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Astronuc

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What is the context.

If one has a function Ae

To have a finite flux, the function describing the spatial distribution of the flux muxt be finite.

Also, if the flux is described by a function proportional to 1/r, then as r-> 0, it would go to infinity, so 1/r cannot be used to describe a flux at r=0.

The flux, which is the number of neutrons passing through some unit area per unit time, usually neutrons/cm

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The section is about determining boundary condition, and to do this, we have to consider of what is physically possible values of flux. By this logic we could deduct that the value of flux must be real, nonnegative and also finite.

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Some more of the excerpt:

"Since the diffusion equation is a partial differential equation, it is necessary to specify certain boundary conditions that must be satisfied by the solution...For example, since a negative or imaginary flux has no meaning, it follows that [tex]\phi[/tex] must be a real, non-negative function.

That being stated, Astronuc is certainly spot on. But, one should be careful to not extend this in the following way:

It is true that the flux being proportional to [tex]1/r[/tex], itself, if [tex]r =0[/tex] is a point in the solution domain, but the caution I wanted to spread was that do not dismiss all functions that simply have this term in it, one must consider the entire term. For instance, in a spherical, bare, reactor one finds the flux is a solution to the diffusion equation in steady-state as,

[tex]\phi (r) = A\frac{\sin (r/L)}{r} + B\frac{\cos (r/L)}{r}[/tex]

where [tex]L[/tex] is the optical (or diffusion) length, and [tex]A[/tex] and [tex]B[/tex] are constants. A suitable boundary condition is that the flux [tex]\phi < \infty [/tex] as [tex]r\rightarrow 0[/tex], enforcing this limit (although both terms involve [tex]1/r[/tex]), it is noted that the term:

[tex]\lim_{r\rightarrow 0}\frac{\sin (r/L)}{r} \rightarrow \frac{1}{L}[/tex]

while the cosine term becomes infinite (implying we require [tex]B=0[/tex]). (One can show that the sine term has a removable singularity by standard methods (e.g. expansion), or you may enforce the symmetry condition that [tex]\lim_{r\rightarrow 0} 4\pi r^2 J(r) = 0[/tex] to retrieve the same result). The moral I wanted to relate was just that do not think that any [tex]1/r[/tex] term is no good, one must take things as a whole. I realize this was posted awhile back, but perhaps it can help future visitors.

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