有限元方法求解Helmholtz方程的边值问题，matlab编程

clc
clear
tic;
% n 是行与列划分的格子数，对整个[0,1]*[0,1]有n^2个划分,can为方程中的参数k
%这里我们用1,2,3按逆时针来表示一个三角形的各个顶点
% s是一个n^2*10的关联矩阵，s(i,1)表示第i个三角形，s(i,2),s(i,3),s(i,4)分别表示第i个三角形的1,2,3所对应的顶点
% s(i,5),s(i,6)；s(i,7)，s(i,8)；s(i,9)，s(i,10)分别表示顶点1,2,3所代表的坐标
%生成关联矩阵s
%A是总刚矩阵
%声明符号变量x y
n=20;
can=20;
s=zeros(2*n^2,10);
h=1/n;
st=1/(2*n^2);
A=zeros((n+1)^2,(n+1)^2);
syms x y;
for  k=1:1:2*n^2
        s(k,1)=k;
        q=fix(k/(2*n));
        r=mod(k,(2*n));
        if (r~=0)
            r=r;
        else r=2*n;q=q-1;
        end
        if (r<=n)
            s(k,2)=q*(n+1)+r;
            s(k,3)=q*(n+1)+r+1;
            s(k,4)=(q+1)*(n+1)+r+1;
            s(k,5)=(r-1)*h;
            s(k,6)=q*h;
            s(k,7)=r*h;
            %%
            % 
            %   for x = 1:10
            %       disp(x)
            %   end
            % 
            s(k,8)=q*h;
            s(k,9)=r*h;
            s(k,10)=(q+1)*h;
        else
            s(k,2)=q*(n+1)+r-n;
            s(k,3)=(q+1)*(n+1)+r-n+1;
            s(k,4)=(q+1)*(n+1)+r-n;
            s(k,5)=(r-n-1)*h;
            s(k,6)=q*h;
            s(k,7)=(r-n)*h;
            s(k,8)=(q+1)*h;
            s(k,9)=(r-n-1)*h;
            s(k,10)=(q+1)*h;
        end
end
%下面生成基函数L(i)表示第i个点顶点的基函数
%生成单刚矩阵d
%生成单刚矩阵并将其加入总纲矩阵
d=zeros(3,3);
B=zeros((n+1)^2,1);
b=zeros(3,1);
%生成A的总刚
for   k=1:1:2*n^2
     L(1)=(1/(2*st))*((s(k,7)*s(k,10)-s(k,9)*s(k,8))+(s(k,8)-s(k,10))*x+(s(k,9)-s(k,7))*y);
     L(2)=(1/(2*st))*((s(k,9)*s(k,6)-s(k,5)*s(k,10))+(s(k,10)-s(k,6))*x+(s(k,5)-s(k,9))*y);
     L(3)=(1/(2*st))*((s(k,5)*s(k,8)-s(k,7)*s(k,6))+(s(k,6)-s(k,8))*x+(s(k,7)-s(k,5))*y);
     for i=1:1:3
        for j=i:3
            d(i,j)=int(int(((((diff(L(i),x))*(diff(L(j),x)))+((diff(L(i),y))*(diff(L(j),y))))-((can^2)*L(i)*L(j))),x,0,1),y,0,1);
            d(j,i)=d(i,j);
        end
     end   
     for i=1:1:3
         for j=1:1:3
             A(s(k,(i+1)),s(k,(j+1)))=A(s(k,(i+1)),s(k,(j+1)))+d(i,j);
         end
     end
     for i=1:1:3
         b(i)=int(int((L(i)),x,0,1),y,0,1);
         B(s(k,(i+1)),1)=B(s(k,(i+1)),1)+b(i);
     end
     
end
%下面根据边界条件来求解有限元方程组Mx=B,齐次边界条件约掉了很多项
M=zeros((n+1)^2,n^2);
j=n^2;
for i=(n^2+n):-1:1
    if ((mod(i,(n+1)))~=1)
        M(:,j)=A(:,i);
        j=j-1;
    else continue
    end
end
%preanswer是未知点的值是（n+1）^2*(n^2)的
preanswer=M\B;
%得到所有节点的值
answer=zeros((n+1)^2,1);
j=1;
for i=1:1:(n^2+n)
    if ((mod(i,(n+1)))~=1)
        answer(i)=preanswer(j);
        j=j+1;
    else answer(i)=0;
    end
end
%%
% 
%   for x = 1:10
% 
%   for x = 1:10
% 
%   for x = 1:10
%       disp(x)
%   end
% 
%       disp(x)
%   end
% 
%       disp(x)
%   end
% 
%生成求解后的图
Z=zeros((n+1),(n+1));
for i=1:1:(n+1)^2
      s=fix(i/(n+1))+1;
      r=mod(i,(n+1));
      if(r==0)
          r=n+1;
          s=s-1;
      else
      end
      Z(r,s)=answer(i);
end
[X,Y]=meshgrid(1:-h:0,0:h:1);
surf(X,Y,Z);



toc;
t=toc;

温馨提示：答案为网友推荐，仅供参考

当前网址：https://verywind.cn/ee/y72f23vry.html