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# partial differential equation

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5577-10.2-10E AID: 595|18/09/2012
RID: 175| 19/10/12
Consider the initial – boundary value problem
( ) ( )
( )
, 0 1, 0
0, 0, 1, 10, 0
,0 10 , 0 1
xx t
u u x t
u t u t t
u x x x
= < < >
= = >
= < <
...... (1)
For the initial boundary value problem
( ) ( )
( ) ( )
2
0 1
, 0 , 0
0, , , , 0
,0 , 0
t xx
u c u x p t
u t T u p t T t
u x f x x p
= < < >
= = >
= < <
...... (2)
the solution is
( ) ( ) ( )
, ,u x t v x t S x= +
...... (3)
Where
( )
2
2
1
, sin
n
c t
p
n
n
n
v x t b x e
p
π
π
÷
=
=
÷
...... (4)
( )
1 0
0
T T
S x T x
p
= +
( ) ( )
( )
0
2
sin , 1, 2,3
p
n
n
b f x S x x dx n
p p
π
= =
÷
L
...... (5)
( )
0f x =
Compare (1) with (2),
1c =
( )
10f x x=
0
0T =
1
10T =
So,
( )
10 0
0
1
10
S x x
x
= +
=
Use
1c =
,
,
( )
10f x x=
,
( )
10S x x=
in (5),
( )
1
0
2
10 10 sin
1 1
0
n
n
b x x x dx
π
=
÷
=
So,
0
n
b =
Also, substitute
1c =
,
,
0
n
b =
in (4),
( ) ( )
2
2
1
1
1
, 0 sin
1
0
n
t
n
n
v x t x e
π
π
÷
=
=
÷
=
Therefore
( )
, 0v x t =
Use this result and
( )
10S x x=
in (3), the solution is
( )
, 10u x t x=

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5577-10.2-10E AID: 595|18/09/2012 RID: 175| 19/10/12Consider the initial - boundary value problem ...... (1)For the initial boundary value problem ...... (2)the solution is ...
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