# Asymptotic behavior of the solution of the high degree differential equation $(x^{2n}y^{(n)})^{(n)}-x^2y=\lambda \; y$

The following differential equation has two independent solutions, one of the two is decreasing exponentially at infinity (k-Bessel function).

$$(x^2y')'-x^2y=\lambda \;y$$

Now for a higher-degree differential equation like:

$$(x^{2n}y^{(n)})^{(n)}-x^2y=\lambda \; y$$

We have $2n$ independent solutions. How can we find information on their asymptotic behavior near infinity? How many linearly independent solutions can we find that will decrease exponentially at infinity? (I guess $n$?) (any book of reference on this subject?)

If you want a quick and dirty way to find the asymptotics of $y(x)$ as $x \to \infty$, you can use the WKB ansatz $y(x) = e^{S(x)}$, with the hypothesis that $S^{(k)}/S' \to 0$ as $x\to \infty$ for all $k>1$. Substituting this form into your equation and keeping only the leading terms at infinity, you find $$x^{2n} (S')^{2n} + x^2 = 0 .$$ The solutions are $S'(x) = (-x^{2-2n})^{1/2n} = (-1)^{1/2n} x^{1/n-1}$ and hence the expected asymptotics for the $2n$ independent solutions are $y_k(x) \sim \exp(e^{i\frac{\pi}{2n}(1+2k)} n x^{1/n})$, $k=0,\ldots,2n-1$. In the case $n=1$, you get the asymptotics $y(x) \sim e^{\pm i x}$, and not $e^{\pm x}$ as your question suggested. You get the $e^{\pm x}$ asymptotics by flipping the sign of the $x^2 y$ term in the equation.
If you want to be more systematic, it helps to apply some transformations to the equation first. If you introduce $z = x^n y^{(n)}$, then the equation is equivalent to $$\begin{bmatrix} x^n y^{(n)} \\ (x^n z)^{(n)} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -x^2+\lambda & 0 \end{bmatrix} \begin{bmatrix} y \\ z \end{bmatrix} .$$ Next, it is convenient to rescale the dependent variables such that the leading power of $x$ on the right-hand side has a diagonalizable matrix as coefficient (if it's actually possible). Here, this looks like $$\begin{bmatrix} x^n (x y)^{(n)} + P(x\frac{d}{dx}) y \\ (x^n z)^{(n)} \end{bmatrix} = x \begin{bmatrix} 0 & 1 \\ -1+\frac{\lambda}{x} & 0 \end{bmatrix} \begin{bmatrix} x y \\ z \end{bmatrix} ,$$ where now $P(x\frac{d}{dx})$ is a constant coefficient polynomial in its argument, of order less than $n$. The last step is to substitute $x=t^{n}$, where the power of $t$ is chosen so that the resulting equation has asymptotically constant coefficients as $t\to \infty$. Since $x\frac{d}{dx} = \frac{1}{n} t\frac{d}{dt}$, the result is $$\begin{bmatrix} (t^n y)^{(n)} + t^{-n} Q(t\frac{d}{dt}) y \\ z^{(n)} + t^{-n} R(t\frac{d}{dt}) z \end{bmatrix} = n^n \begin{bmatrix} 0 & 1 \\ -1+\frac{\lambda}{t^n} & 0 \end{bmatrix} \begin{bmatrix} t^n y \\ z \end{bmatrix} ,$$ where again the constant coefficient polynomials $Q$ and $R$ are of order less than $n$. From here the theory of asymptotics of ODEs at irregular singular points tells us that the independent solutions of your equation will have the asymptotic expansions $y_k(t) = t^{-n} Y_k(t) (1 + O(1/t))$, where the last factor stands for an asyptotic series in inverse powers of $t$ and $Y_k(t)$ solves the asymptotic constant coefficient equation. In this case corresponds to $Y_k(t) = \exp(\alpha_k t)$, where the $\alpha_k$ are the different $n$-th roots of the eigenvalues of matrix $n^n [\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}]$, namely $\alpha_k = e^{i\frac{\pi}{2n}(1+2k)} n$, $k=0,\ldots,2n-1$, which reproduces the form of the $S(x)$ phase function from the WKB solution. But now we have more detailed information about the asymptotics.
• Exact there is a mistake in the sign in front of $x^2y$ to have solutions with an exponential decrease, I will correct it. Thanks for you answer. – Bertrand Oct 1 '17 at 14:24
A good reference on the subject is W. Wasow, Asymptotic expansions for ordinary differential equations. John Wiley & Sons, Inc., New York-London-Sydney 1965. It is not difficult to obtain the answer for small $n$, but computation for arbitrary $n$ may be complicated. Anyway, there is an algorithm of obtaining these asymptotics.