baverloll 发表于 2011-8-30 16:05

请教一下matlab解非线性方程组的问题

本帖最后由 baverloll 于 2011-8-30 16:07 编辑

我在学习振动力学的时候有个问题，描述如下：
非线性方程组是其实是由两个非线性方程加两个不等式限制所组成的。
方程一：
V=sqrt((A1*Wn/sqrt(1-At^2)*exp(-At*Wn*(t-t1))*cos(Wn*sqrt(1-At^2)*t1)+A2*Wn/sqrt(1-At^2)*exp(-At*Wn*(t-t2))*cos(Wn*sqrt(1-At^2)*t2)+A3*Wn/sqrt(1-At^2)*exp(-At*Wn*(t-t3))*cos(Wn*sqrt(1-At^2)*t3)+A4*Wn/sqrt(1-At^2)*exp(-At*Wn*(t-t4))*cos(Wn*sqrt(1-At^2)*t4)+A5*Wn/sqrt(1-At^2)*exp(-At*Wn*(t-t5))*cos(Wn*sqrt(1-At^2)*t5))^2+(A1*Wn/sqrt(1-At^2)*exp(-At*Wn*(t-t1))*sin(Wn*sqrt(1-At^2)*t1)+A2*Wn/sqrt(1-At^2)*exp(-At*Wn*(t-t2))*sin(Wn*sqrt(1-At^2)*t2)+A3*Wn/sqrt(1-At^2)*exp(-At*Wn*(t-t3))*sin(Wn*sqrt(1-At^2)*t3)+A4*Wn/sqrt(1-At^2)*exp(-At*Wn*(t-t4))*sin(Wn*sqrt(1-At^2)*t4)+A5*Wn/sqrt(1-At^2)*exp(-At*Wn*(t-t5))*sin(Wn*sqrt(1-At^2)*t5))^2)=5;
方程二（整个方程是上个方程关于Wn的偏导数）：
1/2/((A1*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t1))*cos(Wn*(1-At^2)^(1/2)*t1)+A2*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t2))*cos(Wn*(1-At^2)^(1/2)*t2)+A3*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t3))*cos(Wn*(1-At^2)^(1/2)*t3)+A4*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t4))*cos(Wn*(1-At^2)^(1/2)*t4)+A5*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t5))*cos(Wn*(1-At^2)^(1/2)*t5))^2+(A1*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t1))*sin(Wn*(1-At^2)^(1/2)*t1)+A2*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t2))*sin(Wn*(1-At^2)^(1/2)*t2)+A3*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t3))*sin(Wn*(1-At^2)^(1/2)*t3)+A4*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t4))*sin(Wn*(1-At^2)^(1/2)*t4)+A5*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t5))*sin(Wn*(1-At^2)^(1/2)*t5))^2)^(1/2)*(2*(A1*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t1))*cos(Wn*(1-At^2)^(1/2)*t1)+A2*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t2))*cos(Wn*(1-At^2)^(1/2)*t2)+A3*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t3))*cos(Wn*(1-At^2)^(1/2)*t3)+A4*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t4))*cos(Wn*(1-At^2)^(1/2)*t4)+A5*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t5))*cos(Wn*(1-At^2)^(1/2)*t5))*(A1/(1-At^2)^(1/2)*exp(-At*Wn*(t-t1))*cos(Wn*(1-At^2)^(1/2)*t1)-A1*Wn/(1-At^2)^(1/2)*At*(t-t1)*exp(-At*Wn*(t-t1))*cos(Wn*(1-At^2)^(1/2)*t1)-A1*Wn*exp(-At*Wn*(t-t1))*sin(Wn*(1-At^2)^(1/2)*t1)*t1+A2/(1-At^2)^(1/2)*exp(-At*Wn*(t-t2))*cos(Wn*(1-At^2)^(1/2)*t2)-A2*Wn/(1-At^2)^(1/2)*At*(t-t2)*exp(-At*Wn*(t-t2))*cos(Wn*(1-At^2)^(1/2)*t2)-A2*Wn*exp(-At*Wn*(t-t2))*sin(Wn*(1-At^2)^(1/2)*t2)*t2+A3/(1-At^2)^(1/2)*exp(-At*Wn*(t-t3))*cos(Wn*(1-At^2)^(1/2)*t3)-A3*Wn/(1-At^2)^(1/2)*At*(t-t3)*exp(-At*Wn*(t-t3))*cos(Wn*(1-At^2)^(1/2)*t3)-A3*Wn*exp(-At*Wn*(t-t3))*sin(Wn*(1-At^2)^(1/2)*t3)*t3+A4/(1-At^2)^(1/2)*exp(-At*Wn*(t-t4))*cos(Wn*(1-At^2)^(1/2)*t4)-A4*Wn/(1-At^2)^(1/2)*At*(t-t4)*exp(-At*Wn*(t-t4))*cos(Wn*(1-At^2)^(1/2)*t4)-A4*Wn*exp(-At*Wn*(t-t4))*sin(Wn*(1-At^2)^(1/2)*t4)*t4+A5/(1-At^2)^(1/2)*exp(-At*Wn*(t-t5))*cos(Wn*(1-At^2)^(1/2)*t5)-A5*Wn/(1-At^2)^(1/2)*At*(t-t5)*exp(-At*Wn*(t-t5))*cos(Wn*(1-At^2)^(1/2)*t5)-A5*Wn*exp(-At*Wn*(t-t5))*sin(Wn*(1-At^2)^(1/2)*t5)*t5)+2*(A1*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t1))*sin(Wn*(1-At^2)^(1/2)*t1)+A2*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t2))*sin(Wn*(1-At^2)^(1/2)*t2)+A3*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t3))*sin(Wn*(1-At^2)^(1/2)*t3)+A4*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t4))*sin(Wn*(1-At^2)^(1/2)*t4)+A5*Wn/(1-At^2)^(1/2)*exp(-At*Wn*(t-t5))*sin(Wn*(1-At^2)^(1/2)*t5))*(A1/(1-At^2)^(1/2)*exp(-At*Wn*(t-t1))*sin(Wn*(1-At^2)^(1/2)*t1)-A1*Wn/(1-At^2)^(1/2)*At*(t-t1)*exp(-At*Wn*(t-t1))*sin(Wn*(1-At^2)^(1/2)*t1)+A1*Wn*exp(-At*Wn*(t-t1))*cos(Wn*(1-At^2)^(1/2)*t1)*t1+A2/(1-At^2)^(1/2)*exp(-At*Wn*(t-t2))*sin(Wn*(1-At^2)^(1/2)*t2)-A2*Wn/(1-At^2)^(1/2)*At*(t-t2)*exp(-At*Wn*(t-t2))*sin(Wn*(1-At^2)^(1/2)*t2)+A2*Wn*exp(-At*Wn*(t-t2))*cos(Wn*(1-At^2)^(1/2)*t2)*t2+A3/(1-At^2)^(1/2)*exp(-At*Wn*(t-t3))*sin(Wn*(1-At^2)^(1/2)*t3)-A3*Wn/(1-At^2)^(1/2)*At*(t-t3)*exp(-At*Wn*(t-t3))*sin(Wn*(1-At^2)^(1/2)*t3)+A3*Wn*exp(-At*Wn*(t-t3))*cos(Wn*(1-At^2)^(1/2)*t3)*t3+A4/(1-At^2)^(1/2)*exp(-At*Wn*(t-t4))*sin(Wn*(1-At^2)^(1/2)*t4)-A4*Wn/(1-At^2)^(1/2)*At*(t-t4)*exp(-At*Wn*(t-t4))*sin(Wn*(1-At^2)^(1/2)*t4)+A4*Wn*exp(-At*Wn*(t-t4))*cos(Wn*(1-At^2)^(1/2)*t4)*t4+A5/(1-At^2)^(1/2)*exp(-At*Wn*(t-t5))*sin(Wn*(1-At^2)^(1/2)*t5)-A5*Wn/(1-At^2)^(1/2)*At*(t-t5)*exp(-At*Wn*(t-t5))*sin(Wn*(1-At^2)^(1/2)*t5)+A5*Wn*exp(-At*Wn*(t-t5))*cos(Wn*(1-At^2)^(1/2)*t5)*t5))=0;
另外还有两个不等式：
不等式一：
sqrt((A1*0.85*Wn/sqrt(1-At^2)*exp(-At*0.85*Wn*(t-t1))*cos(0.85*Wn*sqrt(1-At^2)*t1)+A2*0.85*Wn/sqrt(1-At^2)*exp(-At*0.85*Wn*(t-t2))*cos(0.85*Wn*sqrt(1-At^2)*t2)+A3*0.85*Wn/sqrt(1-At^2)*exp(-At*0.85*Wn*(t-t3))*cos(0.85*Wn*sqrt(1-At^2)*t3)+A4*0.85*Wn/sqrt(1-At^2)*exp(-At*0.85*Wn*(t-t4))*cos(0.85*Wn*sqrt(1-At^2)*t4)+A5*0.85*Wn/sqrt(1-At^2)*exp(-At*0.85*Wn*(t-t5))*cos(0.85*Wn*sqrt(1-At^2)*t5))^2+(A1*0.85*Wn/sqrt(1-At^2)*exp(-At*0.85*Wn*(t-t1))*sin(0.85*Wn*sqrt(1-At^2)*t1)+A2*0.85*Wn/sqrt(1-At^2)*exp(-At*0.85*Wn*(t-t2))*sin(0.85*Wn*sqrt(1-At^2)*t2)+A3*0.85*Wn/sqrt(1-At^2)*exp(-At*0.85*Wn*(t-t3))*sin(0.85*Wn*sqrt(1-At^2)*t3)+A4*0.85*Wn/sqrt(1-At^2)*exp(-At*0.85*Wn*(t-t4))*sin(0.85*Wn*sqrt(1-At^2)*t4)+A5*0.85*Wn/sqrt(1-At^2)*exp(-At*0.85*Wn*(t-t5))*sin(0.85*Wn*sqrt(1-At^2)*t5))^2)<0.005;
不等式二：
sqrt((A1*1.15*Wn/sqrt(1-At^2)*exp(-At*1.15*Wn*(t-t1))*cos(1.15*Wn*sqrt(1-At^2)*t1)+A2*1.15*Wn/sqrt(1-At^2)*exp(-At*1.15*Wn*(t-t2))*cos(1.15*Wn*sqrt(1-At^2)*t2)+A3*1.15*Wn/sqrt(1-At^2)*exp(-At*1.15*Wn*(t-t3))*cos(1.15*Wn*sqrt(1-At^2)*t3)+A4*1.15*Wn/sqrt(1-At^2)*exp(-At*1.15*Wn*(t-t4))*cos(1.15*Wn*sqrt(1-At^2)*t4)+A5*1.15*Wn/sqrt(1-At^2)*exp(-At*1.15*Wn*(t-t5))*cos(1.15*Wn*sqrt(1-At^2)*t5))^2+(A1*1.15*Wn/sqrt(1-At^2)*exp(-At*1.15*Wn*(t-t1))*sin(1.15*Wn*sqrt(1-At^2)*t1)+A2*1.15*Wn/sqrt(1-At^2)*exp(-At*1.15*Wn*(t-t2))*sin(1.15*Wn*sqrt(1-At^2)*t2)+A3*1.15*Wn/sqrt(1-At^2)*exp(-At*1.15*Wn*(t-t3))*sin(1.15*Wn*sqrt(1-At^2)*t3)+A4*1.15*Wn/sqrt(1-At^2)*exp(-At*1.15*Wn*(t-t4))*sin(1.15*Wn*sqrt(1-At^2)*t4)+A5*1.15*Wn/sqrt(1-At^2)*exp(-At*1.15*Wn*(t-t5))*sin(1.15*Wn*sqrt(1-At^2)*t5))^2)<0.005;
我的想法就是用solve函数将上面两个等式的所有解带入到下面那两个不等式中，但是前面两个方程用solve函数解不出来，提示是empty sym，还有后面那两个不等式如果改成等式，全部等于0又怎样解？搞了很长时间不知道哪里有问题，请各位指点一下，谢谢了！！！

baverloll 发表于 2011-8-30 19:41

有人有思路或者经验吗？急啊！！！

hjlovehyuk 发表于 2011-9-4 09:32

求解非线性方程组
http://forum.vibunion.com/forum-viewthread-tid-105234-fromuid-172740.html
第一页的一个帖子，我不太懂，不知道能解决不

ChaChing 发表于 2011-9-4 11:27

:@L式子这样复杂, 不知有多少看官会细看!? 个人水平/时间是有限, 所以...
页: [1]
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