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8/4/2019 Partial Differential Equation Using Laplace
1/15
Solution of PDEs using the Laplace Transform*
A powerful technique for solving ODEs is
to apply the Laplace Transform Converts ODE to algebraic equation that is
often easy to solve
Can we do the same for PDEs? Is it everuseful? Yes to both questions particularly useful for
cases where periodicity cannotbe assumed,thwarting use of Fourier series, henceseparation of variables
*Kreysig, 8th Edn Sec 11.12
8/4/2019 Partial Differential Equation Using Laplace
2/15
Laplace Transform
The key point: we could handle functions that
were discontinuous, in any event not periodic HLT has zillions of useful examples of LT:
)()(
)()(
twsW
sVtv
We ask: can we apply the LT to solvePartialdifferential
equations for cases where separation of variables is ineffective?
8/4/2019 Partial Differential Equation Using Laplace
3/15
Example*: waggling a semi-infinite string
=otherwise
if
0
20sin),0(
tttw
*Kreysig, 8th Edn, page 644
Find the displacement w**(x,t) of an elastic string subject to the followingconditions:
1. The semi-infinite string is initially at rest, on the x-axis
2. For some time t>0the left hand end of the string (x=0) is moved sinusoidally:
3. Furthermore
0,0),(lim =
ttxwx
for
We will tackle this problem using the Laplace Transform; but first, we try a
simpler example
** just in this part of the notes, we use w(x,t) for the
PDE, rather than u(x,t) because u(t) is conventionallyassociated with the step function
8/4/2019 Partial Differential Equation Using Laplace
4/15
A recap on the LT
( ) ( ) ( ) 1)0( ==+ wtutawtw &
We first solve the first order ODE
where u(t) is the unit step:
t=0
1
ssU
1)( =:HLT
Laplace Transform (reach for HLT): ( ) ( )( ) ( )ssaWwssW
10 =+
( )asa
a
assW
+
+=
111Rearranging, and partial fractions
( ) ( ) ( )[ ]at
eatuatw
+= 11
Once more unto HLT, inverse LT:
8/4/2019 Partial Differential Equation Using Laplace
5/15
Next: a beguilingly simple PDE
0=
+
t
wx
x
w
( ) ( ) ttwxw == ,0;00,Subject to boundary conditions:
Applying separation of variables, it is easy to show that
=2
2
),(
xtk
etxw
In fact any function of the form2
),(egsolution,ais,2
22xttxw
xtf =
But runs into problems with boundary conditions
!!all!!all
:conditionsboundarytheApply
tetwxexw
kt
kx
,0),0(,0)0,(
2
====
8/4/2019 Partial Differential Equation Using Laplace
6/15
Try again, this time applying the LT
0=
+
t
wx
x
w
( ) ( ) ttwxw == ,0;00,
Again, consider the simple first order PDE:
Subject to boundary conditions:
Of course, ),( txw is a function ofx and t; but when we compute partialderivative ofw with respect to twe treatx as a constant
0=
+
t
wxL
x
wLFirst, the LT is a linear operator:
sW),x(wsWt
wL ==
0The second term is easy:
but what about the first term?
8/4/2019 Partial Differential Equation Using Laplace
7/15
Back to basics
x
Wdte)t,x(w
xdte
x
w
x
wL
stst
=
=
=
00
Definition of LT Integral of a derivative
= derivative of integral
Gathering the bits together 0=+
xsWx
W
= sxdxWdW
which we solve to get:
( ) ( ) 2/sx2
escs,xW=To find
Note that our separation of variables approach also gave the exponential term; but
with a problematic constant here we see that it is theLaplace variable s
8/4/2019 Partial Differential Equation Using Laplace
8/15
Applying the final boundary condition: ( ) tt,w =0
( )2
10ss,W =Treating this as a function oft, we
can take its LT
And, we found ( ) ( ) 2/sx2
escs,xW=
Evidently, ( ) 21
ssc =
( )
= 22 x
2
1tux
2
1tt,xwConsulting HLT or Kreysig
p296 line 11, we find finally
Note the general form is still a function of2
2xt
But the inclusion of the discontinuous step resolves our earlier difficulties
with boundary conditions
8/4/2019 Partial Differential Equation Using Laplace
9/15
Back to waggling the semi-infinite string
=otherwise
if
0
20sin),0(
tttw
*Kreysig, 8th Edn, page 644
Find the displacement w(x,t) of an elastic string subject to the following conditions:
1. The semi-infinite string is initially at rest, on the x-axis
2. For some time t>0the left hand end of the string (x=0) is moved sinusoidally:
3. Furthermore
0,0),(lim =
ttxwx
for
We will tackle this problem using the Laplace Transform
** just in this part of the notes, we use w(x,t) for the
PDE, rather than u(x,t) because u(t) is conventionallyassociated with the step function
8/4/2019 Partial Differential Equation Using Laplace
10/15
To solve the wave equation
0
,0)0,(
00),(lim)(),0(
0
2
22
2
2
=
===
=
=
t
x
t
w
xw
ttxwtftw
x
wc
t
w
conditionsinitial)(forandtosubject
=
=2
22
0
2 )0,()(
:
x
wLc
t
wxswsWs
t
t
torespectwithLTthetakewe,FIRST
The initial conditions mean that the second and third terms drop out
8/4/2019 Partial Differential Equation Using Laplace
11/15
Recall formula for LT
2
2
2
2
0
2
2
2
2
2
2
0
2
2
)(),(x
WwL
xdttxwe
xx
wL
dtx
we
x
wL
st
st
=
=
=
=
:ationdifferentiandnintegratioofordertheExchanging
c
sx
c
sx
esBesAsxW
W
c
s
x
W
x
WcWs
+=
=
=
)()(),(
02
2
2
2
2
222
so
:thatfollowsIt
8/4/2019 Partial Differential Equation Using Laplace
12/15
Apply the boundary condition
csx
x
stst
xx
esFsxW
sFsBsW
sA
dttxwedttxwesxW
sWtfLsF
/
00
)(),(
)()(),0(
0)(
0),(lim),(lim),(lim
),0())(()(
=
==
=
===
==
soand
:againationdifferenti&nintegratioExchanging
+
8/4/2019 Partial Differential Equation Using Laplace
13/15
Heat equation example usingLaplace Transform
x0
We consider a semi-infinite
insulated bar which is initially at aconstant temperature, then the end
x=0is held at zero temperature.
2
22
x
w
ct
w
=
We are to solve the Diffusion Equation:
Subject to the initial and boundary conditions:
==
xtxw
tw
Txw
as0),(
0),0(
)0,( 0
8/4/2019 Partial Differential Equation Using Laplace
14/15
Applying the Laplace Transform
0
2
22
)0,(
TsW
xwsW
t
wL
x
wLc
=
=
=
2
2
2
2
xW
xwL
=
As usual:
02
22
02
22
TsWx
WcTsW
x
Wc =
=
and so
sTesBesAsxW
s
TsxW
esBesAsxW
xc
s
xc
s
xcsx
cs
0
0
)()(),(
),(
)()(),(
++=
=
+=
isSolution
:PI
:CFhaveWe
Evidently, B(s)=0fromthe third condition
8/4/2019 Partial Differential Equation Using Laplace
15/15
Boundary condition
s
TesAsxWxc
s
0)(),( +=
0)(
0),0(,0),0(
0 =+
==
s
TsA
sWtw
haveweSince
sc
x
es
T
s
TsxW
= 00),(
We have:
and so
= tc
x
erfTtxw 2),( 0The final solution is:
=
=
=
t
a
erft
a
erfcs
e
L
sL
sa
212
11
1
1Inverse Laplace Transforms:
where erfis the error function*
*see http://mathworld.wolfram.com/Erf.html