Mathematical Physics Vol 1

Chapter 7. Partial differential equations

376

For k = 0we obtain

rR ′′ + R ′ = 0 , F ′′ = 0 . with the substitution R ′ = f ( r ) , for the first equation, we obtain d f / dr = − f / r or ln f = − ln r + ln C 1 ⇒ r f = C 1 = r d R d r ⇒

(7.262)

R = C 1 ln r + C 2 .

(7.263)

For the second equation we obtain F = C 3 · ϕ + C 4 . Thus, the solution of the initial equation is: for k̸ = 0 u = C 1 r k + C 2 r − k · ( C

3 cos ( k ϕ )+ C 4 sin ( k ϕ ))

(7.264)

for k = 0

u =( C 1 ln r + C 2 ) · ( C 3 ϕ + C 4 ) . (7.265) We determine the constants from the condition that the solution satisfies the boundary conditions. Namely, we have assumed that on the circular contour the function u has some predefined value. However, as after completing the round of the circle we reach the same point again, this solution must be periodic with period 2 π . This practically means that k must be an integer ± 1 , ± 2 ,... , and C 3 = 0 in (7.265). We would obtain the same solutions if we assumed that k is a natural number, i.e. k = 1 , 2 ,..., n , so we finally obtain u 0 = u | k = 0 = C 4 ( C 1 ln r + C 2 ) , (7.266) u n = u | k = n = C n r n + D n r − n ( a n cos ( n ϕ )+ b n sin ( n ϕ )) , n = 1 , 2 ,... (7.267) Further on, the function u must be continuous, by assumption, at all point inside the region bounded by K , and thus at point r = 0 as well. However, as the functions ln r and r − n are not defined at that point, it follows that the coefficients next to them are equal to zero ( C 1 = D n = 0), and the solution gains the form

a 0 2

n ( a

u 0 ( r , ϕ )=

; u n ( r , ϕ )= r

n cos ( n ϕ )+ b n sin ( n ϕ )) .

(7.268)

Applying the superposition principle, for u we obtain

+ ∞ ∑ n = 1

a 0 2

r n ( a

u ( r , ϕ )=

n cos ( n ϕ )+ b n sin ( n ϕ )) .

(7.269)

+

We have not yet used the condition at the boundary K . Let u ( R , ϕ )= u | r = R = f ( ϕ ) , where f is a known function. From (7.269) we obtain f ( ϕ )= a 0 2 + R n ( a n cos ( n ϕ )+ b n sin ( n ϕ )) .

+ ∞ ∑ n = 1

(7.270)

Thus, a known function should be expanded into a Fourier series, which yields

+ π Z

+ π Z

1 π R n

1 π R n

a n =

f ( t ) cos ( nt ) d t , b n =

f ( t ) sin ( nt ) d t .

(7.271)

− π

− π