Mathematical Physics Vol 1

7.8 Examples

411

Problem 269 Solve the equation

u t = u xx , 0 < x < 1 , t > 0 ,

(7.325)

with the conditions

u ( x , 0 )= x − x 2 , u

x ( 0 , t )= u x ( 1 , t )= 0 .

(7.326)

Solution Starting with the substitution u ( x , t )= X ( x ) · T ( t ) and dividing the initial equation (7.325), we obtain

˙ T T

X ′′ X

= − λ

=

that is

˙ T = − λ T ∧ X ′′ = − λ X .

Integration of the first equation yields

λ t

T = e − (7.327) Note that we did not include here the integration constant, as it is "absorbed" by constants that appear later (because X · T , see 7.324). From X ′′ + λ X = 0 ⇒ X ( x )= c 1 sin kx + c 2 cos kx , and from the initial conditions ( X ′ ( 0 )= X ′ ( 1 )= 0) we obtain c 1 k cos0 − c 2 k sin0 = 0 , c 1 k cos k − c 2 k sin k = 0 , (7.328) which leads to the equation c 1 = 0 , sin k = 0 ⇒ k = 0 , π , 2 π ,..., n π . Substituting into (7.327), for T we obtain T n ( t )= c 3 e − n 2 π 2 t and X n ( x )= c 2 cos ( n π x ) , and thus u ( x , t )= ∞ ∑ n = 0 a n e − n 2 π 2 t cos ( n π x ) , ( a n ≡ c 1 c 3 ) . in order to determine the constants a n we shall use the initial conditions, expanding the given functions ( x − x 2 ) into Fourier series u ( x , 0 )= x − x 2 = ∞ ∑ n = 0 a n cos ( n π x )= .

∞ ∑ n = 1

a 0 2

a n cos ( n π x ) .

=

+