# Peeter Joot's (OLD) Blog.

• ## Archives

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 293,800

## PHY452H1S Basic Statistical Mechanics. Problem Set 2: Generating functions and diffusion

Posted by peeterjoot on January 26, 2013

# Disclaimer

## Question: Diffusion

The usual diffusion equation for the probability density in one dimension is given by

\begin{aligned}\frac{\partial {P}}{\partial {t}}(x, t) = D \frac{\partial^2 {{P}}}{\partial {{x}}^2}(x, t)\end{aligned} \hspace{\stretch{1}}(1.0.1)

where $D$ is the diffusion constant. Define the Fourier components of the probability distribution via

\begin{aligned}P(x, t) = \int_{-\infty}^\infty \frac{dk}{2 \pi} \tilde{P}(k, t) \exp\left( i k x \right)\end{aligned} \hspace{\stretch{1}}(1.0.2a)

\begin{aligned}\tilde{P}(k, t) = \int_{-\infty}^\infty dx P(x, t) \exp\left( -i k x \right)\end{aligned} \hspace{\stretch{1}}(1.0.2b)

This is useful since the diffusion equation is linear in the probability and each Fourier component will evolve independently. Using this, solve the diffusion equation to obtain $P(k,t)$ in Fourier space given the initial $\tilde{P}(k,0)$.

### Assuming an initial Gaussian profile

\begin{aligned}P(x, 0) = \frac{1}{{\sqrt{2 \pi \sigma^2}}} \exp\left(-\frac{x^2}{2 \sigma^2}\right),\end{aligned} \hspace{\stretch{1}}(1.3)

obtain the probability density $P(x,t)$ at a later time $t$. (NB: Fourier transform, get the solution, transform back.) Schematically plot the profile at the initial time and a later time.

### A small modulation on top of a uniform value

Let the probability density be proportional to

\begin{aligned}\frac{1}{{L}} + A \sin(k_0 x)\end{aligned} \hspace{\stretch{1}}(1.4)

at an initial time $t = 0$. Assume this is in a box of large size $L$, but ignore boundary effects except to note that it will help to normalize the constant piece, assuming the oscillating piece integrates to zero. Also note that we have
to assume $A < 1/L$ to ensure that the probability density is positive. Obtain $P(x,t)$ at a later time $t$. Roughly how long does the modulation take to decay away? Schematically plot the profile at the initial time and a later time.

Inserting the transform definitions we have

\begin{aligned}0 &= \left( \frac{\partial {}}{\partial {t}} - D \frac{\partial^2 {{}}}{\partial {{x}}^2} \right) P \\ &= \left( \frac{\partial {}}{\partial {t}} - D \frac{\partial^2 {{}}}{\partial {{x}}^2} \right) \int_{-\infty}^\infty \frac{dk}{2 \pi} \tilde{P}(k, t) \exp\left( i k x \right) \\ &=\int_{-\infty}^\infty \frac{dk}{2 \pi} \left(\frac{\partial {}}{\partial {t}} \tilde{P}(k, t) + k^2 D\tilde{P}(k, t) \right)\exp\left( i k x \right),\end{aligned} \hspace{\stretch{1}}(1.0.5)

We conclude that

\begin{aligned}0 = \tilde{P}(k, t) + k^2 D\tilde{P}(k, t),\end{aligned} \hspace{\stretch{1}}(1.0.6)

or

\begin{aligned}\tilde{P}(k, t) = A(k) e^{-k^2 D t}.\end{aligned} \hspace{\stretch{1}}(1.0.7)

If the Fourier transform of the distribution is constant until time $t$, so that $\tilde{P}(k, t < 0) = \tilde{P}(k, 0)$, we can write

\begin{aligned}\boxed{\tilde{P}(k, t) = \tilde{P}(k, 0) e^{-k^2 D t}.}\end{aligned} \hspace{\stretch{1}}(1.0.8)

The time evolution of the distributions transform just requires multiplication by the decreasing exponential factor $e^{-k^2 D t}$.

Propagator for the diffusion equation

We can also use this to express the explicit time evolution of the distribution

\begin{aligned}P(x, t) &= \int_{-\infty}^\infty \frac{dk}{2 \pi} \tilde{P}(k, 0) e^{-k^2 D t}\exp\left( i k x \right) \\ &= \int_{-\infty}^\infty \frac{dk}{2 \pi} \int_{-\infty}^\infty dx' P(x', 0) \exp\left( -i k x' \right)e^{-k^2 D t}\exp\left( i k x \right) \\ &= \int_{-\infty}^\infty dx' P(x', 0) \int_{-\infty}^\infty \frac{dk}{2 \pi} \exp\left( -k^2 D t + i k (x - x') \right)\end{aligned} \hspace{\stretch{1}}(1.0.9)

Our distribution time evolution is given by convolve with a propagator function

\begin{aligned}P(x, t) = \int dx' P(x', 0) G(x', x)\end{aligned} \hspace{\stretch{1}}(1.0.10a)

\begin{aligned}G(x', x) =\int_{-\infty}^\infty \frac{dk}{2 \pi} \exp\left( -k^2 D t + i k (x - x') \right)\end{aligned} \hspace{\stretch{1}}(1.0.10b)

For $t \ge 0$ we can complete the square, finding that this propagator is

\begin{aligned}G(x', x) &= \int_{-\infty}^\infty \frac{dk}{2 \pi} \exp\left( -k^2 D t + i k (x - x') \right) \\ &= \exp\left( \left(\frac{i (x - x')}{2 \sqrt{D t}} \right)^2 \right)\int_{-\infty}^\infty \frac{dk}{2 \pi} \exp\left( - \left(k \sqrt{D t} + \frac{i (x - x')}{2 \sqrt{D t}} \right)^2 \right)\end{aligned} \hspace{\stretch{1}}(1.0.11)

or

\begin{aligned}\boxed{G(x', x) =\frac{1}{\sqrt{4 \pi D t}} \exp\left(-\frac{(x - x')^2}{4 D t}\right).}\end{aligned} \hspace{\stretch{1}}(1.0.12)

A schematic plot of this function as a function of $t$ for fixed $x - x'$ is plotted in (Fig1).

Fig1: Form of the propagator function for the diffusion equation

### Gaussian

For the Gaussian of 1.3 we compute the initial time Fourier transform

\begin{aligned}\tilde{P}(k) &= \int_{-\infty}^\infty dx \frac{1}{{\sqrt{2 \pi \sigma^2}}} \exp\left(-\frac{x^2}{2 \sigma^2}-i k x \right) \\ &= \frac{1}{{\sqrt{2 \pi \sigma^2}}} \exp\left(-\left( \frac{\sqrt{ 2 \sigma^2}}{2} k i\right)^2\right)\int_{-\infty}^\infty dx \exp\left(-\left( \frac{x}{\sqrt{2 \sigma^2} } + \frac{\sqrt{ 2 \sigma^2}}{2} k i\right)^2\right) \\ &= \exp\left(-\frac{\sigma^2 k^2}{2}\right).\end{aligned} \hspace{\stretch{1}}(1.0.13)

The time evolution of the generating function is

\begin{aligned}\tilde{P}(k, t) = \exp\left(-\frac{\sigma^2 k^2}{2} - D k^2 t\right),\end{aligned} \hspace{\stretch{1}}(1.0.14)

and we can find our time evolved probability density by inverse transforming

\begin{aligned}P(x, t) &= \int_{-\infty}^\infty \frac{dk}{2 \pi} \exp\left(-\frac{\sigma^2 k^2}{2} - D k^2 t + i k x\right) \\ &= \exp\left(i \frac{x}{ 2 \sqrt{\frac{\sigma^2}{2} + D t} }\right)^2\int_{-\infty}^\infty \frac{dk}{2 \pi} \exp\left(-\left(k \sqrt{\frac{\sigma^2}{2} + D t} + i \frac{x}{ 2 \sqrt{\frac{\sigma^2}{2} + D t} }\right)^2\right)\end{aligned} \hspace{\stretch{1}}(1.0.15)

For $t \ge 0$ this is

\begin{aligned}\boxed{P(x, t) =\frac{1}{{\sqrt{2 \pi \left( \sigma^2 + 2 D t \right) } }}\exp\left(-\frac{x^2}{2 \left( \sigma^2 + 2 D t \right) }\right).}\end{aligned} \hspace{\stretch{1}}(1.0.16)

As a check, we see that this reproduces the $t = 0$ value as expected. A further check using Mathematica applying the propagator 1.0.12, also finds the same result as this manual calculation.

This is plotted for $D = \sigma = 1$ in (Fig2) for a couple different times $t \ge 0$.

Fig2: Gaussian probability density time evolution with diffusion

### Boxed constant with small oscillation

The normalization of the distribution depends on the interval boundaries. With the box range given by $x \in [a, a + L]$ we have

\begin{aligned}\int_a^{a + L} dx \left( \frac{1}{{L}} + A \sin( k_0 x) \right) dx=1 - \frac{A}{k_0} \left( \cos( k_0(a + L) ) - \cos( k_0 a ) \right)\end{aligned} \hspace{\stretch{1}}(1.0.17)

With an even range for box $x \in [-L/2, L/2]$ this is unity.

To find the distribution at a later point in time we can utilize the propagator

\begin{aligned}P(x, t) = \int_{-L/2}^{L/2} dx' \frac{1}{{2 \sqrt{\pi D t} }} \left( \frac{1}{{L}} + A \sin( k_0 x' ) \right) \exp\left( - \frac{(x' - x)^2}{2 \sqrt{D t} }\right)\end{aligned} \hspace{\stretch{1}}(1.0.18)

Let’s write this as

\begin{aligned}P(x, t) = P_{\mathrm{rect}}(x, t) + P_{\mathrm{sin}}(x, t)\end{aligned} \hspace{\stretch{1}}(1.0.19a)

\begin{aligned}P_{\mathrm{rect}}(x, t) =\frac{1}{{2 L \sqrt{\pi D t} }} \int_{-L/2}^{L/2} dx' \exp\left( - \frac{(x' - x)^2}{2 \sqrt{D t} }\right)\end{aligned} \hspace{\stretch{1}}(1.0.19b)

\begin{aligned}P_{\mathrm{sin}}(x, t)=\frac{A}{2 \sqrt{\pi D t} } \int_{-L/2}^{L/2} dx' \sin( k_0 x' ) \exp\left( - \frac{(x' - x)^2}{2 \sqrt{D t} }\right)\end{aligned} \hspace{\stretch{1}}(1.0.19c)

Applying a $u = (x' - x)/\sqrt{4 D t}$ change of variables for the first term, we can reduce it to a difference of error functions

\begin{aligned}P_{\mathrm{rect}}(x, t) &= \frac{1}{{L}} \int_{-L/2}^{L/2} dx' \frac{1}{{2 \sqrt{\pi D t} }} \exp\left( - \frac{(x' - x)^2}{2 \sqrt{D t} }\right) \\ &= \frac{1}{{L \sqrt{\pi}}}\int_{-\frac{L/2 +x}{2 \sqrt{D t}}}^{\frac{L/2 - x}{2 \sqrt{Dt}}} du e^{-u^2} \\ &= \frac{1}{{2 L}} \left( \text{erf}\left( \frac{L/2 -x}{2 \sqrt{D t}} \right)-\text{erf}\left( -\frac{L/2 +x}{2 \sqrt{D t}} \right)\right)\end{aligned} \hspace{\stretch{1}}(1.0.20)

Following Mathematica, lets introduce a two argument error function for the difference between two points

\begin{aligned}\text{erf}(z_0, z_1) \equiv \text{erf}(z_1) - \text{erf}(z_0).\end{aligned} \hspace{\stretch{1}}(1.0.21)

Using that our rectangular function’s time evolution can be written

\begin{aligned}P_{\mathrm{rect}}(x, t)=\frac{1}{{2 L}} \text{erf}\left( -\frac{L/2 +x}{2 \sqrt{D t}} ,\frac{L/2 -x}{2 \sqrt{D t}} \right)\end{aligned} \hspace{\stretch{1}}(1.0.22)

For $L = D = 1$, and $t = 10^{-8}$, this is plotted in (Fig3). Somewhat surprisingly, this difference of error functions does appear to result in a rectangular function for small $t$.

Fig3: Rectangular part of the probability distribution for very small t

The time evolution of this non-oscillation part of the probability distribution is plotted as a function of both $t$ and $x$ in (Fig4).

Fig4: Time evolution of the rectangular part of the probability distribution

For the sine piece we can also find a solution in terms of (complex) error functions

\begin{aligned}P_{\mathrm{sin}}(x, t) &= A \int_{-L/2}^{L/2} dx' \frac{1}{{2 \sqrt{\pi D t} }} \sin( k_0 x' ) \exp\left( - \frac{(x' - x)^2}{2 \sqrt{D t} }\right) \\ &= \frac{A}{\sqrt{\pi}}\int_{-\frac{L/2 +x}{2 \sqrt{D t}}}^{\frac{L/2 - x}{2 \sqrt{Dt}}} du \sin( k_0 ( x + 2 u \sqrt{D t} ) ) e^{-u^2} \\ &= \frac{A}{2 i \sqrt{\pi}}\int_{-\frac{L/2 +x}{2 \sqrt{D t}}}^{\frac{L/2 - x}{2 \sqrt{Dt}}} du \left( e^{ i k_0 ( x + 2 u \sqrt{D t} ) } -e^{ -i k_0 ( x + 2 u \sqrt{D t} ) }\right)e^{-u^2} \\ &= \frac{A}{2 i \sqrt{\pi}}\left( e^{ i k_0 x }\int_{-\frac{L/2 +x}{2 \sqrt{D t}}}^{\frac{L/2 - x}{2 \sqrt{Dt}}} du e^{ -u^2 + 2 i k_0 u \sqrt{D t} } -e^{ -i k_0 x }\int_{-\frac{L/2 +x}{2 \sqrt{D t}}}^{\frac{L/2 - x}{2 \sqrt{Dt}}} du e^{ -u^2 -2 i k_0 u \sqrt{D t} }\right) \\ &= \frac{A}{2 i \sqrt{\pi}}e^{ -k_0^2 D t } \left( e^{ i k_0 x }\int_{-\frac{L/2 +x}{2 \sqrt{D t}}}^{\frac{L/2 - x}{2 \sqrt{Dt}}} du e^{ -(u - i k_0 \sqrt{D t})^2 } -e^{ -i k_0 x }\int_{-\frac{L/2 +x}{2 \sqrt{D t}}}^{\frac{L/2 - x}{2 \sqrt{Dt}}} du e^{ -(u + i k_0 \sqrt{D t})^2 }\right) \\ &= \frac{A}{2 i \sqrt{\pi}}e^{ -k_0^2 D t } \left( e^{ i k_0 x }\int_{-\frac{L/2 +x}{2 \sqrt{D t}} - i k_0 \sqrt{D t}}^{\frac{L/2 - x}{2 \sqrt{Dt}} - i k_0 \sqrt{D t}} dv e^{ -v^2 }-e^{ -i k_0 x }\int_{-\frac{L/2 +x}{2 \sqrt{D t}} + i k_0 \sqrt{D t}}^{\frac{L/2 - x}{2 \sqrt{Dt}} + i k_0 \sqrt{D t}} dv e^{ -v^2 }\right) \\ &= \frac{A}{4 i }e^{ -k_0^2 D t } \left( e^{ i k_0 x }\text{erf}\left( -\frac{L/2 +x}{2 \sqrt{D t}} - i k_0 \sqrt{D t}, \frac{L/2 - x}{2 \sqrt{Dt}} - i k_0 \sqrt{D t} \right)-e^{ -i k_0 x }\text{erf}\left( -\frac{L/2 +x}{2 \sqrt{D t}} + i k_0 \sqrt{D t}, \frac{L/2 - x}{2 \sqrt{Dt}} + i k_0 \sqrt{D t}\right)\right)\end{aligned} \hspace{\stretch{1}}(1.0.23)

This is plotted for $A = D = L = 1$, $k_0 = 8 \pi$, and $t = 10^{-8}$ in (Fig5).

Fig5: Verification at t -> 0 that the diffusion result is sine like

The diffusion of this, again for $A = D = L = 1$, $k_0 = 8 \pi$, and $t \in [10^{-5}, 0.01]$ is plotted in (Fig6). Again we see that we have the expected sine for small $t$.

Fig6: Diffusion of the oscillatory term

Putting both the rectangular and the windowed sine portions of the probability distribution together, we have the diffusion result for the entire distribution

\begin{aligned}\boxed{\begin{aligned}P(x, t)&=\frac{1}{{2 L}} \text{erf}\left( -\frac{L/2 +x}{2 \sqrt{D t}} ,\frac{L/2 -x}{2 \sqrt{D t}} \right) \\ &+\frac{A}{4 i }e^{ -k_0^2 D t + i k_0 x }\text{erf}\left( -\frac{L/2 +x}{2 \sqrt{D t}} - i k_0 \sqrt{D t}, \frac{L/2 - x}{2 \sqrt{Dt}} - i k_0 \sqrt{D t} \right) \\ &-\frac{A}{4 i }e^{ -k_0^2 D t -i k_0 x }\text{erf}\left( -\frac{L/2 +x}{2 \sqrt{D t}} + i k_0 \sqrt{D t}, \frac{L/2 - x}{2 \sqrt{Dt}} + i k_0 \sqrt{D t}\right)\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.0.24)

It is certainly ugly looking! We see that the oscillation die off is dependent on the $\exp( -k_0^2 D t)$ term. In time

\begin{aligned}t = \frac{1}{{k_0^2 D}},\end{aligned} \hspace{\stretch{1}}(1.0.25)

that oscillation dies away to $1/e$ of its initial amplitude. This dispersion is plotted at times $t = 10^{-5}$ and $t = 1/(k_0^2 D)$ for $L = D = 1$, $k_0 = 8 \pi$ and $A = 1/2$ in (Fig7).

Fig7: Initial time distribution and dispersion of the oscillatory portion to 1/e of initial amplitude

Similar to the individual plots of $P_{\mathrm{rect}}(x, t)$ and $P_{\mathrm{sin}}(x, t)$ above, we plot the time evolution of the total probability dispersion $P(x, t)$ in (Fig8). We see in the plots above that the rectangular portion of this distribution will also continue to flatten over time after most of the oscillation has also died off.

Fig8: Diffusion of uniform but oscillating probability distribution

An easier solution for the sinusoidal part

After working this problem, talking with classmates about how they solved it (because I was sure I’d done this windowed oscillating distribution the hard way), I now understand what was meant by “ignore boundary effects”. That is, ignore the boundary effects in the sinusoid portion of the distribution. I didn’t see how we could ignore the boundary effects because doing so would make the sine Fourier transform non-convergent. Ignoring pesky ideas like convergence we can “approximate” the Fourier transform of the windowed sine as

\begin{aligned}\tilde{P}_{\mathrm{sin}}(k) &\approx A \int_{-\infty}^\infty \sin (k_0 x) e^{-i k x} dx \\ &= \frac{A }{2 i} \int_{-\infty}^\infty \left(e^{i (k_0 - k) x} -e^{-i (k_0 + k) x} \right)dx \\ &= \frac{A \pi}{i} \left( \delta(k - k_0) - \delta(k + k_0)\right)\end{aligned} \hspace{\stretch{1}}(1.0.26)

Now we can inverse Fourier transform the diffusion result with ease since we’ve got delta functions. That is

\begin{aligned}P_{\mathrm{sin}}(x, t) &\approx \frac{1}{{2 \pi}} \frac{A \pi}{i} \int\left( \delta(k - k_0) - \delta(k + k_0)\right)e^{-D k^2 t} e^{i k x} dk \\ &= e^{-D k_0^2 t} \frac{ e^{i k_0 x} - e^{-i k_0 x}}{2 i} \\ &= e^{-D k_0^2 t} \sin( k_0 x )\end{aligned} \hspace{\stretch{1}}(1.0.27)

## Question: Generating function

The Fourier transform of the probability distribution defined above $\tilde{P}(k)$ is called the “generating function” of the distribution. Show that the $n$-th derivative of this generating function $\partial^n P(k)/\partial k^n$ at the origin $k = 0$ is related to the $n$-th moment of the distribution function defined via $\left\langle{{x^n}}\right\rangle = \int dx P(x) x^n$. We will later see that the “partition function” in statistical mechanics is closely related to this concept of a generating function, and derivatives of this partition function can be related to thermodynamic averages of various observables.

\begin{aligned}{\left.{{ \frac{\partial^n}{\partial k^n} \tilde{P}(k) }}\right\vert}_{{k = 0}} &= {\left.{{ \frac{\partial^n}{\partial k^n} \left( \int_{-\infty}^\infty dx P(x) \exp\left( -i k x \right)\right)}}\right\vert}_{{k = 0}} \\ &= {\left.{{ \left(\int_{-\infty}^\infty dx P(x) (-i x)^n\exp\left( -i k x \right)\right)}}\right\vert}_{{k = 0}} \\ &= (-i)^n \int_{-\infty}^\infty dx P(x) x^n \\ &= (-i)^n \left\langle{{x^n}}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.0.26)