常微分方程式を解く方法 $y^{-1}$ 期間

私の専攻は数学ではありませんが、次の常微分方程式に出くわしました。 $y(x)$：

$$\left(y^3y^{\prime\prime\prime}\right)^\prime+\frac{5}{8}xy^\prime-\frac{1}{2}y+\frac{a}{y}=0,$$ ここで、素数は導関数を示し、 $a$ は正の定数です。 $0<a\le1$。このODEには、次の4つの初期条件があります。$y^\prime(0)=y^{\prime\prime\prime}(0)=0$、および他の2つ $y(0)$ そして $y^{\prime\prime}(0)$ 無限大での境界条件（BC）の狙い撃ち法（方法）によって決定されます。

無限大で目的のBCを見つけるために、遠方界の漸近的振る舞いを理解しようとします。私の問題では、遠方界BCがこのODEの準定常的な振る舞いであることを知っています。

Case 1.

If I drop the first term and solve $$\frac{5}{8}xy^\prime-\frac{1}{2}y+\frac{a}{y}=0,$$ I obtain a general solution $$y=\sqrt{2a+e^{2c}x^{8/5}},$$ where $c$ is a free constant, and the negative one has been dropped due to $y>0$ in my problem. At this point, can I say the far-field asymptotic behavior is $$y\sim e^cx^{4/5}.$$

Case 2. If I drop all the nonlinear terms in the ODE and solve $$\frac{5}{8}xy^\prime-\frac{1}{2}y=0,$$ I got a simpler solution $$y=cx^{4/5}.$$ In this case, I might say the far-field asymptotic behavior is $$y\sim cx^{4/5}.$$

My first question is which case is right? Or both of them are wrong, how can I got it?

Alternatively, if I requires

$$\left(y^3y^{\prime\prime\prime}\right)^\prime \sim 0 \quad as \quad x \rightarrow \infty,$$ and let $y=Ax^\alpha$, by plug it in the above asymptotics, I have

$$A^4x^{4\alpha-3}\alpha(\alpha-1)(\alpha-2)=0$$ which yields $\alpha=0,1,2$. Here, my second question is: could I say the far-field asymptotic behavior is $$y\sim Ax^2,$$ because this is the fastest growing one?

Update Thanks for @Frits Veerman's answer, I understand that the far-field behavior is not affected by the last term $\frac{a}{y}$, which is my "afraid" term. So come to my third questions: whether or not the following derivation is right?

Case 3. Now, an alternative far-field behavior is produced $$\left(y^3y^{\prime\prime\prime}\right)^\prime \sim \frac{1}{2}y-\frac{5}{8}xy^\prime \quad as \quad x \rightarrow \infty,$$ Again let $y=Bx^\alpha$, by plug it in above asymptotics, I have $$\frac{1}{2}(1-\frac{5}{4}\alpha)=B^3\alpha(\alpha-1)(\alpha-2)(4\alpha-3)x^{3\alpha-4}.$$ For this equality, it must have $$3\alpha-4=0,$$ which gives $\alpha=\frac{4}{3}.$ Thus I obtain a different far-field behavior $$y=Bx^{\frac{4}{3}}.$$

Am I right?

I think this question is related to the perturbation solution of ODE. Anyone can help me, or recommend some famous books or journal papers to me? Many thanks!

Regards,

Robin

To be very short: you've obtained the right solution, i.e. the far field behaviour of $y$ is indeed $y \leadsto c\,x^\frac{4}{5}$, for some $c$ to be determined by the boundary conditions in the far field. Now, how to show this 'properly'?

The standard approach is to rescale the $x$-variable, as we're interested in the far field behaviour of the solution, i.e. in large values of $x$. Introducing \begin{equation} x = \frac{\xi}{\varepsilon},\qquad 0<\varepsilon \ll 1, \end{equation} the ODE transforms to \begin{equation} \varepsilon \left( \varepsilon^3 y\, y_{\xi\xi\xi}\right)_\xi +\frac{5}{8} \xi y_\xi - \frac{1}{2}y + \frac{a}{y} = 0. \end{equation} As $\varepsilon^4$ is indeed very small, we can argue that the influence of this term in the ODE is negligible. So, we can look at the reduced ODE \begin{equation} \frac{5}{8} \xi y_\xi - \frac{1}{2}y + \frac{a}{y} = 0, \end{equation} which has a solution $y(\xi) = \sqrt{2 a+ c \xi^{\frac{8}{5}}}$. So, $y$ grows with $\xi$, so for large $\xi$, we see that $y \leadsto c\,\xi^{\frac{4}{5}}$. In hindsight, that means that the term $\frac{a}{y}$ in the ODE also becomes negligible for large values of $y$ -- but that's something we didn't know beforehand.

A very good, albeit somewhat mathematical, source for these techniques is

C. Kuehn, Multiple Time Scale Dynamics, Springer, 2015.

For a more gentle introduction, you could look at

M.H. Holmes, Introduction to Perturbation Methods, Springer, 1995,

or the golden oldie

J. Kevorkian and J.D. Cole, Perturbation Methods in Applied Mathematics, Springer, 1985 (2nd ed).

Note: I skipped over a lot of details in the above: in particular, I neglected the possibility that $y$ could have very steep jumps, which would mean that the term $\varepsilon\left(\varepsilon^3y\,y_{\xi\xi\xi}\right)_\xi$ is not small at those jumps. If you're interested, do read more about multiple time scale analysis and asymptotic techniques.

Addition: To address the second part of your question: the behaviour of $y$ for small values of $x$ is a priori uncorrelated with its far field behaviour, i.e. its behaviour for large values of $x$. So, if you're interested in far field behaviour, it's not really useful to study the behaviour of the ODE as $x \to 0$. However, if you're interested in the behaviour of $y$ near $x=0$, you could indeed try to substitute $y = A x^\alpha$ and choose $A$, $\alpha$ such that the leading order terms (i.e. lowest order in $x$) cancel. My back-of-the-envelope analysis suggests $y \leadsto \left(-\frac{625 a}{24}x^4\right)^{\frac{1}{5}}$ as $x \to 0$.

Actually, when calculating the above, you find that \begin{equation} y(x) = \left(-\frac{625 a}{24}x^4\right)^{\frac{1}{5}} \end{equation} is an exact solution to the ODE. So that's nice, and definitely helpful.

To address the third part of your question: indeed, if you neglect the $\frac{a}{y}$-term and substitute $B x^\beta$, you obtain \begin{equation} y(x) = \left(\frac{27}{56} x^4\right)^{\frac{1}{3}}, \end{equation} again as an exact solution.

The lesson to be learned here? Don't just assume certain terms in the ODE are negligible, because this can (and often will) influence the result. Rescaling your variables (like I did with $\xi = \varepsilon x$) will show you when certain terms are negligible, and what you have to assume to drop them. Otherwise, you're just solving a different ODE, with different results.