# 制御と振動の数学/第一類/Laplace 変換/三角関数の Laplace 変換とその応用

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## §1

(2.18)
${\displaystyle {\frac {d^{2}\sin \beta t}{dt^{2}}}=-\beta ^{2}\sin \beta t}$

を公式

${\displaystyle {\mathcal {L}}[f'']=s^{2}{\mathcal {L}}[f]-sf(0)-f'(0)}$

を用いて Laplace 変換する．

${\displaystyle f(t)=\sin \beta t}$

とおくと，${\displaystyle f(0)=0,f'(0)=\beta }$ であるから，式(2.18)

${\displaystyle s^{2}{\mathcal {L}}[\sin \beta t]-\beta =-\beta ^{2}{\mathcal {L}}[\sin \beta t]}$
${\displaystyle \therefore {\mathcal {L}}[\sin \beta t]={\frac {\beta }{s^{2}+\beta ^{2}}}}$

を得る．

また，

${\displaystyle \cos \beta t={\frac {1}{\beta }}{\frac {d}{dt}}\sin \beta t}$

を Laplace 変換すると，

${\displaystyle {\mathcal {L}}[\cos \beta t]={\frac {s}{\beta }}{\mathcal {L}}[\sin \beta t]={\frac {s}{s^{2}+\beta ^{2}}}}$

となる．

Laplace 変換の定義式から，直接三角関数の Laplace 変換を導け．

${\displaystyle {\mathcal {L}}[\sin \omega t]=\int _{0}^{\infty }\sin \omega t\ e^{-st}dt}$

${\displaystyle ={\frac {1}{s}}\left[\sin \omega t\ e^{-st}\right]_{\infty }^{0}+{\frac {\omega }{s}}\int _{0}^{\infty }\cos \omega t\ e^{-st}dt}$

${\displaystyle ={\frac {1}{s}}\left(0-0\right)+{\frac {\omega }{s}}\int _{0}^{\infty }\cos \omega t\ e^{-st}dt}$

${\displaystyle ={\frac {\omega }{s}}\left\{{\frac {1}{s}}\left[\cos \omega t\ e^{-st}\right]_{\infty }^{0}-{\frac {\omega }{s}}\int _{0}^{\infty }\sin \omega t\ e^{-st}dt\right\}}$

${\displaystyle ={\frac {\omega }{s^{2}}}(1-0)-{\frac {\omega ^{2}}{s^{2}}}{\mathcal {L}}[\sin \omega t]}$

${\displaystyle \therefore (1+{\frac {\omega ^{2}}{s^{2}}}){\mathcal {L}}[\sin \omega t]={\frac {\omega }{s^{2}}}}$

${\displaystyle \therefore {\mathcal {L}}[\sin \omega t]={\frac {\omega }{s^{2}+\omega ^{2}}}}$

また ${\displaystyle {\frac {\omega }{s}}{\mathcal {L}}[\cos \omega t]={\mathcal {L}}[\sin \omega t]}$ より

${\displaystyle {\mathcal {L}}[\cos \omega t]={\frac {s}{\omega }}\cdot {\frac {\omega }{s^{2}+\omega ^{2}}}={\frac {s}{s^{2}+\omega ^{2}}}}$

${\displaystyle \diamondsuit }$

${\displaystyle {\frac {d^{2}}{dt^{2}}}\cos \beta t=-\beta ^{2}\cos \beta t}$ を Laplace 変換することにより上の結果を導け．

${\displaystyle f(t)=\cos \beta t,\quad f\sqsupset F}$ とすると，

${\displaystyle s^{2}F-s\cdot 1-0=-\beta ^{2}F}$

${\displaystyle \therefore {\mathcal {L}}[\cos \beta t]=F={\frac {s}{s^{2}+\beta ^{2}}}}$

${\displaystyle \diamondsuit }$

${\displaystyle \sin \beta t\sqsupset {\frac {\beta }{s^{2}+\beta ^{2}}}}$
(2.19)
${\displaystyle \cos \beta t\sqsupset {\frac {s}{s^{2}+\beta ^{2}}}}$

を得る．

${\displaystyle {\frac {\beta }{s^{2}+\beta ^{2}}}={\frac {\beta }{s^{2}(1+{\frac {\beta ^{2}}{s^{2}}})}}={\frac {\beta }{s^{2}}}-{\frac {\beta ^{3}}{s^{4}}}+{\frac {\beta ^{5}}{s^{6}}}-\cdots }$[1]

この原像を求めると，

${\displaystyle \sin \beta t=\beta t-{\frac {\beta ^{3}t^{3}}{3!}}+{\frac {\beta ^{5}t^{5}}{5!}}-}$

となる．これは ${\displaystyle \sin \beta t}$ Taylor 展開である．

${\displaystyle {\frac {s}{s^{2}+\beta ^{2}}}={\frac {s}{s^{2}(1+{\frac {\beta ^{2}}{s^{2}}})}}}$${\displaystyle ={\frac {1}{s}}-{\frac {\beta ^{2}}{s^{3}}}+{\frac {\beta ^{4}}{s^{5}}}-{\frac {\beta ^{6}}{s^{7}}}+\cdots }$[2]
よってその原像は，
${\displaystyle \cos \beta t=1-{\frac {\beta ^{2}t^{2}}{2!}}+{\frac {\beta ^{4}t^{4}}{4!}}-{\frac {\beta ^{6}t^{6}}{6!}}+\cdots }$

${\displaystyle \diamondsuit }$

1. ^ 初項：${\displaystyle {\frac {\beta }{s^{2}}}}$，公比：${\displaystyle -{\frac {\beta ^{2}}{s^{2}}}}$ の無限等比級数．
2. ^ 初項：${\displaystyle {\frac {1}{s}}}$，公比：${\displaystyle -{\frac {\beta ^{2}}{s^{2}}}}$ の無限等比級数．

${\displaystyle {\frac {d^{2}x}{dt^{2}}}+4x=0;\quad x(0)=2,x'(0)=-3}$

を解け．

${\displaystyle s^{2}{\mathcal {L}}[x]-2s+3+4{\mathcal {L}}[x]=0}$

よって，

${\displaystyle {\mathcal {L}}[x]={\frac {2s}{s^{2}+4}}-{\frac {3}{s^{2}+4}}=2\cdot {\frac {s}{s^{2}+2^{2}}}+{\frac {3}{2}}{\frac {2}{s^{2}+2^{2}}}}$

この原像は，

${\displaystyle x(t)=2\cos 2t-{\frac {3}{2}}\sin 2t}$

${\displaystyle \diamondsuit }$

${\displaystyle {\begin{cases}{\frac {dx}{dt}}&=\beta y\\{\frac {dy}{dt}}&=-\beta x\end{cases}}\quad {\begin{cases}x(0)&=0\\y(0)&=1\end{cases}}}$

を解け．

${\displaystyle {\begin{cases}s{\mathcal {L}}[x]&=\beta {\mathcal {L}}[y]\\s{\mathcal {L}}[y]-1&=-\beta {\mathcal {L}}[x]\end{cases}}}$

これを ${\displaystyle {\mathcal {L}}[x]}$${\displaystyle {\mathcal {L}}[y]}$ について解くと，

${\displaystyle {\begin{cases}{\mathcal {L}}[x]&={\frac {\beta }{s^{2}+\beta ^{2}}}\\{\mathcal {L}}[y]&={\frac {s}{s^{2}+\beta ^{2}}}\end{cases}}}$[1]

となるから，この原像は

${\displaystyle {\begin{cases}x(t)&=\sin \beta t\\y(t)&=\cos \beta t\end{cases}}}$

である．

${\displaystyle \diamondsuit }$

1. ^ ${\displaystyle X\sqsubset x,Y\sqsubset y}$ とすると ${\displaystyle X,Y}$ の連立方程式
${\displaystyle {\begin{cases}sX-\beta Y&=0\\\beta X+sY&=1\end{cases}}}$
${\displaystyle {\begin{pmatrix}s&-\beta \\\beta &s\end{pmatrix}}\left({\begin{array}{c}X\\Y\end{array}}\right)=\left({\begin{array}{c}0\\1\end{array}}\right)}$
を得る．したがって，
${\displaystyle \left({\begin{array}{c}X\\Y\end{array}}\right)={\begin{pmatrix}s&-\beta \\\beta &s\end{pmatrix}}^{-1}\left({\begin{array}{c}0\\1\end{array}}\right)={\frac {1}{s^{2}+\beta ^{2}}}{\begin{pmatrix}s&\beta \\-\beta &s\end{pmatrix}}\left({\begin{array}{c}0\\1\end{array}}\right)={\frac {1}{s^{2}+\beta ^{2}}}\left({\begin{array}{c}\beta \\s\end{array}}\right)}$

バネの振動

${\displaystyle m{\frac {d^{2}x}{dt^{2}}}=-kx}$

の周期を求めてみよう．今，

${\displaystyle {\frac {d^{2}x}{dt^{2}}}+\beta ^{2}x=0,\quad \beta ={\sqrt {\frac {k}{m}}}}$

と変形しておいて Laplace 変換する．

${\displaystyle s^{2}{\mathcal {L}}[x]-sx(0)-x'(0)+\beta ^{2}{\mathcal {L}}[x]=0}$

これを ${\displaystyle {\mathcal {L}}[x]}$ について解くと，

${\displaystyle {\mathcal {L}}[x]={\frac {s}{s^{2}+\beta ^{2}}}x(0)+{\frac {x'(0)}{s^{2}+\beta ^{2}}}}$

この原像を求めると，

${\displaystyle x(t)=x_{0}\cos \beta t+{\frac {v_{0}}{\beta }}\sin \beta t}$

ただし ${\displaystyle x(0)=x_{0},\ \ x'(0)=v_{0}}$ とおいた．次に ${\displaystyle \sin }$${\displaystyle \cos }$ を合成して

${\displaystyle x(t)=A\sin(\beta t+\varphi )}$

ここに，

${\displaystyle A:={\sqrt {x_{0}^{2}+{\frac {v_{0}^{2}}{\beta ^{2}}}}},\quad \varphi :=\tan ^{-1}{\frac {\beta x_{0}}{v_{0}}}}$

と変形すると，周期 ${\displaystyle T}$ は，

${\displaystyle T={\frac {2\pi }{\beta }}=2\pi {\sqrt {\frac {m}{k}}}}$

であることが分かる．さて、バネに錘 ${\displaystyle m}$ をつけたときの伸びを ${\displaystyle \delta }$ とすると， 力の釣り合いの式から，

${\displaystyle k\delta =mg\quad \therefore {\frac {m}{k}}={\frac {\delta }{g}}}$

を得るから，この伸び ${\displaystyle \delta }$ を用いると，

${\displaystyle T=2\pi {\sqrt {\frac {\delta }{g}}}}$

となる．${\displaystyle \delta }$ を静たわみと呼ぶことがある． このバネ振子と振り子とを比べてみると面白い． この振り子の運動方程式，

${\displaystyle ml{\frac {d^{2}\theta }{dt^{2}}}=-mg\sin \theta }$

は，${\displaystyle \theta }$ が小さいときは ${\displaystyle \sin \theta \fallingdotseq \theta }$ であるから

${\displaystyle {\frac {d^{2}\theta }{dt^{2}}}+\beta ^{2}\theta =0,\quad \beta ^{2}={\frac {g}{l}}}$

となる．よってこの振り子の周期は，

${\displaystyle T=2\pi {\sqrt {\frac {l}{g}}}}$

である．さきの静たわみ ${\displaystyle \delta }$ は，この振り子の長さ ${\displaystyle l}$ に相当する．

${\displaystyle {\frac {d^{2}x}{dt^{2}}}+\beta ^{2}x=f(t),\quad x(0)=x_{0},\ \ x'(0)=v_{0}}$

${\displaystyle X\sqsubset x,\ \ F\sqsubset f(t)}$ とおき，与方程式の Laplace 変換をとると，

${\displaystyle s^{2}X-sx_{0}-v_{0}+\beta ^{2}X=F}$

${\displaystyle \therefore X={\frac {x_{0}s+v_{0}}{s^{2}+\beta ^{2}}}+{\frac {1}{\beta }}{\frac {\beta F}{s^{2}+\beta ^{2}}}}$

この原像は，

${\displaystyle x(t)=x_{0}\cos \beta t+{\frac {v_{0}}{\beta }}\sin \beta t+{\frac {1}{\beta }}\int _{0}^{t}f(\tau )\sin \beta (t-\tau )d\tau }$

${\displaystyle \diamondsuit }$

${\displaystyle {\begin{cases}{\frac {dx}{dt}}&=\beta y\\{\frac {dy}{dt}}&=-\beta x\end{cases}},\quad {\begin{cases}x(0)&=x_{0}\\y(0)&=y_{0}\end{cases}}}$

${\displaystyle X\sqsubset x(t),\ \ Y\sqsubset y(t)}$ とおき，与方程式の Laplace 変換をとると，

${\displaystyle {\begin{cases}sX-x_{0}&=\beta Y\\sY-y_{0}&=-\beta X\end{cases}}}$

${\displaystyle {\begin{pmatrix}s&-\beta \\\beta &s\end{pmatrix}}\left({\begin{array}{c}X\\Y\end{array}}\right)=\left({\begin{array}{c}x_{0}\\y_{0}\end{array}}\right)}$

${\displaystyle \left({\begin{array}{c}X\\Y\end{array}}\right)={\begin{pmatrix}s&-\beta \\\beta &s\end{pmatrix}}^{-1}\left({\begin{array}{c}x_{0}\\y_{0}\end{array}}\right)={\frac {1}{s^{2}+\beta ^{2}}}{\begin{pmatrix}s&\beta \\-\beta &s\end{pmatrix}}\left({\begin{array}{c}x_{0}\\y_{0}\end{array}}\right)}$${\displaystyle ={\frac {1}{s^{2}+\beta ^{2}}}\left({\begin{array}{c}sx_{0}+\beta y_{0}\\-\beta x_{0}+sy_{0}\end{array}}\right)}$
この原像は，
${\displaystyle {\begin{cases}x(t)&=x_{0}\cos \beta t+y_{0}\sin \beta t\\y(t)&=y_{0}\cos \beta t-x_{0}\sin \beta t\end{cases}}}$

${\displaystyle \diamondsuit }$

## §2

${\displaystyle f(t)\sqsupset F(s)\Longrightarrow f(t)e^{\alpha t}\sqsupset F(s-\alpha )}$

を式(2.19) に用いると，

${\displaystyle e^{\alpha t}\sin \beta t\sqsupset {\frac {\beta }{(s-\alpha )^{2}+\beta ^{2}}}}$
(2.20)
${\displaystyle e^{\alpha t}\cos \beta t\sqsupset {\frac {s-\alpha }{(s-\alpha )^{2}+\beta ^{2}}}}$

を得る．

${\displaystyle {\frac {d^{2}x}{dt^{2}}}+2{\frac {dx}{dt}}+5x=4,\quad x(0)=2,\quad x'(0)=-4}$

を解け．

${\displaystyle x(t)\sqsupset X(s)}$ とおくと，

${\displaystyle \{s^{2}X-2s+4\}+2\{sX-2\}+5X=0}$

これを ${\displaystyle X(s)}$ について解く．

${\displaystyle X={\frac {2s}{s^{2}+2s+5}}={\frac {2(s+1)-2}{(s+1)^{2}+2^{2}}}}$[1]

この原像を求めると，

${\displaystyle x(t)=e^{-t}(2\cos 2t-\sin 2t)}$[2]

${\displaystyle \diamondsuit }$

${\displaystyle {\frac {d^{2}x}{dt^{2}}}+{\frac {dx}{dt}}+x=f(x)}$
${\displaystyle x(0)=x'(0)=0}$

を解け．

${\displaystyle x(t)\sqsupset X(s),f(t)\sqsupset F(s)}$

とおくと，

${\displaystyle s^{2}X+sX+X=F}$

となる．これを ${\displaystyle X}$ について解くと，

${\displaystyle X={\frac {F}{s^{2}+s+1}}}$

ところで

${\displaystyle {\frac {1}{s^{2}+s+1}}={\frac {1}{\left(s+{\frac {1}{2}}\right)^{2}+\left({\frac {\sqrt {3}}{2}}\right)^{2}}}\sqsubset {\frac {2}{\sqrt {3}}}e^{-{\frac {t}{2}}}\sin {\frac {\sqrt {3}}{2}}t}$

であるから

${\displaystyle x(t)={\frac {2}{\sqrt {3}}}\int _{0}^{t}\left\{e^{-{\frac {1}{2}}(t-\tau )}\sin {\frac {\sqrt {3}}{2}}(t-\tau )\right\}f(\tau )d\tau }$[3]

${\displaystyle \diamondsuit }$

${\displaystyle {\frac {dx^{2}}{dt^{2}}}+{\frac {dx}{dt}}+x=7e^{2t}\quad x(0)=x'(0)=0}$

${\displaystyle x(t)\sqsupset X}$

とおくと，

${\displaystyle s^{2}X+sX+X={\frac {7}{s-2}}}$
${\displaystyle \therefore X={\frac {7}{(s-2)(s^{2}+s+1)}}={\frac {A}{s-2}}+{\frac {Bs+C}{s^{2}+s+1}}}$

とおいて，

${\displaystyle A=1,B=-1,C=-3}$

すなわち

${\displaystyle X={\frac {1}{s-2}}+{\frac {-s-3}{s^{2}+s+1}}}$
${\displaystyle ={\frac {1}{s-2}}+{\frac {-(s+{\frac {1}{2}})}{s^{2}+s+1}}+{\frac {-{\frac {5}{2}}}{s^{2}+s+1}}}$
${\displaystyle ={\frac {1}{s-2}}+}$${\displaystyle {\frac {-(s+{\frac {1}{2}})}{(s+{\frac {1}{2}})^{2}+({\frac {\sqrt {3}}{2}})^{2}}}+{\frac {-5}{\sqrt {3}}}{\frac {\frac {\sqrt {3}}{2}}{(s+{\frac {1}{2}})^{2}+({\frac {\sqrt {3}}{2}})^{2}}}}$

この原像は，

${\displaystyle x(t)=e^{2t}-e^{-{\frac {t}{2}}}\left(\cos {\frac {\sqrt {3}}{2}}t+{\frac {5}{\sqrt {3}}}\sin {\frac {\sqrt {3}}{2}}t\right)}$

${\displaystyle \diamondsuit }$

${\displaystyle x(t)={\frac {2}{\sqrt {3}}}\int _{0}^{t}\left\{e^{-{\frac {1}{2}}(t-\tau )}\sin {\frac {\sqrt {3}}{2}}(t-\tau )\right\}f(\tau )d\tau }$

${\displaystyle f(t)=7e^{2t}}$ を代入すると，

${\displaystyle x(t)={\frac {2}{\sqrt {3}}}\int _{0}^{t}e^{\frac {-t+\tau }{2}}\sin {\frac {\sqrt {3}}{2}}(t-\tau )\cdot 7e^{2\tau }d\tau }$
${\displaystyle ={\frac {14}{\sqrt {3}}}e^{-{\frac {t}{2}}}\int _{0}^{t}e^{\frac {5\tau }{2}}\sin {\frac {\sqrt {3}}{2}}(t-\tau )d\tau }$[4]
${\displaystyle I_{1}=\int _{0}^{t}e^{{\frac {5}{2}}\tau }\sin {\frac {\sqrt {3}}{2}}(t-\tau )d\tau }$ とおいて部分積分を実行すると，
${\displaystyle I_{1}={\frac {2}{\sqrt {3}}}\left[e^{\frac {5\tau }{2}}\cos {\frac {\sqrt {3}}{2}}(t-\tau )\right]_{0}^{t}-{\frac {5}{2}}\cdot {\frac {2}{\sqrt {3}}}\int _{0}^{t}e^{\frac {5\tau }{2}}\cos {\frac {\sqrt {3}}{2}}(t-\tau )d\tau }$
${\displaystyle ={\frac {2}{\sqrt {3}}}\left\{e^{{\frac {5}{2}}t}-\cos {\frac {\sqrt {3}}{2}}t\right\}-{\frac {5}{\sqrt {3}}}\int _{0}^{t}e^{\frac {5\tau }{2}}\cos {\frac {\sqrt {3}}{2}}(t-\tau )d\tau }$
${\displaystyle I_{2}=\int _{0}^{t}e^{\frac {5\tau }{2}}\cos {\frac {\sqrt {3}}{2}}(t-\tau )d\tau }$ とおいて部分積分を実行すると，
${\displaystyle I_{2}={\frac {-2}{\sqrt {3}}}\left[e^{{\frac {5}{2}}\tau }\sin {\frac {\sqrt {3}}{2}}(t-\tau )\right]_{0}^{t}+{\frac {5}{2}}\cdot {\frac {2}{\sqrt {3}}}\int _{0}^{t}e^{{\frac {5}{2}}\tau }\sin {\frac {\sqrt {3}}{2}}(t-\tau )d\tau }$
${\displaystyle ={\frac {-2}{\sqrt {3}}}\left\{0-\sin {\frac {\sqrt {3}}{2}}t\right\}+{\frac {5}{\sqrt {3}}}\int _{0}^{t}e^{{\frac {5}{2}}\tau }\sin {\frac {\sqrt {3}}{2}}(t-\tau )d\tau }$
${\displaystyle ={\frac {2}{\sqrt {3}}}\sin {\frac {\sqrt {3}}{2}}t+{\frac {5}{\sqrt {3}}}I_{1}}$

${\displaystyle I_{1}={\frac {2}{\sqrt {3}}}\left(e^{{\frac {5}{2}}t}-\cos {\frac {\sqrt {3}}{2}}t\right)-{\frac {5}{\sqrt {3}}}\left({\frac {2}{\sqrt {3}}}\sin {\frac {\sqrt {3}}{2}}t+{\frac {5}{\sqrt {3}}}I_{1}\right)}$
${\displaystyle ={\frac {2}{\sqrt {3}}}\left(e^{{\frac {5}{2}}t}-\cos {\frac {\sqrt {3}}{2}}t\right)-{\frac {10}{3}}\sin {\frac {\sqrt {3}}{2}}t-{\frac {25}{3}}I_{1}}$

すなわち，

${\displaystyle \left(1+{\frac {25}{3}}\right)I_{1}={\frac {2}{\sqrt {3}}}\left(e^{{\frac {5}{2}}t}-\cos {\frac {\sqrt {3}}{2}}t\right)-{\frac {10}{3}}\sin {\frac {\sqrt {3}}{2}}t}$
${\displaystyle \therefore I_{1}={\frac {3}{28}}\cdot {\frac {2}{\sqrt {3}}}\left(e^{{\frac {5}{2}}t}-\cos {\frac {\sqrt {3}}{2}}t\right)-{\frac {3}{28}}\cdot {\frac {10}{3}}\sin {\frac {\sqrt {3}}{2}}t}$
${\displaystyle ={\frac {\sqrt {3}}{14}}\left(e^{{\frac {5}{2}}t}-\cos {\frac {\sqrt {3}}{2}}t\right)-{\frac {5}{14}}\sin {\frac {\sqrt {3}}{2}}t}$
${\displaystyle \therefore x(t)={\frac {14}{\sqrt {3}}}e^{-{\frac {t}{2}}}\left\{{\frac {\sqrt {3}}{14}}\left(e^{{\frac {5}{2}}t}-\cos {\frac {\sqrt {3}}{2}}t\right)-{\frac {5}{14}}\sin {\frac {\sqrt {3}}{2}}t\right\}}$
${\displaystyle =e^{2t}-e^{-{\frac {t}{2}}}\left(\cos {\frac {\sqrt {3}}{2}}t+{\frac {5}{\sqrt {3}}}\sin {\frac {\sqrt {3}}{2}}t\right)}$

${\displaystyle \diamondsuit }$

${\displaystyle {\frac {dx^{2}}{dt^{2}}}+{\frac {dx}{dt}}+x=f(t),\quad x(0)=x_{0},x'(0)=v_{0}}$

${\displaystyle x(t)\sqsupset X,f(t)\sqsupset F}$

とおくと

${\displaystyle (s^{2}X-x_{0}s-v_{0})+(sX-x_{0})+X=F}$
${\displaystyle \therefore X={\frac {x_{0}s+v_{0}+x_{0}}{s^{2}+s+1}}+{\frac {F}{s^{2}+s+1}}}$

${\displaystyle {\frac {x_{0}s+v_{0}+x_{0}}{s^{2}+s+1}}={\frac {x_{0}(s+{\frac {1}{2}})+v_{0}+{\frac {x_{0}}{2}}}{(s+{\frac {1}{2}})^{2}+{\frac {3}{4}}}}}$
${\displaystyle =x_{0}\cdot {\frac {s+{\frac {1}{2}}}{(s+{\frac {1}{2}})^{2}+({\frac {\sqrt {3}}{2}})^{2}}}+(v_{0}+{\frac {x_{0}}{2}})\cdot {\frac {2}{\sqrt {3}}}\cdot {\frac {\frac {\sqrt {3}}{2}}{(s+{\frac {1}{2}})^{2}+({\frac {\sqrt {3}}{2}})^{2}}}}$

この原像は，

${\displaystyle u(t)=x_{0}e^{-{\frac {1}{2}}t}\cos {\frac {\sqrt {3}}{2}}t+(v_{0}+{\frac {x_{0}}{2}})\cdot {\frac {2}{\sqrt {3}}}e^{-{\frac {1}{2}}t}\sin {\frac {\sqrt {3}}{2}}t}$

${\displaystyle v(t)={\frac {2}{\sqrt {3}}}\left(e^{-{\frac {1}{2}}t}\sin {\frac {\sqrt {3}}{2}}t\right)*f(t)}$
${\displaystyle ={\frac {2}{\sqrt {3}}}\int _{0}^{t}\left\{e^{-{\frac {1}{2}}(t-\tau )}\sin {\frac {\sqrt {3}}{2}}(t-\tau )\right\}f(\tau )d\tau }$

よって解 ${\displaystyle x(t)}$ は，

${\displaystyle x(t)=u(t)+v(t)=x_{0}e^{-{\frac {1}{2}}t}\cos {\frac {\sqrt {3}}{2}}t+(v_{0}+{\frac {x_{0}}{2}})\cdot {\frac {2}{\sqrt {3}}}e^{-{\frac {1}{2}}t}\sin {\frac {\sqrt {3}}{2}}t+{\frac {2}{\sqrt {3}}}\int _{0}^{t}\left\{e^{-{\frac {1}{2}}(t-\tau )}\sin {\frac {\sqrt {3}}{2}}(t-\tau )\right\}f(\tau )d\tau }$

${\displaystyle \diamondsuit }$

1. ^ これはまず二次式 ${\displaystyle s^{2}+2s+5}$ を平方式 ${\displaystyle (s+1)^{2}-1+5}$ に展開し，分子は一次項 ${\displaystyle (s+1)}$ を含むように適当な定数を足し引きしたもの．
2. ^ ${\displaystyle {\frac {2s}{s^{2}+2s+5}}={\frac {2(s+1)-2}{(s+1)^{2}+2^{2}}}={\frac {2(s+1)}{(s+1)^{2}+2^{2}}}-{\frac {2}{(s+1)^{2}+2^{2}}}}$
${\displaystyle \sqsubset e^{-t}\cdot {\mathcal {L}}^{-1}\left[2\cdot {\frac {s}{s^{2}+2^{2}}}-{\frac {2}{s^{2}+2^{2}}}\right]}$
${\displaystyle =e^{-t}\left(2\cos 2t-\sin 2t\right)}$
3. ^ ${\displaystyle \because {\frac {F}{s^{2}+s+1}}={\frac {F}{\left(s+{\frac {1}{2}}\right)^{2}+\left({\frac {\sqrt {3}}{2}}\right)^{2}}}\sqsubset }$${\displaystyle {\mathcal {L}}^{-1}\left[{\frac {1}{\left(s+{\frac {1}{2}}\right)^{2}+\left({\frac {\sqrt {3}}{2}}\right)^{2}}}\right]*f(t)=\left(e^{-{\frac {1}{2}}t}\cdot {\mathcal {L}}^{-1}\left[{\frac {2}{\sqrt {3}}}{\frac {\frac {\sqrt {3}}{2}}{s^{2}+\left({\frac {\sqrt {3}}{2}}\right)^{2}}}\right]\right)*f(t)}$
${\displaystyle ={\frac {2}{\sqrt {3}}}\left(e^{-{\frac {1}{2}}t}\sin {\frac {\sqrt {3}}{2}}t\right)*f(t)}$
4. ^ さらに加法定理を使いたくなるが，ここは我慢のしどころである…．

## §3

${\displaystyle {\frac {1}{(s^{2}+\beta ^{2})^{2}}}}$ および ${\displaystyle {\frac {s}{(s^{2}+\beta ^{2})^{2}}}}$

の原像を求めよう．

(i)

${\displaystyle {\frac {1}{s^{2}+\beta ^{2}}}\sqsubset {\frac {1}{\beta }}\sin \beta t}$

であったから，

${\displaystyle {\frac {1}{(s^{2}+\beta ^{2})^{2}}}\sqsubset {\frac {1}{\beta }}\sin \beta t*{\frac {1}{\beta }}\sin \beta t={\frac {1}{\beta ^{2}}}\int _{0}^{t}\sin \beta (t-\tau )\sin \beta \tau d\tau }$
${\displaystyle ={\frac {1}{2\beta ^{2}}}\int _{0}^{t}\{\cos \beta (t-2\tau )-\cos \beta t\}d\tau }$[1]
${\displaystyle ={\frac {1}{2\beta ^{2}}}\left[{\frac {\sin \beta (t-2\tau )}{-2\beta }}-\tau \cos \beta t\right]_{0}^{t}}$
${\displaystyle ={\frac {1}{2\beta ^{2}}}\left[{\frac {1}{-2\beta }}\left\{\sin \beta (-t)-\sin \beta t\right\}-t\cos \beta t\right]}$
${\displaystyle ={\frac {1}{2\beta ^{2}}}\left({\frac {1}{\beta }}\sin \beta t-t\cos \beta t\right)}$

(ii)

${\displaystyle {\frac {1}{(s^{2}+\beta ^{2})^{2}}}\sqsubset x(t)}$

とおくと，今求めた ${\displaystyle x={\frac {1}{2\beta ^{2}}}\left({\frac {1}{\beta }}\sin \beta t-t\cos \beta t\right)}$ より ${\displaystyle x(0)=0}$ である．

${\displaystyle {\frac {dx}{dt}}\sqsupset s\cdot {\frac {1}{(s^{2}+\beta ^{2})^{2}}}-sx(0)}$

にて ${\displaystyle x(0)=0}$ より

${\displaystyle {\frac {dx}{dt}}\sqsupset s\cdot {\frac {1}{(s^{2}+\beta ^{2})^{2}}}}$

となるから、上の結果を用いて

${\displaystyle {\frac {s}{(s^{2}+\beta ^{2})}}\sqsubset {\frac {d}{dt}}\left\{{\frac {1}{2\beta ^{2}}}\left({\frac {1}{\beta }}\sin \beta t-t\cos \beta t\right)\right\}={\frac {1}{2\beta ^{2}}}\left({\frac {1}{\beta }}\cdot \beta \cos \beta t-\cos \beta t+t\beta \sin \beta t\right)={\frac {t}{2\beta }}\sin \beta t}$

を得る[2]．以上をまとめると

${\displaystyle {\frac {1}{(s^{2}+\beta ^{2})^{2}}}\sqsubset {\frac {1}{2\beta ^{2}}}\left({\frac {1}{\beta }}\sin \beta t-t\cos \beta t\right)}$
(2.21)
${\displaystyle {\frac {s}{(s^{2}+\beta ^{2})^{2}}}\sqsubset {\frac {t}{2\beta }}\sin \beta t}$

この応用として，外力を伴う単振動を取り扱おう．

${\displaystyle m{\frac {d^{2}x}{dt^{2}}}+kx=F\sin \omega t}$

${\displaystyle t=0}$ では静止していたものとする．[3]いま，

${\displaystyle {\frac {k}{m}}=:\beta ^{2},\quad {\frac {F}{m}}=:K}$

とおくと，上式は，

(2.22)
${\displaystyle {\frac {d^{2}x}{dt^{2}}}+\beta ^{2}x=K\sin \omega t;\quad x(0)=x'(0)=0}$

となる．これを解けばよい．

(i) ${\displaystyle \quad \omega \neq \beta }$ の場合

${\displaystyle (s^{2}+\beta ^{2})X={\frac {K\omega }{s^{2}+\omega ^{2}}}}$
${\displaystyle X={\frac {K\omega }{(s^{2}+\omega ^{2})(s^{2}+\omega ^{2})}}}$
${\displaystyle X={\frac {K\omega }{\omega ^{2}-\beta ^{2}}}\left({\frac {1}{s^{2}+\beta ^{2}}}-{\frac {1}{s^{2}+\omega ^{2}}}\right)}$

この原像は

${\displaystyle x={\frac {K\omega }{\omega ^{2}-\beta ^{2}}}\left({\frac {1}{\beta }}\sin \beta t-{\frac {1}{\omega }}\sin \omega t\right)}$

おもりの位置 ${\displaystyle x}$ に，微分方程式の形に由来する力学系の固有振動の項 ${\displaystyle \sin \beta t}$ の他， 外力による振動の項 ${\displaystyle \sin \omega t}$ が現れていることに注目する．この二つの振動数が近づくほど ${\displaystyle {\frac {K\omega }{\omega ^{2}-\beta ^{2}}}}$ の分母の影響により，${\displaystyle |x|}$ が大きくなることがわかる． 力学系の固有振動数 ${\displaystyle \beta }$ と外力の振動数 ${\displaystyle \omega }$ が同一となると、ついにはこの力学系にて問題を引き起こすのである．

(ii)${\displaystyle \quad \omega =\beta }$ の場合

${\displaystyle X={\frac {K\beta }{(s^{2}+\beta ^{2})^{2}}}}$

この原像は，

${\displaystyle x={\frac {K}{2\beta ^{2}}}(\sin \beta t-\beta t\cdot \cos \beta t)}$

${\displaystyle t\to \infty }$ のとき ${\displaystyle |x(t)|\to \infty }$

となり，建造物の場合などでは破壊が起こる．いわゆる共振現象と呼ばれているものがこれである．

${\displaystyle {\frac {d^{2}x}{dt^{2}}}+9x=6\cos 3t,\quad x(0)=2,x'(0)=0}$

を解け．

${\displaystyle X\sqsubset x}$

とおくと

${\displaystyle s^{2}X-2s-0+9X={\frac {6s}{s^{2}+9}}}$
${\displaystyle X={\frac {2s}{s^{2}+9}}+{\frac {6s}{(s^{2}+9)^{2}}}}$

この原像は，

${\displaystyle x=2\cos 3t+6\cdot {\frac {t}{2\cdot 3}}\sin 3t}$
${\displaystyle =2\cos 3t+t\sin 3t}$

${\displaystyle \diamondsuit }$

(2.21)第一移動定理 を用いると，

${\displaystyle {\frac {1}{[(s-\alpha )^{2}+\beta ^{2}]^{2}}}\sqsubset {\frac {e^{\alpha t}}{2\beta ^{2}}}\left({\frac {1}{\beta }}\sin \beta t-t\cos \beta t\right)}$
(2.23)
${\displaystyle {\frac {s-\alpha }{[(s-\alpha )^{2}+\beta ^{2}]^{2}}}\sqsubset {\frac {e^{\alpha t}}{2\beta }}t\sin \beta t}$

を得る．

${\displaystyle {\frac {d^{2}x}{dt^{2}}}+2{\frac {dx}{dt}}+2x=-2e^{-t}\sin t,\quad x(0)=0,x'(0)=1}$

を解け．

${\displaystyle X\sqsubset x}$

とおくと

${\displaystyle s^{2}X-1+2sX+2X=-2\cdot {\frac {1}{(s+1)^{2}+1}}}$
${\displaystyle (s^{2}+2s+2)X=1-2\cdot {\frac {1}{(s+1)^{2}+1}}}$
${\displaystyle X={\frac {1}{(s+1)^{2}+1}}-2\cdot {\frac {1}{[(s+1)^{2}+1]^{2}}}}$

この原像は，

${\displaystyle x=e^{-t}\sin t-2\cdot {\frac {e^{-t}}{2}}\left({\frac {1}{1}}\sin t-t\cos t\right)}$
${\displaystyle x=e^{-t}\sin t-e^{-t}(\sin t-t\cos t)}$
${\displaystyle x=te^{-t}\cos t}$

${\displaystyle \diamondsuit }$

1. ^ 加法定理より
${\displaystyle \cos(A+B)=\cos A\cos B-\sin A\sin B}$…①
${\displaystyle \cos(A-B)=\cos A\cos B+\sin A\sin B}$…②
② - ① より ${\displaystyle \cos(A-B)-\cos(A+B)=2\sin A\sin B}$
すなわち　${\displaystyle \sin A\sin B={\frac {1}{2}}\left\{\cos(A-B)-\cos(A+B)\right\}}$ （積和の公式）
これを適用する．
2. ^ または，${\displaystyle {\frac {s}{(s^{2}+\beta ^{2})^{2}}}={\frac {s}{s^{2}+\beta ^{2}}}\cdot {\frac {1}{\beta }}{\frac {\beta }{s^{2}+\beta ^{2}}}\sqsubset {\frac {1}{\beta }}\cos \beta t*\sin \beta t}$ とし，これを求める．
${\displaystyle {\frac {1}{\beta }}\cos \beta t*\sin \beta t={\frac {1}{\beta }}\int _{0}^{t}\cos \beta (t-\tau )\sin \beta \tau d\tau }$
加法定理より
${\displaystyle \sin(A+B)=\sin A\cos B+\cos A\sin B}$…①
${\displaystyle \sin(A-B)=\sin A\cos B-\cos A\sin B}$…②
① - ② より ${\displaystyle \sin(A+B)-\sin(A-B)=2\cos A\sin B}$
すなわち　${\displaystyle \cos A\sin B={\frac {1}{2}}\left\{\sin(A+B)-\sin(A-B)\right\}}$ （積和の公式）
これを適用すると，
${\displaystyle {\frac {1}{\beta }}\int _{0}^{t}\cos \beta (t-\tau )\sin \beta \tau d\tau ={\frac {1}{2\beta }}\int _{0}^{t}\{\sin \beta t-\sin \beta (t-2\tau )\}d\tau }$
${\displaystyle ={\frac {1}{2\beta }}\left\{t\sin \beta t+\left[{\frac {\cos \beta (t-2\tau )}{-2\beta }}\right]_{0}^{t}\right\}}$
${\displaystyle ={\frac {1}{2\beta }}\left[t\sin \beta t+{\frac {1}{-2\beta }}\left\{\cos(-t)-\cos(t)\right\}\right]}$
${\displaystyle ={\frac {t}{2\beta }}\sin \beta t}$
3. ^ 水平面上，質量 ${\displaystyle m}$ のおもりと自由長 ${\displaystyle L}$，バネ定数 ${\displaystyle k}$ のバネを結合した系を ${\displaystyle X}$ 軸上に置き、このときのおもりの位置を ${\displaystyle x=0}$， バネのおもりとは反対側の一端（静止した状態では ${\displaystyle X=-L}$) の位置に新しい座標系 ${\displaystyle Y}$ を置いて ${\displaystyle Y}$ の大きさおよび正の向きは ${\displaystyle X}$ と同一とし，このバネの一端の座標軸 ${\displaystyle Y=0}$ に対する向きを含めた変位を ${\displaystyle y}$ とする．今 ${\displaystyle x}$ および ${\displaystyle y}$ が任意の値をとるとき，バネの自由長からの伸びは符号を含めて ${\displaystyle x-y}$．おもりに対する運動方程式を立てると，
${\displaystyle m{\frac {d^{2}x}{dt^{2}}}=-k(x-y)}$
すなわち，
${\displaystyle m{\frac {d^{2}x}{dt^{2}}}+kx=ky}$
いま，${\displaystyle y}$ を強制的に変位させ，それが ${\displaystyle y=A\sin \omega t}$ ならば，
${\displaystyle m{\frac {d^{2}x}{dt^{2}}}+kx=kA\sin \omega t}$
これは ${\displaystyle kA=F}$ とおけば，おもり ${\displaystyle m}$ に遠隔力としての外力 ${\displaystyle F\sin \omega t}$ を与えたことと同じである．