This is a worked example of the sort of optimization problem that is typical of the calculus of variations. This document contains the text of the lecture presentation on this example, and you may follow along the transcript below, if you like.

## Posing the problem

Instructor: Now I'd like to discuss a particular case of interest in this sort of optimization problem. It's called the Zermelo Navigation Problem, discussed by E. Zermelo in 1931.

The physical situation to be modeled is the following one. We have a boat capable of a certain maximum speed and we want to understand how to navigate on a body of water, choosing paths that go from an originating point O to a destination D in the least possible time.

Now, if there is no current or wind to consider, so that it is equally easy to go in any direction, we can just follow the straight line segment from O to D. (We're assuming that the body of water is convex, otherwise, we'll have to take the shape of the coastline into account. This can be done, but I won't do this today, so that we can focus on a different aspect of the problem.)

Now suppose, instead, that there is a current, i.e, that the water is moving. Now, when you point the boat in a particular direction, the actual velocity vector of the boat with respect to the fixed bottom is the sum of the heading velocity and the current velocity. In a steady current, for example, this causes the boat to be dragged sideways.

You might think, at first glance, that the right thing to do is just keep heading directly for D. Such a strategy will result in the black paths in a steady current (drawn in blue), whereas aiming steadily at the appropriate fixed angle to the current will result in the green paths, i.e., straight lines.

It's not hard to see that the green paths have a shorter time traverse. Of course, since we are assuming a steady current, this problem is invariant under translation. Thus, our earlier Theorem on shortest paths with translation symmetry shows that the straight lines must be the shortest paths for time traversal. Actually, it's an interesting calculus exercise to determine the traverse time of the black paths to see this directly. I'll leave this to you. If you want hints, consult the hints page for today's lecture. After you've done this, you might want to think about why the 'head directly for the goal' strategy might still have some good points, even though it's not optimal. After all, there must be some reason that it seems to occur to most people first!

## The Euler-Lagrange equation

Now let's take a look at the mathematical formulation of the problem and see how the Euler Lagrange equations come out:

## Optimal trajectories

Robert L. Bryant <bryant@math.duke.edu>
```email address: bryant@math.duke.edu
office 'phone: (919)-660-2805
office number: 128A Physics
```

Updated: 10 May 1998