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Fredholm Operator

Nov 14, 20242 min read

PDEsFunctionalAnalysis Existence and uniqueness of a weak solution of an Elliptic PDE for the regularized operator, , follows from the Lax-Milgram Theorem for all (the value of is not important, as it is just a tool to get invertibility of ). One can recover a solution to the original PDE in the following manner:

The regularized elliptic operator has a compact inverse, namely and the Fredholm alternative states that

This categorizes the possible solutions to the original PDE: Either

  1. is bijective so that
Misplaced & Lu = f &\iff L_\mu u = \mu u + f \\ &\iff u = L_\mu^{-1}(\mu u + f) \\ &\iff (I - \mu L_\mu^{-1}) u = L_\mu^{-1} f \\ &\iff (I - K) u = L_\mu^{-1} f \\ \end{aligned}$$ which yields a unique weak solution to the inhomogeneous PDE. 2. $\frac{\lambda}{\lambda + \mu}$ is an eigenvalue of $(I - K)$ with $u \neq 0$: $$\begin{aligned} (L-\lambda I)u = 0 &\iff L_\mu u = (\lambda +\mu) u \\ &\iff u = (\lambda +\mu) L_\mu^{-1} u \\ &\iff Ku = \frac{\mu}{\lambda + \mu} u \\ &\iff (I - K) u = \frac{\lambda}{\lambda + \mu} u \end{aligned}$$ which yields a non-trivial solution to the homogenous PDE. Here the $\lambda$'s consist of the eigenvalues of $L$ which is at most countable by the [[Spectral Theorem]] for compact operators. The corresponding eigenfunctions can be thought of intuitively as "resonances" that prevent the equation from being solvable.

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