1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partial differential equation

  1. Jan 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi
    I need some help in proving that [itex]\partial^2 V / \partial x^2[/itex] + [itex]\partial^2 V / \partial y^2[/itex] = 0 , when V = 1/2ln(x^2+y^2)
    2. Relevant equations



    3. The attempt at a solution
    For [itex]\partial V / \partial x[/itex] here's what I get: (1/x^2+y^2) *2x*1/2 = x * (x^2+y^2)^-1
    and for [itex]\partial V / \partial y[/itex]
    (x^2+y^2)^-1 *y ....

    As you can see if i differentiate them once , I don't end up with any terms which may cancel each other to get zero..
    what am I over looking ? PDE is new to me. I know we keep one variable constant in respect to other,etc..
    Thanks
     
  2. jcsd
  3. Jan 14, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The equation has second derivatives of V. Why are you stopping with the first derivative? You were doing fine.
     
  4. Jan 14, 2012 #3
    I should have posted my complete answer which I know is wrong...
    Here is the second derivative for x , keeping y fixed.

    For [itex]\partial V / \partial x[/itex] here's what I get: (1/x^2+y^2) *2x*1/2 = x * (x^2+y^2)^-1

    =>
    -(x^2+y^2)^-1 * (x^2+y^2)^-2 *2x^2...

    Which seems wrong...
     
  5. Jan 14, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It is wrong. But I can see parts of the right answer in it. Did you use the product rule on the first derivative? Can you show your steps?
     
  6. Jan 14, 2012 #5
    Yes, that I have. Sure :
    Let u = x , v = (x^2+y^2)^-1
    Then using product rule :

    f'' (x,v) = u' (v) + (u)v' ( Differentiating w.r.t x , keeping y constant)
    =>
    1* (x^2+y^2)^-1 + x [ -1 ( x^2+y^2)^-2 * 2x) ]


    There's something about this expression which makes me think that it's wrong.. ( (x^2+y^2)^-2 , shouldn't the right version be : -1(x^2)^-2 * 2x ?
     
  7. Jan 14, 2012 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No, it's right. That doesn't look a lot like what you posted before. Were you just being sloppy? Here I'll simplify it a little and translate into TeX:
    [tex]\frac{1}{(x^2+y^2)}-\frac{2 x^2}{(x^2+y^2)^2}[/tex]
     
  8. Jan 14, 2012 #7
    Oh silly me.. ok then in that case my second partial derivative of y is :

    [tex]\frac{1}{(x^2+y^2)}-\frac{2 y^2}{(x^2+y^2)^2}[/tex]
    It still doesn't answer the question.. they want me to prove that the sum of these two derivatives = 0 , am I right ?
     
  9. Jan 14, 2012 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, prove the sum is zero. Add them and do some algebra. Put stuff over a common denominator.
     
  10. Jan 14, 2012 #9
    Oh.. that makes sense.. Thanks for your help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Partial differential equation
Loading...