Ordinary Di erential Equations. Session 7roquesol/Math_308_Fall_2019_Session_7_Print.pdfLaplace...
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Laplace Transform
Ordinary Differential Equations. Session 7
Dr. Marco A Roque Sol
10/08/2019
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
Laplace Transform
Among the tools that are very useful for solving linear differentialequations are integral transforms. An integral transform is arelation of the form
F (s) =
∫ β
αK (s, t)f (t)dt
where K (s, t) is a given function, called the kernel of thetransformation, and the limits of integration α and β are alsogiven. It is possible that α = −∞ or β =∞ or both. The relation,introduced above, transforms the function f into another functionF , which is called the transform of f .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
There are several integral transforms that are useful in appliedmathematics, but we consider only the Laplace Transform (https://en.wikipedia.org/wiki/Pierre-Simon_Laplace )
Laplace Transform
Let f (t) be given for t ≥ 0. Then the Laplace transform of f ,which we will denote by L {f (t)} = F (s), is defined by theequation
L {f (t)} = F (s) =
∫ ∞0
e−st f (t)dt
whenever this improper integral converges.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
The Laplace transform makes use of the kernel K (s, t) = e−st . Inparticular for linear second order differential equations withconstant coeficients is particular useful, since the solutions arebased on the exponential function.
The general idea in using the Laplace transform to solve adifferential equation is as follows:
1. Use the relation L {f (t)} = F (s) to transform an initial valueproblem for an unknown function f in the t-domain (time domain)into an algebraic problem for F in the s-domain (frequencydomain).
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
2. Solve this algebraic problem to find F .
3. Recover the desired function f from its transform F . This laststep is known as inverting the transform (which in general involvecomplex integration) and denoted by L −1{F (s)}(= limω→∞
12πi
∫ σ+iωσ−iω F (s)estds).
OBS The full power of Laplace Transform becomes available onlywhen we regard F (s) as a function of a complex variable. However,for our purposes it will be enough to consider only real values for s.
The Laplace transform F of a function f exists if f satisfies certainconditions:
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
Theorem
Suppose that
1. f is piecewise continuous on the interval 0 ≤ t ≤ A for anypositive A.
2. |f (t)| ≤ Keat when t ≥ M. In this inequality, K , a, and M arereal constants, K and M necessarily positive.
Then the Laplace transform L {f (t)} = F (s), defined by
L {f (t)} = F (s) =
∫ ∞0
e−st f (t)dt
exists for s > a.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
Remember that a function, f (t), is piecewise continuous on theinterval α ≤ t ≤ β if the interval can be partitioned by a finitenumber of points α = t0 < t1 < . . . < tn = β so that
1. f is continuous on each open subinterval ti−1 < t < ti .
2. f approaches a finite limit as the endpoints of each subintervalare approached from within the subinterval.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
Example 62
Find the Laplace transform for f (t) = 1, t ≥ 0
Solution
L {f (t)} = F (s) =
∫ ∞0
e−st f (t)dt
L {1} = F (s) =
∫ ∞0
e−stdt = − limA→∞
e−st
s
∣∣∣A0
=1
s; s > 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
Example 63
Find the Laplace transform for f (t) = eat , t ≥ 0
Solution
L {f (t)} = F (s) =
∫ ∞0
e−st f (t)dt
L {eat} = F (s) =
∫ ∞0
eate−stdt =
∫ ∞0
e−(s−a)tdt =
− limA→∞
e−(s−a)t
s − a
∣∣∣A0
=1
s − a; s > a
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
Example 64
Find the Laplace transform for
f (t) =
1 0 ≤ t < 1k t = 10 t > 1
where k is a constant. In engineering contexts f (t) oftenrepresents a unit pulse, perhaps of force or voltage.
Solution
L {f (t)} = F (s) =
∫ ∞0
f (t)e−stdt =
∫ 1
0e−stdt =
−e−st
s
∣∣∣10
=1− e−s
s
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
Example 65
Find the Laplace transform for f (t) = sin(at), t ≥ 0
Solution
L {f (t)} = F (s) =
∫ ∞0
e−st f (t)dt =
∫ ∞0
sin(at)e−stdt = Int by Parts
F (s) = limA→∞
[−e−stcos(at)
a
∣∣∣A0− s
a
∫ A
0e−stcos(at)dt
]= Int by Parts
F (s) =1
a− s2
a2
∫ ∞0
sin(at)e−stdt =1
a− s2
a2F (s)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
Hence, solving for F(s), we have
F (s) =a
s2 + a2
Now, the Laplace transform is a linear operator, that is, supposethat f1 and f2 are two functions whose Laplace transforms exist fors > a1 and s > a2, respectively. Then, for s > max{a1, a2}
L {c1f1 + c2f2} =
∫ ∞0
e−st{c1f1 + c2f2}dt =
c1
∫ ∞0
e−st f1dt + c2
∫ ∞0
e−st f2dt = c1L {f1}+ c2L {f2}
Thus, we have
L {c1f1 + c2f2} = c1L {f1}+ c2L {f2}Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Definition of The Laplace Transform
OBS
L −1{d1F1 + d2F2} = d1L−1{F1}+ d2L
−1{F2}Example 66
Find the Laplace transform for f (t) = 5e−2t − 3sin(4t), t ≥ 0
Solution
L {5e−2t − 3sin(4t)} = 5L {e−2t} − 3L {sin(4t)} =
L {5e−2t − 3sin(4t)} =5
s + 2− 12
s2 + 16; s > 0
L −1{ 5
s + 2− 12
s2 + 16} = 5L −1{ 1
s + 2} − 3L −1{ 4
s2 + 42} =
5e−2t − 3sin(4t)Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
To see how we can apply the method of Transform Laplace tosolve linear differential equations with constant coefficients. Weestablish the following results.
Theorem
Suppose that f is continuous and f ′ is piecewise continuous on anyinterval 0 ≤ t ≤ A. Suppose further that there exist constantsK , a, and M such that |f (t)| ≤ Keat for t ≥ M. Then
L {f ′} = sL {f } − f (0)
proof
L {f , (t)} =
∫ ∞0
e−st f ′(t)dt = limA→∞
∫ A
0e−st f ′(t)dt =
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
limA→∞
[∫ t1
0e−st f ′(t)dt +
∫ t2
t1
e−st f ′(t)dt +
∫ t3
t2
e−st f ′(t)dt + ...+
∫tn−1
tn = Ae−st f ′(t)dt
]=
and integrating by parts, we have
limA→∞
{e−st f (t)
∣∣∣t10
+ e−st f (t)∣∣∣t2t1
+ ...+ e−st f (t)∣∣∣tn=A
tn−1
+
s
[∫ t1
0e−st f ′(t)dt +
∫ t2
t1
e−st f ′(t)dt + ...+
∫ tn=A
tn−1
e−st f ′(t)dt
]}=
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
limA→∞
[e−sAf (A)− f (0) + s
∫ A
0e−st f (t)dt
]= s
∫ A
0e−st f (t)dt − f (0)
In his way we obtain
L {f ′} = sL {f } − f (0)
As a corollary, we have the following
Corollary
Suppose that the functions f , f ′, ..., f (n−1) are continuous and thatf (n) is piecewise continuous on any interval 0 ≤ t ≤ A.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
Suppose further that there exist constants K , a, and M such that|f (t)| ≤ Keat , |f ′(t)| ≤ Keat , ..., |f (n−1)(t)|Keat for t ≥ M. ThenL {f (n)(t)} exists for s > a and is given by
L {f (n)(t)} = snL {f (t)} − sn−1f (0)
−sn−2f ′(0) . . .− sf (n−2)(0)− f (n−1)(0)
We use the previous results to solve IVP’s using LaplaceTransform. It is most useful for problems involvingnonhomogeneous differential equations. However, just to illustratethe method we will start with a homogeneus case
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
Example 67
Consider the IVP
y ′′ − y ′ − 2y = 0; y(0) = 1, y ′(0) = 0
Solution
Using the traditional method we find that the general solution is
y(x) = c1e−t + c2e
2t
and applying initial conditions we get c1 = 2/3 and c2 = 1/3.Hence, the particular solution is
y(x) =2
3e−t +
1
3e2t
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
Now, let us solve the same problem by using the Laplacetransform. We start off with the differential equation
y ′′ − y ′ − 2y = 0
Applying the Laplace Transform
L {y ′′ − y ′ − 2y = 0}because of linearity
L {y ′′} −L {y ′} − 2L {y} = 0
and using corollary 6.2.1
s2L {y} − sy(0)− y ′(0)− [sL {y} − y(0)]− 2L {y} = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
Taking Y (s) = L {y} and applying initial coditions(y(0) = 1, y ′(0) = 0), we obtain
s2Y (s)− sy(0)− y ′(0)− [sY (s)− y(0)]− 2Y (s) = 0 =⇒
Y (s) =s − 1
s2 − s − 2=
s − 1
(s − 2)(s + 1)
The above can be written, using partial fractions, as
Y (s) =1/3
s − 2+
2/3
s + 1
Now, applying the inverse Laplace transform
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
y(t) = L −1{Y (s)} = L −1{
1/3
s − 2+
2/3
s + 1
}
y(t) = L −1{Y (s)} =1
3L −1
{1
s − 2
}+
2
3L −1
{1
s + 1
}but we know that L {eat} = 1
s−a or equivalently L −1{ 1s−a} = eat
y(t) =1
3e2t +
2
3e−t
The same procedure can be applied to the general second orderlinear equation with constant coefficients
ay ′′ + by ′ + cy = f (t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
obtaining
Y (s) =(as + b)y(0) + ay ′(0)
as2 + bs + c+
F (s)
as2 + bs + c; F (s) = L {f (t)}
The main difficulty that occurs in solving initial value problems bythe transform method lies in the problem of determining thefunction y = y(t), corresponding to the inverse transform of Y (s).
Since we will not deal with the formula for the inverse transform,because it requires complex integration, we will use a table ofcommon Laplace Transform for basic functions (http://www.math.tamu.edu/~roquesol/Laplace_Table.pdf)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
Example 68
Consider the IVP
y ′′ + y = sin(2t); y(0) = 2, y ′(0) = 1
Solution
Let us solve the problem by using the Laplace transform. We startoff with the differential equation
y ′′ + y = sin(2t)
Applying the Laplace Transform
L {y ′′ + y = sin(2t)}
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
because of linearity
L {y ′′}+ L {y} = L {sin(2t)}
and using a previous corollary
s2L {y} − sy(0)− y ′(0) + L {y} =2
s2 + 4
taking Y (s) = L {y} and applying initial coditions(y(0) = 2, y ′(0) = 1), we obtain
s2Y (s)− 2s − 1 + Y (s) =2
s2 + 4=⇒
Y (s) =2s3 + s2 + 8s + 6
(s2 + 1)(s2 + 4)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
The above can be written, using partial fractions, as
Y (s) =as + b
s2 + 1+
cs + d
s2 + 4=
(as + b)(s2 + 4) + (cs + d)(s2 + 1)
(s2 + 1)(s2 + 4)=
(a + c)s3 + (b + d)s2 + (4a + c)s + (4b + d)
(s2 + 1)(s2 + 4)=
2s3 + s2 + 8s + 6
(s2 + 1)(s2 + 4)
Then, comparing coefficients of like powers of s, we have
a + c = 2; b + d = 1;
4a + c = 0; 4b + d = 6;
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
Therefore, a = 2, b = 5/3, c = 0, d = −2/3 and
Y (s) =2s
s2 + 1+
5/3
s2 + 1− 2/3
s2 + 4
Now, taking the inverse, we have
y(t) = 2cos(t) +5
3sin(t)− 1
3sin(2t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
Example 69
Consider the IVP
y (4) − y = 0; y(0) = 0, y ′(0) = 1, y ′′(0) = 0, y ′′′(0) = 0
Solution
Let us solve the problem by using the Laplace transform. We startoff with the differential equation
y (4) − y = 0
Applying the Laplace Transform
L {y (4) − y = 0}
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
because of linearity
L {y (4)} −L {y} = 0
and using corollary 6.2.1
s4L {y} − s3y(0)− s2y ′(0)− sy ′′(0)− y ′′′(0)− Y (s) = 0
taking Y (s) = L {y} and applying initial coditions(y(0) = 0, y ′(0) = 1, y ′′(0) = 0, y ′′′(0) = 0), we obtain
s4L {y} − s2y ′(0)− Y (s) = 0 =⇒
Y (s) =s2
s4 − 1=
s2
(s2 − 1)(s2 + 1)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
The above can be written, using partial fractions, as
Y (s) =as + b
s2 − 1+
cs + d
s2 + 1=
(as + b)(s2 + 1) + (cs + d)(s2 − 1)
(s2 − 1)(s2 + 1)=
(a + c)s3 + (b + d)s2 + (a− c)s + (b − d)
(s2 − 1)(s2 + 1)=
s2
(s2 − 1)(s2 + s)
Then, comparing coefficients of like powers of s, we have
a + c = 0; b + d = 1;
a− c = 0; b − d = 0;
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
therefore, a = 0, b = 1/2, c = 0, and d = 1/2, and
Y (s) =1/2
s2 − 1+
1/2
s2 + 1
and take the inverse, we have
y(t) =sinh(t) + sin(t)
2=
(et − e−t)/2 + sin(t)
2=
et
4− e−t
4+
sin(t)
2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
Example 70
The Laplace transforms of certain functions can be foundconveniently from their Taylor series expansion. Using the Taylorseries for sin(t)
sin(t) =∞∑n=0
(−1)nt2n+1
(2n + 1)!
Let
f (t) =
{sin(t)
t t 6= 01 t = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
Find the Taylor series for f about t = 0. Assuming that theLaplace transform of this function can be computed term by term,determine L {f (t)}
Solution
For t 6= 0 we have that f can written as
f (t) =sin(t)
t=∞∑n=0
(−1)nt2n+1
(2n + 1)!t=∞∑n=0
(−1)nt2n
(2n + 1)!
Applying the Laplace Transform
L {f (t)} = L
{ ∞∑n=0
(−1)nt2n
(2n + 1)!
}
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
because of linearity
L {f (t)} =∞∑n=0
(−1)n
(2n + 1)!L {t2n}
and using the table of Laplace Transforms, L {tm} = m!sm+1
L {f (t)} =∞∑n=0
(−1)n
(2n + 1)!
(2n)!
s2n+1
L {f (t)} =∞∑n=0
(−1)n(2n)!
(2n + 1)!s2n+1=∞∑n=0
(−1)n(2n)!
(2n + 1)(2n)!s2n+1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Solution of Initial Value Problems
L {f (t)} =∞∑n=0
(−1)n
(2n + 1)s2n+1=∞∑n=0
(−1)n(1/s)2n+1
(2n + 1)
but, if we remember
tan−1(x) =∞∑n=0
(−1)nx2n+1
(2n + 1)
Therefore, we have
L {f (t)} = tan−1(1/s)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
To deal effectively with functions having jump discontinuities, it isvery helpful to introduce a function known as the unit stepfunction or Heaviside function. This function will be denoted byuc and is defined by
uc(t) =
{0 t < c1 t ≥ c
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
The step can also be negative. For instance
u(t) = (1− uc(t)) =
{1 t < c0 t ≥ c
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
In fact, any piecewise-defined function can be written as a linearcombination of uc(t)’s functions. For instance consider thefunction
f (t) =
2 0 ≤ t < 11 1 ≤ t < 22 2 ≤ t < 30 3 ≤ t
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
We start with the function f1(t) = 2u0, which agrees with f (t) on[0, 1). To produce the jump down of one unit at t = 1, we add−u1(t) to f1(t), obtaining f2(t) = 2u0 − u1(t), which agrees withf (t) on [1, 2). The jump of one unit at t = 2 corresponds toadding u2(t), which gives f3(t) = 2u0 − u1(t) + u2(t). Thus weobtain
f (t) = f3(t) = 2u0 − u1(t) + u2(t)
The Laplace transform of uc for c ≥ 0 is easily determined:
L {uc} =
∫ ∞0
e−stucdt =
∫ ∞c
e−stdt =e−cs
s, s > 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
For a given function f defined for t ≥ 0, we will often want toconsider the related function g defined by
g(t) = uc f (t − c)
which represents a translation of f a distance c in the positive tdirection.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
Theorem
If F (s) = L {f (t)} exists for 0 ≤ a < s, and if c is a positiveconstant, then
L {uc f (t − c)} = e−csL {f (t)} = e−csF (s),
Conversely, if f (t) = L −1{F (s)}, then
L −1{e−csF (s)} = uc f (t − c)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
Example 71
If the function f is defined
f (t) =
{sin(t) 0 ≤ t < π/4sin(t) + cos(t − π/4) π/4 ≤ t
find L {f (t)}.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
Solution
Note that f (t) = sint + g(t), where
g(t) =
{0 0 ≤ t < π/4cos(t − π/4) π/4 ≤ t
Thus
g(t) = uπ/4cos(t − π/4)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
and
L {f (t)} = L {sin(t)}+ L {uπ/4cos(t − π/4)} =
L {sin(t)}+ e−πs/4L {cos(t)}
and using the table of Laplace Transforms
L {f (t)} =1
s2 + 1+ e−πs/4
s
s2 + 1=
1 + se−πs/4
s2 + 1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
Let’s consider the following theorem
Theorem
If F (s) = L {f (t)} exists for 0 ≤ a < s, and if c is a constant, then
L {ect f (t)} = F (s − c), s > a + c
Conversely, if f (t) = L −1{F (s)}, then
L −1{F (s − c)} = ect f (t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
Example 72
Find the inverse transform of
H(s) =1
s2 − 4s + 5Solution
First of all the polynomial s2 − 4s + 5, has complex roots. Bycompleting the square in the denominator, we can write
H(s) =1
(s − 2)2 + 1= F (s − 2)
where F (s) = (s2 + 1)−1. L −1{F (s)} = sin(t). It follows fromthe previous theorem that
h(t) = L −1{H(s)} = L −1{F (s − 2)} = e2tsin(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7
Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions
Step Functions
Shift in the time − domain (t − domain)
If F (s) = L {f (t)} exists for 0 ≤ a < s, and if c is a positiveconstant, then
L {uc f (t − c)} = e−csF (s) L −1{e−csF (s)} = uc f (t − c)
Shift in the frequency − domain (s − domain)
If F (s) = L {f (t)} exists for 0 ≤ a < s, and if c is a constant, then
L {ect f (t)} = F (s − c), s > a + c L −1{F (s − c)} = ect f (t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7