Formation of first order PDE

Let $\psi_1(x,y,u)$ and $\psi_2(x,y,u)$ be two given functions. Let $F$ be an arbitrary function of $\psi_1$ and $\psi_2$ in the form

$\displaystyle F(\psi_1,\psi_2)=0.$

(1.3)

We can form a differential equation by eliminating the arbitrary function $F$

. Differentiate Eq.(

) w.r.t.

and

gives,

$\begin{displaymath}\begin{split} \frac{\partial F}{\partial \psi_1}\bigg[\frac{\... ...}{\partial u}\frac{\partial u}{\partial y}\bigg]=0. \end{split}\end{displaymath}$

(1.4)

For the existence of solution

$\displaystyle \begin{vmatrix} \frac{\partial \psi_1}{\partial x}+\frac{\partial... ...psi_2}{\partial y}+\frac{\partial \psi_2}{\partial u}u_y \par \end{vmatrix}= 0$

Therefore,

$\displaystyle (\psi_{1,x}+\psi_{1,u}u_x)(\psi_{2,y}+\psi_{2,u}u_y)-(\psi_{2,x}+\psi_{2,u}u_x)(\psi_{1,y}+\psi_{1,u}u_y)=0,$

This simplifies to

$\displaystyle \boxed{ u_x\frac{\partial(\psi_1,\psi_2)}{\partial(y,u)}+ u_y\fra... ...(\psi_1,\psi_2)}{\partial(u,x)}= \frac{\partial(\psi_1,\psi_2)}{\partial(x,y)}}$

(1.5)

This gives rise to

$\displaystyle \boxed{ a(x,y,u)u_x+ b(x,y,u)u_y=c(x,y,u),}$

(1.6)

where

$\displaystyle a=\frac{\partial(\psi_1,\psi_2)}{\partial(y,u)}~b=\frac{\partial(\psi_1,\psi_2)}{\partial(u,x)}~$ and $\displaystyle ~\frac{\partial(\psi_1,\psi_2)}{\partial(x,y)}.$

Example 1.2.1 Form the PDE by eliminating the arbitrary function from

$\displaystyle F(x+y+u,x^2+y^2+u^2)=0.$

Solution Let $\psi_1=x+y+u~$ and $~\psi_2=x^2+y^2+u^2$ and differentiating $F$

w.r.t.

and

, one get

$\begin{displaymath}\begin{split} F_{\psi_1}(1+u_x)+F_{\psi_2}(2x+2uu_x) &=0,\\ F_{\psi_2}(1+u_y)+F_{\psi_2}(2y+2uu_y)&=0,\nonumber \end{split}\end{displaymath}$