Partial Differential Equation Using Laplace

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


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?


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


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:


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

====


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?


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


8/15

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


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


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


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


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

+


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


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


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

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Partial Differential Equation Using Laplace