计算含有偏微分的二重​积分积不出来答案是为​什么,百度也没找到解​决办法,求大佬帮忙看​看!

b=0.0175;h=0.114;h0=0.01;N=4;r0=0.0325;d1=(h-(N+1)*h0)/N;
syms x y n;
U=x*y+symsum(32*b^2*(-1)^((n+1)/2)*sin(n*pi*x/(2*b))*sinh(n*pi*y/(2*b))/(n^3*pi^3*cosh(n*pi*d1*sqrt(d1^2+pi*(2*(r0+b)+b)^2)/(2*b))),n,1,inf);
A=int(int(x^2+y^2+x*diff(U,y)-y*diff(U,x),x,0,sqrt(d1^2+pi*(2*(r0+b)+b)^2)),y,-h0/2,h0/2);
A=vpa(A,10)
结果是:
A =
numeric::int(numeric::int(x^2 + y^2 + x*(x + (1724034232352768*symsum(((-1)^(n/2 + 1/2)*((exp(-(n*x*pi*200i)/7)*1i)/2 - (exp((n*x*pi*200i)/7)*1i)/2)*((100*n*pi*exp(-(200*n*y*pi)/7))/7 + (100*n*pi*exp((200*n*y*pi)/7))/7))/(n^3*(exp(-(7525588108204847*n*pi)/78812993478983680)/2 + exp((7525588108204847*n*pi)/78812993478983680)/2)), n, 1, Inf))/5454681879044716875) - y*(y - (1724034232352768*symsum(((-1)^(n/2 + 1/2)*(exp(-(200*n*y*pi)/7)/2 - exp((200*n*y*pi)/7)/2)*((100*n*pi*exp(-(n*x*pi*200i)/7))/7 + (100*n*pi*exp((n*x*pi*200i)/7))/7))/(n^3*(exp(-(7525588108204847*n*pi)/78812993478983680)/2 + exp((7525588108204847*n*pi)/78812993478983680)/2)), n, 1, Inf))/5454681879044716875), x == 0..7525588108204847/36028797018963968), y == -1/200..1/200)
>>

 采纳的回答

cpxnwm
cpxnwm 2023-5-22

0 个投票

你一直增加这个数计算,如果增大到某个值,得到的结果变化很小,不就可以认为积分到无穷大也是这个值嘛

更多回答(0 个)

类别

帮助中心File Exchange 中查找有关 Simulink Design Optimization 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!