ODE-06

Systems Theory

Linear Systems of ODEs

Learn to solve systems of linear differential equations using matrix exponentials, fundamental matrices, Jordan canonical form, and the powerful Liouville formula.

Learning Objectives

Understand and compute matrix exponentials
Construct fundamental matrices for linear systems
Apply Jordan form to solve systems
Use the Liouville formula for Wronskian evolution
Solve systems with distinct and repeated eigenvalues
Handle complex eigenvalues with real solutions
Apply variation of parameters to nonhomogeneous systems
Understand the solution space structure

1. Matrix Exponential

The matrix exponential is fundamental to solving linear systems of ODEs. It generalizes the scalar exponential $e^,{at}$ to matrices.

Definition 6.1: Matrix Exponential

For a square matrix $A \in M_n(\mathbb,{R}, )$ , the matrix exponential is defined by:

e^A = \sum_{k=0}^{\infty} \frac{A^k}{k!} = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots

This series converges for any matrix $A$ .

Theorem 6.1: Convergence of Matrix Exponential

For any matrix $A \in M_n$ , the series $\sum_{k=0}^{\infty} \frac{A^k}{k!}$ converges absolutely, and $e^,{At}$ converges uniformly on any finite interval.

Proof of Theorem 6.1:

Using any matrix norm $\|\cdot\|$ :

\left\|\sum_{k=0}^{\infty} \frac{A^k}{k!}\right\| \leq \sum_{k=0}^{\infty} \frac{\|A\|^k}{k!} = e^{\|A\|} < \infty

For uniform convergence on $[0, T]$ , we have $\|e^{At}\| \leq e^{\|A\|T}$ . ∎

Theorem 6.2: Properties of Matrix Exponential

1. If $AB = BA$ , then $e^,{A+B}, = e^A e^B$

2. $e^A$ is always invertible with $(e^A)^,{-1}, = e^,{-A}$

3. If $P$ is invertible, then $e^,{PAP^{-1}}, = Pe^A P^,{-1}$

4. $\frac,{d}{dt},e^,{At}, = Ae^,{At}, = e^,{At},A$

Example 6.1: Matrix Exponential of Diagonal Matrix

Problem: Compute $e^,{At}$ for $A = \text,{diag},(a_1, \ldots, a_n)$ .

Solution:

For a diagonal matrix, $A^k = \text,{diag},(a_1^k, \ldots, a_n^k)$ . Therefore:

e^{At} = \sum_{k=0}^{\infty} \frac{A^k t^k}{k!} = \text{diag}\left(\sum_{k=0}^{\infty}\frac{(a_1 t)^k}{k!}, \ldots, \sum_{k=0}^{\infty}\frac{(a_n t)^k}{k!}\right)

= \text{diag}(e^{a_1 t}, \ldots, e^{a_n t})

Example 6.2: Matrix Exponential with Nilpotent Part

Problem: Compute $e^,{At}$ for $A = \begin,{pmatrix}, 2 & 1 \\ 0 & 2 \end,{pmatrix}$ .

Solution:

Write $A = 2I + N$ where $N = \begin,{pmatrix}, 0 & 1 \\ 0 & 0 \end,{pmatrix}$ . Since $N^2 = 0$ (nilpotent):

e^{Nt} = I + Nt = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}

Since $2I$ commutes with $N$ :

e^{At} = e^{2It} e^{Nt} = e^{2t} \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} e^{2t} & te^{2t} \\ 0 & e^{2t} \end{pmatrix}

2. Fundamental Matrix

Definition 6.2: Fundamental Matrix

For the homogeneous system $\frac,{dx}{dt}, = Ax$ , a fundamental matrix $\Phi(t)$ is an $n \times n$ matrix whose columns are $n$ linearly independent solutions.

The standard fundamental matrix satisfies $\Phi(0) = I$ .

Theorem 6.3: Standard Fundamental Matrix

For the constant coefficient system $\frac,{dx}{dt}, = Ax$ , the matrix $\Phi(t) = e^,{At}$ is the standard fundamental matrix.

Proof of Theorem 6.3:

We verify: (1) $\Phi'(t) = Ae^,{At}, = A\Phi(t)$ ✓, and (2) $\Phi(0) = e^0 = I$ ✓.

By the Liouville formula, $\det(\Phi(t)) = \det(I) \cdot e^,{\text{tr}(A)t}, \neq 0$ , so columns are linearly independent. ∎

Theorem 6.4: Liouville Formula

For the system $\frac,{dx}{dt}, = A(t)x$ with fundamental matrix $\Phi(t)$ , the Wronskian $W(t) = \det(\Phi(t))$ satisfies:

W(t) = W(t_0) \exp\left(\int_{t_0}^{t} \text{tr}(A(s))\,ds\right)

Remark:

The Liouville formula shows that the Wronskian is either always zero or never zero. This gives an elegant criterion for linear independence of solutions.

3. Systems with Distinct Eigenvalues

Theorem 6.5: Fundamental Matrix for Distinct Eigenvalues

If $A$ has $n$ distinct eigenvalues $\lambda_1, \ldots, \lambda_n$ with corresponding eigenvectors $r_1, \ldots, r_n$ , then:

\Phi(t) = \begin{pmatrix} e^{\lambda_1 t} r_1 & \cdots & e^{\lambda_n t} r_n \end{pmatrix}

is a fundamental matrix for $\frac,{dx}{dt}, = Ax$ .

Proof of Theorem 6.5:

Each column $e^,lambda_k t, r_k$ satisfies:

\frac{d}{dt}(e^{\lambda_k t} r_k) = \lambda_k e^{\lambda_k t} r_k = A(e^{\lambda_k t} r_k)

since $Ar_k = \lambda_k r_k$ . The columns are linearly independent because eigenvectors corresponding to distinct eigenvalues are independent. ∎

Algorithm 6.1: Solving Systems with Distinct Eigenvalues

Input: System $\frac,{dx}{dt}, = Ax$ where $A$ has distinct eigenvalues

Output: General solution

1. Find eigenvalues $\lambda_1, \ldots, \lambda_n$ from $\det(A - \lambda I) = 0$

2. For each $\lambda_k$ , solve $(A - \lambda_k I)r = 0$ for eigenvector $r_k$

3. General solution: $x(t) = C_1 e^,lambda_1 t, r_1 + \cdots + C_n e^,lambda_n t, r_n$

Example 6.3: System with Distinct Real Eigenvalues

Problem: Solve $\frac,{dx}{dt}, = \begin,{pmatrix}, 6 & -3 \\ 2 & 1 \end,{pmatrix},x$ .

Solution:

Characteristic equation: $\det(A - \lambda I) = (6-\lambda)(1-\lambda) + 6 = \lambda^2 - 7\lambda + 12 = (\lambda - 3)(\lambda - 4) = 0$

Eigenvalues: $\lambda_1 = 3, \lambda_2 = 4$

For $\lambda_1 = 3$ : $(A - 3I)r = 0$ gives $r_1 = (1, 1)^T$

For $\lambda_2 = 4$ : $(A - 4I)r = 0$ gives $r_2 = (3, 2)^T$

General solution:

x(t) = C_1 e^{3t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{4t} \begin{pmatrix} 3 \\ 2 \end{pmatrix}

4. Complex Eigenvalues

Theorem 6.6: Real Solutions from Complex Eigenvalues

If $\lambda = \alpha + i\beta$ is a complex eigenvalue with eigenvector $v = u + iw$ , then two real linearly independent solutions are:

x_1(t) = e^{\alpha t}(u\cos\beta t - w\sin\beta t)

x_2(t) = e^{\alpha t}(u\sin\beta t + w\cos\beta t)

Example 6.4: System with Complex Eigenvalues

Problem: Solve $\frac,{dx}{dt}, = \begin,{pmatrix}, 3 & 5 \\ -5 & 3 \end,{pmatrix},x$ .

Solution:

Characteristic equation: $(3-\lambda)^2 + 25 = 0$ , giving $\lambda = 3 \pm 5i$

For $\lambda_1 = 3 + 5i$ : eigenvector $v = (1, i)^T = (1, 0)^T + i(0, 1)^T$

So $u = (1, 0)^T$ and $w = (0, 1)^T$

Real solutions:

x_1(t) = e^{3t}\begin{pmatrix} \cos 5t \\ -\sin 5t \end{pmatrix}, \quad x_2(t) = e^{3t}\begin{pmatrix} \sin 5t \\ \cos 5t \end{pmatrix}

General solution:

x(t) = e^{3t}\begin{pmatrix} C_1\cos 5t + C_2\sin 5t \\ -C_1\sin 5t + C_2\cos 5t \end{pmatrix}

5. Jordan Form and Repeated Eigenvalues

Definition 6.3: Jordan Block

A Jordan block $J_k(\lambda)$ of size $k$ for eigenvalue $\lambda$ is:

J_k(\lambda) = \begin{pmatrix} \lambda & 1 & & \\ & \lambda & \ddots & \\ & & \ddots & 1 \\ & & & \lambda \end{pmatrix}_{k \times k} = \lambda I_k + N_k

where $N_k$ is nilpotent with $N_k^k = 0$ .

Theorem 6.7: Exponential of Jordan Block

For a Jordan block $J_k(\lambda) = \lambda I + N$ :

e^{J_k(\lambda)t} = e^{\lambda t} \begin{pmatrix} 1 & t & \frac{t^2}{2!} & \cdots & \frac{t^{k-1}}{(k-1)!} \\ & 1 & t & \cdots & \frac{t^{k-2}}{(k-2)!} \\ & & \ddots & \ddots & \vdots \\ & & & 1 & t \\ & & & & 1 \end{pmatrix}

Remark:

When $A$ has repeated eigenvalues but is not diagonalizable, we need generalized eigenvectors. If $A = PJP^,{-1}$ is the Jordan decomposition, then $e^,{At}, = Pe^,{Jt},P^,{-1}$ .

6. Nonhomogeneous Systems

Theorem 6.8: Variation of Parameters for Systems

The general solution of $\frac,{dx}{dt}, = Ax + f(t)$ is:

x(t) = e^{At}C + \int_{t_0}^{t} e^{A(t-s)} f(s)\,ds

where $C$ is an arbitrary constant vector.

Proof of Theorem 6.8:

The homogeneous solution is $x_h = e^,{At},C$ . For a particular solution, set $x_p = e^,{At},C^*(t)$ :

e^{At}\frac{dC^*}{dt} = f(t) \implies \frac{dC^*}{dt} = e^{-At}f(t)

Integrating: $C^*(t) = \int_,{t_0},^t e^,{-As},f(s)ds$ . Thus $x_p = e^,{At},\int_,{t_0},^t e^,{-As},f(s)ds = \int_,{t_0},^t e^,{A(t-s)},f(s)ds$ . ∎

Example 6.5: Nonhomogeneous System

Problem: Solve $\frac,{dx}{dt}, = \begin,{pmatrix}, 1 & 0 \\ 0 & 2 \end,{pmatrix},x + \begin,{pmatrix}, e^t \\ 1 \end,{pmatrix}$ .

Solution:

For this diagonal system: $e^,{At}, = \text,{diag},(e^t, e^,{2t},)$

Particular solution:

x_p = \int_0^t \begin{pmatrix} e^{t-s} & 0 \\ 0 & e^{2(t-s)} \end{pmatrix} \begin{pmatrix} e^s \\ 1 \end{pmatrix} ds = \int_0^t \begin{pmatrix} e^t \\ e^{2(t-s)} \end{pmatrix} ds

= \begin{pmatrix} te^t \\ \frac{1}{2}(e^{2t} - 1) \end{pmatrix}

General solution:

x(t) = \begin{pmatrix} C_1 e^t + te^t \\ C_2 e^{2t} + \frac{1}{2}(e^{2t} - 1) \end{pmatrix}

Practice Quiz

Linear Systems Quiz

Questions

Correct

Accuracy

The matrix exponential

e^A

is defined as:

Easy

Not attempted

A

and

B

commute (

AB = BA

), then:

Easy

Not attempted

The fundamental matrix

\Phi(t)

\frac{dx}{dt} = Ax

satisfies:

Easy

Not attempted

A = \text{diag}(a_1, \ldots, a_n)

, then

e^{At} =

Medium

Not attempted

The Liouville formula states that for

\Phi'(t) = A(t)\Phi(t)

Medium

Not attempted

For a Jordan block

J = \lambda I + N

where

N

is nilpotent,

e^{Jt} =

Medium

Not attempted

A

has distinct eigenvalues

\lambda_1, \ldots, \lambda_n

with eigenvectors

r_1, \ldots, r_n

, then the fundamental matrix is:

Medium

Not attempted

For the nonhomogeneous system

\frac{dx}{dt} = Ax + f(t)

, the particular solution using variation of parameters is:

Medium

Not attempted

For complex eigenvalues

\alpha \pm i\beta

with eigenvector

v = u + iw

, the real fundamental solutions are:

Hard

Not attempted

P^{-1}AP = J

(Jordan form), then

e^{At} =

Hard

Not attempted

Frequently Asked Questions

How do I compute the matrix exponential?

The method depends on the matrix structure: (1) For diagonal matrices, exponentiate each diagonal entry. (2) For diagonalizable matrices, use

e^,{At}, = Pe^,{Dt},P^,{-1}

. (3) For Jordan form, use

e^,{Jt}

block by block. (4) For 2×2 matrices, explicit formulas exist based on eigenvalue type.

What if eigenvalues are repeated but the matrix is diagonalizable?

A

has repeated eigenvalues but is still diagonalizable (has a full set of eigenvectors), treat it like the distinct eigenvalue case. The key is whether there are

n

linearly independent eigenvectors, not whether eigenvalues are distinct.

How do I convert complex solutions to real solutions?

For complex eigenvalue

\alpha + i\beta

with eigenvector

u + iw

, the complex solution

e^,{(\alpha+i\beta)t},(u+iw)

gives real solutions by taking real and imaginary parts:

e^,{\alpha t},(u\cos\beta t - w\sin\beta t)

and

e^,{\alpha t},(u\sin\beta t + w\cos\beta t)

What is the geometric meaning of the Liouville formula?

The Liouville formula describes how volumes evolve under the flow of a linear system. If

\text,{tr},(A) < 0

, volumes contract; if

\text,{tr},(A) > 0

, they expand. This is related to dissipation in physical systems.

When do I need Jordan form vs. just eigenvalues?

You need Jordan form when the matrix is not diagonalizable—i.e., when there are fewer independent eigenvectors than the algebraic multiplicity of an eigenvalue. In practice, this happens with repeated eigenvalues and defective matrices.

How does this connect to the scalar case?

A single

n

-th order ODE is equivalent to an

n

-dimensional first-order system. The characteristic equation of the system's matrix equals the characteristic equation of the scalar ODE. Eigenvalues correspond to characteristic roots.