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Say I have $u=f(x,y)=x^2+y^2$. Then $u-x^2-y^2=0$. We can write $g(u,x,y)=u-x^2-y^2=0$.

As a result, we have $\nabla {g}=(1,-2x,-2y)$.

How do we interpret $\nabla {g}$. If we were to plot $g$ for various values of $x$ and $y$, we'd get $0$ everywhere. But $\nabla {g}$, which could be interpreted as the rate of change of $g$ is non-zero at many points. For example, $g(2,1,1)=0$. Here, $\nabla {g}=(1,-2,-2)$.

Thanks in advance!

Siminore
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  • the triple $(1,-2,-2)$ gives you the direction that standing at $(2,1,1)$ the function $f$ grows faster – janmarqz Mar 01 '14 at 06:23
  • How can something that is $0$ for all values of $x$ and $y$ grow? –  Mar 01 '14 at 06:24
  • $0$ is only the level of the paraboloid $u-x^2-y^2=0$ – janmarqz Mar 01 '14 at 06:26
  • I feel this case is slightly different from a general level surface. If I say $u-x^2-y^2=0$, then I have to determine the values of $u,x,y$ such that this relation is defined. Here, there cannot be any values of $x,y,z$ such that $u-x^2-y^2\neq 0$. Hence, $f(u,x,y)=u-x^2-y^2$ is equal to $0$ for all $u,x,y$. In other words, I'm not trying to draw a level surface. A level surface is all that I'm getting. –  Mar 01 '14 at 06:31
  • anyway along this paraboloid the gradient $[1,-2x,-2y]$ gives you the direction at the position $(u,x,y)$ which $f$ varies more – janmarqz Mar 01 '14 at 06:37

2 Answers2

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You are confusing yourself with poor notation.

If you have $f(x,y) = x^2+y^2$, this defines a real valued function defined everywhere. It takes values in $[0,\infty)$.

If you pick some $u \in [0,\infty)$, and consider the set of $(x,y)$ pairs that satisfy the equation $f(x,y) = u$, that is, $L_u = \{ (x,y) | f(x,y) = u \}$ then you are no longer considering arbitrary pairs of $(x,y)$ pairs, but only considering those that satisfy the equation $f(x,y) = u$.

By reusing $f$ as in $f(u,x,y) = 0$ you are confusing things and the notation police will show up at your door. Let us use $\phi(u,x,y) =u-f(x,y)$ instead.

Now note that $\phi(u,x,y) = 0 $ iff $(x,y) \in L_u$. This does not mean that $\phi$ is zero everywhere. For example, if $\phi(u,x,y) = 0$, then we have $\phi(u+1,x,y) = 1$. The equation $\phi(u,x,y) = 0$ defines a surface $G \subset \mathbb{R}^3$ (in fact, it is the graph of the function $f$).

If you pick a point $(u',x',y') \in G$, then $\nabla \phi(u',x',y')$ is a normal to the surface $G$ at the point $(u',x',y')$.

copper.hat
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  • $u$ is not a constant. It is a function. It is $f(x,y)$. –  Mar 01 '14 at 07:29
  • For context: "we have the quasi-linear equation $a(x,y,u)u_x+b(x,y,u)u_y-c(x,y,u)=0$. We assume the possible solution in the form $u=u(x,y)$." –  Mar 01 '14 at 07:31
  • @copper.hat- Sorry I forgot to address you in the above comments. –  Mar 01 '14 at 07:38
  • Then I am truly confused. You use $f$ above once with two parameters $(x,y)$ and then with three $(u,x,y)$. If you say $u=f(x,y)$ and then write $f(u,x,y)$ then I don't understand what you are writing. – copper.hat Mar 01 '14 at 07:49
  • @copper.hat- Terribly sorry for that. I think I have meaningful notation down now. I'm too used to referring to every function as $f$. –  Mar 01 '14 at 07:52
  • Then you need to write $g(f(x,y),x,y) = 0$, which is a function of $(x,y)$? I am still missing something. – copper.hat Mar 01 '14 at 07:54
  • No you're not. $g(f(x,y),x,y)=g(u,x,y)=u-x^2-y^2$. Remember that $f(x,y)=x^2+y^2$. Now we know that $g(u,x,y)$ is identically $0$ for all $x,y$. So how does a non-zero $\bigtriangledown g$ make sense? –  Mar 01 '14 at 07:58
  • This is a function of $(x,y)$ not $(u,x,y)$, so the gradient with respect to $(u,x,y)$ is a bit meaningless in this context. – copper.hat Mar 01 '14 at 08:01
  • The gradient of $g(u,x,y)$ is used extensively in my book. I'll quote the parts: "we have the quasi-linear equation $a(x,y,u)u_x+b(x,y,u)u_y−c(x,y,u)=0$. We assume the possible solution in the form $u=u(x,y)$. Then we have $g(u,x,y)\equiv u(x,y)-u=0$. Hence, $\bigtriangledown g=(g_x,g_y,g_u)=(u_x,u_y,-1)$. This gradient vector is normal to the solution surface. " –  Mar 01 '14 at 08:06
  • @copper.gat- $g(x,y,u)$ is just the function obtained by bringing $u,x,y$ to the left hand side. Obviously, it will be identically equal to $0$ for all $x,y$ –  Mar 01 '14 at 08:08
  • I think there is some notational abuse here... – copper.hat Mar 01 '14 at 08:10
  • @AyushKhaitan, $g$ is identically zero, when confined to the surface. But it easily extends to the ambient space of all $(u,x,y)$. In this extended space, we consider measuring the gradient, where $g$ sees all the other level surfaces. – Brady Trainor Mar 01 '14 at 08:10
  • You can define a function $g(a,b,c) = a-(b^2+c^2)$. This is defined everywhere and has gradient $(1, -2b, -2c)$. Then you must have $g(f(x,y),x,y) = 0$, where $f$ is defined as above. This is a function of $(x,y)$. Given any $(x,y)$, you can compute the gradient of $g$ at $(f(x,y),x,y)$, and, in general, it will not be zero. – copper.hat Mar 01 '14 at 08:16
  • @BradyTrainor- I think in this cae, there can't be any other level surfaces. Let $u=x^2+y^2$. All we're graphing is $u-x^2-y^2$. Hence, we're graphing $(x^2+y^2)-(x^2+y^2)$. –  Mar 01 '14 at 08:20
  • @AyushKhaitan, consider the point $(u,x,y)=(2,1,1)$. This point is on your surface. But $(u,x,y)=(3,1,1)$ is not on your surface. For the latter point, we have $g(3,1,1)=3-1^2-1^2=1$, thus corresponding to another level set, $g^{-1}(1)={(u,x,y,z):g(u,x,y)=1}$. Every $g^{-1}(c)$ is another level set. – Brady Trainor Mar 01 '14 at 19:26
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You start with the $(x,y)$-plane and there have a function $f(x,y):=x^2+y^2$. Then you look at the graph $$G_f:=\{(x,y,u)\>|\> (x,y)\in{\mathbb R}^2, \ u=f(x,y)\}$$ of this function, which is a surface in three-dimensional $(x,y,u)$-space. This surface can be viewed as solution set of the equation $u-x^2-y^2=0$. The left hand side $g(x,y,u):= x^2+y^2-u$ of this equation is a function of three variables, and our graph $G_f$ is a level surface of this function $g$. It follows that at each point $(x,y,u)\in G_f$ the gradient $\nabla g(x,y,u)=(-2x,-2y,1)$ is orthogonal to the tangent plane of $G_f$ at $(x,y,u)$. If we project this gradient onto the $(x,y)$-plane we obtain $\nabla'g=(-2x,-2y)=-\nabla f(x,y)$. The latter "plane" gradient vector is at each point $(x,y)\in{\mathbb R}^2$ orthogonal to the level line of $f$ through $(x,y)$.

Christian Blatter
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