Implicit ODEs (Examples)

This is a combination of the Section 5 of Chapter 2 of Arnold’s ODE and Section 3 of his Geometrical Methods in the Theory of Ordinary Differential Equations.

Currently working through examples of Clairaut’s equation.

Example 1:

Example 2:

Understanding problem 2 (p. 115):

Question: What does it mean to take p and x as coordinates on E? Does it mean to project the surface E onto the (p, x) plane?

To find the trace of the contact plane () at each point on the surface E (defined by ), we need to determine the tangent plane at that point.

The tangent plane is given by

.

which gives:

Hence, we have:

as the equation of the tangent plane.

The trace of the contact plane is the intersection between the contact plane and the tangent plane. That is, we also have additional relation . Substituting this relation into the equation of the tangent plane gives us the form . Note that this form describes a line at every point (p, x) on the (p, x)-plane. Also note that this form does not vanish at any point on the (p, x)-plane. The resulting integral curves on the (p, x) plane are (Question: are these also the equations describing the integral curves on the surface of E?).

Projecting the integral curves back onto the surface E, we can express them using equations in terms of x and t (because E is described by ):

which is equivalent to:

Problem 3 on p. 116:

Question: Why did Arnold take p and x as coordinates on E in the previous question, but changed to take p and t as coordinates in this question?

Ans: This is due to the form obtained for the traces of the contact plane. In this question, the form is independent of x and hence could directly projected as a line described by this same form on the (p, t)-plane.

Finding the traces of the contact plane follows similar procedure from previous question (being lazy, I am not going to type this out).

Unlike the previous problem, does vanish when t = f’. What does this mean? It means that the contact plane completely coincides with the tangent plane at these points (called the exceptional points).

The set of exceptional points form a curve. On the surface E, it is parameterized by (f'(C)C – f(C), f'(C), C). When projected onto the (t, x)-plane, the curve can be parameterized as stated in Arnold’s solution: t = f'(C) and x = Ct – f(C).

Now, let’s return back to finding the integral curves from the form .