# Explaining Characteristics of a PDE

I have the PDE: $$y\cdot u_x+x\cdot u_y=0$$ I'm trying to find out what the characteristics of this PDE are. Now I know the characteristic equation is given by: $$\frac{\partial y}{\partial x}=\frac{x}{y}$$ So are the Characteristics the parabolas that have $x=y$ and $x=-y$ as asymptotes along with the lines $x=y$ and $x=-y$ themselves, or am I missing some Characteristics?

The method of characteristics consists in setting $u(x,y) = u(x(s),y(s))$, and transform the PDE in $(x,y)$ into an ODE in $s$. Here, $$\frac{\mathrm{d}u}{\mathrm{d}s} = x'\, u_x + y'\, u_y \, .$$ We set the equations of characteristics $$x' = 1 \qquad\text{and}\qquad y' = \frac{x}{y}\, ,$$ such that $\mathrm{d}u/\mathrm{d}s = 0$, i.e. $u$ is constant along the characteristics. Thus, $$\frac{y'}{x'} = \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{y} \, ,$$ which solutions are $y^2 = x^2 + C$, with constant $C$. The value of $u$ is completely determined by the constant $C$, so that the solutions of the PDE are $$u(x,y) = f(y^2 - x^2)\, ,$$ where $f$ is any differentiable function. To go further, this post is related.
• The curves of the form $y^2-x^2=C$ are known as hyperbolas, @FelicioGrande – user357151 Sep 19 '17 at 1:03
• Ah yes, sorry. But these hyperbolas do not intersect the lines $x=y$ and $x=-y$, correct? Unless of course the 2 cases $x=y$ and $x=-y$. @Michelle – Felicio Grande Sep 19 '17 at 1:08
• @FelicioGrande Right, the case $C=0$ is exceptional; we get two lines, which is kind of a degenerate hyperbola. – user357151 Sep 19 '17 at 1:15