Categories: Mathematics.

Fredholm alternative

The Fredholm alternative is a theorem regarding equations involving a linear operator L^\hat{L} on a Hilbert space, and is useful in the context of multiple-scale perturbation theory. It is an alternative because it gives two mutually exclusive options, given here in Dirac notation:

  1. L^∣u⟩=∣f⟩\hat{L} \Ket{u} = \Ket{f} has a unique solution ∣u⟩\Ket{u} for every ∣f⟩\Ket{f}.
  2. L^†∣w⟩=0\hat{L}^\dagger \Ket{w} = 0 has nonzero solutions. Then regarding L^∣u⟩=∣f⟩\hat{L} \Ket{u} = \Ket{f}:
    1. If ⟨w | f⟩=0\Inprod{w}{f} = 0 for all ∣w⟩\Ket{w}, then it has infinitely many solutions ∣u⟩\Ket{u}.
    2. If ⟨w | f⟩≠0\Inprod{w}{f} \neq 0 for any ∣w⟩\Ket{w}, then it has no solutions ∣u⟩\Ket{u}.

Where L^†\hat{L}^\dagger is the adjoint of L^\hat{L}. In other words, L^∣u⟩=∣f⟩\hat{L} \Ket{u} = \Ket{f} has non-trivial solutions if and only if for all ∣w⟩\Ket{w} (including the trivial case ∣w⟩=0\Ket{w} = 0) it holds that ⟨w | f⟩=0\Inprod{w}{f} = 0.

As a specific example, if L^\hat{L} is a matrix and the kets are vectors, this theorem can alternatively be stated as follows using the determinant:

  1. If det(L^)≠0\mathrm{det}(\hat{L}) \neq 0, then L^u⃗=f⃗\hat{L} \vec{u} = \vec{f} has a unique solution u⃗\vec{u} for every f⃗\vec{f}.
  2. If det(L^)=0\mathrm{det}(\hat{L}) = 0, then L^†w⃗=0⃗\hat{L}^\dagger \vec{w} = \vec{0} has nonzero solutions. Then regarding L^u⃗=f⃗\hat{L} \vec{u} = \vec{f}:
    1. If w⃗⋅f⃗=0\vec{w} \cdot \vec{f} = 0 for all w⃗\vec{w}, then it has infinitely many solutions u⃗\vec{u}.
    2. If w⃗⋅f⃗≠0\vec{w} \cdot \vec{f} \neq 0 for any w⃗\vec{w}, then it has no solutions u⃗\vec{u}.

Consequently, the Fredholm alternative is also brought up in the context of eigenvalue problems. Define M^=(L^−λI^)\hat{M} = (\hat{L} - \lambda \hat{I}), where λ\lambda is an eigenvalue of L^\hat{L} if and only if det(M^)=0\mathrm{det}(\hat{M}) = 0. Then for the equation M^∣u⟩=∣f⟩\hat{M} \Ket{u} = \Ket{f}, we can say that:

  1. If λ\lambda is not an eigenvalue, then there is a unique solution ∣u⟩\Ket{u} for each ∣f⟩\Ket{f}.
  2. If λ\lambda is an eigenvalue, then M^†∣w⟩=0\hat{M}^\dagger \Ket{w} = 0 has nonzero solutions. Then:
    1. If ⟨w | f⟩=0\Inprod{w}{f} = 0 for all ∣w⟩\Ket{w}, then there are infinitely many solutions ∣u⟩\Ket{u}.
    2. If ⟨w | f⟩≠0\Inprod{w}{f} \neq 0 for any ∣w⟩\Ket{w}, then there are no solutions ∣u⟩\Ket{u}.

References

  1. O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.