Mathematical Physics Vol 1

Chapter 7. Partial differential equations

368

The constants are determined from the initial conditions

∞ ∑ n = 1

∞ ∑ n = 1

π n l

u ( x , 0 )= ϕ ( x )=

u n ( x , 0 )=

A n sin

x ,

(7.199)

∞ ∑ n = 1

∞ ∑ n = 1

∂ u n ∂ t

π n l

π n l

u t ( x , 0 )= ψ ( x )=

aB n sin

x .

(7.200)

=

t = 0

Thus, the problem is reduced to expanding the known functions ϕ ( x ) and ψ ( x ) into Fourier series.

Example 242 Find the solution of the equation a 2 u

xx = u t , 0 ≤ x ≤ l , 0 ≤ t ≤ t 0 ,

(7.201)

that satisfies the initial

u ( x , 0 )= ϕ ( x ) , 0 ≤ x ≤ l ,

(7.202)

and boundary conditions

u ( 0 , t )= 0 , u ( l , t )= 0 0 ≤ t ≤ t 0 .

(7.203)

Solution Let us first note that, according to 4 ◦ (p. 362), this equation is of parabolic type. We shall look for a solution in the form u ( x , t )= X ( x ) · T ( t ) ⇒ (7.204) u t = X · ˙ T , u x = X ′ · T , u xx = X ′′ · T , (7.205) where X̸ ≡ 0and T̸ ≡ 0, as we are looking for a nontrivial solution. Substituting (7.204) into equation (7.201) we obtain a 2 X ′′ T = X ˙ T , that is X ′′ X = 1 a 2 ˙ T T = − λ . (7.206) Thus, the initial equation (7.201) is now decomposed to two ordinary differential equations: X ′′ + λ X = 0 ∧ ˙ T + a 2 λ T = 0 , (7.207) from where we obtain, as in the previous case, for X ( x ) X n ( x )= ¯ C n sin π n l x , (7.208) while for T ( t ) we obtain T n ( t )= ¯ C n e − a 2 λ n t , (7.209) where ¯ C n is, for now, undetermined.