6.3 Partial Differential Equations 377
Example 1.
∂^2 u
∂x^2=∂u
∂t, 0 <x< 1 , 0 <t,
u( 0 ,t)= 1 , u( 1 ,t)= 1 , 0 <t,
u(x, 0 )= 1 +sin(πx), 0 <x< 1.The partial differential equationand the boundary conditions(that is, every-
thing that is valid fort>0) are transformed, while the initial condition is
incorporated by the transform
d^2 U
dx^2 =sU−(
1 +sin(πx))
, 0 <x< 1 ,U( 0 ,s)=^1
s, U( 1 ,s)=^1
s.
This boundary value problem is solved to obtain
U(x,s)=^1 s+sins+(ππx 2 ).We d i r e c t o u r a t t e n t i o n n o w t oUas a function ofs. Because sin(πx)is a con-
stant with respect tos,tablesmaybeusedtofind
u(x,t)= 1 +sin(πx)exp(
−π^2 t)
.
Example 2.
∂^2 u
∂x^2 =∂^2 u
∂t^2 ,^0 <x<^1 ,^0 <t,
u( 0 ,t)= 0 , u( 1 ,t)= 0 , 0 <t,
u(x, 0 )=sin(πx), 0 <x< 1 ,
∂u
∂t(x, 0 )=−sin(πx), 0 <x< 1.Under transformation the problem becomes
d^2 U
dx^2=s^2 U−ssin(πx)+sin(πx), 0 <x< 1 ,
U( 0 ,s)= 0 , U( 1 ,s)= 0.The functionUis found to be
U(x,s)= s−^1
s^2 +π^2sin(πx),