Linearized system along a trajectory

For a system of first order ODEs $\dot{x} = f (x)$ , let $γ (t)$ be an arbitrary reference trajectory that does not necessarily satisfy the differential equation (so $\dot{γ} (t) \neq f (γ (t))$ ).

Define the error $e (t) = x (t) - γ (t)$ . Its derivative is:

\dot{e} (t) = \dot{x} (t) - \dot{γ} (t) = f (γ (t) + e (t)) - \dot{γ} (t)

Expanding $f$ in a first-order Taylor series around $γ (t)$ :

f (γ (t) + e (t)) \approx f (γ (t)) + d f_{γ (t)} e (t)

Substituting this back gives:

\dot{e} (t) = M (t) e (t) + d (t)

where:

$M (t) = d f_{γ (t)}$ is the Jacobian matrix of $f$ evaluated along the trajectory (time-varying).
$d (t) = f (γ (t)) - \dot{γ} (t)$ is the defect (or drift) vector.

1. If $γ (t)$ is a solution

Then we have $d (t) = 0$ , and

\dot{e} (t) = M (t) e (t)

is called the linearized system, or the variational equation. Its coefficient matrix $M (t) = d f_{γ (t)}$ is the Jacobian of the vector field, not of the flow: see Jacobian of the flow.

2. What Happens When $γ (t)$ Is Not a Solution?

Because $d (t) \neq 0$ , the error system is affine/forced rather than homogeneous:

Driven Error Dynamics: Even if the tracking error starts at zero ( $e (0) = 0$ ), the system immediately develops non-zero error ( $\dot{e} (0) = d (0) \neq 0$ ). The natural vector field pushes the state away from the target curve.
Rapid Breakdown: The linear approximation degrades quickly because $d (t)$ drives $e (t)$ away from zero, violating the assumption that $e$ remains small.

3. Example: linearizing along a solution curve

Consider the 2D autonomous nonlinear system:

(\begin{matrix} {\dot{x}}_{1} \\ {\dot{x}}_{2} \end{matrix}) = (\begin{matrix} x_{1} \\ x_{1}^{2} + x_{2} \end{matrix})

Step 1: Define the Reference Trajectory

To linearize along a trajectory, first select a known nominal solution $γ (t) = (\begin{matrix} γ_{1} (t) \\ γ_{2} (t) \end{matrix})$ .

Given the initial condition $γ (0) = (\begin{matrix} 1 \\ 1 \end{matrix})$ , solve the nonlinear equations:

${\dot{γ}}_{1} = γ_{1} ⟹ γ_{1} (t) = \exp (t)$
${\dot{γ}}_{2} - γ_{2} = (\exp (t))^{2} = \exp (2 t) ⟹ γ_{2} (t) = \exp (2 t)$

The reference trajectory moving through state space is:

γ (t) = (\begin{matrix} \exp (t) \\ \exp (2 t) \end{matrix})

Step 2: Compute the Time-Varying Jacobian $M (t)$

Calculate the general Jacobian matrix $d f_{x}$ for $f (x) = (\begin{matrix} x_{1} \\ x_{1}^{2} + x_{2} \end{matrix})$ :

d f_{x} = (\begin{matrix} \frac{\partial f_{1}}{\partial x_{1}} & \frac{\partial f_{1}}{\partial x_{2}} \\ \frac{\partial f_{2}}{\partial x_{1}} & \frac{\partial f_{2}}{\partial x_{2}} \end{matrix}) = (\begin{matrix} 1 & 0 \\ 2 x_{1} & 1 \end{matrix})

Evaluating $d f_{x}$ along the reference trajectory $γ (t)$ yields the linear time-varying system matrix $M (t)$ :

M (t) = d f_{γ (t)} = (\begin{matrix} 1 & 0 \\ 2 \exp (t) & 1 \end{matrix})

Step 3: Formulate and Solve the Linearized Error System

Let $e (t) = x (t) - γ (t) = (\begin{matrix} e_{1} (t) \\ e_{2} (t) \end{matrix})$ represent a small tracking perturbation. The linearized error dynamics are given by $\dot{e} (t) = M (t) e (t)$ :

(\begin{matrix} {\dot{e}}_{1} \\ {\dot{e}}_{2} \end{matrix}) = (\begin{matrix} 1 & 0 \\ 2 \exp (t) & 1 \end{matrix}) (\begin{matrix} e_{1} \\ e_{2} \end{matrix})

Suppose the system is given a small initial displacement along $x_{1}$ : $e (0) = (\begin{matrix} ϵ \\ 0 \end{matrix})$ .

Solve for $e_{1} (t)$ :

{\dot{e}}_{1} = e_{1} ⟹ e_{1} (t) = ϵ \exp (t)

Solve for $e_{2} (t)$ :

{\dot{e}}_{2} = 2 \exp (t) (ϵ \exp (t)) + e_{2} = 2 ϵ \exp (2 t) + e_{2}

Solving this linear non-homogeneous differential equation with initial condition $e_{2} (0) = 0$ :

e_{2} (t) = 2 ϵ \exp (2 t) - 2 ϵ \exp (t)

The linearized approximation predicts the perturbed trajectory as:

x_{linearized} (t) = γ (t) + e (t) = (\begin{matrix} \exp (t) + ϵ \exp (t) \\ \exp (2 t) + 2 ϵ \exp (2 t) - 2 ϵ \exp (t) \end{matrix})

Step 4: Compare Linearized Prediction to True Nonlinear Trajectory

Solving the full nonlinear system with perturbed initial conditions $x (0) = (\begin{matrix} 1 + ϵ \\ 1 \end{matrix})$ gives the exact solution:

x_{exact} (t) = (\begin{matrix} (1 + ϵ) \exp (t) \\ (1 + 2 ϵ + ϵ^{2}) \exp (2 t) - (2 ϵ + ϵ^{2}) \exp (t) \end{matrix})

Subtracting the reference trajectory $γ (t)$ yields the exact error:

e_{exact} (t) = (\begin{matrix} ϵ \exp (t) \\ (2 ϵ + ϵ^{2}) \exp (2 t) - (2 ϵ + ϵ^{2}) \exp (t) \end{matrix})

Comparing the linearized prediction to the true error:

e_{exact} (t) - e_{linearized} (t) = (\begin{matrix} 0 \\ ϵ^{2} (\exp (2 t) - \exp (t)) \end{matrix}) = O (ϵ^{2})

The linearized system along the reference trajectory successfully captures all primary linear dynamics ( $O (ϵ)$ ), leaving an approximation error proportional only to $ϵ^{2}$ .

4. Example: Linearizing Along a Non-Solution Trajectory

Consider the same nonlinear autonomous system:

(\begin{matrix} {\dot{x}}_{1} \\ {\dot{x}}_{2} \end{matrix}) = (\begin{matrix} x_{1} \\ x_{1}^{2} + x_{2} \end{matrix})

Step 1: Define the Reference Trajectory and Exact Solution

Exact Solution $x (t)$ : For initial condition $x (0) = (\begin{matrix} 1 \\ 1 \end{matrix})$ , the exact nonlinear trajectory is:

x (t) = (\begin{matrix} \exp (t) \\ \exp (2 t) \end{matrix})

Reference Trajectory $γ (t)$ : Choose an arbitrary (non-solution) curve that passes through the same initial point $γ (0) = (\begin{matrix} 1 \\ 1 \end{matrix})$ :

γ (t) = (\begin{matrix} 1 + t^{2} \\ 1 + t^{2} \end{matrix})

Since $\dot{γ} (t) = (\begin{matrix} 2 t \\ 2 t \end{matrix}) \neq f (γ (t))$ , this curve is not a trajectory of the system.

Step 2: Calculate System Matrices $M (t)$ and Defect Vector $d (t)$

Jacobian $M (t) = d f_{γ (t)}$ :

M (t) = (\begin{matrix} 1 & 0 \\ 2 γ_{1} & 1 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 2 (1 + t^{2}) & 1 \end{matrix})

Defect Vector $d (t) = f (γ (t)) - \dot{γ} (t)$ :

d (t) = (\begin{matrix} 1 + t^{2} \\ (1 + t^{2})^{2} + (1 + t^{2}) \end{matrix}) - (\begin{matrix} 2 t \\ 2 t \end{matrix}) = (\begin{matrix} t^{2} - 2 t + 1 \\ t^{4} + 3 t^{2} - 2 t + 2 \end{matrix})

Step 3: Solve the Linearized Error System $\dot{e} (t) = M (t) e (t) + d (t)$

Since $x (0) = γ (0)$ , the initial error is $e (0) = (\begin{matrix} 0 \\ 0 \end{matrix})$ .

1. First Component $e_{1} (t)$

{\dot{e}}_{1} = e_{1} + (t^{2} - 2 t + 1), e_{1} (0) = 0

Using an integrating factor $\exp (- t)$ :

e_{1} (t) = \exp (t) - t^{2} - 1

2. Second Component $e_{2} (t)$

{\dot{e}}_{2} = 2 (1 + t^{2}) e_{1} (t) + e_{2} + (t^{4} + 3 t^{2} - 2 t + 2), e_{2} (0) = 0

Substituting $e_{1} (t) = \exp (t) - t^{2} - 1$ :

{\dot{e}}_{2} - e_{2} = 2 (1 + t^{2}) (\exp (t) - t^{2} - 1) + t^{4} + 3 t^{2} - 2 t + 2 = 2 (1 + t^{2}) \exp (t) - t^{4} + t^{2} - 2 t

Solving this linear first-order differential equation yields:

e_{2} (t) = (\frac{2}{3} t^{3} + 2 t - 28) \exp (t) + t^{4} + 4 t^{3} + 13 t^{2} + 28 t + 28

Step 4: Construct the Approximated Trajectory

Combining $x_{approx} (t) = γ (t) + e (t)$ :

x_{approx} (t) = (\begin{matrix} 1 + t^{2} \\ 1 + t^{2} \end{matrix}) + (\begin{matrix} \exp (t) - t^{2} - 1 \\ (\frac{2}{3} t^{3} + 2 t - 28) \exp (t) + t^{4} + 4 t^{3} + 13 t^{2} + 28 t + 28 \end{matrix})

Simplifying each component:

x_{approx} (t) = (\begin{matrix} \exp (t) \\ (\frac{2}{3} t^{3} + 2 t - 28) \exp (t) + t^{4} + 4 t^{3} + 14 t^{2} + 28 t + 29 \end{matrix})

Step 5: Comparison of Exact $x (t)$ vs. Approximated $γ (t) + e (t)$

Trajectory Component	Exact Solution x(t)	Approximated Solution xapprox(t)
First Component ( $x_{1}$ )	$\exp (t)$	$\exp (t)$
Second Component ( $x_{2}$ )	$\exp (2 t)$	$(\frac{2}{3} t^{3} + 2 t - 28) \exp (t) + t^{4} + 4 t^{3} + 14 t^{2} + 28 t + 29$

Key Analytical Insights

Exact Agreement in $x_{1} (t)$ :

Because the first state equation ${\dot{x}}_{1} = x_{1}$ is purely linear, the linearized error equation recovers the exact solution $x_{1} (t) = \exp (t)$ for all $t$ .
Local Accuracy Around $t = 0$ :

Expanding both $x_{2} (t)$ and $x_{2, approx} (t)$ in a Taylor series around $t = 0$ :
- Exact: $x_{2} (t) = \exp (2 t) = 1 + 2 t + 2 t^{2} + \frac{4}{3} t^{3} + O (t^{4})$
- Approximated: $x_{2, approx} (t) = 1 + 2 t + 2 t^{2} + 0 t^{3} + O (t^{4})$
The approximation matches the exact solution up to order $O (t^{2})$ . The residual error for small times is $x (t) - x_{approx} (t) = (\begin{matrix} 0 \\ \frac{4}{3} t^{3} + O (t^{4}) \end{matrix})$ .
Breakdown Over Time: Because the reference trajectory $γ (t)$ is not a valid solution curve ( $d (t) \neq 0$ ), the forcing term continuously injects artificial error dynamics. As $t$ grows, $\exp (2 t)$ dominates the exact solution, whereas the linear approximation grows like $t^{3} \exp (t)$ , causing rapid divergence for larger $t$ .

5. Error equation in an alternative coframe

The error system above is written in canonical coordinates. To express it in a coframe chosen along the trajectory $γ (t)$ , see error equation in an alternative coframe.