3a. Why is it correct to write this as: d^2Φ/dx^2 + B^2Φ = 0 (an ordinary differential equation)

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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: $\Phi_c/\Phi_{avg} = (\Phi_c/\Phi_{avg})_x (\Phi_c/\Phi_{avg})_y (\Phi_c/\Phi_{avg})_z = (\pi/2)^2 = 3.88$

          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:
$\Phi_c/\Phi_{avg} = (\Phi_c/\Phi_{avg})_x (\Phi_c/\Phi_{avg})_y (\Phi_c/\Phi_{avg})_z = (\pi/2)^2 = 3.88$

3a. Why is it correct to write this as: d^2Φ/dx^2 + B^2Φ = 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 Fz) = (/Φavg)z = π/2.
For a parallelepiped (-Lx/2 < x < Lx/2, -Ly/2 < y < Ly/2, and -Lz/2 < z < Lz/2) the solution
is the product of three 1-D solutions: Φ = cos(π x/Lx)cos(π y/Ly)cos(π z/Lz), and
B^2 = (π/Lx)^2 + (π/Ly)^2 + (π/Lz)^2.
3e. Derive an expression for the peak-to-average flux assuming (for convenience, less work!) a
cube (L = Lx = Ly = Lz). The result is true if not a cube, and from the product solution (of the three
1-D solutions) Φ(x, y, z) = Φ(x)Φ(y)Φ(z), one can immediately write:
/Φavg = (/Φavg)x (/Φavg)y (/Φavg)z = (π/2)^2 = 3.88

Added by Gregorio M.

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