Mathematical Physics Vol 1

Chapter 7. Partial differential equations

404

and find its solution for the following - boundary conditions

u ( 0 , t )= 0 , u ( 2 , t )= 0 ,

- initial condition

2 −

u ( x , 0 )= 2sin

π x

sin ( π x )+ 4sin ( 2 π x )= f ( x ) .

Solution According to 4 ◦ (see p. 362) the equation is of parabolic type. We will look for its solution in the form (method of separation of variables)

u ( x ; t )= X ( x ) T ( t ) ,

and thus teh equation (7.317) becomes

4 X ( x ) ˙ T ( t )= X ′′ ( x ) T ( t ) .

R Note that X̸ ≡ 0 and T̸ ≡ 0, as we are looking for a non-trivial solution. If we now divide this equation by XT , we obtain

˙ T T

X ′′ X

= λ ,

4

(7.318)

=

where λ is a constant. Given that the left hand side is a function of the independent variable t , and the right hand side of the independent variable x , the equal sign is possible only in the case when they are equal to a constant ( λ ). Thus, this PDE is decomposed to two ordinary differential equations 4 ˙ T − λ T = 0 , ∧ X ′′ − λ X = 0 . Boundary conditions now come down to u ( 0 , t )= X ( 0 ) · T ( t )= 0 ⇒ X ( 0 )= 0 , u ( 2 , t )= X ( 2 ) · T ( t )= 0 ⇒ X ( 2 )= 0 . Thus, the first part of the problem is to find values of the constant λ for which we obtain non-trivial solution of the problem: X ′′ + λ X = 0 , X ( 0 )= X ( 2 )= 0 , (7.319)

as well as the corresponding function X ( x ) .