Singular solution

Singular solution

A singular solution "ys"("x") of an ordinary differential equation is a solution that is tangent to every solution from the family of general solutions. By "tangent" we mean that there is a point "x" where "ys"("x") = "yc"("x") and "y's"("x") = "y'c"("x") where "yc" 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

Consider the following Clairaut's equation:

: y(x) = x cdot y' + (y')^2 ,!

where primes denote derivatives with respect to "x". We write "y' = p" and then

: y(x) = x cdot p + (p)^2. ,!

Now, we shall take the differential according to "x":

: p = y' = p + x p' + 2 p p' ,!

which by simple algebra yields

: 0 = ( 2 p + x )p'. ,!

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 is:

: y_c(x) = c cdot x + c^2 ,!

where "c" is determined by the initial value.

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

: y_s(x) = -(1/2)x^2 + (-(1/2)x)^2 = -(1/4) cdot x^2. ,!

Now we shall check whether this is a singular solution.

First condition of tangency: "ys"("x") = "yc"("x"). We solve

: c cdot x + c^2 = y_c(x) = y_s(x) = -(1/4) cdot x^2 ,!

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

Second condition tangency: "y's"("x") = "y'c"("x").

We calculate the derivatives:

: y_c'(-2 cdot c) = c ,!: y_s'(-2 cdot c) = -(1/2) cdot x |_{x = -2 cdot c} = c. ,!

We see that both requirements are satisfied and therefore "ys" is tangent to general solution "yc". Hence,

: y_s(x) = -(1/4) cdot x^2 ,!

is a singular solution for the family of general solutions

: y_c(x) = c cdot x + c^2 ,!

of this Clairaut equation:

: y(x) = x cdot y' + (y')^2. ,!

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

: y(x) = x cdot y' + f(y'). ,!

ee also

* caustic (mathematics).


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