# Parameterization of the initial condition of nonlinear PDE

(NOTE: I am not asking for the solution to this PDE)

I have the PDE, $$u_x u_y + ln(x^2)=0,$$ with the condition that, $$u(x_0,y)=y,$$ where $$x_0$$ is a constant. I am to find the explicit solution (using Charpit's equations). In order to do this I first need to parameterize the initial condition in the form of a curve $$\Gamma (s)$$ in order to solve.

I am used to parameterizing curves but not of this type.

I have written down, $$u=s, x=x_0, y=s,$$ just as a hunch but I have no working to show for it. Is this correct? How do you parameterize equations of this type?

Thanks in advance.

• does $\ln(x^2) = 2\ln(x)$ help at all – phdmba7of12 Apr 6 at 14:46
• @phdmba7of12 maybe; when solving the equation. Not for the parameterization. Solving the PDE becomes fairly trivial once I have the parameterization. – Dan Elsender Apr 6 at 14:54
• what specific text are you referring to describing this method – phdmba7of12 Apr 6 at 14:59
• a precedent exists here on the stack math.stackexchange.com/questions/2674756/… – phdmba7of12 Apr 6 at 15:00
• @phdmba7of12 I'm asking how to parameterize the initial condtion, not how to solve the PDE. – Dan Elsender Apr 6 at 15:05

## 1 Answer

The correct parameterisation is $$u=s, y=s, x=x_0$$ This leads to a solution through Charpit's method of $$u=-2xln(x)+2x+2x_0ln(x_0)-2x_0+y$$