Problem statement:

Consider the PDE: $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=\frac{1}{\sqrt{x^2+y^2}}, (x,y)\neq (0,0) $

Determine all solutions to the equation of the form

$f(x,y)=g(r)$ where $r= \sqrt{x^2+y^2}$

My progress:

Now $f(x,y)=g(r)=g(\sqrt{x^2+y^2})$ and so the partial derivatives are:

$\frac{\partial f}{\partial x}=g'(\sqrt{x^2+y^2})\frac{x}{\sqrt{x^2+y^2}}$

$\frac{\partial f}{\partial y}=g'(\sqrt{x^2+y^2})\frac{y}{\sqrt{x^2+y^2}}$

Now our PDE will become:

$\frac{x^2}{\sqrt{x^2+y^2}}+\frac{y^2}{\sqrt{x^2+y^2}}=\frac{1}{\sqrt{x^2+y^2}} $

And furthermore: $x^2+y^2=1$ which is an equation for a circle at origo. So how did I end up wrong?

  • 1
    $\begingroup$ What did you do with $g'(r)$? The idea of this homework is that you transform a PDE for $f$ into a ODE for $g$. $\endgroup$
    – Fabian
    Jan 23, 2013 at 22:17
  • $\begingroup$ Alright, I will give it a new try but I've red about transform a PDE to ODE but I wasn't paying enough attention apparently. $\endgroup$
    – EricAm
    Jan 23, 2013 at 22:23

2 Answers 2


As the OP has answered the question himself, I just want to add the general solution of this PDE can be obtained via the method of characteristics. For that one converts the PDE in an ODE along the characteristic line. The system of ODE which needs to be solved reads $$\frac{dx}{ds} = x, \frac{dy}{ds} = y, \frac{df}{ds}= \frac{1}{r}; $$ where $s$ parameterizes the characteristic line.

The general solution for $x$ and $y$ is given by $$\tag{1}x= x_0 e^{s},y=y_0 e^{s};$$ thus we have $$\frac{df}{ds} = \frac{e^{-s}}{\sqrt{x_0^2+y_0^2}} $$ which is solved by $$f= f_0(y_0/x_0) - \frac{e^{-s}}{\sqrt{x_0^2+y_0^2}}$$ where $f_0$ depends on the characteristic line via $y_0/x_0$. Using (1) to eliminate $x_0$ and $y_0$, we obtain $$f= f_0(y/x) - \frac{1}{\sqrt{x^2+y^2}}$$ with an arbitrary function $f_0$.


I actually solved it myself:

$\frac{\partial f}{\partial x}=g'(r)\frac{x}{r}$ and

$\frac{\partial f}{\partial y}=g'(r)\frac{y}{r}$

This gives: $g'(r)\frac{x^2+y^2}{r}=\frac{1}{r} \Rightarrow rg'(r)=\frac{1}{r}$

With solution: $g(r)=C-\frac{1}{\sqrt{x^2+y^2}}$


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