The iterate of the Laplace transform: Show that, provided the integrals concerned exist,

7 Laplace transforms

6. The iterate of the Laplace transform: Show that, provided the integrals concerned exist,

L²[f(t)] = _∞

0

ds e^−us _∞

0

dt e^−stf(t) = _∞

0

dt f(t) (t+u). The last integral is the so-called Stieltjes transformof f(t).

LCR series circuit: As another standard example from elementary physics, consider an LCR series circuit under a sinusoidal applied voltage of amplitudeV₀ and (angular) frequency ω. Recall that we have already determined the complex admittance of the system, when discussing linear response and dispersions relations.

The differential equation satisfied by the charge q(t) on the capacitor is Lq¨+Rq˙+ (1/C)q=V₀ sin ωt.

As you know, this equation is precisely the equation of motion of a sinusoidally forced damped simple harmonic oscillator. The damping constant is R/L =γ, which is just the reciprocal of the time constant of an LR circuit. The natural frequency of the circuit in the absence of the resistor 1/√

LC =ω₀. The conditionω₀ > ¹₂γ corresponds to the underdamped case. Let’s consider this case first , for definiteness.²¹ It is convenient, in this case, to work in terms of the shifted frequency

ω_u =

ω²₀− ¹₄γ²1/2

.

7. Suppose the initial conditions are such that both the initial charge and the initial current are equal to zero, i.e.,

q(0) = 0 and q(0) = 0.˙

21The critically damped and overdamped cases will be considered subsequently.

The solution for q(t) can be written as the sum of a transient part and a steady state part,

q(t) =q^tr(t) +q^st(t).

Show that these are given, respectively, by q^tr(t) =

V₀ω L ω_u

(ω²−ω²₀+ ¹₂γ²) sin (ω_ut) +ω_uγ cos (ω_ut) (ω²−ω₀²)²+ω²γ²

e^−γt/2 and

q^st(t) =− V₀

L

(ω²−ω²₀) sin (ωt) +ω γ cos (ωt) (ω²−ω²₀)²+ω²γ²

.

Hint: Take the Laplace transform of both sides of the differential equation. With the initial conditions q(0) = 0 and ˙q(0) = 0, we get

q(s) = (V₀ω/L)

(s²+ω²)(s²+γs+ω₀²).

Resolve the right-hand side into partial fractions, and invert the Laplace transform.

All you need to use is the fact that the inverse Laplace transform of (s+a)⁻¹ ise^−at. Observe that the transient component of the solution, q^tr(t), is characterized by the frequency ω_u that depends on the circuit parameters L, C and R. This part decays to zero exponentially in time, owing to the dissipation in the system. The steady state component q^st(t), on the other hand, oscillates with the same frequency ω as the ap- plied voltage. These statements apply equally to the current in the circuit, given by I(t) = ˙q(t).

8. The remarks just made are, in fact, applicable to all initial conditions. In order to check this out, consider general values q(0) and ˙q(0) =I(0), respectively, of the initial charge on the capacitor and the initial current in the circuit. Show that the complete solution forq(t) is now given by the one already found above, plus an extra term added to q^tr(t), namely,

I(0) + ¹₂γ q(0) ω_u

sin (ω_ut) +q(0) cos (ω_ut)

e^−γt/2.

Complementary function and particular integral: Note, incidentally, that the last expression above is precisely the solution to the homogeneousdifferential equation Lq¨+Rq˙+ (1/C)q = 0 that is satisfied by the charge on the capacitor in theabsenceof any applied voltage. With reference to the inhomogeneous differential equation forq(t), the full solution given above represents the particular integral, while the solution of the homogeneous equation represents the complementary function.

• In the solution of an inhomogeneous differential equation, the purpose of adding

‘the right amount’ of the complementary function to the particular integral is to ensure that the boundary conditions (in this case, the initial conditions) are satisfied by the solution.

The example just considered ought to help you see this quite clearly.

9. By now, it should be obvious to you that the solutions in the critically damped (ω0 = ¹₂γ) and overdamped (ω0 < ¹₂γ) cases may be written down by simple analytic continuation of the solution in the underdamped case.²² Do so.

Laplace transforms and random processes: The master equations for certain Markov processes lead to simple differential equations for the generating functions of these processes. Such equations are solved very easily using Laplace transforms. Here are a few examples.

10. The Poisson process: Radioactive decay of an unstable isotope provides a phys- ical example of a Poisson process. If P_n(t) is the probability that exactly n events of the process take place in a time interval t, and λ >0 is the mean rate at which events take place, then

dP₀(t)

dt =−λ P₀(t) and dP_n(t) dt =λ%

P_n−1(t)−P_n(t)&

, n≥1.

This coupled set of ordinary differential equations is to be solved with the initial con- ditions P₀(0) = 1 and P_n(0) = 0 forn ≥1. It follows at once that P₀(t) = e^−λt. The generating function f(z, t) =_∞

n=1P_n(t)zⁿ satisfies the differential equation²³

∂f

∂t +λ(1−z)f =λz P₀ =λz e^−λt,

with the initial condition f(z,0) = 0. Use Laplace transforms to obtain the solution f(z, t) =e^−λt

e^λzt −1 .

Picking out the coefficient of zⁿ in the power series expansion of the exponential we get the Poisson distribution

P_n(t) = e^−λt(λt)ⁿ n! .

The mean number of events in a time interval t is thereforeλt. Every cumulant of the distribution is also equal to λt.

22Typically, trigonometric functions will become hyperbolic functions.

23Note that the generating function has been defined as a sum fromn= 1 rather thann= 0.

The Poisson process is an example of what is known as abirth process, because the random variable n(t) never decreases as tincreases. The next example shows what happens when the value of the random variable can both increase as well as decrease as time elapses (a birth-and-death process).

11. A biased random walk on a linear lattice: Consider a linear lattice, with its sites labelled by an integer j ∈Z. A walker on this lattice jumps from any site to one of the two neighboring sites with a mean jump rate λ. The probability of a jump from the site j to the site (j+ 1) is p, and that of a jump to the site (j−1) is q = (1−p), where 0 < p < 1. Successive jumps are statistically independent of each other. Let P_j(t) denote the probability that the walker is at sitej at timet. (The random variable in this problem is j). The master equation satisfied by the set of probabilities {P_j(t)}

is dP_j(t)

dt =λ%

p P_j−1(t) +q P_j+1(t)

‘gain’ terms

− P_j(t)

‘loss’ term

&

, j ∈Z.

Without loss of generality,²⁴ we may take the site 0 to be the starting point of the random walk (RW for short). The initial condition is then

P_j(0) =δ_j,0. Define the generating function

f(z, t) = ∞ j=−∞

P_j(t)z^j.

Note that the summation is over all integers j. Hence the equation above is to be regarded as a Laurent series expansion of f(z, t).

(a) From the differential equation satisfied by P_j(t) and its initial condition, show that f(z, t) satisfies the equation

∂f

∂t =λ(pz−1 +qz⁻¹)f , with the initial condition f(z,0) = 1 .

(b) Solve this equation using Laplace transforms, to get f(z, t) =e^−λte^{λt(pz+qz−1)}.

24Because the lattice is ofinfiniteextent in both directions! Finite boundaries would obviously spoil this translation invariance.

(c) The probability we seek, P_j(t), is the coefficient of z^j in the Laurent expansion of f(z, t) in powers of z. It is helpful to write

pz+qz⁻¹ =√ pq

z

p/q+z⁻¹ q/p

.

Now expand the second exponential factor in the solution for f(z, t), and pick out the coefficient of z^j. Consider j ≥0 first. Show that

P_j(t) =e^−λt(p/q)^j/2 ∞ n=0

1

n! (n+j)! (λt√

pq)^2n+j, j ≥0.

The modified Bessel function of the first kind and of order ν may be defined by means of its power series, namely,

I_ν(z) = ∞

n=0

1

Γ(ν+n+ 1)n!

₁

2z2n+ν

.

Compare this series with that for the ordinary Bessel function J_ν(z) given earlier.

Apart from the factor (−1)ⁿ in the summand in the case of J_ν(z), the series are the same.²⁵ Like the Bessel function J_ν(z), the modified Bessel function I_ν(z) is also an entire function ofz. The generating function for the modified Bessel function of integer order is²⁶

e^t(z+z−1)/2 = ∞ j=−∞

I_j(t)z^j.

Owing to the symmetry of the generating function under the exchange z ↔ z⁻¹, it is obvious that I_−j(z) =I_j(z) for every integer value of j.

(d) Compare the series expansion obtained above for P_j(t) with the power series for the modified Bessel function, to obtain

P_j(t) =e^−λt(p/q)^j/2I_j(2λt√ pq).

Verify that exactly the same expression is valid for negative integer values of j as well, using the fact that I_−j =I_j for any integer j.

25The two functions are related to each other according to J_ν(z) = e^iπν/2I_ν(−iz), or I_ν(z) = e^−iπν/2J_ν(iz).

26For completeness, I mention that the generating function of the Bessel functionJ_k is given by

e^{t(z−z−1)/2}=

∞

j=−∞

J_j(t)z^j.

An RW in whichp=qis called abiased random walk. Whenp > q[respectively, q > p], there is a bias to the right [left], and the factor (p/q)^j/2 in the expression for P_j(t) shows that the probability of positive [negative] values of j is enhanced, at any t >0.

In the case of anunbiased random walk,p=q= ¹₂. The probability distribution then simplifies to

P_j(t) =e^−λtI_j(λt) (unbiased RW in 1 dimension).

Since I_j = I_−j, we have in this case P_j(t) = P_−j(t) at any time t, as expected on physical grounds.

The asymptotic or long-time behavior of the RW is of importance. In the problem at hand, the characteristic time scale is set by λ⁻¹. Now, for λt 1, the leading asymptotic behavior of the modified Bessel function is given by

I_j(λt)∼ e^λt

√2πλt,

independent of j. Therefore, in an unbiased RW, P_j(t) has a leading asymptotic time- dependence∼t^−1/2, which is characteristic of purelydiffusive behavior.²⁷ In marked contrast, the leading asymptotic behavior of P_j(t) for a biased random walk is given by P_j(t)∼t^−1/2e^{−λt(1−2√pq)}. Since 1>2√

pq whenp=q, this shows thatP_j(t) decays exponentially with time for all j.

12. Unbiased random walk in d dimensions: Consider an unbiased simple RW