Welcome back, we continue with our preamble into the finite element method. And you will recall that in the previous segment, we had started working with this linear, one dimensional, partial differential equation, which was of elliptic type. Okay? And, we, we, we'd written out this differential equation. We had, also motivated it in the context of elasticity. And also looked at how essentially the same partial differential equation arises in other problems. Right? We'd looked at heat transport as well as mass diffusion. Okay? We'll pick up now, but we'll focus on the PDE itself. All right? So, the topic of this segment is what we call the strong form of the p, of the partial differential equation. Okay? So. So the strong form of a linear PDE, or partial differential equation, of elliptic type. Right? In one dimension. Okay? Now let me write the, the PDE and this is the same PDE that we were working with in the previous segment, but let me put it down now. Okay? So what you want to do is the following. Right? We've set ourselves the task of finding U, function of X, given certain data. Right? We're given Ug, U not. We're given Ug, or we're given t, right? And remember how we qualify the difference between Ug and tUg would be the Dirichlet boundary data, and t would be the Neumann boundary data. Okay? We're given these. We're also given the function f, function of x. And the constitutive relation. And that constitutive relation is sigma equals E. U, X. Right? Given all of this, what we want to do is sorry, we already said that we wanted to find U. So we want to find U given all of this information, such that the spatial gradient of sigma plus f equals 0 in the open interval 0, L. Okay? With boundary conditions, Boundary conditions being u(0) equals U not. And, either U at L equals U given or sigma at L equals T. Okay? This is our specification of of the linear partial differential equation, of elliptic type in 1D. Now, we call this a strong form of the equation, right? That's new terminology that I will introduce. We call it the strong form for two reasons, one is that if you look at the PDE itself, you will observe that the PDE has been specified to hold essentially at every point in the interval of interest, right? When we say that d segment dx plus f equals 0 in the open interval 0,L, we're specifying that that PDE has to hold at every point. Right? So, in a sense, that is a strong condition, we've seen that at every point we want it to hold. Additionally, you observe that when we account for this, constitutive relation, right? What this implies is that, if we substitute that considerative relationship for sigma in the partial differential equation, what we get is the following form. Right? We get d, d, x of sigma. But sigma itself is E du, dx plus f equals 0 in the open integral. Now, if you observe that term, what we see is that there are two derivatives effectively on the field of interest, right? There are two derivatives implied here on the prime of field U. And that is a slightly strong condition, because really for this equation to make sense, classically, we do require that U should be a function of a type that allows us to take two derivatives okay. As you may imagine if U is not a terribly smooth function, if it has discontinuities and so on. If it, if, if even worse, right, if it has delta functions, right? If there are delta functions in it. There are certain difficulties we are going to encounter with taking two derivatives of it. Okay? All right? And so, what, what we're doing here is w, w, we are, we, we require so called strong conditions of what I will loosely call smoothness. Okay? On U. Right? And why is this? Because the strong form, and I will capitalize it here to emphasize that, for now, it's something special to us, right? We require strong conditions of smoothness on U effects, because the strong form has to Spatial derivatives. 'Kay, and there's also this idea that we require the whole point wise, 'kay, so we require The partial differential equation to hold pointwise In the interval of interest. Okay, these are the reasons for which we call this form of the equation the strong form. All right, and this is of course is the classical form in which you have seen partial differential equations posed up to this point, assuming of course that you do not already know something about the mathematical background to finite element methods. Okay, all right, so, so this is the strong form with a PDE of interest and, and, of course, we observed in the previous segment that it, it models elasticity, it models heat transport, mass diffusion and so on. What I'm going to do now in the next few minutes is sketch out for us the approach to an analytic solution, 'kay? And I say just sketch out because I'm not actually going to produce an analytical an, an analytic solution. But I will point us in the direction of of how we would proceed, okay? Of course I'll be using the fact that this is all in 1D, so it makes life a little easier for me, okay? So so let's like now look at an analytic, analytic solution. 'Kay? The approach is very straightforward and you probably already know how I'm going to approach it. It is to simply integrate them, okay? So, we have d/dx of sigma equals minus f, all right, and I did that by just moving f over to the other side, it's just integrate, okay? So we integrate this now. I'm going to integrate this from 0 not all the way up to L, 'kay, but I'm going to integrate this from 0 to say y, okay? And here the the dx, of course. And here I have integral of f. Also over the same limits, right? 0 to y, dx, right? Where y belongs to the open integral, sorry, it belongs now to the closed integral. Right, y belongs to the closed interval, 0 to L. Importantly, we do not necessarily, we do not actually in so, in solving this PDE, we do not want to integrate all the way up to L, right, because we want to maintain a general solution applicable at any point, right, and that's why we don't take the upper limit all the way up to L. Okay this again is completely straight forward, right, so what we, what we know we get from here is we get sigma at y minus sigma at 0 equals minus integral of fdx over the limit 0 to y. F in general is at this point it's, it's not quite unknown. It's data, which we would know if we set out to actually solve the problem. But I want to leave it general. I do not want to assume a particular form of f or anything, right? All right. So we have this. But then we realize that what this implies is, of course this, right, it is that E u, x, evaluated at y minus E, u comma x evaluated at 0 equals minus integral 0 to y, fdx, okay? because that's what it implies for us, okay? And you, and you see how I get this last line by simply making the substitution for the constitutive relation for sigma, all right, at the two limit points. Now in in sa, in what is actually some what of an abuse of notation, let me simplify this to write as let me simplify and write this as E du/dy equals minus integral 0 to y fdx plus E du/dx evaluated at x equals 0, okay? Right? Okay, so and now you know what, what the next step is. Well, we need to do one more integration, right? So what that does for us then is the forage now we'll integrate this again now let's integrate from 0 to z, right? Or 0 to zed depending on which part of the world you come from. Okay, so that becomes E that the, the integral here is E, du/dy dy equals minus integral 0 to y of fdx, and then we're doing another integration, right, 0 to z dy. Okay, and just to make it clear, let's do this, okay? And then we have another term here, which is plus E du/dx evaluated at 0, okay? But this term also gets integrated from 0 to z, dy, all right? Okay again, straightforward enough what, what we get here, now, now note that of course E could be a function of position. Right, so, so, you're, you're allowed to make E a function of position. So E could also depend upon y. All right, so now when you do that integral we know what we get, we, we essentially get to make things simpler, let me forget about that particular dependent, okay? Let me, let me leave things as they are. Okay, so let's suppose E's a constant and we go ahead with the integration what we get here is now E. U at z minus E. Again, u at zero equals everything that we have here on the right hand side, correct? So we have integral zero two z, integral zero two y f d x, d y plus integral zero to z. Well actually this last integral can be made very simple now, right? Right, because it is just E. D u, d x evaluated at zero, right? Times z. Okay. So this is the final form that we get, and since we want to write this all as a, as a result for u of z. We write it as one over e. Of integral zero two z. Integral zero two y, f, d x, d y, plus e, d u, d x. Evaluated at zero times z + E multiplying u at zero and close the brackets, ok? So this is the final, the general form of our equation. Of course I made things even a little simpler for us by saying that lets just assume that E is not a function of position. Okay, and if E were a function of position, of course, would involve a little more. But just for the point of fixing ideas, I want to take this approach. Okay. Now, you observed that we have some unknowns here. Would, we know how to apply them, them, right. We would apply the boundary conditioning, we would apply the differential boundary condition here for u of zero. That is equal to u not, okay? That leaves us however with one more unknown, which is this okay? And this would be obtained by applying the boundary condition that we have at l. Okay? This term which is determined. By applying the given boundary condition. BC for boundary condition at X. Equals L. Right? X equals l or in this case z equals l because we are finally expressing this as, a function of z. Okay? Alright. Now. If the bound recondition that we have z equals l is a displacement boundary condition, if its a directional boundary condition it could be straight forward to have to apply it. What if it were a Neuman boundary condition? Think about it. Straight forward of course its strictly almost if it were a Neuman boundary condition we would simply carry out a derivative. A few right and thereby allow us. That would allow us to apply the boundary condition right. So if boundary condition at x equals l is a Neuman boundary condition, we first calculate, calculate or go to u comma z. Using that form of the function, okay? And that will then allow us to apply the Neuman boundary condition, right? And then apply The boundary condition, e u comma z at l. Okay? That will then allow us to determine this unknown. Okay? I also note now, that I have just 1 sign off in this, slide. I have a minus sign here, which I forgot to carry over. So let me just put it in here. Okay? That takes care of things. So, this would be the. General approach to getting an analytic solution of the strong form of the pde that we are working with. Okay. The trouble with this of course is that we can do this only for very special forms, or for the forcing function f. All right? And furthermore, we did observe during the process that I simplified things by saying that E is a constant. Okay? I'm assuming that E is a constant. It does not always have to be the case. I assumed E were a constant to go through the process. If E is not a constant it makes the process a little more complicated right? And likewise if F were some sort of complicated function. Furthermore this sort of simple minded approach or simplistic approach works only in one dimension right? As we go up to, to higher dimensions integrating up the equations of, of this form. This, this particular linear elliptic PDE. It is not at all a trivial matter, alright? Exact solutions or analytic solutions exist only for very special cases. And what we are trying develop here is really an approach to solve general boundary value problems. All right? With general data, general data in the form of F and E. So that's what we're going to go on to. Shortly, but we'll do that in the next segment, and we'll end this segment here.
