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逆演算子の公式

■逆演算子の公式

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2}D\hspace{2}}F \(x\) =\int F \(x\) dx}$ ⇒導出
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}} \{ cF \(x\) \} =c\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}}F \(x\) }$ ⇒導出
$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}} \{ F \(x\) +G \(x\) \} }$ $3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}}F \(x\) +\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}}G \(x\) }$ ⇒導出
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}}{e}^{ \alpha x }=\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(\alpha \) \hspace{2}}{e{5}\hspace{5}\hspace{5}\hspace{5}\hspace{10} }^{ \alpha x }}$ 　　 $3$\displaystyle{ \( f \(\alpha \) \neq 0 \) }$ ⇒導出
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}}F \(x\) }$ $3$\displaystyle{={e}^{ {\alpha}x }\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \( D+{\alpha} \) \hspace{2}} \[ {e}^{ - {\alpha}x }F \(x\) \] }$ ⇒導出
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} D- {\alpha} \hspace{2}}F \(x\) ={e}^{ {\alpha}x }\int {e}^{ - {\alpha}x } F \(x\) dx}$ ⇒導出
$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}{ \( D- {\alpha} \) }^{n}\hspace{2}}F \(x\) }$ $3$\displaystyle{ ={e}^{ \alpha x }\int \cdot \cdot \cdot \int {e}^{ - \alpha x }F \(x\) dx\cdot \cdot \cdot dx }$ ⇒導出
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}} \[ k{e}^{ i{\alpha}x } \] ={\xi} \(x\) +{\eta} \(x\) i}$ ならば
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}} \[ k\cos {\alpha}x \] ={\xi} \(x\) ,}$ $3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}} \[ k\sin {\alpha}x \] ={\eta} \(x\) }$ ⇒導出
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} {D}^{2}+{a}^{2} \hspace{2}}\cos ax=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2a \hspace{2}}x\sin ax}$ ⇒導出
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}} \[ {e}^{ {\alpha}x }F \(x\) \] ={e}^{ {\alpha}x }\frac{\hspace{2}1\hspace{2}}{\hspace{2} \( D- {\alpha} \) \hspace{2}}F \(x\) }$ ⇒導出
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} {D}^{2}+{a}^{2} \hspace{2}}\sin ax=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2a \hspace{2}}x\cos ax}$ ⇒導出
$3$\displaystyle{ F\left({x}\right) }$ が多項式である定数係数線形微分方程式 $3$\displaystyle{ f\left({D}\right)y=F\left({x}\right) }$ において
$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} f\left({t}\right) \hspace{2}}=\hspace{1}{a}_{0}+\hspace{1}{a}_{1}t+\hspace{1}{a}_{2}{t}^{2}+\cdots }$ $3$\displaystyle{ +\hspace{1}{a}_{r}{t}^{r}+g\left({t}\right){t}^{ r+1 } }$ 　(ただし， $3$\displaystyle{ g\left({t}\right) }$ は多項式)

と表されるとき，逆演算子 $3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} f\left({D}\right) \hspace{2}} }$ は

$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} f\left({D}\right) \hspace{2}}=\hspace{1}{a}_{0}+\hspace{1}{a}_{1}D+\hspace{1}{a}_{2}{D}^{2}+\cdots }$ $3$\displaystyle{ +\hspace{1}{a}_{r}{D}^{r} }$

となる．　⇒導出
$3$\displaystyle{F \(x\) }$ が多項式の場合， $3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} f \(D\) \hspace{2}}F \(x\) }$ を計算するのに次の展開を利用することがある．
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1- aD \hspace{2}}}$ $3$\displaystyle{=1+aD+{a}^{2}{D}^{2}+{a}^{3}{D}^{3}+\cdot \cdot \cdot }$ ⇒導出
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} f\left({D}\right)g\left({D}\right) \hspace{2}}F\left({x}\right)}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} f\left({D}\right) \hspace{2}}\left{{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} g\left({D}\right) \hspace{2}}F\left({x}\right) }\right}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} g\left({D}\right) \hspace{2}}\left{{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} f\left({D}\right) \hspace{2}}F\left({x}\right) }\right}}$ ⇒導出
以下のような逆演算子は，部分分数に分解することが出来る．
$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} \left({ 1- aD }\right)\left({ 1- bD }\right) \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} a- b \hspace{2}}\left{{ \frac{\hspace{2}a\hspace{2}}{\hspace{2} \left({ 1- aD }\right) \hspace{2}}- \frac{\hspace{2}b\hspace{2}}{\hspace{2} \left({ 1- bD }\right) \hspace{2}} }\right}}$ ⇒導出

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最終更新日2024年10月7日