一维有限元法.doc

_实习三、一维问题的有限元方法一）实习问题：设 将原问题的边界条件齐次化二）算法描述：1，单元剖分2，i=1 3,计算数值积分：即得单元上的4，将迭加到总的中5，若il(j) l2=l(i); l(i)=l(j); l(j)=l2; end endend for i=1:n-1 x(i+1)=l(i); end x(1)=0;x(n+1)=1; for i=1:n h(i)=x(i+1)-x(i); end %一次区间元法 %A的求解 for i=1:n a(i,1)=1/h(i)+h(i)/3; a(i,2)=-1/h(i)+h(i)/6; a(i,3)=a(i,2); a(i,4)=a(i,1); end for i=1:n-1 A(i,i)=a(i,4)+a(i+1,1); end for i=1:n-2 A(i,i+1)=a(i+1,2); end for i=2:n-1 A(i,i-1)=a(i,3); end %b的求解 for i=2:n+1 b1(i,i-1)=f1(x(i-1),h(i-1); b1(i,i)=f2(x(i-1),h(i-1); end for i=2:n b2(i)=b1(i,i)+b1(i+1,i); end for i=1:n-1 b(i)=b2(i+1); end u=inv(A)*b ; for i=2:n un(i)=u(i-1); endun(1)=0;un(n+1)=0; %还原原始的u值uz=un+x*(exp(1)-exp(-1);uz%真解的求解for i=1:n+1 u1(i)=exp(x(i)-exp(-x(i)+x(i)*x(i)*exp(x(i)-x(i)*exp(x(i);endu1%误差的计算for i=1:n+1 e(i)=abs(uz(i)-u1(i)/u1(i)*100);ende%作图subplot(1,2,1)plot(x,u1)xlabel(自变量x的范围);ylabel(函数值u的取值);title(真解的图象);gridsubplot(1,2,2)plot(x,uz)xlabel(自变量x的范围);ylabel(函数值u的取值);title(有限元法算得的近似解的图象);grid四）图形显示的计算结果：将区间随机分（利用rand函数）为20份的计算结果： 图1：预测值与真实值的作图比较表1：预测值与真实值的数值比较真实值预测值误差（%）000.00000.01350.01350.07260.03360.03360.07250.18160.18150.06570.34920.34900.05150.41170.41150.04650.54600.54580.03770.66460.66440.03160.73150.73130.02870.82680.82660.02540.87530.87510.02400.87820.87800.02391.09411.09390.01811.32561.32540.01281.35511.35490.01221.52691.52680.00921.82531.82520.00501.99631.99630.00302.02042.02030.00272.15292.15290.00152.35042.35040.0000分析：从图1与表1中可以看出，预测值与真实值极其的接近，其误差最大值是0.0