The first term of the right-hand side of equation (3.18) reads: —div(J(r, v, t)) = Dr2′(r, v, t). The diffusion equation is obtained from the Boltzmann equation when neutrons are assumed to be monoenergetic, or, in other words, to belong to a single group. The integration^ over velocities in equation (3.18) can be dropped, giving

Vi^(r) — ^ ^(Mn S(r, t).

i j ‘

(3.24)

Note that we have replaced ST in equation (3.18) by Xa, since diffusion has no effect on the flux in one-group formalism. Using relation (3.75), equation (3.24) can be written as

= Dr2′(r, t) + ‘(r, t) X Z[j)(r)(ki — 1) + S(r, t). (3.25)

j

Equation (3.25) leads to a few interesting remarks. In an infinite and homogeneous medium, with an evenly distributed neutron source, the

equation should not include derivatives of ‘(r, t), since ‘(r, t) should be independent of r. Thus equation (3.25) simplifies to

‘(t)J2 Zij)(ki — 1) + S(t). (3.26)

v @ t —

Consider first the case that, for t > 0, S(t) = 0 and ‘(0) is finite. Then equation (3.26) has the solution

	(3.27)

which shows that if кж > 1 the flux diverges, while it decreases to 0 for кж < 1. It is time independent only if кж = 1. This condition can never be met in reality. Rather, in critical reactors, a time-dependence of the absorption cross-sections is used, so that кж fluctuates about 1.

Equations (3.26) and (3.27) have a simple interpretation if one defines the neutron lifetime as the mean time separating creation and absorption of the neutron,

then, equation (3.26) reads

@|^= ‘(t) (kl~ 1} + S(t) (3.29)

ot T

omitting the source term. This equation simply expresses that every time a neutron disappears, кж neutrons are re-emitted, the average time between two neutron absorption events being the time T. With the same notations equation (3.27) reads

‘(t)='(0) e^”-1)t/T (3.30)

which shows that the characteristic evolution time of the neutron flux is т/кж — 1|. Other quantities like neutron density n(t), fission rates, specific power W(t), obey the same type of evolution.

Consider, now the case where кж < 1, and S(t) = S0 is time indepen­dent, but positive. The solution of equation (3.26) for stationary states reads

S0

(1—к^)^ .

The number of absorption reactions per second is then

(3.32)

which agrees with equation (3.71).