Consider a stochastic search model with resetting for an unknown stationary target $a\in\mathbb{R}$ with known distribution $\mu$. The searcher begins at the origin and performs Brownian motion with diffusion constant $D$. The searcher is also armed with an exponential clock with spatially dependent rate $r$, so that if it has failed to locate the target by the time the clock rings, then its position is reset to the origin and it continues its search anew from there. Denote the position of the searcher at time $t$ by $X(t)$. Let $E_0^{(r)}$ denote expectations for the process $X(\cdot)$. The search ends at time $T_a=\inf\{t\ge0:X(t)=a\}$. The expected time of the search is then $\int_{\mathbb{R}}(E_0^{(r)}T_a)\thinspace\mu(da)$. Ideally, one would like to minimize this over all resetting rates $r$. We obtain quantitative growth rates for $E_0^{(r)}T_a$ as a function of $a$ in terms of the asymptotic behavior of the rate function $r$, and also a rather precise dichotomy on the asymptotic behavior of the resetting function $r$ to determine whether $E_0^{(r)}T_a$ is finite or infinite. We show generically that if $r(x)$ is on the order $|x|^{2l}$, with $l>-1$, then $\log E_0^{(r)}T_a$ is on the order $|a|^{l+1}$; in particular, the smaller the asymptotic size of $r$, the smaller the asymptotic growth rate of $E_0^{(r)}T_a$. The asymptotic growth rate of $E_0^{(r)}T_a$ continues to decrease when $r(x)\sim \frac{D\lambda}{x^2}$ with $\lambda>1$; now the growth rate of $E_0^{(r)}T_a$ is more or less on the order $|a|^{\frac{1+\sqrt{1+8\lambda}}2}$. However, if $\lambda=1$, then $E_0^{(r)}T_a=\infty$, for $a\neq0$.

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具有空间相关重置的扩散搜索

考虑一个随机搜索模型，对未知的固定目标 $a\in\mathbb{R}$ 进行重置，已知分布 $\mu$。搜索器从原点开始，以扩散常数 $D$ 执行布朗运动。搜索者还配备了一个具有空间相关速率 $r$ 的指数时钟，因此如果它在时钟响起时未能定位目标，则其位置将重置为原点，并从那里继续搜索. 用$X(t)$表示搜索者在时间$t$的位置。让 $E_0^{(r)}$ 表示对过程 $X(\cdot)$ 的期望。搜索在 $T_a=\inf\{t\ge0:X(t)=a\}$ 时间结束。搜索的预期时间为 $\int_{\mathbb{R}}(E_0^{(r)}T_a)\thinspace\mu(da)$。理想情况下，人们希望在所有重置率 $r$ 上最小化这一点。我们获得了 $E_0^{(r)}T_a$ 的定量增长率作为 $a$ 的函数，根据速率函数 $r$ 的渐近行为，以及对重置渐近行为的相当精确的二分法函数 $r$ 来确定 $E_0^{(r)}T_a$ 是有限的还是无限的。我们一般地表明，如果 $r(x)$ 在订单 $|x|^{2l}$ 上，其中 $l>-1$，则 $\log E_0^{(r)}T_a$ 在订单上$|a|^{l+1}$; 特别是$r$的渐近大小越小，$E_0^{(r)}T_a$的渐近增长率越小。当$r(x)\sim \frac{D\lambda}{x^2}$且$\lambda>1$时，$E_0^{(r)}T_a$的渐近增长率继续下降；现在 $E_0^{(r)}T_a$ 的增长率或多或少在 $|a|^{\frac{1+\sqrt{1+8\lambda}}2}$ 的顺序上。然而，如果 $\lambda=1$，则 $E_0^{(r)}T_a=\infty$，对于 $a\neq0$。