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 Variables for Elliptic Integral

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Given the differential equation


    use the substitution of dependent variable

    [tex]u=ve^{ \alpha x + \beta y}[/tex]

    and a scaling change of variables

    [tex]y'= \gamma y[/tex]

    to reduce the differential equation to


    2. Relevant equations
    I have no idea

    3. The attempt at a solution
    I tried a direct substitution of both variables:

    [tex]u_{x}=\alpha ve^{\alpha x+\beta y}[/tex]
    [tex]u_{xx}=\alpha^{2}ve^{\alpha x+\beta y}[/tex]
    [tex]u_{y}=\beta ve^{\alpha x+\beta y}[/tex]
    [tex]u_{yy}=\beta^{2} ve^{\alpha x+\beta y}[/tex]

    Plugging in this gives

    [tex]\alpha^{2} ve^{\alpha x+\beta y}+3\beta^{2}ve^{\alpha x+\beta y}-2\alpha ve^{\alpha x+\beta y}+24\beta ve^{\alpha x+\beta y}+5ve^{\alpha x+\beta y}[/tex]

    You can obviously factor out [itex]ve^{\alpha x+\beta y}[/itex], but that doesn't really do much for you. I also tried doing this with the y' substitution. It also occurred to me that since v is probably supposed to be understood as v(x,y), I tried this set of substitutions:

    [tex]u_{x}=\alpha ve^{\alpha x+\beta y}+v_{x}e^{\alpha x+\beta y}[/tex]
    [tex]u_{xx}=\alpha^{2} ve^{\alpha x+\beta y}+\alpha v_{x}e^{\alpha x+\beta y}+v_{xx}e^{\alpha x+\beta y}[/tex]
    [tex]u_{x}=\beta ve^{\alpha x+\beta y}+v_{y}e^{\alpha x+\beta y}[/tex]
    [tex]u_{xx}=\beta^{2} ve^{\alpha x+\beta y}+\beta v_{y}e^{\alpha x+\beta y}+v_{yy}e^{\alpha x+\beta y}[/tex]

    None of these attempts gave me any insight into the problem.
    Last edited: Sep 21, 2010
  2. jcsd
  3. Sep 21, 2010 #2


    User Avatar
    Homework Helper
    Gold Member

    You need to treat [itex]v[/itex] as a function of [itex]x[/itex] and [itex]y[/itex] and use the product rule to differentiate [itex]u(x,y)=v(x,y)e^{\alpha x+\beta y}[/itex].

    Also, your [itex]\LaTeX[/itex] isn't displaying properly because you aren't puuting spaces between \alpha and x (or \beta and y )
  4. Sep 21, 2010 #3
    I fixed the LaTeX and (hopefully now that it's displaying correctly), you'll see that I used the product rule at the bottom of my attempted solution. However, it is non-obvious to me how making this correction by using the product rule helps me. I will have lots of first order partial derivatives floating around that are not in the desired equation.
  5. Sep 21, 2010 #4


    User Avatar
    Homework Helper
    Gold Member

    You'll end up with a bunch of terms involving [itex]v[/itex] and its partial derivatives, all multyiplied by [itex]e^{\alpha x+\beta y}[/itex], which you can factor out of your DE (since it is never zero, getting rid of this factor doesn't exclude any solutions)... giving you a different DE for [itex]v[/itex]...one that you can simplify by choosing nice values for [itex]\alpha[/itex] and [itex]\beta[/itex].

    Give it a shot and post your attempt.
  6. Sep 21, 2010 #5
    Haha, it did not occur to me to simply pick alpha and beta, even though they are arbitrary parameters introduced in the change of variables.

    I ended up with [itex]v_{xx}+3v_{yy}-31v[/itex], and I'm assuming you can take care of the factor of 3 in front of [itex]3v_{yy}[/itex] by simply correctly setting [itex]\gamma[/itex] when you substitute in y'.
  7. Sep 22, 2010 #6


    User Avatar
    Homework Helper
    Gold Member

    I think the factor of -31 might be a little off (my back of envelope calc gave me a factor of -49), so you'll probably want to double check that. But otherwise, yes...can you see what value of [itex]\gamma[/itex] will get rid of the factor of 3?
  8. Sep 22, 2010 #7
    I used [itex]\gamma = \sqrt{3}[/itex]. This is correct? If not, perhaps [itex]\frac{1}{\sqrt{3}}[/itex].
  9. Sep 22, 2010 #8


    User Avatar
    Homework Helper
    Gold Member

    You tell me....the chain rule tells you that [itex]v_y(x,y'(y))=\frac{dy'}{dy}v_{y'}[/itex], so what value of [itex]\gamma[/itex] makes [itex]v_{y'y'}=3v_{yy}[/itex]?
  10. Sep 22, 2010 #9
    [tex]\frac{d^{2}y'}{dy^{2}}=0 \Rightarrow \gamma v_{y'y'}=v_{yy}[/tex]
    [tex]\gamma = 1/3[/tex]
  11. Sep 22, 2010 #10


    User Avatar
    Homework Helper
    Gold Member

    Ermm... shouldn't you have


  12. Sep 22, 2010 #11
    Yes, I should. Forgot to apply chain rule while using product rule. Bah! Thanks for all the help.

    Also, I rechecked the 31, and I got it right unless I made a mistake in my substitution or if I missed a term while I was gathering terms. I'll double check it from the start before I hand it in.

    EDIT: Upon reviewing the work, 49 is the correct number, not 31.
    Last edited: Sep 22, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook