One of the many good things about my teaching career is that as I teach across the curriculum, I fill in the gaps of my own education.

I got my Ph. D. in topology (low dimensional manifolds; in particular, knot theory) and hadn’t seen much of differential equations beyond my “engineering oriented” undergraduate course.

Therefore, I learned more about existence and uniqueness theorems when I taught differential equations; though I never taught the existence and uniqueness theorems in a class, I learned the proofs just for my own background. In doing so I learned about the Picard iterated integral technique for the first time; how this is used to establish “uniqueness of solution” can be found here.

However I recently discovered (for myself) what thousands of mathematicians already know: the Picard process can be used to yield an interval of existence for a solution for a differential equation, even if we cannot obtain the solution in closed form.

**The situation**

I assigned my numerical methods class to solve with and to produce the graph of from to .

There is a unique solution to this and the solution is valid so long as the and value of the solution curve stays finite; note that

So, is it possible that the values for this solution become unbounded?

Answer: yes.

**What follows are the notes I gave to my class.**

Numeric output seems to indicate this, but numeric output is NOT proof.

To find a proof of this, let’s turn to the Picard iteration technique. We

know that the Picard iterates will converge to the unique solution.

The integrals get pretty ugly around here; I used MATLAB to calculate the

higher order iterates. I’ll show you

where means assorted polynomial terms from order 6 to 11.

Here is one more:

We notice some patterns developing here. First of all, the coefficient of

the term is staying the same for all where

That is tedious to prove. But what is easier to show (and sufficient) is

that the coefficients for the terms for all appear to be

bigger than 1. This is important!

Why? If we can show that this is the case, then our ”limit” solution will have an interval of convergence less than 1. Why? Substitute and see that the sum

diverges because the not only fail to converge to zero, but they

stay greater than 1.

So, can we prove this general pattern?

YES!

Here is the idea: where is a polynomial of order

and is a polynomial whose terms all have order or greater.

Now put into the Picard process:

Note: all of the terms of of degree or higher must come from

the second integral.

Now by induction we can assume that all of the coefficients of the

polynomial are greater than or equal to one.

When we ”square out” the polynomial, the coefficients of the new

polynomial will consist of the sum of positive numbers, each of which is

greater than 1. For the coefficients of the polynomial of

degree or higher: if one is interested in the

coefficient, one has to add at least numbers together, each of which

is bigger than one.

Now when one does the integration on these particular terms, one, of course,

divides by (power rule for integration). But that means that the

coefficient (after integration) is then greater than 1.

Here is a specific example:

Say

Now

Remember that are all greater than or equal to one.

Now

Now when we integrate term by term, we get:

But note that and

Since all of the factors are greater than or equal to 1.

Hence in our new polynomial approximation, the order 4 terms or less all

have coefficients which are greater than or equal to one.

We can make this into a Proposition:

**Proposition**

Suppose where each

If

Then for all

Proof. Of course, and

Let

Then we can calculate: (since all of the are

defined):

If is odd, then

If is even then

The Proposition is proved.

Of course, this possibly fails for where as we would fail to

have a sufficient number of terms in our sum.

Now if one wants a challenge, one can modify the above arguments to show that the coefficients of the approximating polynomial never get ”too big”; that is, the coefficient of the order term is less than, say, .

It isn’t hard to show that where

Then one can compare to the derivative of the geometric series to show that

one gets convergence on an interval up to but not including 1.