04.01. The Pure Dirichlet Problem (Part 1)

Okay.

We will continue with other aspectsof our finite element method.

We're sticking still, with ourlinear one dimensional elliptic PDE.

But what I want to do now, is take the opportunity to go to the case, where we could have a problem, a boundary value problem.

where all the boundary dataare specified as Dirichlet data, 'kay? So, what if there is no Neumann data? Okay.

So, we will call this segment pure Dirichlet problems.

Okay? Or just, Dirichlet problems.

So, in the context of what we are doinghere, the statement is the following.

Find U h belonging to S h, which is a subset of S, where S consists of, all functions u such that u, and 0 equals u not, and u at L equals u given, okay? Now, since, S h is a subset of S, what this implies, of course, for us, forour finite dimensional weak form is that S h consistsof functions u h.

Now, they could belong to H 1, right? Like we have considered forour finite dimensional weak form.

When we consider Dirichlet, and Neumann data, okay? So, u h belongs to H 1, and omega, such that, u h, and 0 equals u 0, and u h, at L equals u g.

Right? That's the kind of u h, we want to find.

Okay.

So, find u h of, of this form such that For all w h belonging to V h, and just for consistency, with whatever the problem, we do this properly.

V h is a subset of V which consists of all functions w, such that, w at 0 equals 0.

And now, w at L, also equals 0, okay? So wherever, we have Dirichlet data, the weighting function satisfiesthe homogenous Dirichlet condition, okay? And then, this implies for us that V h consist of all functions w h, also belonging to the space H 1.

On omega such that w h of 0 equals 0, and w h at L, also equals 0.

Okay? Okay.

So, essentially, we're saying find u hbelonging to S h, where S h consists of functions that have that are H1, and have Dirichlet data on both ends.

Okay? Such that, where all w h belonging to V h, which now, consists of functions satisfyinghomogenous Dirichlet data.

Right? At x equals 0, and x equals L, 'kay? Okay.

What needs to hold? Well, the same old weak form.

Integral over omega w h comma x sigma h A d x equals integral over omega w h f A d x, and we are done.

There is no Neumann data, right? There are, there are no Neumann boundaryconditions in, in, in this problem.

So, we don't have the contributionfrom the traction, right? This is it.

Okay.

Essentially, the formulation is the same, as we've studied for the Dirichlet-Neumann problem.

The only detail is what happens with our homogenous Dirichlet conditions, on the weighting function, okay? So here is the picture actually.

So the physical picture, if you care for it, is this one.

Right?At this end, we know, that we have u not.

Which we know, we are thinking of, as being equal to 0.

At this end, we have not a traction condition.

But we have that displacement, if this is an elasticity problem, is equal to some given value.

Right?Alternately, if this were a, heat conduction ora mass diffusion problem.

It would say, that we would be saying, that we have to the temperature specified, at both ends, right? Or maybe the concentration specified, at both ends.

Right?And we have of course, our forcing.

Right?This is f.

Okay? [INAUDIBLE] Sorry.

All right.

This is our forcing, in this direction.

Okay.

Right?So, so this is the physical picture.

Now, we proceed just as before, right? Let's assume that we havea decomposition of the domain, a partition of the domain, into our elements.

Right?And the subdomains which are elements.

That would be omega 1, that would be omega N e L.

Okay? Now, observe that whatwe are saying is that because of the form of our Dirichlet data, at both boundary nodes.

We are going to do something specialabout the weighting function, basis function, the basis functions forthe weighting function, in elements omega 1, and omega N e L, okay? So, what this implies is that As far as basis functions go.

Here again, let's suppose that weare working with linear polynomials Over You know, these are polynomialsthat compact support, right? Just as before, all of that is the same, right? Over each element omega e, okay? So then, what we're saying is that w h in element e equals 1, is equal to only, basis function 2.

Okay? And, and our basis functions 1, and basis function 2 are exactly, the same as before.

Okay? So, we take only basis function 2, function of c through xtimes c 2 e equals 1.

Okay? And the same sort, well, the similar sort of thing, holds for element e equals N e L.

Okay? This one is, which one, which basis function do we use, for element N e L? Remember, that this element is this one.

Element 1 is this.

Right? And so, what we're saying is that foromega h, we use only that basis function, right? In element 2, sorry, in in element 1.

Element N e L on the other hand, is going to be this one, right? I jumped ahead of myself, by also drawing the basis function, right? But you get the idea.

That is a basis function, the only onethat we will use in element N e L.

Okay? So, this one is just N1, c 1 e equals N e L.

Okay? So really, this is the only thing.

This is really, the only difference.

Now, when we go back, andlook at our the integrals contributing to our weak form, okay? So, we get the following, okay? Integral over omega w h comma x sigma h A d x equals, now, when we work things through, what we will see is that, we get in element 1, we get c 2 1, okay? Multiplying e A over h e Minus 1, 1.

And here the co, the column vector is d 1, for the element 1, d 2, for element 1.

Okay? We get the usual summation over all the L, over the remaining elements, e equals 2 to N e L mi, minus 1.

Okay? I'm sorry, I should Got too close to that one.

Okay.

Right? Note that the sum goes from element2 to element N e L minus 1.

Okay? So, that means, that we are looking atall the elements in between here, right? As contributing, the same sort of term.

Okay? As I'm about to writeout in this summation.

Okay.

So here, we get properly c 1 element e, c 2 element e.

EA over h e.

Our little matrix here, which gives us our element stiffness matrix, would be 1 minus 1 minus 1 1 And multiplying it here, are the degrees of freedoms d 1 e, d 2 e.

And then, added on to this, is the verylast, the very contribution, from the very last element, from element N e L, whichwe have above us here, in the figure.

Right? [INAUDIBLE] You know inthe figure.

And up here we see, that the contribution comes only, from That degree field.

Okay? And in our global numbering system, right? Well, let's look at it, in our localnumbering system, that is c 1 N e L.

Right? Element N e L.

Okay? Just as this, as far as c degrees offreedom were concern, was c 2, element 1.

So, here we get c 1 N e L.

Multiplying.

E A Over h e.

Okay? E A over h e, but now, it, this contribution from the So called stiffness contribution, from thiselement, it takes 1 to form 1 minus 1.

And here, we get [INAUDIBLE] it is multiplied by d 1 for element N e L, and d 2 for element N e L.

Okay? And in a related manner, right.

In a related manner, what we willsee is that, what we obtain is that, as far as our force contributionsare concerned, right? Which come only, from our forcing function, there being no traction condition here, w h f A d x equals.

Okay.

So just as before, we, we get a contribution c 2 1 f A h e over 2, okay? This would be the only contribution, right? From element 1.

From the other elements, right? We will get a contribution, which has the same form as our main sum over all the elements, when we weredoing the Dirichlet-Neumann problem.

Right? This is the sum of eequals 2 to N e L minus 1.

And multi, and here in this sum the, the contributions from the, from the degrees of freedom forweighting function c 1 e, c 2 e F A h e over 2.

Okay? This thing now, will be multiplied by 1, and 1.

And finally, from the very last element, we would get once again c 1 N e L.

We would have f A h e over 2.

Okay? These are the contributions, we would get.

Right? And really this you know, if you go back, and look at our generalDirichlet-Neumann problem.

What you will see is that, we are just accounting for, if you look at this term, right? If you look at how, we obtained This term.

Right? You could go back, to the Dirichlet-Neumann problem, right? And look at what we had, for the very last element.

And in there, simply set c 2 N e L, right? Equal to 0.

Which should be like, setting the weighting function, corresponding to the second nodeof that element equal to 0.

Okay? And this is what, you would be left with.