I was reading ISBN: 9782868836373 which is dealing with a differential equation at page 139 with a source term in 3D but with three different types of source of neutrons (source term):
- A point
- A wire
- A plane
The differential equations are dealt respectively in spherical, cylindrical and cartesian coordinates and take the following forms, respectively:
- $D \left( \frac{d^2 \phi}{d r^2} + \frac{2}{r} \frac{d \phi}{d r} \right) - \Sigma_a \phi + S = 0$
- $D \left( \frac{d^2 \phi}{d \rho^2} + \frac{1}{\rho} \frac{d \phi}{d \rho} \right) - \Sigma_a \phi + S = 0$
- $D \left( \frac{d^2 \phi}{d x^2} \right) - \Sigma_a \phi + S = 0$
One obviously recognize the Laplacian operator, but the thing is that the source term is not the same in all those equations. The idea is to first solve the equation without the source term, which leads to, with the condition that the function cannot blow up at infinity leads to the following family of solutions:
- $\phi (r) = A \frac{\exp{(- \kappa r)}}{r}$
- $\phi (\rho) = A K_0 (\rho \kappa)$, with $K_0$ a Bessel function
- $\phi (x) = A \exp{(- \kappa |x|)}$
All these solutions make sense, as it vanishes ad infinitum. The author proposes two ways to get the constants $A$. The first method is pretty straight forward using Fick's law and a volume whose limit is the source (a rod of radius $\epsilon \rightarrow 0$ for the first case). They lead to the following results:
- $A = S/4\pi D$
- $A = S/2\pi D$
- $A = S/2\kappa D$
Now my problem is with the second method proposed. The French author Paul Reuss says (for the first case) (my translation):
The source can be mathematically represented by $S \delta (\vec{r})$ where $\delta$ is the 3D Dirac distribution; near the origin, the flow is equivalent to A/r; since the exponential is nearly equal to 1: but, we have the formula: $\delta(\vec{r}) = - \frac{1}{4 \pi} \Delta \frac{1}{r}$. By balancing the coefficients of the two Dirac distributions showing up in the equation, we see that A must be equal to $S/4\pi D$.
For the second case:
The Source can be mathematically represented by $S \delta(\vec{\rho})$ where $\delta$ is the 2D Dirac distribution; knowing that, near the origin, the function $K_0 (u)$ is equivalent to $-ln(u)$ and that we have the formula $\delta (\vec{\rho}) = \frac{1}{2\pi} \Delta ln(\rho)$, we find back the value of A by balancing the coefficient of the two Dirac distributions showing up in the equation.
For the third case:
The source can be mathematically represented by $S\delta(x)$ where $\delta$ is the 1D Dirac distribution; using the formula: $\delta(x) = \frac{1}{2} \Delta |x|$, we find back the value of A by balancing the coefficient of the two Dirac distributions showing up in the equation.
At first, the Dirac distribution seems coherent with the physics definition of the source.
Now I have two problems with the second method.
- At first, it seems there are formulas here with the Dirac distribution that I have never used; is it possible to know a bit more about where they come from?
- Secondly I don't understand the steps, once the formulas are admitted as valid, to conclude to the value of $A$. Can I get some help about it?
Edit: In Annex, there is a part about the Dirac distribution function and the formulas of the Dirac function are said to be "definition with the Laplacian operator" without more information.
Disclaimer: I'm an engineer. The derivation below may be not the most rigorous, or the most straightforward, or the most generic. It works, though.
On the Dirac delta in more than one dimensions
The Dirac delta function has different "forms" according to the number of dimensions one is dealing with. In $n$ dimensions, the Dirac delta must satisfy $$ \int_{R^n} \delta^n(\mathbf{x}-\mathbf{x}_0)\, dV = 1. $$ Evaluating the integral in $R^3$ in spherical coordinates, one has $$ \int_0^\infty 4\pi r^2 \delta^3(\mathbf{x}-\mathbf{x}_0)\, dr = 1, $$ and, therefore, we can identify $\delta^3(\mathbf{x}-\mathbf{x}_0) = \delta(r)/4\pi r^2$ (in which $\delta$ is the one-dimensional, regular Dirac delta).
Now, consider the Poisson equation in three dimensions $$ \Delta f=\delta^3(\mathbf{x}). $$ The RHS is zero for all $\|\mathbf{x}\|\neq 0$, yielding $\Delta f=0$ for $\|\mathbf{x}\|>0$. Solving the Laplace equation, one has $$ f = \frac{A}{r}, \ \ \ r > 0. $$ Using the divergence theorem in a sphere of radius $R$, we have $$ \oint \nabla f \cdot d\mathbf{S} = \iiint \Delta f\, dV. $$ The RHS is the integral of the Dirac delta, and the gradient in the LHS is normal to the surface. Then, $$ 4\pi R^2 \left. \frac{df}{dr}\right|_{R} = 1, $$ yielding $A=-1/4\pi$. Therefore, $$ \delta^3 = -\frac{1}{4\pi} \Delta \frac{1}{r}. $$ The same derivation can be worked for two dimensions. Wikipedia and MathWorld list some properties of the Dirac delta in more than one dimension.
Derivating $A$
All your equations can be written as $$ D \Delta \phi - \Sigma_a \phi + \hat{S} \delta^n = 0, $$ where we use $S=\hat{S} \delta^n$, as suggested by the book. For the spherical case, the solution is $$ \phi = A \frac{e^{-\kappa r}}{r}, $$ and we can evaluate its Laplacian in $r=0$ as $$ \lim_{r \to 0} \Delta \phi = \Delta \frac{A}{r} = -4\pi A \delta^3. $$ Substituting in the equation, we have $$ -4\pi A D \delta^3 - \Sigma_a \phi + \hat{S} \delta^3 = 0, $$ and, finally, equating the terms accompanying $\delta^3$ yields $$ A = \frac{\hat{S}}{4\pi D}. $$ Again, the procedure is the same for the two-dimensional and one-dimensional cases.
I'm confident that that is what the book meant with "balancing the coefficients of the two Dirac distributions". I cannot provide rigorous arguments for these manipulations. I can't say, for example, what happens to the $\Sigma_a \phi$ in the equation when we collect the Dirac deltas. I guess it vanishes in comparison to the Dirac deltas.