（ x^2+y^2 )^2 <= C

1

Hello, everyone ! Do you know how to express the constraints in CVX?
x and y are both variable, C is a constant.

 
C1    ( x^2+y^2 )^2 <= C 

C2    ( x^(-2) + y^(-2) ) ^2 <= C

Thank you for your attention very much!

2

hello, try square_pos function for the square outside (i.e., ()^2). Since the square_pos is nondecreasing, it can accept a convex argument, like x^2 + y^2

3

Thank you for your reply. Can I write it like this:
square_pos( pow_pos(x,2) + pow_pos (y,2) )
square_pos( pow_p(x,-2) + pow_p (y,-2) )

4

This usage seems to cut off part of domains of variable x and y, (i.e., when they are negative). Actually, I don’t know if this change will fit the problem u concern.

5

I would always use norm as much as possible and avoid iterated squares of squares etc. if they can be moved to the constant part.

C1:  C^0.25 >= norm([x,y])
C2:  C^0.5 >= pow_p(x,-2) + pow_p(y,-2)

Yes, the second one will cut off part of the domain, but the function is convex only for x,y>0, so there is no other choice anyway.

6

Thank you very much! I think you are right at avoiding the iterated squares of squares.

7

But when the constraint as follows,it seem that we have to iterate squares.

variable : x, y, q, S 

C1:  S  >= ( x^2 +y^2)^2 - 2( q_f - wk )' (q - q_f) - 2( q_f - wb)' (q - q_f)
8

Yes in this case I guess you have no choice since you have to bound all this by a variable. I guess something like square_pos(sum_squares([x,y])) or similar should work.

9

That should be
square_pos(sum_square([x,y]))
I.e., no s at the end of sum_square

Alternatively, you can use square_pos(x^2+y^2) as @pz233 suggests.

help square_pos

square_pos Square of positive part.
square_pos(X) is the square of the postive parts of the elements of X;
i.e., square_pos(X)=MAX(X,0).^2. X must be real.

Disciplined convex programming information:
    square_pos(X) is convex and nondecreasing in X. Thus when used in
    CVX expressions, X must be convex (or affine).
10

Thank you for your reply very much!
I will have a try.

11

Thank you for your reply again !
I will have a try.

12

Thank you for your attention, variable x and y are both positive.