Inverse Laplace Transform:

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Inverse Laplace Transform
If Lf(x) F(p), then f(x) is called an inverse
Laplace transform of F(p), and we write
f(x) L-1F(p).
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or
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Example Find a function f(x) whose transform is
Solution (a)
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Problem 5. Let ? be a positive number and
consider the function f? defined by
We have
Show that
If
and
is seen to be some kind of quasi-function that is
infinite at x0 and zero for x gt 0, and has the
properties
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and
This quasi-function ?(x) is called the Dirac
delta function or unit impulse functionn.
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Solution. We know
Taking limt as ? 0 on both side, we have
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and hence
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Unit Step Function(or Heavisides Unit Function).
For each real number a ? 0, the unit step
function u(x-a) is defined for nonnegative x by
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x
a
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In particular, if a 0, this formally becomes
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Problem 2(b) Find the Laplace Transform f(x)
x, where x denotes the greatest integer x
Solution
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Problem 3. Show explicitly that does
not exist.
Solution
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For any C gt 0 there exist M such that x gt M
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Hence
does not exist
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Solution of Initial Value Problems
In this section we see how the Laplace transform
can be used to solve initial value problems for
linear differential equations with constant
coefficients.
The Laplace transform is useful in solving these
differential equations because the transform of f
' is related in a simple way to the transform of
f, as stated in Theorem
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Theorem Hypothesis

• Let f be a real function that is continuous for
  t ? 0 and of exponential
  order e?x.
• 2. Let f (the derivative of f) be piecewise
  continuous in every finite closed interval
  0 ? x ? b.

Conclusion Then Lf exists for p gt ? and
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We now generalized Theorem and obtain the
following result
Theorem Hypothesis

• Let f be a real function having a continuous
  (n-1)st derivative f(n-1)(and hence f, f, ..,
  f(n-2) are also continuous) t ? 0 and assume
  that f, f, ., f(n-1)are all of exponential
  order e?x.
• 2. Suppose f(n) is piecewise continuous in every
  finite closed interval 0 ? x ? b.

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Conclusion Then Lf(n) exists for p gt ? and
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We wish to find the particular solution of the
differential equation
That satisfy the initial condition y(0)y0 and
y(0) y0
Let us apply the Laplace Transform of both side
of (1)
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By linearity of LT we get
Next step is to express Ly and Ly in termas
of Ly. By above Theorem we know that
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Substitute these values in (2) we get
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Example
Consider the initial value problem
Recall
Thus r1 -2 and r2 -3, and general solution
has the form
Using initial conditions
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We now solve this problem using Laplace
Transforms.
and hence
Letting Y(p) Ly, we have
Substituting in the initial conditions, we obtain
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Thus
Using partial fraction decomposition, Y(p) can be
rewritten
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Thus
Recall
Thus
Recalling Y(p) Ly, we have
and hence
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Further properties of Laplace Transforms First
Shifting Theorem
If
then
Proof
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Example Find the
Solution
By shifting theorem
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Problem 2 page 394. Find the inverse LT of
Solution.
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Example Find the
Solution
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Problem 3(b) Page 394
Solve the initial value problem
Taking Laplace transforms on both sides, we get
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or
Hence
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Problem 6 page 394 Solve
Taking Laplace transforms on both sides, we get

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Hence
Hence
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Problem 3(e) Page394. Solve the DE by the method
of LT
Solution.
Taking the Laplace transform of the differential
equation
Substituting in the initial conditions, we obtain
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Thus
and hence
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Thus
Therefore our solution to the initial value
problem is
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Derivatives and Integrals of Laplace transforms
Theorem
If
then
Proof
Differentiating both sides w.r.t. p, we get
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or
More generally
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Example
Hence
Theorem
If
then
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Corollary
Thus
Letting p ? 0, we get
Application
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Hence
And so
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Example
Find
Solution
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Problem 1 page 397. Show that
and use this result to find
Solution.
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Problem 3(a)page 398. Solve the DE by LT
Taking the Laplace transform of the differential
equation
Solution.
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Let
Substituting in the initial condition, we obtain
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Thus
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and hence, we find that
This equation is obviously linear with
IF is
Its general solution is
and hence
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The convolution product of two functions Definiti
on If f(x), g(x) are two functions, we define
their convolution product f g by the formula
For example if f(x)1, g(x)x, we get
Thus 1 x x2/2 (and not x)
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Theorem The convolution product is
commutative that is, f g g f Proof
put x-t u,
Theorem The convolution product is associative
i.e. (f g) h f (g h).
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Theorem The convolution product is
commutative that is, f g g f Proof
put x-t u,
Theorem The convolution product is associative
i.e. (f g) h f (g h).
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Theorem If f(x), g(x) are piecewise
continuous and are of exponential order,
then Lf g Lf Lg (ordinary
product) Proof
Changing the order of integration, we get
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t
The line t x
x
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put x-t u,
Hence
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Thus
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Example
Let f (x) 1 and g(x) sinx. Recall that the
Laplace Transforms of f and g are
Thus
and
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Therefore for these functions it follows that
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Example
Find the Laplace Transform of the function h
given below.
Solution
Note that f (x) x and g(x) sin2x, with
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Thus by Theorem
Example Find Inverse Transform
Find the inverse Laplace Transform of H(p), given
below
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Solution
Let F(p) 2/p2 and G(p) 1/(p - 2), with
Thus by Theorem
We can integrate to simplify h(x), as follows.
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More about convolutions Suppose we wish to solve
the I.V.P
satisfying the initial conditions
We first solve the above equation when
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where
We denote the solution of (1) with f(t) u(t) as
A(t).
Hence
Taking Laplace transforms on both sides, we get
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or
where
Now consider the general problem
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Taking Laplace transforms on both sides, we get
Hence
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Hence
We used the fact
where
and so
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Hence
Applying Leibnizs rule for differentiating
integrals we get
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Again
gives
Hence
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putting t s u
We get y
Also y
as A(0)0
Thus the general solution y can be expressed in
terms of A(t), the solution to
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Problem 4(a) page 410 Solve
For this, we first solve
where u(t) is the unit step function.
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Solution
Here we have
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Second Method
Thus