In order to proceed now, what I'd like to do is fairly quickly go over how things change if instead of looking at a Dirichlet-Neumann problem, we were to look at another type of boundary value problem that I have sometimes spoken of, for this prob, for, for this case right. Let's look at the Dirichlet Dirichlet problem. Okay, and, and fairly quickly set that one, set that up. All right, so now what we're going to do is consider. The Dirichlet problem. Okay. And we distinguish this from our Dirichlet-Neumann problem by noting that when I say it's just a Dirichlet problem, without any mention of Neumann, what I mean is that all the boundary conditions are Dirichlet boundary conditions. Okay? So, what we would be attempting to do for this sort of problem is the following. We'd say find belonging to Sh, subset of S, where Sh consist of all belonging to, H1 functions on such that at 0 equals u0, as we've been saying. And let's suppose that at L, equals u sub L, okay? So we have Dirichlet boundary conditions at both ends. We do not have a Neumann boundary condition at the right end, right, there are no Neumann conditions in this problem, okay? All right. Find of this sort such that. For all wh belonging to vh, sorry. For all wh belonging to vh. Subset of 3 where vh equals the collection of waiting functions belonging to h1 on omega, such that wh and 0 equals 0, and wh at L also equals 0. Okay? All right. Find belonging to Sh such that for wh of this form, right, where you note that Sh and vh are different looking spaces because they have Dirichlet boundary conditions at both ends. Right? Okay. What do we want to solve here? Okay? What has to hold is the following: integral over Omega e Wh comma x, sigma h, A dx equals integral over omega. Sorry, I've been saying integral over omega e, it's actually integral over the entire domain omega. Integral over omega. Wh fA dx plus, plus nothing. All right? Because we don't have a Neumann condition. That's it. Okay? That is our weak form. Now because of the nature of our waiting functions okay, here's what we have. If this is our domain omega. Okay, and let's suppose that were working with quadratic basis functions. So those are our nodes for element one, and those are our nodes for element NEL, right? So this is element omega one, and that is element omega NEL. Okay? Right? We know that the sort of basis functions we'd be considering inside here are that and. That. On that, on element omega one. Here, we will have that basis function for the mid-side node. And, that basis function, right? For the first node of that element. Observe that in both cases for element omega one there is no use of our basis function from the very first node right, from node A equals 1, global node equals 1. That node does not contribute a basis function to our waiting function. Likewise for the waiting function here, our node A equals 2nel. Plus 1 has no contribution. Righ, okay? However, node a equals 1 and node 2nel plus 1 do have contributions for it's only for wh that they don't have contribution, okay. All right. When we work things through, here's what we will get. The matrix. Vector. Weak form. For this problem? Global. Here it is. We get here, c2 c3. Everything carries on until we come to the last element. We get c2nel minus 1, c2nel. 'Kay, that's it, we don't get c2nel plus 1, because we're simply not using that basis function there. Okay. Here we pick up 2E. We, we are now coming up to the, to work with eventually give rise to our stiffness matrix right? So here we have 2EA over h. Right? We have our matrix here. Okay. Now this matrix is going to have contributions when we add up all the contributions coming from the different elements and accounting for the common degrees of freedom across elements. Right? We are going to have a, contribution that, takes on the following form. It will have a form, 8 over 3 minus 4 over 3, minus 4 over 3, 7 over 3, minus 4 over 3, 1 over 6. Minus 4 over 3, 8 over 3 minus 4 over 3, 1 over 6 minus 4 over 3 7 over 3. And so on. Of course here we know that we get some more terms. Right? And so on, right? Except, that when we come all the way down here for the last element we get 7 over 3 minus 4 over 3. 1 over 6, okay? And here, we get minus 4 over 3. 8 over 3. Okay, and 1 over 6. Sorry not 1 over 6 it's, minus 4 over 3. Right. We get this, that's the last row we get, okay? In particular, what I want to point out is that if compare this with the corresponding metrics at this stage, for the Durtally Noin problem, there is an extra row, for the Durtally Noin problem, which is missing here. Okay? This would now multiply the following matrix, right? It will multiply D 1, D 2, so on, all the way down to d 2 n e l plus 1. Okay, actually I realize that I do need another. Column here. Sorry, let me just put it in here. I do need another column here which is Minus 4 over 3. 1 over 6. That's what I did. Okay. Right. We have this equal to, on the right hand side, we have our c2, c3 vector. We're going all the way now to c 2 n e l c, 2 n e l. Minus 1, and 2 c n e l. Okay? This multiplying. Alright, vector here which is filled with our multiplied by f a h, over 2. Okay. And here we have 4 over 3. 2 over 3, 4 over 3. So on, until the last entry here is one third. All right, actually no. We don't get the one third entry here. Because we have only up to c2 n e l degrees of freedom. The C vector. Right? As a result, the last entry here would also be 4 over 3, okay? 4 over 3 would be the entry from the mid-side node of the last element, okay? The contribution from the last node of the last element, which is the one that I've marked out here Is missing also in the vector that arises in the forcing function. Okay? We have here also then the contribution from traction. Well we don't have a contribution from traction, because there is no traction in this problem. Right? That's it then. Okay, there's actually, I don't even need this parenthesis. Okay, that's it. That's what our right hand side looks like, okay. I can point out something about dimensions. This is matrix is, this vector is 2nel minus 1. This vector as before is 2NEL plus 1. This vector is 2NEL minus 1 just like the C vector on the left hand side because it is the same. The forcing vector now rises only from the uniformity, from the distributed forcing. Which we are resuming to be uniformly distributed and therefore has a constant F. Okay so this is also 2NEL minus 1 in dimension. Okay. This matrix that will eventually give rise to our stiffness matrix, we know has therefore, dimensions 2nel minus 1 times 2nel plus 1 in the d vector here. D1 is known. It is U0. Right. It is this condition. This is also known. It is U sub L. That known condition. We will do what we did in the case of the Dirichlet Riemann Problem. We will note that the d1 degree of freedom multiplies out that column. Whereas, a d. 2 N E L plus 1. Degree of freedom multiplies out, back column. Since D1 and D2NEL plus 1 are known, because U0 and Ul are given to us, we are going to move close to columns that are marked over to the right-hand side, just as we did in the case of the problem, except [INAUDIBLE] problem, we did that only to this column, all right? Okay, let's go ahead and do this now. Right? Redo this. We get C2 C3 up to C2nel-1 C2nel multiplying 2EA over h times this matrix now which is 8 over 3 minus 4 over 3, minus 4 over 3, 7 over 3, minus 4 over 3, 1 over 6, minus 4 over 3, 8 over 3, minus 4 over 3. 1 over 6 minus 4 over 3, 7 over 3, and so on. Now, we come down here and it ends in 7 over 3 minus 4 over 3 in the last element, minus 4 over 3, 8 over 3. Okay? All right? Multiplying In the d vector, we don't have d2 because, sorry, we don't have d1 because we moved it to the to the right-hand side. We come all the way down and we end here at d2nl. Okay? Equal to here, c2, c3, up to c2nel, okay. Multiplying fAh over 2. The vector here from the distributed forcing is four-third, two-third, all the way down until it ends in 4 over 3, at a mid side node of, at the mid side node of the last element. The last two contributions are 2EA over h times d1, right. Multiplying the first column, right, of our matrix on the previous slide, which is minus four-third, one-sixth, zero. All right, zero's all the way down And minus 2EA over h d2nel plus 1, multiplying 0, 0, zeroes all the way down except 'til we come to the last two entries, right, and here we get one-sixth and minus 4 over 3. Okay? We close parentheses, here. And note that d1 is known, because from a Dirichlet condition, that is U not. And d2nel plus 1 is known because from our Dirichlet condition, that is UL, okay? We have here C transpose. That matrix is K. This vector is D. Okay? C transpose has dimensions 2nel minus 1. D also has dimensions 2nel minus 1. K therefore is 2nel minus 1, squared. It is a square matrix. C again is 2nel minus 1, the entire vector here has dimensions 2nel minus 1. Okay? This is our f vector. Everything sitting inside these parentheses is our f vector. This, of course, is C transpose. Okay? Putting it all together. Again, for this Dirichlet-Dirichlet problem we end up with C transpose, K d equals C transpose F, just as we did for the Dirichlet-Neumann problem. Okay? So, for the Dirichlet problem also we have this. Okay? So whether we have the Dirichlet-Neumann problem or the Dirichlet problem we see that C transposed K d minus F equals 0. Now, remember that when we write out our weak form, right, such as in this case at the top of this slide I've written out our weak form fairly briefly and I made a statement on how it has to hold for all WH belonging to the space VH. My question is when we come down to this final version of the weak form, whether for the Dirichlet problem or the Dirichlet-Neumann problem. Where can we see a term that represents the weak the form? Right, it is c transports right? Or, or the c vector. Because that contains the degrees of freedom which are going to be interpolated to form the actual reading function field. All right? They have been interpreted in some basis functions. So if our weak form has to hold for OWH what can we say about this final equation, with regard to C? All right. We can say that this equation must hold for all C belonging to R2nel. This would be for the Dirichlet-Neumann problem. Or, for all C belonging to R2nel minus 1. This would be for the Dirichlet problem. Right. Well, if that is the case, one can show from stand, standard arguments that this implies therefore that K d minus F has to be equal to 0. Okay? Or Kd equals F is our standards form of the finite element equations for linear problems. Okay. And as we saw before, as we did for the case of the linear problem, we say from here d is now equal to K inverse F. How we actually solve this is a different matter which we won't get into in this series of lectures, but nevertheless, this is the definition of our nodal solution. Right? Once we have that, we can go back and regain our fields by using the basis functions. All right, this is a great place to end this segment.