Axisymmetric problem satisfying laplace's equation :

( no f - dependence in the equation : ¶²V/¶ r² + 1/r ¶ V/¶ r + ¶²V/¶ z² = 0 )

 All derivatives evaluated at the center of the cell(r,z). Dr = a and Dz = b .

A taylor series expansion gives

                    V(r+a,z+b) = V(r,z) + a ¶ V/¶ r + b ¶ V/¶ z + ab/2 ¶²V/¶ r¶ z + a²/2 ¶²V/¶ r² + b²/2 ¶²V/¶ z² .

+ higher terms. In different notation we have

                    V(r,z+b) = V(r,z) + b ¶V/¶z + 1/2 b² ¶²V/¶ z² = Vo + b V_z + 1/2 b² V_zz .

                    V(r,z-b) = V(r,z) - b ¶V/¶z + 1/2 b² ¶²V/¶ z² = Vo - b V_z + 1/2 b² V_zz .

Then V_zz = ( V(r,z+b) + V(r,z-b) - 2Vo)/b² .

                    V(r+a,z) = V(r,z) + a ¶V/¶r + 1/2 a² ¶²V/¶ r² + ... = Vo + aV_r + 1/2 a² V_rr + ..

                    V(r-a,z) = V(r,z) - a ¶V/¶r + 1/2 a² ¶²V/¶ r² + ... = Vo - aV_r + 1/2 a² V_rr + .. .

Now we find

                     V_r = (V(r+a,z)-V(r-a,z))/(2a), and

                    V_rr = (V(r+a,z)+V(r-a,z) - 2 Vo)/a² .

Since Laplace'e equation with no phi dependence is   V_rr + 1/r V_r + V_zz = 0, we have

                     0 = (V(r+a,z)+V(r-a,z) -2 Vo)/a² +1/r (V(r+a,z)-V(r-a,z))/(2a) +(V(r,z+b) + V(r,z-b) - 2Vo)/b² .

This can be solved for Vo:

                    Vo( 2/a² + 2/b²)= (V(r+a,z)+V(r-a,z))/a² +1/r (V(r+a,z)-V(r-a,z))/(2a) +( V(r,z+b) +V(r,z-b))/b² .

When we are away from r = 0 and z = 0 we have

                   V(r,z) = C_rr (V(r+a,z)+V(r-a,z)) +C_r/r (V(r+a,z)-V(r-a,z)) + C_zz (V(r,z+b)+V(r,z-b)) ,

where the constants C_rr, C_r, and C_zz are

                    C_rr = 1/2 b²/(a² + b²),             C_r = 1/4 ab²/( a² + b²), and             C_zz = 1/2 a²/( a² + b²) .

 At r=0, catastrophe? If V_r(0) ¹ 0, then at r=0 V_r/r definitely blows up. For Laplace's equation to hold, V_r(0) = 0. Now let s equal the radial distance from the z-axis and taylor-expand Vr: (s<<1)

                    V_r(s) = V_r(0) + s V_rr(0) + higher powers of s,

and since V_r(0) = 0, we have V_r(s) = s V_rr(0) + ... . Laplace's equation at s is

                    V_rr(s) + 1/s( V_r(0) + s V_rr(0)) + V_zz(s) .

Now using V_r(0) = 0 we see that for tiny s near r=0 we are ok because the limit of s->0 gives V_rr(0) + V_rr(0) + V_zz(0) , or

                    2 V_rr + V_zz = 0

Another way to deal with r=0 is to sit on the z-axis and write V_xx + V_yy + V_zz = 0.

Since we are at r=0, an x-derivative is identical to a y-derivative and also identical to an r-derivative. This means that V_xx = V_rr = V_yy. Then Laplace's equation becomes

                  2 V_rr + V_zz = 0 at r=0.

At r=0 we have V(0+a,z) = Vo + 0 + 1/2 a² V_rr, so V_rr = (2/a²) (V(0+a,z)-Vo) .

At r=0 for 2V_rr + V_zz = 0, we find

0 = 2 (2/a²) (V(0+a,z)-Vo) + ( V(0,z+b) + V(0,z-b) - 2Vo)/b² . Then Vo is found to be

Vo = [(4/a²)V(0+a,z) + (V(0,z+b) + V(0,z-b))/b²]/[4/a² + 2/b²] . Or

                    V(0,z) = C_ro V(0+a,z) + C_zo (V(0,z+b) + V(0,z-b)), where

                    C_ro = 2b²/(a²+2b²)             and C_zo = 1/2 a²/(a²+2b²) .

At z = 0. Where we have symmetry in the z-axis, like z=0 in the mid-plane of a cylindrical pipe, V(r,z) = V(r,-z), and V_z(r,0) = 0. Using the taylor expansion for V(r,0+b) and V(r,0-b), we find

 V_zz(r,0) = 2 (V(r,0+b) - Vo)/b² . Therefore at z = 0

                      V(r,0) = C_rr (V(r+a,0)+V(r-a,0)) +C_r/r (V(r+a,0)-V(r-a,0)) + 2C_zz V(r,0+b) .

 At r=0 and z = 0, we solve 2V_rr + V_zz = 0. V(0,0+b) = Vo + 0 + 1/ 2 b² V_zz, and

V_zz = (V(0,0+b)-Vo) (2/b²), and V_rr = (2/a²)(V(0+a,0)-Vo) . Thus

0 = 2 (2/a²)(V(0+a,0)-Vo) + (V(0,0+b)-Vo) (2/b²) . Then

Vo = [ 4/a² V(0+a,0) + 2/b² V(0,0+b) ] / [4/a² + 2/b²] . Therefore

                    V(0,0) = C_ro V(0+a,0) + 2C_zo V(0,0+b) .