How many strict local minima a quartic polynomial in two variables might have?

Question

What is the maximum number $N$ of strict local minima that a degree 4 polynomial $p:\Bbb R^2\to \Bbb R$ can have?

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One dimension: For a single variable quartic polynomial $q:\Bbb R \to \Bbb R$ the answer to this question would be easy: The polynomial has at most two local minima. Indeed, the derivative $q'$ is a cubic polynomial, so by Fundamental Theorem of Analysis $q$ has at most 3 critical points. However, when $q$ has 3 critical points, then all of them are simple roots, and so two consecutive critical points of $q$ can not be both local minima (local minima/maxima are alternating).

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Along the lines: My first thought was to study $p$ along the lines $(x,y)=(a+bt,c+dt)$ as one would do to prove that a quadratic polynomial $p:\Bbb R^2\to \Bbb R$ has no more than one strict local minimum. However, this argument fails in the attempt to show that a quartic polynomial $p$ has at most two strict local minima. The problem is that unlike two points, three points in $\Bbb R^2$ can not in general be interpolated by a line. Three points in $\Bbb R^2$ can be interpolated by a quadratic curve like $(x(t),y(t))=(t,at^2+bt+c)$, but then $t\mapsto p(x(t),y(t))$ could be a polynomial of degree $8$.

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At lest 4: Since the polynomial $q(t)=(t^2-1)^2$ has minima at the two points $t=\pm 1$, the polynomial $p(x,y) = q(x) + q(y)$ has minima at the four points points $(x,y)=(\pm 1,\pm 1)$. Thus $N\geq 4$.

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At most 9: As commented by @GerryMyerson, the partial derivatives $p_x,p_y$ of $p$ are cubic polynomials, and so by Bézout's Theorem the curves $p_x(x,y)=0$ and $p_y(x,y)=0$ intersect at no more than $3\times 3$ isolated points. Hence, $p$ has no more than 9 critical points. The question then is how many critical points $p$ needs to have in order to have $N$ strict local minima:

• For the case of $N=2$ and a general function $f$ this was answered in the post: If a two variable smooth function has two global minima, will it necessarily have a third critical point?
• I asked the related question for a degree 4 polynomial and $N\geq 2:$ Is there a quartic polynomial in two variables that have multiple local minima and no other critical points?

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If you could not determine the exact value of $N$, could you at least give some lower or upper bound, like that $N>4$ or that $N$ is finite?

Answer 1

Answer: The maximum number of strict local minima of a quartic polynomial is $N=5$.

A few examples of such polynomials are provided in the answers to this cross-posting: https://mathoverflow.net/questions/442736, and they are visualized at the end of this post. In what follows will be argued that $N$ can not be greater than 5.

The papers "Counting Critical Points of Real Polynomials in Two Variables" http://www.jstor.org/stable/2324459 and "The Index of grad f (x, y)", Alan H. Durfee, Topology Vol. 37, No. 6, pp. 1339Ð1361, 1998 shows that the number of critial points is bounded by the vector field index $i$ of the gradient vector field of the polynomial p(x,y).

Due to Bezout's theorem we know that we have at most 9 critical points (for an example, see the plot below).

The papers show that for non-degenerate critical points we have: $$i=m+n-s$$ where $i$ is the index, $m$ is the number of local maxima, $n$ is the number of local minima, and $s$ is the number of saddle points.

The paper "The Index of grad f (x, y)" shows that: $$i \leq max(1,d−3),$$ where $d$ is the degree of the polynomial, in this case $d=4$ so that we get the bound $i \leq 1$.

Combining this with the index formula above we have as the most extreme possible case 5 minima, 0 maxima, and 4 saddle points, for a total of 9 critical points:

$$i=5+0-4=1$$

We cannot have more than 5 minima since then we don't get enough saddle points to get the index down to the bound 1. So, 5 is a definite upper bound. This is a polynomial with 4 local minima, 1 local maximum, and 4 saddle points for a total of 9 critical point and index 1:

$$p(x,y)=(8x^4-8x^2+1)+(16y^4-12y^2+1)$$

The OP gave a simpler polynomial on this form but I like this one since it is the sum of two Chebyshev polynomials.

Below is a plot of this polynomial where the red curves are the loci of the zeros of the gradient components. Where they intersect we have a critical point.

There are of course also degree 4 polynomials with 3 minima (and of course 2 and 1). Here is an example with 3 minima and 2 saddle points: $$p(x,y)=(xy + y + x^2 -1)^2+(x^2 + 2xy + y^2 + x-y -1)^2$$

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Finally, the cross-posting: https://mathoverflow.net/questions/442736 features a few polynomials with 5 minima. Below is a visualization of one of the polynomials, discovered by Peter Mueller:

Note: the plot for (0,0) is rotated 180 degrees vs. the larger figure and some of the others are also rotated.

Here is a visualization of a simpler polynomial, found a bit later by Peter Mueller:

Note: The coordinate values are rounded and the orientation of each highlighted minimum has been changed for clarity. In this case the symmetry of the polynomial was used to show the nature of the minima (only 3 types). (The previous polynomial is also symmetric with 3 different types of minima but there all 5 are shown separately.)

Here is a zoomed-in view of the centrally located saddle point (located at the origin):

The visualization below shows yet another of Mueller's polynomials. This one with an interesting minimum at the origin, located close to a saddle point (see also https://mathoverflow.net/questions/442736):

The following visualization shows a polynomial written as a sum of squares with an interesting minimum at the origin (Mueller):

Note: This doesn't look like a minimum but it is! There are shallow saddle points on either side of the minimum.