I have just entered the study of ODEs. However, the professor, without having talked at all about it in class, asked us to solve the following partial differential equation:

$\displaystyle r\frac{\partial v}{\partial r} = av$, where $a$ is a constant.

Here, $v$ is a function of $r$ and $\phi$ (I transformed a harder PDE into this easier one using polar coordinates).

We are supposed to be able to use ODE methods to solve this. I cannot see how. Any help will be much appreciated!

  • $\begingroup$ There needs to be more information. What variables is $v$ a function of? If $v$ is just a function of $r$ then this is just some rather silly notation for a first-order ordinary differential equation. $\endgroup$ – Gyu Eun Lee Jan 14 '13 at 22:39
  • $\begingroup$ I will correct it at once! $\endgroup$ – user44069 Jan 14 '13 at 22:41

By the method of separation of variables, let $v(r, \phi) = f(r) g(\phi)$. This implies that $\displaystyle r \frac{\partial f}{\partial r} = a f(r)$ which is a straightforward first-order linear ordinary differential equation with solution $\displaystyle f(r) = r^a + c$ and consequently $v(r, \phi) = g(\phi) (r^a + c)$


So this basically looks like a standard seperation of variables problem.

I will call theta t from now on.

You can divide both sides by v(r,t), multiply both sides by r, and "multiply" both sides by dr to get the following:

v(r,t) dv = a*r dr

You can now proceed to integrate this expression getting yourself:

1/2 * v(r,t)^2 = 1/2*a*r^2 + f(t)

Thus after rearranging the expression you end up with:

v(r,t) = sqrt(a*r^2 + f(t)) for any arbitrary single variable expression f(t) [where t once again denotes theta].

  • $\begingroup$ v(r,t)= √(ar^2+ f(θ)) (i used microsoft word to format it) $\endgroup$ – frogeyedpeas Jan 14 '13 at 23:00
  • $\begingroup$ whoop it looks like i forgot to divide out r... That should correct itself instead to: $\endgroup$ – frogeyedpeas Jan 15 '13 at 1:56
  • $\begingroup$ v(r,t) = f(t)*r(t)^a $\endgroup$ – frogeyedpeas Jan 15 '13 at 1:59

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