Finding first integrals of PDE

Question

$$x(y+u)u_x-y(x+u)u_y=(x-y)u$$

Not sure how to find the first inegrals of this, I am told that one first integral is: uxy , however I am unsure how to find this and find the other first integral.

Thanks!

Answer 1

Using the method of characteristics:

$$\frac{dx}{dt}=x(y+u)\\ \frac{dy}{dt}=-y(x+u)\\ \frac{du}{dt}=u(x-y)$$

Now we notice that:

$$\frac{d}{dt}(\ln(uxy))=\frac{1}{x}\frac{dx}{dt}+\frac{1}{y}\frac{dy}{dt}+\frac{1}{u}\frac{du}{dt}=y+u-x-u+x-y=0$$

and we conclude that $uxy=C$ is an integral of motion.

Also it is not hard to see either that

$$\frac{d}{dt}(x+y-u)=xy+ux-xy-uy-ux+uy=0$$

and thus $x+y-u=D$ is the second integral of motion.

This allows you to write the general solution of the PDE as :

$$F(u(x,y)xy,x+y-u(x,y))=0$$

where $ F $ is an arbitrary function subject to the initial conditions.