Chrystal's equation

In mathematics, Chrystal's equation is a first order nonlinear ordinary differential equation, named after the mathematician George Chrystal, who discussed the singular solution of this equation in 1896.Chrystal G., "On the p-discriminant of a Differential Equation of the First order and on Certain Points in the General Theory of Envelopes Connected Therewith.", Trans. Roy. Soc. Edin, Vol. 38, 1896, pp. 803–824. The equation reads asDavis, Harold Thayer. Introduction to nonlinear differential and integral equations. Courier Corporation, 1962.Ince, E. L. (1939). Ordinary Differential Equations, London (1927). Google Scholar.

: $\left(\frac{dy}{dx}\right)^2 + Ax \frac{dy}{dx} + By + Cx^2 =0$

where $A,\ B, \ C$ are constants, which upon solving for $dy/dx$ , gives

: $\frac{dy}{dx} = -\frac{A}{2} x \pm \frac{1}{2} (A^2 x^2 - 4By - 4Cx^2)^{1/2}.$

This equation is a generalization particular cases of Clairaut's equation since it reduces to a form of Clairaut's equation under condition as given below.

Solution

Introducing the transformation $4By=(A^2-4C-z^2)x^2$ gives

: $xz\frac{dz}{dx} = A^2 + AB - 4C \pm Bz - z^2.$

Now, the equation is separable, thus

: $\frac{z \, dz}{A^2 + AB - 4C \pm Bz - z^2} = \frac{dx}{x}.$

The denominator on the left hand side can be factorized if we solve the roots of the equation $A^2 + AB - 4C \pm Bz - z^2=0$ and the roots are $a,\ b = \pm \left[ B +\sqrt{(2A+B)^2 - 16C} \right]/2$ , therefore

: $\frac{z \, dz}{-(z-a)(z-b)} = \frac{dx}{x}.$

If $a\neq b$ , the solution is

: $x \frac{(z-a)^{a/(a-b)}}{(z-b)^{b/(a-b)}} = k$

where $k$ is an arbitrary constant. If $a=b$ , ( $(2A+B)^2 - 16C=0$ ) then the solution is

: $x(z-a) \exp \left[\frac a {a-z}\right]=k.$

When one of the roots is zero, the equation reduces to a special-case of Clairaut's equation and a parabolic solution is obtained in this case, $A^2+ AB -4C=0$ and the solution is

: $x(z\pm B)=k, \quad \Rightarrow \quad 4By = - AB x^2 - (k\pm Bx)^2.$

The above family of parabolas are enveloped by the parabola $4By=-ABx^2$ , therefore this enveloping parabola is a singular solution.

References

Category:Eponymous equations of physics

Category:Ordinary differential equations