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注册日期: 2019-11-21
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我不知道这个问题与数学或编程是否相关,我在Matlab中绝对是新手。程序FEM_50在Omega中将有限元方法应用于拉普拉斯方程-Uxx(x,y)-Uyy(x,y)= F(x,y)。如何更改以将FEM应用于方程-Uxx(x,y)-Uyy(x,y)+ U(x,y)= F(x,y)?在此页面上: http : //sc.fsu.edu/~burkardt/m_src/fem_50/fem_50.html其他代码文件(如果需要)。
function fem_50 ( ) %% FEM_50 applies the finite element method to Laplace's equation. % % Discussion: % % FEM_50 is a set of MATLAB routines to apply the finite % element method to solving Laplace's equation in an arbitrary % region, using about 50 lines of MATLAB code. % % FEM_50 is partly a demonstration, to show how little it % takes to implement the finite element method (at least using % every possible MATLAB shortcut.) The user supplies datafiles % that specify the geometry of the region and its arrangement % into triangular and quadrilateral elements, and the location % and type of the boundary conditions, which can be any mixture % of Neumann and Dirichlet. % % The unknown state variable U(x,y) is assumed to satisfy % Laplace's equation: % -Uxx(x,y) - Uyy(x,y) = F(x,y) in Omega % with Dirichlet boundary conditions % U(x,y) = U_D(x,y) on Gamma_D % and Neumann boundary conditions on the outward normal derivative: % Un(x,y) = G(x,y) on Gamma_N % If Gamma designates the boundary of the region Omega, % then we presume that % Gamma = Gamma_D + Gamma_N % but the user is free to determine which boundary conditions to % apply. Note, however, that the problem will generally be singular % unless at least one Dirichlet boundary condition is specified. % % The code uses piecewise linear basis functions for triangular elements, % and piecewise isoparametric bilinear basis functions for quadrilateral % elements. % % The user is required to supply a number of data files and MATLAB % functions that specify the location of nodes, the grouping of nodes % into elements, the location and value of boundary conditions, and % the right hand side function in Laplace's equation. Note that the % fact that the geometry is completely up to the user means that % just about any two dimensional region can be handled, with arbitrary % shape, including holes and islands. % clear % % Read the nodal coordinate data file. % load coordinates.dat; % % Read the triangular element data file. % load elements3.dat; % % Read the quadrilateral element data file. % load elements4.dat; % % Read the Neumann boundary condition data file. % I THINK the purpose of the EVAL command is to create an empty NEUMANN array % if no Neumann file is found. % eval ( 'load neumann.dat;', 'neumann=[];' ); % % Read the Dirichlet boundary condition data file. % load dirichlet.dat; A = sparse ( size(coordinates,1), size(coordinates,1) ); b = sparse ( size(coordinates,1), 1 ); % % Assembly. % for j = 1 : size(elements3,1) A(elements3(j,:),elements3(j,:)) = A(elements3(j,:),elements3(j,:)) ... + stima3(coordinates(elements3(j,:),:)); end for j = 1 : size(elements4,1) A(elements4(j,:),elements4(j,:)) = A(elements4(j,:),elements4(j,:)) ... + stima4(coordinates(elements4(j,:),:)); end % % Volume Forces. % for j = 1 : size(elements3,1) b(elements3(j,:)) = b(elements3(j,:)) ... + det( [1,1,1; coordinates(elements3(j,:),:)'] ) * ... f(sum(coordinates(elements3(j,:),:))/3)/6; end for j = 1 : size(elements4,1) b(elements4(j,:)) = b(elements4(j,:)) ... + det([1,1,1; coordinates(elements4(j,1:3),:)'] ) * ... f(sum(coordinates(elements4(j,:),:))/4)/4; end % % Neumann conditions. % if ( ~isempty(neumann) ) for j = 1 : size(neumann,1) b(neumann(j,:)) = b(neumann(j,:)) + ... norm(coordinates(neumann(j,1),:) - coordinates(neumann(j,2),:)) * ... g(sum(coordinates(neumann(j,:),:))/2)/2; end end % % Determine which nodes are associated with Dirichlet conditions. % Assign the corresponding entries of U, and adjust the right hand side. % u = sparse ( size(coordinates,1), 1 ); BoundNodes = unique ( dirichlet ); u(BoundNodes) = u_d ( coordinates(BoundNodes,:) ); b = b - A * u; % % Compute the solution by solving A * U = B for the remaining unknown values of U. % FreeNodes = setdiff ( 1:size(coordinates,1), BoundNodes ); u(FreeNodes) = A(FreeNodes,FreeNodes) \ b(FreeNodes); % % Graphic representation. % show ( elements3, elements4, coordinates, full ( u ) ); return end 回答: 我不会做你的作业,但是这里有一些提示:
然后,重复调用此函数,在此函数中您可以将体力修改为: New_body_Force = Original_body_force - Solution_from_previous_call 这应该收敛到期望的结果。 第二个选项不如首先介绍的“干净”方式那么优雅,但我只是喜欢定点迭代。 希望这可以帮助。 更多&回答... |
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