Expected Lifetime and Capacity

Abstract

We investigate sharp isoperimetric problems for random walks on weighted graphs. Symmetric weights on edges determine the one step transition probabilities for the random walk, measure of sets and capacity between sets. In that setup one can be interested in the exit time of the random walk from a set, i.e. to find for a fixed starting point the “optimal” set of given volume which maximizes the expected time when the walk leaves the set. A strongly related problem is to find a set of fixed volume which has minimal conductance with respect to a given set. In both problems the answer is less appealing than in the case of Euclidean space. As demonstrated by a simple counterexample, there is no unique optimal set. The Berman-Konsowa principle is used in the search for optimal sets. It allows to construct a new graph on which the calculation of conductance and mean exit time is tractable.

Appendix

Proof of Proposition 4.2

Let us recall that we assume that D is an optimal set and slightly change its boundary along the border crossing edges. We consider the Lagrange function with multiplier $\lambda \in \mathbb{R}:$

$$\displaystyle{ \mathrm{Cap}_{L}^{\mathbb{P}}\left (D_{\xi },F\right ) +\lambda \mu \left (D_{\xi }\right ). }$$

Denote $\xi _{l} =\xi _{l}\left (x,y\right ): w_{l} = \left (\xi _{l},x,y\right ) \in \partial D$ forming the perturbation vector $\xi = \left [\xi _{l}\right ].$ Let $z_{l} = \partial F \cap l$ be the endpoint of the path l at the boundary of F.

$$\displaystyle{ \frac{\partial } {\partial \xi _{l}}\left [\sum _{l}\mathrm{Cap}_{L}^{\mathbb{P}}\left (w_{ l},z_{l}\right ) +\lambda \mu \left (D_{\xi }\right )\right ]. }$$

Setting the derivative zero and using $r_{l} = R^{l}\left (w_{l},z_{l}\right )$ we have that

$$\displaystyle\begin{array}{rcl} 0& =& \frac{\partial } {\partial \xi _{l}}\left [\sum _{l}\mathrm{Cap}_{L}^{\mathbb{P}}\left (w_{ l},z_{l}\right ) +\lambda \mu \left (D_{\xi }\right )\right ] = {}\\ & =& \frac{\partial } {\partial \xi _{l}}\left [ \frac{1} {r_{l}} +\lambda \mu _{l}\right ] {}\\ & =& -\frac{R^{l}\left (x_{l},y_{l}\right )} {r_{l}^{2}} +\lambda \mu ^{l}\left (x_{ l},y_{l}\right ) {}\\ \end{array}$$

for all path l ∈ L and

$$\displaystyle{ \mu ^{l}\left (x_{ l},y_{l}\right )r_{l} =\mathrm{ const} }$$

is a necessary condition for the optimality. □ 

Proof of Theorem 5.3

We consider the variational problem

$$\displaystyle{ \max _{F^{{\prime}}:\,\mu (F^{{\prime}})\leq M}E_{x}\left (F^{{\prime}}\right ). }$$

Assume that F is optimal with a path system L and the probability $\mathbb{P}$ on it. As in the case of the capacity we perturb F in a small neighborhood. The maximal solution should satisfy for a suitable λ and for all path l that

$$\displaystyle\begin{array}{rcl} \frac{\partial } {\partial s_{l}}\left [E_{x}\left (F\right ) +\lambda \mu \left (F\right )\right ]& =& 0 {}\\ \frac{\partial } {\partial s_{l}}\left [R\sum _{p} \frac{e_{p}} {R_{p}} +\lambda \mu \left (F\right )\right ]& =& \frac{\partial } {\partial s_{l}}\left [R\sum _{p}\left ( \frac{e_{p}} {R_{p}} +\lambda \mu _{p}\right )\right ] = 0, {}\\ \end{array}$$

where s _l is the length of l and we use μ _l for the volume of the path l. Let $E =\sum _{} \frac{e_{l}} {R_{l}}$, the density of μ is $\mu \left (z_{l}\right ) = \frac{d\mu } {ds}\vert _{s_{l}},$ where s is the arc length parametrization of $l_{z}: w\left (s_{l}\right ) = z_{l}$. Furthermore, $\mu _{l} =\mu \left (z_{l}\right )$ and the density of resistance is $\rho \left (z_{l}\right ) = 1/\mu (z_{l})$, then the derivative is as follows

$$\displaystyle\begin{array}{rcl} \frac{\partial } {\partial s_{l}}\left [R\sum _{p}\left ( \frac{e_{p}} {R_{p}} +\lambda \mu _{p}\right )\right ] = \left ( \frac{\partial } {\partial s_{l}}R\right )E + R \frac{\partial } {\partial s_{l}}E +\lambda \mu \left (z_{l}\right ).& & {}\\ \end{array}$$

One can find that

$$\displaystyle{ \left ( \frac{\partial } {\partial s_{l}}R\right ) = \frac{\partial } {\partial s_{l}} \frac{1} {\sum _{p} \frac{1} {R_{p}}} = R^{2}\frac{\rho \left (z_{l}\right )} {R_{l}^{2}} }$$

and

$$\displaystyle\begin{array}{rcl} \frac{\partial } {\partial s_{l}}e_{l}& =& \frac{\partial } {\partial s_{l}}\int r\left (w_{s},z_{l}\right )\mu \left (w\left (s\right )\right )ds {}\\ & =& \frac{\partial } {\partial s_{l}}\int _{0}^{s_{l} }\int _{s}^{s_{l} }\rho \left (w\left (t\right )\right )dt\mu \left (w\left (s\right )\right )ds {}\\ & =& \rho \left (z_{l}\right )\mu _{l}, {}\\ \end{array}$$

$$\displaystyle{ \frac{\partial } {\partial s_{l}}E = \frac{\partial } {\partial s_{l}} \frac{e_{l}} {R_{l}} =\rho \left (z_{l}\right )\frac{\mu _{l}R_{l} - e_{l}} {R_{l}^{2}}. }$$

It is trivial that $e_{l} \leq \mu _{l}R_{l}$, so the defined δ _l is nonnegative. Furthermore,

$$\displaystyle{ \frac{\partial } {\partial s_{l}}E = \frac{\rho \left (z_{l}\right )} {R_{l}^{2}} \delta _{l}e_{l}. }$$

$$\displaystyle{ \frac{\partial } {\partial s_{l}}E_{x}\left (F\right ) = R^{2}\frac{\rho \left (z_{l}\right )} {R_{l}^{2}} E + R\frac{\rho \left (z_{l}\right )} {R_{l}^{2}} \delta _{l}e_{l} +\lambda \mu \left (z_{l}\right ) = 0 }$$

$$\displaystyle{ R\frac{\rho \left (z_{l}\right )^{2}} {R_{l}^{2}} E + \frac{\rho \left (z_{l}\right )^{2}} {R_{l}^{2}} \delta _{l}e_{l} =\mathrm{ const}. }$$

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