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Pennsylvania State University

Mathematics

MATH 251

Unknown

Spring

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The Laplace Transform
Definition and properties of Laplace Transform, piecewise continuous
functions, the Laplace Transform method of solving initial value problems
The method of Laplace transforms is a system that relies on algebra (rather
than calculus▯based methods) to solve linear differential equations. While it
might seem to be a somewhat cumbersome method at times, it is a very
powerful tool that enables us to readily deal with linear differential equations
with discontinuous forcing functions.
Definition: Let f(t) be defined for t ≥ 0. The Laplace transform of f(t),
denoted by F(s) or L {f(t)}, is an integral transform given by
∞ −st
L{f(t)} = F(s) = ∫0 e f (t)dt .
Provided that this (improper) integral exists, i.e. that the integral is
convergent.
For functions continuous on [0, ∞), the above transformation is one▯to▯one.
That is, different continuous functions will have different transforms.
© 2008 Zachary S Tseng C▯1 ▯ 1 1
Example: Let f(t) = 1, then F(s) = , s > 0.
s
∞ −st ∞ −st −1 −st∞
L{f(t)} = ∫0 e f( )dt = ∫0 e dt = e 0
s
The integral is divergent whenever s ≤ 0. However, when s > 0, it
converges to
−1 −1 1
(0−e =0) (−1) = = F(s) .
s s s
1
Example: Let f(t) = t, thenF ( ) s2 , s > 0.
[This is left to you as an exercise.]
1
Example: Let f(t) = e , then F(s) = , s > a.
s a
∞ −st at ∞ (a−s)t 1 (a−s)t
L{f(t)} = ∫ e e dt = ∫ e dt = e
0 0 a − s 0
The integral is divergent whenever s ≤ a. However, when s > a, it
converges to
1 (0−e =0) 1 (−1) = 1 = F(s)
a − s a − s s − a .
© 2008 Zachary S Tseng C▯1 ▯ 2 Definition: A function f(t) is called piecewise continuous if it only has
finitely many (or none whatsoever – a continuous function is considered to
be “piecewise continuous”!) discontinuities on anyinterval [a, b], and that
both one▯sided limits exist as t approaches each of those discontinuity from
within the interval. The last part of the definition means that f could have
removable and/or jump discontinuities only; it cannot have any infinity
discontinuity.
Theorem: Suppose that
1. f is piecewise continuous on the interval 0 ≤ t ≤ A for any A > 0.
2. │f(t)│ ≤ Ke when t ≥ M, for any real constant a, and any positive
constants K and M. (This means that f is “of exponential order”, i.e.
its rate of growth is no faster than that of exponential functions.)
Then the Laplace transform, F(s) = L {f (t)}, exists for s > a.
Note: The above theorem gives a sufficient condition for the existence of
Laplace transforms. It is not a necessary condition. A function does not
need to satisfy the two conditions in order to have a Laplace transform.
Examples of such functions that nevertheless have Laplace transforms are
logarithmic functions and the unit impulse function.
© 2008 Zachary S Tseng C▯1 ▯ 3 Some properties of the Laplace Transform
1. L{0} = 0
2. L{f (t) ± g(t)} = L{f (t)} ± L{g(t)}
3. L{cf (t)} = cL{f (t)} , for any constant c.
Properties 2 and 3 together means that the Laplace transform is linear.
4. [The derivative of Laplace transforms]
L{(−t)f (t)} = F′(s) or, equivalently L{t f (t)} = −F′(s)
d 1 − 2 2
Example: L{t } = − (L{t})′ = − 2 = − 3 = 3
ds s s s
In general, the derivatives of Laplace transforms satisfy
L{(−t) f (t)} = F (s)) or, equivalentlyL{t f (t)} = (−1) F (s)n)
Warning: The Laplace transform, while a linear operation, is not
multiplicative. That is, in general
L{f (t)g(t)} ≠ L{f (t)}L{g(t)}.
at 1
Exercise: (a) Use property 4 above, and the fact thL{e } = s −a ,
at 1 2 at
to deduce thatL{te } = 2. (b) What will L{t e } be?
(s −a)
© 2008 Zachary S Tseng C▯1 ▯ 4 Exercises C▯1.1:
1 – 5 Use the (integral transformation) definition of the Laplace transform
to find the Laplace transform of each function below.
1. t2 2. te 6t
−t
3. cos 3t 4. e sin 2t
iαt
5.* e , where i and α are constants, i = −1 .
6 – 8 Each function F(s) below is defined by a definite integral. Without
integrating, find an explicit expression for each F(s).
[Hint: each expression is the Laplace transform of a certain function. Use
your knowledge of Laplace Transformation, or with the help of a table of
common Laplace transforms to find the answer.]
∞ −(s+7)t ∞ 2 −(s−3)t
6. ∫0 e dt 7. ∫0 t e dt
∞ −st
8. ∫0 4e sin 6t dt
Answers C▯1.1:
2 1
1. 3 2. 2
s (s −6)
s 2
3. 2 4. 2
s +9 s +2s +5
s α
5. s +α 2 +i s +α 2
iαt
Note: Since the Euler’s formula says that e = cos αt + isin αt, therefore,
L {e iα} = L {cos αt + isin αt} . That is, the real part of its Laplace transform
corresponds to that of cos αt, the imaginary part corresponds to that of sin αt.
(Check it for yourself!)
1 2 24
6. 7. 3 8. 2
s +7 (s −3) s +36
© 2008 Zachary S Tseng C▯1 ▯ 5 Solution of Initial Value Problems
We now shall meet “ the new System”: how the Laplace transforms can be
used to solve linear differential equations algebraically.
Theorem: [Laplace transform of derivatives] Suppose f is of exponential
order, and that f is continuous and f ′ is piecewise continuous on any interval
0 ≤ t ≤ A. Then
L{f ′(t)} = sL{f (t)} − f (0)
Applying the theorem multiple times yields:
2
L{f ″(t)} = s L{f (t)} − sf (0) − f ′(0),
3 2
L{f ″′(t)} = s L{f (t)} − s f (0) − sf ′(0) − f ″(0),
:
:
(n) n n−1 n−2
L{f 2 (n −3)= s L{f ((n −2)s (n −1)− s f ′(0) − …
− s f (0) − sf (0) − f (0).
This is an extremely useful aspect of the Laplace transform: that it changes
differentiation with respect to t into multiplication by s (and, as seen a little
earlier, differentiation with respect to s into multiplication by −t, on the other
hand). Equally importantly, it says that the Laplace transform, when applied
to a differential equation, would change derivatives into algebraic
expressions in terms of s and (the transform of) the dependent variable itself.
Thus, it can transform a differential equation into an algebraic equation.
We are now ready to see how the Laplace transform can be used to solve
differentiation equations.
© 2008 Zachary S Tseng C▯1 ▯ 6 Solving initial value problems using the method of
Laplace transforms
To solve a linear differential equation using Laplace transforms, there are
only 3 basic steps:
1. Take the Laplace transforms of both sides of an equation.
2. Simplify algebraically the result to solve forL{y} = Y(s) in terms
of s.
3. Find the inverse transform of Y(s). (Or, rather, find a function y(t)
whose Laplace transform matches the expression ofY(s).) This
inverse transform, y(t), is the solution of the given differential
equation.
The nice thing is that the same 3▯step procedure works whether or not the
differential equation is homogeneous or nonhomogeneous. The first two
steps in the procedure are rather mechanical. The last step is the heart of the
process, and it will take some practice. Let’s get started.
© 2008 Zachary S Tseng C▯1 ▯ 7 © 2008 Zachary S Tseng C▯1 ▯ 8 Example: y″ − 6y′ + 5y = 0, y(0) = 1, y′(0) = −3
[Step 1] Transform both sides
L{y″ − 6y′ + 5y } =L{0}
2
s L{y} − sy(0) − y′(0) ) − 6sL{y} − y(0)) + 5L{y} = 0
[Step 2] Simplify to fiY(s) = L{y}
s L{y} − s −(−3)) − 6(s L{y} − 1) + 5L{y} = 0
2
(s − 6 s + 5)L{y} − s + 9 = 0
(s − 6 s + 5)L{y} = s − 9
s −9
L{y} = s −6s +5
[Step 3] Find the inverse transfy(t)
Use partial fractions to simplify,
s −9 a b
L{y} = 2 = +
s − 6s + s−1 s −5
s −9 a(s −5) b(s −1)
2 = +
s −6s +5 (s −1)(s −5) (s −5)(s −1)
s −9 = a(s −5) +b(s −1) = (a +b)s + (−5a −b)
© 2008 Zachary S Tseng C▯1 ▯ 9 Equating the corresponding coefficients:
1 = a + b a = 2
−9 = −5a − b b = −1
Hence,
s −9 2 1
L{y} = s −6s +5 = s −1 − s −5 .
The last expression corresponds to the Laplace transform of
2e t− e . Therefore, it must be that
t 5t
y(t) = 2e − e .
Many of the observant students no doubt have noticed an interesting aspect
(out of many) of the method of Laplace transform: that it finds the particular
solution of an initial value problem directly, without solving for the general
solution first. Indeed, it usually takes more effort to find the general solution
of an equation than it takes to find a particular solution!
The Laplace Transform method can be used to solve linear differential
equations of any order, rather than just second order equations as in the
previous example. The method will also solve a nonhomogeneous linear
differential equation directly, using the exact same three basic steps, without
having to separately solve for the complementary and particular solutions.
These points are illustrated in the next two examples.
© 2008 Zachary S Tseng C▯1 ▯ 10 Example: y′ + 2y = 4te ,t y(0) = −3.
[Step 1] Transform both sides
L{y′ + 2y } =L{4te }−2t
−2t 4
(sL{y} − y(0)) + 2L{y} = L{4te } = 2
(s +2)
[Step 2] Simplify to finY(s) = L{y}
4
(sL{y} − (−3)) + 2L{y} = 2
(s+2)
4
(s + 2)L{y} + 3 = 2
(s +2)
(s + 2)L{y} = 4 − 3
(s+2) 2
2 2
4 3 4−3(s + 2) −3s −12s −8
L{y} = 3 − = 3 = 3
(s + 2) s + 2 (s + 2) (s + 2)
[Step 3] Find the inverse transfoy(t)
By partial fractions,
2
−3s −12s −8 a b c
L{y} = ( + 2) 3 = ( + 2) 3+ ( + 2) 2 + s + 2 .

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