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 Z ∞ ∞ −st −st e ut (x, t) dt = e u(x, t) 0 + s L(ut (x, t) =
∞
e−st u(x, t) dt = sU (x, s) − u(x, 0)
0
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 . U1 (x, s) U2 (x, s) √ = −2 s W (U1 , U2 ) = 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)