Mathematical Physics Vol 1

Chapter 7. Partial differential equations

416

Finally, for function v ( x , t ) , we obtain

∞ ∑ n = 1

( 1 − ( − 1 ) n ) n 3

n π 3

32 π 3

π 2 n 2

9 t sin

e −

v ( x , t )=

x ,

and the solution of the initial equation is u = v + x ,

∞ ∑ n = 1

( 1 − ( − 1 ) n ) n 3

n π 3

32 π 3

π 2 n 2

9 t sin

e −

u ( x , t )= x +

x .

(7.343)

R In this case also an analysis of the expression 1 − ( − 1 ) n can be performed, from which it can be seen that the even members are equal to zero, and that the function u is expressed in terms of odd members only, which we leave to the reader.

Problem 272 Solve the PDE

u t = u xx , 0 < x < 1 , t > 0 ,

(7.344)

with the conditions

u ( x , 0 )= 0 | {z } initial ,

u x ( 0 , t )= − 1 , u x ( 1 , t )= 0 | {z } boundary .

(7.345)

Solution The problem is, conceptually, similar to the previous, but in this case the substitution u = v + ax + b does not lead to suitable boundary conditions, because u x = v x + a , and thus

u x ( 0 , t )= v x ( 0 , t )+ a = − 1 ⇒ a = − 1 , ( for v x ( 0 , t )= 0 ) u x ( 1 , t )= v x ( 1 , t )+ a = 0 ⇒ a = 0 ( for v x ( 1 , t )= 0 )

which is contradictory.

We shall try with u = v + ax 2 + bx . This yields

u t = u xx ⇒ v t = v xx + 2 a ,

(7.346)

and we do not obtain the same partial equation. Let us try the following transformation

u = v + a ( x 2 + 2 t )+ bx

(7.347)

which yields the equation

= v t + 2 a ) ⇒

u xx = v xx + 2 a , u t

v xx = v t .

(7.348)