Miscellaneous Exercises 285
to get a rough graphical solution of the potential equation for hydrodynamics
problems.
Where a physical boundary is formed by an impervious wall, the velocity
vectorVmust be parallel to the boundary. This fact leads to two boundary
conditions. First, the wall must coincide with a streamline; thusψ=constant
along a boundary. Second, the component ofVthat is normal to the wall must
be zero there, because no fluid passes through it; thus the normal derivative of
φis zero,∂φ/∂n=0, at a boundary. See Miscellaneous Exercises 30–32.
Chapter Review
See the CD for Review Questions.
Miscellaneous Exercises
1.Solve the potential equation in the rectangle 0<x<a,0<y<bwith
the boundary conditionsu( 0 ,y)= 1 , u(a,y)= 0 , 0 <y<b,
u(x, 0 )= 0 , u(x,b)= 0 , 0 <x<a.2.Ifa=bin Exercise 1, thenu(a/ 2 ,a/ 2 )= 1 /4. Use symmetry to explain
this fact.
3.Solve the potential equation on the rectangle 0<x<a,0<y<bwith
the boundary conditionsu( 0 ,y)= 1 , u(a,y)= 1 , 0 <y<b,
∂u
∂y(x,^0 )=^0 ,∂u
∂y(x,b)=^0 ,^0 <x<a.4.Same as Exercise 3, but the boundary conditions areu( 0 ,y)= 1 ,∂u
∂x(a,y)=^0 ,^0 <y<b,u(x, 0 )= 1 ,∂u
∂y(x,b)=^0 ,^0 <x<a.