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Find general solution to Bessel's equation with substitution

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]81x^{2}y'' + 27xy' + (9x^{\frac{2}{3}}+8)y = 0 [/tex]
    Hint: y = x1/3u
    x1/3 = z

    2. The attempt at a solution
    Change of variables gives:
    [tex] \frac{d^{2}y}{dx^{2}} = x^{\frac{1}{3}}\frac{d^{2}u}{dx^{2}}+\frac{2}{3}x^{-\frac{2}{3}}\frac{du}{dx} - \frac{2}{9}x^{-\frac{5}{3}}u [/tex]

    Plugging into original equation:
    [tex]81x^{\frac{7}{3}}u'' + 81x^{\frac{4}{3}}u'+(9x-x^{\frac{1}{3}})u = 0[/tex]

    Then using x1/3 = z, z' = 1/3 x-2/3
    [tex]\frac{d^{2}u}{dx^{2}} = -\frac{2}{9}x^{-\frac{5}{3}}\frac{du}{dz}+\frac{1}{3}x^{-\frac{2}{3}}\frac{d^{2}u}{dz^{2}}[/tex]

    Plugging in:
    [tex]3z^{4}u'' + zu' + (z^{2} - \frac{1}{9})u = 0 [/tex]

    The Bessel equation does not have a coeffcient in front of the first term and it is also of the second order:
    x2y'' + xy' + (x2 - n2)y = 0
    Can anyone help me get my equation into the Bessel equation form please?
     
  2. jcsd
  3. Oct 6, 2009 #2
    [tex]81x^{\frac{7}{3}}u'' + 81x^{\frac{4}{3}}u'+(9x-x^{\frac{1}{3}})u = 0[/tex]

    Double check the power of x in the term 9xu.


    [tex]\frac{d^{2}u}{dx^{2}} = -\frac{2}{9}x^{-\frac{5}{3}}\frac{du}{dz}+\frac{1}{3}x^{-\frac{2}{3}}\frac{d^{2}u}{dz^{2}}[/tex]

    Redo your calculation of [tex]\frac{d^{2}u}{dx^{2}}[/tex] because the term [tex]\frac{1}{3}x^{-\frac{2}{3}}\frac{d^{2}u}{dz^{2}}[/tex] is not right (missed a chain rule along the way?)
     
  4. Oct 6, 2009 #3
    Yup, found it. Thanks
     
  5. Jun 7, 2010 #4
    can you show me why the term [tex]\frac{1}{3}x^{-\frac{2}{3}}\frac{d^{2}u}{dz^{2}}[/tex] is wrong???

    (I have already followed and checked the chain rule step by step but I still got this term.)
     
    Last edited: Jun 7, 2010
  6. Jun 7, 2010 #5

    vela

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    You have

    [tex]\frac{du}{dx} = \frac{1}{3}x^{-2/3} \frac{du}{dz}[/tex]

    Then you differentiate again, applying the product rule, to get

    [tex]\frac{d^2u}{dx^2} = -\frac{2}{9}x^{-5/3}\frac{du}{dz} + \frac{1}{3}x^{-2/3}\frac{d}{dx}\left(\frac{du}{dz}\right)[/tex]

    The mistake you're probably making is saying

    [tex]\frac{d}{dx}\left(\frac{du}{dz}\right) \rightarrow \frac{d^2u}{dz^2}[/tex]

    which results in the term you got, but you have to remember you're differentiating with respect to x, not z, so you have to apply the chain rule one more time.
     
  7. Jun 7, 2010 #6
    I've tried to apply the chain rule again as you have told me to do so . I got these following results.

    as from du/dx = {(1/3)[x^(-2/3)]}du/dz
    thus du/dz = [3x^(-2/3)]du/dx
    and then [d(du/dz)]/dx = d[(3x^2/3)(du/dx)]/dx
    using product rule; [d(du/dz)]/dx = [2x^(-1/3)]du/dx + [3x^(2/3)]{[(d^2)u]/dx^2}

    so i couldn't move on again.
     
  8. Jun 8, 2010 #7

    vela

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    Why did you solve for du/dz? Remember, you're trying to replace du/dx and d2u/dx2 in the original equation.
     
  9. Jun 8, 2010 #8
    thx for your advice. Ok, this is my last question. I want you to show me in detail how to get d^2u/dx^2 . please dont't blame me whether i didn't try it myself because actually I did but I stuck with this kind of problems for 3 days .(I have the limit of time before my test!!!) thx a lot
     
  10. Jun 8, 2010 #9

    vela

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    You have [itex]z=x^{1/3}[/itex] so that

    [tex]\frac{dz}{dx} = \frac{1}{3} x^{-2/3} = \frac{1}{3z^2}[/tex]

    The chain rule tells you, for an arbitrary function f,

    [tex]\frac{df}{dx} = \frac{dz}{dx} \frac{df}{dz} = \frac{1}{3z^2} \frac{df}{dz}[/tex]

    Differentiating with respect to x is the same thing as differentiating with respect to z and dividing by 3z2.

    So to find du/dx, you'd set f=u and you'd get

    [tex]\frac{du}{dx} = \frac{1}{3z^2} \frac{du}{dz}[/tex]

    and to find d2u/dx2, set f=du/dx and do the same thing:

    [tex]\frac{d^2u}{dx^2} = \frac{1}{3z^2} \frac{d}{dz} \left(\frac{du}{dx}\right) = \frac{1}{3z^2} \frac{d}{dz} \left(\frac{1}{3z^2} \frac{du}{dz}\right)[/tex]

    I'll leave it to you to work out the rest.
     
  11. Jun 8, 2010 #10
    Thank you very much , I got it!!!
     
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