A one dimensional example of divergence: Mystery

[GreenWombat]

TL;DR Summary

When I draw the vector field of y = -1 + x^2 it seems different from the calculated divergence.

I am trying to understand “divergence” by considering a one-dimensional example of the vector y defined by:
. the parabola: y = -1 + x^2

The direction of the vector y will either be to the right ( R) when y is positive or to the Left (L).

The gradient = dy/dx = Divergence = Div y = 2 x

x	-3	-2	-1	-0.5	0	0.5	1	2	3
x^2	9	4	1	0.25	0	0.25	1	4	9
y	8	3	0	-0.75	-1	-0.75	0	3	8
Arrow direction	R	R	-	L	L	L	-	R	R
Div y = 2x	-6	-4	-2	-1	0	1	2	4	6

To see the vector field of y, I drew an arrow for each value of y, in the direction indicated by the sign of y, at the corresponding value of x on a grid.
As suggested by the above table:
. The arrows at x = -1 and -2 point right.
. There is no arrow at x = 1 as y = 0
. The arrows at x = -0.5, 0 and 0.5 point left.
It seems to me that there is a sink at x = 1

Similarly is seems that there is a source at x = 1.

However, Div y indicates:
. for x < 0 all points are sinks and
. for x > 0 all points are sources.

I think I have something wrong here. Can anyone help?

I am trying to create a one-dimensional explanation along the lines of the two-dimensional examples given here.
https://www.khanacademy.org/math/mu...ves/divergence-and-curl-articles/a/divergence

 

[Stephen Tashi]

Imagine the vector field plotted on the x-axis. All the arrows would be overlapping each other, but imagine you can distinguish them. For $h > 0$ the line segment $[x, x + h]$ has an arrow at the lower end and an arrow at the upper end. The sizes of these arrows are $-1 + x^2$ and $-1 + (x+h)^2$.

There is a net source in the line segment if the arrow at the upper end is more positive (or less negative) than the arrow at the lower end. The average density of the source in the interval is $\frac{ (-1 + (x+h)^2) - (-1 + x^2)}{h}$. To compute the divergence, you take the limit as $h \rightarrow 0$ and you can see that this amounts to computing the derivative of $-1 + x^2$.

So for $x < 0$ you need a negative divergence in order to make the arrow on the right of an interval less positive or more negative that the arrow on the left. The fact that arrows may change signs over an interval doesn't, by itself, determine the sign of the divergence. In the case they do change signs, what matters is whether the change from left to right is from positive to negative or vice versa.

If we think of the arrow at the left end of the interval as "given", we can imagine the divergence as some physical phenomenon that causes a change to that arrow as it "passes through" the interval.

 

[GreenWombat]

In a mystery, everything becomes suspect and your reply straightened out a few things. Thanks

Various people suggest that you can understand divergence by thinking of the vector field to as the velocity field of flowing gas. I am trying to understand this for myself, and so I can explain it to my daughter, by considering one-dimensional examples.

******* For y = x - c
Div y = 1

For c = 1 we have

. x	-3	-2	-1	0	1	2	3
. y	-4	-3	-2	-1	0	1	2
Arrow direction	L	L	L	L	-	R	R
Div y	1	1	1	1	1	1	1

This is uniformly divergent.

The value of c sets the value of x away from which the flow arrows diverge.
The apparent point of divergence is a source, as is every point in the whole field - as shown by div y. (everything expanding - like the big bang)

****** For y = -x + c
Div y = -1

For c = 1 we have y = -x + 1

. x	-3	-2	-1	0	1	2	3
. y	4	3	2	1	0	-1	-2
Arrow direction	R	R	R	R	-	L	L
Div y	-1	-1	-1	-1	-1	-1	-1

This is uniformly “negatively divergent” = convergent.

The value of c sets the value of x towards which the flow arrows converge.
The apparent point of convergence is a sink, as is every point in the whole field - as shown by Div y.

******* For y = x^2
Div y = 2x

. x	-3	-2	-1	0	1	2	3
. y	9	4	1	0	1	4	9
Arrow direction	R	R	R	-	R	R	R
Div y	-6	-4	-2	0	2	4	6

Here, as x increases, the divergence consistently increases from -6 (convergent) to +6 (divergent).These one-dimensional examples (two linear and the two parabolic) would seem to be as simple as you can get.

Understanding divergence via these “flow arrows” is becoming clearer for me in the linear examples - but the parabolic examples are still confusing.

Is there some better way of understanding these flow arrows, or are there better examples.

I have been considering divergence because of Maxwell’s laws. I’ve been hoping for a single simple equation in which there is one point of divergence surrounded by a zero divergence. (This is the opposite of the above examples in which there are points of zero-divergence surrounded by non-zero divergence.) It seems that this is a vain search as the magnitude of an electric field around an electric charge is E = kQ/r^2 and even this equation only applies “around the electric charge”, and not where r = 0, where div E is non-zero.