 Hello students, in modeling stochastic phenomena it is a very rewarding to obtain analytical solutions wherever possible or else to obtain some asymptotic approximations. This has been the motivating factor for introducing several tools and techniques. In the last few lectures we heard about characteristic functions introduced via Fourier transforms and then generating functions. The generating function that we introduced is basically a kind of an integral transform. It transforms the function from the variable space to a reciprocal or a conjugate space in a special case when the function f x is defined along the positive real line the generating function assumes a special significance. In such a context reducing the generating function to the form of what is called the Laplace transform is very beneficial. For this purpose we study the Laplace transformation some of its properties and some of its possible applications in the stochastic processes. Another interesting insight why Laplace transform occurs quite frequently in modeling stochastic phenomena is that the stochastic processes always have an indexed set. So, the distribution functions generally march in the direction of time and therefore, have the range of the time variable defined in the positive real line. That suits the definition of Fourier Laplace transforms very well and is very helpful in actually solving many problems. So, let us define the Laplace transform over a function let us say f of t where t is greater than or equal to 0 and the function f t is such that f t equal to 0 if t is less than 0. So, we basically now talk of function which looks like something which is defined over only the positive side, but identically 0 on the negative axis. For such functions the Laplace transform sometimes written as L of the function f t t mapped onto a reciprocal variable s is defined as an integral over the positive real line that is t varying from 0 to infinity of the function e to the power minus st with respect to the given function f t. So, you may note that in the previous generating function that we defined we had used e to the power plus st as the notation. So, it is conceptually just about reversing the sign of the reciprocal variable. In some conventions the Laplace transform of a function f is often denoted by a symbol f tilde of s also which is e to the power minus st f t dt. Let us calculate some Laplace transforms using this definition supposing we have a function f t equal to 0 t less than 0 and equal to some constant lambda e to the power minus lambda t. Then its Laplace transform f tilde of s will be by definition integral 0 to infinity e to the power minus st into lambda e to the power minus lambda t dt which is equal to lambda integral 0 to infinity e to the power s plus lambda t dt which is going to be lambda by lambda plus s because this integral by the by the choice of the variables the value of the integrand will tend to 0 as the argument as the variable goes to infinity. So, we have a important result that for an exponential function the Laplace transform is a reciprocal function of the conjugate variable. In general if f t for example is some function lambda into lambda t to the power n e to the power minus lambda t it is actually a combination of an algebraic function at an exponential. Then following the prescription of the Laplace transform f tilde of s is going to be lambda into 0 to infinity lambda t to the power n e to the power minus lambda plus st dt and one can quickly identify this is actually a gamma function or a factorial function from what we had heard in our previous lectures. So, accordingly one can integrate this very easily and we get the result lambda to the power n plus 1 divided by lambda plus s to the power n plus 1 into gamma n plus 1 and if n is integer it is in general it is gamma n plus 1 if n is integer it will be n factorial, but if is n is not an integer this integral is defined for all n greater than minus 1 it is not defined for n less than minus 1. So, one can thus calculate or determine the Laplace transform of a large class of function and today we have lot of results available with respect to these transforms. Most of this information is in available in tables of Laplace transforms which lists. For example, some of the some of those results can be listed here for us some of them we might require for our future use and if f t is given the corresponding f tilde s is going to be for example, if f t is 1 f tilde s is going to be 1 by s because it is e to the power minus st dt only which is 1 by s of course, it is valid for all s positive greater than 0. The domain of validity has to be determined after the integration is done depending on the singularity. So, similarly for example, if f t is e to the power alpha t then this will be e to the power alpha t e to the power minus st dt and you are going to get 1 by s minus alpha. We can see that this will be now have a pole at s equal to alpha. Similarly, if my function is t to the power n then it is the case of putting lambda equal to 1 in the previous example. So, it is going to be gamma n plus 1 divided by s to the power n plus 1. If you have a periodic function like cos omega t defined only in the positive side the function may be 0 on the negative side. So, it may be a part oscillating function. So, for like for example, this may corresponds to if this is 0 function is 0, but it may be a function like this like cos omega t. So, its Laplace transform is going to be s by s square plus omega square and if my function is delta t minus t naught then its Laplace transform will be simply e to the power minus st naught etcetera. Many functions even the advanced functions and special functions also have Laplace transforms which are available. In problems when we are dealing if we arrive at a situation where Laplace transform is such that its inversion is not available then one has to follow some other techniques. And this inversion is a very elaborated complex process it is carried out through what is called as Laplace inversion theorem or via Bromwich integral that is if f s is given then the function itself is obtained by carrying out integration over a line of the function e to the power now plus st and f tilde s ds where c is a constant such that the line of integration is parallel to the imaginary axis and is located to the right of all singularities f tilde s. So, what basically says is after having done the Laplace transform we consider s now as a complex variable in the complex plane. So, we construct an s plane. So, we have now the function f s investigated in its s plane f tilde s is investigated in the s plane then that function multiplied by e to the power st function is integrated over a line parallel to the imaginary axis from minus i infinity to plus i infinity and that line being to the right of all singularities. The property of our integration conducted during generation of Laplace transform if transform is such that most all the singularities would lie somewhere towards the left or for the negative values of s or values of x real parts are likely to be negative even if they exist there will be always a line beyond the to the right of which there will be no singularities. So, that is why this integration line is defined. The proof of the Bromwich integral is a little complicated and we do not attempt it here unless it is required specifically we will we will not be using the Bromwich integration. So, this is a general and useful information for you to know. One important application of Laplace transform is what is called as the convolution theorem. This theorem has a physical basis like this. Imagine that I have a system which is a black box I give some input signal to it the input input signal could be f t time varying function. So, in input like for example, force applied on a particle which is in a viscous medium. And the system is a black box there is some delay after which the effect of this force is imparted on to the receiving system. So, here there will be an output function which we can call it as let us say phi of t. The system response to it or the system transfer mechanism from f to phi could be by a delayed action. It may take some time let us say tau so that it sends a sends the impact of f t to phi t with a certain delay by a known kind of a retardation function or a response function. So, basically there is a concept of a response function R. So, under a linear response kind of a model one can then say that my phi function that is my output that I am receiving at time t is basically a superposition of all the impulses that have been input into the system that is f t at a time prior to t controlled by the retardation function R. So, if a force f acts on a system at time t prime and that force is a transmitted to the receiving end as an output signal to phi by a delay through a response function then the force that is acting on t prime is transmitted at time t. So, R t minus t prime implies that what the action time t prime the signal is transmitted to time t and this can happen at all times. So, all the times prior to t causality demands that any force that is impressed after the time t cannot provide a output provide a response. So, that is why the integration is limited to 0 to t in some some situations one can use minus infinity t, but we always begin our time at time 0. So, in many experimentation one could have measured data or information on the output response like for example, velocity of a particle. There may be theoretical information on the response function also the delay function is also known from that one might have to reconstruct the impulse function which has been the cause of it. So, it is a cause effect relationship we have to delineate the cause from the information obtained from the effect. As you can see from this integral it is a simple linear superposition, but still it is not directly possible because there is a convolution. So, this is called as a convolution integral often sometimes this integral is indicated by a notation f cross g well in the present case it is f cross r. So, one could write phi as f cross r this is just a definition or another notation for the same integral. It is here that Laplace transforms becomes a very powerful and useful tool to invert the problem and obtain information on the impulse function from the information that is contained in the stress and the output function. Now, we show that this is called the convolution theorem. The convolution theorem simply states that if we take the Laplace transform with respect to t that is the phi tilde of s s is the conjugate variable it is simply the product of the Laplace transform of f with respect to the same variable multiplied by the Laplace transform of the response function with respect to the same variable this is called the convolution theorem. We try to prove it let us define the function once again phi of t is integral 0 to t f of t prime response r of t minus t prime dt prime is definition. Now, multiply this by e to the power minus s t and integrate over 0 to infinity ds both sides. So, we will have for example, this is going to be not ds it should be dt we are integrating over time. So, when we accomplish it on the left hand side we are going to have 0 to infinity e to the power minus s t phi t dt and the right hand side we write as 0 to infinity is an integration over t and we already have an integration over t prime integration over t prime up to 0 to t and the overall functions then will be e to the power minus s t f t prime r t minus t prime dt dt prime. So, it is now an integration over 2 variables and and we may note that it is not cannot be immediately implemented because the function r contains both t and t prime together. So, one cannot directly interchange integrals. So, one has to first integrate over t prime and then over t. So, presently therefore, the integration procedure that is adopted in the way we have written is that we have a space of t and t prime. So, integration is done over an area in this space because it is dt dt prime is an area in the space of t and t prime and as it is written here the procedure of integration is that the inner integral varies from 0 to t. So, you fix it where t value then make t prime vary from 0 to t. So, 0 to t is a line t prime t equal to t prime is the line at 45 degree angle it should be and then the inner integral is carried out So, first for a fixed t you integrate t prime from 0 t equal to t t prime equal to 0 to t prime equal to t then you vary t this is the second integral is doing that work. So, this way we expand the entire space to the positive side of t and to the lower half of the area bounded by t equal to t prime line. So, the function at every point at t t prime is now basically integrated or added up in by this rule go up vertically on each line and then go shift the line parallely with an increasing order in the t direction. We can interchange the integration process. So, here in the previous one first t prime integration then t integration. Now, we interchange to achieve first t integration then t prime integration let us see how to do that. So, as you can see we can cover this area also first by performing fixing the value of t prime like for example, here and varying t from t prime to infinity then along this line we make t or t prime vary from 0 to infinity. So, if I want to take t prime outside same double integral will now assume the form of t prime varying from 0 to infinity and t varying from t prime to infinity of the same function minus t prime dt dt prime. So, we have now interchanged the integral by judiciously taking care to see that the inner integral now varies from t prime to infinity. Now integration is simple we make another variable transformation we say t prime t minus t prime if you call as a new variable we can do the t integration first then dt will be du when t is t prime u will be 0 the limit of u the bounds of u will be u varying from 0 to infinity. So, we now therefore, have the Laplace transform of phi S will be the t prime integral varying from 0 to infinity and the u integral also varying from 0 to infinity and the function f t prime. Now, r t minus t prime will be u by this definition and e to the power minus S u plus t prime will appear and again it will be integrated over both du and dt prime. Now, you can see that the functions are separated because e to the power minus S u this will be e to the power minus S u into e to the power minus S t you can take out then it just becomes two different integrals 0 to infinity e to the power minus S t prime f t prime dt prime to 0 to infinity e to the power minus S u r u du which is f tilde of S into r tilde of S proved.