如何求解如下优化问题

smileqq

f=min(abs(0.5*ay_tp1)+abs(0.5*ay_tp2));
s.t.
0<p<10,-10<q<0,q^2-4*p<=0;2*pi/(4*p-q^2)^(1/2)-tf=0;
# o, ?7 v' X1 m8 i3 q%%其中：
w=3.75;
9 @! H, A$ @! L+ S9 |' ktf=5;
% l, k5 h: |/ Gtp1=0;
tp2=2*(pi-atan(q*(4*p-q^2)^(1/2)/(q^2-4*p)))/((4*p-q^2)^(1/2));
$ j9 P9 x5 Y; ]6 h" c8 a  @A1=(4*p/(4*p-q^2))^(1/2);
A2=((4*p-q^2)/2)^(1/2);
A3=q/((4*p-q^2)^(1/2));
- d( j$ h; U2 S( H6 x
: W! l+ Y! c, Hay_tp1=(w*p*exp(q*tp1/2)*A1*sin(((A2/(2^(1/2)))*tp1)+1/A3))/(1-exp(q*tf/2)*A1*sin(A2*tf-atan(A3)));
0 O) ~9 l3 Q% A: k5 r6 h0 r8 ?ay_tp2=(w*p*exp(q*tp2/2)*A1*sin(((A2/(2^(1/2)))*tp2)+1/A3))/(1-exp(q*tf/2)*A1*sin(A2*tf-atan(A3)));
6 q, z$ K% m1 c5 f使用fmincon函数求解时显示 相邻两次迭代点的变化小于预先给定的容忍度。
求助大佬。感谢

SsaaM7

路过，帮顶一下

ExxNEN

看不懂啊，有心无力哦

haidaowang

下面结果参考下：
Objective Function (Min.): 1.28973902645915E-5
p: 0.394784176495481
! F- [  j6 B4 s% `7 fq: -4.3001505065871E-5
) I& n. _7 s3 ~: k3 r$ D) z) W6 c
Constrained Functions:
# C; V, j2 X! W% A2 |' g) q1: q^2-4*p-0 = -1.5791367041328
- s/ _  [$ V" c7 h2 t% s" @2: 2*pi/(4*p-q^2)^(1/2)-5-0 = 6.57038867757365E-11