Ordinary Differential Equations/Preliminaries from calculus

In this section, we shall do some preparations that will come in handy later, when we need them in order to prove existence/uniqueness theorems. This is since those do rely heavily on some techniques from calculus, which may not usually be taught within a calculus course. Hence this section.

We shall begin with very useful estimation inequalities, called Gronwall's inequalities or inequalities of Gronwall type. These allow us, if we are given one type of estimation (involving an integral with a product of functions), to conclude another type of estimation (involving the exponential function).

Gronwall's inequalitiesEdit

Theorem 2.1 (Right Gronwall's inequality):

Let   and let   such that for all  


Then for all  , also



We define a new function by


By the fundamental theorem of calculus, we immediately obtain


where the inequality follows from the assumption on  . From this follows that


We may now multiply both sides of the equation by   and use the equation

  (by the product and chain rules)

to justify


Hence, the function


is non-increasing. Furthermore, if we set   in that function, we obtain




From   (assumption) follows the claim. 

This result was for functions extending from   to the right. An analogous result holds for functions extending from   to the left:

Theorem 2.2 (Left Gronwall's inequality):

Let   and   such that for all  


then for all  


Note that this time we are not integrating from   to  , but from   to  . This is more natural either, since this means we are integrating in positive direction.

Proof 1:

We rewrite the proof of theorem 12.1 for our purposes.

This time, we set


reversing the order of integration in contrast to the last proof.

Once again, we get  . This time we use


and multiply   by   to obtain


which is why


is non-decreasing. Now inserting   in the thus defined function gives


and thus for  


Proof 2:

We prove the theorem from theorem 12.1. Indeed, for   we set   and  . Then we have


by the substitution  . Hence, we obtain by theorem 12.1, that


for  . Therefore, if now  ,


The Arzelà–Ascoli theoremEdit

Theorem 2.3 (Arzelà–Ascoli):

Let   be a sequence of functions defined on an interval   which is

  • equicontinuous (that is, for any   there exists   such that  ) and
  • uniformly bounded (that is, there exists   such that  ).

Then   contains a uniformly convergent subsequence.


Let   be an enumeration of the set  . The set   is bounded, and hence has a convergent subsequence   due to the Heine–Borel theorem. Now the sequence   also has a convergent subsequence  , and successively we may define   in that way.

Set   for all  . We claim that the sequence   is uniformly convergent. Indeed, let   be arbitrary and let   such that  .

Let   be sufficiently large that if we order   ascendingly, the maximum difference between successive elements is less than   (possible since   is dense in  ).

Let   be sufficiently large that for all   and    .

Set  , and let  . Let   be arbitrary. Choose   such that   (possible due to the choice of  ). Due to the choice of  , the choice of   and the triangle inequality we get


Hence, we have a Cauchy sequence, which converges due to the completeness of  . 

Convergence considerationsEdit

In this section, we shall prove two or three more or less elementary results from analysis, which aren't particular exciting, but useful preparations for the work to come.

Theorem 2.4:

Let   be a sequence of functions defined on an interval  , whose image is contained within a compact set  , let   be a continuous function, and assume further that   uniformly. Then


Proof: Let   be arbitrary. Since   is a continuous function defined on a compact set, it is even uniformly continuous (this is due to the Heine–Cantor theorem). This means that we may pick   such that   for all  . Since   uniformly, we may pick   such that for all   and  ,  . Then we have for   and   that


The next result is very similar; it is an extension of the former theorem making   time-dependent.

Theorem 2.5:

Let   be a sequence of functions defined on an interval  , whose image is contained within a compact set   such that   uniformly, and let this time   be a function from   to  . Then


uniformly in  .


First, we note that the set   is compact. This can be seen either by noting that this set is still bounded and closed, or by noting that for a sequence in this space, we may first choose a convergent subsequence of the "induced" sequence of   and then a convergent subsequence of what's left in   (or the other way round).

Thus, the function   is uniformly continuous as before. Hence, we may choose   such that   implies   (note that   is a norm on   and since this space is still finite-dimensional, all norms there are equivalent; at least to the norm with respect to which continuity is measured).

Since   uniformly, we may pick   such that for all   and  ,  . Then for   and all  , we have


Banach's fixed-point theoremEdit

We shall later give two proofs of the Picard-Lindelöf existence of solutions theorem; one can be given using the machinery above, whereas a different one rests upon the following result by Stefan Banach.

Theorem 2.6:

Let   be a complete metric space, and let   be a strict contraction; that is, there exists a constant   such that


Then   has a unique fixed point, which means that there is a unique   such that  . Furthermore, if we start with a completely arbitrary point  , then the sequence


converges to  .


First, we prove uniqueness of the fixed point. Assume   are both fixed points. Then


Since  , this implies  .

Now we prove existence and simultaneously the claim about the convergence of the sequence  . For notation, we thus set   and if   is already defined, we set  . Then the sequence   is nothing else but the sequence  .

Let  . We claim that


Indeed, this follows by induction on  . The case   is trivial, and if the claim is true for  , then  .

Hence, by the triangle inequality,


The latter expression goes to zero as   and hence we are dealing with a Cauchy sequence. As we are in a complete metric space, it converges to a limit  . This limit further is a fixed point, as the continuity of   (  is Lipschitz continuous with constant  ) implies