偏微分方程的有限元法求解

1、 %對(duì)于d2u/dx2=f的FEM解算器，其中f=x*(1-x)%邊界條件u(0)=0, u(1)=0.%精確解用以比對(duì)xx=linspace(0,1,101);%產(chǎn)生0-1之間的均分指令，101為元素個(gè)數(shù)uex=(1/6).*xx.3-(1/12).*xx.4-(1/12).*xx;%對(duì)力項(xiàng)設(shè)置高斯點(diǎn)的數(shù)目NGf=2;if (NGf=2)xiGf=-1/sqrt(3);1/sqrt(3);%1、2的值aGf=1 1;else,NGf=1;xiGf=0.0;aGf=2.0;end%單元數(shù)目Ne=5;%建立網(wǎng)格節(jié)點(diǎn)x=linspace(0,1,Ne+1);%零剛性矩陣K=zeros(Ne+1,N

2、e+1);b=zeros(Ne+1,1);%對(duì)所有單元循環(huán)計(jì)算剛性和殘差for ii=1:Ne,kn1=ii;kn2=ii+1;x1=x(kn1);x2=x(kn2);dx=x2-x1;%每一個(gè)單元的長度dxidx=2/dx;%d/dxdxdxi=1/dxidx;%dx/ddN1dxi=-1/2;%d1/ddN2dxi=1/2;%d2/ddN1dx=dN1dxi*dxidx;%-1/(xj-xj-1)dN2dx=dN2dxi*dxidx;%1/(xj-xj-1)K(kn1,kn1)=K(kn1,kn1)-2*dN1dx*dN1dx*dxdxi;%Rj的第二項(xiàng)K(kn1,kn2)=K(kn1,k

3、n2)-2*dN1dx*dN2dx*dxdxi;K(kn2,kn1)=K(kn2,kn1)-2*dN2dx*dN1dx*dxdxi;K(kn2,kn2)=K(kn2,kn2)-2*dN2dx*dN2dx*dxdxi;%用高斯積分估計(jì)力項(xiàng)的積分for nn=1:NGf%NGf=2xiG=xiGf(nn);%得到高斯點(diǎn)的N1=0.5*(1-xiG);%求N1和N2(即在xiG的權(quán)重/插值) 形狀函數(shù)在的值N2=0.5*(1+xiG);%值fG=xiG*(1-xiG);%對(duì)點(diǎn)求fgG1=N1*fG*dxdxi;%在節(jié)點(diǎn)處估計(jì)權(quán)函數(shù)在高斯點(diǎn)的被積函數(shù)gG2=N2*fG*dxdxi;%估計(jì)是個(gè)積分值b(kn1)=b(kn1)+aGf(nn)*gG1;% aGf為1b(kn2)=b(kn1)+aGf(nn)*gG2;endend%在x=0處設(shè)置Dirichlet條件kn1=1;K(kn1,:)=zeros(size(1,Ne+1);K(kn1,kn1)=1;b(kn1)=0;%在x=1處設(shè)置Dirichlet條件kn1=1;K(kn1,:)=zeros(size(1,Ne+1);K(kn1,kn1)=1;b(kn1)=0;%求解方程v=Kb;%v為Kx=b的解plot(x,v,