-1
$\begingroup$

Is there a general solution for first-order partial differential equations of the form

$$m(x) \partial_x f(x,y) = n(y) \partial_y f(x,y)$$

for given $m(x),n(y)$ and reasonable boundary conditions (e.g. $f(x,0)=0$ etc.)?

$\endgroup$

1 Answer 1

2
$\begingroup$

Change the variables in $x$ and $y$, i.e., $\tilde x =\tilde x(x)$ and $\tilde y = \tilde y(y)$ to make $m(x) = n(y) = 1$. Then equation can be written as

$ \partial_{\tilde x}f =\partial_{\tilde y}f. $

The equation above is equivalent to that $f= f(\tilde x+ \tilde y)$.

$\endgroup$
2
  • $\begingroup$ This won't work if $m$ and/or $n$ vanish at some points of the domain, and the OP may be concerned about these cases. $\endgroup$ Aug 25, 2014 at 0:54
  • $\begingroup$ Robert, thanks for your comment! This is the first step I would like to try. Assuming that the zeros of $m(x)$ and $n(y)$ are discrete, it will give solutions at different pieces. I am not sure there is any general statement I can make given no information of $m$ and $n$. $\endgroup$
    – BewSMA
    Aug 25, 2014 at 1:55

Not the answer you're looking for? Browse other questions tagged or ask your own question.