Mathematical Physics Vol 1

Chapter 7. Partial differential equations

422

Now, substituting this relation, we obtain

R m ( r )= J 0 ( k m r )= J 0

r .

α m r 0

(7.386)

Given that

λ 2 = c 2 k 2 = c 2 k 2

2 m ,

m = λ

(7.387)

the equation (7.378) becomes ¨ T + λ 2

m T = 0 ⇒ T m ( t )= a m cos λ m t + b m sin λ m t ,

(7.388)

From where we obtain

u m = T m ( t ) R m ( r )=( a m cos λ m t + b m sin λ m t ) J 0 α m r 0 r .

(7.389)

From here we further obtain for solution u

r .

( a m cos λ m t + b m sin λ m t ) J 0

∞ ∑ m = 1

∞ ∑ m = 1

α m r 0

u =

u m =

(7.390)

Furthermore, the constants a m and b m must be determined. For determining them, we have the unused initial conditions available. From the first condition (7.374) we obtain

a m J 0

r = f ( r ) ,

∞ ∑ m = 1

α m r 0

u ( r , 0 )=

(7.391)

from where we can determine the coefficients a m as well as the coefficients of the Fourier-Bessel series, which represents the expansion of the known function f ( r ) , and thus we obtain

r f ( r ) J 0

r d r , m = 1 , 2 ,...

r 0 Z 0

α m r 0

2

a m =

(7.392)

r 2

2 1 ( α m )

0 J

Similarly, we obtain also the coefficients b m from the remaining condition (7.375) ∂ u ∂ t t = 0 = g ( r ) . (7.393)

Problem 275 The electric potential of a point source in a homogeneous isotropic medium satisfies the Laplace equation, which in the cylindrical coordinate system has the form ∂ 2 U ∂ r 2 + 1 r ∂ U ∂ r + ∂ 2 U ∂ z 2 = 0 . Solve this equation, with two boundary conditions - the potential along the boundary planes of the medium is constant, and - the current flux in the direction perpendicular to the boundary plane is constant.