search

# partial differential equation

Exam Practice

### Rating

Showing Page:
1/1
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 =

### Unformatted Attachment Preview

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 ...
Purchase document to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.
Review
Review

Anonymous
Very useful material for studying!

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4