京都大学 情報学研究科 数理工学専攻 2023年8月実施 常微分方程式

Author

Casablanca

Description

日本語版

\(a(t), b(t) ≢ 0\) を \(t\) の多項式として次の微分方程式を考える．

\[ \begin{align} \frac{d^2 x}{d t^2} + a(t) \frac{dx}{dt} + b(t)x = 0 \tag{1} \end{align} \]

\(k\) をある自然数として \(x = t^k\) が解であるものとする.このとき，以下の問いに答えよ.

(i) \(k\) を定めよ.

(ii) \(a(t)\) を \(b(t)\) を用いて表わせ.

(iii) 式 (1) は \(x = t^k\) と線形独立な有理関数解をもたないことを示せ.

English Version

Let \(a(t), b(t) ≢ 0\) be polynomials of \(t\) and consider the differential equation

\[ \begin{align} \frac{d^2 x}{d t^2} + a(t) \frac{dx}{dt} + b(t)x = 0 \tag{1} \end{align} \]

Assume that \(x = t^k\) is a solution, where \(k\) is a positive integer. Answer the following questions.

(i) Determine \(k\).

(ii) Express \(a(t)\) in terms of \(b(t)\).

(iii) Show that Eq. (1) has no rational function solution that is linearly independent of \(x = t^k\).

Kai

(i)

if \(k\geq 2\), plug \(x = t^k\) in,

\[ k(k-1)t^{k-2} + kt^{k-1}a(t) + t^kb(t) = 0 \]

thus

\[ k(k-1) + kta(t)+t^2b(t) = 0 \]

since \(a(t)\), \(b(t)\) are both polynomials of t, \(kta(t)+t^2b(t)\) has no constant term. Thus \(k(k-1)=0\), which is in conflict with \(k \geq 2\). Therefore \(k=1\)

(ii)

\(a(t) = -tb(t)\)

(iii)

Let \(x(t) = t u(t)\), we have

\[ x'(t) = u(t) + tu'(t), x''(t) = 2u'(t) + tu''(t) \]

and obtain:

\[ tu''(t) + (2-t^2b(t))u'(t) = 0 \]

Let \(v(t) = u'(t)\)

\[ t\frac{dv(t)}{dt} + (2-t^2b(t))v(t) = 0 \]

since \(x(t)\) is a rational function, we can easily see that \(u(t)\) is a rational function and \(v(t)\) is a rational function. Let \(v(t) = \frac{p(t)}{q(t)}\),

\[ t(p'(t)q(t) - p(t)q'(t)) + (2-t^2b(t))p(t)q(t) = 0 \]

if \(p(t)q(t) \neq 0\), the times of \((2-t^2b(t))p(t)q(t)\) is greater than the times of \(t(p'(t)q(t) - p(t)q'(t))\). Thus \(p(t)q(t) \equiv 0\), \(v(t) = 0\), \(v(t) = C\), \(x(t) = Ct\) is the only ration function solution.