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北京石油化工学院2026年研究生招生接收调剂公告
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ghtyr

铁虫 (小有名气)

[求助] Matlab利用fsolve函数求解非线性方程组,报错,不知道哪里出问题了

错误如下,谢谢大家了。
[sol,fval]=fsolve(@myfun4,[1,1,1,1])
未定义与 'function_handle' 类型的输入参数相对应的运算符 '*'。

出错 myfun4 (line 61)
F(1)=fa2*cos(alpha1*180/pi)+fb2*cos(alpha2*180/pi)-fa11*sin(alpha1*180/pi)-fb1*sin(alpha2*180/pi);

下面是程序:
function F=myfun4(x)
L=2.5;
B=1.92;
alpha2 =10*pi/180;
temp1 = B/(L/2)+1/tan(alpha2);
alpha1 = atan(1/temp1)*180/pi;
alpha2 = alpha2*180/pi;

b = 0.28;
rs = 0.175;
k=0.0175;
M = 1800;
g = 9.8;
Fz1 = M*g/4/1000;
fai = 13/180*pi;
c=4.14;
l=0.7;
p1 = Fz1/b./l;p2 = p1;

D1= x(3)*sin(alpha1*pi/180)+x(4)*cos(alpha1*pi/180);
D2= x(3)*sin(alpha2*pi/180)+x(4)*cos(alpha2*pi/180);

Rs1=(L/2+D1*cos(alpha1*pi/180)-x(4))/sin(alpha1*pi/180);
Rs2=(L/2+D2*cos(alpha2*pi/180)-x(4))/sin(alpha2*pi/180);

v = 2;
R = L/2/tan(alpha2*pi/180)+B/2;
w = v/sqrt((R+x(3))^2+x(4)^2);
w1=w*(Rs1+x(1))/rs;
w2=w*(Rs2+x(2))/rs;

jx1= (Rs1+x1)*(cos((l/2-D1-y1)*w/rs/w1)-1)-y1*sin((l/2-D1-y1)*w/rs/w1);
jy1=  (Rs1+x1)*sin((l/2-D1-y1)*w/rs/w1)+y1*cos((l/2-D1-y1)*w/rs/w1)-(l/2-D1);
sin1=    -x(1)/sqrt(x(1)^2+y1^2);
cos1=     -y1/sqrt(x(1)^2+y1^2);

funa1=@(x1,y1) (c+p1*tan(fai))*(1-exp(-sqrt(jx1^2+jy1^2)/k))*sin1;
fa1=  integral2(funa1,x(1)-b/2,x(1)+b/2,-(l/2+D1),l/2-D1);

funa2=@(x1,y1) (c+p1*tan(fai))*(1-exp(-sqrt(jx1^2+jy1^2)/k))*cos1;
fa2= integral2(funa2,x(1)-b./2,x(1)+b./2,-(l/2+D1),l/2-D1);

funa3=@(x1,y1) (c+p1*tan(fai))*(1-exp(-sqrt(jx1^2+jy1^2)/k))*(y1^2-x(1)*x1)/sqrt(x(1)^2+y1^2);
fa3=integral2(funa3,x(1)-b/2,x(1)+b/2,-(l/2+D1),l/2-D1);


jx2= (Rs2+x2)*(cos((l/2-D2-y2)*w/rs/w2)-1)-y2*sin((l/2-D2-y2)*w/rs/w2);
jy2= (Rs2+x2)*sin((l/2-D2-y2)*w/rs/w2)+y2*cos((l/2-D2-y2)*w/rs/w2)-(l/2-D2);
sin2=   -x(2)/sqrt(x(2)^2+y2^2);
cos2=   -y2/sqrt(x(2)^2+y2^2);

funb1=@(x2,y2) (c+p2*tan(fai))*(1-exp(-sqrt(jx2^2+jy2^2)/k))*sin2;
fb1=  integral2(funb1,x(2)-b/2,x(2)+b/2,-(l/2+D2),l/2-D2);

funb2=@(x2,y2) (c+p2*tan(fai))*(1-exp(-sqrt(jx2^2+jy2^2)/k))*cos2;
fb2=  integral2(funb2,x(2)-b/2,x(2)+b/2,-(l/2+D2),l/2-D2);

funb3=@(x2,y2) (c+p2*tan(fai))*(1-exp(-sqrt(jx2^2+jy2^2)/k))*(y2^2-x(2)*x2)/sqrt(x(2)^2+y2^2);
fb3= integral2(funb3,x(2)-b/2,x(2)+b/2,-(l/2+D2),l/2-D2);

F(1)=fa2*cos(alpha1*180/pi)+fb2*cos(alpha2*180/pi)-fa11*sin(alpha1*180/pi)-fb1*sin(alpha2*180/pi);
F(2)=fa2*sin(alpha1*180/pi)+fb2*sin(alpha2*180/pi)-fa11*cos(alpha1*180/pi)-fb1*cos(alpha2*180/pi);
F(3)=fa1*Rs1+fb1*Rs2-fa3-fb3;
F(4)=fa1-fb1;
end
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