I think I am. Yeah yeah yeah what is that So if everything goes way back when and I'll just say monthly report we'll wrap up all the new stuff today. And Thursday will be a review. So I will prayer some stock but I will not prepare anywhere close to 75 minutes worth of stuff. So I encourage you to think back look back. Take a little time. I know it's crazy tiny semester. A little time to look back and stuff that in retrospect don't understand with management. And the best way to get at it would be some specific exercise. Ok so you can ask like what's the deal with the Laplace transform I can't answer that right You know specifically what it is that you have tremor but. And then I can work through those as much time as we got. Are there any questions people are going to go over very very long on all of it are. So. Corrupt bottom line is that if there is it will be on the kinds of things done primarily from just Yeah so I would say well with what we talked about we have written questions yet but I'd say there are maybe two or three from the newest stuff since the second exam. And then somewhere between three to five from the other you remember the previous. Sometimes in the course of doing one of the new problems you've got old stuff to do. So it's not easy to always breaking up. But that's sort of what we're doing. Anything else. We are working on an occasion equation or a place which is one partial derivative P equals positive number times. Second partial with respect to a. And then in addition to that you've got two boundary conditions has not been spent yet And one initially Japan will use omega alpha squared is that. Okay and then the wonderfully strategy because there is separation. So we find solution. In particular for a function x times z and plug that in and then do a very small amount juggling. And you end up with one over alpha square times t double prime over t t is a function of one variable primes though regular derivatives role as her eggs. And since one side depends on t only depends on x only neither side can move at all. So they are both constant constant lambda. And if we did this base first. Name is rearrange that x double prime plus lambda. So this has the form of an eigenvalue problem. But has with every eigenvalue problem we don't really know what to do until we know the boundary conditions. So we need the result from beard. Depending on what the boundary conditions on. Two major cases two cases I've already described. The first year And that means me 0 value and say at the end 0 and pi. That's the easiest case. As before we can shift that to a different interval. We want to just by transforming the variable. Does that more ego without blueberry pie Well this is the problem we did before. So the eigenvalues of L square. And with that we have eigenfunctions and also in next slide. Right so if lambda takes this what are these guys then there are non-0 solutions in this or any constant. Ok. So now well you don't get anyone boundary all that matters is elsewhere so positive themselves. All right so now we go back to this equation. That has two double prime. I'm sorry not to just keep row 100. And that equals y as alpha square lambda. So that's a first order equation linear who spent a lot of time doing that and this is as easy as it is exponential some constant times e to the minus alpha square So if we chose an eigenvalue then we Wednesday index everything. And so those go diving function. So at each eigenvalue solution file. So it's just a part of capital x capital T. So there's sometimes that is allowed times the easy part first square alpha squared and t times the space part. So we found not just a solution was found in bunches with. If we're thinking about like Ebola on the subway or everything in one study they separated. We're yeah the next big question what do we do with all these which one comes up or how do you know which one you well the answer is because the original their partial equation here and has no forcing term is homogenous. And that means linear combinations of solutions is that they are partially diverted. That diffusion equation with linear combinations of solutions are also. Sometimes this is giving a fancier name superposition like that word because this means So that's just add them all up. And one minus alpha square team side. So if you think about it go back to where it started. We don't just take my word for it right if you took the partial derivative of this sum well partial derivative just inside the zone right and then going would've equity derivatives only gonna affect this term here. So what's the theater but he is going to pull now on this where. And then we just do. So that's just cuz we wear CAN BE. But if we put together just three days minus lambda n times sine and that's because they're eigenfunctions that's equal to the second derivative sine. Or it's just bloody obvious because you differentiate twice you get minus n squared. So this whole thing is some alpha square cn times the exponential secondary site index. And that's that derivative comes back to his attention elsewhere as well So that song as the one we started with that is it's like everything else has been his last. If I had written down the solution to this. And yet that have gone through the motions and check that pretty easily. But much more interesting to find out where it came from because that can be generalized to other kinds of problems. Okay. So we have this solution. I'm an infinite sum but I'll tell you what gear extra notation anymore. So everything in there and determined except for these ECNs. So on one hand it's a solution for any CN. But we have one more condition that we still had an unusual u at x and time 0 is some given function gives a solution at every point in space and so on. One hand ax is equal to mu. On the other hand It's equal to the sum from n equals one to infinity. C n times e to the 0 line. Right that thing it's just a Fourier sine series. Which means we already know how to hold him. He ends before its size every two over pi integral 0 to pi f of x times sine x We finally close the loop on this long digression would be mad to differential equations and this is how Fourier came up with the series in the first place by thinking about it. We were going from 0 to L and sends here. And I know that happens this is n pi x over L. And then this becomes two over l out here. Because our next plus CE Right I mean how useful is well let's see. So let's say the initial condition so that it makes sense that f should go to 0 at the ends because it has to satisfy that condition than the initial condition onto it as well you can get away without doing. So this is the rod. Well as mathematics helped me and comes back. All plus sign up for the final. We're going to give you a little interval table are things like x plus x squared Okay and then a lot simplifies so at both end is one sine of pi you're always integer n cosine n i minus one to the n. So let's see when n is two over pi. I mean it's out at the upper right or the upper. This is 0. So I just get minus two minus 13 and this is all the lower boundary. And the lower boundary this one is 0 minus negative two times minus 10. So if n is even then this is just the both of these. So again as I said we get 42. So the only odd n from song. So n is equal to one we get y is equal to y is just one over 1 q times E to the minus alpha squared times one squared times sine of x which is just That's n equals one. Skip n equals to 0. So that's one over three cubed times e to the minus alpha squared n squared is going to be nine squared times the sine of 3x. Skip n equals four. N equals five. You get one over five E to the minus five squared alpha squared sine of five etcetera. Well what can we learn Well I guess because all the exponentials here have negative x right so they're all doing it. So now if you think about heat a glass rod or no I guess. And I put both ends and ice which I'm him infinite huge amount of items. Then no matter what the thing was initially. I just let it sit around. The whole thing just goes to 0 degrees. So that makes sense physically. If the if the end points are held fixed and everybody is going to hold a smooth out until everybody. Because any agree agree flow. That's that's how we derived the equation. So not only that we can say little bit one day they all go to 0 but they don't all go to 0. So minus x squared It's faster and larger values of e to the minus 20. For it's a lot faster than e to the minus t. And so the solution ends up looking like a shrinking multiple site Now to get more detailed event it's difficult just working by hand. But he's got these formulas are very easy to do that. Okay so here is a dividing my initial division. And I threw it at x squared times pi minus x just because it won't be that symmetric parabolic slightly more interesting. Alright and then I'm setting this constant two over pi and reason. And then I am. If you're computing by the Fourier sign. Okay so the first thing is to look at how well that reconstructs the original function. So there's still a 5-0 I referred to reconstruct F by taking the sum and Papa FM Islam on the scene. So you can see that it's not so great. Yeah. He's kind of good but there's no reason to settle for just lovely approximation. One foot long do. So let's just crank it up to 50. Were able visions. And now as a tip whenever you're trying to compare two things and you get a graph like this you say no the same. It's not very useful right because How similar argument. So let's usually better difference. Now you see oh okay so they agree to about three digit 3.5 which is which is fine least. Okay so here's the actual solutions to the equation severe. So all I want to do it that same reconstruction and instead of just the CK I get I get e to the minus n squared. Alright I said I'll be equal to one times sine There's the initial again and that's that's represented well by the initial theory. And then we start vetted before you see and all types of pores is 0. And again that's the boundary. And the thing is going to 0. But it's also getting more and more like one assignment. So at this point you'd be hard pressed to tell the difference between that red curve and just art assignment you'd have a different disease. And it just engineers didn't shrink exponentially. As the liquid. Yes. Mm. The place I got. Yeah so this one because you put this on the other side it's got a negative coefficient of a backward integration goes up exponentially with the actually this is the loss of the equation. This is the diffusion equation because the foreigner So I promised you two solutions are two different problems than the other one is boiling conditions partial derivative with respect to x. Well if we're back to so we had to rewind all the way back to here. Okay so now back to this situation. But you put in if you have all x t then this is just x prime times capital T with the X parameter. So that we have 0. So we've got that and we got ODE. So yeah it's easiest if we break it into three pieces lambda positive lambda square plus mu square 0. So from there the general solution is y cosine of u plus c2 sine of boundary conditions tell us 00 which is. So that converts Well. So we just gave you from our heart. And then zeros and ones pi which is minus one. Psi was used Okay and we can look at this as a linear system or the Cs 0 values in the system. So a x equals 0 that kind of thing. And we could say where does a singular or were you just look at and say look the first equation tells us that since I assume that lambda is positive. So birth which tells us that c2 is 0. And so that tells us that just 0 is equal to U C1 sine. So if I'm going to avoid everything in sight being 0 the whole solution for x being 0. I'm going to avoid that. I have that C1 is nonzero only u is non-zero. So non-zero resolution requires that this last part b. So that means using it. So lambda sub n squared. Whereas the negatives don't Jane swearing isn't giving anything new but I have the robot 0 I haven't done. So this is all under case a where is that lambda is greater than 0. But these are definitely I recognize. And the solutions that they go with it for C1s. So there's allows C1 to be non-zero rows and I need to get the postman. Makes sense writing assignments is I wanted to go to 0 at cosines are flat. So the derivatives. Okay so that was a that can be your own capital n thermal finding your own. General solution is one lets us. Tell us. Personally tells us that c2 is Europe. Second tells us that c2 is 0 right so C1 is free. So any constant work for every process that has 0 derivative of yet. So lambdas at 00 is an eigenvalue. Eigenfunction is just 1. Third possibility lambda negative views were double-prime minus use where there's some general solution is c1 e to the minus mu plus. Now under conditions tell us 00 sorry contributions from both I get minus c1 plus c2 times than on the other hand. High mass right minus mu c y e to the minus. And I want to give too. So a matrix has determinant u mu u nu. So you square both times e to the plus or minus this one was the determinant of the matrix and the system here. So one thing that he singular Singular if this equals 0 to mu one or real values of real non-zero values of u. This is impossible. Remember this whole case started with assuming that lambda is negative number and it's negative news where you'd better be ready. So there are no more. So to summarize. The spatial part now. Lambda n is equal to n where n starts at 0. And yeah eigenfunctions that cosine sine. Notice that the case n equals 0 that's right handedness. I think the time part goes exactly like before. So we get a solution in the form of a linear combination of all. So now n starts at 0 e to the minus alpha square where cosine. So I just switch cosine or sine and started using the initial condition that x equal to 0 or set t equal to 0 and I didn't alphabets. And I get some cosine of x. That's just a cosine. Remember the cosine series the 0 term is tricky usually revise 1.501. But in any case we know how to compute the A's If you gave me an F of X I could give you which tells me to Zn tells me. One thing is different this time. As t goes to infinity all the exponentials set. So you limit will approach. Just Caesar. Constant. C. Zed is 1.5. So one over pi integral f of x times one. Anybody remember count one. What is this I will try. So if I insulate the ends of the rod and nobody can get in around. So gotta conservation situation. So the solution still us smooth out all the derivatives to do that is to approach a constant but that constant has to preserve the energy so it will be ok.. And then the next most important term would be cosine x i and the decaying exponential. Pretty cool. Alright. For this last thing. Down shift a little bit. Let's go back to these Forget all that partial stop Fourier theories like the first one. So a and B are pulses. Now when you get a problem based on the physical world trying to fall. Generally speaking all of these things that go with y is measured in terms of ln concentration or whatever it is. And Apple at the time. And if you want to actually compute the solution to that particular problem you need to know all those things. But to actually get answers to the problem or to think about the problem. You don't necessarily need to think about all those things independently. I mean after all the fundamental mathematics doesn't change whether you're using kilograms or my group. Doesn't care whether you're using standards around the numbers. But the concept remedy. So what we're gonna do is we're going to define new dimensionless capital Y and capital T is nothing new to. You then have to go in. And I'm going to say that little town little y is equal to eta. Where now eat at our dimension. To be determined So if you go back to the little y is equal to. And if you apply the chain rule says the science says is in the denominator the top of that. So now I can divide if you make the leading coefficient of one. So they cancel out your i tau. Tau And I'd like to have one impossible for me to find this one. So that means tau must be a over B. Inverse. I said these are being determined. So there's my determination tau. I can go ahead and let this be one as well. So that tells me that b must be one. So all that I end up with following the y plus y equals capital. T. Capital T is just little q. Okay so now we could just say this is as easy as we can get these in turn there's no more AB. So we can reason about this problem. And then if we wanted to answer the original problem just use our definition around. What did we get back to the dimensional problem the other thing. Here's our dance. All constants being pauses. We've already no problem in terms of all these parameters and B k and capital. But wouldn't it be nice if we had fewer parameters. So dizzying and see where it takes us. So on the left here. I get the derivative of eta times. Here In the denominator into derivative twice tau squared o b and d y t over tau. Well I guess I'll triggered my little t for faculty and lived through to make a leading coefficient of one. Here I'm sorry. There was a house over. Okay. Now the way this works is that the problem gave me the end of the day. Email is that I get to choose. So by looking at these first two I can't make them both equal to one was extraordinarily lucky. We have how to play the Jews. This is the one our energies to me. Because that makes the natural frequency equal to one. So that is the tau. It's the square root of one over what we call the natural frequency. And then I do have the ability. I think that's what he wanted. He didn't want people to remember this. And that tells me that it's a house where And that's true. So what I'm left with the one I have a mass in the middle. Yeah. I had one why third order term and I have a one. So or they've locked yet I change it to be what I want. But I do know that this has no dimension because everyday or every other year was engineered to do them. Define it to be some other number. And it turned out to be convenient to put a factor of two in there. So I'm going to have zeta. One sound that's used for. I did that because now two zeta here. So now the characteristic polynomial is S square plus two zeta S plus one. So S is negative zeta plus or minus the square root Theta squared sites so now it becomes easy to say well it's critically it's underdamped if zeta is less than one it's critically damped zeta equal to one over day after day. That's why I did it with pay attention to. So everything on the no part of the problem boils down to just knowing zeta which is his magic number. And then everything on this side is just omega over omega n. Okay So everything in terms of just your brand zeta omega n. So even if you don't care about units this can be a useful thing to do. Instead of having to explore a parameter space by lambda explores Dubois lived. Every problem I had prepared. Aren't the only five minutes to finish up on Thursday. Review I guess if you have questions about personal works with leadership
math305-010-20190514-093001.mp4
From Tobin Driscoll May 14, 2019
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