1

Week 10

5. Applications of the LT to PDEs (continued)

Example 1:

,0for02

22

2

2

xx

uc

t

u

,0at)( xtfu

.0at0,0

tt

uu

This problem describes propagation of a signal generated at the end of a semi-infinite string.

(2)

(3)

(1)

Solve the following initial-boundary-value problem:

2

Solution:

Take the LT of Eq. (1) and use IC (3):

,02

222

x

UcUs

),(),0( sFsU

.exp)(exp)(),(

c

sxsB

c

sxsAsxU

hence,

(5)

(4)

hence, (4) yields

Take the LT of BC (2):

).()()( sFsBsA

Where do we get another condition to determine A(s) and B(s)?...

3

It can be seen from (4), that U(x, s) may grow as x → +∞, so we should make sure that it doesn’t!

,0)( sA

Evidently, the behaviour of (4) as x → +∞ depends on the sign of Re s... so, what should we assume it to be?

after which (4)-(5) yield

.exp)(),(

c

sxsFsxU

Given that the path of integration in the inverse LT can be moved arbitrarily to the right, we can safely assume that Re s > 0. Hence, (4) is bounded as x → +∞ only if

Take the inverse LT...

4

,exp)(),( 1

c

sxsFtxu L

hence, using the 2nd Shifting Theorem with a = x/c,

),/u()]([),( /1 cxtsFtxu cxtt

Lhence,

)./u()/(),( cxtcxtftxu

Example 2:

,0for02

2

xx

u

t

u

Solve the following initial-boundary-value problem:

5

),(),0( tftu

.0,0at0 xtu

This problem describes spreading of heat in a half-space from a source at the boundary.

Solution:

The usual routine yields

,e)(),( xssFsxU

hence,

.e)( xssW

)],()([),( 1 sWsFtxu Lwhere

(6)

6

Rearranging (6) using the convolution theorem, we obtain

),()()]()([),( 1 twtfsWsFtxu L

Comment:

where f(t) is a given function (the BC) and

.4

exp4

][e2

3

1

t

x

t

xxsL

We shall use a formula from Q4c of TS6:

].[e)( 1 xstw L

(8)

(7)

7

Summarising (7)-(8), we obtain

.d)(4

exp)(4

)(),(0

2

3

t

t

x

t

xftxu (9)

8

).u(4

exp4

),(2

3t

t

x

t

xtxG

Comment:

,d),()(),(0

txGftxu

where

Re-write (9) in the form (observe the highlighted parts)

,d)u()(4

exp)(4

)(),(0

2

3

tt

x

t

xftxu

or, equivalently,

G(x, t) is called the Green’s function of this problem.

9

Comment:

Generally, a Green’s function G provides means to represent the solution of a problem by a convolution integral of G with the function describing the boundary or initial condition. In the latter case, the solution has the form

,d),()(),( b

a

txGftxu

where (a, b) is the domain where the problem is to be solved.