3a. Why is it correct to write this as: $d^2\Phi/dx^2 + B^2\Phi = 0$ (an ordinary differential equation)?
(for 3b thru 3e) The following are steps you should likely do in checking a differential equation and
a particular solution.
3b. Given D has units of length [L], show the (above) diffusion equation is dimensionally correct.
Having done 3b, you (should) have confidence/be willing to invest the time to do the following.
3c. Demonstrate the above solution is correct. Substitute into the diffusion equation, and show you
get an identity (e.g., 0 = 0); and verify it satisfies the boundary conditions.
3d. Verify the peak-to-average flux (the axial peaking factor, or $F_z$) = $(\Phi_c/\Phi_{avg})_z = \pi/2$.
For a parallelepiped ($-L_x/2 < x < L_x/2$, $-L_y/2 < y < L_y/2$, and $-L_z/2 < z < L_z/2$) the solution
is the product of three 1-D solutions: $\Phi = \Phi_c \cos(\pi x/L_x)\cos(\pi y/L_y)\cos(\pi z/L_z)$, and
$B^2 = (\pi/L_x)^2 + (\pi/L_y)^2 + (\pi/L_z)^2$.
3e. Derive an expression for the peak-to-average flux assuming (for convenience, less work!) a
cube ($L = L_x = L_y = L_z$). The result is true if not a cube, and from the product solution (of the three
1-D solutions) $\Phi(x, y, z) = \Phi(x)\Phi(y)\Phi(z)$, one can immediately write:
$\Phi_c/\Phi_{avg} = (\Phi_c/\Phi_{avg})_x (\Phi_c/\Phi_{avg})_y (\Phi_c/\Phi_{avg})_z = (\pi/2)^2 = 3.88$
