1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

0.4 Singular Boundary Value Problems 39

radiuscmay be described in polar(r,θ)coordinates as occupying the region
0 ≤r≤c. The origin, atr=0, is a mathematical boundary, yet physically this
point is in the interior of the disk.
At a singular point, one cannot specify a value foru(x 0 ), the solution of
the differential equation, or for its derivative. However, it is usually necessary
to require that bothu(x 0 )andu′(x 0 )be finite, or bounded. Tacitly, wealways
require that the solution and its derivative be finite ateverypoint of the interval
where we are solving a differential equation. But when a singular point is a
boundary point of that interval, we enforce the condition explicitly. In the
example that follows we shall see how these conditions act so as to make the
solution of a boundary value problem unique.

Example: Radial Heat Flow.
Suppose a long cylindrical bar, surrounded by a medium at temperatureT,
carries an electrical current. If heat flows in the radial direction much faster
than in the axial direction, the temperatureu(r)in the rod may be described
by the problem

1

r

d

dr

(

rdu

dr

)

=−H, 0 ≤r<c, (1)

u(c)=T. (2)

Here,cis the radius of the rod,ris a polar coordinate, andH(constant) is
proportional to the electrical power being converted into heat.
In this problem, only the physical boundary condition has been noted. The
mathematical boundaryr=0 is a singular point, as is clear from the differen-
tial equation in the form

d^2 u

dr^2 +

1

r

du

dr=−H.

Thus, at this point we will require thatuanddu/drbe finite:

u( 0 ), u′( 0 ) finite. (3)

Now the differential equation (1) is easy to solve. Multiply through byrand
integrate once to find that

rdu

dr

=−Hr

2

2

+c 1.

Divide through this equation byrand integrate once more to determine that

u(r)=−Hr

2

4

+c 1 ln(r)+c 2.