Question

Find the critical points of the following function, and classify it as degener-
ate or non-degenerate. If the latter, further classify it as local maxima, local
minima, or as saddle points.
f(x, y) = x4 + y4 - 4xy + 1.

Answer 1

Critical points are the points where the gradient is undefined or the gradient is zero (see https://en.wikipedia.org/wiki/Critical_point_(mathematics)). In our case

$\dfrac{\partial f}{\partial x} = 4x^3-4y = 0; \; \dfrac{\partial f}{\partial y} = 4y^3-4x = 0;$

therefore, $x^3=y \; \text{and} \; y^3=x$ , so $(x^3)^3 = x$ , $x^9-x = 0$ . Therefore, $x(x^8-1) = 0.$ So the critical points are the following:

$x=0, y=0^3=0, f(x,y) = 1; \\ x =1, y=1^3=1, f(x,y) = -1; \\ x=-1, y=(-1)^3=-1, f(x,y)=-1.$

To determine whether the point is critical or not, we should calculate determinant of matrix comprised of partial derivatives of 2nd order.

Let $A = \dfrac{\partial^2 f}{\partial x^2}\,, \; C = \dfrac{\partial^2 f}{\partial y^2}\,, \; B = \dfrac{\partial^2 f}{\partial x\partial y}.$ In our case $A=12x^2\, \; C = 12y^2, \; B = -4.$

We should calculate $\Delta = \begin{vmatrix} A & B \\ B & C \end{vmatrix}= AC-B^2.$

1) $x = 0,\; y=0.$

$A=12\cdot0^2 = 0,\; C = 12\cdot0^2 = 0, \; B = -4.$

$\Delta = AC-B^2 = 0\cdot0-(-4)^2 = -16 < 0,$ so there is no extremum, it is a saddle point.

2) $x = 1,\; y=1.$

$A=12\cdot1^2 = 12,\; C = 12\cdot1^2 = 12, \; B = -4.$

$\Delta = AC-B^2 = 12\cdot12-(-4)^2 = 128 > 0,$ so there is an extremum. $A>0,$ therefore it is local minimum. The matrix is indefinite, so this is a saddle point.

3) $x = -1,\; y=-1.$

$A=12\cdot(-1)^2 = 12,\; C = 12\cdot(-1)^2 = 12, \; B = -4.$

$\Delta = AC-B^2 = 12\cdot12-(-4)^2 = 128 > 0,$ so there is an extremum. $A>0,$ therefore it is local minimum.