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?


1 Answer 1


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.

  • $\begingroup$ Thanks! I'm mostly looking for a word description or a way to describe what each characteristic looks like however @Harry49 $\endgroup$ Sep 19, 2017 at 0:28
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    $\begingroup$ The curves of the form $y^2-x^2=C$ are known as hyperbolas, @FelicioGrande $\endgroup$
    – user357151
    Sep 19, 2017 at 1:03
  • $\begingroup$ 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 $\endgroup$ Sep 19, 2017 at 1:08
  • $\begingroup$ @FelicioGrande Right, the case $C=0$ is exceptional; we get two lines, which is kind of a degenerate hyperbola. $\endgroup$
    – user357151
    Sep 19, 2017 at 1:15

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