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partial differential equation

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5577-10.2-6E AID: 595|18/09/2012
RID: 175| 19/10/12
Consider the initial – boundary value problem
( ) ( )
( )
, 0 2, 0
0, 2, 0, 0
, 0 1
,0 , 0 2
2 , 1 x<2
xx t
u u x t
u t u t t
x x
u x x
x
= < < >
= = >
<
= < <
...... (1)
For the initial - boundary value problem
( ) ( )
( ) ( )
2
, 0 , 0
0, 0, , 0, 0
,0 , 0
t xx
u c u x p t
u t u p t t
u x f x x p
= < < >
= = >
= < <
, ...... (2)
the solution is
( )
2
2
1
, sin
n
c t
p
n
n
n
u x t b x e
p
π
π
÷
=
=
÷
...... (3)
Where
( )
0
2
sin , 1, 2,3,
p
n
n
b f x x dx n
p p
π
= =
÷
L
...... (4)
Compare (1) with (2),
1c =
2p =
( )
, 0 1
2 , 1 2
x x
f x
x x
<
=
<
Substitute
1c =
,
2p =
,
( )
, 0 1
2 , 1 2
x x
f x
x x
<
=
<
in (4), it follows
( )
( ) ( )
( )
2
0
1 2
0 1
1 2
0 1
1 2 2
0 1 1
2
sin
2 2
sin sin
2 2
sin 2 sin
2 2
sin 2sin sin
2 2 2
n
n
b f x x dx
n n
f x x dx f x x dx
n n
x x dx x x dx
n n n
x x dx x dx x x dx
π
π π
π π
π π π
=
÷
= +
÷ ÷
= +
÷ ÷
= +
÷ ÷ ÷
Continue
So,
0
n
b =
Substitute
1c =
,
2p =
,
0
n
b =
in (3),
( )
2
2
2
1
1
, 0 sin
2
0
n
t
n
n
u x t x e
π
π
÷
=
=
÷
÷
=
Therefore the solution is
( )
, 0u x t =

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5577-10.2-6E 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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