Review / Part I — Explicit & Qualitative Methods / Ch 7

Linear Systems via Eigenvectors

Ch 7 — Part I — Explicit & Qualitative Methods

A homogeneous linear system \(\mathbf{x}' = A\mathbf{x}\) (with \(A\) an \(n\times n\) constant matrix) has solutions constructed from eigenvalues and eigenvectors of \(A\).

Building solutions from eigenpairs

If \(A\mathbf{v} = \lambda \mathbf{v}\), then \(\mathbf{x}(t) = e^{\lambda t}\mathbf{v}\) solves \(\mathbf{x}' = A\mathbf{x}\). If \(A\) has \(n\) linearly independent eigenvectors \(\mathbf{v}_1, \ldots, \mathbf{v}_n\) with eigenvalues \(\lambda_1, \ldots, \lambda_n\), the general solution is $$\mathbf{x}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + \cdots + c_n e^{\lambda_n t}\mathbf{v}_n.$$

Finding eigenpairs

Solve \(\det(A - \lambda I) = 0\) for \(\lambda\) (characteristic equation of the matrix), then solve \((A - \lambda I)\mathbf{v} = 0\) for a nonzero \(\mathbf{v}\).

Complex eigenvalues

If \(\lambda = r + is\) has eigenvector \(\mathbf{v}\), the real-valued general solution is $$c_1 \operatorname{Re}(e^{\lambda t}\mathbf{v}) + c_2 \operatorname{Im}(e^{\lambda t}\mathbf{v}).$$

Companion system

Any higher-order linear ODE or system of ODEs can be rewritten as a first-order system \(\mathbf{x}' = A\mathbf{x} + \mathbf{f}(t)\); introduce variables for each derivative.

Practice Problems

Problem 1medium
Find the general solution of \(\mathbf{x}' = \begin{pmatrix}1 & 2\\2 & 1\end{pmatrix}\mathbf{x}.\)
Hint
Eigenvalues: trace=2, det=-3, so char poly \(\lambda^2-2\lambda-3\).
Solution

Roots \(\lambda = 3, -1\). For \(\lambda = 3\): \((A-3I)\mathbf{v}=0\) gives \(\mathbf{v} = (1,1)^T\). For \(\lambda = -1\): \(\mathbf{v} = (1,-1)^T\).

Answer: \(\mathbf{x}(t) = c_1 e^{3t}\binom{1}{1} + c_2 e^{-t}\binom{1}{-1}.\)

Problem 2medium
Solve \(\mathbf{x}' = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\mathbf{x},\; \mathbf{x}(0)=\binom{1}{0}.\)
Hint
This is a rotation matrix. Eigenvalues are \(\pm i\).
Solution

Take the eigenpair \(\lambda = i,\; \mathbf{v} = \binom{1}{-i}\). Then \(e^{it}\mathbf{v} = \binom{\cos t + i\sin t}{\sin t - i\cos t}\). Its real and imaginary parts give two real solutions: \(\binom{\cos t}{\sin t}\) and \(\binom{\sin t}{-\cos t}\). So \(\mathbf{x}(t) = c_1\binom{\cos t}{\sin t} + c_2\binom{\sin t}{-\cos t}\). The IC \((1,0)\) gives \(c_1 = 1,\; c_2 = 0\).

Answer: \(\mathbf{x}(t) = \binom{\cos t}{\sin t}.\)

Problem 3easy
Rewrite \(y'' + 3y' + 2y = 0\) as a first-order system and identify its matrix.
Hint
Let \(x_1 = y,\; x_2 = y'\).
Solution

Then \(x_1' = x_2\) and \(x_2' = -2x_1 - 3x_2\). So \(A = \begin{pmatrix}0 & 1 \\ -2 & -3\end{pmatrix}\).

Answer: See solution.