{"id":833,"date":"2018-11-22T05:28:51","date_gmt":"2018-11-22T05:28:51","guid":{"rendered":"http:\/\/msp01.epgpbooks.inflibnet.ac.in\/?post_type=chapter&p=833"},"modified":"2019-04-30T11:59:29","modified_gmt":"2019-04-30T11:59:29","slug":"laplace-transforms","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/chapter\/laplace-transforms\/","title":{"rendered":"Laplace transforms"},"content":{"raw":"
\"epgp<\/a>\r\n<\/span><\/div>\r\n
\r\n\r\nTABLE OF CONTENTS<\/strong>\r\n

1. Definition of Laplace transform<\/p>\r\n

1.1 Linearity of Laplace transform<\/p>\r\n

1.2 The first shifting theorem<\/p>\r\n

2. Existence of Laplace transforms<\/p>\r\n

2.1 Transforms of derivatives<\/p>\r\n

2.2 Transforms of integrals<\/p>\r\n

3. The inverse Laplace transforms<\/p>\r\n

3.1 Linearity property of the inverse transforms<\/p>\r\n

3.2 Inverse transform of rational functions<\/p>\r\n

4. Solution of initial value problem<\/p>\r\n

5. The step function<\/p>\r\n

5.1 Representation using step function<\/p>\r\n

5.2 The second shifting theorem<\/p>\r\n

6. Initial value problem with piecewise continuous inhomogeneous term<\/p>\r\n

6.1 The step function method<\/p>\r\n

7. Convolution<\/p>\r\n

7.1 Definition of convolution<\/p>\r\n

7.2 Convolution Theorem<\/p>\r\n

7.3 Solution of initial value problem revisited<\/p>\r\n\r\n<\/div>\r\n

<\/div>\r\n
\r\n\r\n \r\n\r\nLEARNING OBJECTIVES<\/strong>\r\n

1. Laplace transforms are defined and the first shifting theorem proved.<\/p>\r\n

2. Conditions for existence of Laplace transforms is stated and formulae for Laplace transform of the derivative and integral of a function derived.<\/p>\r\n

3. The inverse Laplace transforms are defined and method of finding inverse transform of a rational function is described.<\/p>\r\n

4. Solution of initial value problem using Laplace transforms is described.<\/p>\r\n

5. Step function is defined and representation of a piecewise continuous function in terms of step function is provided. Second shifting theorem about inverse transforms is proved.<\/p>\r\n

6. Initial value problem with piecewise continuous inhomogeneous term is solved both without and with the use of step functions.<\/p>\r\n

7. Convolution of two functions is defined; convolution theorem for Laplace transforms is proved and used in solution of initial value problems.<\/p>\r\n\r\n<\/div>\r\n \r\n

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L a p l a c e\u00a0 t r a n s f o r m s<\/strong><\/p>\r\n \r\n\r\n1. Definition of Laplace transforms<\/strong><\/span>\r\n\r\n \r\n

Laplace transforms provide a very important method of solving initial value problems. The basic premise is to transform a difficult problem into a comparatively easier one; solve the easier problem and then use it to find the solution to the original problem. The major advantage of this method is that we can find the solution of an initial value problem without any need to find the general solution first. The method is particularly handy for differential equations involving constant coefficients. Though many of the problems can be solved by other methods that we have discussed earlier, but for some the method of Laplace transforms is far easier. This is particularly true for problems in which the inhomogeneous term has a discontinuity.<\/p>\r\n\"\"\r\n\r\nIn that case the integral may not exist even if the function\u00a0f(x)<\/em> is continuous. The improper integral is defined through a limiting process. Thus for example the improper integral of f<\/em> over a<\/em> to \u221e is defined as\r\n\r\n\"\"\r\n

If the limit is finite we say the <\/span>improper integral exists<\/em> or is <\/span>convergent<\/em>; otherwise we say that the improper integral <\/span>does not exist<\/em> or is <\/span>divergent<\/em>. Unless stated otherwise we will assume that the improper integrals that we come across in this chapter do exist.<\/span><\/p>\r\n

We define the Laplace transform<\/em> of a function f<\/em> as follows: if f<\/em> is a function defined for t \u2265 0 and s<\/em> is any real number, then the Laplace transform of f<\/em> is defined by the integral<\/p>\r\n\"\"\r\n\r\nprovided the improper integral converges.\r\n\r\n\"\"\r\n\r\n<\/div>\r\n \r\n\r\n1.1 Linearity of Laplace transform<\/span>\r\n\r\n\"\"\r\n\r\n \r\n\r\nProof<\/span><\/span>\r\n\r\nLet <\/span>a<\/em> and <\/span>b<\/em> be constants and <\/span>f<\/em> and <\/span>g<\/em> two functions.\u00a0 Then<\/span>\r\n\r\n\"\"\r\n\r\nSome simple examples<\/span>\r\n\r\n\"\"\r\n

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1.2 The first shifting theorem<\/strong><\/span><\/p>\r\n

We now prove a very useful theorem on Laplace transforms which helps us find Laplace transform of a function from that of other known transforms. The first shifting theorem<\/em> state that if<\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\nProof<\/span>\r\n\r\n\"\"\r\n\r\nWe can write this theorem as<\/span>\r\n\r\n\"\"\r\n\r\nExample:<\/span>\r\n\r\n\"\"\r\n\r\nIn the above theorem put<\/span>\r\n\r\n\"\"\r\n\r\nThen<\/span>\r\n\r\n\"\"\r\n\r\n \r\n\r\n<\/div>\r\n

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2. Existence of Laplace transforms<\/strong><\/span><\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nExample:<\/span>\r\n\r\n\"\"\r\n

This is an example of a piecewise continuous function. The Laplace transform for this function is<\/p>\r\n\"\"\r\n\r\nHence\r\n\r\n\"\"\r\n\r\n<\/div>\r\n

<\/div>\r\n \r\n
\r\n\r\n2.1 Transforms of derivatives<\/span>\r\n

In view of application to solution of differential equations, we need to know the transform of derivative of a function. This is provided by the follow theorem:<\/p>\r\n \r\n\r\nTheorem<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\n2.2 Transforms of integrals<\/span>\r\n\r\n\"\"\r\n\r\nTheorem<\/span>\r\n\r\n\"\"\r\n\r\nProof<\/span>\r\n\r\n\"\"\r\n\r\n \r\n\r\n3. The inverse Laplace transforms<\/strong><\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n

then <\/span>f<\/em> is the <\/span>inverse Laplace transform<\/em> of <\/span>F<\/em>.\u00a0 We write it as<\/span><\/p>\r\n\"\"\r\n

We are familiar with inverse transforms from the theory of Fourier transforms. For Fourier transforms evaluation of the inverse is rather straightforward. However for Laplace transforms evaluation of the inverse is usually much more difficult as it involves evaluation of a contour integral. <\/span><\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nFor <\/span>t<\/em> > 0, close the contour from the left, then from residue theorem<\/span>\r\n\r\n\"\"\r\n\r\n \r\n\r\nExample<\/span><\/span>\r\n\r\nLet us look at a particularly simple example to illustrate the method:<\/span>\r\n\r\n\"\"\r\n\r\nThis function has a pole at <\/span>s<\/em> = 1, so <\/span>\u03b3<\/em> > 1.\u00a0 Hence from the above formula (10), we have<\/span>\r\n\r\n\"\"\r\n\r\nThis result is expected since we have already seen that<\/span>\r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n3.1 Linearity property of the inverse transforms<\/span>\r\n\r\nThe linearity property of the inverse transform follows from that of the Laplace transform.\u00a0 Thus we have\r\n\r\n\"\"\r\n\r\nExample<\/span>\r\n\r\n\"\"\r\n\r\n \r\n\r\n3.2 Inverse transform of rational functions<\/span><\/span>\r\n\r\n\"\"\r\n\r\nExample<\/span><\/span>\r\n\r\nLet us find inverse transform of<\/span>\r\n\r\n\"\"\r\n\r\nWe make the partial fractions of the above function:<\/span>\r\n\r\n\"\"\r\n\r\nHence<\/span>\r\n\r\n\"\"\r\n\r\n \r\n\r\n4.\u00a0 Solution of initial value problem<\/span><\/strong><\/span>\r\n

We will now apply Laplace transforms to find the solution of initial value problem for linear second order equations with constant coefficients. For this we will use the theorem on the Laplace transforms of the derivatives of functions. We have the initial value problem<\/span><\/p>\r\n\"\"\r\n\r\nTo solve the initial value problem we take the Laplace transform of both the sides:<\/span>\r\n\r\n\"\"\r\n

Here <\/span>F<\/em> is the Laplace transform of <\/span>f<\/em>.\u00a0 Now we use the result on the Laplace transform of derivatives, equations (5)\u00a0<\/span>and (6), and the initial values given in equation (12):<\/span><\/p>\r\n\"\"\r\n\r\nHence<\/span>\r\n\r\n\"\"\r\n\r\nOr<\/span>\r\n\r\n\"\"\r\n\r\nOn taking the inverse Laplace transform we have the required solution\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nLet us illustrate the procedure by an example.<\/span>\r\n\r\nExample<\/span><\/span>\r\n\r\nLet us solve the initial value problem<\/span>\r\n\r\n\"\"\r\n\r\nWe take the Laplace transform of both sides of the above equation:<\/span>\r\n\r\n\"\"\r\n\r\nOr<\/span>\r\n\r\n\"\"\r\n\r\nNow we use the result on the Laplace transform of derivatives and the initial values<\/span>\r\n\r\n\"\"\r\n\r\nHence<\/span>\r\n\r\n\"\"\r\n\r\nOn taking the inverse Laplace transform, we obtain the solution<\/span>\r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n5.\u00a0 The step function<\/strong><\/span>\r\n

We now wish to find solution of initial value problem for a differential equation in which the inhomogeneous term is a piecewise continuous function. In fact the method of Laplace transforms is most useful for such problems. We have to first develop a method of finding the Laplace transform of such a function. Let us consider the case of a function with one discontinuity. The procedure can obviously be generalized. So consider the function<\/p>\r\n\"\"\r\n\r\nThe Laplace transform of this function is<\/span>\r\n\r\n\"\"\r\n\r\nIn the second integral make the transformation t = u + a<\/em> so that<\/span>\r\n\r\n\"\"\r\n

Since the variable of integration has no effect in a definite integral, we replace it by t<\/em> so that<\/p>\r\n\"\"\r\n\r\nExample<\/span>\r\n\r\n\"\"\r\n\r\nFrom the above formula (15)<\/span>\r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n5.1 Representation using step function<\/strong><\/span>\r\n

The piecewise continuous function can be dealt with in a more convenient form by using the unit step function<\/em> defined by<\/p>\r\n\"\"\r\n\r\nUsing this step function we can rewrite the piecewise continuous function (14) as<\/span>\r\n\r\n\"\"\r\n

To make use of this relation in finding the Laplace transform of a piecewise continuous function we have the following theorem:<\/p>\r\n \r\n\r\nTheorem<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nProof<\/span>\r\n\r\nBy definition of Laplace transform<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nReplace the variable <\/span>u<\/em> by <\/span>t<\/em>, and we have the desired result:<\/span>\r\n\r\n\"\"\r\n\r\nExample-1<\/span><\/span>\r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\nExample-2<\/span>\r\n\r\nFind the Laplace transform of f(t)<\/em> given by\r\n\r\n\"\"\r\n\r\nUsing step function we can write this function as\r\n\r\n\"\"\r\n\r\nHence\r\n\r\n\"\"\r\n\r\n \r\n\r\n5.2 The second shifting theorem<\/span><\/span>\r\n\r\nIf in theorem (18), we replace <\/span>t<\/em> by <\/span>t<\/em> \u2013 <\/span>a<\/em> in <\/span>f<\/em>(<\/span>a<\/em>), we obtain<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nExample<\/span><\/span>\r\n\r\nFind<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\n<\/div>\r\n
\r\n

6. Initial value problem with piecewise continuous inhomogeneous term<\/strong><\/span><\/p>\r\n

Let us consider the initial value problem<\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n6.1 The step function method<\/span>\r\n

We now solve the same problem using step function. We illustrate it by the same example as above. The problem is now written as<\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\n7. Convolution<\/strong><\/span>\r\n

We next wish to find a means of calculating the inverse Laplace transform of a product of two functions. As we have seen in many examples, solving of initial value problem leads ultimately to finding inverse Laplace\u00a0transform of a product of two functions, as, for example, in the problem above. For this purpose let us consider the equation<\/span><\/p>\r\n\"\"\r\n\r\n \r\n\r\n7.1 Definition of convolution<\/span><\/span>\r\n\r\n\"\"\r\n\r\n7.2 Convolution Theorem<\/span><\/span>\r\n\r\n\"\"\r\n\r\nProof<\/span><\/span>\r\n\r\nBy definition of Laplace transform<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\n7.3 Solution of initial value problem revisited<\/span><\/span>\r\n

We can now find the solution of the initial value problem of linear second order differential equation with constant coefficients by using the convolution theorem. As we have already seen, solution of the initial value problem<\/span><\/p>\r\n\"\"\r\n\r\nby the Laplace transform method reduces to<\/span>\r\n\r\n\"\"\r\n\r\nThe second can be evaluated by partial fraction and the first by the convolution theorem.\u00a0 Let<\/span>\r\n\r\n\"\"\r\n\r\n \r\n\r\n<\/div>\r\nSUMMARY<\/strong>\r\n