CN107092581A - The transport equation response matrix Block Diagonalization method discussed based on symmetric group - Google Patents
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Abstract
Description
技术领域technical field
本发明涉及核反应堆堆芯设计和安全技术领域,具体涉及一种基于对称群论的输运方程响应矩阵分块对角化方法。The invention relates to the technical field of nuclear reactor core design and safety, in particular to a block diagonalization method of a transport equation response matrix based on symmetric group theory.
背景技术Background technique
全堆芯输运节块计算在中子学的计算中具有重要的意义。在中子输运方程的求解中,若能够提前构造节块内部边界处中子出入射流的响应关系,则可根据该响应关系,通过节块的扫描迭代直至收敛,即可获得节块的中子泄漏,从而求解出各个节块中的中子通量分布。这种方法称为响应矩阵方法。输运方程的角度项常常采用球谐函数进行离散。若角度项展开取前N阶的球谐函数,并将输运方程在前N阶的球谐函数上进行投影,即可获得(N+1)2个线性代数方程组,即PN近似方程组。由于节块内部未知量的总数与球谐函数的个数是正相关的,因此,未知量的总数随着角度展开阶数呈平方量级增长。The calculation of the whole core transport segment is of great significance in the calculation of neutronics. In the solution of the neutron transport equation, if the response relation of the neutron entry and exit jet at the inner boundary of the nodal block can be constructed in advance, then the midpoint of the nodal block can be obtained by scanning and iterating the nodal block until convergence according to the response relation. neutron leakage, so as to solve the neutron flux distribution in each node. This approach is called the response matrix approach. The angle term of the transport equation is usually discretized by using spherical harmonics. If the angle term is expanded to take the first N-order spherical harmonic functions, and the transport equation is projected on the first N-order spherical harmonic functions, (N+1) 2 linear algebraic equations can be obtained, that is, the PN approximate equations . Since the total number of unknowns inside a block is positively correlated with the number of spherical harmonics, the total number of unknowns increases quadratically with the order of angle expansion.
响应矩阵的构建通常涉及矩阵的乘法以及求逆运算,这些浮点运算量大致与未知量个数的三次方呈正比关系。因此,当未知量的个数较多时,计算内存以及浮点运算总量将会大大增加。The construction of the response matrix usually involves matrix multiplication and inversion operations, and the amount of these floating-point operations is roughly proportional to the cube of the number of unknowns. Therefore, when the number of unknowns is large, the calculation memory and the total amount of floating-point operations will be greatly increased.
在实际堆芯计算中,对于泄漏较强、具有强烈各向异性散射效应的小型快中子反应堆,若想得到更为精确的数值计算结果,通常要采用较高的角度阶数进行逼近。此外,若进行燃耗、瞬态等计算,各个节块的截面不同,导致各个节块需要单独计算出各自的响应矩阵,这对于计算内存与计算效率造成了极大的挑战。In the actual core calculation, for small fast neutron reactors with strong leakage and strong anisotropic scattering effect, if you want to obtain more accurate numerical calculation results, you usually need to use a higher angle order for approximation. In addition, if fuel consumption, transient, etc. are calculated, the cross-sections of each block are different, so that each block needs to calculate its own response matrix separately, which poses a great challenge to computing memory and computing efficiency.
发明内容Contents of the invention
为了克服上述现有技术存在的问题,本发明的目的是提供一种基于对称群论的输运方程响应矩阵分块对角化方法,该方法基于对称群论,采用不可约对称化的基函数,将输运方程的求解简化为若干个互相解耦的子问题,从而将响应矩阵分解为一系列互相解耦的位于原矩阵的对角线位置的子块。在矩阵乘法以及求逆运算中,可以对子块矩阵单独进行乘法及求逆运算,从而大大降低矩阵的浮点运算量。此外,由于矩阵进行了分块对角化处理,与原始矩阵相比,新矩阵的非零元素大大减小,存储响应矩阵所需的计算内存也大大降低。In order to overcome the problems in the above-mentioned prior art, the object of the present invention is to provide a method for block diagonalization of the transport equation response matrix based on symmetric group theory, which is based on symmetric group theory and uses irreducible symmetrized basis functions , the solution of the transport equation is simplified into several mutually decoupled sub-problems, and thus the response matrix is decomposed into a series of mutually decoupled sub-blocks located on the diagonal positions of the original matrix. In the matrix multiplication and inversion operations, the multiplication and inversion operations can be performed on the sub-block matrix independently, thereby greatly reducing the amount of floating-point operations of the matrix. In addition, due to the block-diagonalization of the matrix, the new matrix has significantly fewer non-zero elements compared to the original matrix, and the computational memory required to store the response matrix is also greatly reduced.
为了实现上述目的,本发明采取了以下技术方案予以实施:In order to achieve the above object, the present invention has taken the following technical solutions to implement:
基于对称群论的输运方程响应矩阵分块对角化方法,该方法包括以下步骤:A block diagonalization method of the response matrix of the transport equation based on symmetric group theory, the method includes the following steps:
步骤一,对于整个堆芯求解区域按照规则的几何进行划分,从而得到规则的节块,基于变分节块方法,对中子输运方程建立泛函,构造输运方程的弱解形式;采用球谐函数以及正交多项式对中子角通量密度进行离散,球谐函数与空间正交多项式共同构成了通量的基函数,如公式(8)所示Step 1: Divide the entire core solution area according to regular geometry to obtain regular nodal blocks. Based on the variational nodal method, a functional is established for the neutron transport equation to construct a weak solution form of the transport equation; Spherical harmonic functions and orthogonal polynomials discretize the neutron angular flux density, and spherical harmonic functions and spatial orthogonal polynomials together constitute the basis function of the flux, as shown in formula (8)
其中ψ(r,Ω)为偶通量密度;J(r,Ω)为节块边界处的拉格朗日乘子项;g(Ω)=[Y0,0,Y2,-2,Y2,-1,Y2,0,Y2,1,Y2,2...]T为偶阶的归一化的球谐函数;f(r)=[f1,f2,f3...]T为定义在节块v内部的正交多项式;h(r)=[h1,h2...]T为定义在节块v边界处的正交多项式;表示位于1号边界面处的边界基函数,表示位于2号边界面处的边界基函数;以及ξ为分别为偶中子角通量密度展开矩以及中子流密度展开矩;T为转置符号;where ψ (r,Ω) is the even flux density; J(r,Ω) is the Lagrangian multiplier term at the block boundary; g(Ω)=[Y 0,0 ,Y 2,-2 , Y 2,-1 ,Y 2,0 ,Y 2,1 ,Y 2,2 ...] T is the even order normalized spherical harmonic function; f(r)=[f 1 ,f 2 ,f 3 ...] T is an orthogonal polynomial defined inside the node v; h(r)=[h 1 , h 2 ...] T is an orthogonal polynomial defined at the boundary of the node v; Indicates the boundary basis function located at the No. 1 boundary surface, Indicates the boundary basis function located at the No. 2 boundary surface; And ξ is the expansion moment of the even neutron angular flux density and the expansion moment of the neutron current density respectively; T is the transpose symbol;
将离散形式的通量以及拉格朗日乘子项带入泛函中整理可得响应矩阵方程:Bring the flux in discrete form and the Lagrangian multiplier term into the functional to sort out the response matrix equation:
j+=Bs+Rj- j + =Bs+Rj -
(16) (16)
其中为偶中子角通量密度展开矩;j+为出射偏中子流密度展开矩;j-为入射偏中子流密度展开矩;s为源项展开矩;R,H,C,B为变分节块方法的四个响应矩阵;in j + is the expanded moment of the outgoing partial neutron flux density; j - is the expanded moment of the incident partial neutron flux density; s is the expanded moment of the source term; R, H, C, B are Four response matrices for the variational block method;
步骤二,根据步骤一中的节块的几何形状,确定节块所有的对称性变换构成的对称群;利用对称群的投影算符对步骤一中的基函数进行投影,从而构造出不可约对称化的基函数;Step 2: According to the geometric shape of the nodal block in step 1, determine the symmetry group formed by all the symmetry transformations of the nodal block; use the projection operator of the symmetry group to project the basis functions in step 1, thereby constructing an irreducible symmetry The basis function of the transformation;
投影算符由对称算符组成projection operator composed of symmetric operators
式中In the formula
n——群的阶数,即有限群中的群元个数; n - the order of the group, that is, the number of group elements in the finite group;
sα——第α个不可约表示矩阵的维数;s α ——the dimension of the αth irreducible representation matrix;
R——群元素;R - group element;
PR——与R群元对应的对称性算符;P R ——the symmetry operator corresponding to the R group element;
G——群的名称;G - the name of the group;
——群元素R对应的第α个不可约表示矩阵的j行第j列元素的复数共轭; ——the complex conjugate of the α-th irreducible representation matrix element R corresponding to the j-th row and j-th column element;
使用投影算符对基函数进行投影即获得基函数在不同的对称性空间下的不可约分量;若边界面基函数组成基函数向量使用投影算符对函数向量进行投影可得:Use the projection operator to project the basis functions to obtain the irreducible components of the basis functions in different symmetry spaces; if the basis functions of the boundary surface form the basis function vector use projection operator Projecting the function vector gives:
投影算符由对称算符组成,对称算符作用于基函数向量等同于某一矩阵UR乘以函数向量,即:Projection operators consist of symmetric operators that operate on vectors of basis functions It is equivalent to multiplying a certain matrix U R by a function vector, namely:
从而可得:Thus available:
其中矩阵 where matrix
函数向量的函数是线性相关的,这时需要对矩阵进行高斯消去,从而得到新的满秩矩阵其大小为 为矩阵的秩;function vector The functions of are linearly related, and the matrix needs to be Perform Gaussian elimination to obtain a new full-rank matrix its size is for the matrix rank;
得到矩阵之后从而获得第α个不可约表示的第j列的基函数:get the matrix Then the basis function of the jth column of the αth irreducible representation is obtained:
重复上述方法,利用其他的投影算符,最终求得所有的不可约对称化的基函数;Repeat the above method and use other projection operators to finally obtain all irreducible symmetrization basis functions;
步骤三,将不可约对称化的基函数替换步骤一中的球谐函数以及空间正交多项式,Step 3, replace the spherical harmonic functions and spatial orthogonal polynomials in step 1 with irreducibly symmetrized basis functions,
其中K(r,Ω)与H(r,Ω)为不可约对称化的基函数;采用新的基函数对变量进行离散,可以得到新的响应关系:Among them, K(r, Ω) and H(r, Ω) are the basis functions of irreducible symmetry; using the new basis function to discretize the variables, a new response relationship can be obtained:
新的响应矩阵为对角分块的矩阵。new response matrix is a diagonally block matrix.
与现有技术相比,本发明有如下突出优点:Compared with the prior art, the present invention has the following outstanding advantages:
1.利用节块的对称性可以将复杂的输运问题分解为一系列解耦的子问题,从而简化问题的求解。1. Using the symmetry of the nodal block, the complex transport problem can be decomposed into a series of decoupled sub-problems, thus simplifying the solution of the problem.
2.将基函数进行对称化处理后,即可得到对角分块的响应矩阵,从而大大的减少响应矩阵的计算时间。2. After symmetrizing the basis function, the response matrix of diagonal blocks can be obtained, thereby greatly reducing the calculation time of the response matrix.
3.大幅度的减少了响应矩阵的非零元素的个数,从而较大程度的降低了响应矩阵所需的计算内存。3. The number of non-zero elements of the response matrix is greatly reduced, thereby greatly reducing the computational memory required by the response matrix.
附图说明Description of drawings
图1六角形节块对称性示意图。Figure 1 Schematic diagram of the symmetry of a hexagonal segment.
图2响应矩阵非零元素位置示意图。Figure 2 Response Matrix Schematic diagram of the location of non-zero elements.
图3采用不可约对称化的基函数后的响应矩阵非零元素位置示意图。Figure 3 The response matrix after using irreducibly symmetric basis functions Schematic diagram of the location of non-zero elements.
具体实施方式detailed description
下面结合附图和具体实施方式对本发明作进一步详细说明:Below in conjunction with accompanying drawing and specific embodiment the present invention is described in further detail:
本发明方法将基于对称群理论,采用不可约对称化的基函数,将输运方程的求解简化为若干个互相解耦的子问题,从而将响应矩阵分解为一系列互相解耦的位于原矩阵的对角线位置的子块。在矩阵乘法以及求逆运算中,可以对子块矩阵单独进行乘法及求逆运算,从而大大降低矩阵的浮点运算量。此外,由于矩阵进行了分块对角化处理,与原始矩阵相比,新矩阵的非零元素大大减小,存储响应矩阵所需的计算内存也大大降低。The method of the present invention will be based on the theory of symmetry groups, and adopt irreducible symmetrization basis functions to simplify the solution of the transport equation into several mutually decoupled sub-problems, thereby decomposing the response matrix into a series of mutually decoupled sub-problems located in the original matrix The subblocks at the diagonal positions of . In the matrix multiplication and inversion operations, the multiplication and inversion operations can be performed on the sub-block matrix independently, thereby greatly reducing the amount of floating-point operations of the matrix. In addition, due to the block-diagonalization of the matrix, the new matrix has significantly fewer non-zero elements compared to the original matrix, and the computational memory required to store the response matrix is also greatly reduced.
整个计算具体包括以下步骤:The whole calculation specifically includes the following steps:
步骤一,基于变分节块方法,对中子输运方程建立泛函,构造输运方程的弱解形式,奇偶通量的角度项与空间项分别采用球谐函数以及正交多项式进行离散。Step 1. Based on the variational nodal method, a functional is established for the neutron transport equation, and the weak solution form of the transport equation is constructed. The angle and space terms of the odd and even fluxes are discretized by spherical harmonics and orthogonal polynomials, respectively.
单能的二阶偶对称形式的中子输运方程可写成如下形式:The neutron transport equation in the unienergetic second-order even symmetric form can be written as follows:
r——空间位置变量;r - space position variable;
Ω——角度变量;Ω——angle variable;
ψg——第g能群中子角通量密度;ψ g — neutron angular flux density of energy group g;
Φg——中子标通量密度;Φ g — neutron standard flux density;
g——能群标识;g—energy group identification;
Σt,g——第g能群的总截面;Σ t,g ——the total cross-section of the gth energy group;
——第g能群的自散射截面; ——the self-scattering cross section of the gth energy group;
Sg——第g能群的中子输运方程源项,包含群间散射源项以及裂变源项,S g ——the source term of the neutron transport equation of the gth energy group, including the intergroup scattering source term and the fission source term,
这里假设散射为各项同性的,第g群的奇偶通量定义如下Here it is assumed that the scattering is isotropic, and the odd and even fluxes of the gth group are defined as follows
ψg(r,Ω)——第g群的偶通量密度;ψ g (r,Ω)——the even flux density of the gth group;
χg(r,Ω)——第g群的奇通量密度;χ g (r,Ω)——the odd flux density of the gth group;
令make
Jγ(r,Ω)=nγ·Ωχγ(r,Ω) (3)J γ (r,Ω)=n γ ·Ωχ γ (r,Ω) (3)
nγ——为节块γ边界上的外法线方向;n γ —— is the outer normal direction on the boundary of the node γ;
χγ——为节块γ边界上的奇通量;χ γ —— is the odd flux on the boundary of the node γ;
对所求解的区域建立全局泛函,全局泛函为各个节块局部泛函的加和形式。A global functional is established for the area to be solved, and the global functional is the sum of the local functionals of each node.
局部泛函为The local functional is
其中I[ψ,J]为边界积分项,对于内部边界where I[ψ,J] is the boundary integral term, for the internal boundary
对于真空边界For the vacuum boundary
上式中Γi为内部边界,Γj为真空边界。In the above formula, Γ i is the internal boundary, and Γ j is the vacuum boundary.
于是,求解中子输运方程的问题则转化为了求解全局泛函极小值的变分问题,而变分问题的解即为二阶奇偶方程的弱形式解。Therefore, the problem of solving the neutron transport equation is transformed into the variational problem of solving the global functional minimum, and the solution of the variational problem is the weak form solution of the second-order odd-even equation.
(5)式中J(r,Ω)为拉格朗日乘子项,其作用是消除边界条件带来的约束,通过拉格朗日乘子项,泛函中加入了边界条件,从而实现了节块间的耦合。In the formula (5), J(r, Ω) is the Lagrange multiplier term, its function is to eliminate the constraints brought by the boundary conditions, through the Lagrange multiplier term, the boundary conditions are added to the functional, so as to realize coupling between nodes.
将偶通量与拉格朗日乘子项的角度项展开成球谐函数的形式,空间项展开成正交基函数的形式:Expand the even flux and the angle term of the Lagrange multiplier term into the form of spherical harmonic functions, and the space term into the form of orthogonal basis functions:
其中g(Ω)=[Y0,0,Y2,-2,Y2,-1,Y2,0,Y2,1,Y2,2...]T为偶阶的归一化的球谐函数;f(r)=[f1,f2,f3...]T为定义在节块v内部的正交多项式,h(r)=[h1,h2...]T为定义在节块v边界处的正交多项式;表示位于1号边界面处的边界基函数,表示位于2号边界面处的边界基函数;以及ξ为分别为偶中子角通量密度展开矩以及中子流密度展开矩;T为转置符号。Where g(Ω)=[Y 0,0 ,Y 2,-2 ,Y 2,-1 ,Y 2,0 ,Y 2,1 ,Y 2,2 ...] T is the normalization of even order spherical harmonic function of ; f(r)=[f 1 ,f 2 ,f 3 ...] T is an orthogonal polynomial defined inside the nodal block v, h(r)=[h 1 ,h 2 ... ] T is an orthogonal polynomial defined at the boundary of the node block v; Indicates the boundary basis function located at the No. 1 boundary surface, Indicates the boundary basis function located at the No. 2 boundary surface; and ξ are the expansion moments of the even neutron angular flux density and the expansion moments of the neutron current density respectively; T is the transpose symbol.
把(8)式带入到泛函(5)中即可得到传统变分节块方法的响应矩阵。Bring (8) into the functional (5) to get the response matrix of the traditional variational block method.
其中in
M=[M1 M2 ... Mγ ...] (10)M=[M 1 M 2 ... M γ ...] (10)
泛函关于通量展开矩的导数为零,可得偶通量展开矩与拉格朗日乘子项的关系:The derivative of the functional with respect to the flux expansion moment is zero, and the relationship between the even flux expansion moment and the Lagrangian multiplier term can be obtained:
定义γ面的分流展开矩:Define the split flow expansion moment of the γ surface:
其中 in
在求解展开矩的策略上采用四色棋盘扫描以及响应矩阵的计算方法,响应矩阵公式如下:In the strategy of solving the expansion moments, the four-color checkerboard scanning and the calculation method of the response matrix are adopted. The formula of the response matrix is as follows:
j+=Bs+Rj- (16)j + =Bs+Rj - (16)
其中in
B=[G+I]-1CT (18)B=[G+I] -1 C T (18)
R=[G+I]-1[G-I] (19)R=[G+I] -1 [GI] (19)
H=A-1 (22)H=A -1 (22)
步骤二,根据节块的几何形状,确定节块所有的对称性变换构成的对称群。二维六角形的所有对称性变换共有12个,如图1所示。这些对称性变换包括6个旋转变换算符他们分别代表逆时针旋转0度、60度、120度、180度、240度、300度。其中也可记作单位算符E。6个反射变换算符为{σv1,σv2,σv3,σd1,σd2,σd3},他们为关于各自对称轴的反射变换。所有的对称算符构成了C6v群。对于六棱柱节块还需要考虑关于xoy平面的对称性,因此共有24个对称算符,构成群D6h。Step 2: Determine the symmetry group formed by all the symmetry transformations of the nodal block according to the geometric shape of the nodal block. There are 12 total symmetry transformations of a two-dimensional hexagon, as shown in Figure 1. These symmetry transformations include 6 rotation transformation operators They represent counterclockwise rotation of 0 degrees, 60 degrees, 120 degrees, 180 degrees, 240 degrees, and 300 degrees, respectively. in It can also be recorded as the unit operator E. The six reflection transformation operators are {σ v1 ,σ v2 ,σ v3 ,σ d1 ,σ d2 ,σ d3 }, which are reflection transformations about their respective symmetry axes. All symmetric operators form the C 6v group. For the hexagonal prism block, it is also necessary to consider the symmetry about the xoy plane, so there are 24 symmetry operators in total, Constitute group D 6h .
利用对称群的投影算符对基函数进行投影,从而构造出不可约对称化的基函数。The basis functions are projected by using the projection operator of the symmetry group, so as to construct the irreducible symmetrization basis functions.
不妨假设所有的对称性变换构成群G,群G有若干个投影算符,其中为第α个不可约表示矩阵的第j列的投影算符。It may be assumed that all symmetric transformations form a group G, and the group G has several projection operators, among which is the projection operator for the jth column of the αth irreducible representation matrix.
投影算符由对称性算符组成projection operator Consists of symmetry operators
式中In the formula
n——群的阶数,即有限群中的群元个数;n——the order of the group, that is, the number of group elements in the finite group;
sα——第α个不可约表示矩阵的维数;s α ——the dimension of the αth irreducible representation matrix;
R——群元素;R - group element;
PR——与R群元对应的对称性算符;P R ——the symmetry operator corresponding to the R group element;
G——群的名称;G - the name of the group;
——群元素R对应的第α个不可约表示矩阵的j行第j列元素的复数共轭; ——the complex conjugate of the α-th irreducible representation matrix element R corresponding to the j-th row and j-th column element;
以二维六角形节块为例,根据群论可知,群C6v共有8个投影算符,他们分别为Taking the two-dimensional hexagonal block as an example, according to the group theory, there are 8 projection operators in the group C 6v , and they are respectively
其中in
——不可约表示矩阵A1的投影算符; ——the projection operator of the irreducible representation matrix A 1 ;
——不可约表示矩阵A2的投影算符; ——the projection operator of the irreducible representation matrix A 2 ;
——不可约表示矩阵B1的投影算符; ——the projection operator of the irreducible representation matrix B1;
——不可约表示矩阵B2的投影算符; ——the projection operator of the irreducible representation matrix B2 ;
——不可约表示矩阵E1的第一列投影算符; - the projection operator of the first column of the irreducible representation matrix E1;
——不可约表示矩阵E1的第二列投影算符; - the projection operator of the second column of the irreducible representation matrix E1;
——不可约表示矩阵E2的第一列投影算符; - the projection operator of the first column of the irreducible representation matrix E2 ;
——不可约表示矩阵E2的第二列投影算符; - the projection operator of the second column of the irreducible representation matrix E2;
公式右端的算符为步骤二中的的对称性变换算符。The operator at the right end of the formula is the symmetry transformation operator in step 2.
使用投影算符对基函数进行投影即可获得基函数在不同的对称性空间下的不可约分量。若边界面基函数组成基函数向量不妨设使用对函数向量进行投影可得:The irreducible components of the basis functions in different symmetry spaces can be obtained by using the projection operator to project the basis functions. If the basis functions of the boundary surface form the basis function vector May wish to set use Projecting the function vector gives:
投影算符由对称算符组成,而某一对称算符作用于基函数向量可以得到一组新的基函数。若基函数空间对于对称算符是封闭的,那么得到的新的基函数仍然属于原来的函数空间。因此对称算符作用于基函数向量等同于某一矩阵UR乘以函数向量,即:Projection operators consist of symmetric operators, and a symmetric operator acts on the basis function vector A new set of basis functions can be obtained. If the basis function space is closed to the symmetric operators, then the new basis functions obtained still belong to the original function space. So the symmetric operator acts on the basis function vector It is equivalent to multiplying a certain matrix U R by a function vector, namely:
其中矩阵UR可以通过下式获得where the matrix U R can be obtained by
根据公式(32)以及公式(34)可得According to formula (32) and formula (34), we can get
其中矩阵 where matrix
函数向量的长度与原来基函数向量的长度式相等的。而中的函数仅仅式的第α个不可约表示的第j列的分量,因此可以看出的函数是线性相关的。这时需要对矩阵进行高斯消去,从而得到新的满秩矩阵其大小为 为矩阵的秩。若对进行所有不可约表示空间以及不同的列进行求和可得:function vector The length of the original basis function vector are equal in length. and The function in is only of the form The component of the jth column of the αth irreducible representation of , so it can be seen that functions are linearly related. At this time, the matrix needs to be Perform Gaussian elimination to obtain a new full-rank matrix its size is for the matrix rank. If yes Summing over all irreducible representation spaces and distinct columns gives:
得到矩阵之后从而可以获得第α个不可约表示的第j列的基函数:get the matrix Then the basis function of the jth column of the αth irreducible representation can be obtained:
重复上述方法,利用其他的投影算符,最终求得所有的不可约基函数。Repeat the above method and use other projection operators to finally obtain all irreducible basis functions.
对于二维六角形节块,可根据公式(24)至公式(31)中的投影算符构造出C6v群的不可约基函数。而对于三维六棱柱节块,由于因此可以先生成C6v的不可约基函数,再使用C1h的投影算符对C6v的不可约基函数进一步投影,最终可获得群D6h的不可约对称化的基函数。For two-dimensional hexagonal blocks, the irreducible basis functions of the C 6v group can be constructed according to the projection operators in formula (24) to formula (31). And for the three-dimensional hexagonal prism block, due to Therefore, the irreducible basis functions of C 6v can be generated first, and then the projection operator of C 1h can be used to further project the irreducible basis functions of C 6v , and finally the irreducible symmetrization basis functions of group D 6h can be obtained.
步骤三,将不可约对称化的基函数替换公式(8)中的球谐函数以及正交多项式:Step 3, replace the spherical harmonics and orthogonal polynomials in formula (8) with irreducible symmetrized basis functions:
不可约基函数与原基函数的关系如下所示:The relationship between the irreducible basis function and the original basis function is as follows:
若采用不可约对称化的基函数,则可以得到新的响应关系式:If the basis functions of irreducible symmetry are adopted, a new response relation can be obtained:
由于采用了不可约对称化的基函数,利用了节块的对称性,新的响应矩阵为对角分块矩阵,如图3所示。Due to the use of irreducibly symmetric basis functions and the use of the symmetry of nodal blocks, the new response matrix is a diagonal block matrix, as shown in Figure 3.
以三维六角形节块的响应矩阵G(公式(20))为例,若采用原基函数,矩阵如图2所示。该计算采用了P3的角度近似,边界出的空间多项式采用2阶展开,体积内部的多项式采用6阶展开。图2中矩阵G共有11504个非零元素。而若采用不可约基函数,可以得到新的矩阵这两个矩阵的关系为:Taking the response matrix G (formula (20)) of the three-dimensional hexagonal block as an example, if the original basis function is used, the matrix is shown in Fig. 2 . The calculation adopts the angle approximation of P3, the space polynomial out of the boundary adopts the 2nd order expansion, and the polynomial inside the volume adopts the 6th order expansion. The matrix G in Figure 2 has a total of 11504 non-zero elements. However, if irreducible basis functions are used, a new matrix can be obtained The relationship between these two matrices is:
矩阵如图3所示。从图中可以看出,矩阵具有16个位于对角线位置的子块。在矩阵进行求逆以及乘法运算时,可以对子块单独进行求逆乘法运算,这将大大节省矩阵计算时间。同时矩阵的非零元素也大大减小,从11504减少至1454,因此,较大程度的减少了计算所需的内存。计算TAKETA问题采用不同的角度展开阶数的总时间的比较如表1所示。对于P7的角度展开,计算加速比可达到15倍。matrix As shown in Figure 3. It can be seen from the figure that the matrix There are 16 sub-blocks located diagonally. When matrix inversion and multiplication are performed, sub-blocks can be independently inverted and multiplied, which will greatly save matrix calculation time. At the same time, the non-zero elements of the matrix are also greatly reduced, from 11504 to 1454, so the memory required for calculation is greatly reduced. Table 1 shows the comparison of the total time for calculating the TAKETA problem using different angle expansion orders. For the angle unwrapping of P7, the calculation speedup can reach 15 times.
表1不同角度展开阶数问题的计算总时间比较Table 1 Comparison of the total calculation time of the expansion order problem from different angles
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