I'm confused about how $\nabla\cdot ( x, y)= 2$ at each point in the vector field.

Question

I'm confused about how the divergence at each point in the vector field described by $( x, y ) $ is equivalent to 2. For instance, take the point $( -2, 1 )$. At this point, the $x$-component is decreasing, since it's becoming more negative. At the 2nd, 3rd, and 4th quadrants of the $xy$ plane, the $x$- or $y$-components are decreasing since they're getting more negative. Perhaps, I'm confused about the geometric meaning of summing the components of the gradient vector. That's become I'm also confused about how the divergence having a positive value at a specific point means that its pointing away from the origin. Any help would be appreciated!

Answer 1

First, let me address the things that are not related to the divergence of $\langle x,y\rangle$:

1. In quadrants that are not the 1st quadrant, the $x$- and/or $y$- components of the vector field would decrease (become more negative) if you followed the vector field. But the rate of change in the direction of the vector field is not what divergence (or anything else I'm aware of) measures.
2. We are not summing the components of the gradient vector, because a gradient field does not have a gradient vector. A scalar field $f(x,y)$ has a gradient vector $\langle f_x(x,y), f_y(x,y)\rangle$. A vector field $\langle M(x,y), N(x,y)\rangle$ instead has a gradient matrix $$\begin{bmatrix}M_x(x,y) & M_y(x,y) \\ N_x(x,y) & N_y(x,y)\end{bmatrix}$$ and the divergence is only looking at two components of that matrix (the diagonal ones).
3. The vector field is pointing away from the origin everywhere, but this doesn't necessarily say anything about the divergence at a specific point. A good related example is $\langle x,y\rangle/(x^2+y^2)^p$ for various powers $p$; if $p$ is large, then the divergence will actually become negative sufficiently far from the origin, even though the vectors have the same direction.

Instead, the divergence is only about the rate of change of the components of $\langle x,y\rangle$. Here are a couple of ways to think about that.

First, going exactly from the definition: no matter which point $(x,y)$ you're at, the $x$-component of $\langle x,y\rangle$ will become a bit larger if you increase $x$, and the $y$-component will become a bit larger if you increase $y$. This is the mechanical reason the divergence is positive.

Second, appealing to physics: if you think of $\langle x,y\rangle$ as representing the velocity of gas at point $(x,y)$, then the divergence measures how much the gas is expanding at that point. For the vector field $\langle x,y\rangle$, it is very easy to see that it is expanding at $(0,0)$ from the picture. That's because at $(0,0)$, the gas is not doing anything other than expanding. At another point $(a,b)$, the gas is moving very quickly in the direction $\langle a,b\rangle$, and also expanding. It's expanding because, relative to the particles at $(a,b)$, the particles closer to the origin are moving a bit slower, and the particles further from the origin are moving a bit faster. All in all, the particles near $(a,b)$ are spreading out.

Answer 2

Let $\mathbf{F}(x,y) = x \mathbf{i} + y \mathbf{j}$ be the vector field in question (as comments say, $\langle x, y \rangle$ is an inner product, not a vector field).

As you found, the divergence of this particular vector field is $\nabla \cdot \mathbf{F}(x,y) = \frac{\partial\mathbf{F}}{\partial x} + \frac{\partial\mathbf{F}}{\partial y} = 1+1=2$

I think your confusion is around the interpretation of the value $2$, and the divergence more generally. There is a good explanation on Wikipedia for the general concept, but to keep this answer self-contained I'll summarize the intuition here.

A good physical model for vector fields is a fluid, because the mathematical properties of vector fields map nicely to commonsense notions about fluids. In particular, the divergence of the vector field at a point tells us if the field is "expanding" (divergence > 0), "compressing" (divergence < 0), or neither (divergence = 0).

I think one of your confusions is mistaking the direction of the vectors themselves (i.e., $(-2,1)$) with the divergence, which is about the net rate of "flow" into the point $(-2,1)$ vs the point itself. Imagine a small square centered on your point $(-2,1)$ with dimensions $2\delta \times 2\delta$. We want to calculate the flux or "flow" $\phi(x,y;\delta)$ of fluid out of this box (centered at our point $x,y$):

$$ \phi(x,y;\delta) = \int_{x-\delta}^{x+\delta} \mathbf{F}(z,y-\delta)\cdot \mathbf{-j}\;dz+ \int_{x-\delta}^{x+\delta} \mathbf{F}(z,y+\delta)\cdot \mathbf{j}\;dz $$ $$+\int_{y-\delta}^{y+\delta} \mathbf{F}(x+\delta,z)\cdot \mathbf{i}\;dz +\int_{y-\delta}^{y+\delta} \mathbf{F}(x-\delta,z)\cdot \mathbf{-i}\;dz$$

Substituting in the defintion of $\mathbf F (x,y)$ we get:

$$ \phi(x,y;\delta) = \int_{x-\delta}^{x+\delta} [z\mathbf i + (y-\delta)\mathbf j]\cdot \mathbf{-j}\;dz+ \int_{x-\delta}^{x+\delta} [z\mathbf i + (y+\delta)\mathbf j]\cdot \mathbf{j}\;dz $$ $$+\int_{y-\delta}^{y+\delta} [(x+\delta)\mathbf i + z\mathbf j]\cdot \mathbf{i}\;dz +\int_{y-\delta}^{y+\delta} [(x-\delta)\mathbf i + z\mathbf j]\cdot \mathbf{-i}\;dz$$

In each term, we see that we are taking the dot product with one of the standard unit vectors (since we defined the square so its sides are aligned to our coordinate system) which greatly simplifies the calculation:

$$ \phi(x,y;\delta) = \int_{x-\delta}^{x+\delta} -(y-\delta)\;dz+ \int_{x-\delta}^{x+\delta} (y+\delta)\;dz +\int_{y-\delta}^{y+\delta} (x+\delta)\;dz +\int_{y-\delta}^{y+\delta} -(x-\delta)\;dz =$$

$$2\delta \left[ -(y-\delta) + (y+\delta) + -(x-\delta) + (x+\delta)\right] = 8\delta^2$$

All we're doing above is calculating the component of the vector field that is perpendicular to the edge of the box we are focusing on (hence four terms) and pointing out of the box.

Note that if the flow into the box (i.e., the integral is negative for that term) is balanced by flow out of the box (i.e., positive integral for a particular term), then we have zero net flow and zero divergence. A good example of this is water or other incompressible fluid: if you pump water into one side of a full pipe, the same volume must come out the other end, so the net flow into the pipe is zero. The same is not true of a gas -- think of filling up a tire -- we can pump a lot of air into the tire without any of it coming out (unless there's a leak ;)

To get divergence, we want to look at the flow rate per unit area of the square. The area of the square is $2\delta \times 2\delta = 4\delta^2$ so dividing by the area we arrive at our divergence:

$$\nabla \cdot \mathbf F = 2 = \frac{8\delta^2}{4 \delta ^2}$$

This worked out nicely since we are dealing with a linear vector field in this case. In the more general case we need to take the limit as $\delta \to 0$ but we get the same outcome.

So, a divergence of 2 means that every point in this vector fields is experiencing the same rate of expansion (e.g., uniformly expanding gas). This means that if you put a drop of ink at $(x,y)$ it will "smear out" as it flows away from the origin (the expansion caused by the divergence).

Note that a positive divergence doesn't mean all vectors point away from the origin -- we can always "shift" the origin of the vector field. For example:

$$\mathbf F(x,y) = (x-a)\mathbf i + (y-b) \mathbf j$$