matlab语言编程w0=1u0=0.4u=[0.7 0.4 0.1]v0=0.2v=[0.1 0.3 0.2]i=3j

matlab语言编程w0=1u0=0.4u=[0.7 0.4 0.1]v0=0.2v=[0.1 0.3 0.2]i=3j=mod(i,3)+1syms ux vyR=sqrt(((u0-ux)^2)+((v0-vy)^2)+(w0)^2)f=(3*((u0-ux)^2)/(R^5))-1/(R^3)I=int(int(f,vy,((v(3)-v(2))/(u(3)-u(2)))*ux+v(2)-((v(3)-v(2))/(u(3)-u(2)))*u(2),((v(1)-v(2))/(u(1)-u(2)))*ux+v(1)-((v(1)-v(2))/(u(1)-u(2)))*u(2)),ux,u(2),u(1))+int(int(f,vy,((v(3)-v(2))/(u(3)-u(2)))*ux+v(2)-((v(3)-v(2))/(u(3)-u(2)))*u(2),((v(1)-v(3))/(u(1)-u(3)))*ux+v(1)-((v(1)-v(3))/(u(1)-u(3)))*u(1)),ux,u(1),u(3))w0=0u0=0.4u=[0.7 0.4 0.1]v0=0.2v=[0.1 0.3 0.2]i=3j=mod(i,3)+1a(i)=(u(i)-u0)*(v(j)-v(i))-(v(i)-v0)*(u(j)-u(i))b(i)=(u(j)-u(i))*(u(i)-u0)+(v(j)-v(i))*(v(i)-v0)c(i)=(u(j)-u(i))*(u(j)-u0)+(v(j)-v(i))*(v(j)-v0)l(i)=sqrt((u(j)-u(i))^2+(v(j)-v(i))^2)r(i)=sqrt((u0-u(i))^2+(v0-v(i))^2+w0^2)R(j)=sqrt((u0-u(j))^2+(v0-v(j))^2+w0^2)%g(j)=sqrt((l(i))^2+2*((b(i))^2)+(r(i))^2)s1(i)=(u(j)-u(i))*(v(j)-v(i))*(r(i)-R(j))/(r(i)*R(j)*((l(i))^2))s2(i)=a(i)*(c(i)*r(i)-b(i)*R(j))/(r(i)*R(j)*(((a(i))^2)+((l(i))^2)*(w0^2)))t1(i)=(-(u(i)-u0)*a(i)-(v(j)-v(i))*w0^2)/((u(j)-u(i))*r(i)*w0)t2(i)=(-(u(j)-u0)*a(i)-(v(j)-v(i))*w0^2)/((u(j)-u(i))*R(j)*w0)t3(i)=(-(v(i)-v0)*a(i)+(u(j)-u(i))*w0^2)/((v(j)-v(i))*r(i)*w0)t4(i)=(-(v(j)-v0)*a(i)+(u(j)-u(i))*w0^2)/((v(j)-v(i))*R(j)*w0)d(i)=sqrt((u0-u(i))^2+(v0-v(i))^2)h1(i)=((u(j)-u(i))*(v(j)-v(i))*(d(i)-d(j)))/(d(i)*d(j)*(l(i))^2)h2(i)=(c(i)*r(i)-b(i)*r(j))/(a(i)*d(i)*d(j))%f1(i)=(1/w0)*(atan(t1(i))-atan(t2(i)))+(2/(3*w0))*(-atan(t3(i))+atan(t4(i)))+(1/3)*(s1(i)-(s2(i)/l(i))*((v(j)-v(i))^2))%f_1=f1(1)+f1(2)+f1(3)%f9(i)=(1/(2*w0))*(atan(t1(i))-atan(t2(i))+atan(t3(i))-atan(t4(i)))%f_9=f9(1)+f9(2)+f9(3)IL1(i)=h1(i)-(h2(i)/l(i))*((v(j)-v(i))^2)I_L1=IL1(1)+IL1(2)+IL1(3)%I_1=3*f_1-f_9在这上面还要再添加上上面东西可以实现这个功能 就是对这个程序还要怎么对其进行处理麻烦知道的速度解决下还好.这是两种积分方法的比较问题