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## Main Question or Discussion Point

Hi all!

I attempted to solve Airy's equation

http://en.wikipedia.org/wiki/Airy_function

[tex]\ddot y = \lambda ty[/tex]

in a rather nontraditional way which I highly doubt to be correct for obvious reasons, but I can't find my mistake. Here's my attempt:

first, rewrite the second order ode as a system of two 1st order ode's:

define [tex]\dot y = z[/tex]. Then we have:

[tex] \left(\begin{array}{c}\dot y \\ \dot z\end{array}\right) = \left(\begin{array}{cc}0&1 \\ \lambda t&0\end{array}\right)\left(\begin{array}{c} y \\ z\end{array}\right) [/tex]

Here comes the step I suspect: I simply exponentiate the matrix with the antiderivatives of the original entries, to get:

[tex] \left(\begin{array}{c} y\\ z\end{array}\right)=\exp(\left(\begin{array}{cc}0&t\\ \frac{\lambda}{2}t^2 & 0\end{array}\right))\left(\begin{array}{c} y_0\\ z_0\end{array}\right) [/tex]

since the value of the exponent at t=0 is identity.

To back up this step, I decompose the matrix in the exponential in a trivial way:

[tex]\left(\begin{array}{cc}0&t\\ \frac{\lambda}{2}t^2 & 0\end{array}\right) = t\left(\begin{array}{cc}0&1\\ 0 & 0\end{array}\right)+\frac{\lambda}{2}t^2\left(\begin{array}{cc}0&0\\ 1 & 0\end{array}\right) =: a e_2+b e_4 [/tex]

and set it in the series of exp by definition. Then I take the time derivative, to ?confirm? the validity of the solution:

[tex]\partial_t\exp (a e_2+b e_4) = \partial_t\big\sum_n\frac{(a e_2+b e_4)^n}{n!}= \big\sum_n\frac{\partial_t(a e_2+b e_4)^n}{n!} =(\dot a e_2+\dot b e_4)\big\sum_n\frac{(a e_2+b e_4)^{n-1}}{(n-1)!}=(\dot a e_2+\dot b e_4) \exp (a e_2+b e_4) [/tex]

thereby establishing the initial system of ode's (since a' = 1 and b' = \lambda t by definition)

Further it's quite simple since one can change basis in order to diagonalise (here it's doable, otherwise just take the jordan normal form) the exponent and perform the exponential, and then go back to the initial basis constructing the final solution for y(t).

The big problem is whatever you do, you won't get in any way any of the Airy functions Ai or Bi, hence my suspicion for this solution.

It could be, after all, that the obtained solution is some special linear combination of the linearly independent Ai and Bi, but I doubt it.

I would be very thankful, if someone points out where the above calculations break down :)

regards, marin

MISTAKE FOUND: matrices do not commute, so the product rule breaks down :)

I attempted to solve Airy's equation

http://en.wikipedia.org/wiki/Airy_function

[tex]\ddot y = \lambda ty[/tex]

in a rather nontraditional way which I highly doubt to be correct for obvious reasons, but I can't find my mistake. Here's my attempt:

first, rewrite the second order ode as a system of two 1st order ode's:

define [tex]\dot y = z[/tex]. Then we have:

[tex] \left(\begin{array}{c}\dot y \\ \dot z\end{array}\right) = \left(\begin{array}{cc}0&1 \\ \lambda t&0\end{array}\right)\left(\begin{array}{c} y \\ z\end{array}\right) [/tex]

Here comes the step I suspect: I simply exponentiate the matrix with the antiderivatives of the original entries, to get:

[tex] \left(\begin{array}{c} y\\ z\end{array}\right)=\exp(\left(\begin{array}{cc}0&t\\ \frac{\lambda}{2}t^2 & 0\end{array}\right))\left(\begin{array}{c} y_0\\ z_0\end{array}\right) [/tex]

since the value of the exponent at t=0 is identity.

To back up this step, I decompose the matrix in the exponential in a trivial way:

[tex]\left(\begin{array}{cc}0&t\\ \frac{\lambda}{2}t^2 & 0\end{array}\right) = t\left(\begin{array}{cc}0&1\\ 0 & 0\end{array}\right)+\frac{\lambda}{2}t^2\left(\begin{array}{cc}0&0\\ 1 & 0\end{array}\right) =: a e_2+b e_4 [/tex]

and set it in the series of exp by definition. Then I take the time derivative, to ?confirm? the validity of the solution:

[tex]\partial_t\exp (a e_2+b e_4) = \partial_t\big\sum_n\frac{(a e_2+b e_4)^n}{n!}= \big\sum_n\frac{\partial_t(a e_2+b e_4)^n}{n!} =(\dot a e_2+\dot b e_4)\big\sum_n\frac{(a e_2+b e_4)^{n-1}}{(n-1)!}=(\dot a e_2+\dot b e_4) \exp (a e_2+b e_4) [/tex]

thereby establishing the initial system of ode's (since a' = 1 and b' = \lambda t by definition)

Further it's quite simple since one can change basis in order to diagonalise (here it's doable, otherwise just take the jordan normal form) the exponent and perform the exponential, and then go back to the initial basis constructing the final solution for y(t).

The big problem is whatever you do, you won't get in any way any of the Airy functions Ai or Bi, hence my suspicion for this solution.

It could be, after all, that the obtained solution is some special linear combination of the linearly independent Ai and Bi, but I doubt it.

I would be very thankful, if someone points out where the above calculations break down :)

regards, marin

MISTAKE FOUND: matrices do not commute, so the product rule breaks down :)

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