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Changing Independent Variable in the Bessel Equation

  1. Nov 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Given the bessel equation $$x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} -(1-x)y=0$$ show that when changing the variable to ##u = 2\sqrt{x}## the equation becomes $$u^2\frac{d^2y}{du^2}+u\frac{dy}{du}+(u^2-4)y = 0$$

    2. Relevant equations

    3. The attempt at a solution
    ##u=2\sqrt{x}##, ##du/dx=\frac{1}{\sqrt{x}}=\frac{2}{u}##, ##x = u^2/4## and ##x^2=u^4/16##, since ##u=u(x)## we have $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\frac{2}{u}\frac{dy}{du}$$ and obtain the operator ##\frac{d}{dx} = \frac{2}{u}\frac{d}{du}## then $$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\left(\frac{2}{u}\frac{d}{du}\right)\left(\frac{2}{u}\frac{dy}{du}\right)=\frac{4}{u^2}\frac{d^2y}{du^2}$$ plugging into the original now we obtain $$\frac{u^4}{16}\frac{4}{u^2}\frac{d^2y}{du^2}+\frac{u^2}{4}\frac{2}{u}\frac{dy}{du}-\left(1-\frac{u^2}{4}\right)y = \frac{u^2}{4}\frac{d^2y}{du^2}+\frac{u}{2}\frac{dy}{du}+\left(\frac{u^2}{4}-1\right)y=0$$ multiplying by 4 we obtain $$u^2\frac{d^2y}{du^2}+2u\frac{dy}{du}+(u^2-4)y=0$$ where I somehow have an additional factor of 2 on the ##dy/du## term despite getting the correct coefficients for the first and last term.
     
  2. jcsd
  3. Nov 18, 2016 #2

    Ssnow

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    Gold Member

    Hi, attention ##\left(\frac{2}{u}\frac{d}{du}\right)\left(\frac{2}{u}\frac{dy}{du}\right)\not=\frac{4}{u^2}\frac{d^2y}{du^2}##, you must apply the product rule for derivation! ...
     
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