Functions of Several Variables: Level Curves, Continuity

By a function $f (x_{1}, x_{2}, \dots, x_{n})$ of $n$ variables, we mean of course a rule which assigns to each point $(x_{1}^{(0)}, x_{2}^{(0)}, \dots, x_{n}^{(0)})$ of some region in $n$ -space a single number, which we denote by $f (x_{1}^{(0)}, x_{2}^{(0)}, \dots, x_{n}^{(0)})$ . Let us for the moment consider functions of two variables $f (x, y)$ . It would be possible to represent this function graphically in three dimensions, by plotting the set of points $(x, y, f (x, y))$ for $(x, y)$ in the domain of the function. But since attempts to represent 3-space on the blackboard are awkward at best, we choose instead to draw "contour map" for the function; that is, we plot on the blackboard the set of points $(x, y)$ where $f (x, y) = 0$ , the set where $f (x, y) = 7$ , or in general, the set of points $(x, y)$ where $f (x, y) = a$ , where $a$ is any number in the range of $f (x, y)$ . In general, these sets of points will be curves, and are customarily called the level curves of $f (x, y)$ , or less precisely, the level lines of the contour map. These curves represent paths which lie at constant height on the surface representing the function in 3-space.

Example 1.

Let us consider the contour map for the function

f(x, y) = \begin{cases} \frac{xy^2}{x^2 + y^4}, & (x, y) \neq (0, 0) \\ 0, & (x, y) = (0, 0) \end{cases}

Consider the parabola $x = α y^{2}$ . For $(x, y)$ on this curve, $α \neq 0, y \neq 0$ , we have

f (x, y) = \frac{α y^{4}}{(α^{2} + 1) y^{4}} = \frac{α}{α^{2} + 1},

or $f (x, y)$ is a constant. By solving for $x$ in terms of $y$ in the equation

\frac{x y^{2}}{x^{2} + y^{4}} = β, β \neq 0,

we see that each level curve of height $β \neq 0$ , is just a parabola $x = α y^{2}$ with the point $(0, 0)$ omitted. Thus

β = \frac{α}{α^{2} + 1} \cdot

Now what numbers can have the form

\frac{α}{α^{2} + 1} ?

To answer this question, we investigate the function

g (α) = \frac{α}{α^{2} + 1}

for maxima and minima, finding that it has a maximum when $α = 1$ , a minimum when $α = - 1$ . Thus

- \frac{1}{2} \leq \frac{α}{α^{2} + 1} \leq \frac{1}{2} \cdot

The level curve of minimum height $(- \frac{1}{2})$ is the parabola $y = - x^{2}$ , that of maximum height $(+ \frac{1}{2})$ is the parabola $y = x^{2}$ (always omitting $(0, 0)$ ). Thus the contour map looks like the one on the following page.

Illustration for Functions of Several Variables

Observe that for any fixed value of $x$ , this function $f (x, y)$ is continuous in $y$ , and that it is also continuous in $x$ when $y$ is held fixed. It is also continuous along any line; that is, if $(x_{0}, y_{0})$ is a point on a line, then for $(x, y)$ near to $(x_{0}, y_{0})$ and on the line, $f (x, y)$ is near $f (x_{0}, y_{0})$ . However, we can not call this function continuous at $(0, 0)$ . For that would mean that whenever $(x, y)$ was near to $(0, 0)$ , $f (x, y)$ would be near $f (0, 0) = 0$ . But there are points on the parabola $y = x^{2}$ as near as we wish to $(0, 0)$ , and at these $f (x, y) = \frac{1}{2} \cdot$ The surface has a ridge along $y = x^{2}, x \neq 0$ ; a trough along $y = - x^{2}, x \neq 0$ ; a shallow trough along $y = 0$ for $x > 0$ ; and a mild ridge along $y = 0$ for $x < 0$ . It has a discontinuity at $(0, 0)$ .

Let us make the definition of continuity more precise for these functions so that we can discuss it intelligently:

$f (x, y)$ is called continuous at $(x_{0}, y_{0})$ if for any positive number $ϵ$ , we can find a neighborhood of $(x_{0}, y_{0})$ such that whenever $(x, y)$ is in this neighborhood, then

| f (x, y) - f (x_{0}, y_{0}) | < ϵ \cdot

A neighborhood may be regarded as a circular disk with $(x_{0}, y_{0})$ as center or as a square with $(x_{0}, y_{0})$ as center. If the function discussed above were to be continuous at $(0, 0)$ , then presumably we could start with $ϵ = \frac{1}{4}$ and find a circle about $(0, 0)$ such that for all points $(x, y)$ inside it,

| f (x, y) - 0 | < \frac{1}{4} \cdot

But inside every such circle we can find points (other than $(0, 0)$ ) of the parabola $x = y^{2}$ , where $f (x, y) = \frac{1}{2} \cdot$ Then we should have to have

\frac{1}{2} = | \frac{1}{2} - 0 | < \frac{1}{4},

which is nonsense.

Example 2.

As a second example, about which we shall say more a little later, let

f(x, y) = \begin{cases} \dfrac{4xy(x^2 - y^2)}{x^2 + y^2}, & (x, y) \neq (0, 0) \\ 0, & (x, y) = (0, 0) \end{cases} \cdot

(In terms of polar coordinates $(r, θ)$ , where $x = r \cos θ, y = r \sin θ, f (x, y) = r^{2} \sin 4 θ$ .) Though we shall not describe the level curves in detail, the contour map is roughly as follows:

The $+$ signs in the contour map indicate sectors like the one with a $+$ sign in which the contour lines are sketched, and similarly for the $-$ signs. Thus if we stand at $(0, 0)$ , we see four valleys sloping away, alternating with four mountains. This function is continuous everywhere.

EXERCISES

Exercise 1.

Draw a contour map and discuss continuity for each of the following functions:
a) $f (x, y) = x y$ ,
b) f(x, y) = \begin{cases} \frac{2x(x^2 + y^2)}{x^2 + (x^2 + y^2)^2}, & (x, y) \neq (0, 0) \\ 0, & (x, y) = (0, 0) \end{cases} \cdot

Functions of Several Variables

EXERCISES

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