by Jazzbo » Wed Mar 27, 2013 3:48 pm
I agree that it depends on the definition of emptiness. If emptiness is some measure of how much space the graph takes compared to the whole, like the area of the curve over the area of the whole, then the answer depends on your perspective. Ideally, the graph is 1 dimensional and thus has no area. So it would be zero the whole way.
If you consider the actual physical artifact instead of the idealized concept it represents, then the curve, of course, is not actually 1-dimensional but 2-d since any real-world line has thickness. And if you think not in terms of the area of the whole plane but instead the area of the rectangle before you and consider the ratio of those, then it is not zero. If you consider a straight line, it's area scales with length assuming constant thickness (an idealization). If you merely change the width of the graph and not the height, then it's area also scales by the length. And so you have a horizontal line again, just not at zero. That constant is dependent on the thickness of the line and the height of the graph.
Now I was wondering if one could have a curve as a solution instead of a horizontal line. Before I do that, I will briefly explore why a line has to be horizontal from a different point of view.
First consider a diagonal line with thickness w and length l at an angle theta going across a graph of with width x and height h:
h w
| /
|l/
|/theta
|___ x
x/l = cos theta
thus l = x / (cos theta)
Area of line over area of graph is lw/(hx) = wx/(hx cos theta) = w/(h cos theta)
Since w, h, and theta are constant, this is a constant which means it must be a horizontal line as argued before. The graph would be y = w/h.
Now let us look at a general curve. We are proposing that the area of the thick line after horizontal distance x over the area of the rectangle of the graph up to width x equals the value of the function at x.
(integral x'=0->x [w*dl])/(hx) = y, where h is height of the graph and x is the current width of the graph and dl is the differential length of the curve (construed here has a function of x) and w is the width of the curve.
So (w/h) integral x'=0->x [dl] = xy
Take derivative of each side by x.
(w/h)dl/dx = x*dy/dx + y
(w/h)sqrt( (dy/dx)^2 + 1) = x*dy/dx + y
Now, let us consider boundary conditions. The highest ratio possible between the areas is 1. Of course it is going to be less than one. The graph will not take up the whole area, so there must be a horizontal asymptote. That means that as x goes to infinity, dy/dx goes to zero. dy/dx should go to zero much faster than x goes to infinity, so the above will give you,
w/h = y when x approaches infinity.
In other words, the asymptote is the horizontal line discussed earlier. We also know that any deviation of the curve from a horizontal line lengthens the line and hence increases the area. So the curve cannot go down. It can only go horizontally or up. Thus if there is a solution to the differential equation that is not a horizontal line, it must start below the asymptote. But the initial behavior of the curve must mirror the behavior of the tangent line. Any line will have a value of w/(h cos theta) which is always greater than or equal to w/h. Since it cannot be greater, the horizontal line is the only possible solution.