Mathematical Physics Vol 1

7.8 Examples

405

Assume that the solution of the equation (7.319) is of the form

α x α 2

+ λ = 0 √ − λ x .

α x ⇒ α 2 Ce α x

α x

+ λ Ce

X ( x )= Ce

= 0 ⇒ Ce

⇒ α = ± p − λ ⇒ X ( x )= C 1 e α 1 x

√

α 2 x

− λ x

+ C 2 e −

+ C 2 e

= C 1 e

The analysis (discussion with respect to value λ , namely, is it: positive, 0 or negative) (see discussion on p. 366) shows that the only interesting case is the one where

µ xi

µ xi = C 1 ( cos µ x + i sin µ x )+ C 2 ( cos µ x − i sin µ x ) ,

+ C 2 e −

X ( x )= C 1 e

A cos µ x +( C 1 − C 2 ) i | {z }

B sin µ x =

X ( x )=( C 1 + C 2 ) | {z } X ( x )= A cos µ x + B sin µ x .

From boundary conditions it follows that

X ( 0 )= A ∧ X ( 2 )= B sin2 µ = 0 . From the last condition we conclude that B̸ = 0, because otherwise we would obtain the trivial solution ( A = 0 , B = 0 ⇒ u ( x , t )= 0 for each x and t ), and thus, because B̸ = 0, the following must be true sin2 µ = 0 ⇒ 2 µ = n π ⇒ µ = n π 2 , n = 1 , 2 ,..., λ = − µ 2 = − n π 2 2 Finally, for the first part of the solution, we obtain

n π 2

X n ( x )= B sin

x .

Now the second equation (when λ is determined) becomes

1 4

2 n 2 π 2 16 2

λ n 4

n π

d T T

4 ˙ T = λ T ⇒

= −

⇒

=

n 2 π 2 16

t

t ⇒ T = e −

ln T = −

,

and thus

x · exp −

t

n 2 π 2 16

n π 2

u n = X n ( x ) · T n ( t )= B n sin

that is (property 1. - superposition, on p. 356)

exp −

t

b n sin

2

∞ ∑ n = 1

n 2 π 2 16

n π x

u ( x , t )=

where b n are constants. The constants b n can be determined from the initial conditions, i.e.

b n sin

2

∞ ∑ n = 1

n π x

u ( x , 0 )= ϕ ( x )=

(7.320)

.