Miscellaneous Exercises 287
12.Apply the following formula (see Section 4.4, Exercise 16) for the solu-
tion of the potential problem in the upper half-plane if the boundary
condition isu(x, 0 )=f(x),where
f(x)={ 0 , x<0,
1 , x>0,u(x,y)=π^1∫∞
−∞f(x′)y (^2) +(xy−x′) 2 dx′.
13.Apply the formula in Exercise 12 to the case wheref(x)=1,−∞<x
<∞. The solution of the problem should beu(x,y)≡1.
- a. Find the separation-of-variables solution of the potential problem in
a disk of radius 1 if the boundary condition isu( 1 ,θ)=f(θ ),where
f(θ )={−π−θ, −π<θ<0,
π−θ, 0 <θ <π.b.Show that the function given in polar and Cartesian coordinates byu(r,θ)=2tan−^1( rsin(θ )
1 −rcos(θ ))
=2tan−^1( y
1 −x)
satisfies the potential equation (use the Cartesian coordinates) and
the boundary condition. The following identity is useful:sin(θ )
1 −cos(θ )=tan(π−θ
2)
.
c. Sketch some level curves of the solution inside the circle of radius 1.15.Solve the potential equation in a disk of radiuscwith boundary condi-
tions
u(c,θ)=
{ 1 , 0 <θ <π,
0 , −π<θ<0.16.What is the value ofuat the center of the disk in Exercise 15?
17.Same as Exercise 15, but the boundary condition is
u(c,θ)=∣∣
sin(θ )