如何求解如下优化问题

smileqq

3 N. ~( |! P4 I0 nf=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;
%%其中：
, ?8 n, Y8 c" b  e- g# V) G* Ww=3.75;
tf=5;
tp1=0;
: S) F# H8 q0 f7 ltp2=2*(pi-atan(q*(4*p-q^2)^(1/2)/(q^2-4*p)))/((4*p-q^2)^(1/2));
A1=(4*p/(4*p-q^2))^(1/2);
4 ~% V* M$ J$ k( z/ p5 sA2=((4*p-q^2)/2)^(1/2);
A3=q/((4*p-q^2)^(1/2));
2 R1 J. e/ L6 {8 Z
ay_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)));
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)));
1 I/ F. \( ~. \+ Z! p1 h5 k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。
; F* c" Y5 p2 |  Y0 T( y. j0 @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
/ S: Y) t/ \; |5 Y  Q! S& Up: 0.394784176495481
" k) `! R2 h& m5 f3 u) D, W: _; jq: -4.3001505065871E-5

( C! s4 G$ P3 q  P2 ZConstrained Functions:
% [+ \. _+ _! y" ?* L1: q^2-4*p-0 = -1.5791367041328
2: 2*pi/(4*p-q^2)^(1/2)-5-0 = 6.57038867757365E-11