 So we're now going to take a look at an example problem involving the potential function. And what we're going to do, we're going to solve for the potential function for the same velocity field that we looked at when we looked at the stream function, and that was one where we had this constant a, and so this is the vector field that we have for velocity, and a equals a constant. And we're told to find the potential function. So we're looking for the velocity potential, and that is phi. So the way that we're going to do this, we'll use a similar manner to what we did with the stream function. We begin by looking at the two components of velocity, but for the stream function u is partial phi by partial x, and what we can do, looking at our velocity relationship, that is that component there. So that is equal to ax, and if I go ahead and integrate that, what I then get is phi is equal to ax squared divided by 2 plus some function of y. It could be a constant or a function of y. We leave it arbitrary at this point. I am then going to take this, and I'm going to plug it into the relationship for velocity as d phi by dy, and then I'm going to equate that to minus ay, which is from the velocity that we know. So let's go ahead and do that. And when we do that, the first term, the a over 2x squared is 0, and then we have f of y, and so all we have is the derivative of f with respect to y, and that is then equal to the v component of velocity for the velocity field we have, which is minus ay. And so with that, what I can do is I can integrate to solve for f, and once we integrate for f, we get this plus, and in this case, it's going to be some arbitrary constant that we don't know what it is. So we can take this here and plug it into that, and that gives us our velocity potential. So that would be the velocity potential function. What I'm going to do now is I'm going to show you contour plots that have been generated with a equals 1 and the constant equal to 0. And so the velocity potential function, when you do that, looks something like this. So this is for phi equals a, actually it would be 1 over 2 because I've substituted 1 for a. So we have 1 over 2x squared minus 1 over 2y squared and plus 0 for our constant. And so that is what the velocity potential looks like for the velocity field, and that was a velocity field that was xi minus yj. So for that velocity field, that is the potential function that we obtain. And you can see it looks quite a bit different from the stream function if you refer back to the example problem where we solve for the stream, or for the stream function of that velocity.