You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
author:
- 'D.Chakraborty [^1]'
title: Persistence in Random Walk in Composite Media
---

The phenomenon of persistence in various stochastic processes has been well documented over the past decade[@1]-[@13]. Even the most simple of all stochastic processes, a random walker in a homogeneous and infinite media, exhibits the phenomenon of persistence and the non trivial persistence exponent $\theta$ has the value $1/2$ [@9]. In an experimental setup finite boundaries become important and we have recently investigated the effect of finite boundaries on the survival probability of a random walker in an homogeneous system[@16]. Its then a natural question to ask as to how the survival probability behaves for a random walker in an heterogeneous system. Although random walk in spatially disordered media has been already studied [@17]-[@19] and in few cases exact results are known for a similar quantity ,the first passage time[@20]-[@23], little is known about the persistence probability. We shall consider one such class of heterogeneous system which is known as composite media and is often encountered in experimental science. A composite system essentially comprises of segments of different homogeneous media which differ in their macroscopic properties, such as diffusion coefficients. Redner has already investigated the first passage properties for a diffusive process in such a composite system[@24]. He considers a linear chain of $N$ blocks each of length $L_i$ having diffusivities $D_i$. The mean first passage time in such a system is calculated to be $$\label{1}
\langle t \rangle =\frac{1}{2} \sum_{i=1}^{N} \frac{L_i^2}{D_i} + \sum_{i<j}^{N} \frac{L_i L_j}{D_j}.$$

While the first passage probability is simply the probability that the particle escapes from one of the boundaries, the persistence probability is different and is defined as the probability that the random walker has not crossed the origin up to time $t$. If the system was homogeneous then the persistence probability of a random walker would be simply $p(t) \sim t^{-\theta}$ with $\theta=1/2$. A composite system is slightly different in the sense that near the boundaries the difference in diffusivities tend to give a bias to random walker.

The simplest of composite media that can be constructed is the one with two homogeneous segments with different diffusivities. We shall first derive the result for such a system and later generalize this result for different types of composite media.

Consider two homogeneous media of diffusivities $D_1$, henceforth called medium 1 and $D_2$, medium 2. A slab of medium 1 is placed between $-L$ and $+L$ and the rest of the space is filled with medium 2 as shown in Fig. 1. Since the diffusion coefficients are different, it follows that the stochastic noise correlator will be different and in particular they are $$\label{2}
\langle \eta_i(t_1) \eta_j(t_2) \rangle =2D_i \delta_{ij} \delta(t_1-t_2),$$ where $i,j$ are the medium indices $1$ and $2$ and $t_1>t_2$. For a random walker the the probability that the walker is at $(x,t)$ starting from $(x_0,0)$ simply obeys the diffusion equation in two segments as

\[3\] =D\_1 -Lx L\
=D\_2 -L &gt; x &gt; L

The exact dynamics of the problem can solved by considering the Laplace transform of Eq.(\[3a\]) and Eq.(\[3b\]) in which case the solution to the equations become

\[4\] P(x,s)=A\_1 (-|x|)\
+A\_2 (|x|) -L x L\
P(x,s)=A\_3 (-|x|)

The coefficients $A_1, A_2, A_3$ are found from the boundary conditions that the probability $P(x,t)$ and the current $-D\frac{\partial P}{\partial x}$ is continuous across the boundary. Finally, the third unknown coefficient is found from the normalization of the probability. The resulting expression, however, is complicated and it is difficult to extract any information from it.

We, instead, take a different approach to derive our result. The equation of motion for the random walker is not changed in spite of the heterogeneity of the system and is simply

\[5\] = \_1(t) -L x L,\
= \_2(t) -L &gt; x &gt; L.

Let the time required for a random walker to reach the boundary $x=\pm L$ be $\tau$. In which case we can write down the solution for the equation of motion in the two different regions. For $-L<x<L$ the solution is simply $$\label{6}
    x(t)=\int_{0}^{t} \mathrm{d} t' \eta_1(t'),$$ whereas for $-L>x>L$, when the particle is in medium 2 the solution for $x(t)$ becomes $$\label{7}
    x(t)= \int_{0}^{\tau} \mathrm{d} t' \eta_1(t') +\int_{\tau}^{t} 
\mathrm{d} t' \eta_2(t').$$ Eq.(\[7\]) simply states the fact the random walker has spent time $\tau$ in medium 1 and the rest of the time in medium 2. Both Eq.(\[7\]) and Eq.(\[8\]) are valid when the walker is deep inside medium 1 or medium 2, since none of the equations considers the hopping across the boundary. When deep inside either media the multiple boundary crossings are rare events whereas when the walker is near the boundary multiple crossings are frequent and it is due to these multiple crossing events the crossover is not sharp and there will be two crossover time scales in the problem.

The correlation $\langle x(t_1) x(t_2) \rangle$ can now be worked out carefully. First consider the case $-L<x(t_1)<L$, $-L<x(t_2)<L$ and $t_1>t_2$. In this case the correlator becomes $$\label{8}
\langle x(t_1) x(t_2) \rangle=\int_{0}^{t_1}\mathrm{d}t'_1\int_{0}^{t_2}
\mathrm{d}t'_2 \quad \langle \eta_1(t'_1) \eta(t'_2) \rangle=2D_1 t_2\\$$ where we have used Eq.(\[2\]) for the noise correlator. If, however, $-L>x(t_1)>L$, $-L<x(t_2)<L$ and $t_1>t_2$ then we have $$\begin{aligned}
\label{9}
\nonumber
\langle x(t_1) x(t_2) \rangle = \langle\left[\int_{0}^{t_2} \mathrm{d} t'_2 \eta_1(t'_2) \right]\times \\
\left[ \int_{0}^{\tau} \mathrm{d} t'_1 \eta_1(t'_1) +\int_{\tau}^{t} \mathrm{d} t'_1 \eta_2(t'_1) \right]\rangle\end{aligned}$$ Since $\langle \eta_1(t)\eta_2(t) \rangle=0$, the above expression simplifies to $$\label{10}
\langle x(t_1) x(t_2) \rangle =\int_{0}^{\tau}\mathrm{d} t'_1 \int_{0}^{t_2} \mathrm{d} t'_2 \langle \eta_1(t'_2)  \eta_1(t'_1) \rangle$$ As $\tau >t_2$ the correlation $\langle x(t_1) x(t_2) \rangle$ becomes $$\label{11}
\langle x(t_1) x(t_2) \rangle = 2D_1 t_2$$ Finally, for $-L>x(t_1)>L$ and $-L>x(t_2)>L$, $t_1>t_2 $, the correlator becomes $$\begin{aligned}
\label{12}
\nonumber
\langle x(t_1) x(t_2) \rangle =\langle\left[\int_{0}^{\tau} \mathrm{d} t'_1 \eta_1(t'_1) +\int_{\tau}^{t} \mathrm{d} t'_1 \eta_2(t'_1) \right]\times\\
\left[ \int_{0}^{\tau} \mathrm{d} t'_2 \eta_1(t'_2) +\int_{\tau}^{t} \mathrm{d} t'_2 \eta_2(t'_2)\right]\rangle\end{aligned}$$ Since the cross-correlation of the noise is zero we arrive at $$\begin{aligned}
\label{13}
\nonumber
\langle x(t_1) x(t_2) \rangle = \int_{0}^{\tau} \mathrm{d}t'_1 \int_{0}^{\tau} \mathrm{d}t'_2 \langle \eta_1(t'_1) \eta_1(t'_2) \rangle\\
+ \int_{\tau}^{t} \mathrm{d} t'_1 \int_{\tau}^{t} \mathrm{d} t'_2 \langle\eta_2(t'_1) \eta_2(t'_2)\rangle\end{aligned}$$ The first term in Eq.(\[13\]) is easy to calculate and gives us $$\label{14}
\int_{0}^{\tau} \mathrm{d}t'_1 \int_{0}^{\tau} \mathrm{d}t'_2 \langle \eta_1(t'_1) \eta_1(t'_2)=2D_1\tau$$ To evaluate the second term we make a transformation of variable $t''=t'-\tau$ and we have $$\label{15}
\int_{0}^{t_1-\tau} \mathrm{d} t''_1 \int_{0}^{t_2-\tau} \mathrm{d} t''_2 2D_2 \delta(t''_1-t''_2) = 2D_2 (t_2-\tau)$$ Hence, the complete correlator is $$\label{16}
\langle x(t_1)x(t_2)\rangle = 2(D_1-D_2) \tau+2D_2t_2$$

Of all the quantities in Eq.(\[16\]) the only unknown is $\tau$. Since Eq.(\[16\]) gives us noise averaged quantities we might as well replace $\tau$ by an average value, which is simply $\tau=\frac{L^2}{2D_1}$, the average time for a random walker to reach $x=\pm L$.

It is clear from Eq.(\[11\]) and Eq.(\[16\]) that there are two relevant time scales in the problem. The first one is $\tau_1=\left(\frac{D_1-D_2}{D_1}\right)\frac{L^2}{2D_1}$ whereas the second one is $\tau_2=\left(\frac{D_1-D_2}{D_1}\right)\frac{L^2}{2D_2}$. It is between these two time scales when the random walker undertakes multiple hoppings across the boundary, as a result of which the temporal regime $\tau_1<t<\tau_2$ gives the crossover region in the system.

The complete correlator now takes the form $$\begin{aligned}
\label{17}
\nonumber
\langle x(t_1)x(t_2)\rangle &=& 2 D_1 t_2 \phantom{2 cm} \textrm{for}\quad t_2<\tau_1 ,\\
\nonumber
&=& \left(\frac{D_1-D_2}{D_1}\right)L^2+2D_2 t_2 \quad \textrm{for}\quad 
t_2>\tau_2,\\\end{aligned}$$

In the limit $D_1=D_2=D$ we get the correct result for a homogeneous medium.

A numerical simulation of the system for two different values of $L$ and two different sets of $D_1$ and $D_2$ has been done. Simulation result for the mean square displacement $\langle x^2(t)\rangle$ is shown in Fig.2 and Fig.3. Configuration averaging of $10^4$ has been done for both the systems to obtain the result.

To calculate the survival probability in the two regimes we use Eq.(\[17\]) and with suitable transformations both in $x$ and $t$ convert the process to a Gaussian stationary process. Define a normalized variable $\bar{X}(t)=x(t)/\sqrt{\langle x^2(t) \rangle}$. The correlator in this normalized variable, $\langle \bar{X}(t_1)\bar{X}(t_2)\rangle$, is then given by $$\begin{aligned}
\label{18}
\nonumber
\langle \bar{X}(t_1)\bar{X}(t_2)\rangle &=& \sqrt{\frac{t_2}{t_1}} \quad \textrm{for}\quad t_2<\tau_1 \\ 
\nonumber
&=&\sqrt{\frac{\beta L^2+2D_2 t_2}{\beta L^2+2D_2 t_1}}\quad \textrm{for}\quad t_2>\tau_2 ,\\\end{aligned}$$ with $\beta=\frac{D_1-D_2}{D_1}$. For $t<\tau_1$ we define the usual transformation in time, $T=\ln t$ and the two time correlation function in the new time variable becomes $$\label{19}
\langle \bar{X}(T_1) \bar{X}(T_2)\rangle = e^{-1/2(T_1-T_2)},$$ and the survival probability $p(t)$ for this temporal regime, in real time, is then $$\label{20}
    p(t)\sim t^{-1/2}.$$

For $t>\tau_2$ we define a new time variable $T$ as $$\label{21}
e^{T}=\beta L^2+2D_2 t.$$ The correlation function $\langle \bar{X}(T_1) \bar{X}(T_2) \rangle$ takes the form of Eq(\[19\]), except that the time transformations are different. Since the process is a Gaussian stationary process and the correlator is exponentially decaying, the survival probability in the new time variable, $P(T)$, is $$\label{22}
    P(T)=e^{-T/2}.$$ In real time the survival probability $p(t)$ takes the form $$\label{23}
    p(t)\sim \frac{1}{\sqrt{\beta L^2+2D_2 t}}.$$

A plot of the survival probability, Eq.(\[20\]) and Eq.(\[23\]) for the two time regimes is shown in Fig. 4 and Fig. 5. The crossover timescales $\tau_1$, $\tau_2$ and the crossover regimes are also indicated in the figures. Configuration averaging of $10^6$ has been done to obtain the numerical results of Fig. 4 and Fig 5. Theoretical and numerical values of $\tau_1$ and $\tau_2$ are presented in Table A for two different set of values of $L$, $D_1$ and $D_2$.

**Table A**

   Parameter Values   $\tau_1^{th}$   $\tau_1^{nu}$   $\tau_2^{th}$   $\tau_2^{nu}$
  ------------------ --------------- --------------- --------------- ---------------
        $L=20$                                                       
      $2D_1=20$            18            17.161            180           179.988
      $2D_2=2.0$                                                     
        $L=30$                                                       
      $2D_1=30$            27            27.228            270           271.899
      $2D_2=3.0$                                                     

\

In this section we consider a composite system that is made of three homogeneous media with diffusivities $D_1$, $D_2$ and $D_3$. The medium with diffusivity $D_1$ is placed between $\pm L_1$ while the second medium with diffusivity $D_2$ is placed symmetrically between $-(L_2+L_1)<x<-L_1$ and $L_1<x<(L_1+L_2)$ as shown in the figure.

For a random walker in region I the average time to reach the boundary $\pm L_1$ is $\tau=\frac{L_1^2}{2D_1}$. When the random walker is in region II the average time to cross a region the length $L_2$ is once again $\tau'=\frac{L_2^2}{2D_2}$. Thus, for $t<\tau$ the particle spends its time in region I, for $\tau<t<\tau'$ the walker is in region II while for $t>\tau'$ the walker escapes to region III. The equation of motion in all the three region are $$\label{25}
\frac{\mathrm{d}x}{\mathrm{d}t}=\eta_i(t) \quad \textrm{with i=1,2,3 for three media,}$$ with the noise correlator $$\label{26}
\langle \eta_i(t) \eta_j(t') \rangle =2 D_i \delta_{ij} \delta(t-t'),$$ where $i,j$ are the medium indices running from $1$ to $3$.

The solutions to Eq(\[25\]) for the three regions are respectively $$\label{27}
x(t)=\int_0^t \mathrm{d}t'\eta_1(t')\quad \textrm{for $-L_1<x(t)<L_1$}\\$$

$$\label{28}
x(t)=\int_0^{\tau} \mathrm{d}t'\eta_1(t')
 +\int_{\tau}^t \mathrm{d}t'\eta_2(t')$$

for $L_1<x(t)<L_2$ and $-L_2<x(t)<-L_1$,

$$\begin{aligned}
\label{29}
\nonumber
x(t)=\int_0^{\tau} \mathrm{d}t'\eta_1(t')
 +\int_{\tau}^{\tau+\tau'}\mathrm{d}t'\eta_2(t')+\int_{\tau+\tau'}^t \mathrm{d} t' 
\eta_3(t')\\
\nonumber 
\textrm{for $x(t)>L_2$ and $x(t)<-L_2$}.\\\end{aligned}$$

The two time correlation function $\langle x(t_1) x(t_2) \rangle$ can be worked out carefully and for both $x(t_1)$ and $x(t_2)$ lying in region I, with $t_1>t_2$ is $$\label{30}
\langle x(t_1) x(t_2) \rangle =2 D_1 t_2$$

For $x(t_1)$ and $x(t_2)$ lying in region II, $\langle x(t_1) x(t_2) \rangle$ takes the form $$\label{31}
\langle x(t_1) x(t_2) \rangle =\beta L_1^2 +2D_2 t_2$$

while for $x(t_1)$ and $x(t_1)$ lying in region III, using the fact that the cross correlations of the noise is zero, the correlator becomes $$\begin{aligned}
\label{32}
\nonumber
\langle x(t_1) x(t_2) \rangle=\int_{0}^{\tau} \mathrm{d}t'_1 \int_{0}^{\tau} \mathrm{d}t'_2 \langle \eta_1(t'_1) \eta_1(t'_2) \rangle\\
\nonumber
+ \int_{\tau}^{\tau+\tau'} \mathrm{d} t'_1 \int_{\tau}^{\tau+\tau'} \mathrm{d} t'_2 \langle\eta_2(t'_1) \eta_2(t'_2)\rangle \\
+\int_{\tau+\tau'}^{t_1} \mathrm{d} t'_1 \int_{\tau+\tau'}^{t_2} \mathrm{d} t'_2 \langle\eta_3(t'_1) \eta_3(t'_2)\rangle.\end{aligned}$$

The first integral is simply $2D_1\tau$. The second integral is performed by making use of the transformation $t''=t'-\tau$ and the integral reduces to $$\nonumber
\int_{0}^{\tau'}\mathrm{d} t'_1 \int_{0}^{\tau'} \mathrm{d} t'_2 \langle\eta_2(t'_1) \eta_2(t'_2)\rangle =2D_2\tau'$$

while for the third integral we use the transformation $t''=t'-(\tau+\tau')$ and the integral is evaluated to be $2D_3 (t_2-\tau-\tau')$. Hence the correlator becomes $$\begin{aligned}
\label{33}
\nonumber
\langle x(t_1) x(t_2) \rangle&=&2D_1 \tau +2D_2 \tau' +2D_3 (t_2-\tau-\tau')\\
\nonumber
&=& \beta_1 L_1^2+\beta_2 L_2^2 +2D_3 t_2 ,\\\end{aligned}$$ with $\beta_1=(D_1-D_3)/D_1$ and $\beta_2=(D_2-D_3)/D_2$. It is clear from Eq.(\[30\]), Eq.(\[31\]) and Eq.(\[33\]) that there four relevant time scales in the problem. The first one is obviously $\tau_1=\beta \frac{L_1^2}{2D_1}$. The second one is $\tau_2=\beta \frac{L_1^2}{2D_2}$. The temporal regime $\tau_1<t<\tau_2$ represents the crossover regime from region I to region II, when the walker feels the effect of the inhomogeneity. Similarly, the third time scale is $\tau_3=\frac{1}{2D_2}(\beta_1 L_1^2+\beta_2 L_2^2)$ and the fourth time scale is $\tau_4=\frac{1}{2D_3}(\beta_1 L_1^2+\beta_2 L_2^2)$. $\tau_3<t<\tau_4$ is the crossover regime from region II to region III and it is during this time when the walker spends most of its time near the boundary of region II and region III. Thus the proper time scales for which Eq.(\[30\]), Eq.(\[31\]) and Eq.(\[33\]) are valid are respectively $0<t<\tau_1$, $\tau_2<t<\tau_3$ and $t>\tau_4$ while the time intervals $\tau_1<t<\tau_2$ and $\tau_3<t<\tau_4$ represents the two crossover regimes.

The mean square displacement $\langle x^2(t) \rangle$ is then $$\begin{aligned}
\label{34}
\nonumber
\langle x^2(t) \rangle&=&2D_1t \quad\textrm{for $0<t<\tau_1$}\\
\nonumber
&=& \beta L_1^2 + 2D_2 t_2 \quad \textrm{for $\tau_2<t<\tau_3$}\\
\nonumber
&=& \beta_1 L_1^2+\beta_2 L_2^2+2D_3 t \\
\quad \textrm{for $t>\tau_4$}.\end{aligned}$$

Note that for $D_2=D_3$ we recover the first case, that is a composite media with two homogeneous components while for $D_1=D_2=D_3=D$ we recover the case for a homogeneous system.

To obtain the survival probability from Eq.(\[30\]), Eq.(\[31\]) and Eq.(\[33\]) we follow the usual procedure of defining suitable transformations in space and time as in the earlier section. Thus, we define a normalized variable $\bar{X}(t)$ as $\bar{X}(t)=\frac{x(t)}{\sqrt{\langle x^2(t) \rangle}}$ and the correlator in the normalized variable becomes $$\begin{aligned}
\label{35}
\nonumber
\langle \bar{X}(t_1)\bar{X}(t_2)\rangle &=&\sqrt{\frac{t_2}{t_1}} \quad \textrm{for $0<t<\tau_1$}\\
\nonumber
&=& \sqrt{\frac{\beta L_1^2+2D_2 t_2}{\beta L_1^2+2D_2 t_1}}\quad \textrm{for $\tau_2<t<\tau_3$}\\
\nonumber
&=& \sqrt{\frac{\beta_1 L_1^2+\beta_2 L_2^2+2D_3 t_2}{\beta_1 L_1^2+\beta_2 L_2^2+2D_3 t_1}} \quad \textrm{for $t>\tau_4$} \\\end{aligned}$$

The time transformation $t\rightarrow T$ for the three different regimes are defined in the following way $$\begin{aligned}
\label{36}
\nonumber
e^{T} &=& t \quad \textrm{for $0<t<\tau_1$} \\
\nonumber
&=& \beta L_1^2 + 2D_2 t \quad \textrm{for $\tau_2<t<\tau_3$}\\
\nonumber
&=& \beta_1 L_1^2+\beta_2 L_2^2 + 2D_3 t \quad \textrm{for $t>\tau_4$}\\\end{aligned}$$

and the correlator $\langle \bar{X}(T_1) \bar{X}(T_2) \rangle$ becomes $$\begin{aligned}
\label{37}
\langle \bar{X}(T_1) \bar{X}(T_2) = e^{-\frac{1}{2}(T_1-T_2)}\end{aligned}$$ for all the three temporal regimes, the difference being that the time transformations are different in the three regimes. The process is now a Gaussian stationary process.

Since the correlator is exponentially decaying, the survival probability in the transformed time variable is simply $$\label{38}
 P(T)=e^{-T/2}.$$ In real time, using Eq.(\[38\]), the survival probability becomes $$\begin{aligned}
\label{39}
\nonumber
 p(t)&\sim&t^{-1/2} \quad \textrm{for $0<t<\tau_1$}\\
 \nonumber
&\sim& \frac{1}{\sqrt{\beta L_1^2+2D_2t}} \quad \textrm{for $\tau_2<t<\tau_3$}\\
&\sim& \frac{1}{\sqrt{\beta_1 L_1^2+\beta_2 L_2^2+2D_3t}} \\
\nonumber
\quad \textrm{for $t>\tau_4$}.\end{aligned}$$

![Plot of survival probability $p(t)$ vs time in log-log scale for $L_1=30$, $L_2=210$, $2D_1=100$, $2D_2=10$ and $2D_3=5$. The crossover time scales and the crossover regime is indicated in the figure. The circular points are actual data from numerical simulation and the solid lines are fit of Eq.(\[41\]).](fig8.EPS){height="7cm" width="8cm"}

A plot of the survival probability for two different set of values of $L_1$, $L_2$, $D_1$, $D_2$ and $D_3$ is shown in Fig. 5 and Fig. 6. Configuration averaging of $10^6$ has been done to obtain the numerical results of Fig. 5 and Fig. 6. Theoretical and numerical values of the time scales are presented in Table D.

**Table D**

[|c|c|]{} Parameter Values & Time Scales\
$L_1=30$ & $\tau_1^{th}=8.1$, $\tau_1^{nu}=8.68$\
$L_2=210$& $\tau_2^{th}=81$, $\tau_2^{nu}=80.4393$\
$2D_1=100$& $\tau_3^{th}=1705.5$, $\tau_3^{nu}=1702.58$\
$2D_2=10$, $2D_3=5$ & $\tau_4^{th}=3411$, $\tau_4^{nu}=3421.27$\
$L_1=60$ & $\tau_1^{th}=16.2$, $\tau_1^{nu}=16.01$\
$L_2=360$& $\tau_2^{th}=162$, $\tau_2^{nu}=162.096$\
$2D_1=200$& $\tau_3^{th}=4228.2$, $\tau_3^{nu}=4245$\
$2D_2=20$, $2D_3=2$ & $\tau_4^{th}=42282$, $\tau_4^{nu}=42329.8$\

To conclude, we have investigated the phenomenon of persistence for the case of a random walker in a composite media with two and three homogeneous components. We have presented a very simplified theory to explain the survival probability of a random walker in such inhomogeneous systems. For the two component system, analytical results show that there are two relevant time scales in the problem and this time interval is the crossover regime for the problem. Similarly, for the three component systems there are four relevant time scales and two crossover regimes. The fact that the crossover regimes are not sharp is due to the multiple hoppings that a random walker undergoes near the boundary.

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