider dij for all vertices. We divide the vertices of Ln into the following four categories.

    Case 1. Vertex 4i2(i=1,2,,n) of Ln:

    f4i2=103n(56n224n+37).

    Case 2. Vertex 4i1(i=1,2,,n) of Ln:

    f4i1=103n(152n248n+29).

    Case 3. Vertex 4i(i=1,2,,n) of Ln:

    f4i=103n(140n248n+43).

    Case 4. Vertex 4i3(i=1,2,,n) of Ln:

    f4i3=103n(136n26n+71).

    According to Eq.(1.3), the Gutman index of Ln is

    Gut(Ln)=f4i+f4i1+f4i2+f4i32=10n3(242n263n+61).

    Therefore, combining with Kf(Ln) and Gut(Ln), we have

    limnKf(Ln)Gut(Ln)=14.

    The result as desired.

    Finally, we want to know the complexity of Ln.

    Theorem 4.5. For the graph Ln, we have

    τ(Ln)=23n+233n2.

    Proof. Based on Lemma 2.4, we can get

    8ni=1di4ni=2αi4nj=1βi=2(19n4)τ(Ln).

    Note that

    8ni=1di=3442n56n4.
    4ni=2αi=259(38n8)(4125)n.
    4nj=1βi=(43)2(65)3n2.

    Hence,

    τ(Ln)=23n+233n2.

    The proof is over.

    Thus, we can get the complexity of Ln from W1 to W8 which are listed in Table 3.

    Table 3.  The complexity of W1,W2W8.
    G τ(G) G τ(G)
    W1 96 W5 208971104256
    W2 20736 W6 45137758519296
    W3 4478976 W7 9749755840167936
    W4 967458816 W8 2105947261476274176

     | Show Table
    DownLoad: CSV

    In this paper, the linear chain network with n hexagons and 2n1 squares is considered. We have devoted ourselves to calculating the (multiplicative degree) Kirchhoff index, Wiener index, Gutman index, and complexity. In the meantime, we deduced that the ratio of the (multiplicative degree) Kirchhoff index to the (Gutman) Wiener index is nearly a quarter when n tends to infinity. These rules also apply to some other graphs.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This research was funded by the Anhui Provincial Natural Science Research Major Project (No. 2022AH040317) and the Anhui Provincial 2023 Action Project for Cultivating Young and Middle Aged Teachers in Universities (No. DTR2023095).

    No potential conflicts of interest were reported by the authors.



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