Let's look at Hamilton's principle function again. Now we said this was true assuming we'd have Holonomic constraints on the system, this would work and we have this variational. We said is going to be the extremum. You often find people say that this is the action integral and it's nature's equations of motion for this system fall out as a minimum of this action integral, which is plain wrong, it's not a minimum. It is an extremum. Del S equal to zero just means you found the flat point in the curve, but it could be an inflection point, it could be a hyperbolic. It could be a minimum, it could be a maximum. It just depends. All you've said is you found a flat spot and extremum point of that. Let's just keep that in mind. We will go through one example now to illustrate this. Super simple. Spring mass, you could do this in your sleep. Hamilton's extended principal is, this sledgehammer approach on a fly. But it's something we can quickly do and understand the structure of what happens with this stuff. We have initial conditions. Immediately this makes it a fixed end point problem. We're trying to figure out what's happening. We know what the exact solution is. Given these initial conditions, you can write this out as a sine term and an amplitude. We've done this. Now we're going to look at variations. If this is the optimal answer, I'm going to now come up with variations and prove to you depending on some very arbitrary things in the system, the solution is either a minimum, a maximum, or a saddle point. It can be either just depends on how you familiarize your variations. Just to drive a point. We don't get minimums, we get extremums of the stuff. My variations are done like this. You notice at t is zero and t final, this will give you zero again, because I have no uncertainty in my initial conditions and also the end conditions where I'm going to be, fixed end point problems. This variation is perfectly allowable. In between it just has to be C2, twice differentiable continuous and this function it's a second-order polynomial. It is by inspection. Therefore, my very path is the optimal, the true path plus these variations. You can just plug in the algebra. I'm going to outline this. You don't have to do this by hand. We say interesting. Let's look at what happens under these conditions. The action integral was the integral of 0 to t_f over the Lagrangian. The Lagrangian was kinetic energy minus potential energy so m over 2x dot squared minus k over 2x squared and I pulled out the 1/2 just to the left. Now, the action integral of the very path instead of x, I just plugging in x totally, which is x plus Delta x that I defined. You do that and you expand it, you will still get the reference part plus everything that depends on it, and because it's quadratic, there's first-order and second-order Delta x and Delta x dots showing up. This is just algebra you could do that. If you trust Mathematica that's what I used to solve all this, because I just did a little quicker. We're there. Now what happens to the variations that the variation between this and this, and if you take that, this expression minus the reference, this all drops out, you're just left with this and this integral because I have Delta x carefully defined here, it's a simple quadratic polynomial. Again, you can do these integrals. I'd use Mathematica, but it's just tedious, it has a bunch. You do all that and I can actually evaluate all of this. All these temporal integrals from t_0 to t_f with that particular variations and maybe take its derivatives and integrate all those and you do all this algebra and you end up with this expression. That shows you that neighboring path, the mc is just a constant 60t_f another constant for the problem, 10 minus this. Now it depends. How did we pick t_f? If t_f squared is equal to 10 over Omega square root of the system, then the variation is actually equal to the extrema. That's like a saddle point. There is an answer, but there's also neighboring answers that look exactly the same and they all would have. If you plug in this into my original variation, this stuff will all vanish in the end, and it goes back to where it was. But if t_f is greater than that, you have one sign. If t_f is greater than this other one, then this is less than or greater than depending on the sign of that. You control by simply where t_f is an arbitrary number. I'm integrating for five seconds or 15 seconds or 30 seconds. All of a sudden some answers are a minimum because you found a local minimum other ones are a local maximum. It just shows in all cases you found an extremum, but you can never argue you found the minimum of the integrated Lagrangian, it's just an extremum. This is a simple counter-example that just cannot always be a minimum. There is something you said up there. Any questions on this one? What does that physically represent in the system? Gosh, this is why we do abstract thinking so we don't have to put it back to reality. I don't know. It's a good question. What does this physically mean because these extended principle, it's basically del t plus del w. The virtual work and the virtual energy integrated over time must be zero. How did we get there? That's the action integral that we found. Integrating the Lagrangian like this, we found an extremum of that expression essentially. Nature ends up being the equations of motion of whatever a solid pen or if this pen has ink in it that's flowing back and forth. Just like a spacecraft with fuel, it turns out it all satisfies that, that there is a functional that is being [inaudible]. It's just not minimized or maximized. That's as deep as I can go. The rest of it is math. I trust the math that comes out of these thought experiments.