Mathematical Physics Vol 1

7.8 Examples

415

Let us now apply the method of separation X ( x ) · T ( t ) : v t = v xx ⇒ X ′′ · T = ˙ T · X = 0 . From here, dividing by XT , we obtain X ′′ X = ˙ T T = − k 2 ,

that is a system of two ordinary DE:

X ′′ + k 2 X = 0 ∧ ˙ T + k 2 T = 0 . The boundary conditions for the first equation are given by equations (7.342). The solution of the first equation is of the form (as already demonstrated several times) X ( x = c 1 sin kx + c 2 cos kx . ) From (7.342) we obtain: X ( 0 )= c 2 = 0 , X ( 3 )= c 1 sin3 x = 0 ⇒ k = n π 3 , X = c 1 sin n π 3 x . From the second of equation we obtain ˙ T + k 2 T = 0 T = c 3 e − n 2 π 2 9 t . Note that these solutions depend on n = 1 , 2 ,... , and thus on of the solutions is v n = X n · T n = c 1 c 3 |{z} b n e − n 2 π 2 9 t · sin n π 3 x . Applying the principle of superposition, we obtain

∞ ∑ n = 1

n π 3

n 2 π 2

9 t sin

b n e −

v ( x , t )=

x .

The constants b n are obtained from the initial condition

∞ ∑ n = 1

n π 3

v ( x , 0 )= x = 3 x − x 2 expanding the function 3 x − x 2 into the Fourier sine series, i.e. b n = 2 3 Z 3 0 ( 3 x − x 2 ) sin n π 3 x d x = = − 6 ( 2 x − 3 ) n 2 π 2 sin n π 3 x + 3 ( n 2 π 2 x 2 − 3 n 2 π 2 x − 18 ) n 3 π 3 cos n π 3 x 3 0 = = 32 ( 1 − ( − 1 ) n ) n 3 π 3 . b n sin