I want to minimize a double integral

Can I use the Matlab optimization toolbox to minimize

integral of f(u)^2 over a square D in the space of smooth functions u = (u1,u2) such that u1 = u2 = 0 on the boundary of D. f is a smooth real valued function. There are no other constraints.

[Walter Roberson]

If there no other parameters, then assuming real valued f(u), the solution is f(u) = 0.

If f(u) is given and arbitrary, then what parameters are to be adjusted to try to minimize it ?

[Guojun Liao]

Hi Walter, Thank you for your comment. The function f(u) in my mind is a differential operator and f(u) = 0 is a differential equation with u = 0 on boundary. My theoretical conjecture is "f(u) = 0 implies u = 0 on D". I do not have a proof in the general case yet. I'd like to conduct numerical experiments to see if there is a counter example: whether or not there exists u (not equal zero in D) such that the integral is zero (thus f(u) = 0). I want to know if Matlab optimization toolbox could help in the numerical experiments.

Answer 1

[Walter Roberson]

To be sure that we are talking about the same thing: if I understand correctly, you want u(x,y) such that

D(u(x,y),x)(0,y) = 0
D(u(x,y),x)(D,y) = 0
D(u(x,y),y)(x,0) = 0
D(u(x,y),y)(x,D) = 0
minimize( int(u(x,y)^2, x=0..D, y=0..D) )

We know that u(x,y) = 0 satisfies that the integral is 0. We also know that u(x,y)^2 can never be negative if u(x,y) is strictly real valued. We know that the integral of an expression that is everywhere non-negative can never be negative, so we deduce that we cannot get an integral that is less than the integral of the all-zero function.

Your question then appears to be whether there is a different function u(x,y) that is non-zero at some point, but the integral for which is at least as minimal as the integral of the all-zero function. But with u(x,y) being real valued, u(x,y)^2 is non-negative. If there is a point (x,y) at which u(x,y)^2 is positive (some point that is non-zero, and must be positive because square of a non-zero real value is positive), then there cannot be some other u(x,y)^2 negative to drag the integral down to zero again. Therefore if any real-valued u(x,y) is non-negative the integral of u(x,y)^2 must be non-negative, and so would be a larger integral than the all-zero integral. Therefore over real-valued functions, u(x,y) = 0 is the only one that has the minimal integral of 0.

If you do not restrict u(x,y) to real values, then you can get arbitrary negative values. Consider

 T*cos(2*x*Pi/D)*cos(2*y*Pi/D)*j

with j = sqrt(-1)

then this satisfies the boundary conditions that the differentials are 0 at the boundaries 0 an D, but the square of it is non-positive everywhere, and the integral of u(x,y)^2 would be -(1/4)*T^2*D^2 . Since T can be made arbitrarily large, you could get arbitrarily negative integrals.