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Note: 1–2 lectures, §1.6 in [EP], §2.6 in [BD]

Another type of equation that comes up quite often in physics and engineering is an exact equation. Suppose is a function of two variables, which we call the potential function. The naming should suggest potential energy, or electric potential. Exact equations and potential functions appear when there is a conservation law at play, such as conservation of energy. Let us make up a simple example. Let

We are interested in the lines of constant energy, that is lines where the energy is conserved; we want curves where , for some constant . In our example, the curves are circles. See Figure 1.15.

We take the total derivative of :

For convenience, we will make use of the notation of and . In our example,

We apply the total derivative to , to ﬁnd the diﬀerential equation . The diﬀerential equation we obtain in such a way has the form

An equation of this form is called exact if it was obtained as for some potential function . In our toy example, we obtain the equation

Since we obtained this equation by diﬀerentiating , the equation is exact. We often wish to solve for in terms of . In our example,

An interpretation of the setup is that at each point is a vector in the plane, that is, a direction and a magnitude. As and are functions of , we have a vector ﬁeld. The particular vector ﬁeld that comes from an exact equation is a so-called conservative vector ﬁeld, that is, a vector ﬁeld that comes with a potential function , such that

Let be a path in the plane starting at and ending at . If we think of as force, then the work required to move along is

That is, the work done only depends on endpoints, that is where we start and where we end. For example, suppose is gravitational potential. The derivative of given by is the gravitational force. What we are saying is that the work required to move a heavy box from the ground ﬂoor to the roof, only depends on the change in potential energy. That is, the work done is the same no matter what path we took; if we took the stairs or the elevator. Although if we took the elevator, the elevator is doing the work for us. The curves are those where no work need be done, such as the heavy box sliding along without accelerating or breaking on a perfectly ﬂat roof, on a cart with incredibly well oiled wheels.

An exact equation is a conservative vector ﬁeld, and the implicit solution of this equation is the potential function.

Now you, the reader, should ask: Where did we solve a diﬀerential equation? Well, in applications we generally know and , but we do not know . That is, we may have just started with , or perhaps even

It is up to us to ﬁnd some potential that works. Many diﬀerent will work; adding a constant to does not change the equation. Once we have a potential function , the equation gives an implicit solution of the ODE.

Example 1.8.1: Let us ﬁnd the general solution to . Forget we knew what was.

If we know that this is an exact equation, we start looking for a potential function . We have and . If exists, it must be such that . Integrate in the variable to ﬁnd

(1.5) |

for some function . The function is the “constant of integration”, though it is only constant as far as is concerned, and may still depend on . Now diﬀerentiate (1.5) in and set it equal to , which is what is supposed to be:

Integrating, we ﬁnd . We could add a constant of integration if we wanted to, but there is no need. We found . Next for a constant , we solve

for in terms of . In this case, we obtain as we did before.

Exercise 1.8.1: Why did we not need to add a constant of integration when integrating ? Add a constant of integration, say , and see what you get. What is the diﬀerence from what we got above, and why does it not matter?

The procedure, once we know that the equation is exact, is:

- (i)
- Integrate in resulting in .
- (ii)
- Diﬀerentiate this in , and set that equal to , so that we may ﬁnd by integration.

The procedure can also be done by ﬁrst integrating in and then diﬀerentiating in . Pretty easy huh? Let’s try this again.

OK, so and . We try to proceed as before. Suppose exists. Then . We integrate:

for some function . Diﬀerentiate in and set equal to :

But there is no way to satisfy this requirement! The function cannot be written as plus a function of . The equation is not exact; no potential function exists.

Is there an easier way to check for the existence of , other than failing in trying to ﬁnd it? Turns out there is. Suppose and . Then as long as the second derivatives are continuous,

Let us state it as a theorem. Usually this is called the Poincarè Lemma^{5}.

Theorem 1.8.1 (Poincarè). If and are continuously diﬀerentiable functions of , and . Then near any point there is a function such that and .

The theorem doesn’t give us a global deﬁned everywhere. In general, we can only ﬁnd the potential locally, near some initial point. By this time, we have come to expect this from diﬀerential equations.

Let us return to the example above where and . Notice and , which are clearly not equal. The equation is not exact.

We write the equation as

so and . Then

The equation is exact. Integrating in , we ﬁnd

Diﬀerentiating in and setting to , we ﬁnd

So , and will work. Take . We wish to solve . First let us ﬁnd . As then . Therefore , so . Now we solve for to get

We leave to the reader to check that .

This vector ﬁeld is not conservative if considered as a vector ﬁeld of the entire plane minus the origin. The problem is that if a curve was a circle around the origin, say starting at and ending at going counterclockwise, then if existed we would expect

That is nonsense! We leave the computation of the path integral to the interested reader, or you can consult your multivariable calculus textbook. So there is no potential function deﬁned everywhere outside the origin .

If we think back to the theorem, it did not guarantee such a function anyway. It only guaranteed a potential function locally. As we start at the point . Considering and integrating in or in we ﬁnd

So the implicit solution is . Solving, . That is, the solution is a straight line. Solving gives us that , and so is the desired solution. See Figure 1.16, and note that the solution only exists for .

The reader should check that this equation is exact. Let and . We follow the procdure for exact equations

and

Therefore or and . We try to solve . We easily solve for and then just take the square root:

When , the term in front of vanishes. You can also see that our solution is not valid in that case. However, one could in that case try to solve for in terms of starting from the implicit solution . In this case the solution is somewhat messy and we leave it as implicit.

Sometimes an equation is not exact, but it can be made exact by multiplying with a function . That is, perhaps for some nonzero function ,

is exact. Any solution to this new equation is also a solution to .

In fact, a linear equation

is always such an equation. Let be the integrating factor for a linear equation. Multiply the equation by and write it in the form of .

Then , so , while , so . In other words, we have an exact equation. So integrating factors for linear functions are just a special case of integrating factors for exact equations.

But how do we ﬁnd the integrating factor ? Well, given an equation

should be a function such that

Therefore,

At ﬁrst it may seem we replaced one diﬀerential equation by another one. True, but hope is not lost.

A strategy that often works is to look for a that is a function of alone, or a function of alone. If is a function of alone, that is , then we write instead of , and is just zero. Then

In particular, ought to be a function of alone (not depend on ). If so, then we have a linear equation

Letting , we solve using the standard integrating factor method, to ﬁnd . The constant in the solution is not relevant, we need any nonzero solution, we take . So is the integrating factor.

Similarly we could try a function of the form . Then

In particular ought to be a function of alone. If so, then we have a linear equation

Letting we ﬁnd , and we can take . So is the integrating factor.

Let and . Compute

As this is not zero, the equation is not exact. We notice

is a function of alone. We compute the integrating factor

We multiply our given equation by to obtain

which is an exact equation that we solved in Example 1.8.5. The solution was

First compute

As this is not zero, the equation is not exact. We observe

is a function of alone. We compute the integrating factor

Therefore we look at the exact equation

The reader should double check that this equation is exact. We follow the procdure for exact equations

and

(1.6) |

Consequently or . Thus . It is not possible to solve for in terms of elementary functions, so let us be content with the implicit solution:

We are looking for the general solution and we divided by above. We should check what happens when , as the equation itself makes perfect sense in that case. We plug in to ﬁnd the equation is satisﬁed. So is also a solution.

Exercise 1.8.2: Solve the following exact equations, implicit general solutions will suﬃce:

a)

b)

c)

d)

Exercise 1.8.3: Find the integrating factor for the following equations making them into exact equations:

a)

b)

c)

d)

Exercise 1.8.4: Suppose you have an equation of the form: .

a) Show it is exact.

b) Find the form of the potential function in terms of and .

Exercise 1.8.5: Suppose that we have the equation .

a) Is this equation exact?

b) Find the general solution using a deﬁnite integral.

Exercise 1.8.6: Find the potential function of the exact equation in two diﬀerent ways.

a) Integrate in terms of and then diﬀerentiate in and set to .

b) Integrate in terms of and then diﬀerentiate in and set to .

Exercise 1.8.7: A function is said to be harmonic function if .

a) Show that is an exact equation. Therefore there exists (at least locally) the so-called harmonic conjugate function such that and .

Verify that the following are harmonic and ﬁnd the corresponding harmonic conjugates :

b)

c)

d)

Exercise 1.8.101: Solve the following exact equations, implicit general solutions will suﬃce:

a)

b)

c)

d)

Exercise 1.8.102: Find the integrating factor for the following equations making them into exact equations:

a)

b)

c)

d)

Exercise 1.8.103: a) Show that every separable equation can be written as an exact equation, and verify that it is indeed exact. b) Using this rewrite as an exact equation, solve it and verify that the solution is the same as it was in Example 1.3.1.

^{5}Named for the French polymath Jules Henri Poincarè (1854–1912).