0

According to (hope my calculation below is correct) https://en.wikipedia.org/wiki/Quadratic_function

a bivariate quadratic function is a second-degree polynomial of the form $$ f(\xi,\eta)=\mathcal{A}\, \xi^2 + \mathcal{B}\, \eta^2 + \mathcal{C}\, \xi + \mathcal{D}\, \eta + \mathcal{E}\, \xi\, \eta + \mathcal{F}, $$ where $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$, $\mathcal{D}$, $\mathcal{E}$, and $\mathcal{F}$ are constants,

(i) achieves its minimum $f(\xi_0,\eta_0)=\frac{\mathcal{D} (\mathcal{C}\,\mathcal{E}-\mathcal{A}\, \mathcal{D})-\mathcal{B}\, \mathcal{C}^2}{4\, \mathcal{A}\,\mathcal{B}-\mathcal{E}^2}+\mathcal{F}$ at $(\xi_0,\eta_0)=\big(\frac{-2\,\mathcal{B}\,\mathcal{C}+\mathcal{D}\,\mathcal{E}}{4\,\mathcal{A}\,\mathcal{B}-\mathcal{E}^2},\frac{-2\,\mathcal{A}\,\mathcal{D}+\mathcal{C}\,\mathcal{E}}{4\,\mathcal{A}\,\mathcal{B}-\mathcal{E}^2}\big)$ provided that that $4\,\mathcal{A}\,\mathcal{B}-\mathcal{E}^2 >0$ and $\mathcal{A}>0$;

(ii) achieves its minimum at $(\xi_0,\eta_0)=\big(\xi_0,-\frac{1}{\mathcal{E}}(\mathcal{C}+2\,\mathcal{A}\,\xi_0)\big)$ with $f(\xi_0,\eta_0)=-\frac{\mathcal{C}^2}{4\,\mathcal{A}}+\mathcal{F}=-\frac{\mathcal{D}^2}{4\,\mathcal{B}}+\mathcal{F}$ for any $\xi_0\in\mathbb{R}$ provided that $4\,\mathcal{A}\,\mathcal{B}-\mathcal{E}^2=0$, $\mathcal{A}>0$, and $-2\,\mathcal{B}\,\mathcal{C}+\mathcal{D}\,\mathcal{E}=-2\,\mathcal{A}\,\mathcal{D}+\mathcal{C}\,\mathcal{E}=0$.

My question is: Consider $g(u,v)=\mathcal{A}\, u^2 + \mathcal{B}\, v^2 + \mathcal{C}\, w^2+\mathcal{D}\, u + \mathcal{E}\, v + \mathcal{F}\, w+ \mathcal{G}\, u\,v +\mathcal{H}\, v\,w +\mathcal{I}\, u\,w + \mathcal{J},$. What is the condition under which the minimum of $g(u,v)$ exists?

Any reference or idea is welcome. Thanks!

LCH
  • 815

1 Answers1

1

You can write the function in a matrix form:

$$g(u,v)=[u,v]^T {\sf A} [u,v]+\text{linear and constant terms}$$

Notice that if you are looking for a minimum only in $(u,v)$ space, you can pack $w$ into the constant and linear terms. This is a quadratic form... linear terms can be packed into the quadratic terms by substitution $u\to u-u_0$ (same for $v$), and finding $u_0,v_0$ for which the linear terms vanish. That already gives you the center of this quadratic form, which is always an extremum (but may be a maximum, minimum or a saddle point). It's minimum, if matrix ${\sf A}$ has two positive eigenvalues, it's a saddle, if eigenvalues are of opposite signs and maximum if it has two negative eigenvalues. In your parameterization, it's $${\sf A}=\begin{bmatrix}A&G/2\\G/2&B\end{bmatrix}$$ For 2×2 matrix, we know that sum of eigenvalues is the trace (A+B>0) and product (which also has to be positive to ensure the eigenvalues are the same sign) equals the determinant, therefore $AB-G^2/4>0$.

Notice how $w$ does nothing at all for the condition... only shifts the minimum. If you want a minimum in (u,v,w) space, just pack this into a 3×3 matrix and require 3 positive eigenvalues.

orion
  • 15,781