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Lab 05

Matrix Boundary Lab

Can an integral be evaluated by solving one small linear system instead of integrating by parts?

Pick a basis closed under differentiation, see differentiation become the matrix Ω, and evaluate the truncated integral of e^(bx) f(x) with the resolvent M(b) = (bI + Ωᵀ)⁻¹, in six visible steps. The same resolvent solves y′ − by = f with y(0) = 0. Both are compared with quadrature.

  • Computational Demonstration
  • Book Framework
  • Classical Foundation
0100002000030000Φ′ = ΩΦM(b) = (bI + Ωᵀ)⁻¹[e^(bx) Φᵀ M u]₀ᵃ

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Scope of this instrument

Supported

  • Bases {1,x,…,xm−1}\{1,x,\dots,x^{m-1}\} (2≤m≤62 \le m \le 6), {cos⁡ωx,sin⁡ωx}\{\cos\omega x,\sin\omega x\}, {eαx}\{e^{\alpha x}\}, {eαxcos⁡ωx,eαxsin⁡ωx}\{e^{\alpha x}\cos\omega x, e^{\alpha x}\sin\omega x\}, and {xsin⁡ωx,xcos⁡ωx,sin⁡ωx,cos⁡ωx}\{x\sin\omega x, x\cos\omega x, \sin\omega x, \cos\omega x\}.
  • The book's convention Φ′=ΩΦ\Phi'=\Omega\Phi, f=uTΦf=u^{\mathsf T}\Phi, and ∫0aebxf(x) dx=[ebxΦ(x)TM(b) u]0a\int_0^a e^{bx}f(x)\,dx=\big[e^{bx}\Phi(x)^{\mathsf T}M(b)\,u\big]_0^a.
  • The first-order ODE y′−by=fy'-by=f, y(0)=0y(0)=0, through M(−b)M(-b) (book eq. 5.1).
  • Explicit messages when bI+ΩTbI+\Omega^{\mathsf T} or ΩT−bI\Omega^{\mathsf T}-bI is singular.

Not supported

  • Functions outside the span of the chosen basis.
  • Resonant cases (singular bI+ΩTbI+\Omega^{\mathsf T}), which the book handles by limits; the lab reports them instead.
  • Products of bases (the Kronecker-state extension), the matrix-exponential variant, and higher-order ODEs.
  • ω=0\omega = 0 for trigonometric bases, where the functions are not independent.

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