Keywords
[1] If \(Ax=\lambda x\) then \(u(t) = e^{\lambda t}x\) will solve \(\frac{du}{dt}=Au\). Each \(\lambda\) and \(x\) give a solution \(e^{\lambda t}x\)
[2] If \(A = X \Lambda X^{-1}\) then \(u(t)=e^{At}u(0)=Xe^{\Lambda t}X^{-1}u(0)=c_1 e^{\lambda_1t}x_1 + \cdots + c_n e^{\lambda_n t}x_n\)
[3] \(A\) is stable and \(u(t) \rightarrow 0\) and \(e^{At}\rightarrow 0\) when all eigenvalues of \(A\) have real part \(<0\).
[4] Matrix exponential \(e^{At}=I + At + \cdots + (At)^n / n! + \cdots = X e^{\Lambda t}X^{-1}\) if \(A\) is diagonalizable
[5] Second order equation First order system: \(u'' + Bu' + Cu = 0\) is quivalent to \(\begin{bmatrix} u \\\ u' \end{bmatrix}' = \begin{bmatrix} 0 & 1 \\\ -C & -B \end{bmatrix} \begin{bmatrix} u \\\ u' \end{bmatrix}\)
Differential Equations
Eigenvalues and eigenvectors are perfect for \(\frac{du}{dt}=Au\). As a prior knowledge, the derivative of \(e^{\lambda t}\) is \(\lambda e^{\lambda t}\).
The main point is to convert constant-coefficient differential equations into linear algebra.
The ordinary equations \(\frac{du}{dt}=u\) and \(\frac{du}{dt}=\lambda u\) are solved by exponentials:
\(\frac{du}{dt}=u\) products \(u(t)=Ce^t\)
\(\frac{du}{dt}=\lambda u\) produces \(u(t)=Ce^{\lambda t}\).
When \(t=0\), we have \(u(0)=C\). The solutions that start from \(u(0)\) at \(t=0\) are
\[u(t) = u(0)e^t\] \[u(t)=u(0)e^{\lambda t}\]This is the 1 by 1 case. Let’s generalize to n by n. The unknown is a vector \(u\). We start from the initial vector \(u(0)\) and the \(n\) equations contain a square matrix \(A\). We expect \(n\) exponents \(e^{\lambda t}\) in \(u(t)\), from \(n\) \(\lambda's\)
\[\frac{du}{dt}=Au,\ \ u(0)=\begin{bmatrix} u_1(0) \\\ \cdots \\\ u_n(0) \end{bmatrix}\ \text{at}\ t=0\]These differential equations are linear which means if \(u(t)\) and \(v(t)\) are solutions, so is \(Cu(t)+Dv(t)\). Thus, the first job is to find n “pure exponential solutions” \(u=e^{\lambda t}x\) by using \(Ax=\lambda x\).
Notice that \(A\) is a constant matrix. So \(\frac{du}{dt}=Au\) is “linear with constant coefficients”.
Solve linear constant coefficient equations by exponentials \(e^{\lambda t}x\), when \(Ax=\lambda x\)
Solution of \(\frac{du}{dt}=Au\)
Pure exponential solution
The pure exponential solution is \(e^{\lambda x}\) times a fixed vector \(x\). \(\lambda\) is an eigenvalue of \(A\) and \(x\) is the eigenvector.
Choose \(u=e^{\lambda t}x\) when \(Ax=\lambda x\). Then, \(\frac{du}{dt}=\lambda e^{\lambda t}x\) agrees with \(Au = A e^{\lambda t} x\)
After we fine the pure exponential solutions, the complete solution is the linear combination of those pure solutions
Complete solution
\[u(t) = Cu_1(t) + Du_2(t)\]Summary: Three steps
[1] Write \(u(0)\) as a combination \(c_1x_1 + \cdots + c_n x_n\) of the eigenvectors of A
[2] Multiply each eigenvector \(x_i\) by its growth factor \(e^{\lambda_i t}\).
[3] The solution is the same combination of those pure solutions \(e^{\lambda t}x\)
\[\frac{du}{dt}=Au\] \[u(t) = c_1 e^{\lambda_1 t}x_1 + \cdots + c_n e^{\lambda_n t}x_n\]References
[1] Introduction to Linear Algebra, 5th edition