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

Author

Casablanca

Description

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

Kai

(i)

If $k = 1$, then $a(t) = tb(t) = 0.

If $k \geq 2$, then $k(k-1) + kta(t) + t^2b(t) = 0$.

Easy to see $k(k-1) + kta(t) + t^2b(t) = 0$ is neccessary and sufficient.

(ii)

Let $\Phi(t) = u(t)t^k, u(t) !\equiv \text{Constant}$,

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

\[ \Phi''(t) = t^ku''(t) + 2kt^{k-1}u'(t) + k(k-1)t^{k-2}u(t) \]

then

\[ p(t) = t^{k-1}u'(t), p'(t) = (k-1)t^{k-2}u'(t) + t^{k-1}u''(t) \]

And we can obtain:

\[ t u''(t) + (2k + a(t)t)u'(t) = 0 \]

Therefore

\[ a(t) = -\frac{u''(t)}{u'(t)} - \frac{2k}{t} = -\frac{(3k-1)p'(t)}{tp(t)} \]

\[ b(t) = \frac{2k^2}{t^2} - \frac{kp'(t)}{tp(t)} \]

(iii)

\[ a(t) = -\frac{1}{t}, b(t) = \frac{1}{t^2} \]

(iv)

Let $x_1$, $x_2$ be 2 independent particular solutions, then

\[ x_1'' = -a(t)x_1' - b(t)x_1 , x_2'' = -a(t)x_2' - b(t)x_2 \]

and we have

\[ \begin{aligned} (x_1'x_2 - x_1 x_2')' &= x_1''x_2 - x_1 x_2''\\ &= -a(t)x_1'x_2 + a(t)x_1x_2'\\ &= -a(t)(x_1'x_2 - x_1 x_2') \end{aligned} \]

so we get

\[ x_1'x_2 - x_1x_2' = C e^{\int a(t)dt} \]

Let $A(t)$ denote $\int a(t)dt$. Since $a(t)$ is a polynomial, and $Ce^{-A(t)}$ is a polynomial for $x_1'x_2 - x_1x_2'$ is a polynomial, we obtain $a(t) \equiv 0$.

Then, consider

\[ \frac{d^2x}{dt^2} + b(t)x = 0 \]

If $x_1, x_2$ are independent polynomials, and $b(t)$ is polynomials, suppose that $x(t)$ is $m$ times and $b(t)$ is $n$ times, then $b(t) \equiv 0$. Thus $a(t), b(t)$ can't be polynomials.

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