Mathematical Physics Vol 1

Chapter 7. Partial differential equations

418

R From the last two problems the question arises, is it possible to transform the PDE u t = u xx , 0 < x < L , t > 0 , (7.356) with conditions u ( x , 0 )= f ( x ) , u ( 0 , t )= p ( t ) , u ( L , t )= q ( t ) , (7.357) so that we obtain the standard boundary conditions ( X ( a )= 0, X ( b )= 0)? The answer is YES. The transformation is u = v + A ( t ) x + B ( t ) (7.358) and the conditions are u ( L , t )= v ( L , t |{z} ) = 0 + A ( t ) · L + B ( t ) ⇒ u ( L , t )= A ( t ) · L + B ( t )= q ( t ) , u ( 0 , t )= v ( 0 , t |{z} ) = 0 + A ( t ) · 0 + B ( t ) ⇒ B ( t )= p ( t ) . From these equations it follows that A ( t )= q − p L . The partial differential equation, for v is ˙ v = v xx − ˙ q ( t ) x L + ˙ p ( L − x ) L , and the initial condition

p ( 0 ) − q ( 0 ) L

v ( x , 0 )= f ( x )+

x − p ( 0 ) .

Problem 273 Determine the PDE type and find its solution (conduction of heat equation - with source) u t = u xx + Q ( x ) , 0 < x < L , t > 0 , (7.359) that satisfies the following conditions

u ( 0 , t )= 0 , u ( L , t )= 0 , u ( x , 0 )= f ( x ) .

(7.360)

Solution When solving this problem, we will first analyze the special case when L = 2, f ( x )= 2 x − x 2 and Q ( x )= 1 −| x − 1 | . If we observe this problem without the source, the solution is of the form u ( x , t )= ∞ ∑ n = 1 T n ( t ) sin n π x 2 , (7.361) where T n ( t )= 16 ( 1 − ( − 1 ) n ) n 3 π 3 e − n 2 π 2 4 t (7.362) The function Q ( x ) should now be represented by a Fourier sine series, i.e.

∞ ∑ n = 1

n π x 2

Q ( x )=

q n sin

(7.363)

,