# RECONSTRUCTING A FUNCTION FROM ITS GRADIENT

This section has three parts. In Part 1 we show how to find given its gradient

In Part 2 we show that, although all gradients are expressions of the form

set and , not all such expressions are gradients. In Part 3 we tackle the problem of recognizing which expressions are actually gradients.

## Part 1

Problem1. Find given that

Solution. Since

we have

Integrating with respect to , we have

where is independent of but may depend on . Differentiation with respect to gives

The two equations for can be reconciled only by having

This means that

We could have started by integrating with respect to and then differentiating with respect to . This procedure would have yielded the same result.

Problem 2. Find given that

Solution. Here we have

We will proceed as in the first problem. Integrating with respect to , we have

with independent of . Differentiation with respect to gives

The two equations for can be reconciled only by having

This means that

The function

is the general solution of the vector differential equation

Each particular solution can be obtained by assigning a particular value to the constant of integration .

## Part 2

The next problem shows that not all linear combinations are gradients.

Problem 3. Show that is not a gradient.

Solution. Suppose on the contrary that it is a gradient. Then there exists a function such that

Obviously

and thus

This contradicts the mixed partial derivative theorem: the four partial derivatives under consideration are everywhere continuous and thus, according to the mixed partial derivative theorem, we must have

## Part 3

We come now to the problem of recognizing which linear combinations

are actually gradients. But first we need to review some ideas and establish some new terminology.

An open set (in the plane or in three-space) is said to be connected if any two points of the set can be joined by a polygonal path that lies entirely within the set. An open connected set is called an open region. A curve

is said to be closed if it begins and ends at the same point:

It is said to be simple if it does not intersect itself:

As is intuitively clear, a simple closed curve in the plane separates the plane into two disjoint open connected sets: a bounded inner region consisting of all points surrounded by the curve and an unbounded outer region consisting of all points not surrounded by the curve. (Jordan Curve Theorem.)

Finally, we come to the notion we need in our work with gradients:

Let be an open region of the plane, and let be an arbitrary simple closed curve. is said to be simply connected if is in implies the inner region of is in

THEOREM Let and be functions of two variables, each continuously differentiable on a simply connected open region . The linear combination is a gradient on if and only if

for all .

Proof. A complete proof of this theorem is given in an advanced calculus course. We will prove the result under the additional assumption that has the form of an open rectangle with sides parallel to the coordinate axes. Suppose that is a gradient on this open rectangle , say

Since

we have

Since and have continuous first partials, has continuous second partials. Thus, according to the mixed partials theorem we have

Conversely, suppose that

We must show that is a gradient on . To do this, we choose a point from and form the function

The first integral is independent of . Hence it follows from the fundamental theorem of calculus that

Differentiating with respect to we have

The first term is since once again we are differentiating with respect to the upper limit. In the second term the variable appears under the sign of integration. It can be shown that, since and are continuous,

Anticipating this result, we have

We have now shown that

It follows that is the gradient of on

Example 4. The vector functions

and

are both everywhere defined. The first vector function is the gradient of a function everywhere defined:

The second is not a gradient:

and so

Example 5. The vector function defined on the punctured disc by setting

satisfying

We will see later that is not a gradient on that set. The set is not simply connected so the theorem does not apply.

# Double Integral over a Rectangle

Let be a function continuous on the rectangle

Our object is to define the double integral

Recall that to define the integral

we first introduced some auxiliary notions: partition , upper sum , and lower sum We were then able to define

as the unique number that satisfies the inequality

We will follow exactly the same procedure to define the double integral

First we explain what we mean by a partition of the rectangle . To do this, we begin with a partition

and a partition

The set

is called a partition of . consists of all the grid points

The partition breaks up into nonoverlapping rectangles where for each fixed

On each rectangle , the function takes on a maximum value , and a minimum value The sum of all the products

is called the upper sum for :

The sum of all the products

is called the lower sum for :

Example 1. Consider the function

on the rectangle

As a partition of take

and as a partition of take

The partition then breaks up the initial rectangle into the six rectangles.

On each rectangle , the function takes on its maximum value at the point , the corner farthest from the origin:

Thus

(area of ) +(area of ) + (area of )

+ (area of ) + (area of ) + (area of )

On each rectangle , takes on its minimum value at the point , the corner closest to the origin:

Thus

(area of ) +(area of ) + (area of )

+ (area of ) + (area of ) + (area of )

We return now to the general situation. As in the one-variable case, it can be shown that, if is continuous, then there exists one and only one number that satisfies the inequality

Definition of the double integral over a rectangle

The unique number that satisfies the inequality

is called the double integral of over and is denoted by

# The Double Integral as a Volume

If is nonnegative on the rectangle , the equation

represents a surface that lies above . In this case the double integral

gives the volume of the solid that is bounded below by and bounded above by the surface

To see this, consider a partition of . breaks up into subrectangles ; and thus the entire solid into parts . Since contains a rectangular solid with base and height , we must have

(area of ) volume of .

Since is contained in a rectangular solid with base and height , we must have

volume of (area of ).

In short, for each pair of indices and , we must have

(area of ) volume of (area of ).

Adding up these inequalities, we are forced to conclude that

volume of .

Since is arbitrary, the volume of must be the double integral:

volume of =

The double integral

gives the volume of a solid of constant height erected over . In square units this is just the area of :

# Some Computations

Double integrals are generally computed by techniques that we will take up later. It is possible, however, to evaluate simple double integrals directly from the definition.

Problem 2. Evaluate

where is the rectangle

Solution. Here

We begin with a partition

and a partition

This gives

as an arbitrary partition of On each rectangle , has constant value . This gives and throughout. Thus

Similarly

Since

and was chosen arbitrarily, the inequality must hold for all partitions of . This means that

Remark. If

gives the volume of the rectangular solid of constant height erected over the rectangle .

Problem 3. Evaluate

where

Solution. With with a partition

and a partition

we have

as an arbitrary partition of On each rectangle the function

has a maximum

and a minimum

Thus

and

For each pair of indices and ,

This means that for arbitrary we have

The middle term can be written

The first double sum reduces to

The second double sum reduces to

The sum of these two numbers

satisfies the inequality

The integral is therefore :

Remark. This last integral should not be interpreted as a volume. The expression does not keep a constant sign on

This document created by Scientific WorkPlace 4.0.