1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

256 Chapter 4 The Potential Equation

it is clear that the temperature cannot be greater at one point than at all other
nearby points. For if such were the case, heat would flow away from the hot
point to cooler points nearby, thus reducing the temperature at the hot point.
But then the temperature would not be unchanging with time. We return to
this matter in Section 4.
A complete boundary value problem consists of the potential equation in a
region plus boundary conditions. These may be of any of the three types

ugiven, ∂u

∂n

given, or αu+β∂u

∂n

given

along any section of the boundary. (By∂u/∂n,wemeanthedirectionalderiv-
ative in the direction normal, or perpendicular, to the boundary.) Whenuis
specified along the whole boundary, the problem is calledDirichlet’s problem;
if∂u/∂nis specified along the whole boundary, it isNeumann’s problem.The
solutions of Neumann’s problem are not unique, for ifuis a solution, so isu
plus a constant.
It is often useful to consider the potential equation in other coordinate sys-
tems. One of the most important is the polar coordinate system, in which the
variables are

r=

√

x^2 +y^2 ,θ=tan−^1

(y

x

)

,

x=rcos(θ ), y=rsin(θ ).

By convention we requirer≥0. We shall define

u(x,y)=u

(

rcos(θ ),rsin(θ )

)

=v(r,θ)

and find an expression for the Laplacian ofu,

∇^2 u=

∂^2 u

∂x^2 +

∂^2 u

∂y^2 ,

in terms ofvand its derivatives by using the chain rule. The calculations are
elementary but tedious. (See Exercise 7.) The results are

∂^2 u

∂x^2

=cos^2 (θ )∂

(^2) v
∂r^2
−2sin(θ )cos(θ )
r
∂^2 v
∂θ∂r
+sin
(^2) (θ )
r^2
∂^2 v
∂θ^2
+sin
(^2) (θ )
r
∂v
∂r
+2sin(θ )sin(θ )
r^2
∂v
∂θ

,

∂^2 u

∂y^2

=sin^2 (θ )∂

(^2) v
∂r^2
+2sin(θ )cos(θ )
r
∂^2 v
∂θ∂r
+cos
(^2) (θ )
r^2
∂^2 v
∂θ^2
+cos
(^2) (θ )
r
∂v
∂r−
2sin(θ )sin(θ )
r^2
∂v
∂θ.