Singular Solution - Further Example of Failure of Uniqueness

Further Example of Failure of Uniqueness

The previous example might give the erroneous impression that failure of uniqueness is directly related to . Failure of uniqueness can also be seen in the following example of a Clairaut's equation:

We write y' = p and then

Now, we shall take the differential according to x:

which by simple algebra yields

This condition is solved if 2p+x=0 or if p'=0.

If p' = 0 it means that y' = p = c = constant, and the general solution of this new equation is:

where c is determined by the initial value.

If x + 2p = 0 than we get that p = −(1/2)x and substituting in the ODE gives

Now we shall check when these solutions are singular solutions. If two solutions intersect each other, that is, they both go through the same point (x,y), then there is a failure of uniqueness for a first-order ordinary differential equation. Thus, there will be a failure of uniqueness if a solution of the first form intersects the second solution.

The condition of intersection is : y_s(x) = y_c(x). We solve

to find the intersection point, which is .

We can verify that the curves are tangent at this point y'_s(x) = y'_c(x). We calculate the derivatives:

Hence,

is tangent to every member of the one-parameter family of solutions

of this Clairaut equation: