Singular solution

Consider the homogeneous linear ordinary differential equation

${\displaystyle xy'(x)+2y(x)=0,\,\!}$

where primes denote derivatives with respect to x. The general solution to this equation is

${\displaystyle y(x)=Cx^{-2}.\,\!}$

For a given ${\displaystyle C}$, this solution is smooth except at ${\displaystyle x=0}$ where the solution is divergent. Furthermore, for a given ${\displaystyle x\not =0}$, this is the unique solution going through ${\displaystyle (x,y(x))}$.

Consider the differential equation

${\displaystyle y'(x)^{2}=4y(x).\,\!}$

A one-parameter family of solutions to this equation is given by

${\displaystyle y_{c}(x)=(x-c)^{2}.\,\!}$

Another solution is given by

${\displaystyle y_{s}(x)=0.\,\!}$

Since the equation being studied is a first-order equation, the initial conditions are the initial x and y values. By considering the two sets of solutions above, one can see that the solution fails to be unique when ${\displaystyle y=0}$. (It can be shown that for ${\displaystyle y>0}$ if a single branch of the square root is chosen, then there is a local solution which is unique using the Picard–Lindelöf theorem.) Thus, the solutions above are all singular solutions, in the sense that solution fails to be unique in a neighbourhood of one or more points. (Commonly, we say "uniqueness fails" at these points.) For the first set of solutions, uniqueness fails at one point, ${\displaystyle x=c}$, and for the second solution, uniqueness fails at every value of ${\displaystyle x}$. Thus, the solution ${\displaystyle y_{s}}$ is a singular solution in the stronger sense that uniqueness fails at every value of x. However, it is not a singular function since it and all its derivatives are continuous.

In this example, the solution ${\displaystyle y_{s}(x)=0}$ is the envelope of the family of solutions ${\displaystyle y_{c}(x)=(x-c)^{2}}$. The solution ${\displaystyle y_{s}}$ is tangent to every curve ${\displaystyle y_{c}(x)}$ at the point ${\displaystyle (c,0)}$.

The failure of uniqueness can be used to construct more solutions. These can be found by taking two constant ${\displaystyle c_{1}<c_{2}}$ and defining a solution ${\displaystyle y(x)}$ to be ${\displaystyle (x-c_{1})^{2}}$ when ${\displaystyle x<c_{1}}$, to be ${\displaystyle 0}$ when ${\displaystyle c_{1}\leq x\leq c_{2}}$, and to be ${\displaystyle (x-c_{2})^{2}}$ when ${\displaystyle x>c_{2}}$. Direct calculation shows that this is a solution of the differential equation at every point, including ${\displaystyle x=c_{1}}$ and ${\displaystyle x=c_{2}}$. Uniqueness fails for these solutions on the interval ${\displaystyle c_{1}\leq x\leq c_{2}}$, and the solutions are singular, in the sense that the second derivative fails to exist, at ${\displaystyle x=c_{1}}$ and ${\displaystyle x=c_{2}}$.

The previous example might give the erroneous impression that failure of uniqueness is directly related to ${\displaystyle y(x)=0}$. Failure of uniqueness can also be seen in the following example of a Clairaut's equation:

${\displaystyle y(x)=x\cdot y'+(y')^{2}\,\!}$

We write y' = p and then

${\displaystyle y(x)=x\cdot p+(p)^{2}.\,\!}$

Now, we shall take the differential according to x:

${\displaystyle p=y'=p+xp'+2pp'\,\!}$

which by simple algebra yields

${\displaystyle 0=(2p+x)p'.\,\!}$

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 of this new equation is:

${\displaystyle y_{c}(x)=c\cdot x+c^{2}\,\!}$

where c is determined by the initial value.

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

${\displaystyle y_{s}(x)=-(1/2)x^{2}+(-(1/2)x)^{2}=-(1/4)\cdot x^{2}.\,\!}$

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

${\displaystyle c\cdot x+c^{2}=y_{c}(x)=y_{s}(x)=-(1/4)\cdot x^{2}\,\!}$

to find the intersection point, which is ${\displaystyle (-2c,-c^{2})}$.

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

${\displaystyle y_{c}'(-2\cdot c)=c\,\!}$

${\displaystyle y_{s}'(-2\cdot c)=-(1/2)\cdot x|_{x=-2\cdot c}=c.\,\!}$

Hence,

${\displaystyle y_{s}(x)=-(1/4)\cdot x^{2}\,\!}$

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

${\displaystyle y_{c}(x)=c\cdot x+c^{2}\,\!}$

of this Clairaut equation:

${\displaystyle y(x)=x\cdot y'+(y')^{2}.\,\!}$

• Chandrasekhar equation
• Chrystal's equation
• Caustic (mathematics)
• Envelope (mathematics)
• Initial value problem
• Picard–Lindelöf theorem

• Rozov, N.Kh. (2001) [1994], "Singular solution", Encyclopedia of Mathematics, EMS Press