Singular Solutions and Envelopes

What is a Singular Solution and Why is it Special

When solving an ODE $y' = f(x, y)$, we usually find the general solution—a family of curves depending on a constant $C$. It seems that by selecting the appropriate $C$, any solution can be obtained. But this is not always the case.

A singular solution is a solution that cannot be obtained from the general solution for any specific value of the constant $C$. It cannot be "fit" into the family, though it satisfies the equation. Such solutions have fundamental physical and geometric significance.

Geometrically, a singular solution is the envelope of a family of integral curves: a curve that, at each of its points, is tangent to at least one integral curve from the general family. At the points of tangency, a branch "splits off" from the general solution—the singular solution.

How to Find Singular Solutions

$p$-discriminant Method: Suppose the general solution is written as $F(x, y, C) = 0$. We seek singular solutions by eliminating $C$ from the system:

$ F(x, y, C) = 0 \quad \text{and} \quad \frac{\partial F}{\partial C} = 0. $

Physical meaning: $\frac{\partial F}{\partial C} = 0$ means that, at this point, the "neighboring" curves of the family converge—these are precisely the points of tangency with the envelope.

$y$-discriminant (or $C$-discriminant) Method: For an equation of the form $y' = p$, we write $g(x, y, p) = 0$ (an equation expressing $p$ in terms of $x$ and $y$). The system $g = 0$ and $\frac{\partial g}{\partial p} = 0$ yields candidates for singular solutions.

Example: Clairaut’s Equation and Its Envelope

Let us return to the equation $y = x y' + (y')^2$. The general solution is $y = Cx + C^2$ (a family of straight lines). Let's write it as $F(x, y, C) = y - Cx - C^2 = 0$.

Condition $\frac{\partial F}{\partial C} = 0$: $-x - 2C = 0$, from which $C = -x/2$.

Substitute into $F = 0$: $y - (-x/2)x - (-x/2)^2 = y + x^2/2 - x^2/4 = y + x^2/4 = 0$.

The singular solution: $y = -x^2/4$, a parabola.

Let’s check that it is indeed a solution of the equation: $y' = -x/2$. Substitute: $x y' + (y')^2 = x(-x/2) + x^2/4 = -x^2/2 + x^2/4 = -x^2/4 = y$. ✓

Geometrically: each straight line $y = Cx + C^2$ touches the parabola $y = -x^2/4$ at exactly one point. The parabola is the envelope of the family of lines.

A Remarkable Example: The Problem of Light

Problem: Find the curve whose normal bisects the distance from the origin to the point of tangency with the $x$-axis. This leads to a Clairaut equation, and the envelope turns out to be an astroid $x^{2/3} + y^{2/3} = R^{2/3}$. An astroid is a curve traced by a point on a circle rolling inside a larger circle.

Caution When Finding Singular Solutions

Important: Not every solution of the system $F = 0$, $\frac{\partial F}{\partial C} = 0$ is a singular solution! One must verify:

1. The curve must be a solution of the ODE (check by substitution).
2. The curve must not coincide with any curve of the general family.
3. Check whether the singular solution is an envelope (i.e., whether every point of the singular solution is indeed a point of tangency with some curve of the general family).

It can happen that the system $F = 0$, $\frac{\partial F}{\partial C} = 0$ yields "extra" curves that are neither singular solutions nor envelopes (for example, return points of the family).

Orthogonal Trajectories

An orthogonal trajectory for a family of curves is a curve that intersects each curve of the family at a right angle. If the family is given by $y' = f(x, y)$, then the orthogonal trajectories satisfy $y' = -1/f(x, y)$.

Application in Electrostatics: The field lines of the electric field are orthogonal to equipotential surfaces. If the equipotentials are known (for example, concentric circles around a point charge), the field lines are found as orthogonal trajectories.

Example: The family of circles $x^2 + y^2 = C$ (equipotentials of a point charge). Differentiate: $2x + 2y y' = 0$ $\rightarrow$ $y' = -x/y$. Orthogonal trajectories: $y' = y/x \rightarrow dy/y = dx/x \rightarrow \ln|y| = \ln|x| + \text{const} \rightarrow y = kx$. These are straight lines through the origin—the field lines of the Coulomb field!

Example from Hydrodynamics: Streamlines of an incompressible fluid flow around an obstacle are orthogonal to potential lines. Finding orthogonal trajectories is a standard problem in hydrodynamics.

Singular Solutions in Mechanics: The Tractrix Problem

A boat on a rope of length $a$ is towed along the shore (the $x$-axis). The boat initially is at the point $(0, a)$. The trajectory of the boat—the tractrix—is determined by a Clairaut equation.

Equation: $y' = -\left(\frac{y}{\sqrt{a^2 - y^2}}\right)$. Solution: $x = a \ln\left(\frac{a + \sqrt{a^2 - y^2}}{y}\right) - \sqrt{a^2 - y^2} + C$.

The tractrix has the remarkable property that the length of the tangent segment from the point of tangency to its intersection with the $x$-axis is constant and equal to $a$. The surface of revolution of the tractrix—the pseudosphere—is a surface of constant negative Gaussian curvature (the analogue of a sphere in Lobachevskian geometry).

Question for reflection: Singular solutions violate the uniqueness in Picard's theorem. This means that through a point on a singular solution, more than one solution of the Cauchy problem passes. What do you think happens physically to the system at such a point?