Vai's Uni Notes

Fundamental Matrix

4 min read

Building every solution of the homogeneous linear state equation from 𝑛 known solutions.

The Solution Problem

  • Return to the linear system:
    • ̇𝐱(𝑡)=𝐴(𝑡)𝐱(𝑡)+𝐵(𝑡)𝐮(𝑡),𝐱(𝑡0)=𝐱0
  • Problem: given any 𝐱0∈ℝ𝑛, 𝐮(𝑡)∈ℝ𝑚 (𝑡≥𝑡0), find 𝐱(𝑡), 𝑡≥𝑡0
  • Start with the homogeneous eq’n:
    • ̇𝐱(𝑡)=𝐴(𝑡)𝐱(𝑡),𝐱(𝑡0)=𝐱0
  • Basic idea: if we know any 𝑛 solutions from linearly independent ICs, we know all solutions

Fundamental Solutions

  • Suppose we know, or can easily find, 𝑛 solutions 𝛙𝑘(𝑡) from an arbitrary set of ICs {𝛙𝑘(𝑡0)}𝑛𝑘=1 which are linearly independent, i.e.
    • ̇𝛙𝑘(𝑡)=𝐴(𝑡)𝛙𝑘(𝑡)𝛙𝑘(𝑡0)=𝛙𝑘,0𝑘=1,…,𝑛
    • with 𝛙𝑘,0 arbitrary, but linearly independent
  • Any such set {𝛙𝑘(𝑡)}𝑛𝑘=1 are known as fundamental solutions

Fundamental Matrix

Fundamental Matrix

  • Ψ(𝑡)=[𝛙1(𝑡),𝛙2(𝑡),…,𝛙𝑛(𝑡)]
  • is known as a fundamental matrix
  • Note:
    • ̇Ψ(𝑡)=[̇𝛙1(𝑡),…,̇𝛙𝑛(𝑡)]=[(𝐴𝛙1(𝑡)),…,(𝐴𝛙𝑛(𝑡))]=𝐴Ψ(𝑡)
  • i.e.
    • ̇Ψ(𝑡)=𝐴(𝑡)Ψ(𝑡)withΨ(𝑡0)=[𝛙1(𝑡0),…,𝛙𝑛(𝑡0)]nonsingular
    • A matrix version of the homogeneous state eq’n, with nonsingular B/C

Solutions as Combinations of Fundamental Solutions

  • Claim: for any 𝐱0=𝐱(𝑡0)∈ℝ𝑛, and any set of fundamental solutions {𝛙𝑘(𝑡)}, the resulting 𝐱(𝑡) can be written as a linear combination of {𝛙𝑘(𝑡)}, i.e.
    • 𝐱(𝑡)=∑𝑛𝑘=1𝛼𝑘𝛙𝑘(𝑡)∀𝑡≥𝑡0
  • Pf: by definition, {𝛙𝑘(𝑡0)} form a basis for ℝ𝑛
    • ⇒ arbitrary 𝐱0 admits the expansion
    • 𝐱0=∑𝑛𝑘=1𝛼𝑘𝛙𝑘(𝑡0)
    • so it’s true for 𝑡=𝑡0. What about 𝑡>𝑡0? See if this assumption is consistent:
    • ̇𝐱(𝑡)=∑𝑛𝑘=1𝛼𝑘̇𝛙𝑘(𝑡)=∑𝑛𝑘=1𝛼𝑘(𝐴(𝑡)𝛙𝑘(𝑡))=𝐴(𝑡)∑𝑛𝑘=1𝛼𝑘𝛙𝑘(𝑡)=𝐴(𝑡)𝐱(𝑡)✓
    • So 𝐱(𝑡) satisfies the DE and B/C. By uniqueness it is the solution

Matrix Form

  • Note: can write 𝐱(𝑡) as
    • 𝐱(𝑡)=Ψ(𝑡)𝛂where𝛂=[𝛼1⋮𝛼𝑛]
  • where 𝛂 also satisfies
    • 𝐱0=Ψ(𝑡0)𝛂⇒𝛂=Ψ(𝑡0)−1𝐱0
    • valid since Ψ(𝑡0) is nonsingular
  • ⇒𝛂=Ψ−1(𝑡)𝐱(𝑡)∀𝑡≥𝑡0
    • which would imply that 𝛂 is a “conserved” quantity of the homogeneous dynamics (provided Ψ(𝑡) stays invertible: see the lemmas below)
  • Thus

Nonsingularity of the Fundamental Matrix

Lemma 1

  • Any fundamental matrix satisfies:
  • dd𝑡detΨ(𝑡)=trace{𝐴(𝑡)}detΨ(𝑡)
  • trace{𝐴(𝑡)}=∑𝑛𝑘=1𝑎𝑘𝑘(𝑡) (sum of diagonal elements)

Lemma 2

  • If ∫𝑡𝑡0trace{𝐴(𝜏)}d𝜏>−∞ for any finite 𝑡≥𝑡0, then Ψ(𝑡) is nonsingular for any finite 𝑡≥𝑡0
  • Pf: let 𝑑(𝑡)=detΨ(𝑡) and 𝑎(𝑡)=trace{𝐴(𝑡)}. Then Lemma 1 states
    • ̇𝑑(𝑡)=𝑎(𝑡)𝑑(𝑡)
    • a scalar linear DE, with explicit solution
    • 𝑑(𝑡)=𝑑(𝑡0)𝑒∫𝑡𝑡0𝑎(𝜏)d𝜏
    • 𝑑(𝑡)=0⇒ either 𝑑(𝑡0)=0 or 𝑒∫(⋅)=0
    • But 𝑑(𝑡0)=detΨ(𝑡0)≠0 by assumption
    • The 2nd requires
    • ∫𝑡𝑡0𝑎(𝜏)d𝜏=−∞
    • Assuming this is not true, then 𝑑(𝑡)≠0 for any finite 𝑡≥𝑡0
  • ⇒ all “realistic” 𝐴(𝑡) will satisfy the required condition
  • Hence:
    • detΨ(𝑡)≠0 ∀𝑡≥𝑡0
    • Ψ−1(𝑡) exists ∀𝑡≥𝑡0
    • {𝛙𝑘(𝑡)} linearly independent ∀𝑡≥𝑡0

Derivative of the Inverse

  • When 𝐮(𝑡)=𝟎: dd𝑡[𝛂]=𝟎=dd𝑡[Ψ−1(𝑡)𝐱(𝑡)] (𝛂 conserved)
  • What if 𝐮(𝑡)≠𝟎? Is 𝛂 still “conserved”? We need dd𝑡Ψ−1:
  • Differentiate ΨΨ−1=𝐼:
    • [dd𝑡Ψ]Ψ−1+Ψ[dd𝑡Ψ−1]=0withdd𝑡Ψ=𝐴Ψ
    • ⇒dd𝑡Ψ−1=−Ψ−1[𝐴Ψ]Ψ−1

Derivative of the Inverse Fundamental Matrix

  • dd𝑡Ψ−1=−Ψ−1𝐴

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