Mathematical Physics Vol 1

Chapter 7. Partial differential equations

414

- boundary

u ( 0 , t )= 0 , u ( 3 , t )= 3 .

(7.341)

Solution Let us look for the solution of the problem by applying the method of separation of variables, i.e. u ( x , t )= X ( x ) T ( t ) . Boundary conditions, for function X ( x ) yield u ( 0 , t )= X ( 0 ) · T ( t )= 0 ⇒ X ( 0 )= 0 , u ( 3 , t )= X ( 3 ) · T ( t )= 3 . The second condition does not separate the boundary conditions, and it is thus not suitable for solving by this method. The question arises: is it possible to find a transformation

u ( x , t )= f ( v ( x , t ))

to separate both the PDE and the conditions? Let us try with a linear transformation

u ( x , t )= v ( x , t )+ ax + b .

The constants a and b should be determined so that the initial PDE is transformed into an equation of the same form, i.e. v t = v xx , and that the conditions are separated, i.e. in our case: v ( 0 , t )= 0 and v ( 3 , t )= 0. In our case we obtain: u xx = v xx + 0 + 0 = v xx , u t = v t + 0 + 0 = v t ⇒ v xx = v t .

The boundary conditions are

u ( 0 , t )= v ( 0 , t )+ a · 0 + b = 0 ⇒ v ( 0 , t )= b = 0 , u ( 3 , t )= v ( 3 , t )+ a · 3 + b = 0 ⇒ v ( 0 , t )= 3 + 3 a = 0 ⇒ a = − 1 , u = v + x ,

v ( 0 , t )= 0 , v ( 3 , t )= 0 .

The former shows that it is possible to find a suitable transformation! The problem now comes down to solving the equation

v t = v xx

with boundary conditions

v ( 0 , t )= 0 , v ( 3 , t )= 0 ,

(7.342)

and initial condition

v ( x , 0 )= u ( x , 0 ) − x = 3 x − x 2 .