Singular solution

A singular solution of a differential equation is a solution that satisfies the following conditions:

1. It solves the original differential equation.
2. It is tangent to every solution from the family of general solutions of the ODE. By tangent we mean that there is a point x where y_s(x) = y_c(x) and y'_s(x) = y'_c(x) where y_c is any general solution.

Usually, singular solutions appear in differential equations when there is a need to divide in a term that might be equal to zero. Therefore, when one is solving a differential equation and using division one must check what happens if the term is equal to zero, and whether it leads to a singular solution.

Example

Clairaut's equation:

<math> y(x) = x \cdot y' + (y')^2 <math>

We write y' = p and then

<math> y(x) = x \cdot p + (p)^2 <math>

Now, we shall take the differential according to x:

<math> p dx = dy = p ( dx ) + x ( dp ) + 2 p ( dp ) <math>

which by simple algebra yields

<math> 0 = ( 2 p + x )dp <math>

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

If dp=0 it means that y' = p = c = Const and the general solution is:

<math> y_c(x) = c \cdot x + c^2 <math>

where c is determined by the initial value.

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

<math> y_s(x) = -(1/2)x^2 + (-(1/2)x)^2 = -(1/4) \cdot x^2 <math>

Now we shall check whether this a singular solution.

First condition of tangency: y_s(x) = y_c(x) . We solve

<math> c \cdot x + c^2 = y_c(x) = y_s(x) = -(1/4) \cdot x^2 <math>

to find the intersection point, which is (-2c, -c).

Second condition tangency: y'_s(x) = y'_c(x) .
We calculate the derivatives:

<math> y_c'(-2 \cdot c) = c <math>

<math> y_s'(-2 \cdot c) = -(1/2) \cdot x |_{x = -2 \cdot c} = c <math>

We see that both requirements are satisfied and therefore y_s is tangent to general solution y_c. Hence,

<math> y_s(x) = -(1/4) \cdot x^2 <math>

is a singular solution for the family of general solutions

<math> y_c(x) = c \cdot x + c^2 <math>

of this Clairaut equation:

<math> y(x) = x \cdot y' + (y')^2 <math>

Note: The method shown here can be used as general algorithm to solve any Clairaut's equation, i.e. first order ODE of the form

<math> y(x) = x \cdot y' + f(y'). <math>

See also: caustic (mathematics).