연립상미분방정식Ch4(Systems of ODEs)

-이전 글

physical-world.tistory.com/24

Variation of Parameters(매개변수 변환법)Ch2.10

연립상미분방정식Ch4(Systems of ODEs)

연립상미분방정식을 행렬을 이용해서 해를 찾아간다.
그리고, 연립상미분방정식을 벡터 방정식으로 나타낼 수 있다.

 

행렬을 사용하기 위해선 행렬의 성질과 활용을 알아야한다.

• 행렬의 계산(행렬 덧셈, 행렬 곱)

• 전치행렬(transpose) : A행렬의 transpose=$A^T$

• 역행렬(inverse of matrix) : A행렬의 역행렬=$A^{-1}$

• 단위행렬(unit matrix) : $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

• 행렬곱(determinant) : A의 행렬곱=$|A|$

• 고유값, 고유벡터(eigenvalue, eigenvector) : $\mathbf{Ax}=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}\lambda x_1\\ \lambda x_2\end{array}\right]=\lambda \mathbf{x}$

  에서 $$\lambda :eigen-value,\mathbf{x}:eigen-vector$$ 

연립상미분방정식을 벡터방정식으로 나타낼 수 있다(systems of ODEs as vector equation). 즉, 행렬을 벡터로 볼 수 있다.

$$\{\begin{array}{l}{y}_1^{\prime }=a_{11}{y}_1+a_{12}{y}_2\\ y_2^{\prime }=a_{21}{y}_1+a_{22}{y}_2\end{array}$$

$$\mathbf{y'} = $$\left[\begin{array}{c}{y}_1^{\prime }\\ y_2^{\prime }\end{array}\right]$$ =
$$\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$$
\cdot $$\left[\begin{array}{c}{y}_1\\ y_2\end{array}\right]$$=\mathbf{Ay}$$

 

일반적인 방법론도 중요하지만, 바로 예시로부터 연립상미분방정식의 해를 알아보자.

• 1계 상수계수 연립상미분방정식(초기값 문제)

$$\{\begin{array}{l}{y}_1(t{)}^{\prime }=-0.02y_1+0.02y_2\\ y_2(t{)}^{\prime }=0.02y_1-0.02y_2\end{array}\mathrm{\&}\{\begin{array}{l}{y}_1(0)=0\\ y_2(0)=150\end{array}$$

• Step1  : 벡터방정식 꼴로 바꾸기.

$$\Rightarrow $$\left[\begin{array}{c}{y}_1(t{)}^{\prime }\\ y_2(t{)}^{\prime }\end{array}\right]$$ = 
$$\left[\begin{array}{cc}-0.02& 0.02\\ 0.02& -0.02\end{array}\right]$$ 
\cdot $$\left[\begin{array}{c}{y}_1\\ y_2\end{array}\right]$$$$

• Step2 : $\mathbf{y}\equiv e^{\lambda t}\mathbf{x}\Rightarrow \mathbf{Ax}=\lambda \mathbf{x}\Rightarrow det(\mathbf{A}-\lambda \mathbf{I})=0$

proof)

$${\mathbf{y}}^{\prime }=\mathbf{Ay}\Rightarrow \lambda e^{\lambda t}\mathbf{x}=e^{\lambda t}\mathbf{Ax}$$

$$\Rightarrow (\mathbf{Ax}-\lambda \mathbf{x})e^{\lambda t}$$

$$\mathbf{Ax}=\lambda \mathbf{x}$$

 

문제를 마저 풀면,

$$det(\mathbf{A}-\lambda \mathbf{I})=\left|\begin{array}{cc}-0.02-\lambda & 0.02\\ 0.02& -0.02-\lambda \end{array}\right|=0$$

$$\Rightarrow (0.02+\lambda {)}^2-{0.02}^2=0$$

$${\lambda }_1=0,{\mathbf{x}}^{(1)}=\left[\begin{array}{c}1\\ 1\end{array}\right]$$

$${\lambda }_2=-0.04,{\mathbf{x}}^{(2)}=\left[\begin{array}{c}1\\ -1\end{array}\right]$$

일반해:

$$\mathbf{y}=c_1{\mathbf{x}}^{(1)}{e}^{{\lambda }_{1}t}+c_2{\mathbf{x}}^{(2)}{e}^{{\lambda }_{2}t}$$

$$=c_1\left[\begin{array}{c}1\\ 1\end{array}\right]+c_2\left[\begin{array}{c}1\\ -1\end{array}\right]e^{-0.04t}$$

 

• Step3 : 초기값 대입.

$$\Rightarrow c_1=75,c_2=-75$$

$$\Rightarrow \mathbf{y}=75\left[\begin{array}{c}1\\ 1\end{array}\right]-75\left[\begin{array}{c}1\\ -1\end{array}\right]e^{-0.04t}$$

$$\Rightarrow \{\begin{array}{l}{y}_1=75-75e^{-0.04t}\\ y_2=75+75e^{0.04t}\end{array}$$

 

-다음 글

physical-world.tistory.com/28

비제차 선형 연립상미분방정식Ch4.6(Nonhomogeneous Linear Systems of ODEs)