8. Some Additional Examples
In addition to the Fourier transform and eigenfunction expansions, it is sometimes
convenient to have the use of the Laplace transform for solving certain problems in partial
differential equations. We will quickly develop a few properties of the Laplace transform and
use them in solving some example problems.
Laplace Transform
Definition of the Transform
Starting with a given function of t, fÝtÞ, we can define a new function !fÝsÞ of the variable s.
This new function will have several properties which will turn out to be convenient for
purposes of solving linear constant coefficient ODE’s and PDE’s. The definition of !fÝsÞ is as
follows:
Definition Let fÝtÞ be defined for t ³ 0 and let the Laplace transform of fÝtÞ be defined by,
K
LßfÝtÞà = X e ?st fÝtÞ dt = !fÝsÞ
0
For example:
fÝtÞ = 1, -t ³ 0,
fÝtÞ = e bt , -t ³ 0,
?st
K
1
!
Lß1à = X e ?st dt = e?s | t=K
t=0 = s = fÝsÞ for s > 0
0
?Ýb?sÞt
K
Lße bt à = X e ?Ýb?sÞt dt = e
| t=K = 1 = !fÝsÞ, for s > b.
0
s?b
Ýs ? bÞ t=0
The Laplace transform is defined for all functions of exponential type. That is, any function
fÝtÞ which is
(a) piecewise continuous = has at most finitely many finite jump discontinuities on any
interval of finite length
(b) has exponential growth: for some positive constants M and k
|fÝtÞ| ² M e kt for all t ³ 0, .
Properties of the Laplace Transform
The Laplace transform has the following general properties:
1. Linearity
2. Homogeneity
LßC 1 fÝtÞ + C 2 gÝtÞà = C 1 !fÝsÞ + C 2 ĝÝsÞ
LßfÝatÞà = 1a !f as
3. Transform of the Derivative
for
a>0
Lßf v ÝtÞà = s !fÝsÞ ? fÝ0Þ
Lßf vv ÝtÞà = s 2 !fÝsÞ ? sfÝ0Þ ? f v Ý0Þ etc
1
4. Derivative of the Transform
Lßt fÝtÞà = ? !f v ÝsÞ
Lßt 2 fÝtÞà = Ý?1Þ 2 !f vv ÝsÞ etc
Some Special Transforms
There are some transform pairs that are useful in solving problems involving the heat
equation. The derivations are given in an appendix.
k e ?k 2 /4t , t > 0
4^t 3
1 e ?k 2 /4t , t > 0
^t
ÝS.1Þ
fÝtÞ =
ÝS.2Þ
fÝtÞ =
ÝS.3Þ
fÝtÞ = erf c
k
2 t
k³0
LßfÝtÞà = 1s e ?k s ,
k>0
erfÝzÞ =
and
k>0
1 e ?k s ,
s
LßfÝtÞà =
, t>0
erf cÝzÞ = 1 ? erfÝzÞ
Here
LßfÝtÞà = e ?k s ,
2 X z e ?x 2 dx.
^ 0
Additional Properties of the Transform
Let fÝtÞ be a function of exponential type and suppose that for some b > 0,
0
hÝtÞ =
if 0 < t < b
fÝt ? bÞ
if
t>b
Then hÝtÞ is just the function fÝtÞ, delayed by the amount b. Then
K
K
0
b
LßhÝtÞà = X hÝtÞ e ?st dt = X fÝt ? bÞ e ?st dt
Let z = t ? b so that
K
K
0
0
LßhÝtÞà = X fÝzÞ e ?sÝz+bÞ dz = e ?bs X fÝzÞ e ?sz dz = e ?bs !fÝsÞ.
If we define
HÝt ? bÞ =
0 if 0 < t < b
1 if t > b
then
hÝtÞ = HÝt ? bÞ fÝt ? bÞ
and we find
5. Transform of a Delay
LßHÝt ? bÞ fÝt ? bÞà = e ?bs !fÝsÞ, for b > 0.
2
A related results is the following
K
K
Lße bt fÝtÞà = X e bt fÝtÞ e ?st dt = X fÝtÞ e ?Ýs?bÞt dt = !fÝs ? bÞ.
0
0
i.e.,
Lße bt fÝtÞà = !fÝs ? bÞ
6. Delay of a Transform
Results 5 and 6 assert that a delay in the function induces an exponential multiplier in the
transform and, conversely, a delay in the transform is associated with an exponential
multiplier for the function.
A final property of the Laplace transform asserts that
LßÝf D gÞÝtÞà = !fÝsÞ ĝÝsÞ
t
Ýf D gÞÝtÞ := X fÝt ? bÞ gÝbÞ db
7. Inverse of a Product
where
0
The product, Ýf D gÞÝtÞ , is called the convolution product of f and g. Life would be simpler
if the inverse Laplace transform of !fÝsÞ ĝÝsÞ was the pointwise product fÝtÞ gÝtÞ, but it isn’t, it
is the convolution product. The convolution product has some of the same properties as the
pointwise product, namely
Ýf D gÞÝtÞ = Ýg D fÞÝtÞ
and
Ýh D Ýf D gÞÞÝtÞ = ÝÝh D fÞ D gÞÝtÞ.
We will not give the proof of the result 7 but will make use of it nevertheless.
Applications to PDE’s
1. Consider the IBVP
x > 0, t > 0,
/ t uÝx, tÞ = k / xx uÝx, tÞ
uÝx, 0Þ = 0,
x > 0,
uÝ0, tÞ = fÝtÞ,
fÝ0Þ = 0,
t > 0.
Let
ÛÝx, sÞ = LßuÝx, tÞà
Laplace transform in t.
Then
sÛÝx, sÞ ? 0 = k U ”Ýx, sÞ,
ÛÝ0, sÞ = !fÝsÞ.
x > 0,
2
U ”Ýx, sÞ = d 2 ÛÝx, sÞ
dx
Solving this ODE in x, we find
ÛÝx, sÞ = Ae ?x
s/k
+ Be x
s/k
x > 0.
We want ÛÝx, sÞ to remain bounded for all positive x, which requires that B = 0. Then the
boundary condition at x=0 leads to
3
ÛÝx, sÞ = !fÝsÞ e ?x
s/k
x > 0.
,
Then ÝS.1Þ together with property 7 of the Laplace transform, gives
uÝx, tÞ = f D KÝx, 6Þ = X
2
x
e ?x /4kÝt?bÞ fÝbÞ db.
4^kÝt ? bÞ 3
t
0
as the unique solution of the IBVP. Suppose now that we wish to compute the flux through
x=0,
Flux at 0 = ?k/ x uÝ0, tÞ.
Differentiating the integral expression for u does not seem like a pleasant prospect.
However, note that
Lß?k/ x uÝx, tÞà = ?k/ x ÛÝx, sÞ =
ks !fÝsÞ e ?x
Flux at 0 = ?k/ x uÝ0, tÞ = L ?1
ks !fÝsÞ .
and
s/k
,
In order to use our inversion formulas, we write
ks !fÝsÞ =
k !
s fÝsÞ
s
and, recalling that fÝ0Þ = 0, we have s !fÝsÞ = Lßf v ÝtÞà. In addition, ÝS.2Þ with k=0, gives
L ?1
1
s
1
^t
=
,
and so,
Flux at 0 = L ?1
k !
s fÝsÞ
s
=
k t
X
^ 0
f v ÝbÞ
db.
Ýt ? bÞ
Note that if fÝtÞ = At, then
Flux at 0 =
Problem 1
k t
X
^ 0
A
db = 2A kt
^ .
Ýt ? bÞ
Show that if vÝx, tÞ solves
/ t vÝx, tÞ = k / xx vÝx, tÞ
vÝx, 0Þ = 0,
?k/ x vÝ0, tÞ = gÝtÞ,
x > 0, t > 0,
x > 0,
t > 0,
then
4
vÝ0, tÞ = X
t
0
and if gÝtÞ = B, then
2. Consider the IBVP
gÝbÞ
db,
^kÝt ? bÞ
vÝ0, tÞ = 2B
t > 0,
t .
^k
x > 0, t > 0,
/ tt uÝx, tÞ = a 2 / xx uÝx, tÞ
x > 0,
uÝx, 0Þ = / t uÝx, 0Þ = 0,
uÝ0, tÞ = fÝtÞ,
t > 0.
Let
ÛÝx, sÞ = LßuÝx, tÞà
Laplace transform in t.
Then
s 2 ÛÝx, sÞ ? 0 = k U ”Ýx, sÞ,
ÛÝ0, sÞ = !fÝsÞ.
x > 0,
Solving this ODE in x, we find a general solution of the form,
x
x
ÛÝx, sÞ = Ae a s + Be ? a s ,
and both
x
ÛÝx, sÞ = !fÝsÞ e a s
x
ÛÝx, sÞ = !fÝsÞ e ? a s
solve the transformed equation and the boundary condition at x=0. To see how to choose
x
the correct solution, recall that for x > 0, property 5 implies that !fÝsÞ e ? a s is the transform of
x
fÝtÞ delayed by the amount ax > 0. On the other hand, !fÝsÞ e a s is the transform of fÝtÞ
advanced in time by the amount ax > 0. Another way to say this is to say
x
uÝx, tÞ = L ?1 !fÝsÞ e ? a s
= f t ? ax H t ? ax ,
represents the wave form fÝ6Þ propagating from L to R into the region áx > 0â while
x
uÝx, tÞ = L ?1 !fÝsÞ e a s
= f t + ax H t + ax
represents the wave form fÝ6Þ propagating from R to L out of the region áx > 0â. Then the
solution that is relevant for our problem is the wave that travels from L to R into the region
áx > 0â.
3. Consider the IBVP
/ t uÝx, tÞ = K / xx uÝx, tÞ
uÝx, 0Þ = 0,
uÝ0, tÞ = fÝtÞ
/ x uÝ1, tÞ = 0
0 < x < 1, t > 0,
0 < x < 1,
t > 0.
5
Here, we may use the Laplace transform, or if we prefer, we can use eigenfunction
expansion after a suitable modification of the problem. We will solve the problem first by
this means. Since the boundary conditions are not homogeneous, the method does not
apply directly but if we let
vÝx, tÞ = uÝx, tÞ ? fÝtÞ
then
/ t vÝx, tÞ = / t uÝx, tÞ ? f v ÝtÞ
= K / xx uÝx, tÞ ? f v ÝtÞ
= K / xx vÝx, tÞ ? f v ÝtÞ,
vÝx, 0Þ = uÝx, 0Þ ? fÝ0Þ = 0,
vÝ0, tÞ = uÝ0, tÞ ? fÝtÞ = 0,
/ x vÝ1, tÞ = / x uÝ1, tÞ = 0.
Now the method of eigenfunction expansion applies directly to the problem for vÝx, tÞ since
the boundary conditions are homogeneous. Since vÝ0, tÞ = / x vÝ1, tÞ = 0, it follows that the
eigenfunctions are the eigenfunctions of example 6.3, namely
Vn =
We write
n?
1
2
2
^ 2 = W 2n ,
K
vÝx, tÞ = > v n ÝtÞ d n ÝxÞ
d n ÝxÞ = sin n ?
and
n=1
1
2
^x = sin W n x,
n = 1, 2, ...
K
f v ÝtÞ = f v ÝtÞ 1 = f v ÝtÞ > C n d n ÝxÞ,
n=1
where
1
X sin n ?
Ý1, d n Þ
Cn =
= 10
Ýd n , d n Þ
X sin 2 n ?
0
1
2
1
2
^xdx
^xdx
=
4
= W2n .
Ý2n ? 1Þ^
Substituting these expansions into the IBVP for vÝx, tÞ, we conclude that
v vn ÝtÞ + KW 2n v n ÝtÞ = ?f v ÝtÞ C n ,
and
t
v n Ý0Þ = 0,
-n.
v n ÝtÞ = ?C n X e ?KW n Ýt?sÞ f v ÝsÞ ds.
2
0
For example, if fÝtÞ = mt, then
t
2
2
v n ÝtÞ = ?mC n X e ?KW n Ýt?sÞ ds. = mC2n e ?KW n t ? 1
0
KW n
and
K
2
uÝx, tÞ = fÝtÞ + > mC2n e ?KW n t ? 1 d n ÝxÞ,
n=1 KW n
K
2
= mt + 2m > 13 e ?KW n t ? 1 d n ÝxÞ.
K n=1 W n
Alternatively, we may use the Laplace transform to solve this same problem. Let ûÝx, sÞ
denote the Laplace transform of uÝx, tÞ. Then,
6
ûÝ0, sÞ = !fÝsÞ, û v Ý1, sÞ = 0.
s ûÝx, sÞ ? 0 = Kû”Ýx, sÞ,
The general solution of the equation may be written as
ûÝx, sÞ = A exp ?x s/K
+ B exp x s/K ,
and then the boundary conditions lead to
A=
exp
exp ? s/K
s/K
+ exp
!fÝsÞ,
s/K
B=
exp ? s/K
exp ? s/K
+ exp
s/K
!fÝsÞ.
Then
exp ?Ýx ? 1Þ s/K
ûÝx, sÞ =
exp ? s/K
+ exp
exp ?x s/K
=
exp ?2 s/K
exp ?Ý1 ? xÞ s/K
!fÝsÞ +
s/K
exp ? s/K
+ exp
s/K
!fÝsÞ,
!fÝsÞ + exp ?Ý2 ? xÞ s/K !fÝsÞ,
+1
exp ?2 s/K + 1
and since the formula for the sum of a geometric series implies that,
K
1
exp ?2 s/K
this becomes
+1
= >Ý?1Þ n exp ?2n s/K ,
n=0
K
ûÝx, sÞ = !fÝsÞ >Ý?1Þ n exp ?Ýx + 2nÞ s/K
n=0
K
+ !fÝsÞ >Ý?1Þ n exp Ý2n + 2 ? xÞ s/K .
n=0
It follows from (S.1) that
L ?1 exp ? x
K
s
=
2
x
exp ? x
4Kt
4^Kt 3
t > 0,
:= GÝx, KtÞ,
Then, using property 7, we find
K
K
t
t
uÝx, tÞ = >Ý?1Þ n X GÝx + 2n, KÝt ? bÞÞ fÝbÞ db + >Ý?1Þ n X GÝ2n + 2 ? x, KÝt ? bÞÞ fÝbÞ db
n=0
0
n=0
0
t
t
t
0
0
0
= X GÝx, KÝt ? bÞÞ fÝbÞ db + X GÝ2 ? x, KÝt ? bÞÞ fÝbÞ db ? X GÝx + 2, KÝt ? bÞÞ fÝbÞ db
Notice that this representation for the solution looks nothing at all like the eigenfunction
7
expansion obtained previously. However, this problem has a unique solution so the two
representations must produce identical results. What can, in fact, be seen is that since each
representation involves an infinite series of which only a finite number of terms can actually
be computed, each representation produces only an approximate solution. Moreover, it can
be shown that the Laplace transform approximation is most accurate at small values of t
while the eigenfunction expansion is more accurate as t grows large.
Appendix
Some Laplace Transform Formulas
We will derive some Laplace transform formulas which are useful in solving problems
involving the heat equation.
fÝtÞ = L ?1 !fÝsÞ , then
Lemma 1 If
L ?1 !fÝ s Þ
=
K
1
X 0 z e ?z 2 /4t fÝzÞ dz.
4^t 3
Proof- if the lemma holds, then
!fÝ s Þ = X K e ?st
0
=X
K
0
K
1
X 0 z e ?z 2 /4t fÝzÞ dz dt
4^t 3
1 fÝzÞ X K z e ?z 2 /4t e ?st dt dz.
0
^
4t 3
z
then
db = ? 12 z dt
4t
4t 3
2
t = z 2 and as t goes from 0 to K, b goes from K to 0.
4b
b=
Let
Then
!fÝ s Þ = X K 1 fÝzÞ X K 2e ?z 2 s/4b 2 e ?b 2 db dz
0
0
^
=
2 X K fÝzÞ
^ 0
^ ?z
e
2
s
K
dz = X fÝzÞe ?z s dz.
0
Here we used the following result,
K
X 0 e ?z 2 s/4b 2 e ?b 2 db =
^ ?z s
e
.n
2
Applications of the Lemma
We have that
for fÝtÞ = NÝt ? aÞ
!fÝsÞ = e ?as .
Then, by lemma 1,
8
L ?1 !fÝsÞ
= L ?1 e ?a
L ?1 e ?a
s
K
1
X 0 z e ?z 2 /4t NÝz ? aÞ dz
4^t 3
=
s
1 a e ?a 2 /4t
4^t 3
=
for a ³ 0.
(A.1)
Now, if we integrate both sides of (A.1) with respect to the parameter a from b to K,
K
L ?1 X e ?a s da
K
1
X b a e ?a 2 /4t da,
4^t 3
=
b
we get
L ?1
1 e ?b
s
=
s
1 e ?b 2 /4t ,
^t
b ³ 0.
(A.2)
In particular, for b = 0,
1
s
L ?1
=
1 .
^t
Integrate both sides of (A.2) with respect to the parameter b from a to K,
L ?1 X
Let
K
1 e ?b s db
s
a
L ?1 X
so that
K
a
1 e ?b s db
s
erfcÝxÞ =
If we define
1 X K e ?b 2 /4t db.
^t a
dY = db
4t
b
4t
Y=
=
= L ?1 1s e ?a
s
=
2
2 XK
a e ?Y dY.
^
4t
2 X K e ?Y 2 dY,
^ x
then
L ?1 1s e ?a
s
= erfc
a
4t
(A.3)
Proceeding, we integrate both sides of (A.3) with respect to the parameter a from b to K,
L ?1 X
K
b
1 e ?a s da
s
K
= X erfc
b
a
4t
da.
9
Y=
We let
a
4t
dY = da
4t
and then,
L ?1 X
K
b
1 e ?a s da
s
= L ?1
s
1 e ?b
3/2
K
4t X b erfcÝYÞ dY.
4t
s
=
=
4t ierfc
That is,
L ?1
Here
1 e ?b
s 3/2
s
b
4t
(A.4)
K
ierfcÝxÞ = X erfcÝYÞ dY
x
denotes the so called, iterated complementary error function. It can be shown that
ierfcÝxÞ =
1 e ?x 2 ? x erfcÝxÞ.
^
so the iterated complementary error function can be expressed in terms of other functions.
Finally, there are two Laplace transform pairs that are obtainable by elementary means:
L ?1
1
sÝs + cÞ
= 1c Ý1 ? e ?ct Þ
L ?1
e ?as
sÝs + cÞ
= 1c Ý1 ? e ?cÝt?aÞ ÞHÝt ? aÞ := F a ÝtÞ.
and
Then lemma 1 implies
L ?1
e ?a s
s Ý s + cÞ
=
=
K
1
X 0 z e ?z 2 /4t F a ÝzÞ dz.
4^t 3
K
1
X a z e ?z 2 /4t 1c Ý1 ? e ?cÝz?aÞ Þ dz.
4^t 3
Then integration by parts leads to
K
K
X a z e ?z 2 /4t 1c Ý1 ? e ?cÝz?aÞ Þ dz = 2te ac X a e ?cz e ?z 2 /4t dz
so
L ?1
e ?a s
s Ý s + cÞ
ac
K
2
= e X e ?cz e ?z /4t dz,
^t a
10
ac
K
2
2 2
2
= e X e ?Ýz +4zct+4c t Þ/4t e ?c t dz,
a
^t
2
ac+ct
K
X a e ?Ýz+2ctÞ 2 /4t dz,
= e
^t
=
2 e ac+ct 2 X K
a
^
4t
2
+c t
e ?Y dY
i.e.,
L ?1
e ?a s
s Ý s + cÞ
2
= e ac+ct erfc
a +c t
4t
.
11