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4.3 Inverse Laplace Transforms and Initial-Value Problems
In Section 4.2 we saw that the Laplace transform of the solution y = y(x) of the initial-value problem y′′^ + ay′^ + by = f (x); y(0) = α, y′(0) = β
is given by
L[y(x)] = Y (s) = F^ (s) + s (^2) +αs as^ + +^ β b+^ aα
where F (s) = L[f (x)] is the Laplace transform of f.
Now that we know L[y(x)], the obvious question is: What is y(x)? The general problem of finding a function with a given Laplace transform is called the inversion problem. The inversion problem and its application to solving initial-value problems is the topic of this section.
If f is continuous on [0, ∞), and if the Laplace transform, L[f (x)] = F (s) exists for s > λ, then the function F is uniquely determined by f ; that is, the operator L is itself a function. Our first result states that L is a one-to-one function. A proof of this result is beyond the scope of this introductory treatment.
THEOREM 1. If f and g are continuous functions on [0, ∞), and if L[f (x)] = L[g(x)], then f ≡ g; that is f (x) = g(x) for all x ∈ [0, ∞).
The following definition gives the terminology and notation used in treating the inversion problem.
DEFINITION If F (s) is a given transform and if the function f , continuous on [0, ∞), has the property that L[f (x)] = F (s), then f is called the inverse Laplace transform of F (s), and is denoted by f (x) = L−^1 [F (s)].
The operator L−^1 is called the inverse operator of L.
There is a general formula for the inverse operator L−^1 corresponding to (1), Section 4.1, but use of the formula requires a knowledge of complex-valued functions, a topic which is treated in more advanced courses.
The relationship between L and L−^1 is given by the following equations: L−^1 {L[f (x)]} = f (x)
L
L−^1 [F (s)]
= F (s)
for all functions f , continuous on [0, ∞), such that L[f (x)] = F (s).
For convenience here, we reproduce the table of Laplace transforms given at the end of Section 4.1.
Table of Laplace Transforms
f (x) F (s) = L[f (x)]
s , s > 0
eαx^1 s − α , s > α
cos βx s s^2 + β^2 , s > 0
sin βx β s^2 + β^2 ,^ s >^0 eαx^ cos βx s^ −^ α (s − α)^2 + β^2 , s > α
eαx^ sin βx β (s − α)^2 + β^2 ,^ s > α xn, n = 1, 2 ,... n! sn+^ , s > 0
xn^ erx, n = 1, 2 ,... (^) (s −n r!)n+1 , s > r
x cos βx s
(^2) − β 2 (s^2 + β^2 )^2 , s > 0
x sin βx 2 βs (s^2 + β^2 )^2 ,^ s >^0
A simple way to interpret Theorem 1 is that the table can be read either from left to right or from right to left. That is, the table is simultaneously a table of Laplace transforms and of inverse Laplace transforms.
Example 1. (a) If L[f (x)] = F (s) = 1 s − 4 , then f (x) = e^4 x.
(b) If L[f (x)] = F (s) = s s^2 + 9 ,^ then^ f^ (x) = cos 3x.
(c) If L[f (x)] = F (s) = s + 2 s^2 + 4s + 13 =^
s + 2 (s + 2)^2 + 9 =^
s − (−2) [s − (−2)]^2 + 9 ,^ then^ f^ (x) = e−^2 x^ cos 3x.
The properties of the Laplace transform operator L can be used to derive corresponding properties of its inverse operator L−^1. For our purposes, the most important property is that of linearity.
THEOREM 2. The operator L−^1 is linear; that is
L−^1 [F (s) + G(s)] = L−^1 [F (s)] + L−^1 [G(s)], and
To put 1 s^2 − 2 s + 10 in a form in the table, we complete the square in the denominator and “adjust” the numerator:
1 s^2 − 2 s + 10
s^2 − 2 s + 1 + 9
(s − 1)^2 + 9
=^1
(s − 1)^2 + 9
From the table,
L−^1
[
s^2 − 2 s + 10
]
3 L
− 1
[
(s − 1)^2 + 9
]
= 13 ex^ sin 3x.
Putting the two results together, we have
L−^1
[
(s − 3)^2 +^
s^2 − 2 s + 10
]
= 4xe^3 x^ + 13 ex^ sin 3x.
Example 4. Find L−^1 [F (s)] if
F (s) = 2 s^ + 1 s^2 − 2 s − 8
SOLUTION By factoring the denominator, we can write
F (s) = (^) s (^2 2) −s 2 + 1s − 8 = (^) (s + 2)(^2 s^ + 1s − 4).
Now, by partial fraction decomposition,
2 s + 1 (s + 2)(s − 4) =
3 2 s − 4 +
1 2 s + 2.
Therefore
L−^1
[
2 s + 1 s^2 − 2 s − 8
]
=^3
L−^1
[
s − 4
]
+^1
L−^1
[
s + 2
]
= 32 e^4 x^ + 12 e−^2 x.
Example 5. Find L−^1 [F (s)] if
F (s) = 2 s^ + 4 (s − 2)(s^2 − 4 s + 8)
SOLUTION The quadratic factor in the denominator cannot be factored into linear factors.
By partial fraction decomposition
F (s) = 2 s^ + 4 (s − 2)(s^2 − 4 s + 8)
s − 2
Next, we complete the square in the denominator of the second term: 2 s − 2
s − 2
- −^2 s^ + 6 s^2 − 4 s + 4 + 4
s − 2
Finally, we “adjust” the numerator of the second term so that we can use the Table: 2 s − 2 +^
− 2 s + 6 (s − 2)^2 + 4 =^
s − 2 +^
−2(s − 2) + 2 (s − 2)^2 + 4 =^
s − 2 +^
−2(s − 2) (s − 2)^2 + 4 +^
(s − 2)^2 + 4.
Now
L−^1
[
2 s + 4 (s − 2)(s^2 − 4 s + 8)
]
= L−^1
[
s − 2 − 2 s^ −^2 (s − 2)^2 + 4
(s − 2)^2 + 4
]
= 2 L−^1
[
s − 2
]
− 2 L−^1
[
s − 2 (s − 2)^2 + 4
]
+ L−^1
[
(s − 2)^2 + 4
]
= 2 e^2 x^ − 2 e^2 x^ cos 2x + e^2 x^ sin 2x.
Solution of Initial-Value Problems
Here we complete the application of Laplace transforms to the solution of initial-value problems. For our first example, we finish Example 2, Section 4.2.
Example 6. Find the solution of the initial-value problem
y′^ − 2 y = 2e−^3 x; y(0) = − 2.
SOLUTION From the Example, if y = y(x) is the solution, then
L[y′(x) − 2 y(x)] = L[2e−^3 x] = 2L[e−^3 x] =
s + 3
L[y′(x)] − 2 L[y(x)] = 2 s + 3
sL[y(x)] − y(0) − 2 L[y(x)] = 2 s + 3
(s − 2)L[y(x)] + 2 = (^) s + 3^2
(s − 2)L[y(x)] =
s + 3 −^2 Therefore, L[y(x)] = Y (s) = 2 (s − 2)(s + 3)
s − 2
Now, by partial fraction decomposition 2 (s − 2)(s + 3) =^
s − 2 −^
s + 3.
Therefore,
Y (s) =
s − 2 −^
s + 3 −^
s − 2 =^
s − 2 −^
s + 3 and y(x) = L−^1
[
s − 2
− 2 /^5
s + 3
]
= − 85 e^2 x^ − 25 e−^3 x.
SOLUTION If y = y(x) is the solution, then
L[y′′(x) − 5 y′(x) + 6y(x)] = L[x − 1] = L[x] − L[1] =^1 s^2
s L[y′′(x)] − 5 L[y′(x)] + 6L[y(x)] = 1 −^ s s^2
s^2 L[y(x)] − sy(0) − y′(0) − 5 {sL[y(x)] − y(0)} + 6L[y(x)] = 1 − s s^2
(s^2 − 5 s + 6)L[y(x)] − 1 = 1 − s s^2
(s^2 − 5 s + 6)L[y(x)] = 1 − s s^2 + 1.
Therefore,
L[y(x)] = Y (s) = 1 − s s^2 (s^2 − 5 s + 6) +^
s^2 − 5 s + 6 =^
1 − s s^2 (s − 2)(s − 3) +^
(s − 2)(s − 3).
Now, by partial fraction decomposition, 1 − s s^2 (s − 2)(s − 3) =^ −^
s
s^2
s − 2
s − 3
and 1 (s − 2)(s − 3) =^
s − 2 +^
s − 3.
Therefore,
Y (s) = − 1 36
s
+^1
s^2
s − 2
+^7
s − 3
and y(x) = − 361 + 16 x − 34 e^2 x^ + 79 e^3 x.
In many applications of differential equations it is not required to determine the solutions explicitly. Instead what is needed is information about the solutions. Often such information can be obtained by analyzing their Laplace transforms. The next example illustrates this type of application.
Example 9. Consider the differential equation
y′′^ − y′^ − 6 y = 2e−x^ on [0, ∞)
together with the condition y(0) = −1. From Chapter 3, we know that the general solution of the differential equation has the form
y(x) = C 1 e−^2 x^ + C 2 e^3 x^ + Ae−x^ (∗)
where C 1 , C 2 are arbitrary constants and A is a constant which can be determined
The question we want to examine is: Can we choose a value for α = y′(0) so that solution of the resulting initial value-problem
y′′^ − y′^ − 6 y = 2e−x; y(0) = − 1 , y′(0) = α
has limit 0 as x → ∞? Since e−^2 x^ → 0 and e−x^ → 0 as x → ∞, we want to choose α so that the coefficient of the e^3 x^ term is 0.
If y = y(x) is the solution of the initial-value problem, then
L[y′′(x) − y′(x) − 6 y(x)] = L[2e−x] =
s + 1
L[y′′(x)] − L[y′(x)] − 6 L[y(x)] = 2 s + 1
s^2 L[y(x)] − sy(0) − y′(0) − {sL[y(x)] − y(0)} − 6 L[y(x)] = 2 s + 1
(s^2 − s − 6)L[y(x)] + s − α − 1 = 2 s + 1
(s^2 − s − 6)L[y(x)] =
s + 1 + 1 +^ α^ −^ s.
Therefore,
L[y(x)] = Y (s) = 2 (s + 1)(s^2 − s − 6)
=
(s + 1)(s + 2)(s − 3) +^
1 + α − s (s + 2)(s − 3).
Now, by partial fraction decomposition, 2 (s + 1)(s + 2)(s − 3) =^
A
s + 1 +^
B
s + 2 +^
C
s − 3 =^
s + 1 +
2 5 s + 2 +
1 10 s − 3
and 1 + α − s (s + 2)(s − 3)
= D
s − 2
+ E
s − 3
− α+3 5 s + 2
α− 2 5 s − 3
Combining these results, we have
Y (s) = − α^ + 1 5
s + 2
+^2 α^ −^3 10
s − 3
s + 1
Clearly, if 2α − 3 = 0, that is, if α = 3/2, then the e^3 x^ term in (∗) is eliminated. The resulting solution is: y(x) = − 12 e−^2 x^ − 12 e−x.
This solution has initial values: y(0) = − 1 , y′(0) = 32 , and (^) xlim→∞ y(x) = 0.
- y′^ − 4 y = 0; y(0) = −2.
- y′^ + 2y = ex; y(0) = 1.
- y′^ − 3 y = e^2 x; y(0) = 2.
- y′^ + y = sin x; y(0) = 1.
- y′′^ + 4y = 0; y(0) = 2, y′(0) = 2.
- y′′^ − 2 y′^ + 2y = 0; y(0) = 0, y′(0) = 1.
- y′′^ − y = sin x; y(0) = 1, y′(0) = 1.
- y′′^ − y = ex; y(0) = 1, y′(0) = 0.
- y′′^ − y′^ − 2 y = sin 2x; y(0) = 1, y′(0) = 1.
- y′′^ + 2y′^ + y = x + ex; y(0) = − 74 , y′(0) = 94.
- y′′^ + 2y′^ + y = 4e−x; y(0) = 2, y′(0) = −1.
- y′′^ + 3y′^ + 2y = 6ex; y(0) = 2, y′(0) = −1.
- y′′^ − 2 y′^ + 5y = 3e−^2 x; y(0) = 1, y′(0) = 1.
- Given the initial-value problem
y′′^ − y′^ − 6 y = 2e−x; y(0) = α, y′(0) = − 1. (See Example 9)
What value should be assigned to α so that the resulting solution will have limit 0 as x → ∞?
- What initial conditions should be assigned with the differential equation
y′′^ + y = e−x
so that (^) xlim→∞ y(x) = 0 where y = y(x) is the solution.
- Consider the differential equation: y′′^ + y′^ − 2 y = 3 sin 2x together with the initial value y(0) = 2. For what value(s) of β = y′(0) will the resulting solution(s) be bounded?
- Consider the differential equation: y′′^ + 3y′^ + 2y = x together with the initial value y′(0) = −2. For what value(s) of α = y(0) will the resulting solution(s) be bounded?
The Laplace transform method applies to initial-value problems in which the initial values are specified at x = 0. Actually, the method can be applied when the initial
conditions are specified at some point a 6 = 0. All that is required is a simple change of independent variable; a translation. For example, the initial-value problem d^2 y dx^2 − 3 dy dx
- 2y = x; y(1) = 0, y′(1) = 1
is transformed into d^2 y dt^2 −^3
dy dt + 2y^ =^ t^ + 1,^ y(0) = 0, y
by setting t = x−1. With this transformation, we have: t = 0 when x = 1; x = t+1; and dy dx =^
dy dt
dt dx =^
dy dt ·^ 1 =^
dy dt d^2 y dx^2 =^
d dx
[
dy dx
]
d dt
[
dy dx
]
dt dx =^
d dt
[
dy dt
]
d^2 y dt^2.
- Find the solution of the initial-value problem: y′′^ − 3 y′^ + 2y = x; y(1) = 0, y′(1) = 1 by first solving the transformed problem.
- Use the Laplace transform method to solve the initial-value problem
y′^ − 2 y = 1 + x; y(2) = − 1.
