Lectures on the calculus of variations
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Where the heck am I gonna put that minus sign?
Right in there. Then everybody knows that there's also a boundary term, right? So I have to squeeze somewhere in this boundary term. So now that I've done an integration by parts, the boundary term will be the v times the cu prime at the ends of the integral. I think that's right. What we hope is that that goes away. Well, of course, if it doesn't, then we have to -- that there's a good reason that we don't want it to, but here it's nice if it goes away. You see that it does, because we just decided that the boundary conditions on v were zero at both ends. So, v being zero at both ends kills that term.
So now, do you see what I have here? I could write it a little more cleanly. The whole point is that the v -- I now have v there and I have v there so I can factor v out of this. Just put it there. That was a good move.
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This minus sign is still in here. So I now have the integral of some function times v is zero. That's what I'm looking for. The integral of some function, some stuff, times v, and v can be anything -- is zero. What happens now? This integral has to be zero for every v.
So if this stuff had a little bump up, I could take a v to have the same bump and the integral wouldn't be zero. So this stuff can't bump up, it can't bump down, it can't do anything. It has to be zero and that's the strong form. So the strong form is with this minus sign in there, minus the derivative -- see, an extra derivative came onto the cu prime because it came off the v, and the f was just sitting there in the linear, in the no derivative. So, do you see that pattern? You may have seen it before, but calculus variations have sort of disappeared as a subject to teach in advanced calculus.
It used to be here in courses that Professor Hildebrand taught. But actually it comes back because we so much need the weak form in finite elements and other methods. What I wrote over here is the discrete equivalence.
I can't resist, looking at the matrix form for two reasons. First, it's simpler and it copies this. Do you see how this is a copy of that? That matrix form is supposed to be exactly analogous to this continuous form.
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Why is that? Because u transpose f, that's an inner product of u with f -- that's what this is. That integral. So if I forget u prime and think of it as a matrix problem, that's my minimum problem for matrix. I want to find an equation for the winning u. In the end, this is going to be the equation.
Lectures on the Calculus of Variations and Optimal Control Theory
That's the equation that minimizes that. Half of was about this problem. While I concentrated in on this one, because minimum principles are just that little bit trickier. So that's Then in between, people seldom write about but, of course, it's going to work, is that I change u to u plus v, I multiply it out, I subtract, I look at the term linear in v, and that's it.
That would be if I make that just a minus sign and put it all together the way I did here, that's the same thing. So this is the weak form. This is the minimum form, this is the weak form, and this is the strong form. You see that weak form? Somehow in the discrete case it's pretty clear that -- let's see.
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I could write this as -- you see, it's u transpose, a transpose, cav equal f transpose v. The conclusion is if this holds for every v, then this is the same as this. If two things have the same inner product with every vector v, they're the same, and that's the strong form. You'd have to transpose the whole thing, but no problem. So now I guess I've tried to give the main sequence of logic in the continuous case, and it's parallel in the discrete case for this example.
For this specific example because it's the easiest. Let me do what Euler and Lagrange did by extending to a larger class of examples. So now our minimization is still an integral -- I'll still say in 1d, I'll still keep these boundary conditions, but I'm going to allow some more general expression here.
Lectures on the Calculus of Variations and Optimal Control Theory: Second Edition
Instead of that pure quadratic, this could be whatever. Now I'm going to do calculus of variation. Fill in 1d. Calculus of variation, minimize the integral of some function of u and u prime with the boundary conditions, and I'll keep those nice so that integral's still zero to 1 and I'll keep these nice boundary conditions just to make my life easy. So that will lead to v of zero being zero, and v of 1 being zero. What do I have to do? Again, I have to plug in u plus v and compare this result with the same thing having u plus v. So essentially I've got to compare f at u plus v, u plus u prime plus v prime with f of u and u prime.
I have to find the leading term in the difference. So I'll just find out leading term, and then will come the integral. But the first job is really the leading term, and it's calculus, of course. Now can we do that one? It's pure calculus.