LAPLACE
TRANSFORMS
CHAPTER 3$
, Value Problems
Boundary
Laplace transforms
↳
So far differential equations involved the derivative with respect to
one variable It A BVP is differential equation that involves the
.
derivative with respect to two variables x , y or x. t or
y ,
t .
We call it a
partial differential equation .
When solve these pole 's get constants and find
.
we we we
these constants conditions often called
using given
BOUNDARY conditions .
-
Heat .
wave , Laplace
'
Suppose u is a function of exit ) and we are told to solve the
using Laplace transforms I
BVP an
-
L Iu Hit ) ) U Cx s )
=
,
L ( If but ) ) )
=
sulks ) -
u Ix ,
o
L ( 2¥ lait ) ) UH Et la )
o
5 s) su be ol ,
= -
-
, ,
L ( but ) ) Ex Ex = U bust
L ( 3¥ Hit ) ) =
II .
UH . s )
find UH s ) o I as before )
-
-
,
, example :
* just treat x as a constant
SUCKS ) ULK.si
kf÷
-
UH ,
a) =
,
* I
D :=
SUCKS ) -
( I
-
x ) =
kD2ulx , s ) 2x
(k s )
'
D -
Ulk , s ) =
a
-
I
( D2 I ) complementary
*
-
UH , s )
=
O
function
ID Fr ) ( -
Dt Fa ) UH ,ss=0
" FE "
A B Css e-
yo
:
esse
=
+
-
T.se functions of S
particular integral :
ik
F.
)eFk"
Its
' '
Acs Base
-
pz÷
U l " )
'
s =
I ) Ucas ) -
t
La
.
x
.
.
-
t
but lim UGC , s ) =D
I
)
=
(
th
K
-
t
s sa
-
-
-
-
I ( I
-
Foz ) '
Als ) =D
.
-
= 1- .
11¥
that ulo.tt O t o
given
-
>
¥132 S ,
I
-
UCO 's ) o
)
=
2
D2 ) ( x )
I [ (
I
EDY
= -
It ( t Es
)
-
E ' " i
=
I i D= Ot Bcs e
S
: .
Bls ) = -
Is
TRANSFORMS
CHAPTER 3$
, Value Problems
Boundary
Laplace transforms
↳
So far differential equations involved the derivative with respect to
one variable It A BVP is differential equation that involves the
.
derivative with respect to two variables x , y or x. t or
y ,
t .
We call it a
partial differential equation .
When solve these pole 's get constants and find
.
we we we
these constants conditions often called
using given
BOUNDARY conditions .
-
Heat .
wave , Laplace
'
Suppose u is a function of exit ) and we are told to solve the
using Laplace transforms I
BVP an
-
L Iu Hit ) ) U Cx s )
=
,
L ( If but ) ) )
=
sulks ) -
u Ix ,
o
L ( 2¥ lait ) ) UH Et la )
o
5 s) su be ol ,
= -
-
, ,
L ( but ) ) Ex Ex = U bust
L ( 3¥ Hit ) ) =
II .
UH . s )
find UH s ) o I as before )
-
-
,
, example :
* just treat x as a constant
SUCKS ) ULK.si
kf÷
-
UH ,
a) =
,
* I
D :=
SUCKS ) -
( I
-
x ) =
kD2ulx , s ) 2x
(k s )
'
D -
Ulk , s ) =
a
-
I
( D2 I ) complementary
*
-
UH , s )
=
O
function
ID Fr ) ( -
Dt Fa ) UH ,ss=0
" FE "
A B Css e-
yo
:
esse
=
+
-
T.se functions of S
particular integral :
ik
F.
)eFk"
Its
' '
Acs Base
-
pz÷
U l " )
'
s =
I ) Ucas ) -
t
La
.
x
.
.
-
t
but lim UGC , s ) =D
I
)
=
(
th
K
-
t
s sa
-
-
-
-
I ( I
-
Foz ) '
Als ) =D
.
-
= 1- .
11¥
that ulo.tt O t o
given
-
>
¥132 S ,
I
-
UCO 's ) o
)
=
2
D2 ) ( x )
I [ (
I
EDY
= -
It ( t Es
)
-
E ' " i
=
I i D= Ot Bcs e
S
: .
Bls ) = -
Is