Homogeneous linear systems differential equation with constant coefficients

Solution

We want to solve ${\mathbf{x}}^{\prime }=\mathbf{Ax}$. where $\mathbf{x}=\mathbf{x}(t)=\left(\begin{array}{c}{x}_1(t)\\ ⋮\\ x_n(t)\end{array}\right)$, $\mathbf{A}=\left(\begin{array}{ccc}{a}_{11}& \dots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{n1}& \dots & a_{nn}\end{array}\right)$

Try the form of the solution $\mathbf{x}$ for general n, $\mathbf{x}(t)=\xi e^{\lambda t}$, where $\xi$ is constant vector ${\mathbf{x}}^{\prime }=\lambda \xi e^{\lambda t}=\mathbf{A}\xi e^{\lambda t}\Rightarrow \mathbf{A}\xi =\lambda \xi$^[eigenvalue problem]

Real Eigenvalues ${\lambda }_1,\dots ,{\lambda }_n\to {\xi }^{(1)},\dots ,{\xi }^{(n)}$

Solutions: ${\mathbf{x}}^{(1)}={\xi }^{(1)}{e}^{{\lambda }_{1}t},\dots ,{\mathbf{x}}^{(n)}={\xi }^{(n)}{e}^{{\lambda }_{n}t}$

$W({\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)})=\left|\begin{array}{ccc}{\mathbf{x}}^{(1)}& \dots & {\mathbf{x}}^{(n)}\end{array}\right|=\left|\begin{array}{ccc}{\xi }^{(1)}& \dots & {\xi }^{(n)}\end{array}\right|e^{({\lambda }_1+\cdots +{\lambda }_n)t}\ne 0$ So, ${\xi }^{(1)},\dots ,{\xi }^{(n)}$ are linearly independent

General solution: $c_1{\xi }^{(1)}{e}^{{\lambda }_{1}t}+\cdots +c_n{\xi }^{(n)}{e}^{{\lambda }_{n}t},c_1,\dots ,c_n\in \mathbb{R}$

Complex Eigenvalues: ${\lambda }_1=\gamma +i\omega \to {\xi }^1,{\lambda }_2=\gamma -i\omega \to {\xi }^2$

Solutions: ${\xi }^{(1)}=\mathbf{a}+i\mathbf{b},{\xi }^{(2)}=\mathbf{a}-i\mathbf{b},\mathbf{a},\mathbf{b}$ are real vectors

${\mathbf{x}}^{(1)}={\xi }^{(1)}{e}^{{\lambda }_{1}t}=(\mathbf{a}+i\mathbf{b})e^{\gamma t}(\cos \omega t+i\sin \omega t)$ $=\underset{\mathbf{u}(t)}{\underbrace{e^{\gamma t}(\mathbf{a}\cos \omega t+\mathbf{b}\sin \omega t)}}+i\underset{\mathbf{v}(t)}{\underbrace{e^{\gamma t}(\mathbf{a}\sin \omega t+\mathbf{b}\cos \omega t)}}=\mathbf{u}(t)+i\mathbf{v}(t)$

$W(\mathbf{u},\mathbf{v})=\left|\begin{array}{cc}\mathbf{a}\cos \omega t+\mathbf{b}\sin \omega t& \mathbf{a}\sin \omega t+\mathbf{b}\cos \omega t\end{array}\right|e^{2\gamma t}$ $\Rightarrow W(\mathbf{u},\mathbf{v})(0)=\left|\begin{array}{cc}\mathbf{a}& \mathbf{b}\end{array}\right|\ne 0$

Repeated (real) root with less eigenvectors: ${\lambda }_1=\cdots ={\lambda }_n,{\xi }^{(1)},\dots ,{\xi }^{(k)},k<n$

Let ${\lambda }_1={\lambda }_2=\lambda ,\xi$ is only eigen vector, then ${\mathbf{x}}^{(1)}=\xi e^{\lambda t}$ (with $\mathbf{A}\xi =\lambda \xi$)

Try ${\mathbf{x}}^{(2)}=\xi t{e}^{\lambda t}+\eta e^{\lambda t}$, where generalized eigenvector $\eta$ is to be determined Then $({\mathbf{x}}^{(2)}{)}^{\prime }=\xi e^{\lambda t}+\xi \lambda t{e}^{\lambda t}+\eta \lambda e^{\lambda t}=\mathbf{A}{\mathbf{x}}^{(2)}=\mathbf{A}(\xi t{e}^{\lambda t}+\eta e^{\lambda t})$ $\Rightarrow (\mathbf{A}-\lambda I)\eta =\xi \Rightarrow (\mathbf{A}-\lambda I{)}^2\eta =\underset{0}{\underbrace{\mathbf{A}\xi -\lambda I\xi }}\Rightarrow (\mathbf{A}-\lambda I{)}^2=0$

Triple roots: ${\mathbf{x}}^{(3)}=\xi \frac{t^2}{2}{e}^{\lambda t}+\eta t{e}^{\lambda t}+\zeta e^{\lambda t}$, where $\zeta$ is to be determined