The Principle of Duality

There exists a certain transformation, due to Legendre, by which a dual relationship can be set up between one equation of the first order and another of the same order. Let $X$ and $Y$ be new variables defined by the relations

$$X=p,Y=xp-y.$$

and let

$$P=\frac{dY}{dX}.$$

Now, assuming that $\frac{dp}{dx}\ne 0$,

$$dX=dp,dY=xdp+pdx-dy=xdp,$$

and therefore

$$P=x.$$

Also

$$y=xp-Y=XP-Y.$$

Thus the transformation

$$X=p,Y=xp-y$$

is equivalent to

$$x=P,y=XP-Y.$$

They are therefore reciprocally related to one another.¹

By means of this substitution, either of the equations

$$F(x,y,p)=0,F(P,XP-Y,X)=0$$

may be transformed into the other, and in this sense a dual relationship exists between them. When one of the equations is integrable, the other may be integrated by purely algebraical processes.

For instance, let

$$\varphi (X,Y)=0$$

be a solution of the second equation, then on differentiating with respect to $X$,

$$\frac{\partial \varphi }{\partial X}+\frac{\partial \varphi }{\partial Y}P=0.$$

Now $X,Y,P$ may be eliminated between these two equations and

$$x=P,y=XP-Y,$$

thus giving a solution of the equation

$$F(x,y,p)=0.$$

In particular, an equation of the form

$$\mathrm{\Phi }(xp-y)=x\mathrm{\Psi }(p)$$

would become

$$\mathrm{\Phi }(Y)=P\mathrm{\Psi }(X).$$

The variables $X$ and $Y$ are now separable, and the equation is integrable by quadratures.

NOTE. In the case of the Clairaut equation the condition that $\frac{dp}{dx}\ne 0$ is violated for the general solution; this method therefore leads only to the singular solution.

Footnotes

1. If $(x,y)$ and $(X,Y)$ are regarded as points in the plane of the variables $u,v$, the locus of $(x,y)$ is the polar reciprocal of the locus of $(X,Y)$ with respect to the parabola $u^2=2v$, and conversely. ↩