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I come across a confusing point.

Considering the ODE for the function $y(x)$ $$\frac{1}{a^2} y'' + axy = 0$$ I consider the change of variable $ u = ax$ and, using the chain rule for a linear change of variables $$ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{d}^2 y}{\mathrm{d}u^2} \left( \frac{\mathrm{d} u}{\mathrm{d}x} \right)^2 $$

I get the ODE

$$ y'' + uy = 0$$

If I on the other hand define $u = -ax$, I get the ODE $$ y'' - uy = 0$$

The solutions of the equations should be then identical upon changing the sign of the dependent variable. If I check (see http://mathworld.wolfram.com/AiryDifferentialEquation.html). I find the solution of the first equation is given by

$$ y(x) = \frac 1 3 \sqrt{x} \left[A I_{-1/3} \left(\frac 2 3 k x^{3/2}\right) - B I_{1/3} \left(\frac 2 3 k x^{3/2}\right) \right]$$ where $I$ stands for the modified Bessel function of the first kind, while the solution for the second is given by

$$y(x) = \frac 1 3 \sqrt{x} \left[A J_{-1/3} \left(\frac 2 3 k x^{3/2}\right) - B J_{1/3} \left(\frac 2 3 k x^{3/2}\right) \right]$$ where $J$ stands for the modified Bessel function of the first kind. I know that $$I_\beta(x) = i^{-\beta} J_\beta(ix),$$ but that does not imply

$$ I_\beta(x) = J_\beta(-x) $$ So how come by two different changes of variables I end up with different solutions? Thanks a lot

Michael Hardy
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An aedonist
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1 Answers1

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From the first change of variable, you obtained : $$ y(x) = \frac 1 3 \sqrt{x} \left[A_1 I_{-1/3} \left(\frac 2 3 k x^{3/2}\right) - B_1 I_{1/3} \left(\frac 2 3 k x^{3/2}\right) \right]$$ and from the second change of variable : $$y(x) = \frac 1 3 \sqrt{x} \left[A_2 J_{-1/3} \left(\frac 2 3 k x^{3/2}\right) - B_2 J_{1/3} \left(\frac 2 3 k x^{3/2}\right) \right]$$

Do not use the same symbols $A$ and $B$ for the two forms of solution, because $A_1\neq A_2$ and $B_1\neq B_2$. Of course they are arbitrary constants until boundary conditions be specified. But this doesn't imply that they are respectively equal for any boundary conditions. This only implies that $A_2$ is related to $A_1$ and that $B_2$ is related to $B_1$ (with arbitrary $A_1$ and arbitrary $B_1$).

So, from this point, in supposing that $A_1=A_2=A$ and $B_1=B_2=B$ , all that was derived is not correct.

JJacquelin
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