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!

Change of Variable With Legendre Equation

  1. Oct 9, 2015 #1
    1. The problem statement, all variables and given/known data

    Change the independent variable from x to θ by x=cosθ and show that the Legendre equation

    (1-x^2)(d^2y/dx^2)-2x(dy/dx)+2y=0

    becomes

    (d^2/dθ^2)+cotθ(dy/dθ)+2y=0

    2. Relevant equations


    3. The attempt at a solution

    I did get the exact form of what the equation should become, but I had a 2 in front of the cotθ term. I was wondering if the answer that the book provides is missing a factor of 2 in front of the cotθ term?
     
  2. jcsd
  3. Oct 9, 2015 #2
    No, there shouldn't be a "2" factor there.
     
  4. Oct 10, 2015 #3
    Thank you very much for your reply! I have spend many an hour trying to figure out where the 2 goes, but that was to to avail. I have the following:

    (dy/dx)=(dy/dθ)(-1/sqrt (1-x^2))=(dy/dθ)(-1)(sin(θ))^(-1)

    Can you please direct me to where I went wrong on the above statement?

    Thanks!
     
  5. Oct 10, 2015 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The problem lies elsewhere. Show your calculation of ##(1-x^2)y''##.
     
  6. Oct 10, 2015 #5
    Thanks for your reply! Here is my calculation for (1-x^2)y'':

    (d^2y/dx^2) = (d^2y/dθ^2)(1-x^2)^(-1) = (d^2y/dθ^2)(1-(cosθ)^2)^(-1)

    (1-x^2)y'' = (1-(cosθ)^2)(d^2y/dθ^2)(1-(cosθ)^2)^(-1) = (d^2y/dθ^2)
     
  7. Oct 10, 2015 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Differentiate the result you got for y'.
    $$\frac{d^2}{dx^2}y = \frac{d}{dx} y' = -\csc\theta \frac{d}{d\theta} y' = -\csc\theta \frac{d}{d\theta} \left(-\csc\theta \frac{dy}{d\theta}\right)$$ You need to use the product rule.
     
  8. Oct 10, 2015 #7
    I do have that in my solution and it leads to the earlier reply I posted. It also leads to the correct first term of the right answer. Is the following correct?

    -csc(θ)(d/dθ)[y] = (cscθ)^2(d^2y/dθ^2)
     
  9. Oct 10, 2015 #8
    I do have that in my solution and it leads to the earlier reply I posted. It also leads to the correct first term of the right answer. Is the following correct?

    -csc(θ)(d/dθ)[y] = (cscθ)^2(d^2y/dθ^2)
     
  10. Oct 10, 2015 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    No, that's not correct. As I noted earlier, you have to use the product rule. The ##d/d\theta## on the left acts on everything to its right.
     
  11. Oct 12, 2015 #10
    Thank you very much! Sorry for the late reply, but I had to sleep on it. It worked perfectly fine! I was inventing my own incorrect rules for calculus, and that did not work very well!

    P.S. I was wondering how you were able to type those derivatives very clearly. It is much more clear than having slashes and asterisks?
     
  12. Oct 12, 2015 #11
  13. Oct 12, 2015 #12
    Thanks a lot!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Change of Variable With Legendre Equation
Loading...