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2. Directions: (i) Work independently and not discuss the exam with anyone. (ii) You may quote formulae from 2005 18.366
lecture notes and problem sets, but otherwise all steps in your solutions should be derived in detail. (iii) You may consult
the required and recommended books, but no other resources (books, web sites, etc.). Total exam time: 46.5 hours.
1. Discrete versus continuous steps in a random walk.
(a) Consider the probability distribution
for displacements Δx in a random walk on the integers on the integers, and find its generating function
Choose C(a) to set P(1) = 1. Let PN,n be the probability distribution for the position after N IID steps. Use P(z) to
compute the diffusivity D2(a) which arises in the FokkerPlanck equation,
for the continuum approximation
(b) Now consider continuous displacements with the PDF,
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Problems

3. and find its Fourier transform,
Let PN (x) be the PDF for the position after N IID steps. Use pˆ(k) to obtain all the coefficients D2m(b) arising in
the (modified) Kramers-Moyall expansion,
for the continuum approximation
(c) Make a log-linear plot of D2(a) and D2(b(a)) versus a, and explain in what limit they are the same.
2. First passage of N random walks in two dimensions. Consider continuous diffusion processes
approximating random walks in the plane starting at (x, y) = (0, a).
(a) Let G(x,t|x0 = 0) be the Green function for the x component (marginal probability density) and f(t|a) be the first
passage time density for each walker. Compute
and explain why this is the PDF of the location of first passage X (which was derived in lecture by conformal
mapping). Note that
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4. (b) Now suppose N independent random walkers are released from (0, a) at the same time, and compute1 the PDF
εN (x|a) of the location XN where, for the first time, one of these walkers hits the line. What is the critical number of
walkers Nc such that
3. First passage to a circle. Using conformal mapping2 or the method of images, derive the PDF of the angle of
first passage ε(θ) to the unit circle from the point (a, 0) by continuous diffusion in the plane
What is the ratio of hitting probabilities for the nearest and farthest points,
Extra credit: Give a simple geometrical intepretation of ε(θ).
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5. 1 Discrete versus continuous steps in a random walk
Solutions
1.1 Finding a generating function and D2(a)
For pn = Ca|n| , the probability generating function is
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6. we finally obtain the diffusivity,
since the time step in the continuum approximation is τ = 1.
1.2 Continuum approximation
Now we consider the continuum approximation,
which has the same exponential decay as pn for The Fourier transform should look familiar
(from problem set 2), but it’s easy enough to work out again:
The cumulant generating function is
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7. 1.3 Log-linear plot
The two diffusivities are
In the limit a → 1 the width of the distribution b = −1/ log a becomes much larger than the lattice spacing, and thus
the continuum approximation should become exact, D2 ∼ D¯2, which is easily verified. In the opposite limit, a → 0,
the decay length for the distribution is much less than the lattice spacing, and the two models should be very
different. In fact,
These limits are also clear in figure 1.
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8. 2 First passage of N random walks in two dimensions
2.1 First passage position of a single walker
Since the diffusivity is a scalar (isotropic process), the Green function for the x-component (marginal probability
density, after integrating out the y-component) will describe a one-dimensional x diffusion process with the same
D,
By symmetry, the x process plays no role in determining the first passage time, whose (Smirnov1) probability
density will be same as for a one-dimensional y diffusion process with the same D,
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9. since this is an integral over all times t of the probability that the x-component is x given that first passage occurs
at time t. As noted above, these events are independent, so the integrand is just a
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10. Figure 2: Plots of εN (x|a) for several different values of N, for the case when a = D = 1. The case of N = 1 is the
Cauchy distribution, while the cases N > 1 are from numerical integration of equation 1. The substitution 1/t = − log α
was employed to change the integral into one over a finite range 0 < α < 1, and the resulting expression was
evaluated using Simpson’s rule.
product of the two probabilities. The integral is easily evaluated:
which is the same Cauchy distribution we obtained in lecture by conformal mapping. Note that
It does not seem that this integral can be performed analytically, so some numerical integrations are
shown in figure 2.
The variance of the hitting position is given by
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11. For t → ∞ we have f(t|a) ∝ t −3/2, S(t) ∝ t −1/2, fN (t|a) ∝ t −1−N/2, and thus the integrand decays like tfN (t|a) ∝ t
N/2. Therefore, the variance is finite only if N/2 > 1 ⇒ N > 2 ⇒ N ≥ 3. So, once again Nc = 3 is the magic number
of walkers such that the first one will hit in a region of finite variance in space. This should come as no surprise,
because this is the same critical number needed to have a finite mean first passage time for the first walker, as
shown in lecture.
3 First passage to a circle
In the physical z plane, the walker is released at (x = a, y = 0) and hits the unit circle. We would like to map
this domain with w = f(z) to the interior of the unit circle with the source at the origin in the mathematical w
plane, where we know the complex potential is
We could choose a M¨obius transformation with the constraints, f(a) = 0, f(1) = −1, f(−1) = 1, which yields
The complex potential in the z plane is therefore
which is clearly the sum of the source term and an image sink at (a−1, 0) (and a constant). The hitting probability
density is given by the normal electric field on the circle:
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12. which is nine for source at twice the radius (a = 2).
The geometrical interpretation follows from the cumulative distribution function
where γ is the angle formed at a point on the circle by drawing lines to the “charge” at z = a and its “image” at z =
a−1. The probability of hitting between angle θ1 and θ2 on the circles is just the difference of two such angles,
subtended from each of the points to z = a and z = a−1:
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