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author | Prefetch | 2022-10-14 23:25:28 +0200 |
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committer | Prefetch | 2022-10-14 23:25:28 +0200 |
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diff --git a/source/know/concept/fredholm-alternative/index.md b/source/know/concept/fredholm-alternative/index.md new file mode 100644 index 0000000..c813fb4 --- /dev/null +++ b/source/know/concept/fredholm-alternative/index.md @@ -0,0 +1,61 @@ +--- +title: "Fredholm alternative" +date: 2021-05-29 +categories: +- Mathematics +layout: "concept" +--- + +The **Fredholm alternative** is a theorem regarding equations involving +a linear operator $\hat{L}$ on a [Hilbert space](/know/concept/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](/know/concept/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 $\Inprod{w}{f} = 0$ for all $\Ket{w}$, then it has infinitely many solutions $\Ket{u}$. + 2. If $\Inprod{w}{f} \neq 0$ 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 $\Inprod{w}{f} = 0$. + +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 $\Inprod{w}{f} = 0$ for all $\Ket{w}$, then there are + infinitely many solutions $\Ket{u}$. + 2. If $\Inprod{w}{f} \neq 0$ for any $\Ket{w}$, then there are no + solutions $\Ket{u}$. + + + +## References +1. O. Bang, + *Nonlinear mathematical physics: lecture notes*, 2020, + unpublished. |