複素数と複素平面複素数の極形式オイラーの公式、指数関数、微分方程式の解、初期条件

学習環境

複素関数演習〈理工系の数学入門コース/演習新装版〉 (表実(著)、迫田誠治(著)、岩波書店)の第1章(複素数と複素平面)、1-3(複素数の極形式)、問題6の解答を求めてみる。

\frac{d}{dx} f_{\pm k} (x) = \frac{d}{dx} (\cos (\pm k x) + i \sin (\pm k x))

= \mp k \sin (\pm k x) \pm k i \cos (\pm k x)

= \pm i k (- \frac{1}{i} \sin (\pm k x) + \cos (\pm k x))

= \pm i k (\cos (\pm k x) + i \sin (\pm k x))

= \pm i k e^{\pm k i x}

\frac{d^{2}}{{dx}^{2}} f_{\pm k} (x) = \pm i k \frac{d}{dx} e^{\pm k i x} = {(\pm i k)}^{2} e^{\pm k i x} = - k^{2} e^{\pm i k x}

f (x) = e^{i k x}

f (x) = e^{- i k x}

iii

f (x) = a e^{i k x} + b e^{- i k x}

\begin{array}{l} f (0) = a + b = 0 \\ f' (0) = a i k - b i k = k \\ b = - a \\ 2 a i = 1 \\ a = \frac{1}{2 i} \\ y = - \frac{1}{2 i} \end{array}

f (x) = \frac{1}{2 i} e^{i k x} - \frac{1}{2 i} e^{- i k x}

\begin{array}{l} f (0) = a + b = 1 \\ f' (0) a i k - b i k = 0 \\ a = b = \frac{1}{2} \end{array}

f (x) = \frac{1}{2} e^{i k x} + \frac{1}{2} e^{- i k x}

コード(Wolfram Language, Jupyter)

DSolve[{f''[x] == -k^2f[x], f[0] == 1, f'[0] == I k}, f[x], x]

DSolveValue[{f''[x] == -k^2f[x], f[0] == 1, f'[0] == I k}, f[x], x]

% == Exp[I k x]

Simplify[%]

DSolveValue[{f''[x] == -k^2f[x], f[0] == 1, f'[0] == -I k}, f[x], x]

% == Exp[-I k x] // Simplify

DSolveValue[{f''[x] == -k^2f[x], f[0] == 0, f'[0] == k}, f[x], x]

% == (Exp[I k x] - Exp[-I k x]) / (2I) // Simplify

DSolveValue[{f''[x] == -k^2f[x], f[0] == 1, f'[0] == 0}, f[x], x]

% == (Exp[I k x] + Exp[-I k x]) / 2 // Simplify