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partial differential equation

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5577-10.2-9E AID: 595|18/09/2012
RID: 175| 19/10/12
Consider the initial – boundary value problem
( ) ( )
( )
, 0 1, 0
0, 20, 1, 10, 0
,0 0, 0 1
xx t
u u x t
u t u t t
u 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)
Observe that
( )
0f x =
Compare (1) with (2),
1c =
( )
0f x =
0
0T =
1
10T =
So,
( )
10 20
20
1
20 10
S x x
x
= +
=
Substitute
1c
=
,
,
( )
0f x =
,
( )
20 10S x x=
in (5), it follows
( )
( ) ( )
( ) ( )
( )
( )
1
0
1 1
0 0
1
1
2
0
0
2
0 20 10 sin
1 1
20 sin 40 sin
cos sin cos
20 40
cos cos 1
20 0 40
n
n
b x x dx
x n x dx n x dx
x n x n x n x
n n
n
n n
n n n
π
π π
π π π
π π
π
π π
π π π
= +
÷
=
= +
= + +
[ ]
( ) ( )
20
cos 2
20
1 2 Since cos 1 , 1, 2,
n n
n
n
n n
n
π
π
π
π
=
= = =
L
So,
( )
20
1 2
n
n
b
n
π
=
Also, use
1c =
,
,
( )
20
1 2
n
n
b
n
π
=
in (4), it follows that
( ) ( )
( ) ( )
2
2
2 2
1
1
1
1
20
, 1 2 sin
1
20
1 2 sin
n
t
n
n
n
n t
n
n
v x t x e
n
n x e
n
π
π
π
π
π
π
÷
=
=
=
÷ ÷
=
So,
( ) ( ) ( )
2 2
1
20
, 1 2 sin
n
n t
n
v x t n x e
n
π
π
π
=
=
Use this result and
( )
10S x x=
in (3), the solution is
( ) ( ) ( )
2 2
1
20
, 1 2 sin 10
n
n t
n
u x t n x e x
n
π
π
π
=
= +

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5577-10.2-9E 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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