Solving PDEs using Laplace Transforms, Chapter 15
Given a function u(x, t) defined for all t > 0 and assumed to be bounded we can apply the
Laplace transform in t considering x as a parameter.
Z ∞
e−st u(x, t) dt ≡ U (x, s)
L(u(x, t)) =
0
In applications to PDEs we need the following:
Z ∞
e−st ut (x, t) dt = e−st u(x, t)
L(ut (x, t) =
0
∞
0
∞
Z
e−st u(x, t) dt = sU (x, s) − u(x, 0)
+s
0
so we have
L(ut (x, t) = sU (x, s) − u(x, 0)
In exactly the same way we obtain
L(utt (x, t) = s2 U (x, s) − su(x, 0) − ut (x, 0).
We also need the corresponding transforms of the x derivatives:
Z ∞
e−st ux (x, t) dt = Ux (x, s)
L(ux (x, t)) =
0
Z
∞
L(uxx (x, t)) =
e−st uxx (x, t) dt = Uxx (x, s)
0
Consider the following examples.
Example 1.
∂u ∂u
+
= x,
∂x
∂t
with boundary and initial condition
u(0, t) = 0 t > 0,
x > 0, t > 0,
and u(x, 0) = 0, x > 0.
As above we use the notation U (x, s) = L(u(x, t))(s) for the Laplace transform of u.
Then applying the Laplace transform to this equation we have
x
dU
(x, s) + sU (x, s) − u(x, 0) =
dx
s
dU
x
(x, s) + sU (x, s) = .
dx
s
⇒
This is a constant coefficient first order ODE. We solve it by finding the integrating factor
R
µ=e
Thus we have
sdx
= esx
d sx
x
[e U (x, s)] = esx .
dx
s
We integrate both sides to get
e−sx
U (x, s) =
s
Z
1
e r dr + Ce−sx .
sr
We can use integration by parts to evaluate the integral:
Z
Z sx 0
e
sx
x dx
e x dx =
s
Z sx xesx
e
=
−
dx
s
s
xesx esx
− 2.
s
s
So we have
e−sx
U (x, s) =
s
xesx esx
− 2
s
s
+ Ce−sx =
x
1
− 3 + Ce−sx .
2
s
s
We can evaluate the constant C using the boundary condition
0 = U (0, s) = −
so we have
1
+C
s3
⇒ C=
1
s3
1
e−sx
x
−
+
.
s2 s3
s3
U (x, s) =
Taking the inverse Laplace transform we have
u(x, t) = xt −
(t − x)2
t2
+ H(t − x)
2
2
where H is the unit step function (or Heaviside function)
(
0, x < 0
H(x) =
.
1, x ≥ 0
Example 2.
∂u ∂u
+
+ u = 0,
∂x
∂t
with boundary and initial condition
u(0, t) = 0 t > 0,
x > 0, t > 0,
and u(x, 0) = sin(x), x > 0.
As above we use the notation U (x, s) = L(u(x, t))(s) for the Laplace transform of u.
Then applying the Laplace transform to this equation we have
dU
(x, s) + sU (x, s) − u(x, 0) + U (x, s) = 0 ⇒
dx
dU
(x, s) + (s + 1)U (x, s) = sin(x).
dx
This is a constant coefficient first order linear ODE. We solve it by finding the integrating factor
µ=e
Thus we have
R
(s+1)dx
= e(s+1)x
d (s+1)x
e
U (x, s) = e(s+1)x sin(x).
dx
We integrate both sides to get
−(s+1)x
U (x, s) = e
Z
e
(s+1)r
2
sin(r) dr + Ce−(s+1)x .
We can use integration by parts to evaluate the integral:
Z x
(s + 1) sin(x) − cos(x)
(s+1)r
−(s+1)x
e
sin(r) dr =
.
e
s2 + 2s + 2
0
So we have
(s + 1) sin(x) − cos(x)
+ Ce−(s+1)x .
s2 + 2s + 2
We can evaluate the constant C using the boundary condition
U (x, s) =
0 = U (0, s) =
s2
−1
+C
+ 2s + 2
⇒ C=
s2
1
.
+ 2s + 2
So we have
(s + 1) sin(x) − cos(x) + e−(s+1)x)
.
s2 + 2s + 2
Taking the inverse Laplace transform we have
U (x, s) =
u(x, t) = e−t cos(t) sin(x) − e−t sin(t) cos(t) + e−t H(t − x) sin(t − x)
This can be written as
u(x, t) = e−t [sin(x − t) + H(t − x) sin(t − x)] .
Example 3.
∂ 2u
∂u
(x, t) =
(x, t), 0 < x < 2, t > 0,
∂t
∂x2
u(0, t) = 0, u(2, t) = 0
u(x, 0) = 3 sin(2πx).
Take the Laplace transform and apply the initial condition
d2 U
(x, s) = sU (x, s) − u(x, 0) = sU (x, s) − 3 sin(2πx).
dx2
We write this equation as a non-homogeneous, second order linear constant coefficient equation for
which we can apply the methods from Math 3354.
d2 U
(x, s) − sU (x, s) = −3 sin(2πx).
dx2
The general solution can be written as
U (x, s) = Uh (x, s) + Up (x, s)
where Uh (x, s) is the general solution of the homogeneous problem
√
Uh (x, s) = c1 e
sx
+ c2 e−
√
sx
and Up (x, s) is any particular solution of the non-homogeneous problem
Up (x, s) = A cos(2πx) + B sin(2πx).
3
We first use the method of undetermined coefficients to find A and B. To this end we have
d
Up (x, s) = −2πA sin(2πx) + 2πB cos(2πx),
dx
d2
Up (x, s) = −(2π)2 A cos(2πx) + (2π)2 B sin(2πx).
2
dx
Therefore
d2
Up (x, s) − sUp (x, s)
dx2
= (−(2π)2 − s)[A cos(2πx) + B sin(2πx)]
= −3 sin(2πx).
From this we conclude that
−(s + (2π)2 )A = 0,
and
so that
A = 0,
B=
− (s + (2π)2 )B = −3,
3
.
s + 4π 2
Now we have the general solution
√
U (x, s) = c1 e
sx
√
+ c2 e −
sx
+
3
sin(2πx)
(s + 4π 2 )
We note the the Laplace transforms of the boundary conditions give
u(0, t) = 0 ⇒ U (0, s) = 0,
and u(2, t) = 0 ⇒ U (2, s) = 0
So we have
√
0 = U (0, s) = c1 + c2 ,
0 = U (2, s) = c1 e
s2
+ c2 e−
√
s2
which gives c1 = c2 = 0 and we have
U (x, s) =
3
sin(2πx).
(s + 4π 2 )
To find our solution we apply the inverse Laplace transform
3
2
−1
u(x, t) = L
sin(2πx) = 3e−4π t sin(2πx).
2
(s + 4π )
Just as we would have obtained using eigenfunction expansion methods.
Example 4. Next we consider a similar problem for the 1D wave equation.
2
∂ 2u
2∂ u
(x, t) = c
(x, t) + sin(πx),
∂t2
∂x2
u(x, 0) = 0, ut (x, 0) = 0
u(0, t) = 0 u(1, t) = 0.
0 < x < 1, t > 0,
Taking the Laplace transform and applying the initial conditions we obtain
d2 U
sin(πx)
sin(πx)
2
2
(x,
s)
=
s
U
(x,
s)
−
su(x,
0)
−
u
(x,
0)
−
=
s
U
(x,
s)
−
.
t
dx2
s
s
4
We need to solve the constant coefficient non-homogeneous ODE
d2 U
sin(πx)
(x, s) − s2 U (x, s) = −
2
dx
s
Once again we know that
U (x, s) = Uh (x, s) + Up (x, s)
where Uh (x, s) is the general solution of the homogeneous problem
Uh (x, s) = c1 esx + c2 e−sx
and Up (x, s) is any particular solution of the non-homogeneous problem
Up (x, s) = A cos(πx) + B sin(πx).
We apply the method of undetermined coefficients to find A and B. To this end we have
d
Up (x, s) = −πA sin(πx) + πB cos(πx),
dx
d2
Up (x, s) = −π 2 A cos(πx) + π 2 B sin(πx).
dx2
Therefore
d2
Up (x, s) − s2 Up (x, s)
2
dx
= (−π 2 − s2 )[A cos(πx) + B sin(πx)]
sin(πx)
.
=−
s
From this we conclude that
−(s2 + π 2 )A = 0,
and
1
− (s2 + π 2 )B = − ,
s
so that
A = 0,
B=
So we have
Up (x, s) =
s(s2
1
.
+ π2)
sin(πx)
s(s2 + π 2 )
and
U (x, s) = c1 esx + c2 e−sx +
sin(πx)
.
s(s2 + π 2 )
Next we apply the BCs to find c1 and c2 .
0 = U (0, s) = c1 + c2 ,
and 0 = U (1, s) = c1 es + c2 e−s
which implies c1 = 0 and c2 = 0. So we arrive at
U (x, s) =
sin(πx)
.
s(s2 + π 2 )
5
Finally we apply the inverse Laplace transform to obtain
1
−1
−1
sin(πx)
u(x, t) = L (U (x, s)) = L
s(s2 + π 2 )
1 −1 1
s
= 2 L
−
sin(πx)
π
s (s2 + π 2 )
1
= 2 (1 − cos(πt)) sin(πx).
π
Here we have done partial fractions
1
a
bs + c
1
= + 2
= 2
2
2
2
s(s + π )
s (s + π )
π
1
s
− 2
s (s + π 2 )
.
Example 5. This example shows the real use of Laplace transforms in solving a problem we could
not have solved with our earlier work.
∂ 2u
∂u
(x, t) =
(x, t),
∂t
∂x2
u(x, 0) = f (x)
u(x, t) bounded.
−∞ < x < ∞, t > 0,
Under the assumption that u(x, t) is bounded we know that the Laplace transform exists and,
indeed, we have
Z ∞
Z ∞
M
−st
e−st dt =
e |u(x, t)| dt ≤ M
|u(x, t)| ≤ M ⇒ |U (x, s)| ≤
.
s
0
0
Applying the Laplace transform we obtain
d2 U
(x, s) = sU (x, s) − u(x, 0) = sU (x, s) − f (x).
dx2
We write this equation as a non-homogeneous, second order linear constant coefficient equation.
d2 U
(x, s) − sU (x, s) = −f (x).
dx2
The general solution can be written as
U (x, s) = Uh (x, s) + Up (x, s)
where Uh (x, s) is the general solution of the homogeneous problem
√
Uh (x, s) = c1 e
sx
+ c2 e−
√
sx
and Up (x, s) is any particular solution of the non-homogeneous problem. We find
it using √the
√
sx
method of variation of parameters from Math 3354. For this method we use U1 = e , U2 = e− sx .
W (U1 , U2 ) =
√
U1 (x, s) U2 (x, s)
= −2 s
0
0
U1 (x, s) U2 (x, s)
6
Z
x
[−U1 (x, s)U2 (ξ, s) + U2 (x, s)U1 (ξ, s])(−f (ξ))
dξ
W (ξ, s)
0
Z xh √
√
√
√ i
1
sx − sξ
− sx
−e e
= √
+e
e sξ f (ξ) dξ
2 s 0
√
√
Z
Z
e sx x −√sξ
e− sx x √sξ
=− √
e
e f (ξ) dξ
f (ξ) dξ + √
2 s 0
2 s 0
Up (x, s) =
So the general solution can be written as
Z x √
Z x √
√
√
1
1
− sξ
sx
sξ
e
e f (ξ) dξ e− sx .
U (x, s) = c1 − √
f (ξ) dξ e
+ c2 + √
2 s 0
2 s 0
Recall our assumption that u(x, t) be bounded for all −∞ < x < ∞ implies that U (x, s) is also
bounded for all −∞ < x < ∞ for any fixed s > 0.
Now in order that the first term in the general solution stays bounded as x → ∞ we need
Z x √
1
− sξ
f (ξ) dξ = 0
e
lim c1 − √
x→∞
2 s 0
which implies
1
c1 = √
2 s
Z
∞
√
e−
sξ
f (ξ) dξ.
0
In exactly the same way we must have
Z x √
1
sξ
lim c2 + √
e f (ξ) dξ = 0
x→−∞
2 s 0
which implies
−∞ √
Z
1
c2 = − √
2 s
e
0
sξ
1
f (ξ) dξ = √
2 s
Z
Z
x
0
e
√
sξ
f (ξ) dξ.
−∞
Thus
U (x, s) =
1
√
2 s
+
Z
∞
√
− sξ
e
0
1
√
2 s
Z
0
√
− sξ
e
−∞
1
f (ξ) dξ − √
2 s
e
√
− sξ
√
f (ξ) dξ e
sx
0
1
f (ξ) dξ + √
2 s
Z
x √
e
sξ
√
f (ξ) dξ e−
0
√
−√sx Z x
Z
√
e sx ∞ −√sξ
e
sξ
√
√
e
f (ξ) dξ +
e f (ξ) dξ
=
2 s x
2 s −∞
Z ∞ √
1
= √
e− s|x−ξ| f (ξ) dξ
2 s −∞
We want to find the inverse Laplace transform
−√s|x−ξ| e
−1
√
L
.
2 s
7
sx
From our table we have
L
−1
√
2
e−a /(4t)
= √
πt
e−a s
√
s
e−|x−ξ| /(4t)
√
=
≡ K(|x − ξ|, t).
4πt
and if we set a = |x − ξ| then we have
−1
L
√
e− s|x−ξ|
√
2 s
2
So we have
−1
u(x, t) = L (U (x, s)) = L
Z
∞
=
L
−∞
−1
√
−1
e− s|x−ξ|
√
2 s
1
√
2 s
Z
f (ξ) dξ
Z ∞
1
2
=√
e−|x−ξ| /(4t) f (ξ) dξ
4πt −∞
Z ∞
=
K(|x − ξ|, t) f (ξ) dξ
−∞
2
e−x /(4t)
K(x, t) = √
4πt
is called the “Fundamental Heat Kernel”.
8
√
− s|x−ξ|
e
−∞
The function
∞
f (ξ) dξ
Table of Laplace Transforms
Z
f (t) for t ≥ 0
∞
fb = L(f ) =
e−st f (t) dt
0
1
1
s
eat
1
s−a
n!
tn
sn+1
(n = 0, 1, . . .)
ta
Γ(a + 1)
(a > 0)
sa+1
sin bt
b
s 2 + b2
cos bt
sinh bt
s2
s
+ b2
s2
a
− b2
cosh bt
s
s 2 − b2
f 0 (t)
sL(f ) − f (0)
f 00 (t)
s2 L(f ) − sf (0) − f 0 (0)
(−1)n
tn f (t)
dn F
(s)
dsn
eat f (t)



 0 t≤a
u(t − a) =


 1 t>a
L(f )(s − a)
u(t − a)f (t − a)
e−as L(f )(s)
u(t − a)g(t)
e−as L(g(t + a))(s)
δ(t − a)
e−as
Z
(f ∗ g)(t) =
e−as
s
t
f (t − τ )g(τ ) dτ
0
9
L(f ∗ g) = L(f )L(g)
The “error function” denoted by erf(x) is given by
Z x
2
2
erf(x) = √
e−x dx.
π 0
Notice that we can use the properties of integrals to deduce that
erf(−x) = − erf(x).
The complementary error function erfc(x) defined by
Z ∞
2
2
erfc (x) = √
e−s ds.
π x
Notice that
Z
2
erf(x) + erfc(x) = √
π
x
−x2
e
Z
∞
dx +
0
e
−s2
ds = 1.
x
Additional Laplace Transforms
√
2
e−a /(4t)
√
πt
e−a s
√
s
2
ae−a /(4t)
√
2 πt3
√
e−a
es
erf (t)
erfc
r
2
e
b2 t+ab
−e
a
√
2 t
2 /4
√
e−a
s
t −a2 /(4t)
a
e
− a erfc √
π
2 t
b2 t+ab
s
√
a
erfc b t + √
2 t
√
a
a
erfc b t + √
+ erfc √
2 t
2 t
10
erfc(s/2)
s
s
√
e−a s
√
s s
√
e−a s
√ √
s( s + b)
√
be−a s
√
s( s + b)