# 制御と振動の数学/第一類/連立微分方程式の解法/連立微分方程式の解法/行列表示と解法

### (1)

(5.9)
${\displaystyle {\begin{cases}{\frac {d^{n}x}{dt^{n}}}+a_{1}{\frac {d^{n-1}x}{dt^{n-1}}}+a_{2}{\frac {d^{n-2}x}{dt^{n-2}}}+\cdots +a_{n-1}{\frac {dx}{dt}}+a_{n}x=f(t)\\x(0)=\xi _{1},\quad x'(0)=\xi _{2},\cdots ,x^{(n-1)}(0)=\xi _{n}\end{cases}}}$

は，次のような変数を選べば，連立微分方程式とみなすことができる．すなわち，

${\displaystyle x_{1}=x,\quad x_{2}:=x',\quad x_{3}:=x'',\cdots ,x_{n}=x^{(n-1)}}$

とおけば，

${\displaystyle {\begin{cases}{\frac {dx_{1}}{dt}}=x_{2}\\{\frac {dx_{2}}{dt}}=x_{3}\\{\frac {dx_{3}}{dt}}=x_{4}\\\vdots \\{\frac {dx_{n-1}}{dt}}=x_{n}\\{\frac {dx_{n}}{dt}}=-a_{n}x_{1}-a_{n-1}x_{2}-a_{n-2}x_{3}-\cdots -a_{2}x_{n-1}-a_{1}x_{n}+f(t)\end{cases}}}$

となり，また初期条件は，

${\displaystyle x_{1}(0)=\xi _{1},\quad x_{2}(0)=\xi _{2},\quad x_{3}(0)=\xi _{3},\cdots ,x_{n}(0)=\xi _{n}}$

となる．

そこで，この節では，もう少し一般化した定数係数の連立 1 階線形微分方程式，

(5.10)
${\displaystyle {\begin{cases}{\frac {x_{1}}{dt}}=a_{11}x_{1}+a_{12}x_{2}+a_{13}x_{3}+\cdots +a_{1(n-1)}x_{n-1}+a_{1n}x_{n}+f_{1}(t)\\{\frac {x_{2}}{dt}}=a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}+\cdots +a_{2(n-1)}x_{n-1}+a_{2n}x_{n}+f_{2}(t)\\{\frac {x_{3}}{dt}}=a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}+\cdots +a_{3(n-1)}x_{n-1}+a_{3n}x_{n}+f_{3}(t)\\\vdots \\{\frac {x_{n}}{dt}}=a_{n1}x_{1}+a_{n2}x_{2}+a_{n3}x_{3}+\cdots +a_{n(n-1)}x_{n-1}+a_{nn}x_{n}+f_{n}(t)\end{cases}}}$

および初期条件

${\displaystyle x_{1}(0)=\xi _{1},\quad x_{2}(0)=\xi _{2},\quad x_{3}(0)=\xi _{3},\cdots ,x_{n}(0)=\xi _{n}}$

を取り扱うことにする．ここで，

${\displaystyle A:={\begin{pmatrix}a_{11}&a_{12}&a_{13}&\cdots &a_{1n}\\a_{21}&a_{22}&a_{23}&\cdots &a_{2n}\\a_{31}&a_{32}&a_{33}&\cdots &a_{3n}\\\vdots &\vdots &\vdots &&\vdots \\a_{n1}&a_{n2}&a_{n3}&\cdots &a_{nn}\\\end{pmatrix}}}$

および，

${\displaystyle {\boldsymbol {x}}:={\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\\vdots \\x_{n}\end{pmatrix}},\quad {\boldsymbol {f}}={\begin{pmatrix}f_{1}\\f_{2}\\f_{3}\\\vdots \\f_{n}\end{pmatrix}},\quad {\boldsymbol {\xi }}={\begin{pmatrix}\xi _{1}\\\xi _{2}\\\xi _{3}\\\vdots \\\xi _{n}\end{pmatrix}}}$

とおけば，式 (5.10) と初期条件は，

(5.11)
${\displaystyle {\frac {d{\boldsymbol {x}}}{dt}}=A{\boldsymbol {x}}+{\boldsymbol {f}},\quad {\boldsymbol {x}}(0)={\boldsymbol {\xi }}}$

と簡潔に表示できる．ここに，

${\displaystyle {\frac {d{\boldsymbol {x}}}{dt}}={\begin{pmatrix}{\frac {dx_{1}}{dt}}\\{\frac {dx_{2}}{dt}}\\{\frac {dx_{3}}{dt}}\\\vdots \\{\frac {dx_{n}}{dt}}\end{pmatrix}}}$

である．

### (2)

ここで少し記号の約束をしておこう． 関数を成分とする行列，

${\displaystyle B(t):={\begin{pmatrix}b_{11}(t)&b_{12}(t)&b_{13}(t)&\cdots &b_{1n}(t)\\b_{21}(t)&b_{22}(t)&b_{23}(t)&\cdots &b_{2n}(t)\\b_{31}(t)&b_{32}(t)&b_{33}(t)&\cdots &b_{3n}(t)\\\vdots &\vdots &\vdots &&\vdots \\b_{m1}(t)&b_{m2}(t)&b_{m3}(t)&\cdots &b_{mn}(t)\end{pmatrix}}}$

に対して，この行列の微分あるいは積分を，その成分の微分あるいは積分を成分とする行列と定義する．すなわち，

${\displaystyle {\frac {dB(t)}{dt}}:={\begin{pmatrix}{\frac {db_{11}(t)}{dt}}&{\frac {db_{12}(t)}{dt}}&{\frac {db_{13}(t)}{dt}}&\cdots &{\frac {db_{1n}(t)}{dt}}\\{\frac {db_{21}(t)}{dt}}&{\frac {db_{22}(t)}{dt}}&{\frac {db_{23}(t)}{dt}}&\cdots &{\frac {db_{2n}(t)}{dt}}\\{\frac {db_{31}(t)}{dt}}&{\frac {db_{32}(t)}{dt}}&{\frac {db_{33}(t)}{dt}}&\cdots &{\frac {db_{3n}(t)}{dt}}\\\vdots &\vdots &\vdots &&\vdots \\{\frac {db_{m1}(t)}{dt}}&{\frac {db_{m2}(t)}{dt}}&{\frac {db_{m3}(t)}{dt}}&\cdots &{\frac {db_{mn}(t)}{dt}}\end{pmatrix}}}$

あるいは，

${\displaystyle \int _{a}^{b}B(t)dt:={\begin{pmatrix}\int _{a}^{b}b_{11}(t)dt&\int _{a}^{b}b_{12}(t)dt&\int _{a}^{b}b_{13}(t)dt&\cdots &\int _{a}^{b}b_{1n}(t)dt\\\int _{a}^{b}b_{21}(t)dt&\int _{a}^{b}b_{22}(t)dt&\int _{a}^{b}b_{23}(t)dt&\cdots &\int _{a}^{b}b_{2n}(t)dt\\\int _{a}^{b}b_{31}(t)dt&\int _{a}^{b}b_{32}(t)dt&\int _{a}^{b}b_{33}(t)dt&\cdots &\int _{a}^{b}b_{3n}(t)dt\\\vdots &\vdots &\vdots &&\vdots \\\int _{a}^{b}b_{m1}(t)dt&\int _{a}^{b}b_{m2}(t)dt&\int _{a}^{b}b_{m3}(t)dt&\cdots &\int _{a}^{b}b_{mn}(t)dt\end{pmatrix}}}$

と約束する．この約束に従えば，

${\displaystyle {\mathcal {L}}[B(t)]:={\begin{pmatrix}{\mathcal {L}}[b_{11}(t)]&{\mathcal {L}}[b_{12}(t)]&{\mathcal {L}}[b_{13}(t)]&\cdots &{\mathcal {L}}[b_{1n}(t)]\\{\mathcal {L}}[b_{21}(t)]&{\mathcal {L}}[b_{22}(t)]&{\mathcal {L}}[b_{23}(t)]&\cdots &{\mathcal {L}}[b_{2n}(t)]\\{\mathcal {L}}[b_{31}(t)]&{\mathcal {L}}[b_{32}(t)]&{\mathcal {L}}[b_{33}(t)]&\cdots &{\mathcal {L}}[b_{3n}(t)]\\\vdots &\vdots &\vdots &&\vdots \\{\mathcal {L}}[b_{m1}(t)]&{\mathcal {L}}[b_{m2}(t)]&{\mathcal {L}}[b_{m3}(t)]&\cdots &{\mathcal {L}}[b_{mn}(t)]\end{pmatrix}}}$

${\displaystyle {\mathcal {L}}^{-1}[B(t)]:={\begin{pmatrix}{\mathcal {L}}^{-1}[b_{11}(t)]&{\mathcal {L}}^{-1}[b_{12}(t)]&{\mathcal {L}}^{-1}[b_{13}(t)]&\cdots &{\mathcal {L}}^{-1}[b_{1n}(t)]\\{\mathcal {L}}^{-1}[b_{21}(t)]&{\mathcal {L}}^{-1}[b_{22}(t)]&{\mathcal {L}}^{-1}[b_{23}(t)]&\cdots &{\mathcal {L}}^{-1}[b_{2n}(t)]\\{\mathcal {L}}^{-1}[b_{31}(t)]&{\mathcal {L}}^{-1}[b_{32}(t)]&{\mathcal {L}}^{-1}[b_{33}(t)]&\cdots &{\mathcal {L}}^{-1}[b_{3n}(t)]\\\vdots &\vdots &\vdots &&\vdots \\{\mathcal {L}}^{-1}[b_{m1}(t)]&{\mathcal {L}}^{-1}[b_{m2}(t)]&{\mathcal {L}}^{-1}[b_{m3}(t)]&\cdots &{\mathcal {L}}^{-1}[b_{mn}(t)]\end{pmatrix}}}$

は必然である[1]． また ${\displaystyle A(t)}$ を関数を成分とする行列とし，積 ${\displaystyle A(t)\ B(t)}$ が定義できるものとすれば，

(5.11a)
${\displaystyle {\frac {dAB}{dt}}={\frac {dA}{dt}}B+A{\frac {dB}{dt}}}$
(5.11b)
${\displaystyle \int _{a}^{b}{\frac {dB}{dt}}dt=B(b)-B(a)}$

などは明らかであろう．

(1) 式 (5.11a)の証明．

${\displaystyle (AB)_{ij}=\sum _{k}a_{ik}b_{kj}}$

よって，

${\displaystyle \left({\frac {d}{dt}}AB\right)_{ij}=\sum _{k}\left({\frac {d}{dt}}a_{ik}\cdot b_{kj}+a_{ik}\cdot {\frac {d}{dt}}b_{kj}\right)}$
${\displaystyle =\sum _{k}{\frac {d\ a_{ik}}{dt}}b_{kj}+\sum _{k}a_{ik}{\frac {d\ b_{kj}}{dt}}}$
${\displaystyle =\sum _{k}\left({\frac {dA}{dt}}\right)_{ik}\cdot b_{kj}+\sum _{k}a_{ik}\cdot \left({\frac {dB}{dt}}\right)_{kj}}$
${\displaystyle =\left({\frac {dA}{dt}}B\right)_{ij}+\left(A{\frac {dB}{dt}}\right)_{ij}}$
${\displaystyle =\left({\frac {dA}{dt}}B+A{\frac {dB}{dt}}\right)_{ij}}$

(2)式 (5.11b)の証明．

${\displaystyle \int _{a}^{b}{\frac {dB}{dt}}dt=\int _{a}^{b}{\begin{pmatrix}{\frac {d}{dt}}b_{11}&\cdots &{\frac {d}{dt}}b_{1n}\\\vdots &&\vdots \\{\frac {d}{dt}}b_{m1}&\cdots &{\frac {d}{dt}}b_{mn}\end{pmatrix}}dt}$

${\displaystyle ={\begin{pmatrix}\int _{a}^{b}{\frac {d}{dt}}b_{11}dt&\cdots &\int _{a}^{b}{\frac {d}{dt}}b_{1n}dt\\\vdots &&\vdots \\\int _{a}^{b}{\frac {d}{dt}}b_{m1}dt&\cdots &\int _{a}^{b}{\frac {d}{dt}}b_{mn}dt\end{pmatrix}}}$

ここで ${\displaystyle \int _{a}^{b}{\frac {df}{dt}}dt=f(b)-f(a)}$ より，

${\displaystyle \int _{a}^{b}{\frac {dB}{dt}}dt={\begin{pmatrix}b_{11}(b)-b_{11}(a)&\cdots &b_{1n}(b)-b_{1n}(a)\\\vdots &&\vdots \\b_{m1}(b)-b_{m1}(a)&\cdots &b_{mn}(b)-b_{mn}(a)\end{pmatrix}}}$

${\displaystyle ={\begin{pmatrix}b_{11}(b)&\cdots &b_{1n}(b)\\\vdots &&\vdots \\b_{m1}(b)&\cdots &b_{mn}(b)\end{pmatrix}}-{\begin{pmatrix}b_{11}(a)&\cdots &b_{1n}(a)\\\vdots &&\vdots \\b_{m1}(a)&\cdots &b_{mn}(a)\end{pmatrix}}}$

${\displaystyle =B(b)-B(a)}$

${\displaystyle \diamondsuit }$

1. ^
${\displaystyle {\begin{pmatrix}\int _{0}^{\infty }b_{11}e^{-st}dt&\cdots &\int _{0}^{\infty }b_{1n}e^{-st}dt\\\vdots &&\vdots \\\int _{0}^{\infty }b_{m1}e^{-st}dt&\cdots &\int _{0}^{\infty }b_{mn}e^{-st}dt\end{pmatrix}}=\int _{0}^{\infty }{\begin{pmatrix}b_{11}&\cdots &b_{1n}\\\vdots &&\vdots \\b_{m1}&\cdots &b_{mn}\end{pmatrix}}e^{-st}dt}$

また，${\displaystyle b_{11}\sqsubset B_{11},\cdots ,b_{mn}\sqsubset B_{mn}}$ とおけば，

${\displaystyle {\mathcal {L}}{\begin{pmatrix}b_{11}&\cdots &b_{1n}\\\vdots &&\vdots \\b_{m1}&\cdots &b_{mn}\end{pmatrix}}={\begin{pmatrix}B_{11}&\cdots &B_{1n}\\\vdots &&\vdots \\B_{m1}&\cdots &B_{mn}\end{pmatrix}}}$

${\displaystyle \therefore {\mathcal {L}}{\begin{pmatrix}{\mathcal {L}}^{-1}B_{11}&\cdots &{\mathcal {L}}^{-1}B_{1n}\\\vdots &&\vdots \\{\mathcal {L}}^{-1}B_{m1}&\cdots &{\mathcal {L}}^{-1}B_{mn}\end{pmatrix}}={\begin{pmatrix}B_{11}&\cdots &B_{1n}\\\vdots &&\vdots \\B_{m1}&\cdots &B_{mn}\end{pmatrix}}}$

両辺に左から ${\displaystyle {\mathcal {L}}^{-1}}$ を働かせて，

${\displaystyle {\mathcal {L}}^{-1}\cdot {\mathcal {L}}{\begin{pmatrix}{\mathcal {L}}^{-1}B_{11}&\cdots &{\mathcal {L}}^{-1}B_{1n}\\\vdots &&\vdots \\{\mathcal {L}}^{-1}B_{m1}&\cdots &{\mathcal {L}}^{-1}B_{mn}\end{pmatrix}}={\mathcal {L}}^{-1}{\begin{pmatrix}B_{11}&\cdots &B_{1n}\\\vdots &&\vdots \\B_{m1}&\cdots &B_{mn}\end{pmatrix}}}$

${\displaystyle \therefore {\begin{pmatrix}{\mathcal {L}}^{-1}B_{11}&\cdots &{\mathcal {L}}^{-1}B_{1n}\\\vdots &&\vdots \\{\mathcal {L}}^{-1}B_{m1}&\cdots &{\mathcal {L}}^{-1}B_{mn}\end{pmatrix}}={\mathcal {L}}^{-1}{\begin{pmatrix}B_{11}&\cdots &B_{1n}\\\vdots &&\vdots \\B_{m1}&\cdots &B_{mn}\end{pmatrix}}}$

したがって，

(5.12)
${\displaystyle {\mathcal {L}}\left[{\frac {d{\boldsymbol {x}}}{dt}}\right]=s{\mathcal {L}}[{\boldsymbol {x}}]-{\boldsymbol {x}}(0)}$
(5.13)
${\displaystyle {\mathcal {L}}[A{\boldsymbol {x}}]=A{\mathcal {L}}[{\boldsymbol {x}}]}$

などの計算が許される．ただし 式 (5.13) ${\displaystyle A}$ は定数行列とする． たとえば式 (5.12) の証明は次のとおりである。

${\displaystyle {\boldsymbol {x}}=(x_{i})}$

と略記すると，

${\displaystyle {\mathcal {L}}[{\boldsymbol {x}}']={\mathcal {L}}[(x_{i})']=({\mathcal {L}}[x_{i}'])=(s{\mathcal {L}}[x_{i}]-x_{i}(0))=s({\mathcal {L}}[x_{i}])-(x_{i}(0))=s{\mathcal {L}}[{\boldsymbol {x}}]-{\boldsymbol {x}}(0)}$

ただ定義に従って変形していくだけでよい[1]

1. ^ 式 (5.13) の証明は次のとおり．
まず，${\displaystyle A{\boldsymbol {x}}}$${\displaystyle n}$ 行 1 列のベクトルになることに注意して，
上述のとおり，行列 ${\displaystyle M}$ の各成分を ${\displaystyle m_{ij}=(M)_{ij}}$ と，また，ベクトル ${\displaystyle {\boldsymbol {f}}}$ の第 ${\displaystyle i}$ 成分を ${\displaystyle f_{i}=({\boldsymbol {f}})_{i}}$ と表記するものとすると，
${\displaystyle (A{\boldsymbol {x}})_{i}=\sum _{k}a_{ik}x_{k}}$．…①
${\displaystyle {\mathcal {L}}[(A{\boldsymbol {x}})_{i}]={\mathcal {L}}[\sum _{k}a_{ik}x_{k}]}$
${\displaystyle =\sum _{k}a_{ik}{\mathcal {L}}[x_{k}]}$
${\displaystyle =\sum _{k}(A)_{ik}({\mathcal {L}}[{\boldsymbol {x}}])_{k}}$
${\displaystyle =(A{\mathcal {L}}[{\boldsymbol {x}}])_{i}\quad \because \ }$①．
${\displaystyle \therefore {\mathcal {L}}[A{\boldsymbol {x}}]=A{\mathcal {L}}[{\boldsymbol {x}}]}$

### (3)

さて以上の準備の下に，

${\displaystyle {\frac {d{\boldsymbol {x}}}{dt}}=A{\boldsymbol {x}}+{\boldsymbol {f}}}$

は次のように解くことができる． この式を Laplace 変換すると，

${\displaystyle s{\mathcal {L}}[{\boldsymbol {x}}]-{\boldsymbol {x}}(0)=A{\mathcal {L}}[{\boldsymbol {x}}]+{\mathcal {L}}[{\boldsymbol {f}}]}$

すなわち，

${\displaystyle (sI-A){\mathcal {L}}[{\boldsymbol {x}}]={\boldsymbol {x}}(0)+{\mathcal {L}}[{\boldsymbol {f}}]}$

となる．ここに ${\displaystyle I}$${\displaystyle n}$ 次の単位行列である． ${\displaystyle (sI-A)^{-1}}$${\displaystyle (sI-A)}$ の逆行列とすれば，

${\displaystyle {\mathcal {L}}[{\boldsymbol {x}}]=(sI-A)^{-1}{\boldsymbol {x}}(0)+(sI-A)^{-1}{\mathcal {L}}[{\boldsymbol {f}}]}$

となる．いま，

(5.14)
${\displaystyle (sI-A)^{-1}\sqsubset {\mathit {\Phi }}}$

とおけば，

(5.14b)
${\displaystyle {\boldsymbol {x}}(t)={\mathit {\Phi }}(t){\boldsymbol {x}}(0)+\int _{0}^{t}{\mathit {\Phi }}(t-\tau ){\boldsymbol {f}}(\tau )d\tau }$

となる．以上は例題を通して考察したことの繰り返しに過ぎない． これからわかるように，連立微分方程式を解くことの中心は式 (5.14) を計算することである．