Help {negative constant} ./ {convex}

1

I can prove my model is convex.how can i solve this problem?
QQ截图20220315214636
QQ截图20220315220258
ut=0.9;
ur=1-ut;
omiga=rand(1);
rita_t=5;
rita_r=5;
c_r=rita_r/omiga;
c_t=rita_t/omiga;
alpha=2.2;
d=50;
p0=-30;
cvx_begin
variables Dr Dt D
maximize(D)

%%

pt=Dt^alpha;
pr=Dr^alpha;
p=d^alpha;
Pt=sqrt(p0/pt);%%Cannot perform the operation: {negative constant} ./ {convex}
Pr=sqrt(p0/pr);
P=sqrt(p0/p);
g=gP;
cr=norm(rr.diag(g),1)^2;
ct=norm(rt.diag(g),1)^2;
a=sigema/(p0ct);
b=sigema/(p0cr);
% rr=rrPr;
% rt=rtPt;
subject to
Dr>=urD;Dr>=1;
Dt>=utD;Dt>=1;
wt+wr<=Pmax;
omiga(2^c_r-1)sigemapr/(p0creb1)-Pmaxeb1<=0;
omiga(2^c_t-1)sigemapt/(p0cteb2)-Pmaxeb2<=0;
% (2^rita_t-1)(apt/eb2+pr)<=pt;
% (2^rita_r-1)bpr/eb1<=pr;
% eb1<=2^rita_t(2^rita_r-1)/(2^rita_t-1)*eb2;
cvx_end

2

Do you mean how to express (y^2.2)/x ? It can be done by adding
\begin{array}{l} \frac{{{z^2}}}{x} \le ....,\\ {y^{1.1}} \le z, \end{array}
the first constraint is quad_over_lin(z,x)<=....., the second is pow_p(y,1.1)<=z.

3

I am so appreciate your answer！But Pt is also an variable about Dt,It is not a constraint.

4

No Dt found in your picture. Please prove the convexity and explain what you want to express.

5

Dk refer to Dr and Dt,D0 equal to Dr plus Dt.