I’ve called this section Intervals of Validity because all
of the examples will involve them.
However, there is a lot more to this section. We will see a couple of theorems that will
tell us when we can solve a differential equation. We will also see some of the differences
between linear and nonlinear differential equations.
First let's take a look at a theorem about linear first
order differential equations. This is a
very important theorem although we’re not going to really use it for its most
Consider the following IVP.
If p(t) and g(t) are continuous functions on an
open interval and the interval contains to, then there is a unique
solution to the IVP on that interval.
So, just what does this theorem tell us? First, it tells us
that for nice enough linear first order differential equations solutions are
guaranteed to exist and more importantly the solution will be unique. We may
not be able to find the solution, but do know that it exists and that there
will only be one of them. This is the
very important aspect of this theorem.
Knowing that a differential equation has a unique solution is sometimes
more important than actually having the solution itself!
Next, if the interval in the theorem is the largest possible
interval on which p(t) and g(t) are continuous then the interval is
the interval of validity for the solution. This means, that for linear first
order differential equations, we won't need to actually solve the differential
equation in order to find the interval of validity. Notice as well that the
interval of validity will depend only partially on the initial condition. The
interval must contain to,
but the value of yo, has
no effect on the interval of validity.
Let’s take a look at an example.
Example 1 Without
solving, determine the interval of validity for the following initial value
First, in order to use the theorem to find the interval of
validity we must write the differential equation in the proper form given in
the theorem. So we will need to divide out by the coefficient of the
Next, we need to identify where the two functions are not
continuous. This will allow us to find all possible intervals of validity for
the differential equation. So, p(t)
will be discontinuous at since these points will give a division by
zero. Likewise, g(t)
will also be discontinuous at as well as t = 5 since at this point
we will have the natural logarithm of zero. Note that in this case we won't
have to worry about natural log of negative numbers because of the absolute
Now, with these points in hand we can break up the real
number line into four intervals where both p(t) and g(t) will be continuous. These four intervals
The endpoints of each of the intervals are points where at
least one of the two functions is discontinuous. This will guarantee that
both functions are continuous everywhere in each interval.
Finally, let's identify the actual interval of validity
for the initial value problem. The actual interval of validity is the interval
that will contain to =
4. So, the interval of validity for the initial value problem
In this last example we need to be careful to not jump to
the conclusion that the other three intervals cannot be intervals of
validity. By changing the initial
condition, in particular the value of to,
we can make any of the four intervals the interval of validity.
The first theorem required a linear differential
equation. There is a similar theorem for
non-linear first order differential equations.
This theorem is not as useful for finding intervals of validity as the
first theorem was so we won’t be doing all that much with it.
Here is the theorem.
That’s it. Unlike the
first theorem, this one cannot really be used to find an interval of
validity. So, we will know that a unique
solution exists if the conditions of the theorem are met, but we will actually
need the solution in order to determine its interval of validity. Note as well that for non-linear differential
equations it appears that the value of y0
may affect the interval of validity.
Here is an example of the problems that can arise when the
conditions of this theorem aren’t met.
Example 2 Determine
all possible solutions to the following IVP.
First, notice that this differential equation does NOT
satisfy the conditions of the theorem.
So, the function is continuous on any interval, but the
derivative is not continuous at y =
0 and so will not be continuous at any interval containing y = 0.
In order to use the theorem both must be continuous on an interval
that contains yo = 0 and
this is problem for us since we do have yo
Now, let’s actually work the problem. This differential equation is separable and
is fairly simple to solve.
Applying the initial condition gives c = 0 and so the solution is.
So, we’ve got two possible solutions here, both of which
satisfy the differential equation and the initial condition. There is also a third solution to the
IVP. y(t) = 0 is also a solution to the differential equation and
satisfies the initial condition.
In this last example we had a very simple IVP and it only
violated one of the conditions of the theorem, yet it had three different
solutions. All the examples we’ve worked
in the previous sections satisfied the conditions of this theorem and had a
single unique solution to the IVP. This
example is a useful reminder of the fact that, in the field of differential
equations, things don’t always behave nicely.
It’s easy to forget this as most of the problems that are worked in a
differential equations class are nice and behave in a nice, predictable manner.
Let’s work one final example that will illustrate one of the
differences between linear and non-linear differential equations.
Example 3 Determine
the interval of validity for the initial value problem below and give its
dependence on the value of yo
Before proceeding in this problem, we should note that the
differential equation is non-linear and meets both conditions of the Theorem
2 and so there will be a unique solution to the IVP for each possible value
Also, note that the problem asks for any dependence of the
interval of validity on the value of yo. This immediately illustrates a difference
between linear and non-linear differential equations. Intervals of validity for linear
differential equations do not depend on the value of yo. Intervals
of validity for non-linear differential can depend on the value of yo as we pointed out after
the second theorem.
So, let’s solve the IVP and get some intervals of
First note that if yo
= 0 then y(t)
= 0 is the solution and this has an interval of validity of
So for the rest of the problem let's assume that . Now, the differential equation is separable
so let's solve it and get a general solution.
Applying the initial condition gives
The solution is then.
Now that we have a solution to the initial value problem we
can start finding intervals of validity.
From the solution we can see that the only problem that we’ll have is
division by zero at
This leads to two possible intervals of validity.
That actual interval of validity will be the interval that
contains to = 0. This however, depends on the value of yo. If yo
< 0 then and so the second interval will contain to = 0. Likewise if yo > 0 then and in this case the first interval will
contain to = 0.
This leads to the following possible intervals of
validity, depending on the value of yo.
On a side note, notice that the solution, in its final
form, will also work if yo
So what did this example show us about the difference
between linear and non-linear differential equations?
First, as pointed out in the solution to the example,
intervals of validity for non-linear differential equations can depend on the
value of yo, whereas
intervals of validity for linear differential equations don’t.
Second, intervals of validity for linear differential
equations can be found from the differential equation with no knowledge of the
solution. This is definitely not the
case with non-linear differential equations.
It would be very difficult to see how any of these intervals in the last
example could be found from the differential equation. Knowledge of the solution was required in
order for us to find the interval of validity.