# 1.6 Systems of ODE's and Eigenvalue Stability

## 1.6.2 Linear Constant Coefficient Systems

Measurable Outcome 1.1, Measurable Outcome 1.2, Measurable Outcome 1.3, Measurable Outcome 1.4, Measurable Outcome 1.9

The analysis of numerical methods applied to linear, constant coefficient systems can provide significant insight into the behavior of numerical methods for nonlinear problems. Consider the following problem,

\[u_ t = A u, \label{equ:ODElinear_ sys}\] | (1.90)

where \(A\) is a \(d \times d\) matrix. Assuming that a complete set of eigenvectors exists, the matrix \(A\) can be decomposed as,

\[A = R\Lambda R^{-1}, \qquad \Lambda = \mbox{diag}(\lambda _1, \lambda _2, \cdots , \lambda _ d), \qquad R = \left(\begin{array}{c|c|c|c|c}r_1 & r_2 & r_3 & \cdots & r_ d \\ \end{array}\right) \label{equ:eigen}\] | (1.91)

The solution to Equation 1.90 can be derived as follows,

| \(\displaystyle u_ t\) | \(\displaystyle =\) | \(\displaystyle A u\) |  | (1.92)
| \(\displaystyle u_ t\) | \(\displaystyle =\) | \(\displaystyle R\Lambda R^{-1} u\) |  | (1.93)
| \(\displaystyle R^{-1} u_ t\) | \(\displaystyle =\) | \(\displaystyle \Lambda R^{-1} u\) |  | (1.94)

Then, defining \(w = R^{-1} u\),

\[w_ t = \Lambda w.\] | (1.95)

Since \(\Lambda\) is a diagonal matrix, this system of equations is actually uncoupled from each other, so that each of the eigenmodes has its own independent evolution equation,

\[(w_ j)_ t = \lambda _ j w_ j, \qquad \mbox{for each } j = 1 \mbox{ to } d\] | (1.96)

Since each of the eigenmodes has a solution \(w_ j(t) = w_ j(0)\exp (\lambda _ j t)\), then the solution for \(u(t)\) can be written as,

\[u(t) = \sum _{j=1}^ d w_ j(0) r_ j e^{\lambda _ j t}. \label{equ:ODElinear_ sys_ solution}\] | (1.97)

Note that the eigenvalues are in general complex, \(\lambda _ j = {\lambda _ j}_ r + i {\lambda _ j}_ i\). The imaginary part of the eigenvalues determines the frequency of oscillations, and the real part of the eigenvalues determines the growth or decay rate. Specifically,

\[e^{\lambda t} = e^{(\lambda _ r + i \lambda _ i)t} = \left(\cos \lambda _ i t + i\sin \lambda _ i t\right) e^{ \lambda _ r t}.\] | (1.98)

Thus, when \(\lambda _ r > 0\), the solution will grow unbounded as \(t \rightarrow \infty\).