Categories: Mathematics.

# Fredholm alternative

The Fredholm alternative is a theorem regarding equations involving a linear operator $$\hat{L}$$ 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. $$\hat{L} \ket{u} = \ket{f}$$ has a unique solution $$\ket{u}$$ for every $$\ket{f}$$.
2. $$\hat{L}^\dagger \ket{w} = 0$$ has non-zero solutions. Then regarding $$\hat{L} \ket{u} = \ket{f}$$:
1. If $$\braket{w}{f} = 0$$ for all $$\ket{w}$$, then it has infinitely many solutions $$\ket{u}$$.
2. If $$\braket{w}{f} \neq 0$$ for any $$\ket{w}$$, then it has no solutions $$\ket{u}$$.

Where $$\hat{L}^\dagger$$ is the adjoint of $$\hat{L}$$. In other words, $$\hat{L} \ket{u} = \ket{f}$$ has non-trivial solutions if and only if for all $$\ket{w}$$ (including the trivial case $$\ket{w} = 0$$) it holds that $$\braket{w}{f} = 0$$.

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

1. If $$\mathrm{det}(\hat{L}) \neq 0$$, then $$\hat{L} \vec{u} = \vec{f}$$ has a unique solution $$\vec{u}$$ for every $$\vec{f}$$.
2. If $$\mathrm{det}(\hat{L}) = 0$$, then $$\hat{L}^\dagger \vec{w} = \vec{0}$$ has non-zero solutions. Then regarding $$\hat{L} \vec{u} = \vec{f}$$:
1. If $$\vec{w} \cdot \vec{f} = 0$$ for all $$\vec{w}$$, then it has infinitely many solutions $$\vec{u}$$.
2. If $$\vec{w} \cdot \vec{f} \neq 0$$ for any $$\vec{w}$$, then it has no solutions $$\vec{u}$$.

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

1. If $$\lambda$$ is not an eigenvalue, then there is a unique solution $$\ket{u}$$ for each $$\ket{f}$$.
2. If $$\lambda$$ is an eigenvalue, then $$\hat{M}^\dagger \ket{w} = 0$$ has non-zero solutions. Then:
1. If $$\braket{w}{f} = 0$$ for all $$\ket{w}$$, then there are infinitely many solutions $$\ket{u}$$.
2. If $$\braket{w}{f} \neq 0$$ for any $$\ket{w}$$, then there are no solutions $$\ket{u}$$.
1. O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.