mathacademytutoring Home



Laplace Transformation

by | Jun 15, 2022 | Math Learning

1. History⟶

Integral transformation data back to the Work of Leonard Euler (1763 & 1769), who considered them essentially in the form of the inverse Laplace transform in solving second-order, linear ordinary differential equations. Even Laplace, in his great work, “Theorie Analytique Des Probabilities”(1812), credits Euler with introducing integral transforms. It is Spitzer (1878) who attached the name of Laplace to the expression

Y=(s)ds

Employed by Euler. In this form, it is substituted into the differential equation where y is the unknown function of the variable x.

In the late 19th century, the Laplace transform was extended to his complex form by Poincare and Pincherle, Redis covered by Petzval, and extended to two variables by Picard, with the further investigation conducted by Abel and many others.

The first application of the modern Laplace transformations equations arose of Batemer (1910), who was working on radioactive decay

= -

By setting

P(x) = P(t) dt

and obtaining the transformed equation. Bernstein (1920) used the expression

f(c) = ,

called it the Laplace transformation, in his work on theta functions. The modern approach was given particular impetus by Doetsch in the 1920s and 30s; he applied the Laplace transform to differential, integral, and integro-differential equations. This body of work culminated in his foundational 1937 text, “Theorie and Anwendungender Laplace Transformation.”

No account of the Laplace transformation would be complete without mention of the work of Oliver Heaviside, who produced a vast body of what is termed the “Operational Calculus”. The material is scattered throughout his three volumes, Electromagnetic Theory (1894, 1899, 1912), and bears many similarities to the Laplace transform method. Although Heaviside’s calculus was not the theory of Laplace Transformation, he did find favour with electrical engineers as a useful technique for solving their problems. Considerable research went into trying to make the Heaviside calculus rigorous and connecting it with the Laplace transform. One such effort was that of Bromwich, who, among others, discovered the Inverse Laplace Transformation!

Where lying to the right of all singularities of the function x.

2. Introduction⟶

In this article, we shall consider theoretic aspects of the Laplace transform, reserving for the application of the subject to the solution of linear differential equations. This transform is defined by the equation

PIECE-WISE OR SECTIONAL CONTINUITY

A function  is called sectionally continuous or piece-wise continuous in any interval if it is continuous and has finite left and right-hand limits in every subinterval  as shown in the graph of the function

FUNCTION OF EXPONENTIAL ORDER

A function  is said to be of exponential order as x ⟶ ∞

if

i.e., if given a positive integer, there exists a real number M > 0 s.t.

Or

Briefly we say that f(x) is of exponential order.

Sometimes we write,

For example. is of exponential order as being any positive integer.

So we are going to do L’Hospital Rule where we differentiate numerator and denominator individually,
hence,

(by L. Hospital’s rule)

=

Since.
Hence we can say that is an exponential order as.

Another example.   is not of exponential order as t ⟶ ∞.

Taking,       e- at  =  e t (t -a) = ∞.

Therefore,  e- at F (t) = ∞, if a>0.

∴ F(t) is not of exponential order

DEFINITION THE  TRANSFORM  CONCEPT

An improper integral of the form is called Integral Transform of F(t) if it is convergent. Sometimes it is denoted by  or  Thus

The function K(s, t) appearing in the integrand is called Kernel of the transform. Here’s a parameter and is independent of t, s may be real or complex number.

If we take

Then (1) becomes

This transform is known as Laplace transform.

Some examples of well-known transformations

Hankel Transform of F (t) is

Mellin Transform of F (t) is

Fourier transform of F (t) is

DEFINITION. LAPLACE TRANSFORM

Suppose  is a real valued function defined over the interval (- ∞, ∞) s. t.

The Laplace Transform of, denoted by, is defined as

We also write

Here L is called Laplace transformation operator. The parameter s is a real or complex number. In general, the parameter s is taken to be a real positive number. Sometimes we use symbol p for parameter s.

The Laplace transform is said to exist if the integral (1) is convergent for some value of s.

The operation of multiplying  by e- st and integrating from 0 to ∞ is called Laplace Transformation.

Now we are going to see the existence of Laplace transformation on a function

Existence Theorem of Laplace Transform.

Statement: – If  is a function of class A, then Laplace transform of F(t) exists.

Or,

Suppose  is piece-wise continuous in every finite interval and is of exponential order a as  Then  exists ∀ s > a, i.e., Laplace transform exists.

Proof. Let F(t) be piece-wise continuous in every finite interval and of exponential order as t ⟶ ∞.

To show that f(s) = L [F(t)] exists  s > a.

Let t0 > 0. Then

Continuity of F(t) in the finite interval (0, t0) implies that

exists. It remains to show that

exists  s > a. F(t) is of exponential order a as t ⟶ ∞ implies

is finite, i.e., given a number t0, there exists a real number M > 0 st.

t \ge  t0

t \ge  t0

i.e. ,

can be made as small as we please choosing t0 sufficiently large. Hence  exists  s > a.

Fact: The condition given in this part is sufficient for the existence of, but are not the necessary conditions.

LAPLACE TRANSFORMS OF STANDARD FUNCTION

Result 1.

Proof.

Result 2.   if n + 1 > 0.

Proof.

Also  if n is positive integer.

Result 3.  if n + 1 > 0.

For

Result 4.

Proof.

⇒ L(cos at - i sin at) =

⇒ L(cos at) = , L(sin at) =

⇒ L(cos at) = , L(sin at) = .

Result 5.  Find L {sinh (at)}, L {cosh (at)}.

Proof.

L {sinh (at)} = L=  =
(ii) L {cosh (at)} = L=  =

Linear Property.

Suppose f1(s) and f2(s) are Laplace forms of F1(t) and F2(t) respectively. Then

L {c1F1(t) + c2F2(t)} = c1L {F1(t)} + c2L {F2(t)}.

Where c1 and c2 are any constants.

First Shifting Theorem (First Translation)

Proof.  Let L {F(t)} = f(s) = .

Then, L {eat F(t)} =

=

=, where u = s - a > 0

= f (u) = f(s - a)

Fact: This theorem can also be restated as:

If f(s) is the Laplace transform of F (t) and a is any real, or complex number, then f(s + a) is the Laplace transform of). That is to say,

Some examples related with the previous properties and we are going to know how to use these properties to solve a problem

Example.  Find L(t5 e3t).

Solution.   L (t5) =  if n = 5  ⇒             L(t5) =  =  ⇒             L(t5 e3t) =   as L {eat F(t)} = f(s - a).

Example. L(e - t (3 sinh 2t - 5 cosh 2t)}.

Solution. We know that,

if  L {F(t)} = f(s), then L {eat F(t) } = f(s - a).  Also L {sinh 2t} = , L {cosh 2t} =  Therefor L{e- t (3 sinh 2t - 5 cosh 2t)}  = 3.  = .(Ans.)

Example.  Let’s we are going to prove,

Solution. We know that L {cos (at)} =

L {sin (at)} =

This ⇒        L (cos 4t + 3 sin 4t) =  +  =

Using the result,  L {eat F(t)} = f (p - a), we get

Theorem 4. Second shifting Theorem (Second Translatation).

Statement:- If L {F(t)} = f(s) and G(t) =

then        L {G(t)} = e- as f(s)

Or L {F(t - a) H(t - a)} = e- as  f(s)

And  G(t) =

L {G(t)} =  G(t) dt  =  G(t) dt +  G(t) dt  = . 0 dt +  F(t - a) dt  =  +  . F(t - a) dt.

Put t - a = x so that dt = dx.  If t = a, then x = t - a = a - a = 0.  If t = ∞, then x = ∞ - a = ∞.

∴                      L {G(t)} =  F(x) dx  = e- sa  F(x) dx =  f(s).                          Proved.

Example . We are going to find Laplace transform of F(t), where

F(t) =

Solution. Let a = , G(t) = cos t, then

L(G) =  = g(p), as L(cos at) = .

Also                         F(t) =

By second shifting theorem, L(F) = e- ap g(p)

⇒ L(F) = .

Theorem 5. Change of Scale Property.

If              L{f(t)} = f(s), then   L{f(at)} =  f ().

Proof.  Let            L{F(t)} = f(s), then

L{f(at)} =  F(at) dt,

=, at = x

=

=

= . (Proved)

Laplace Transform of Derivatives:

Theorem 6. If L {F(t)} = f(s),  then L {F(n)(t)} = sn f(s) - sn - 1  F(0) - sn - 2 F' (0) …- sF(n - 2) (0) - F(n - 1) (0),  where F(n) (t) stands for .

An Alternate form of Theorem 6.

To express the Laplace transform of the function  in terms of the Laplace transform of the function F(x).

Proof. Let    L{F(t)} = f(s), then  L{F ' (t)} =  =  = - F(0) + sf(s).

∴              L{F ' (t)} = sf(s) - f (0). ………………(1)

In view of this L{F ' (t)} = sg(s) - G(0) = sL {G(t)} - G(0).  Taking G(t) = F ' (t), we get  L{F ' (t)} = sL {F ' (t)} - F ' (0) = s[s  f(s) - F (0)] - F ' (0)  L{F '' (t)} = s2 f(s) - s F(0) - F ' (0).  …………………………(2)

Generalising the results (1) and (2), we get the required result.

Theorem 7. Laplace Transform of Integral.

If  L{F(t)} = f(s), then         f(s) = L.

Proof. Let L {F(t)} = f(s) and

G(t) = .                                                                                                … (1)

We are prove that L =.

From (1) it is clear that

G(0)  = 0  G' (t) =   = F(t).  By theorem 6,  L {G' (t)} =  sL {G (t)} - G (0)  i.e.,          L {F (t)} =  sL {G (t)} - 0 = sL {G (t)}  or                 = L {G (t)} = L.                                                       Proved.

Multiplication by Powers of t:

Theorem 8. If L {F(t)} = f(s),  then                          L {tn F(t)} = (- 1)n  for n = 1, 2, 3, …  Proof. Let        L {F(t)} = f(s) and .  Then                            f(s) =  =  =  =  or                                     (- 1)  =  or                                  (- 1) f' (s) = L {t F(t)}  This proves that the theorem is true for n = 1.

Let us suppose that the theorem is true for n = r so that  (- 1)r f (r) (s) = L {tr F(t)}  =  ⇒       (- 1)r  =  =  (- 1)r + 1 f (r + 1) (s) =  = L {tr + 1 F(t)}.

This proves that the theorem is true for n = r + 1 if it is true for n = r. But we have already shown that the theorem is true for n = 1 and hence for n = 1 + 1 = 2 and therefore for n = 2 + 1 = 3 and so on. Hence it is true for any positive integer n.

Division by t:  Theorem 9.   If L {F(t)} = {f(s)}, then  L =  = ,  provided the integral exists.  Proof. Let L {= L {t F(t)} = f(s). Then  f(s) = .  ∴ a and t are independent variables and hence order of integration in the repeated integral exists.  ∴  =  =  = .

Example. If , then  .  Solution. According to, we get  = .  Dividing by 2, we get  .  Example on Theorem 7 and Theorem 9:  Example. We are going to find Laplace transforms of  (a)  (b)  Solution. If L {F(t) = f(s)], then  L {1 - e- 2t} = L {1} - L {e- 2t} =  ∴                  L=  =  =  =  =  = .  ⇒   L  = . Ans.  (b) L(et - cos t) =  ⇒   L  =  =  = =  =  Example. .  Solution. We know that  =  f(s),  And                        L (sin at) = .  Hence        L (sin t) = .  =  =  =  =  or, L  = cos - 1 s.  ∴ L  cos - 1 s. Ans.

Periodic Functions :

Theorem 10. Let a function F(t) Be period 𝜔 so that  F(t + n𝜔) = F(t)  for n = 1, 2, 3, …  Then              L{F(t)} =  Proof. L{F(t)} =  =  = , where t = x + n𝜔  =  [… F(x + n𝜔) = F(x) for n = 1, 2, 3, …]  =  =  Hence,              L {F(t)} =   (… < 1)  Example. If F(t) = t2, 0 < t < 2 and F(t + 2) = F(t), then we are going to find L{F(t)}.  Solution. Evidently F(t) is a periodic function with period 𝜔 = 2.  L {F(t)} =  (by Th. 10).  = .                                                                  … (i)  Now,    =  =  =  = .  Now (1) becomes  L {F(t)} = .Ans.  Standard Rules of Laplace Transform:  F (t)  L [F(t)]= f(p)  tn  if n + 1 > 0  1  1/p  Cos (at)  p/(p2 + a2)  Sin (at)  a/(p2 + a2)  Cosh (at)  p/(p2 - a2)  Sinh (at)  a/(p2 - a2)  eat  1/(p - a)  J0(at  1/  eat F(t)  f (p - a)  F(at)  F' (t)  p f(p) - F(0)  tn F(t)  (1-)n     tan - 1 (a / p)  Some problems related with the given chart  Problem 1. (a) Find the Laplace transform of t2e - at.  Solution. Since L[tn F(t)] = (- 1)n  F(p) = L[F(t)] = L (e - at) = .  ∴ L[ t2e - at] = (- 1)2  = .  Problem 2. We are going to find the Laplace transform of the followings :  (i) 2e5t           (ii) te2t       (iii) sin (at)          (iv) L(sin 2t . sin 3t)  Solution. We know that L(eat) =                                                                … (1)  and                                          L[tn F(t)] = (- 1)n                                                   … (2)  L(2e5t) = 2 . L(e5t) = L(te2t) = (Le2t) =  = . L(sin at) = L  =  =  = .Ans.  L(sin 2t, sin 3t) = [cos (3t - 2t) - cos (3t + 2t)]  =  [cos t - cos 5t] = . Ans.  Problem 3. Find (i) L[F(t)] if F(t) = 1  (ii) L(5t - 2)  Solition. (i)  =  =  = .Ans.  (ii) Since            L(tn) = ,     L(1) = .  Hence          L(5t - 2) = 5L(t) - 2L(t) = 5.  = .Ans.  Problem 4. Find  L{cos2 t},  (ii)        L{sin2 t}.  Solution. We know that  L(cos at) = ,  and                                L(sin at) = .  L(cos2 t) = =  = . L(sin2 t) = =  = .     Problem 5. We are going to show that,                                                                          L{(1 + te- t)3} =  +  +  + .  Solution. We know that  L{tn} =  if n = 0, 1, 2, 3, ….  Then L {tn eat} = .                                                                                                 … (1)  For L {eat F(t)} = s(s - a) if i{F(t)} = f(s).  In view of equation (1), we shall calculate the given transform  L{(1 + te- t)3} = L {1 + t3 e-3t + 3te-t + 3t2 e-2t}  = L {1} + 3L {te-t} + 3L {t2 e-2t} + 3L {t3 e-3t}  =  = Proved.  Problem 6.  Evaluate  Solution. Let                 F(t) = e-t - e-3t.  Recall that                 L {eat} =  = L {e-t - e-3t} =  ∵                              L  = , where f(s) = L {F(t)}.  This gives,  L  =  =  =  or                 = .  Putting s = 0, we get  = . For log 1 = 0  or                = log 3.                                                                       Ans. log 3.     Problem 7. Prove that  = .  Solution. L cos (at) =  ⇒           L  =  =  In view of this  =  =  =  =  = . Proved.  THE LAPLACE TRANSFORM  Part II - The Inverse Laplace Transform  DEFINITION  If the Laplace transform of a function F(t) is f(s) i.e., L {F(t)} = f(s) then F(t) is called an Inverse Laplace transform of f(s).  We also write     F(t) = L- 1 {f(s)}.  L- 1 is called inverse Laplace transformation operator.  Example.             L{tn} = .  ∴                                   tn = L- 1  .  NULL FUNCTION  Let N(t) be a function of t such that  Then N(c) is called a null function.  Since Laplace transform of all function is zero and hence if  L {F(t) = f(s)}, then  L {F(t) + N(t)} = f(s).  Consequently       L- 1 {f(s)} = F(t),  L- 1 {f(s)} = F(t) + N(t).  This implies that we can have two different functions, with the same Laplace transform. As a result of which the inverse Laplace transform of a function is not unique, if we allow null functions. Hence the inverse Laplace transform of a function is unique if we do not allow null functions which do not appear in case of physical interest.  Table of Laplace Transform  F (t)     f(s) = L   {F(t)}     1  tn     , n = 0, 1, 2, …     eat        cos at        sin at        cosh at        sinh at  Linear Property :  Theorem. If f1(s) and f2(s) are Laplace transform of F1(t) and F2(t) respectively, then  L- 1 {c1 f(s) + c2 f(s)} = c1L- 1 { f1 (s)} + c2L- 2 {f2(s)}  where c1 and c2 are any constants.  Proof. Prove as in theorem 2 of Chapter 1, Part 1, page 10  L {c1 F1(t) + c2 F2(t)} = c1L { F1 (t)} + c2L {F2(t)}  = c1  f1 (s) + c2 f2(s)  or                c1  F1 (t) + c2 F2(t) = L- 1 {c1 f1(s) + c2 f2(s)}  or        c1 L- 1 {f1(s)} + c2  L- 1 {f2(s)} = L- 1 {c1 f1(s) + c2 f2(s)}.                                Proved.  Example . Find L- 1 .  Solution. L- 1 ­  = L- 1  + L- 1  + L- 1  = e2t + 2e- 5t + t3.  Example . Find L- 1 .  Solution. L- 1  = L- 1 + L- 1 + L- 1  = e2t L- 1 + e- 4t  L- 1 + e- 2t  L- 1  = e2t cos 5t + e- 4t cos 9t  +   sin 3t.                                                                             Ans.  Example . Evaluate L- 1 .  Solution. Recall that (tn) = .  ∴ L  =  or  = L- 1  .  Now L- 1  = L- 1  + L- 1   + L- 1  = e4t L- 1  + e2t L- 1   + e- 3t L- 1  = e4t  + e2t sin 5t + e- 3t cos 6t.                                                                                         Ans.  Example. Prove that  L- 1 =  (sin t) (sinh t).  Solution.                 L- 1  =  L- 1  =  p ⟶ p + 1         p ⟶ p - 1  =  =  [et sin t - e -t sin t] =  (sin t) (et - e -t)  = .  Example. Find  Solution.  =  =  =  =  =                                                                                            Ans.  Example Find inverse Laplace transform of  Solution.  =  =  =  =  = .                                                  Ans.  Theorem .  Second Shifting Property. If ,  Then, where  G (t) =  Proof.  =  +  =  +  = 0 +  =  = =  = .  Thus           .  ∴                           = .  Note: This result is also expressible as  =  or  = F(t - a) H(t - a).  Where H (t - a) is Heaviside unit step function.  Example . Find the inverse Laplace transform of .  Solution.  say.  By second shifting theorem,  =  =  =     Example. Evaluate .  Solution. Let L {F(t)} = f(s) = .  Then,   F(t) =  =  =  .  By second shifting theorem,  =  = .                                                    Ans.  Definition Unit Step Function :  The function G(t) such that  G(t) = F(t - a)  where               F(t - a) = 1,      t > a  F(t - a) = 0,       t < a  is called unit step function and is generally denoted by u(t - a).  Theorem.  Change of scale property. If  f(s) = L{f(s)} denotes the Laplace transform of the function F(t), prove that  =, a > 0.  Proof. ∵     , at = x  =  .  For       .  ∴            = .                                                                      Proved.  Example. If  cos t, then  Find  Solution.           cos t.  Writing as for s,  = .  Again writing a = 3,  = .                                                          Ans.  Example.  If  sin t, then  Find                 .  Solution. Here we use  F .  ∴                    .  Put a = 4,     ⇒                .                                                              Ans.  Theorem . Inverse Laplace transform of derivatives. If  ,  then  Proof. By theorem 8 of chapter 1, Part 1, Page 16.  (Prove this)  ∴  or        .                                          Proved.  Example. Evaluate .  Solution. Let  Then  .  But  or  .                                                                Ans.     Theorem.  The Convolution Theorem.  If                        then     = .  Proof.      We have to show that  L  Let                      .  Aim.                                          L(H) = f(s) g(s)  or                                                                L(H) =  =  or                                                                L(H) = .       … (1)  To change the order  Limits are : t = 0, t = ∞ and u = 0, u = t.  Range of integration is OAB.  We want du dt. Draw an elementary strip parallel to T-axis. One end of strip lies on t = u and the other end on t = ∞. For this strip u varies from u = 0 to u = ∞.  ∴  =                     [Put t - u = y, dt = dy]  =  =  =  = .  THE LAPLACE TRANSFORM  Part III - The Applications to Differential Equations  1.0. ARTICE  Consider the differential equation  … (1)  Case I. When a, b are constant  Here we use Theorem 6, Page 15 namely  Taking Laplace transform of (1),  .  Let                                                            .  Then      .  Required solution is obtained by taking the inverse Laplace transform of.  Case II. When a, b are functions of t, i.e., (1) is of the type  … (2)  Here we use Theorem 8, Page 16, i.e.,  = L {tm x(n) (t)}  = .  Taking the Laplace transform (2),  The required solution is obtained by taking inverse Laplace transform of .  WORKED EXAMPLE  Problem. Solve by using Laplace transformation :  (D2 + 9) y = cos (2t) if y(0) = 1, y  = - 1.  Solution. Taking Laplace transform, we get  p2  - py(0) - y'(0) + 9 =  or         (p2 + 9)  = p + a +  where y'(0) = a  or                       =  +  +  or                       =  +  +  or                       =   +  Taking inverse Laplace transform, we get  .  This ⇒  ∴                      .                                                  Ans.  Problem. Solve ,  y(0) = 1, y'(0) = 1.  Solution. Taking Laplace transform of the equation  (2D2 + 5D + 2) y = e- 2t,  we get .  Putting y(0) = 1 = y'(0), we get     ⇒  ⇒       =  ⇒       =                                                                     …(1)  =  = .                              (2)  But                 =  = .  ∴  ∴  ∴                                                                 … (3)  and  ⇒ .                                                                 … (4)  Taking inverse Laplace Transform of (1) and putting values from (2), (3) and (4), we get  y =  +  = . Ans.

 

Problem . Using Laplace transform,

solve yn + y = cos x, where y(0) = 0 = y' (0).

Solution. Taking Laplace transform of given question,

Putting y(0) = 0 = y'(0), we get  =  ⇒  =  ⇒                   =                                                                   … (1)  =  ⇒  =  ⇒            =                                                                                 … (2)  And (2) ⇒      .

A few worked examples should convince the reader that the Laplace transform furnishes a useful technique for solving liner differential equations. Lastly thanks to all of you and hope you all enjoy this article. We will meet next time with a new article.

Thank You.$

References/Bibliography ⟶

This is merely a brief selection. An extensive list is given in reference –

Laplace Transforms and their applications to Differential equations. – N. W. MCLACHLAN
The Laplace Transform: Theory and Applications – Joel L. Schiff

(iii)       Ordinary & Partial Differential

(iv)       Advanced Calculus – Davis V. Widder

Share This