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Math Help - Bessel's equation and second order DE

  1. #1
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    Bessel's equation and second order DE

    Hi all!
    I've hit a wall on the question I'm working on. Can someone look over my work and see where I may have possible gone wrong?

    The question is:
    Show that if y(x) is a solution of Bessel's equation of order p, then, w(z)=z^{-c}y(az^b) is a solution of z^2w''+(2c+1)zw'+[a^2b^2z^{2b}+(c^2-p^2b^2)]w=0.

    Here is my working.
    Considering that w(z)=z^{-c}y(az^b) is a solution, then:
    w'(z)=-cz^{-c-1}y(az^b)+abz^{b-1}y'(az^b)z^{-c} and
    w''(z)=(c^2+c)z^{-c-2}y(az^b)-2cz^{-c-1}abz^{-b-1}y'(az^b)+z^{-c}a^2b^2z^{b-2}y''(az^b)

    Subing into L[w],
    L[w]=z^2(c^2+c)z^{-c-2}y(az^b)-2cz^{-c-1}abz^{-b-1}y'(az^b)+z^{-c}a^2b^2z^{b-2}y''(az^b)+ (2c+1)z(-cz^{-c-1}y(az^b)+abz^{b-1}y'(az^b)z^{-c})+[a^2b^2z^{2b}+(c^2-p^2b^2)]z^{-c}y(az^b).

    I simplify to find:
    bz^{-c}(a^2z^{2b}y''(az^b)+abz^by'(az^b)+b(a^2z^2b-p^2)y(az^b)).
    As L[w]=0, that means a^2z^{2b}y''(az^b)+abz^by'(az^b)+b(a^2z^2b-p^2)y(az^b)=0.
    Making the substitution of x=az^b
    I get the following:
    x^2y''(x)+bxy'(x)+b(x^2-p^2)y(x)=0.
    That equation is soo close to Bessel's equation. I go through my working and I can't find why I have those b's are there.

    Thank you very much for your time.
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  2. #2
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    Quote Originally Posted by Silverflow View Post
    Hi all!
    I've hit a wall on the question I'm working on. Can someone look over my work and see where I may have possible gone wrong?

    The question is:
    Show that if y(x) is a solution of Bessel's equation of order p, then, w(z)=z^{-c}y(az^b) is a solution of z^2w''+(2c+1)zw'+[a^2b^2z^{2b}+(c^2-p^2b^2)]w=0.

    Here is my working.
    Considering that w(z)=z^{-c}y(az^b) is a solution, then:
    w'(z)=-cz^{-c-1}y(az^b)+abz^{b-1}y'(az^b)z^{-c} and
    w''(z)=(c^2+c)z^{-c-2}y(az^b)-2cz^{-c-1}abz^{-b-1}y'(az^b)+z^{-c}a^2b^2z^{b-2}y''(az^b)

    Subing into L[w],
    L[w]=z^2(c^2+c)z^{-c-2}y(az^b)-2cz^{-c-1}abz^{-b-1}y'(az^b)+z^{-c}a^2b^2z^{b-2}y''(az^b)+ (2c+1)z(-cz^{-c-1}y(az^b)+abz^{b-1}y'(az^b)z^{-c})+[a^2b^2z^{2b}+(c^2-p^2b^2)]z^{-c}y(az^b).

    I simplify to find:
    bz^{-c}(a^2z^{2b}y''(az^b)+abz^by'(az^b)+b(a^2z^2b-p^2)y(az^b)).
    As L[w]=0, that means a^2z^{2b}y''(az^b)+abz^by'(az^b)+b(a^2z^2b-p^2)y(az^b)=0.
    Making the substitution of x=az^b
    I get the following:
    x^2y''(x)+bxy'(x)+b(x^2-p^2)y(x)=0.
    That equation is soo close to Bessel's equation. I go through my working and I can't find why I have those b's are there.

    Thank you very much for your time.
    You're missing a term in the second derivative. There should be two y' terms - you only have one. I got (when combined)

     <br />
ab(b-2c-1)z^{b-c-2} y'\left(a z^b\right)<br />
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  3. #3
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    It works!! Thank you very much!
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  4. #4
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    The next part of the question is to use what it given above, and find a general solution for

    x^2y''+(1/8+x^4)y=0

    I have no idea how to do this. I can try to relate coefficients of this DE to the one I mentioned in my first thread, but that doesn't work. Substituting x=az^b doesn't seem to work either...
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  5. #5
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    Quote Originally Posted by Silverflow View Post
    The next part of the question is to use what it given above, and find a general solution for

    x^2y''+(1/8+x^4)y=0

    I have no idea how to do this. I can try to relate coefficients of this DE to the one I mentioned in my first thread, but that doesn't work. Substituting x=az^b doesn't seem to work either...
    Choose a, b, \; \text{and}\; c in

    <br />
z^2w''+(2c+1)zw'+[a^2b^2z^{2b}+(c^2-p^2b^2)]w=0

    so that you obtain

    x^2y''+(1/8+x^4)y=0.

    This then will give you

     <br />
w = z^{-c} y(az^b).<br />
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  6. #6
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    Thanks very much.
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