## Qualitative study of a differential equation

Consider the following Cauchy problem:

Prove that any solution of the above problem cannot be bounded on .

**Solution:** First note that is increasing, since for every . If we neglect the initial condition, we may wonder if the differential equation has some constant solutions. Plugging we get

Define and note that , whcih is strictly decreasing with a unique zero at . It is easy to see that and . This implies that has exactly two zeros and since we have . Thus the only constant solutions to the differential equation are and .

Let’s come back to the problem with initial condition . Suppose that is bounded as . This means that there exists . The differential equation then tells us that there exists . Using the mean value theorem on we find a such that . Taking we get , so we must have

This coupled with shows that .

We have obtained that and for . Let’s now use the particularities of this problem, and note that if we differentiate the two members of the differential equation we get

Using the study made above, we can see that as we have

which implies in particular that for every . Define . On one hand we have and on the other . So we have

since . This contradiction implies that the assumption that was bounded towards is false. We could make a similar reasoning towards .

The same reasoning could be applied to any problem of the type , where is a function such that is decreasing with a unique zero, and .