Mathematical Physics Vol 1

7.8 Examples

417

Thus, the equation is the same, for u ( x , t )= v ( x , t )+ 1 2 ( x It remains to determine the boundary conditions

2 + 2 t ) − x .

u x ( 0 , t )= v x ( 0 , t ) 2 + ax x = 0 + b = − 1 ⇒ b = − 1 , u x ( 1 , t )= v x ( 1 , t ) 2 ax x = 1 + b = − 1 ⇒ a = 1 2 .

(7.349)

The initial condition becomes

1 2

x 2 .

v ( x , 0 )= x −

(7.350)

Separating the variables v = XT yields the equations X ′′ = − k 2 X and T ′ = − k 2 T from where we obtain X = c 1 sin kx + c 2 cos kx , X ′ = c 1 cos kx − c 2 sin kx , (7.351) while the boundary conditions (7.349) yield c 1 = 0 , k = n π , (7.352) and thus X n ( x )= c 2 cos n π x , (7.353) T n ( t )= c 3 e − n 2 π 2 t , (7.354) and v n ( x , t )= c 2 c 3 e − n 2 π 2 t cos n π x . Note that T 0 = c 3 and X 0 = c 2 , so a constant exists for the member X 0 · T 0 = c 2 c 3 . Finally, using the principle of superposition, we obtain for v ( x , t )

∞ ∑ n = 1

n 2 π 2 t cos n π x .

a n e −

v ( x , t )= c 2 c 3 |{z} a 0 2

+

The constants a n are determined from the initial condition (7.350), by expanding the known function x − 1 2 x 2 into a Fourier cosine series, i.e.

1 0 1 0

1 2 x 2 d x = x 2 − 1 2 x 2 cos n π x d x = 1 3

2 1 Z 2 1 Z

1

2 3

x 3

a 0 =

x −

=

,

0

a n =

x −

cos n π x −

2 n 3 π 3

1

( 2 x − x 2 ) n π

2 ( x − 1 ) n 2 π 2

= −

sin n π x

+

=

0

2 n 2 π 2

= −

,

and thus v is of the form

∞ ∑ n = 1

1 3 −

2 π 2

1 n 2

e − n 2 π 2 t cos n π x

v ( x , t )=

and the solution of the initial PDE is

∞ ∑ n = 1

1 2

1 3 −

2 π 2

1 n 2

e − n 2 π 2 t cos n π x

( x 2 + 2 t ) − x +

u ( x , t )=

(7.355)