Finding extreme values of $ f :\mathbb R^2\rightarrow \mathbb R$, when the determinant $\Delta = AC - B^2 = 0$.

Question

We generally rely on the result:

[Let $ f$ be a real valued function from $\mathbb R^2$ with continuous partial derivatives at a stationary point $\vec a$ in $\mathbb R^2$. Let

$A = D_{11}f(\vec a)$,

$ B = D_{12}f(\vec a) = D_{21}f(\vec a) $

$C = D_{22}f(\vec a)$,

and let $\Delta = AC - B^2$ . Then,

(a) If $\Delta>0 $ and $A>0$, $f$ has a relative minimum at $\vec a$

(b) If $\Delta>0 $ and $A<0$, $f$ has a relative maximum at $\vec a$

(c) If $\Delta<0 $ , $f$ has a saddle point at $\vec a$ ]

But when $\Delta = 0, f$ can have a local minimum of a local maximum or a saddle point at $\vec a$.

What are the other ways to proceed when we have $\Delta = 0$ or $A = 0$?

Answer 1

Consider the problem:

Find and classify the extreme values (if any) of the function: $$f(x,y) = y^2+x^2y+x^4$$

Here we have,

$D_1f = \partial f/\partial x = 2xy +4x^3$

$D_2f = \partial f/\partial y = 2y +x^2$

$D_{11}f = \partial^2 f/\partial x^2 =2y +12x^2$

$D_{12}f = \partial^2 f/\partial x \partial y =2x$

$D_{22}f = \partial^2 f/\partial y^2 =2$

For a stationary point (x,y) we want $D_1f(x,y) =0$, and $ D_2f(x,y)=0$. Then,

$D_1f(x,y) =0 \Rightarrow2xy +4x^3 = 0 \Rightarrow x(2y+4x^2)=0$ (equation 1)

$D_2f(x,y) =0 \Rightarrow2y +x^2 = 0 \Rightarrow 2y=-x^2$ (equation 2)

Substituting the value of $2y$ in equation 1, we get, $3x^3=0 \Rightarrow x=0 \Rightarrow y=0$ and so the origin $(0,0)$ is the only stationary point for $\vec f$.

But for this point,

$A = D_{11}f((0,0)) = 2(0)+12(0)^2 = 0$

$B = D_{12}f((0,0)) = 2(0) = 0$

$C = D_{22}f((0,0)) = 2$

i.e. both $A = 0$ and $\Delta = AC - B^2 = 0$. Thus, no direct conclusion can be drawn.

However (using completing the square method),

$f(x,y) - f(0,0) = [y^2+x^2y+x^4] -[(0)^2+(0)^2(0)+(0)^4]$

$= (y)^2+x^2(y)+x^4 $

$= (y)^2+2(x^2/2)(y)+x^4 $

$= (y)^2+2(x^2/2)(y)+ (x^2/2)^2 - (x^2/2)^2 + x^4 $

$= [(y)^2+2(x^2/2)(y)+ (x^2/2)^2] + [x^4 - (x^4/4)] $

$= [y + (x^2/2)]^2 +(3/4)x^4 $

$\geq 0$ , for any $(x,y) \in \mathbb R^2$.

Therefore $f(x,y) \geq f(0,0)$ for all $(x,y) \in \mathbb R^2$ where $(0,0)$ is the only stationary point of f. Thus it is the global minima of f and the minimum value of f is $0 (=f(0,0))$.

The inference here is that in such cases one must always try to analyze not the function $f(\vec x)$ in particular, but the difference $f(\vec x) - f(\vec a)$ and see if something meaningful can be inferred from there.

Thank You!