# Homework Help: Significance of nonvanishing gradient

1. Dec 9, 2016

### Syrus

1. The problem statement, all variables and given/known data
I am working in "Intro to PDEs with Applications" on page 6. Gradients come up in discussions of surfaces expressed as F(x,y,z). In discussing such matters, the buildup includes the assumption that grad F is not equal to the zero vector. A later line reads, "Under the assumption [above], the set of points (x,y,z) in the domain which satisfy the equation F(x,y,z) = c for some appropriate value of c, is a surface is the domain.

2. Relevant equations

3. The attempt at a solution

My question is, what precautions does the gradient not being zero entail? The only answer I've been able to come up with so far is that they mean smooth surface when they simply say surface (since points where grad F = 0) seem to correspond to 'corners' or non-differentiate points. Any deeper or more accurate insight?

2. Dec 9, 2016

If you have a gradient of zero over an extensive region, it allows for $F(x,y,z)$ being constant throughout that region. In this case setting $F(x,y,z)=c$ and finding the points $(x,y,z)$ will not define a "infinetesimally thin" surface, but rather the set of points could consist of something of finite thickness in places where the gradient is zero.

3. Dec 9, 2016

### TJGilb

All a gradient does is point in the direction of max increase for a given function. Therefore, a function that has no point anywhere with a gradient equal to zero simply means there is no peak, no local or absolute maximum in other words.

4. Dec 9, 2016

### Syrus

None of what you guys said attests as to why you couldn't have a "surface" with those properties. Link- I've never heard of "thickness" as a consideration of surfaces- but then again- how does this take away from the SURFACE aspect of it? Gilb- who cares if there's no local or absolute minima or maxima? A surface doesn't need any (e.g. a horizontal plane).

5. Dec 9, 2016

### TJGilb

I misunderstood your question then. Think about it like this. Define the axis for your surface in Cartesian coordinates. Now try thinking up any finite, non-infinite, surface that never switches from having an upward curve to downward curve. One that never peaks. The only possibility is a flat horizontal surface.

6. Dec 9, 2016

### TJGilb

Also what are you taking the gradient of? Density? Height? Temperature? I sort of assumed height in my prior post.

7. Dec 9, 2016

### Syrus

It's just the gradient of an arbitrary surface F(,x,y,z). Not necessarily a function.

8. Dec 9, 2016

### TJGilb

You can only take the gradient of a scalar field, therefore a function. It has to be a function of something. In the example of F(x,y,z) what you would typically be taking the gradient of is z by defining a new function (if you were trying to just take the gradient of the surface itself, think of a contour map in this case).

9. Dec 9, 2016

### TJGilb

Without actually seeing the full context of the text, what this seems to be referring to is that for the gradient of a function of three variables, every set of points that shares the same gradient value forms a surface (just like for a function of two variables, every set of points that shares a gradient forms a contour). Is this what you were asking for clarification on?

10. Dec 9, 2016

I think the thing that defines the surface here is the set of points $(x,y,z)$ such that $F(x,y,z)=c$ where $c$ is a constant. The gradient $\nabla F(x,y,z)$ points normal to the surface at any point and will normally vary in amplitude and direction over the whole surface.

11. Dec 9, 2016

### Syrus

I guess what I'm trying to ask is why is the assumption that grad(F) is nonzero necessary in the ensuing discussions? The text goes on to discuss topics such as using grad to find equations of tangent planes to surfaces and (eventually) to prove that a curve can result as the intersection of two surfaces (by saying grad F1 X grad F2 is also nonzero.

12. Dec 9, 2016

As I previously responded, the gradient being non-zero is a sufficient (probably not necessary, but sufficient) condition to create a well-defined surface. If the surface has a finite thickness, then where is the surface precisely located, i.e. what is the z for a given x and y? It would not be precisely defined if the thickness was finite.

13. Dec 9, 2016

### TJGilb

Well, a gradient of zero $\nabla f = f_x\hat x +f_y\hat y +f_z\hat z = 0$ means that each component is zero, ie the zero vector. If you were to take the cross product of two zero vectors, or any vector with a zero vector for that matter, what do you get? Zero. It defeats the purpose of a cross product in this case, which is to get a new vector orthogonal to both.

14. Dec 9, 2016

### Syrus

Charles Link- I think I understand what you're saying, but let's consider the cone given by F(X,y,z) = z^2-x^2-y^2 = 0. Grad F = 0 at (0,0,0). But this seems to have nothing with the concept you're talking about- or does it?

15. Dec 10, 2016

### TJGilb

It all depends on what you're trying to do with the gradient. There's nothing inherently preventing the gradient from being zero, but it's just not a very valuable for the types of applications it looks like your book is talking about. After all, there is only one point on that surface with a zero gradient, which certainly isn't a curve, and you won't be finding any tangent planes with it.

16. Dec 10, 2016

### Syrus

I'm also not finding any sources on the gradient being sufficient to have a well-defined surface. Any references for this?

17. Dec 10, 2016

### TJGilb

That depends on what function you're taking the gradient of. The gradient of a function of three variables can be divided into a set of surfaces, where each surface represents every point that shares the same gradient magnitude. If it was a function of two variables, instead of being divided into a set of surfaces, it would be a set of curves.

18. Dec 10, 2016

### TJGilb

Oh and the gradient can serve as a normal vector with which to find a tangent plane at any point (assuming it's nonzero).

19. Dec 10, 2016

### Syrus

Well put, I assume that is why they made that assumption- so the ensuing discussion applied to all points considered (as long as grad F is nonzero).

20. Dec 10, 2016

### Ray Vickson

Look up the "Implicit Function Theorem".

Having a nonzero gradient vector at $(x_0,y_0,z_0)$ means that at least one of the partial derivatives is nonzero. For example, if $F_z(x_0,y_0,z_0) \neq 0$, then $dF = 0 = F_x dx + F_y dy + F_z dz$ can be solved for $dz$ in terms of $dx$ and $dy$ (because we are allowed to divide by $F_z \neq 0$):
$$dz = -\frac{F_x}{F_z} dx - \frac{F_y}{F_z} dy.$$
That implies that for $(x,y)$ near $(x_0,y_0)$ we have $z = f(x,y)$ for some smooth function $f(x,y)$. In other words, if $F_z \neq 0$ the equation $F(x,y,z) = 0$ leads to a unique expression for $z$ as a smooth function of $(x,y)$, and the function is unique, at least near $(x_0,y_0,z_0)$.

If the gradient vanishes, we may have a non-smooth and non-unique function (that is, two or more functions $f_1(x,y)$ and $f_2(x,y)$ which can also be non-smooth), or sometimes we can have several different smooth functions $z = f_i(x,y)$ that all satisfy $0 = F(x,y,f_i(x,y))$. Alternatively, we may have no such functions at all; it all depends on details about the nature of the original function $F$.

21. Dec 10, 2016

The cone is a well-defined surface. An isolated point where the gradient is zero will have little effect. I don't have a whole lot more to provide other than to say in the case of the gradient being zero over an extended range would make the calculations your textbook does non-applicable. In an advanced calculus course quite a number of years ago the mathematics professor did tell us there are entire books written about how surfaces are defined. For most practical applications, it is not necessary to study this concept in such depth, but just a common sense approach should give you most of what you need.

22. Dec 10, 2016

### Ray Vickson

For $F(x,y,z) = z^2-x^2-y^2,$ the fact that $\nabla F = (0,0,0)$ at $(x,y,z) = (0,0,0)$ means that the standard theorems about existence of a unique smooth surface $z = f(x,y)$ through $(0,0,0)$ are violated. There are two different surfaces $z = \sqrt{x^2+y^2}$ and $z = - \sqrt{x^2+y^2}$, and both of them happen to be non-smooth (lacking continuous derivatives at $(x,y) = (0,0)$).

23. Dec 10, 2016

### Syrus

Ray- you're absolutely correct- the implicit function theorem is in fact discussed and I believe both your points apply here