RECONSTRUCTING A FUNCTION FROM ITS GRADIENT

This section has three parts. In Part 1 we show how to find $f(x,y)$ given its gradient
MATH
In Part 2 we show that, although all gradients $\nabla f(x,y)$ are expressions of the form
MATH
$($set MATH and MATH, not all such expressions are gradients. In Part 3 we tackle the problem of recognizing which expressions $P(x,y)i+Q(x,y)j$ are actually gradients.

Part 1

Problem1. Find $f$ given that
MATH

Solution. Since
MATH
we have
MATH
Integrating MATH with respect to $x$, we have
MATH
where $\phi (y)$ is independent of $x$ but may depend on $y$. Differentiation with respect to $y$ gives
MATH
The two equations for MATH can be reconciled only by having
MATH
This means that
MATH

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

Problem 2. Find $f$ given that
MATH

Solution. Here we have
MATH
We will proceed as in the first problem. Integrating MATH with respect to $x$, we have


MATH
with $\phi (y)$ independent of $x$. Differentiation with respect to $y$ gives
MATH
The two equations for MATH can be reconciled only by having
MATH
This means that
MATH

The function
MATH
is the general solution of the vector differential equation
MATH
Each particular solution can be obtained by assigning a particular value to the constant of integration $C$.

Part 2

The next problem shows that not all linear combinations $P(x,y)i+Q(x,y)j$ are gradients.

Problem 3. Show that $yi-xj$ is not a gradient.

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

MATH
and thus
MATH
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
MATH
This contradiction shows that $yi-xj$ is not a gradient.

Part 3

We come now to the problem of recognizing which linear combinations
MATH
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


MATH
is said to be closed if it begins and ends at the same point:
MATH
It is said to be simple if it does not intersect itself:
MATH

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

THEOREM Let $P$ and $Q$ be functions of two variables, each continuously differentiable on a simply connected open region $\Omega $. The linear combination $P(x,y)i+Q(x,y)j$ is a gradient on $\Omega $ if and only if
MATH
for all $(x,y)\in \Omega $.

Proof. A complete proof of this theorem is given in an advanced calculus course. We will prove the result under the additional assumption that $\Omega $ has the form of an open rectangle with sides parallel to the coordinate axes. Suppose that $P(x,y)i+Q(x,y)j$ is a gradient on this open rectangle $\Omega $, say
MATH
Since
MATH
we have
MATH
Since $P$ and $Q$ have continuous first partials, $f$ has continuous second partials. Thus, according to the mixed partials theorem we have
MATH

Conversely, suppose that
MATH
We must show that $P(x,y)i+Q(x,y)j$ is a gradient on $\Omega $. To do this, we choose a point $(x_0,y_0)$ from $\Omega $ and form the function
MATH

The first integral is independent of $y$. Hence it follows from the fundamental theorem of calculus that
MATH
Differentiating $f$ with respect to $x$ we have
MATH
The first term is $P(x,y_0)$ since once again we are differentiating with respect to the upper limit. In the second term the variable $x$ appears under the sign of integration. It can be shown that, since $Q$ and MATH are continuous,
MATH

Anticipating this result, we have
MATH

MATH

MATH
We have now shown that
MATH
It follows that $P(x,y)i+Q(x,y)j$ is the gradient of $f$ on $\Omega .$

Example 4. The vector functions

MATH and MATH

are both everywhere defined. The first vector function is the gradient of a function everywhere defined:
MATH
The second is not a gradient:
MATH
and so
MATH

Example 5. The vector function defined on the punctured disc $0<x^2+y^2<1$ by setting
MATH
satisfying
MATH
We will see later that $F$ 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 $f$ be a function continuous on the rectangle
MATH

Our object is to define the double integral
MATH
Recall that to define the integral
MATH
we first introduced some auxiliary notions: partition $P$, upper sum $U_f(P)$, and lower sum $L_f(P).$ We were then able to define
MATH
as the unique number $I$ that satisfies the inequality
MATH
We will follow exactly the same procedure to define the double integral
MATH

First we explain what we mean by a partition of the rectangle $R$. To do this, we begin with a partition
MATH
and a partition
MATH
The set
MATH
is called a partition of $R$. $P$ consists of all the grid points $(x_i,y_j).$

The partition $P$ breaks up $R$ into $m\times n$ nonoverlapping rectangles MATH where for each fixed $i,j,$
MATH
On each rectangle $R_{ij}$, the function $f$ takes on a maximum value $M_{ij} $, and a minimum value $m_{ij}.$ The sum of all the products
MATH
is called the $P$ upper sum for $f$:
MATH
The sum of all the products
MATH
is called the $P$ lower sum for $f$:
MATH

Example 1. Consider the function
MATH
on the rectangle
MATH
As a partition of $[1,4]$ take
MATH
and as a partition of $[1,3]$ take
MATH
The partition $P=P_1\times P_2$ then breaks up the initial rectangle into the six rectangles.

On each rectangle $R_{ij}$, the function $f$ takes on its maximum value $M_{ij}$ at the point $(x_i,y_j)$, the corner farthest from the origin:
MATH
Thus

$U_f(P)=M_{11}$(area of $R_{11}$) +MATH(area of $R_{12}$) + $M_{21}$ (area of $R_{21}$)

+ $M_{22}$(area of $R_{22}$) + $M_{31}$ (area of $R_{31}$ ) + $M_{32}$(area of $R_{32}$)
MATH
On each rectangle $R_{ij}$, $f$ takes on its minimum value $m_{ij}$ at the point $(x_{i-1},y_{j-1})$, the corner closest to the origin:
MATH
Thus

$L_f(P)=m_{11}$(area of $R_{11}$) +MATH(area of $R_{12}$) + $m_{21}$ (area of $R_{21}$)

+ $m_{22}$(area of $R_{22}$) + $m_{31}$ (area of $R_{31}$ ) + $m_{32}$(area of $R_{32}$)
MATH

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

Definition of the double integral over a rectangle $R$

The unique number $I$ that satisfies the inequality
MATH
is called the double integral of $f$ over $R$ and is denoted by
MATH

The Double Integral as a Volume

If $f$ is nonnegative on the rectangle $R$, the equation
MATH
represents a surface that lies above $R$. In this case the double integral
MATH
gives the volume of the solid that is bounded below by $R$ and bounded above by the surface $z=f(x,y).$

To see this, consider a partition $P$ of $R$. $P$ breaks up $R$ into subrectangles $R_{ij}$; and thus the entire solid $T$ into parts $T_{ij}$. Since $T_{ij}$ contains a rectangular solid with base $R_{ij}$ and height $m_{ij}$ , we must have

$m_{ij}$(area of $R_{ij}$) $\le $ volume of $T_{ij}$.

Since $T_{ij}$ is contained in a rectangular solid with base $R_{ij}$ and height $M_{ij}$, we must have

volume of $T_{ij}\le M_{ij}$(area of $R_{ij}$).

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

$m_{ij}$(area of $R_{ij}$) $\le $ volume of $T_{ij}\le M_{ij}$(area of $R_{ij}$).

Adding up these inequalities, we are forced to conclude that

$L_f(P)\le $ volume of $T$ $\le $ $U_f(P)$.

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

volume of $T$ =MATH

The double integral
MATH
gives the volume of a solid of constant height $1$ erected over $R$. In square units this is just the area of $R$:
MATH

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
MATH
where $R$ is the rectangle $[a,b]\times [c,d].$

Solution. Here
MATH
We begin with a partition
MATH
and a partition
MATH
This gives
MATH
as an arbitrary partition of $R.$ On each rectangle $R_{ij}$, $f$ has constant value $\alpha $. This gives $M_{ij}=\alpha $ and $m_{ij}=\alpha $ throughout. Thus
MATH

MATH
Similarly
MATH
Since
MATH
and $P$ was chosen arbitrarily, the inequality must hold for all partitions $P$ of $R$. This means that
MATH

Remark. If $\alpha >0,$
MATH
gives the volume of the rectangular solid of constant height $\alpha $ erected over the rectangle $R$.

Problem 3. Evaluate
MATH
where MATH

Solution. With with a partition
MATH
and a partition
MATH

we have
MATH
as an arbitrary partition of $R.$ On each rectangle $R_{ij}$ the function
MATH
has a maximum
MATH
and a minimum
MATH
Thus
MATH
and
MATH
For each pair of indices $i$ and $j$,
MATH
This means that for arbitrary $P$ we have
MATH

The middle term can be written
MATH

The first double sum reduces to
MATH

MATH
The second double sum reduces to
MATH

MATH
The sum of these two numbers
MATH
satisfies the inequality
MATH
The integral is therefore $2$:
MATH

Remark. This last integral should not be interpreted as a volume. The expression $y-2x$ does not keep a constant sign on MATH

This document created by Scientific WorkPlace 4.0.