Forced Systems and Variation of Constants

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The idea

For $\mathbf x'=A\mathbf x+\mathbf b(t)$ , input continuously contributes new state. Each contribution then evolves under the homogeneous system. This gives a matrix version of convolution and keeps the initial state separate from forcing.

Visual guide

VISUAL GUIDEA cascade responds in two stages

First stage x
Second stage y

For x′ = −x + 1 and y′ = x − y, both initially zero, x rises as 1 − e⁻ᵗ. The second state y = 1 − (1 + t)e⁻ᵗ initially has zero slope and lags behind because it is driven by x rather than directly by the step.

Method and assumptions

For constant $A$ , variation of constants gives $\mathbf x(t)=e^{At}\mathbf x_0+\int_0^t e^{A(t-\tau)}\mathbf b(\tau)d\tau$ . Differentiate the expression to verify both the equation and initial data. For constant input and invertible $A$ , an equilibrium is $\mathbf x_*=-A^{-1}\mathbf b$ .

Worked example: independent driven modes

Let $A=\operatorname{diag}(-1,-2)$ , $\mathbf b=(1,2)$ and $\mathbf x_0=(0,0)$ . Solving each coordinate gives $(1-e^{-t},1-e^{-2t})$ . Both approach the equilibrium $(1,1)$ at different rates.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

For x′=Ax+g(t), which term accounts for forcing when x(0)=0?

∫₀ᵗ eᴬ⁽ᵗ⁻ˢ⁾g(s) ds

eᴬᵗx(0) only

A⁻¹g(t) for every input

Hint 1 · Find a starting point

Each input at time s must propagate from s to t.

Hint 2 · Take the next step

The propagation factor is eᴬ⁽ᵗ⁻ˢ⁾.

Show the reasoning

Answer: ∫₀ᵗ eᴬ⁽ᵗ⁻ˢ⁾g(s) ds

Variation of constants sums those propagated contributions over 0≤s≤t.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: a coupled cascade

For $x'=-x+1,y'=x-y$ with zero initial state, $x=1-e^{-t}$ . Multiplying the second equation by $e^t$ gives $(e^ty)'=e^t-1$ , so $y=1-(1+t)e^{-t}$ . The second stage responds more slowly because its input first passes through the first stage.

Interpreting the result

An equilibrium can exist even if it is unstable. Long-time convergence requires stability of the homogeneous dynamics, not only successful solution of $A\mathbf x_*+\mathbf b=0$ .

Practice

Find the equilibrium of $x'=-2x+6$ .
Find the equilibrium of $x'=x+1$ and assess it.
What happens to the forcing integral at $t=0$ ?

Show worked solutions

$x_*=3$ , attracting.
$x_*=-1$ , unstable because the homogeneous mode is $e^t$ .
Its integration interval is empty, so it is zero and the initial condition remains $\mathbf x_0$ .

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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