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Transfer Matrix MethodG.Eric Moorhouse,University of WyomingReference:Transfer Matrix MethodI.M.Gessel and R.P.Stanley,‘Algebraic Enu-meration’,in Handbook of Combinatorics Vol.2, ed.R.L.Graham et al.,Elsevier,1995,pp.1021–1061.References:Dimensions of CodesN.Hamada,‘The rank of the incidence matrixof points and d-flats infinite geometries’,J. Sci.Hiroshima Univ.Ser.A-I32(1968),381–396.M.Bardoe and P.Sin,‘The permutation mod-ules for GL(n+1,q)acting on P n(q)and F n+1q’,to appear in JLMS./~sin/preprints/hamada.dviG.E.Moorhouse,‘Dimensions of Codes from Finite Projective Spaces’(as html and as Maple worksheet)/~moorhous/src/hamada.html /~moorhous/src/hamada.mwsProblem1Let S k be the set of‘words’of length k consist-ing of‘a’s and‘b’s,with no two consecutive ‘b’s.Determine F k=|S k|.F0=1‘’F1=2‘a’‘b’F2=3‘aa’‘ab’‘ba’F3=5‘aaa’‘aab’‘aba’‘baa’‘bab’F4=8‘aaaa’‘aaab’‘aaba’‘abaa’‘abab’‘baaa’‘baab’‘baba’etc.This gives all but thefirst term of the Fibonacci sequence1,1,2,3,5,8,13,21,34,55,89,144,...To find a formula for F k ,we work instead with the generating function∞k =0F k t k =1+2t +3t 2+5t 3+8t 4+13t 5+···Observe that words w ∈S k correspond to paths of length k ,starting at vertex 1in the digraph12..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................append ‘a’append ‘a’append ‘b’Words not ending in ‘b’Words ending in ‘b’Agenda1.Motivating Problem1(above)2.Counting Walks by the Transfer Matrix Method3.Application to Problem14.Counting Closed Walks5.Counting Weighted Walks in Digraphs withWeighted Edges6.MAPLE Worksheet for Problem17.Application to Coding TheoryThe Transfer Matrix Method Let D be a digraph (directed graph),possibly with loops,having vertices 1,2,3,...,n .Let A =[a ij :1≤i,j ≤n ]be the adjacency matrix of D ;in other words,a ij = 1,if (i,j )is an edge of D ;0,otherwise.A walk of length k in D is a sequence......................................................................................................................................................................................................................................................................................................................................···◦◦◦◦i 0i 1i 2i k→→→→of (not necessarily distinct)vertices such that each ...........................................................................................◦◦i r −1i r →is an edge of D .Counting Walks from i to jLet w ij(k)be the number of walks of length k from vertex i to vertex j in D.Then w ij(k)is the(i,j)-entry of A k.This is readily computed by reading offthe coefficient of t k in the gen-erating function k≥0w ij(k)t k which in turn is the(i,j)-entry of(I−tA)−1=I+tA+t2A2+t3A3+···. Since the(i,j)-entry of(I−tA)−1is of the formpoly.in t of degree≤n−1det(I−tA),w ij(k)satisfies a linear recurrencew ij(k+n)=n−1r=0c r w ij(k+r)for all k≥0where det(I−tA)=1−c n−1t−c n−2t2−···−c0t n.The initial conditions w ij(0),w ij(1),..., w ij(n−1)depend on i and j but the recurrence does not.Counting All WalksLet w(k)= n i=1 n j=1w ij(k),the total num-ber of walks of length k.This is the coefficient of t k in the sum of the entries of(I−tA)−1.In particular w(k)satisfies the same recurrence as the w ij(k)’s:w(k+n)=n−1r=0c r w(k+r)for all k≥0but with different initial conditions.Counting Closed WalksLet w closed(k)= n i=1w ii(k),the total number of closed walks of length k(i.e.starting and ending at the same vertex).This is the coef-ficient of t k in trace((I−tA)−1).In particular w closed(k)satisfies the same linear recurrence as the w ij(k)’s and w(k),but again with different initial conditions.Here we assumed the initial/final vertex to be distinguished,i.e.the walks(i0,i1,i2,...,i k)and (i1,i2,...,i k,i0)are counted as distinct unless all i0=i1=···=i k.ExampleLet F k be the number of‘words’of length k consisting of‘a’s and‘b’s,with no two con-secutive‘b’s.F0=1‘’F1=2‘a’‘b’F2=3‘aa’‘ab’‘ba’F3=5‘aaa’‘aab’‘aba’‘baa’‘bab’F4=8‘aaaa’‘aaab’‘aaba’‘abaa’‘abab’‘baaa’‘baab’‘baba’etc.This gives all but thefirst term of the Fibonacci sequence1,1,2,3,5,8,13,21,34,55,89,144,...Observe that F k is the number of paths of length k ,starting at vertex 1in the digraph12.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................append ‘a’append ‘a’append ‘b’Words not ending in ‘b’Words ending in ‘b’A = 111(I −tA )−1=11−t −t 2 1t t 1−tk ≥0F k t k =sum of (1,1)-and (1,2)-entries of (I −tA )−1=1+t 1−t −t 2=1√5 α21−αt −β21−βt=1√5 k ≥0(αk +2−βk +2)t k where α=(1+√5)/2,β=(1−√5)/2.From (1−t −t 2) k ≥0F k t k =1+t we obtainF k = 1,if k =0;2,if k =1;F k −1+F k −2,if k ≥2so by induction,F k is the (k +1)st Fibonacci number.From the series expansion we obtain the explicit formulaF k =αk +2−βk +2√5for k ≥0.Wraparound VersionLet L k(for k≥0)be the number of‘words’of length k consisting of‘a’s and‘b’s with no consecutive‘b’s,and which do not both start and end with‘b’.For technical reasons we will take L0=2.For k≥2,we are simply counting necklaces with a mber and b lack beads having no two consecutive black beads;however,each neck-lace has a distinguished starting point(a knot in its cord)and a distinguished direction(clock-wise or counter-clockwise).L1=1‘a’L2=3‘aa’‘ab’‘ba’L3=4‘aaa’‘aab’‘aba’‘baa’L4=7‘aaaa’‘aaab’‘aaba’‘abaa’‘abab’‘baaa’‘baba’These are the familiar Lucas numbers which satisfy the same recurrence relation as the Fi-bonacci numbers,but a different initial condi-tion.Note that L k is the number of closed walks of length k in our digraph.k ≥0L k t k =trace ((I −tA )−1)=2−t 1−t −t 2=11−αt +11−βt=k ≥0(αk +βk )t kFrom (1−t −t 2) k ≥0L k t k =2−t we obtainL k = 2,if k =0;1,if k =1;L k −1+L k −2,if k ≥2From the series expansion we obtain the ex-plicit formulaL k=αk+βk for k≥0.Counting Walks with Weighted Edges As before,D is a digraph (directed graph),pos-sibly with loops,having vertices 1,2,3,...,n .Assign a weight to each edge:...........................................................................................◦◦i j →a ij (Non-edges have weight zero.)Define the weight of a walk......................................................................................................................................................................................................................................................................................................................................···◦◦◦◦i 0i 1i 2i k →→→→a i 0i 1a i 1i 2a i 2i 3a i k −1i k of length k to be the producta i 0i 1a i 1i 2a i 2i 3···a i k −1i k .Let A =[a ij :1≤i,j ≤n ].Thenw ij(k):=The sum of all weightsof walks in D of length k from vertex i to vertex j=(i,j)-entry of A khas generating function k≥0w ij(k)t k equal to the(i,j)-entry of(I−tA)−1=I+tA+t2A2+t3A3+···as before.ExampleWe have determined the number F k of words of length k consisting of‘a’s and‘b’s,with no two consecutive‘b’s.How many such words contain r‘a’s and(therefore)k−r‘b’s?12.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................a ab A = a b a0 (I −tA )−1=11−at −abt 2 1bt at 1−atThe sum of the (1,1)-and (1,2)-entries is1+bt1−at −abt 2=1+(a +b )t +(a 2+2ab )t 2+(a 3+3a 2b +ab 2)t 3+(a 4+4a 3b +3a 2b 2)t 4+···Thus,for example,among the F 4=8words of length 4,1has 4‘a’s and 0‘b’s;4have 3‘a’s and 1‘b’;3have 2‘a’s and 2‘b’s.Codes from Finite GeometryConsider the projective plane of order 2:•••••••a bc d e f g ..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................The binary code of this geometry is the sub-space C ≤F 7(where F ={0,1}mod 2)spanned by the lines:a b c d e f g a b c d e f g C ={0000000,1111111,1101000,0010111,0110100,1001011,0011010,1100101,0001101,1110010,1000110,0111001,0100011,1011100,1010001,0101110}|C|=24;dim C =4The code above is the 1-error correcting binary Hamming code of length 7.The projective plane is constructed from F 3by taking as points and lines the 1-and 2-dimensional subspaces of F 3.•••••••001010100011110111101.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Codes of Finite Projective SpacesLet F be thefield of order p e,p prime.Pro-jective n-space over F has as its points,lines, etc.the subspaces of F n+1of dimension1,2, etc.Problem:Compute the dimension of the code C=C n,p,e,k spanned by the subspaces of codi-mension k.Solution by Hamada’s Formula(the follow-ing theorem)is usually computationally infea-sible.Solution by the Transfer Matrix MethodTheorem(Bardoe and Sin,1999)Define M(t)=(1+t+t2+···+t p−1)n+1.Let D=D n,p,e,k be the digraph with vertices 1,2,...,k,and the edge from vertex i to vertex j has weight equal to the coefficient of t pj−i in M(t).Thendim C n,p,e,k=1+sum of weights of closed walks of length ein D=1+ coeff.of t ein tr[(I−tA)−1]where A is the k×k matrix whose(i,j)-entry is the weight of edge(i,j)(defined above).Example:Projective Plane of Order2C=binary code spanned by the seven lines (subspaces of codimension k=1)M(t)=(1+t)3=1+3t+3t2+t3A=[3](coefficient of t1in M(t))(I−tA)−1= 1 1−3ttr[(I−tA)−1]=11−3t=1+3t+9t2+27t3+···dim C=1+3=4。
传递矩阵法是研究转子系统动力学问题的有效手段。
传递矩阵法还具有其它方法(如摄动有限元素法)无法比拟的优点，例如，在做转子系统的临界转速、阻尼固有频率和稳定性计算分析时，由于流体密封交叉刚度、油膜轴承、阻尼项往往是不对称的，再加上陀螺力矩的影响；这样，用随机有限元素法形成的单元刚度矩阵和系统总体刚度矩矩阵往往也是不对称的，阻尼也不可以简单地以小阻尼或比例阻尼系统来替代，求解这样一个非对称系统的复特征值问题，目前还没有一个较为理想的方法。
而传递矩阵法没有随机有限元法在求解这些的问题时带来的这些困难。
因此，传递矩阵法在转子系统动力学问题的研究中占有主导的地位。
传递矩阵法matlab程序传递矩阵法是一种用于计算机程序中传递和操作矩阵的方法，在Matlab中，它被广泛应用于矩阵运算和数据处理等领域。
本文将介绍传递矩阵法的原理和在Matlab中的具体实现。
传递矩阵法是一种通过矩阵传递来操作数据的方法。
它的基本原理是将需要进行操作的数据存储在矩阵中，然后通过矩阵的传递，实现对数据的处理和计算。
这种方法的优势在于可以利用矩阵的高效运算能力，简化程序的编写和调试过程。
在Matlab中，可以使用矩阵操作函数来实现传递矩阵法。
例如，可以使用矩阵的乘法运算来实现矩阵的传递。
假设我们有两个矩阵A 和B，我们希望将矩阵A的数据传递给矩阵B，可以使用如下的Matlab代码实现：```B = A;```这样，矩阵B就完全复制了矩阵A的数据。
通过这种方式，我们可以在程序中传递矩阵，进行各种操作和计算。
除了简单的传递，传递矩阵法还可以实现更复杂的操作。
例如，可以通过传递矩阵进行矩阵的相加、相减、相乘等运算。
假设我们有两个矩阵A和B，我们希望将它们相加得到矩阵C，可以使用如下的Matlab代码实现：```C = A + B;```这样，矩阵C的每个元素都等于矩阵A和矩阵B对应元素的和。
通过传递矩阵法，我们可以很方便地实现这样的矩阵运算。
除了矩阵的运算，传递矩阵法还可以用于数据处理和分析。
例如，可以通过传递矩阵来实现数据的转置、截取、排序等操作。
假设我们有一个矩阵A，我们希望将它的每一列按照从大到小的顺序进行排序，可以使用如下的Matlab代码实现：```B = sort(A,'descend');```这样，矩阵B的每一列都按照从大到小的顺序进行了排序。
通过传递矩阵法，我们可以在Matlab中进行各种复杂的数据处理和分析。
传递矩阵法在Matlab中的应用非常广泛。
无论是矩阵运算、数据处理还是图像处理，都可以通过传递矩阵法来实现。
它不仅提高了程序的效率和可读性，还简化了程序的编写和调试过程。
利用传递矩阵法和Riccati 传递矩阵法分析转子临界转速一、所需求解转子参数将转子简化为如下所示：三个盘的参数为：1232221232221230.0160.050.0160.0120.0250.012P P P d d d I kg m I kg m I kg m I kg m I kg m I kg m ⎪=⋅=⋅=⋅⎨⎪=⋅=⋅=⋅⎩ 另，阶梯轴的三段轴的截面惯性矩分别为： 414243 1.73.20.9J cm J cm J cm ⎧=⎪=⎨⎪=⎩三段轴的单位长度轴段的质量分别为：1232.45/3.063/1.587/m kg m m kg m m kg m =⎧⎪=⎨⎪=⎩二、 试算转轴的传递矩阵取试算转速1200/p rad s ω== ; 则，各轴段的传递矩阵分别为： 第1段840.061.7102.45/l m J m m kg m -=⎧⎪=⨯⎨⎪=⎩1 1.0006e+000 6.0007e-002 5.2943e-007 1.0588e-008 3.7356e-002 1.0006e+000 1.7649e-005 5.2943e-007 6.3506e+003 1.2701e+002 1.0006e+000 6.0007e-002 2.1170e+005 6.3506e+003 3.7356e-002 H = 1.0006e+000⎧⎪⎪⎨⎪⎪⎩ 第2段840.153.2103.063/l m J m m kg m -=⎧⎪=⨯⎨⎪=⎩2 1.0145e+000 1.5044e-001 1.7595e-006 8.7927e-008 3.8782e-001 1.0145e+000 2.3506e-005 1.7595e-006 4.9669e+004 2.4821e+003 1.0145e+000 1.5044e-001 6.6353e+005 4.9669e+004 3.8782e-001 H = 1.0145e+000⎧⎪⎪⎨⎪⎪⎩ 第3段840.053.2103.063/l m J m m kg m -=⎧⎪=⨯⎨⎪=⎩3 1.0002e+000 5.0002e-002 1.9531e-007 3.2552e-009 1.4358e-002 1.0002e+000 7.8128e-006 1.9531e-007 5.5135e+003 9.1890e+001 1.0002e+000 5.0002e-002 2.2054e+005 5.5135e+003 1.4358e-002 H = 1.0002e+000⎧⎪⎪⎨⎪⎪⎩ 第4段840.033.2103.063/l m J m m kg m -=⎧⎪=⨯⎨⎪=⎩4 1.0000e+000 3.0000e-002 7.0313e-008 7.0313e-010 3.1013e-003 1.0000e+000 4.6875e-006 7.0313e-008 1.9848e+003 1.9848e+001 1.0000e+000 3.0000e-002 1.3232e+005 1.9848e+003 3.1013e-003 H = 1.0000e+000⎧⎪⎪⎨⎪⎪⎩ 第5段840.10.9101.587/l m J m m kg m -=⎧⎪=⨯⎨⎪=⎩5 1.0053e+000 1.0011e-001 2.7788e-006 9.2607e-008 2.1163e-001 1.0053e+000 5.5614e-005 2.7788e-006 1.1430e+004 3.8094e+002 1.0053e+000 1.0011e-001 2.2877e+005 1.1430e+004 2.1163e-001 H = 1.0053e+000⎧⎪⎪⎨⎪⎪⎩ 第6段840.060.9101.587/l m J m m kg m -=⎧⎪=⨯⎨⎪=⎩6 1.0007e+000 6.0008e-002 1.0000e-006 2.0000e-008 4.5706e-002 1.0007e+000 3.3338e-005 1.0000e-006 4.1137e+003 8.2272e+001 1.0007e+000 6.0008e-002 1.3714e+005 4.1137e+003 4.5706e-002 H = 1.0007e+000⎧⎪⎪⎨⎪⎪⎩ 此6段传递矩阵均采用MATLAB 编程求解，MATLAB 的源文件为H.m 三、采用传递矩阵法进行各段轴的状态参数的传递初始参数列阵为:0101010120101012011Pd d X X I I p M p I Q mp x θθωθ⎛⎫⎛⎫ ⎪⎪ ⎪ ⎪ ⎪=⎛⎫ ⎪-- ⎪⎪ ⎪ ⎪⎝⎭⎝⎭ ⎪⎝⎭令011X =，则初始矩阵可化为：010*******.046e θθ⎛⎫⎪⎪ ⎪ ⎪⎝⎭以初始矩阵乘第一轴段的传递矩阵，则可得第一段轴的终端状态参数：1011011011010.06306+ 1.0541.102 +5890 3.0885 26566.0 5..7062556k k k k e e X M Q θθθθθ⎛⎫⎛⎫⎪ ⎪ ⎪ ⎪= ⎪ ⎪ ⎪ ++⎪⎝⎭⎝⎭由于考虑支座的支撑刚度系数变化从5101*101*10，先取51*10，那么100001000010001KK k⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥-⎣⎦，此处510k =，则可得支座A 后第2段的起始端参数阵为： 020102010201560201 0.06306 + 1.0541.102 + 5890.0 3.088*10260.2 5.149*2.01076X M Q θθθθθ⎛⎫⎛⎫ ⎪ ⎪ ⎪ ⎪= ⎪ ⎪⎪ ++⎪⎝⎭⎝⎭用第2段的传递矩阵乘此矩阵，可得第2段终端参数：20120120120166 0.2402+ 2.4721.282 + 19.11900.0 1.147*1099133.0 6.170477*1k k k k X M Q θθθθθ⎛⎫⎛⎫ ⎪ ⎪ ⎪ ⎪= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭++用中间圆盘的传递矩阵乘第2段终端参数阵，即可得第3段起始端参数：03010306103016307010.2402 + 2.472 1.282 + 1958022.0 1.848*102.52*10 3.11*10.47X M Q θθθθθ+⎛⎫⎛⎫ ⎪ ⎪ ⎪ ⎪= ⎪ ⎪⎪ ⎭⎝⎭+⎪⎝ 用第3段传递矩阵乘其始端参数矩阵：'013'013'013'6356017 0.3238 + 3.9092.231+ 401.855*10 3.419*102.581*10 3.178*10.02k k k k X M Q θθθθθ⎛⎫⎛⎫ ⎪ ⎪⎪ ⎪= ⎪ ⎪⎪ ⎪⎪⎝⎭++⎝⎭ 用上式乘以支座刚度矩阵，得其终端参数：01305667130130130.3238 + 3.9092.231 + 40.01.855*10 3.419*102.549*10 3.1392*10k k k k X M Q θθθθθ⎛⎫⎛⎫⎪ ⎪⎪ ⎪= ⎪ ⎪⎪ ⎪ ⎝⎭⎭+⎪+⎝ 则，根据可得： ,则可得支座B 后第4段的起始端参数阵为：560104010401040104670.3238 + 3.9092.231 + 40.01.855*10 3.419*102.549*10 3.139*102X M Q θθθθθ⎛⎫⎛⎫ ⎪ ⎪⎪ ⎪= ⎪ ⎪⎪ ⎪⎪⎝⎭⎝++⎭ 同上，用此段轴的传递函数乘其起始端的状态参数，可得：4014014014015667 0.4056 + 5.372 3.281 + 582.626*10 4.369*102.597*10 3..172*20k k k k X M Q θθθθθ+⎛⎫⎛⎫⎪ ⎪ ⎪ ⎪= ⎪ ⎪⎪ ⎪⎝⎭⎝⎭+则，根据40k M =可得：01-16.64θ= 则，可得第5段的起始参数矩阵：050750505-1.3753.69601.12*10X M Q θ⎛⎫⎛⎫ ⎪ ⎪ ⎪ ⎪= ⎪ ⎪ ⎪ ⎭⎝-⎪⎝⎭ 其中，5θ为铰链处的转角。
多体系统传递矩阵法及其应用多体系统是由许多互相作用的体组成的复杂体系，如分子、原子、晶体等。
传递矩阵法是一种处理多体系统的方法，它能够高效地计算多体系统在空间中的相互作用关系，是现代物理学研究中不可或缺的重要手段。
传递矩阵法最早应用于固体物理领域中的声子传输问题。
其基本思想是通过计算相邻体间的相互作用关系，得出整个体系中体与体之间的传递矩阵。
具体来说，传递矩阵法假设每个体以弹性球体为模型，并将每个弹性球体中储存的平面波能量互相传递，形成整个体系中的传递矩阵。
这种方法在研究固体中光声声子的传输、声子光子的散射等问题中具有重要的应用价值。
除了固体物理领域，传递矩阵法还广泛应用于原子、分子的电子结构计算以及化学反应机理的模拟等领域。
计算和分析分子/团簇数据在化学特异性中的作用是大量分子和聚集体计算化学和物理学领域的重要问题。
基于传递矩阵法可以对分子结构、物理特性以及从催化到切削的各种机械和电子反应进行分析和预测。
在实际应用中，传递矩阵法的计算和建模过程也面临着许多挑战，如有限的计算能力、模型精度等问题。
为了解决这些问题，一些改进的传递矩阵法，如多重散射和Greens函数方法等也被提出，以提高计算精度和效率。
同时，也有越来越多的科研工作者尝试将传递矩
阵法与机器学习等前沿技术相结合，从而拓展传递矩阵法的应用范围和精度，实现更加智能化的计算和数据分析。
总之，传递矩阵法在物理、化学、材料学、和计算机科学等领域都扮演着重要的角色。
通过该方法，我们可以更加深入地理解多体系统内部的相互作用关系，进而更好地预测和优化系统的性质和行为，为理论和实践应用提供了新的思路和创新性解决方案。
