Circle near a wall: (radius a)
This solution satisfies the boundary condition on the
wall (
), and the degree it
satisfies the boundary condition of no flux through the circle
boundary increases as the ratio ,
i.e., the velocity due
to the image dipole small on the real circle for .
For a
2D dipole,
- 1.
- Question: verify the boundary condition of no flux at the wall.
- (a)
- Hint: evaluate the component of the velocity vector in the
direction.
- (b)
- Hint: set
in the expression for the velocity vector in the
direction.
Homework 1: At the circle boundary the no flux boundary condition should be satisfied. Check to which degree is the boundary condition of no flux at the circle boundary satisfied.
- More than one wall: