Cumulant generating function and largest eigenvalue of operator

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I am working on a recent paper (arXiv:1805.02887) about an application of large deviation theory to the statistical mechanics of active matter and am a bit bewildered by a result dropped in appendix F (p.9, right column), without any reference or indication.

With $q$ such that $$ \frac{\text{d}q}{\text{d}t} = D_r \left[\frac{1}{2q} - q\right] + \sqrt{D_r} \xi, $$ where $\xi$ is a Gaussian white noise with zero mean and unit variance, and $D_r$ is a constant, they claim that the cumulant generating function of the time-averaged value of $q$, $$ f(k) = \lim_{t \rightarrow \infty} \frac{1}{t} \, \log \left\langle \exp\left(- k \int_0^t \text{d}\tau \, q(\tau)\right)\right\rangle, $$ is the largest eigenvalue of the following operator: $$ L_k[\cdot] = -\frac{\partial}{\partial q} \left[D_r \left(\frac{1}{2q} - q\right) \cdot \right] + \frac{D_r}{2} \frac{\partial^2}{\partial q^2}[\cdot] - kq. $$

How can this be? Where does this result come from?

Thank you for your help!

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Reformulating a large deviation problem of an integrated dynamical quantity as an eigenproblem is a common and useful method.

Consider $$ \epsilon(t) = \int_0^t \text{d}\tau \, q(\tau), $$ such that $$ \dot{\epsilon} = q, $$ we then have that the joint probability distribution of $q$ and $\epsilon$ satisfies the Fokker-Planck equation $$ \frac{\partial}{\partial t} P(q, \epsilon) = \mathcal{L} P(q, \epsilon) - \frac{\partial}{\partial \epsilon} (q P(q, \epsilon)), $$ where $$ \mathcal{L} : \bullet \mapsto - \frac{\partial}{\partial q} \left[D_r \left(\frac{1}{2q} - q\right) \bullet\right] + \frac{D_r}{2} \frac{\partial^2}{\partial q^2} \bullet, $$ is the Fokker-Planck operator inferred from the Langevin equation satisfied by $q$ (see Wikipedia).

We introduce the biased measure $$ P_k(q) = \int \text{d}\epsilon \, \exp(- k \epsilon) P(q, \epsilon) = \left<\exp\left(-k \int_0^t \text{d}\tau \, q(\tau)\right)\right>, $$ then we have from the previous Fokker-Planck equation $$ \begin{aligned} \frac{\partial}{\partial t} P_k(q) &= \frac{\partial}{\partial t} \int \text{d}\epsilon \, \exp(- k \epsilon) P(q, \epsilon)\\ &= \mathcal{L} P_k(q) - q \int \text{d}\epsilon \, \exp(-k\epsilon) \frac{\partial}{\partial \epsilon} P(q, \epsilon)\\ &= (\mathcal{L} - k q) P_k(q), \end{aligned} $$ where we have performed an integration by parts to get to the last line.

We call the operator $$ \mathscr{W}_{k, q} = \mathcal{L} - k q : \bullet \mapsto - \frac{\partial}{\partial q} \left[D_r \left(\frac{1}{2q} - q\right) \bullet\right] + \frac{D_r}{2} \frac{\partial^2}{\partial q^2} \bullet - k q \bullet $$ the tilted generator. What can then be shown is that the SCGF $$ f(k) = \lim_{t \rightarrow \infty} \frac{1}{t} \log\left<\exp\left(-k \int_0^t \text{d}\tau \, q(\tau)\right)\right> $$ is the largest eigenvalue of the following eigenproblem (see Touchette, 2017, arXiv:1705.06492 and Jack, 2019, arXiv:1910.09883 (§ II.E)) $$ \mathscr{W}_{k,q} P_k = \Lambda P_k, $$ which is the assertion of the question.