Clairaut's differential equation

The Clairaut's equation is a nonlinear ordinary differential equation of first order of the form

${\displaystyle y(x)=x\cdot y'(x)+f(y'(x))}$

and is thus a special case of the d'Alembert differential equation . It is named after the French mathematician Alexis-Claude Clairaut .

Determination of some solutions

There are two main types of solutions to Clairaut's differential equation, which are described below. In general, however, these are not all solutions to this differential equation. If there are two different solutions with and , then the function is ${\displaystyle y_{1},y_{2}}$ ${\displaystyle y_{1}(x_{0})=y_{2}(x_{0})}$ ${\displaystyle y_{1}'(x_{0})=y_{2}'(x_{0})}$

${\displaystyle y(x):=\left\{{\begin{array}{ll}y_{1}(x)\ ,&x\leq x_{0}\ ,\\y_{2}(x)\ ,&x>x_{0}\ ,\\\end{array}}\right.}$

Another solution that does not fall into either of the following two solution classes.

Trivial straight line solutions

For each in the domain of are the straight lines ${\displaystyle c}$ ${\displaystyle f}$

${\displaystyle \ y(x):=cx+f(c)}$

Solutions of Clairaut's differential equation.

Non-trivial solutions

Let be differentiable as well as a differentiable function, which of the implicit equation ${\displaystyle f}$ ${\displaystyle c}$

${\displaystyle f\;'(c(x))+x=0}$

enough. Then

${\displaystyle \ y(x):=xc(x)+f(c(x))}$

a solution of Clairaut's differential equation.

proof

The following applies to the straight line ${\displaystyle y'(x)\equiv c}$

${\displaystyle xy'(x)+f(y'(x))=cx+f(c)=y(x)\ .}$

In the case of the nontrivial solutions, the following applies

${\displaystyle {\begin{array}{lll}xy'(x)+f(y'(x))&=&x[xc'(x)+c(x)+f'(c(x))c'(x)]+f(xc'(x)+c(x)+f'(c(x))c'(x))\\&=&xc(x)+f(c(x))=y(x)\ .\\\end{array}}}$

${\displaystyle \Box }$

Relationship between the two types of solutions

The tangents of the nontrivial solutions are trivial straight line solutions.

proof

The tangent of the nontrivial solution through the point is given by the equation ${\displaystyle T}$ ${\displaystyle y(x)=xc(x)+f(c(x))}$ ${\displaystyle (x_{0}\ |\ y(x_{0}))}$

${\displaystyle {\begin{array}{lll}T(x)&=&y'(x_{0})(x-x_{0})+y(x_{0})\\&=&[x_{0}c'(x_{0})+c(x_{0})+f'(c(x_{0}))c'(x_{0})](x-x_{0})+[x_{0}c(x_{0})+f(c(x_{0}))]\\&=&c(x_{0})(x-x_{0})+c(x_{0})x_{0}+f(c(x_{0}))=c(x_{0})x+f(c(x_{0}))\ \\\end{array}}}$

${\displaystyle \Box }$

given. If the nontrivial solution is strictly convex or strictly concave, it divides the plane into an area in which two straight line solutions run through each point and an area that is free of solutions; it is then called the envelope. Solutions are then not only the envelope itself and the line solutions, but also solution curves that run piece by piece on straight lines and piece by piece on the envelope.

literature

• Wolfgang Walter: Ordinary differential equations. 7th edition, Springer Verlag, Berlin 2000, ISBN 3-540-67642-2 , § 4.