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Laplace Transform
So, last class, we were looking at we just started looking at Laplace transforms with the intention of applying them to solve the ordinary differential equation that we had derived the constitutive relationship. So let us, we have not finished with the Laplace recap of the Laplace transform. So, I will just quickly go ahead and recap for a few other things. So, with this definition, you can apply Laplace to the Laplace operator to lot of different functions and you can show what the Laplace looks like.(Refer Slide Time: 1:07)So, some commonly used Laplace transforms are, so if we have a function f(t), this is I amjust going to list the common Laplacians and if you want you can just go over them and youcan derive them for yourself. So, for example,1L(1)s= . 21L(t)s= , 232!L(t )s= .1( ) at L es a=−. The couple of more. So, for example, 2 2 (cos( ))sL ts=+and2 2 L(sin( t))s=+.(Refer Slide Time: 2:45)Some other properties that are important, let me just divide this space here. So,L( f ) = sL( f ) − f (0) , 2 L( f ) = s L( f ) − sf (0) − f (0) , { ( ) ( )} ( ) at L f t a H t a e f s − − − = and( ( )) as L  t a e− − = .(Refer Slide Time: 4:04)And finally, if you have a function h(t), which is defined as the convolution of two functions,so let us say you have function0( ) ( * )( ) ( ) ( )th t = f g t =  f  g t − d , this * is the convolutionoperator. h(s) = f (s)g(s)(Refer Slide Time: 5:05)So, now with this quick recap we are ready to solve the particular ODE that we wanted to,what is that ODE in question? It was this, we have 0 1 0 1 p  p  q  q • •+ = + , we want to solvethis and at the beginning we will not care what 0p , 1p , etc. are. Once we solve this, then wewill insert these values to get our result.(Refer Slide Time: 5:39)So, if you take the Laplace of this, you will. So maybe I will rewrite that equation we have.So, our problem is problem is solve this equation, 0 1 0 1 p  p  q  q • •+ = + . So, let us takeLaplace of both sides and if you do that, you will end up with0 1 0 1 p  (s) p [s (s)  (0 )] q  (s) q [s (s)  (0 )] + + + − = + − .Now, I had already said that this is from the jump condition which we did not derive and thatwas left as a homework for you. We can take this quantity out. So, we have left with asimpler situation, we have 0 1 0 1  (s)( p + p s) = (s)(q + q s) . So, now, we have to take, so nowwe have to decide what we are solving for. So, this is very general Laplace one, the Laplaceequivalent for that above equation. So, let us say we are solving the stress for the stressrelaxation response.(Refer Slide Time: 7:59)So, case one stress relaxation response. So, in this case, we know that ( ) ( ) o  t = H t . So, ifyou take the Laplace of this, we know that ( ) o ss = . Now, we can replace, so here we didnot know what  (s) was and now we have a functional form for that. So, what we will do? Isjust we will replace it, second.So, let us go ahead and replace, so you have 0 10 1( ) o q q sss p p s+=+. So here we have, now, ifyou take the Laplace inverse, you will be able to compute what the stress is, but this is a sortof rational fraction format, if we have to simplify this a little bit before we can easily take theLaplace. So, let us go ahead and try to simplify your right hand side.So, left hand side you do not have to do anything anymore, the left hand side you have to, theright hand side has to be simplified. So, now I am going to try and get to a simple rationalfunction format for this and we get010 111( ) [ ] oqsq qss p psp+= +(Refer Slide Time: 10:13)So, now this is equivalent to o  , this s I am going to take it inside and I am going to open thebracket here and I will see what I get. We have 11 01q 1p psp+, this s and this s it gets cancelledfor this first case. In the second one we have 1 01 1 01q q 1 1p q s psp  +. So, this part I have alreadygot into a form which I can easily take the Laplace inverse of.1 1 01 0 1 1 01 11 1 1( ) [ ] oq q qsp p p q s ps sp p =  +   + +If you look at the tables of the Laplace function the general Laplace functions that you knowthat we can easily take the Laplace inverse of this. So, we have, so this first part, we willleave it as it as it is, we still have to simplify the last part a little bit before we can take theLaplace inverse. So, what we will do is? This part, we are going to write it as two otherfunctions 01qq, sorry I have to this will be 10pp.(Refer Slide Time: 12:48)So, now if you take the two, so now you have these two are common. So you can combinethese two, and then you have rational function of the form1s. So, you have basically twoseparate rational functions, one is 001 qps . So, then you have the other one part which is1 00 1 011( )q qp p psp−+.0 1 00 0 1 011 1( ) [ ( )] oq q qsp s p p psp =  + −+So, this implies if you now take the Laplace inverse of both sides, you will get on the lefthand side you will get the stress sorry on the right hand side you on the left hand side youwill get the stress on the right hand side you will recover these terms. So, we know that the1L(1)s= . So, this is just the fraction and on here you have 0 1 0 ( 0 / 1 )0 1 0( ) [ ( ) ] p p toq q qt ep p p  − = + − .(Refer Slide Time: 14:40)So, if I just divide these two, you have()ot , which is a stress relaxation function, right. So,we are seeking that anyway that we get and now that will have two parts, so this part which isa constant and then you have this part minus again the same fraction here. So, if I decide tocall this G , then you have the function can be written as some( 0 / 1 )0( )( ) ( ) p p totG t G G G e−  = = + −(Refer Slide Time: 15:43)What is 01pp? Let us quickly go back and check. So, 01p= and 11pE= . So, this is back toour ,E.(Refer Slide Time: 15:55)So, this quantity I can replace and I can or I can just say that / 00 ( ) ( ) t G t G G G e −   = + − .Actually, what we will do is? We will call this 0 and we will see why that has to be there,where 0 E = . Now, convince yourself that this quantity is in fact positive.So, what is 1 01 0q qpp− ? Can we quickly calculate this? So, if you calculate this, this shouldcome out to 1 0 1 11 0 11(1 )q q E EE Ep p E p− = + − = = . So, this fraction is greater than 0. So, isactually a positive number, which implies at your Go > G͚.(Refer Slide Time: 17:44)So, if you wanted to plot this curve, what it will look like? It will be some exponentialfunction, some constant, plus an exponential decay will be there, so the exponential decaywill be will not go down to 0, because you always have this constant G that will be left. So,this is your()ot , then this will be ()Gt , maybe it is not legible, let me rewrite this sorry.So, this was your plotting that this is your G . So, G is a nonzero value, so, your stressrelaxation function does not decay down to 0, which is a viscoelastic solid type of response.And that is what we had said in the very-very beginning itself that this by intuition are simplephysics-based arguments.(Refer Slide Time: 19:16)We had said that this model has to behave like a solid body, because the displacement and thesystem, the springs displacement is arresting the total displacement of the system. So, this hasto be a solid like situation and that is what exactly what we have found out that this behaviourdoes satisfy that of a simple solid or a viscoelastic that this response is in line with what wewould expect from viscoelastic solid.(Refer Slide Time: 19:56)So, similarly, to what we have done, we can now also look at Creep response. So, in Creepresponse you have your ( ) ( ) o  t = H t and this implies again that if you take the Laplace ofboth sides end up with ( ) o ss = . So, previously we saw and we will just rewrite thatequation because otherwise it becomes difficult to solve it without looking at it.So, what we saw was, 0 1 0 1 p  p  q  q • •+ = + . There is one more thing I want to point outthat there are four coefficients here in the ordinary differential equation, but your originalmodel only has three constants E1, E and  . So, the 4 coefficients here are not independent,there are actually in reality only 3 coefficients and the fourth one is a dependent one.(Refer Slide Time: 21:25)So, now, we again we are this part is something that we can still reuse. So, we will just writethat, 0 1 0 1  (s)( p + p s) = (s)(q + q s) , so, from the constitutive equation you have, this isgenerally true, for this ordinary differential equation, we have not applied anything till now.So, now apply now, replace ( ) o ss = , and what I want you to do for the next class? Is Iwould like you to give it a try, where you replace this and you try to solve the equation allover again.And now, you are instead of solving for the stress relaxation function, you are going to solvefor the Creep function. So, I would like you to do that and I would like you to convinceyourself that the system response is still in line with the idea of a viscoelastic solid, if it isnot, then we have a problem. So, for today, what we will do is we will stop here, and I would like you to do this and next class I will work it out by myself. So, we will stop here today, thanks.