Miscellaneous Exercises 291
b.Find the gradient ofuand plot some vectorsV=−grad(u)near the
origin.u(x,y)=−tan−^1(y
x)
.
Theflowfield(seeExercise30)givenbythisfunctioniscalledan
irrotational vortex.35.SametasksasinExercise34.Theflowfieldgivenbythisfunctioniscalled
a source at the origin.
u(x,y)=−ln
(√
x^2 +y^2)
.
36.Solve this potential problem in a half-annulus (sketch the region). At
some point, it may be useful to make the substitutions=ln(r).
1
r
∂
∂r(
r∂u
∂r)
+^1
r^2∂^2 u
∂θ^2= 0 , 1 <r<e, 0 <θ <π,u( 1 ,θ)= 0 , u(e,θ)= 0 , 0 <θ <π,
u(r, 0 )= 0 , u(r,π)= 1 , 1 <r<e.37.Solve the Poisson equation,∇^2 u=−f, in polar coordinates by finding a
function that depends only onrfor:
a. f(r,θ)=1;
b.f(r,θ)=1
r^2.
38.In “An improved transmission line structure for contact resistivity mea-
surements” [L.P. Floyd et al.,Solid-State Electronics, 37 (1994): 1579–
1584], a strip of conducting material is carrying a current in the direc-
tion of its length. A second, long conducting strip of widthLis placed at
right angles to the first, forming a cross. A voltage is to be measured by
a probe on the second strip some distance from the first. In the second
strip, the voltageV(x,y)satisfies the boundary value problem
∂^2 V
∂x^2
+∂
(^2) V
∂y^2
= 0 , 0 <x<L, 0 <y,
∂V
∂x
( 0 ,y)= 0 , ∂V
∂x
(L,y)= 0 , 0 <y,
V(x, 0 )=f(x), 0 <x<L.
In this problem,xis in the direction of current flow in the lower strip;
yis in the direction of the length of the second strip;y=0attheedge