]> Stationary points

### Chapter 3: Many Variables; Section 1: Differentiation; page 5

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# Stationary Points

## Overview

In the case of functions with a single variable, we classified the stationary points as maximum, minimum and inflection. Actually an inflection point does not have to be stationary, and is an independent class. In the case of a function of two variables, we already saw examples of stationary points, that are a minimum, a maximum or a saddle. A function of a single variable does not have saddle points, and a saddle point has to be stationary. With an increase in the number of variables, there are also more classes of stationary points. Notice that inflection and saddle points are not related, and it is a mistake to consider them as the manifestation of the same class, as often students do.

On this page we are going to focus on functions with two variables, which are simple to understand and to visualize. We'll need first to deepen our understanding of derivatives, in order to use them for the classification of the stationary points.

The differentiation of a function of two variables,

 $z=z\left(x,y\right)$ (3.1.5.1)

with respect to one or the other of the variables, yields in principle two different values (the same value is coincidental). The reason is that the function is incremented in two different directions: x and y. Generally speaking, if we want to study the characteristics of a function at a given point, we have to increment the function in all possible directions, not only in x and y. As we saw previously, the translation of coordinates (3.1.4.10), does not affect the derivatives. On the other hand, the rotation (3.1.4.13) does.

We already defined a stationary point, as a location where all the first derivatives with respect to the variables vanish (with an additional requirement for continuity, which we'll assume as satisfied). We can prove now this statement, by the use of (3.1.4.13), which is reproduced here:

 $\begin{array}{l}\frac{\partial f}{\partial x\text{'}}=\mathrm{cos}\Delta \phi \frac{\partial f}{\partial x}+\mathrm{sin}\Delta \phi \frac{\partial f}{\partial y}\\ \frac{\partial f}{\partial y\text{'}}=-\mathrm{sin}\Delta \phi \frac{\partial f}{\partial x}+\mathrm{cos}\Delta \phi \frac{\partial f}{\partial y}\end{array}\right\}$ (3.1.5.2)

where x' and y' are the coordinates of the rotated system by an angle of Δφ. According to (3.1.5.2)

 $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0\text{ }\text{yields}\text{ }\frac{\partial f}{\partial x\text{'}}=\frac{\partial f}{\partial y\text{'}}=0$ (3.1.5.3)

for any possible angle of rotation of the coordinates Δφ.

## Sufficient conditions

We already know, that for a function of one variable, a sufficient condition for a stationary point to be at minimum(maximum), is a positive(negative) second derivative. We can translate this condition to the requirement of the sign of the second derivative $\frac{{\partial }^{2}f}{\partial {\left(x\text{'}\right)}^{2}}$ for any possible direction of the coordinate x'.

The rotation of the coordinate system from (x,y) to (x',y') by an angle Δφ according to (3.1.4.11) yields

 $\begin{array}{l}\frac{\partial x}{\partial x\text{'}}=\mathrm{cos}\Delta \phi \text{ }\frac{\partial y}{\partial x\text{'}}=\mathrm{sin}\Delta \phi \\ \frac{\partial x}{\partial y\text{'}}=-\mathrm{sin}\Delta \phi \text{ }\frac{\partial y}{\partial y\text{'}}=\mathrm{cos}\Delta \phi \end{array}\right\}$ (3.1.5.4)

From (3.1.5.2) and (3.1.5.4) it follows, that

 $\begin{array}{l}\frac{{\partial }^{2}f}{\partial {\left(x\text{'}\right)}^{2}}=\frac{\partial }{\partial x\text{'}}\left(\mathrm{cos}\Delta \phi \frac{\partial f}{\partial x}+\mathrm{sin}\Delta \phi \frac{\partial f}{\partial y}\right)=\\ =\left\{\begin{array}{l}{\text{cos}}^{\text{2}}\Delta \phi \frac{{\partial }^{2}f}{\partial {x}^{2}}+\mathrm{cos}\Delta \phi \mathrm{sin}\Delta \phi \frac{{\partial }^{2}f}{\partial y\partial x}+\\ +\mathrm{sin}\Delta \phi \mathrm{cos}\Delta \phi \frac{{\partial }^{2}f}{\partial x\partial y}+{\text{sin}}^{2}\Delta \phi \frac{{\partial }^{2}f}{\partial {y}^{2}}\end{array}\right\}=\\ ={\text{cos}}^{2}\Delta \phi \frac{{\partial }^{2}f}{\partial {x}^{2}}+2\mathrm{sin}\Delta \phi \mathrm{cos}\Delta \phi \frac{{\partial }^{2}f}{\partial x\partial y}+{\text{sin}}^{2}\Delta \phi \frac{{\partial }^{2}f}{\partial {y}^{2}}\end{array}\right\}$ (3.1.5.5)

This expression of $\frac{{\partial }^{2}f}{\partial {\left(x\text{'}\right)}^{2}}$ becomes equal to $\frac{{\partial }^{2}f}{\partial {x}^{2}}$ in the case of $\Delta \phi =0\text{\hspace{0.28em}}\text{or}\text{\hspace{0.28em}}\pi$ , and to $\frac{{\partial }^{2}f}{\partial {y}^{2}}$ in the case of $\Delta \phi =±\frac{\pi }{2}$ , as one could expect. If the expression (3.1.5.5) of $\frac{{\partial }^{2}f}{\partial {\left(x\text{'}\right)}^{2}}$ remains positive(negative) for any possible value of Δφ, then this is a sufficient condition for the stationary point to be at minimum(maximum). The sign of  $\frac{{\partial }^{2}f}{\partial {x}^{2}}$ and of  $\frac{{\partial }^{2}f}{\partial {y}^{2}}$ should also be positive(negative), but this is just a partial requirement.

The requirement for a minimum (3.1.5.5) can be rewritten as

 $\begin{array}{l}\frac{{\partial }^{2}f}{\partial {\left(x\text{'}\right)}^{2}}={\text{cos}}^{\text{2}}\Delta \phi \left(\frac{{\partial }^{2}f}{\partial {x}^{2}}+2t\frac{{\partial }^{2}f}{\partial x\partial y}+{t}^{2}\frac{{\partial }^{2}f}{\partial {y}^{2}}\right)>0\\ t=\mathrm{tan}\Delta \phi \end{array}\right\}$ (3.1.5.6)

The condition (3.1.5.6) becomes

 $F\left(t\right)={t}^{2}\frac{{\partial }^{2}f}{\partial {y}^{2}}+2t\frac{{\partial }^{2}f}{\partial x\partial y}+\frac{{\partial }^{2}f}{\partial {x}^{2}}>0$ (3.1.5.7)

where F is a quadratic (parabolic) function of the parameter t. The function  F(t)  remains positive for any value of  t , and therefore does not have any roots. What are the conditions for such a behaviour? We know that the roots  t  of  F(t) = 0 , if they were existing, are calculable by

 $t=\frac{-\frac{{\partial }^{2}f}{\partial x\partial y}±\sqrt{{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^{2}-\frac{{\partial }^{2}f}{\partial {y}^{2}}\frac{{\partial }^{2}f}{\partial {x}^{2}}}}{\frac{{\partial }^{2}f}{\partial {y}^{2}}}$ (3.1.5.8)

There are no roots, if the expression under the square root of (3.1.5.8) (the discriminant) is negative, or if

 $D=\frac{{\partial }^{2}f}{\partial {x}^{2}}\frac{{\partial }^{2}f}{\partial {y}^{2}}-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^{2}>0$ (3.1.5.9)

This condition holds, only if both  $\frac{{\partial }^{2}f}{\partial {x}^{2}}$ and  $\frac{{\partial }^{2}f}{\partial {y}^{2}}$ have the same sign (and do not vanish). Therefore we should require also, that $\frac{{\partial }^{2}f}{\partial {x}^{2}}$ or  $\frac{{\partial }^{2}f}{\partial {y}^{2}}$ is positive.

By the same reasoning, the sufficient condition for a maximum is (3.1.5.9), with the addition, that  $\frac{{\partial }^{2}f}{\partial {x}^{2}}$ or  $\frac{{\partial }^{2}f}{\partial {y}^{2}}$ is negative.

In the case of

 $D=\frac{{\partial }^{2}f}{\partial {x}^{2}}\frac{{\partial }^{2}f}{\partial {y}^{2}}-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^{2}<0$ (3.1.5.10)

there are two roots, and (3.1.5.7) has opposite signs between and outside of the roots. This means that the stationary points behave as a minimum and as a maximum, depending on the direction of the differentiation. This is exactly the case of a saddle point, and makes (3.1.5.10) a sufficient condition for a saddle point . However if one or both of the derivatives $\frac{{\partial }^{2}f}{\partial {x}^{2}}$ and  $\frac{{\partial }^{2}f}{\partial {y}^{2}}$ vanish,  F(t) (3.1.5.7) does not have to be a parabolic function, and the proof that (3.1.5.10) is also a sufficient condition in this case, is presented below as exercise 1.

The following table summarizes the sufficient conditions, for specifying the classes of a stationary point. If  D = 0 , the class of the point cannot be specified without doing further investigations

 $D=\frac{{\partial }^{2}f}{\partial {x}^{2}}\frac{{\partial }^{2}f}{\partial {y}^{2}}-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^{2}$ $\frac{{\partial }^{2}f}{\partial {x}^{2}}\text{ }\text{or}\text{ }\frac{{\partial }^{2}f}{\partial {y}^{2}}$ Class + − Maximum + + Minimum − any Saddle 0 any Undetermined

A graphical study of these sufficient conditions, by use of the shorthand notation for the partial derivatives, is presented in Fig. Sufficient conditions.

## Applications

The saddle point of the function

 $z=\frac{xy}{2}$ (3.1.5.11)

was already studied graphically in Fig. Level lines (3) and in Fig. Saddle point . We are going to use this example for testing the condition (3.1.5.10). The stationary point of the function is at the coordinate's origin:

 $\begin{array}{l}\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left(\frac{xy}{2}\right)=\frac{y}{2}\text{ }\text{and}\text{ }\frac{\partial z}{\partial y}=\frac{x}{2}\\ \text{yielding}\text{ }x=y=0\text{ }\text{for}\text{ }\frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=0\end{array}\right\}$ (3.1.5.12)

From (3.1.5.12), we obtain

 $\frac{{\partial }^{2}z}{\partial {x}^{2}}=\frac{{\partial }^{2}z}{\partial {y}^{2}}=0\text{ }\text{and}\text{ }\frac{{\partial }^{2}z}{\partial x\partial y}=\frac{1}{2}$ (3.1.5.13)

and the condition (3.1.5.10) is satisfied. Notice, that (3.1.5.13) is exact for any point (x, y) of the function, and the inequality (3.1.5.10) is met everywhere, but only at the stationary point there is a saddle point.

The rotation of the (x, y) coordinates by an angle $\Delta \phi =\frac{\pi }{4}$ , transforms the function (3.1.5.11) in the form of  $z=\frac{{x}^{2}-{y}^{2}}{4}$ , as can be obtained from (3.1.4.11). The stationary point remains at the origin, and the derivatives are

 $\begin{array}{l}\frac{\partial z}{\partial x}=\frac{x}{2}\text{ }\frac{\partial z}{\partial y}=-\frac{y}{2}\\ \frac{{\partial }^{2}z}{\partial {x}^{2}}=\frac{1}{2}\text{ }\frac{{\partial }^{2}z}{\partial {y}^{2}}=-\frac{1}{2}\\ \frac{{\partial }^{2}z}{\partial x\partial y}=0\end{array}\right\}$ (3.1.5.14)

again satisfying (3.1.5.10).

As an example for minimum we'll use a physical problem. There are  n  physical points with masses  mi  located at a plane with fixed coordinated  (x, y)i = (ai, bi) , where  i = 1,2,....,n . What is the point (x, y), such that the quantity  W  is minimal, where

 $W=\sum _{i=1}^{n}{m}_{i}\left[{\left(x-{a}_{i}\right)}^{2}+{\left(y-{b}_{i}\right)}^{2}\right]$ (3.1.5.15)

The stationary point corresponds to

 $\begin{array}{l}\frac{\partial W}{\partial x}=2\sum _{i=1}^{n}{m}_{i}\left(x-{a}_{i}\right)=2\left(x\sum _{i=1}^{n}{m}_{i}-\sum _{i=1}^{n}{m}_{i}{a}_{i}\right)=0\\ \frac{\partial W}{\partial y}=2\sum _{i=1}^{n}{m}_{i}\left(y-{b}_{i}\right)=2\left(y\sum _{i=1}^{n}{m}_{i}-\sum _{i=1}^{n}{m}_{i}{b}_{i}\right)=0\\ \text{yielding}\text{ }x=\frac{\sum _{i=1}^{n}\left({m}_{i}{a}_{i}\right)}{\sum _{i=1}^{n}\left({m}_{i}\right)}\text{ }\text{and}\text{ }y=\frac{\sum _{i=1}^{n}\left({m}_{i}{b}_{i}\right)}{\sum _{i=1}^{n}\left({m}_{i}\right)}\end{array}\right\}$ (3.1.5.16)

One has to test if this result corresponds to a minimum.

 $\begin{array}{l}\frac{{\partial }^{2}W}{\partial {x}^{2}}=2\sum _{i=1}^{n}{m}_{i}=\frac{{\partial }^{2}W}{\partial {y}^{2}}>0\text{ },\text{ }\frac{{\partial }^{2}W}{\partial x\partial y}=0\\ D=\frac{{\partial }^{2}W}{\partial {x}^{2}}\frac{{\partial }^{2}W}{\partial {y}^{2}}-\frac{{\partial }^{2}W}{\partial x\partial y}=4{\left(\sum _{i=1}^{n}{m}_{i}\right)}^{2}>0\end{array}\right\}$ (3.1.5.17)

Indeed, since the mass is by definition positive, the point corresponds to the minimal W. This point (3.1.5.16) is called the centre of mass (or centre of gravity).

## More classes

We know so far about three classes of stationary points. Are there any more? The answer is yes, and we are going to see a couple of examples.

In the following example, there is a stationary point at the origin, which is an inflection along the x and a minimum along the y axes, without being classified so far.

 $z={x}^{3}+{y}^{2}$ (3.1.5.18)

yielding

 $\begin{array}{l}\left\{\begin{array}{l}\frac{\partial z}{\partial x}=3{x}^{2}\\ \frac{\partial z}{\partial y}=2y\end{array}\right\}\\ \left\{\begin{array}{l}\frac{{\partial }^{2}z}{\partial {x}^{2}}=6x\\ \frac{{\partial }^{2}z}{\partial {y}^{2}}=2\\ \frac{{\partial }^{2}z}{\partial x\partial y}=0\end{array}\right\}\end{array}\right\}⇒\left\{\begin{array}{l}\text{stationary}\text{\hspace{0.17em}}\text{point:}\text{ }x=y=0\\ D=\frac{{\partial }^{2}z}{\partial {x}^{2}}\frac{{\partial }^{2}z}{\partial {y}^{2}}-{\left(\frac{{\partial }^{2}z}{\partial x\partial y}\right)}^{2}=0\end{array}\right\}$ (3.1.5.19)

In the case of a function with one variable, a stationary point was really a point, except for the trivial case of  $f\left(x\right)=\text{constant}$ . In the case of two (or more) variables, a stationary value, could become a path or a region. The following example will clarify this statement. The level lines of the function

 $z={\left({x}^{2}+{y}^{2}-2\right)}^{2}$ (3.1.5.20)

are shown in Fig. Level lines of a valley. The function is stationary along the circular path

 ${x}^{2}+{y}^{2}=2$ (3.1.5.21)

of constant  z = 0 , limited by rising slopes reminiscent of a valley. This is not a local minimum, since along the path the value of the function remains constant. On the other hand it is an absolute minimum. The starionary values of this function (3.1.5.20) are summarized at (3.1.5.22).

 $\begin{array}{l}\left\{\begin{array}{l}\frac{\partial z}{\partial x}=4x\left({x}^{2}+{y}^{2}-2\right)\\ \frac{\partial z}{\partial y}=4y\left({x}^{2}+{y}^{2}-2\right)\end{array}\right\}\\ \left\{\begin{array}{l}\frac{{\partial }^{2}z}{\partial {x}^{2}}=4\left(3{x}^{2}+{y}^{2}-2\right)\\ \frac{{\partial }^{2}z}{\partial {y}^{2}}=4\left({x}^{2}+3{y}^{2}-2\right)\\ \frac{{\partial }^{2}z}{\partial x\partial y}=8xy\end{array}\right\}\end{array}\right\}⇒\left\{\begin{array}{l}\left\{\begin{array}{l}\text{stationary:}\text{ }x=y=0\\ D=\left(-8\right)\left(-8\right)-0>0\\ \frac{{\partial }^{2}z}{\partial {x}^{2}}=-8<0\\ \text{point}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{maximum}\end{array}\right\}\\ \left\{\begin{array}{l}\text{stationary:}\text{\hspace{0.17em}}{x}^{2}+{y}^{2}-2=0\\ D=8{x}^{2}8{y}^{2}-{\left(8xy\right)}^{2}=0\\ \frac{{\partial }^{2}z}{\partial {x}^{2}}=8{x}^{2}\ge 0\text{ }\frac{{\partial }^{2}z}{\partial {y}^{2}}\ge 0\\ \text{bottom}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{valley}\end{array}\right\}\end{array}\right\}$ (3.1.5.22)

Since the function (3.1.5.20) can be represented also as a function of one variable r, namely

 ${r}^{2}={x}^{2}+{y}^{2}$ (3.1.5.23)

the analysis shows that at  r² = 2  there is a minimum.

Any additional stationary class in the case of two variables obeys

 $D=\frac{{\partial }^{2}z}{\partial {x}^{2}}\frac{{\partial }^{2}z}{\partial {y}^{2}}-{\left(\frac{{\partial }^{2}z}{\partial x\partial y}\right)}^{2}=0$ (3.1.5.24)

## Exercises

Exercise 1. Prove that (3.1.5.10) is a sufficient condition for a saddle point in the following cases:

1.
$\frac{{\partial }^{2}f}{\partial {y}^{2}}=0\text{ }\text{and}\text{ }\frac{{\partial }^{2}f}{\partial {x}^{2}}\ne 0$
2.
$\frac{{\partial }^{2}f}{\partial {y}^{2}}=0\text{ }\text{and}\text{ }\frac{{\partial }^{2}f}{\partial {x}^{2}}=0$

Exercise 2.

1. Find and classify the stationary points of the function

$f\left(x,y\right)={x}^{3}-{y}^{3}-2xy+2$

2. What are the values of the function for these points?

Exercise 3. Find and classify the stationary points of the function

$\begin{array}{l}f\left(x,y\right)=\left(a{x}^{2}+b{y}^{2}\right)\mathrm{exp}\left(-{x}^{2}-{y}^{2}\right)\\ \text{where the constants hold}\text{ }0

Exercise 4. Use the results of exercise 3 in order to answer what will be the classification in the following cases:

1.
$0
2.
$0

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