Math 246, Fall 2009, Professor David Levermore
7.
Exact Differential Forms and Integrating Factors
Let us ask the following question. Given a firstorder ordinary equation in the form
(7.1)
d
y
d
x
=
f
(
x, y
)
,
when do its solutions satisfy a relation of the form
H
(
x, y
) =
c
where
c
is an arbitrary constant ?
Such an
H
(
x, y
) is called an
integral
of (7.1).
This question is easily answered if we assume that all functions involved are as differntiable
as we need. Suppose that such an
H
(
x, y
) exists, and that
y
=
Y
(
x
) is a solution of differential
equation (7.1). Then
H
(
x, Y
(
x
)) =
H
(
x
I
, Y
(
x
I
))
,
where
x
I
is any point in the interval of definition of
Y
. By differentiating this equation with
respect to
x
we find that
∂
x
H
(
x, Y
(
x
)) +
Y
′
(
x
)
∂
y
H
(
x, Y
(
x
)) = 0
.
Therefore, wherever
∂
y
H
(
x, Y
(
x
))
negationslash
= 0 we see that
Y
′
(
x
) =
−
∂
x
H
(
x, Y
(
x
))
∂
y
H
(
x, Y
(
x
))
.
For this to hold for every solution of (7.1), we must have
d
y
d
x
=
−
∂
x
H
(
x, y
)
∂
y
H
(
x, y
)
,
or equivalently
(7.2)
f
(
x, y
) =
−
∂
x
H
(
x, y
)
∂
y
H
(
x, y
)
,
wherever
∂
y
H
(
x, y
)
negationslash
= 0. The question then arises as to whether we can find an
H
(
x, y
) such
that (7.2) holds for any given
f
(
x, y
)? It turns out that this cannot always be done. In this
section we explore how to seek such an
H
(
x, y
).
7.1.
Exact Differential Forms.
The starting point is to write equation (7.1) in a socalled
differential form
(7.3)
M
(
x, y
) d
x
+
N
(
x, y
) d
y
= 0
,
where
f
(
x, y
) =
−
M
(
x, y
)
N
(
x, y
)
.
There is not a unique way to do this. Just pick one that looks natural. If you are lucky then
there will exist a function
H
(
x, y
) such that
(7.4)
∂
x
H
(
x, y
) =
M
(
x, y
)
,
∂
y
H
(
x, y
) =
N
(
x, y
)
.
When this is the case the differential form (7.3) is said to be
exact
.
1
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It turns out that there is test you can easily apply to find out if you are lucky.
It de
rives from the fact that “mixed partials commute” — namely, the fact that for any twice
differentiable
H
(
x, y
) one has
∂
y
(
∂
x
H
(
x, y
)
)
=
∂
x
(
∂
y
H
(
x, y
)
)
.
This fact implies that if (7.4) holds for some twice differentiable
H
(
x, y
) then
M
(
x, y
) and
N
(
x, y
) satisfy
∂
y
M
(
x, y
) =
∂
y
(
∂
x
H
(
x, y
)
)
=
∂
x
(
∂
y
H
(
x, y
)
)
=
∂
x
N
(
x, y
)
.
In other words, if the differential form (7.3) is exact then
M
(
x, y
) and
N
(
x, y
) satisfy
(7.5)
∂
y
M
(
x, y
) =
∂
x
N
(
x, y
)
.
The remarkable fact is that the converse holds too.
Namely, if the differential form (7.3)
satisfies (7.5) for every (
x, y
) then it is exact — i.e. there exists an
H
(
x, y
) such that (7.4)
holds. Moreover, the problem of finding
H
(
x, y
) is reduced to evaluating two integrals. We
illustrate this fact with examples.
Example:
Solve the initialvalue problem
d
y
d
x
+
e
x
y
+ 2
x
2
y
+
e
x
= 0
,
y
(0) = 0
.
Solution:
Express this equation in the differential form
(
e
x
y
+ 2
x
) d
x
+ (2
y
+
e
x
) d
y
= 0
.
Because
∂
y
(
e
x
y
+ 2
x
) =
e
x
=
∂
x
(2
y
+
e
x
) =
e
x
,
this differential form satisfies (7.5) and is thereby
exact
. We can therefore find
H
(
x, y
) such
that
(7.6)
∂
x
H
(
x, y
) =
e
x
y
+ 2
x ,
∂
y
H
(
x, y
) = 2
y
+
e
x
.
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 Spring '10
 LEVERMORE
 Derivative, Partial differential equation, Differential form, De Rham cohomology

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