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solving partial differential equation

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5577-10.2-21E AID: 595|18/09/2012
RID: 175| 19/10/12
Consider the initial – boundary value problem
( ) ( )
( )
, 0 , 0
0, , 0, 0
,0 1, 0
xx t
x x
u u x t
u t u t t
u x x
π
π
π
= < < >
= = >
= < <
...... (1)
For the initial - boundary value problem with homogeneous boundary conditions with
insulted ends,
( ) ( )
( ) ( )
2
, 0 , 0
0, , 0, 0
,0 , 0
t xx
x x
u c u x p t
u t u p t t
u x f x x p
= < < >
= = >
= < <
...... (2)
the solution is
( )
2
0
1
, cos
c t
n
n
n
u x t A a x e
p
λ
π
=
= +
÷
...... (3)
Where
( )
0
0
1
p
A f x dx
p
=
( )
0
2
cos
p
n
n
a f x x dx
p p
π
=
÷
Compare (1) with (2),
1c =
p
π
=
( )
1f x =
Now
Also,
[ ]
0
0
0
2
1 cos
2
cos
2 sin
2
sin sin 0
n
n
a x dx
nxdx
nx
n
n
n
π
π
π
π
π π
π
π
π
π
=
÷
÷
=
=
=
0, 1n=
Substitute
1c =
,
p
π
=
,
0
1A =
,
0
n
a =
in (3), it follows that
( )
2
1
1
, 1 0 cos
1 0
1
t
n
n
u x t x e
λ
π
π
=
= +
÷
= +
=
Therefore the solution is
( )
, 1u x t =

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5577-10.2-21E AID: 595|18/09/2012 RID: 175| 19/10/12Consider the initial - boundary value problem ...... (1)For the initial - boundary value problem with homogeneous boundary conditions with insulted ends, ...
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