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# partial differential equation

Exam Practice

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

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