An interesting multivariable calculus example

Earlier this semester in my Multivariable Calculus course we were discussing the second derivative test. Recall the pesky condition that if {(a,b) } is a critical point and {D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-(f_{xy}(a,b))^{2}=0}, then the test fails.

A student emailed me after class and asked the following question. Suppose a function {f} has a critical point at {(0,0)} and {D(0,0)=0}. Moreover, suppose that as we approach {(0,0)} along {x=0} we have {f_{xx}(0,y)>0} when {y<0} and {f_{xx}(0,y)<0} when {y>0}. Is that enough to say that the critical point is a not a maximum or a minimum? His thought process was that if we look at slices {y=k} we get curves that are concave up when {k<0} and curves that are concave down when {k>0}—surely that could not happen at a maximum or minimum.

I understood his intuition, but I was skeptical. Indeed, after a little playing around I came up with the following counterexample. The function is

{\displaystyle f(x,y)=\begin{cases} x^{4}+y^{2}e^{-x^{2}} & y\ge 0\\ x^{4}+x^{2}y^{2}+y^{2} & y<0. \end{cases}}

The first partial derivatives are

{\displaystyle f_{x}(x,y)=\begin{cases} 4x^{3}-2xy^{2}e^{-x^{2}} & y\ge 0\\ 4x^{3}+2xy^{2} & y<0, \end{cases}}

{\displaystyle f_{y}(x,y)=\begin{cases} 2ye^{-x^{2}} & y\ge 0\\ 2x^{2}y+2y & y<0. \end{cases}}

Clearly {(0,0)} is a critical point. The second partial derivatives are

{\displaystyle f_{xx}(x,y)=\begin{cases} 12x^{2}-2y^{2}e^{-x^{2}}+4x^{2}y^{2}e^{-x^{2}} & y\ge 0\\ 12x^{2}+2y^{2} & y<0, \end{cases}}

{\displaystyle f_{yy}(x,y)=\begin{cases} 2e^{-x^{2}} & y\ge 0\\ 2x^{2}+2 & y<0, \end{cases}}

{\displaystyle f_{xy}(x,y)=f_{yx}(x,y)=\begin{cases} -4xye^{-x^{2}} & y\ge 0\\ 4xy & y<0. \end{cases}}

Thus {D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-(f_{xy}(0,0))^{2}=0\cdot 2-0^{2}=0}. So the second derivative test fails. But observe that when {x=0} we have

{\displaystyle f_{xx}(0,y)=\begin{cases} -2y^{2} & y\ge 0\\ 2y^{2} & y<0. \end{cases}}

So {f_{xx}(0,y)>0} when {y<0} and {f_{xx}(0,y)<0} when {y>0}. Yet it is easy to see that {(0,0)} is a minimum: {f(0,0)=0} and {f(x,y)>0} for all {(x,y)\ne (0,0)}. A graph of the function is shown below. You can see the concave down cross sections for x=0 and y>0.



  1. Jan Van lent says:

    I think the functions (y-x^2)^4 + x^4 and ((y-x^2)^2 + x^2)^2 also work. They are inspired by the pitchfork bifurcation (compare with the shape of the contour f_x=0).

    1. Jan Van lent says:

      I actually meant (y-x^2)^4 + y^4 and ((y-x^2)^2 + y^2)^2, but the other two also work. From the contour plots for all these functions it is clear that they use the same idea.

      1. Great! Thanks. I haven’t checked the derivatives, but it looks good on Grapher.

    2. Jan Van lent says:

      I just realised that the examples I gave are closely related to the Rosenbrock function [1].


      1. Wow. Cool. Thanks for sharing that!

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