京都大学 情報学研究科 数理工学専攻 2021年8月実施 力学系数学

标签：

• Kyoto-University
• Second-Order-Linear-Ordinary-Differential-Equation
• Polynomial-Solutions-of-Ordinary-Differential-Equation
• Wronskian

Author

Casablanca, 祭音Myyura

Description

大学公表の原題

日本語版

$a(t), b(t)$ を $t$ のある有理式として次の実微分方程式を考える．

$$\frac{d^2 x}{dt} + a(t) \frac{dx}{dt} + b(t)x = 0$$

以下の問いに答えよ．

(i) $k \geqq 1$ をある整数として，$x = t^k$ が式 (1) の解であるための $a(t), b(t)$ に関する必要十分条件を求めよ．

以下では，ある整数 $k \geqq 1$ に対して (i) で求めた条件が成り立つものとし，$\phi(t)$ を $t^k$ と線形独立な解として，

$$p(t) = t\frac{d \phi}{dt} (t) - k \phi(t)$$

とおく．

(ii) $a(t), b(t)$ を $p(t)$ を用いて表わせ．

(iii) $p(t) = t$ のとき $a(t), b(t)$ を定めよ．

(iv) 式 (1) のすべての解が定数でない多項式のとき，$a(t), b(t)$ は多項式でないことを示せ．

English Version

题目描述

设 $a(t),b(t)$ 为关于 $t$ 的有理函数，考虑实微分方程

$$\frac{d^2x}{dt^2}+a(t)\frac{dx}{dt}+b(t)x=0. \tag{1}$$

回答：

1. 对给定整数 $k\ge1$，求 $x=t^k$ 为 (1) 的解时 $a(t),b(t)$ 所满足的充要条件。

以下假设该条件对某个 $k\ge1$ 成立。令 $\phi(t)$ 为与 $t^k$ 线性无关的另一解，并定义

$$p(t)=t\frac{d\phi}{dt}(t)-k\phi(t).$$

继续回答：

2. 用 $p(t)$ 表示 $a(t),b(t)$。
3. 当 $p(t)=t$ 时确定 $a(t),b(t)$。
4. 若 (1) 的每一个解都是非常值多项式，证明 $a(t),b(t)$ 不是多项式。

Kai

(i)

Substitution of $x=t^k$ gives the necessary and sufficient condition

$$k(k-1)+kt\,a(t)+t^2b(t)=0.$$

For $k=1$, this reduces to $a(t)+tb(t)=0$.

(ii)

Let $\phi(t)=u(t)t^k$, where $u$ is not constant. Then

$$\phi'(t) = kt^{k-1}u(t) + t^ku'(t)$$

$$\phi''(t) = t^ku''(t) + 2kt^{k-1}u'(t) + k(k-1)t^{k-2}u(t)$$

then

$$p(t)=t\phi'(t)-k\phi(t)=t^{k+1}u'(t).$$

Substitution gives

$$t u''(t) + (2k + a(t)t)u'(t) = 0$$

Therefore

$$a(t)=-\frac{u''(t)}{u'(t)}-\frac{2k}{t} =\frac{1-k}{t}-\frac{p'(t)}{p(t)},$$

$$b(t)=-\frac{k(k-1)}{t^2}-\frac{k}{t}a(t) =\frac{kp'(t)}{tp(t)}.$$

(iii)

$$a(t)=-\frac{k}{t},\qquad b(t)=\frac{k}{t^2}.$$

(iv)

The original wording says that every solution is a nonconstant polynomial. A homogeneous equation always has the zero solution, so we interpret this as saying that every nonzero solution is a nonconstant polynomial. Under this interpretation, both $a$ and $b$ are individually nonpolynomial; merely showing that they cannot both be polynomials would be weaker.

By the hypothesis, $\phi$ is a polynomial. Thus $p=t\phi'-k\phi$ is a polynomial, and it is nonzero because $\phi$ and $t^k$ are independent. Write $d=\deg p\ge0$. From (ii), as $t\to\infty$,

$$a(t)=\frac{1-k-d}{t}+O(t^{-2}), \qquad b(t)=\frac{kd}{t^2}+O(t^{-3}).$$

If $b$ were a polynomial, its limit $0$ would force $b\equiv0$. Then $x=1$ would be a nonzero constant solution, contrary to the hypothesis. Therefore $b$ is not a polynomial.

If $a$ were a polynomial, the same argument would force $a\equiv0$. Equation (ii) would then give

$$tp'(t)=(1-k)p(t).$$

Comparing leading coefficients gives $d=1-k$. Since $d\ge0$ and $k\ge1$, this implies $k=1$ and $d=0$, hence $p'=0$ and $b\equiv0$, the same contradiction. Thus $a$ is not a polynomial either.