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N Maxwell Model
Welcome back to one more lecture on Introduction to Soft Matter.(Refer Slide Time: 00:37)Last time we were discussing, the specific case of the generalization to N-Maxwell models in parallel. And we had left of where we had discussed that the force balance equation would look very similar to what we had before.(Refer Slide Time: 00:51)And we had also remarked that the geometric constraint remains identical to what it would remain in the case of elements in parallel, which is something that you would expect obviously. And then we had also remark that for each of the individual elements the original equation that we had derived that still holds. Given that, you are using the right force on the, in this particular equation, and the right, the corresponding deformation in the body.(Refer Slide Time: 1:20)So, let us try to see how we can solve these, the set of equations. So, I am just going to rewrite this and I am going to say that this basically corresponds to a set of differential equations such that you have, so this is for first element than you would have one for the second element, I am just expanding the equation that I wrote before.So, this process will go on, till I start writing the equation for the Nth element. So, we had said capital N number of elements are there. So here the last term look something like this, sorry this is not small n this is capital N. So, these are the set of equations, so you have n equations that we have written. Now from the geometry constraint we had said that the deformations are going to be the same in all cases.So, what we can do is we can just delete this, we can drop the subscript from all these different terms, because they are all the same. So, I am just going to do that, I am just go on and delete all the different subscripts. But for the force the subscripts cannot be dropped. So, I have to figure out some other way in which, some other way has to be figured out.And the way to do this would be I want to add all the different forces, so, I just want the same multiplier with all of them, but the multipliers are different. So, I can resolve this problem by multiplying both sides of these two equations by this combination. So, see you already have D/E2, so I am going to multiply this with a set such that terms here and you multiply both sides of this equation.So, I am writing only one side but it is implied that it is also on this side. Basically, this multiplier is something that we have to multiply both sides of equation here. So similarly, what you do for equation 2, equation 2 you already have this E2. So, you start with then the term D/E1 and then you drop the second term, so you multiply it with 3 and like this and you keep on going till you reach the nth multiplier.And you can multiply this entire equation with that. So similarly, by the time you reach the nth equation but you going to have is you are going to multiply it with 1 and this will keep on going, but here you will stop at the (N-1)th multiplier. And in this please recall remember that because the equations are so cumbersome, I am not writing this multiplier both sides but they are on both sides.So, when you multiply this you will finally end up with the same multiplier for all the Fs. And that multiplier would be, so adding all these different terms here and what I will end up having is and this dot dot dot signifies that all the other terms are there and this goes on till N number of terms. And then we have F, so all this will add up and this will give me F and on this side you will have a lot of different terms.Now what you will end up having is you are going to multiply every x with some D andthis kind of multiplier. D is an operator but when you are factoring, when you are multiplyingthis you can also use it as almost as if it were algebraic quantity. So, you will see that whatwill happen is for them for x you will have all the quantities or the term here we will besuch that they will powers of D.21 21 11 1( ).........( ) ( ) ( ..... ) NNN ND DF t q D q D q D xE  E + + = + + + So, this with different coefficients, so what you will end up having is a polynomial n D suchthat where q1, qi’s are constants, some constant which have to be determined. So, you can dothis cumbersome addition and you can determine these constant, so these constants are fullydetermined, but we are not working out here.(Refer Slide Time: 7:29)So, this suggests, suggests a constative equation of the form,21 21 11 1( ).........( ) ( ..... ) NNN ND Dq D q D q DE E  + + = + + + . Now note one thing that thissystem here, this is also result in powers of N, powers of D all the way till N but it will thispolynomial will also have a constant term.So, this is also effectively a polynomial in for the operated D with the highest power beingcapital N again in this case. So, the highest powers on both sides for the operated D are thesame. So, what we can say is that this is actually some polynomial P(D) and this is somepolynomial Q(D). So, this is sort of abbreviation for this entire polynomial.And this set of equations is usually written as, so usually we say written in short form as( ) ( ) P D Q D= . So, this is interesting and now you can probably see the logic behind whywe were writing some of the equations earlier in this particular form.(Refer Slide Time: 10:17)Remember we were write trying to always write it in some P0, P1, P2 and the rational for thatis because once we generalized these basically you end up having a polynomial on this sideand the polynomial on the other side. What the polynomial will look like? Depends on theparticular model whether the powers or will be the same is not guaranteed it again depends onthe particular model that you have choose whether or not there will be a constant term also or again depend on the particular model.So, in this particular case for the N-Maxwell model we have a situation where we have polynomials in D on both sides where and the powers are the exactly the same. That was not the case if you recall for the Jeffrey’s fluid where you had a mismatch in the higher powers and the powers of operator D in the two cases.(Refer Slide Time: 11:15)So, so now let us try to take some special cases, so let us analyse this further. So, this entire N-Maxwell model is going to behave as if this is appropriate for a viscoelastic fluid. But what happens so this is by the way so I just make a note here I think I said that in last time appropriate for viscoelastic fluids. Special cases so what if you want to, I will just quickly go back the diagram so if you want to have solid like response. What you would do here?One thing you could do is take, for example, the last or any of the dashpots. And make this quantity tend to infinity. So, if I take the last spring and I make the viscosity there tend to infinity basically what we have effectively is a spring. So, all of the deformations will be taken by the spring and it will be determined from there.Similarly, if you want to get rid of the initial instantaneous elastic response then the, if you take one of this springs and you can take any of them because I already took N here, I am taking another case where if you take this term and you make it tend to infinity then your initial elastic response will be gone and it will be determined by the viscous effects.(Refer Slide Time: 13:52)So, there are different cases we can do so special case. So, let us say, I am not sure how manycases we will be able to discuss. So, let say the special case. So, let N tend to infinity. So, letus see, what really ends up happening. So, if you take N  to infinity you can redo this entireN calculation for that case or you can take this final equation and you impose the conditionthat is tending to infinity.So one of the things that will happen is, basically for the last term here this quantity willdisappear 1/ N  and you will have the derivate, so the differential operator acting on a F andon this side you will have again at the differential operator acting straight on x  . A littlenaïve way of thinking would be the differential operators can be actually cancelled but that isnot the case what basically it implies is; now you have the simple algebraic relationshipbetween the forces.So, you can actually drop one of these operators, when you drop one of these operators. So,this term gets drop. If you, if this term gets dropped then the power on this side will go downby one. So at least first thing I can do is I can rewrite the left hand side and we can say thatyour left hand side would become and you have all the other terms. But this is now going toend at, N-1.And what happen on this other side, the highest power if you try to redo this what you will isthe highest power of the differential operator on the other side also decreases by 1 so it is N-1. But now you will also have a constant term because one of these, so the way themultiplication and the addition will occur is then you will have a term that does not have thedifferential operator.And your because you had an another E and you still have that if you multiply this numberwill change so essentially what we have here on the other side if you rework is will you havea new set of coefficients and a new polynomial. And I am just going to use keep on using qbut I may be just put a bar here to distinguish it from the previous q, sorry, this will be nowbe 0 and then you will have 1q the and all the way and now the highest power will end at N-1.So, this is interesting and one reason this is interesting is because this side also has highestpowers of N, so maybe I can even write this one as some 10 1 1 ( ..... ) NN p p D P D −− + + . So, this isvery a nice ODE because the polynomials on the both sides are very similar, it is only that theco-efficient are now going to be different. The powers are the same on both sides the highestpowers you have all the different terms and then you also have this other constant term.(Refer Slide Time: 17:16)So, another case, if you want, so let instead of the viscosity tending to infinity what happen isthe elasticity tend to infinity. Be the elasticity tends to infinity and the last term here thisparticular quantity now disappears and you have 1/ N , FN and that will be equal to thisderivative. So basically, if you add up all these, then we will again have 1 powerless on theleft hand side.So, then your this multiplier basically tends to this term become negligible. So, your equationbecome if you rework this whole thing the equation now becomes, if you retain the factoredform again you will end up with the last differential operator will be of the (N-1)th now. Thisis multiplied by F and on this side you will retain more or less what you had. So you canactually rewrite that as, so the coefficients will now be different than the previous case theoriginal case.So maybe I just put a dash here and I am just writing this so that here it looks as if it has thesame powers. Or I mean here you can write it in terms of force or the strain and both weknow that they are equivalent so we can do that to this is strain, this is stress and that is strain.Okay. So, you will that here now there is a mismatch of the highest power, on the left handyour highest power of the operator D is N-1, whereas on this side you have N.(Refer Slide Time: 20:20)So, let us take this type of a case, so let us consider, an equation of the type,0 1 0 1 ( ..... ) ( ......... ) N NN N p + p D+ P D  = q + q D+ q D  . So, we are just considering this type ofequation and we have to let say solve them. Now there is a, the jump condition issue willapply here and you have to redo the jump condition the same way that we have done itbefore.And that is very big, quite cumbersome here so we will not go in details of how the jumpconditions will work out but it will be very-very similar to what we had seen it before. Andthe way we will apply it is when we take Laplace of to both sides and that what we want todo just like the previous case. You remember if you recall, the last Laplacian.So, remember this particular problem when we are looking at solving this particular form and that why you see you can see that we are just recasting this previous equation in this particular way. When you take Laplace of both hand sides it is such, its, the initial conditions were such that they cancelled out.In our and then you are the whole system become such that you are basically taking the Laplace of this particular equation without having to consider the initial conditions is almost equivalent to that, and so you here the derivative and whenever you have derivative you just have a equivalent S multiplied by that and that gives you the Laplacian.(Refer Slide Time: 22:44)So even though we are not going in to the details of initial conditions in this particular casethen you do take the Laplace of both on the both sides then you have, system such that youend up having all those initial conditions cancel on both sides. So, when you do take theLaplace so let say taking Laplace both sides you have system says that you have 0p plus forevery differential operator you basically have one S.And this is going to be very nice for us because what you inessentially end up having is forthe nth power of the differential operator you will have S to the power of N sigma, now this isin the Laplace domain so I will write sigma bar S and on this side you have 0q plus q1 againS the same way to the N qD sorry, not D that differential operator will become S to the powerN and you will have  (s) .0 1 0 1 ( ..... ) ( ) ( ......... ) ( ) N NN N p + p s + P s  s = q + q s + q s  sAnd this is such that this particular, so remember that the left hand side can be easily factoredand this is basically or Laplacian of the polynomial now and I can represent it like this, andthis is basically equal to1 11 1( ).........( )N Ns sE  E + + and on the right hand side this is basicallyequivalent to, this is the Laplacian of the polynomial.Now cube bar this particular polynomial may or may not easily factorable but it still apolynomial so it will have the n number of roots. So, whether or not it is easy to find the rootsit is a different question in this case the polynomial P can be easily factors so the roots arevery well known.P(s) (s) =Q(s) (s)So basically what you have is, so this form is going be to very helpful for us in solving for thedifferent cases, so this form is very similar to what we had done before, so I will just quicklywant to so you the similarity with what we had done before and that should help youunderstand how the solution process will work. So, remember this is what we previously sawright in this particular case it was simple because the polynomial had only one power of theof the differential operator and then things become simple for us.And the same process that we have done can now be repeated for the case of the polynomialwith the nth power. Okay. So, what we do is we will end here and the next class we are goingto take a look at a solution process and what are the different functional form so the stressrelaxation function and the creep compliance is that something that we will see in the nextclass. So, we will stop here today thanks.