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

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).

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 $8$.

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 $N\geq 4$.

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:

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?

  • 1
    $\begingroup$ Here's some rough-and-ready reasoning that suggests an answer and an approach (without actually proving anything). We expect that at isolated local extreme points both partial derivatives vanish. The partial derivatives are cubics. By Bezout's Theorem, we expect the two cubics to have $3\times3=9$ common zeros. $\endgroup$ Commented Jan 18, 2023 at 3:37
  • 1
    $\begingroup$ @GerryMyerson Great thanks for this observation! On the real line the roots of $f'$, if all being simple roots, are alternatively local minima and local maxima. This question gives an example of a function that has two minima and no other critical point: math.stackexchange.com/q/4024737/1134951. Finding a function with 9 local minima and no maximum would perhaps be harder. $\endgroup$ Commented Jan 18, 2023 at 4:50
  • $\begingroup$ I think for $p$ where $p\rightarrow \infty$ in all directions, you should get critical points in addition to minima. If you consider two minima, and a curve that passes between them, we know the derivative of $p$ changes sign along that curve... At least, it makes the approach in that linked answer no longer work. $\endgroup$
    – Alex K
    Commented Jan 27, 2023 at 13:41
  • $\begingroup$ Typo in the “at most 9“ part: you mean cubic, not quadratic (cf. the referenced comment) $\endgroup$ Commented Jan 31, 2023 at 5:27
  • $\begingroup$ @HagenvonEitzen Thanks! Corrected. $\endgroup$ Commented Jan 31, 2023 at 16:38

1 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:


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:


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.

enter image description here

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: $$p(x,y)=(xy + y + x^2 -1)^2+(x^2 + 2xy + y^2 + x-y -1)^2$$

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:

Polynomial with 5 minima and 4 saddle points.

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:

Polynomial with 5 minima and 4 saddle points, irrational coords.

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):

Central saddle point.

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):

Polynomial with interesting minimum next to a saddle point.

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

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

  • 1
    $\begingroup$ Beautiful visualizations! $\endgroup$ Commented Mar 20, 2023 at 8:06
  • $\begingroup$ @PeterMueller Thanks. $\endgroup$
    – Jap88
    Commented Mar 21, 2023 at 1:33
  • $\begingroup$ @Jap88 Thank you for this amazing answer! Can I ask you what program did you use to create these impressive visualizations? $\endgroup$ Commented May 15, 2023 at 14:39
  • $\begingroup$ @PeterMueller Thanks for solving the associated question posed by Jap88 and finding nice examples of quartic polynomials with 5 strict local minima! $\endgroup$ Commented May 15, 2023 at 14:41
  • $\begingroup$ @Jap88 I slightly edited the beginning of your answer in order to make it more clear, I hope that I did not change the meaning accidentally. $\endgroup$ Commented May 15, 2023 at 15:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .