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Homework Help: 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

    [tex]u_{xx}+3u_{yy}-2u_{x}+24u_{y}+5u=0[/tex]

    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

    [tex]v_{xx}+v_{yy}+cv=0[/tex]


    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

    gabbagabbahey

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    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

    gabbagabbahey

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    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

    gabbagabbahey

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    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

    gabbagabbahey

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    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}}v_{y'}+\frac{dy'}{dy}v_{y'y'}=v_{yy}[/tex]
    [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

    gabbagabbahey

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    Ermm... shouldn't you have

    [tex]v_{yy}=\left(\frac{dy'}{dy}\right)^2v_{y'y'}[/tex]

    ??:wink:
     
  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
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