some vector calculus frequently occurs in the study of PDE

Question

As I work through PDE by L.C. Evans, as well as reading papers in the literature of linear/nonlinear PDE. The terms "inward pointing normal" and "outward pointing normal"frequently occur. For instance, in Evans, inward normal is used most often. Whereas in the research of Naviers-STokes equations, outward normal is more often used. As a beginner, I am quite vague about when to use which. Could anyone explain to me and give me some examples that help demonstrating the difference so that we can use the right term in the right situation?

Thanks!

Answer 1

I am going to assume you are talking about normal vectors to surfaces in 3d-space. A similar discussion holds for curves in the plane. Firstly, there it is not always obvious what is meant by "outward" versus "inward". Consider a plane, for example. Even though this surface has two different "sides", it is not clear which should be called the "outside" and which the "inside". In general, the choice of which side of a surface to call the "outside" is called "choosing an orientation". (There are some surfaces which are not orientable, in the sense that they only have "one side", eg. the Mobius strip and Klein bottle but I assume you do not need to worry about those for now)

If, however, you are talking about normal vectors to something like a sphere or an ellipsoid in 3d-space, then there is a natural intuitive choice of which direction to call "outward", and this should be apparent as soon as you draw a picture of the surface.

However, I do not think it is true that you will automatically find the "outward" vector whenever you try to find the normal.

For example, consider the unit sphere in 3d-space. This surface can be described by the equation $x^2 + y^2 + z^2 = 1$. A common way to find a normal vector at, say, the point $(1,0,0)$, is to find the gradient of the function $F(x,y,z)= x^2 + y^2 + z^2$ at that point. This gives you the vector $(2,0,0)$, and indeed this looks like an "outward" normal vector, since from the point $(1,0,0)$ this vector points toward the "outside" of the sphere. However, the same surface is described by the equation $-x^2 -y^2 -z^2 = -1$. If we use the same process to find a normal at (1,0,0), we get the vector $(-2,0,0)$, which points "inwards".

How do you know whether you've found the outward or the inward normal? Here's one recipe: I assume you're looking at a surface defined by an equation like $F(x,y,z)=c$, where c is a constant and F is a (differentiable) function of three variables. There's a similar recipe for parametric surfaces or curves, which I'd be happy to elaborate on if necessary.

• Sketch the surface; if there's a natural choice of what "outward" means, then it should be obvious from your sketch.
• Find a formula for a normal at an arbitrary point $(x,y,z)$ using the gradient of the function F.
• Choose a point on your surface (if possible, a nice one like the origin, or a point on a coordinate axis, to simplify the calculations). Plug the coordinates of that point into the formula for the normal that you found in second step.
• On your sketch, draw the vector that you found in third step
• based at the point that you chose. If it points in the "outward" direction, good. If not, go back to the formula you found in second step and multiply it by $-1$. (As you say, this is how you go from "outward" to "inward", and vice versa)
• As long as your surface is "nice"---and if you're in, say, vector calculus stuff, almost all surfaces you see will fall into this category---then the formula you now have will give you an outward normal at every point on the surface, not just at the one you tested in third step.

I hope this helps.