This question: "Can a real quartic polynomial in two variables have at most 4 isolated local minima?" came up in this post on Math SE but with no answer so far.

Finding examples of 4 isolated minima to a degree 4 polynomial is trivial (see posting) but proving that you cannot have 5 isolated minima seems hard. I cannot find any proof after searching for quite a while.

The only relevant results I can find are in the papers:

Durfee shows that the number of critical points is bounded by the index of the polynomial’s gradient field. A consequence of this seems to be that you can have at most have 5 minima. However, there seems to be no known example with 5 minima, only 4, leading me to believe 4 is the most you can have.

  • $\begingroup$ Since the answer shows that five are possible, I would suggest changing the title to something like "Can a real quartic polynomial in two variables have at most 4 isolated local minima?" (as in the body of the question). Otherwise people who only glance at titles will get the wrong impression of how it turned out... $\endgroup$ Mar 16 at 22:23
  • $\begingroup$ @JukkaKohonen Good suggestion, done. $\endgroup$ Mar 16 at 22:27
  • $\begingroup$ Thanks for making that change. $\endgroup$
    – Jap88
    Mar 16 at 22:31

1 Answer 1


Recent addition: Inspired by DimaPasechnik's and Matt F.'s comments about sum of squares decompositions, I tried the following very natural idea: Try to find $f$ of the form $f(x,y)=A(x,y)^2+B(x,y)^2$, where $A$ and $B$ are relatively prime quadratic polynomials whose zero sets intersect in $4$ points, which then of course are local minima, and attempt to get a fifth local minimum. By a shift we may look for it in $(0,0)$. And indeed, there are examples which can be checked instantly by hand: One possibility is \begin{align} A &= x^2 - y - 6\\ B &= 8x^2 - y^2 - 3y + 2. \end{align} The curves $A=0$ and $B=0$ intersect in the four points \begin{equation} (\pm1, -5)\text{ and } (\pm4, 10). \end{equation} As \begin{equation} f(x, y) = 40 + 20x^2 + 6y^2 + \text{terms of degree }\ge3, \end{equation} we see that $(0, 0)$ is a fifth local minimum.

Original answer: The quartic \begin{equation} f = 3x^4 - 5x^2y^2 + 5y^4 + 20x^2y - 32y^3 + x^2 + 2y^2, \end{equation} which is even in $x$, has $5$ isolated local minima in \begin{equation} (0, 0), (\pm2, -1), (\pm2, 5). \end{equation} In order to prove the claim, one checks that the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ vanish and that the Hessian is positive definite in these points, where the latter is equivalent to the traces and the determinants being positive.

If one carries out the calculation by hand, then the symmetry $f(x,y)=f(-x,y)$ requires only to check $3$ points. Alternatively, the following Sage code verifies the example:

R.<x, y> = QQ[]
f = 3*x^4-5*x^2*y^2+5*y^4+20*x^2*y-32*y^3+x^2+2*y^2
fx, fy = f.derivative(x), f.derivative(y)
fxx = fx.derivative(x)
fxy = fx.derivative(y)
fyy = fy.derivative(y)
tra, det = fxx+fyy, fxx*fyy-fxy^2
cps = ideal([fx, fy]).variety()
minima = [p for p in cps if tra.subs(p) > 0 and det.subs(p) > 0]
print(len(minima) == 5)

Remark 1: As pointed out by Jap88 in the comments (see also his nice visualizations at https://math.stackexchange.com/questions/4620663), previous examples are difficult to visualize. This holds true for this one too. The reason is that some of the saddles points tend to be very close to the minima. The red and blue lines are the curves given by $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$, and the green dots are the minima. As the first picture doesn't show well the situation around $(0,0)$, the second one zooms into this area:

Remark 2: Another rather short example is \begin{equation} f = x^4 - 3x^2y^2 + 3y^4 - 2x^2y + 3y^3 + x^2 - 2y^2. \end{equation} Here, the $5$ local minima are irrational, so the exact algebraic verification by hand is a little more involved. Or, if one relies on Sage again, then one simply replaces the first two lines of the code above with

R.<x, y> = AA[]
f = x^4 - 3*x^2*y^2 + 3*y^4 - 2*x^2*y + 3*y^3 + x^2 - 2*y^2

Here AA is the (exact) field of real algebraic numbers. The image (notation as above) is enter image description here

Remark 3: Conjecture 5.4 in this paper by Durfee et.al predicts that a quartic has at most $4$ isolated local minima. So this is a counterexample.

Remark 4: Theorem 3.1.6. of the paper Critical points of real polynomials ... by Shustin seems to claim that $5$ minima are possible. However, the "proof" given in Section 3.13 merely says "Assume that $d \le 4$. Then all the index distributions except [...] can be easily realized as it was explained above". I don't know what the author means by "above", as all the previous cases handle degrees $d\ge5$.

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    $\begingroup$ Great. I didn't believe such a polynomial existed. $\endgroup$
    – Jap88
    Mar 16 at 14:14
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    $\begingroup$ Using your answer as a starting point, I found the simpler case $$f=26 x^2 - 472 x^3 + 341 x^4 + 624 x^2 y - 624 x^3 y + 156 y^2 + 5268 x y^2 + 960 x^2 y^2 - 8748 y^3 - 5268 x y^3 + 6483 y^4$$ which has local minima at $$(0,0),\ (0,1),\ (1,0),\ (3,2),\ (-2,-1)$$ $\endgroup$
    – user44143
    Mar 16 at 15:23
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    $\begingroup$ @DimaPasechnik, yes, $f+64$ is non-negative and therefore a sum of squares. What I want to know is: Can we write $f$ in a way which makes it obvious that it has exactly five local minima? $\endgroup$
    – user44143
    Mar 16 at 21:26
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    $\begingroup$ I added a visualization of Peter Mueller's polynomial to the posting: math.stackexchange.com/questions/4620663 in case someone is interested. The polynomial is a bit tricky to visualize due to its "multiscale" nature. $\endgroup$
    – Jap88
    Mar 17 at 3:39
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    $\begingroup$ By the way, I noticed that for this polynomial, when one traces the loci of $f_x=0$ and $f_y=0$ there is always one saddle point between two minima. $\endgroup$
    – Jap88
    Mar 17 at 3:53

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