Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with simplifying a 2nd order pde

  1. Sep 7, 2011 #1
    I was given the equation
    dp/ds = 4 + 1/e*d/de(e*dp/de)

    The derivatives in the equation are partial derivatives

    the values of p,s,e are dimensionless numbers.

    I am to assume that the solution is separable and then use finite difference method to solve for p, the finite difference method is not a problem. This is where i am having problems. What will the equation be after the assumption is made and the equation is simplified.

    I have attempted the question:

    I equated the right hand side = 0:

    4 + 1/e*d/de(e*dp/de) = 0

    and made the partial derivatives total derivatives and then applied the chain rule:

    4 + 1/e*dp/de + d/de(dp/de). Is this correct????????
  2. jcsd
  3. Sep 7, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm not sure why you have set the right hand side to zero. If you have a separable solution, this will not be the case for non-trivial solutions. The method of separation of variables should be applied as follows.

    Let [itex]p(e,s) = q(e)r(s)[/itex]. Then we have

    [tex]\frac{\partial}{\partial s}q(e)r(s) = 4 + \frac{1}{e}\frac{\partial}{\partial e}\left(e\frac{\partial}{\partial e} q(e)r(s)\right)\;.[/tex]

    Can you take it from here?
  4. Sep 7, 2011 #3
    I have attempted it again:

    using the equation given i ended up with:

    q*dr/ds = 4 + r(1/e*dq/de+d/de(dq/de))

    what happens to the 4 then??
  5. Sep 7, 2011 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Let's re-write it so that we can see what's happening more clearly:

    [tex]q(e)r^\prime(s) = 4 + r(s)\left[\frac{q^\prime(e)}{e} + q^{\prime\prime}(e)\right][/tex]

    This is still a separable equation, the fact that you have an additional constant term doesn't matter. If you're struggling to see it, let [itex]R(s) = 4+r(s)[/itex]. Then [itex]R^\prime(s) = r^\prime(s)[/itex] leading to

    [tex]q(e)R^\prime(s) = R(s)\left[\frac{q^\prime(e)}{e} + q^{\prime\prime}(e)\right][/tex]

    Can you now continue?
  6. Sep 7, 2011 #5
    Yes thank you very much
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook