一阶常系数微分方程组的笔记

文章目录

一、一阶线性常系数齐次微分方程组
- 1.1 基本定义
- 1.2 解的存在性
- 1.3 解的唯一性
- 1.4 结论
二、一阶线性常系数非齐次微分方程组
- 2.1 基本定义
- 2.2 解的存在性
三、例题

一、一阶线性常系数齐次微分方程组

1.1 基本定义

{ d x 1 ( t ) d t = a 11 x 1 ( t ) + a 12 x 2 ( t ) + ⋯ + a 1 n x n ( t ) d x 2 ( t ) d t = a 21 x 1 ( t ) + a 22 x 2 ( t ) + ⋯ + a 2 n x n ( t ) ⋮ d x n ( t ) d t = a n 1 x 1 ( t ) + a n 2 x 2 ( t ) + ⋯ + a n n x n ( t ) \left\{\begin{array}{c} \frac{\mathbf{d} x_{1}(t)}{\mathbf{d} t}=a_{11} x_{1}(t)+a_{12} x_{2}(t)+\cdots+a_{1 n} x_{n}(t) \\ \frac{\mathbf{d} x_{2}(t)}{\mathbf{d} t}=a_{21} x_{1}(t)+a_{22} x_{2}(t)+\cdots+a_{2 n} x_{n}(t) \\ \vdots \\ \frac{\mathbf{d} x_{n}(t)}{\mathbf{d} t}=a_{n 1} x_{1}(t)+a_{n 2} x_{2}(t)+\cdots+a_{n n} x_{n}(t) \end{array}\right. ⎩ ⎨ ⎧dtdx1(t)=a11x1(t)+a12x2(t)+⋯+a1nxn(t)dtdx2(t)=a21x1(t)+a22x2(t)+⋯+a2nxn(t)⋮dtdxn(t)=an1x1(t)+an2x2(t)+⋯+annxn(t)

给定初始条件： x i ( 0 ) , ( i = 1 , 2 , ⋯ , n ) x_{i}(0),(i=1,2, \cdots, n) xi(0),(i=1,2,⋯,n)。

记 A = ( a i j ) ∈ C n × n , X ( 0 ) = ( x 1 ( 0 ) , x 2 ( 0 ) , ⋯ , x n ( 0 ) ) T X ( t ) = ( x 1 ( t ) , x 2 ( t ) , … , x n ( t ) ) T \begin{array}{l} \mathbf{A}=\left(a_{i j}\right) \in \mathbf{C}^{n \times n}, \\ \boldsymbol{X}(0)=\left(x_{1}(0), x_{2}(0), \cdots, x_{n}(0)\right)^{T} \\ \boldsymbol{X}(t)=\left(x_{1}(t), x_{2}(t), \ldots, x_{n}(t)\right)^{T} \end{array} A=(aij)∈Cn×n,X(0)=(x1(0),x2(0),⋯,xn(0))TX(t)=(x1(t),x2(t),…,xn(t))T

则上述微分方程组可以写成：
{ X ′ ( t ) = A X ( t ) X ( 0 ) = ( x 1 ( 0 ) , x 2 ( 0 ) , … , x n ( 0 ) ) T \left\{\begin{array}{l} \boldsymbol{X}^{\prime}(t)=\boldsymbol{A} \boldsymbol{X}(t) \\ \boldsymbol{X}(0)=\left(x_{1}(0), x_{2}(0), \ldots, x_{n}(0)\right)^{T} \end{array}\right. {X′(t)=AX(t)X(0)=(x1(0),x2(0),…,xn(0))T

1.2 解的存在性

利用矩阵微分性质有
( e − A t X ( t ) ) ′ = − e − A t A ⋅ X ( t ) + e − A t X ′ ( t ) = e − A t ( X ′ ( t ) − A X ( t ) ) \left(e^{-\boldsymbol{A} t} \boldsymbol{X}(t)\right)^{\prime}=-e^{-\boldsymbol{A} t} \mathbf{A} \cdot \boldsymbol{X}(t)+e^{-\mathbf{A} t} \boldsymbol{X}^{\prime}(t)=e^{-\mathbf{A} t}\left(\boldsymbol{X}^{\prime}(t)-\mathbf{A} \boldsymbol{X}(t)\right) (e−AtX(t))′=−e−AtA⋅X(t)+e−AtX′(t)=e−At(X′(t)−AX(t))

故 ( e − A t X ( t ) ) ′ = 0 \left(e^{-\boldsymbol{A} t} \boldsymbol{X}(t)\right)^{\prime}=0 (e−AtX(t))′=0，因此 X ( t ) = e A t C \boldsymbol{X}(\mathrm{t})=e^{A t} \boldsymbol{C} X(t)=eAtC

由初始条件， C = X ( 0 ) \boldsymbol{C}=\boldsymbol{X}(0) C=X(0)，其中 C \boldsymbol{C} C 为常数向量。

1.3 解的唯一性

如果定解问题有两个解 X 1 ( t ) \boldsymbol{X}_1(t) X1(t)， X 2 ( t ) \boldsymbol{X}_2(t) X2(t)，则令

Y ( t ) = X 1 ( t ) − X 2 ( t ) \boldsymbol{Y}(t)=\boldsymbol{X}_1(t)-\boldsymbol{X}_2(t) Y(t)=X1(t)−X2(t)

满足

{ Y ′ ( t ) = X 1 ′ ( t ) − X 2 ′ ( t ) = A X 1 ( t ) − A X 2 ( t ) = A Y ( t ) Y ( 0 ) = X 1 ( 0 ) − X 2 ( 0 ) = 0 \left\{\begin{array}{l} \boldsymbol{Y}^{\prime}(t)=\boldsymbol{X}_{1}^{\prime}(t)-\boldsymbol{X}_{2}^{\prime}(t)=A \boldsymbol{X}_{1}(t)-A \boldsymbol{X}_{2}(t)=\boldsymbol{AY}(t) \\ \boldsymbol{Y}(0)=\boldsymbol{X}_{1}(0)-\boldsymbol{X}_{2}(0)=0 \end{array}\right. {Y′(t)=X1′(t)−X2′(t)=AX1(t)−AX2(t)=AY(t)Y(0)=X1(0)−X2(0)=0

类似推导可知， Y ( t ) = e A t Y ( 0 ) = 0 \boldsymbol{Y}(t)=e^{\mathbf{A} t} \boldsymbol{Y}(0)=\boldsymbol{0} Y(t)=eAtY(0)=0，即 X 1 ( t ) = X 2 ( t ) \boldsymbol{X}_1(t)=\boldsymbol{X}_2(t) X1(t)=X2(t)。

1.4 结论

一阶线性常系数齐次微分方程组的定解问题有唯一解
X ( t ) = e A t X ( 0 ) \boldsymbol{X}(t)=e^{\boldsymbol{A}t}\boldsymbol{X}(0) X(t)=eAtX(0)

二、一阶线性常系数非齐次微分方程组

2.1 基本定义

{ X ′ ( t ) = A X ( t ) + F ( t ) X ( t 0 ) = ( x 1 ( t 0 ) , x 2 ( t 0 ) , … , x n ( t 0 ) ) T \left\{\begin{array}{l} \boldsymbol{X}^{\prime}(t)=\boldsymbol{A} \boldsymbol{X}(t)+\boldsymbol{F}(t) \\ \boldsymbol{X}\left(t_{0}\right)=\left(x_{1}\left(t_{0}\right), x_{2}\left(t_{0}\right), \ldots, x_{n}\left(t_{0}\right)\right)^{T} \end{array}\right. {X′(t)=AX(t)+F(t)X(t0)=(x1(t0),x2(t0),…,xn(t0))T

这里 F ( t ) = ( f 1 ( t ) , f 2 ( t ) , ⋯ , f n ( t ) ) T \boldsymbol{F}(t)=\left(f_{1}(t), f_{2}(t), \cdots, f_{n}(t)\right)^{T} F(t)=(f1(t),f2(t),⋯,fn(t))T 是已知的向量函数， A \boldsymbol{A} A 和 X \boldsymbol{X} X 意义同前。

2.2 解的存在性

( e − A t X ( t ) ) ′ = e − A t ( X ′ ( t ) − A X ( t ) ) = e − A t F ( t ) \left(e^{-\mathbf{A} t} \boldsymbol{X}(t)\right)^{\prime}=e^{-\mathbf{A} t}\left(\boldsymbol{X}^{\prime}(t)-\boldsymbol{A} \boldsymbol{X}(t)\right)=e^{-\mathbf{A} t} \boldsymbol{F}(t) (e−AtX(t))′=e−At(X′(t)−AX(t))=e−AtF(t)

对此方程在 [ t 0 , t ] [t_0,t] [t0,t] 上进行积分，可得
e − A t X ( t ) − e − A t 0 X ( t 0 ) = ∫ t 0 t e − A τ F ( τ ) d τ e^{-\mathbf{A} t} \boldsymbol{X}(t)-e^{-\mathbf{A} t_{0}} \boldsymbol{X}\left(t_{0}\right)=\int_{t_{0}}^{t} e^{-\mathbf{A} \tau} \boldsymbol{F}(\tau) \mathrm{d} \tau e−AtX(t)−e−At0X(t0)=∫t0te−AτF(τ)dτ

上述定解问题的解
X ( t ) = e A ( t − t 0 ) X ( t 0 ) + ∫ t 0 t e A ( t − τ ) F ( τ ) d τ \boldsymbol{X}(t)=e^{\mathbf{A}\left(t-t_{0}\right)} \boldsymbol{X}\left(t_{0}\right)+\int_{t_{0}}^{t} e^{\mathbf{A}(t-\tau)} \boldsymbol{F}(\tau) \mathrm{d} \tau X(t)=eA(t−t0)X(t0)+∫t0teA(t−τ)F(τ)dτ

解答的唯一性和结论已经忽略。

三、例题

例1 求定解问题 { X ′ ( t ) = A X ( t ) X ( 0 ) = ( 1 , 1 , 1 ) T \left\{\begin{array}{l} \boldsymbol{X}^{\prime}(t)=\mathbf{A} \boldsymbol{X}(t) \\ \boldsymbol{X}(0)=(1,1,1)^{T} \end{array}\right. {X′(t)=AX(t)X(0)=(1,1,1)T，其中 A = ( 3 − 1 1 2 0 − 1 1 − 1 2 ) A=\left(\begin{array}{ccc} 3 & -1 & 1 \\ 2 & 0 & -1 \\ 1 & -1 & 2 \end{array}\right) A= 321−10−11−12

解 det ⁡ ( λ I − A ) = λ ( λ − 2 ) ( λ − 3 ) \operatorname{det}(\lambda \mathbf{I}-\mathbf{A})=\lambda(\lambda-2)(\lambda-3) det(λI−A)=λ(λ−2)(λ−3)

特征根 λ 1 = 0 , λ 2 = 2 , λ 3 = 3 \lambda_{1}=0, \lambda_{2}=2, \lambda_{3}=3 λ1=0,λ2=2,λ3=3，相应的三个线性无关的特征向量分别为：
X 1 = ( 1 , 5 , 2 ) T , X 2 = ( 1 , 1 , 0 ) T , X 3 = ( 2 , 1 , 1 ) T \boldsymbol{X}_{1}=(1,5,2)^{T}, \quad \boldsymbol{X}_{2}=(1,1,0)^{T}, \quad \boldsymbol{X}_{3}=(2,1,1)^{T} X1=(1,5,2)T,X2=(1,1,0)T,X3=(2,1,1)T

故
T = ( 1 1 2 5 1 1 2 0 1 ) , T − 1 = − 1 6 ( 1 − 1 − 1 − 3 − 3 9 − 2 2 − 4 ) \boldsymbol{T}=\left(\begin{array}{lll} 1 & 1 & 2 \\ 5 & 1 & 1 \\ 2 & 0 & 1 \end{array}\right), \quad \boldsymbol{T}^{-1}=-\frac{1}{6}\left(\begin{array}{ccc} 1 & -1 & -1 \\ -3 & -3 & 9 \\ -2 & 2 & -4 \end{array}\right) T= 152110211 ,T−1=−61 1−3−2−1−32−19−4

所求的解为
X = e A t X ( 0 ) = T ( 1 e 2 t e 3 t ) T − 1 X ( 0 ) = − 1 6 ( − 1 + 3 e 2 t − 8 e 3 t − 5 + 3 e 2 t − 4 e 3 t − 2 − 4 e 3 t ) \begin{array}{l} \boldsymbol{X}=e^{\mathbf{A t}} \boldsymbol{X}(0)=\boldsymbol{T}\left(\begin{array}{lll} 1 & & \\ & e^{2 t} & \\ & & e^{3 t} \end{array}\right) \boldsymbol{T}^{-1} \boldsymbol{X}(0)=-\frac{1}{6}\left(\begin{array}{l} -1+3 e^{2 t}-8 e^{3 t} \\ -5+3 e^{2 t}-4 e^{3 t} \\ -2-4 e^{3 t} \end{array}\right) \end{array} X=eAtX(0)=T 1e2te3t T−1X(0)=−61 −1+3e2t−8e3t−5+3e2t−4e3t−2−4e3t

例2 求定解问题 { d X ( t ) d t = A X ( t ) + F ( t ) X ( 0 ) = ( 1 , 1 , 1 ) T \left\{\begin{array}{l} \frac{\mathrm{d} \boldsymbol{X}(t)}{\mathrm{d} t}=\mathbf{A} \boldsymbol{X}(t)+\boldsymbol{F}(t) \\ \boldsymbol{X}(0)=(1,1,1)^{T} \end{array}\right. {dtdX(t)=AX(t)+F(t)X(0)=(1,1,1)T 的解，其中 A A A 如上例， F ( t ) = ( 0 , 0 , e 2 t ) T \boldsymbol{F}(t)=\left(0,0, e^{2 t}\right)^{T} F(t)=(0,0,e2t)T。

解该问题的定解为 X ( t ) = e A t X ( 0 ) + ∫ 0 t e A ( t − τ ) F ( τ ) d τ \boldsymbol{X}(t)=e^{\mathbf{A} t} \boldsymbol{X}(0)+\int_{0}^{t} e^{\mathbf{A}(t-\tau)} \boldsymbol{F}(\tau) \mathrm{d} \tau X(t)=eAtX(0)+∫0teA(t−τ)F(τ)dτ

e A ( t − τ ) F ( τ ) = T e [ J ( t − τ ) ] T − 1 F ( τ ) = ( 1 1 2 5 1 1 2 0 1 ) ( 1 e 2 ( t − τ ) e 3 ( t − τ ) ) ( − 1 6 ) ( 1 − 1 − 1 − 3 − 3 9 − 2 2 − 4 ) ( 0 0 e 2 τ ) = − 1 6 ( − e 2 τ + 9 e 2 t − 8 e 3 t − τ − 5 e 2 τ + 9 e 2 t − 4 e 3 t − τ − 2 e 2 τ − 4 e 3 t − τ ) 。 \begin{aligned} e^{\mathbf{A}(t-\tau)} \boldsymbol{F}(\tau) & =\boldsymbol{T} e^{[J(t-\tau)]} \boldsymbol{T}^{-1} \boldsymbol{F}(\tau) \\ & =\left(\begin{array}{lll} 1 & 1 & 2 \\ 5 & 1 & 1 \\ 2 & 0 & 1 \end{array}\right)\left(\begin{array}{lll} 1 & & \\ & e^{2(t-\tau)} & \\ & & e^{3(t-\tau)} \end{array}\right)\left(\begin{array}{l} -\frac{1}{6} \\ \end{array}\right)\left(\begin{array}{ccc} 1 & -1 & -1 \\ -3 & -3 & 9 \\ -2 & 2 & -4 \end{array}\right)\left(\begin{array}{l} 0 \\ 0 \\ e^{2 \tau} \end{array}\right)=-\frac{1}{6}\left(\begin{array}{l} -e^{2 \tau}+9 e^{2 t}-8 e^{3 t-\tau} \\ -5 e^{2 \tau}+9 e^{2 t}-4 e^{3 t-\tau} \\ -2 e^{2 \tau}-4 e^{3 t-\tau} \end{array}\right) 。 \end{aligned} eA(t−τ)F(τ)=Te[J(t−τ)]T−1F(τ)= 152110211 1e2(t−τ)e3(t−τ) (−61) 1−3−2−1−32−19−4 00e2τ =−61 −e2τ+9e2t−8e3t−τ−5e2τ+9e2t−4e3t−τ−2e2τ−4e3t−τ 。

对变量 τ \tau τ 从 0 到 t t t 进行积分，即得
P = − 1 6 ( 1 2 + ( 9 t + 15 2 ) e 2 t − 8 e 3 t 5 2 + ( 9 t + 3 2 ) e 2 t − 4 e 3 t 1 + 3 e 2 t − 4 e 3 t ) \boldsymbol{P}=-\frac{1}{6}\left(\begin{array}{l} \frac{1}{2}+\left(9 t+\frac{15}{2}\right) e^{2 t}-8 e^{3 t} \\ \frac{5}{2}+\left(9 t+\frac{3}{2}\right) e^{2 t}-4 e^{3 t} \\ 1+3 e^{2 t}-4 e^{3 t} \end{array}\right) P=−61 21+(9t+215)e2t−8e3t25+(9t+23)e2t−4e3t1+3e2t−4e3t
因此 X ( t ) = e A t X ( 0 ) + P \boldsymbol{X}(t)=e^{\mathbf{A} t} \boldsymbol{X}(0)+\boldsymbol{P} X(t)=eAtX(0)+P

X ( t ) = − 1 6 ( − 1 2 + ( 9 t + 21 2 ) e 2 t − 16 e 3 t − 5 2 + ( 9 t + 9 2 ) e 2 t − 8 e 3 t − 1 + 3 e 2 t − 8 e 3 t ) X(t)=-\frac{1}{6}\left(\begin{array}{l} -\frac{1}{2}+\left(9 t+\frac{21}{2}\right) e^{2 t}-16 e^{3 t} \\ -\frac{5}{2}+\left(9 t+\frac{9}{2}\right) e^{2 t}-8 e^{3 t} \\ -1+3 e^{2 t}-8 e^{3 t} \end{array}\right) X(t)=−61 −21+(9t+221)e2t−16e3t−25+(9t+29)e2t−8e3t−1+3e2t−8e3t