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Operators - Discussion

1. What is an operator? A function whose domain is a function. When we do something to functions, rather than call it a function of functions, we call it an operator on functions.
   • The derivative $\frac{df}{dx}$ operates on a function $f(x)$ to produce a new function $f'$.
   • The partial derivative $\frac{\partial f}{\partial x}$ operates on a function $f(x,y)$ to produce a new function $f_x(x,y)$.
   • The integral $\int_a^b f(x) dx$ operates on a function $f(x)$ to produce a number (the area under $f$ from $a$ to $b$).
   • The gradient $\vec \nabla f$ operates on a function $f(x,y,z)$ to produce a vector field $(f_x,f_y, f_z)$.
   • The del operator $\vec \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right)$ or $\vec \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$ can be used in a variety of ways to produce an operator.
2. Three new quantities (A great book: "Div, Grad, Curl, and all that" by H. M. Schey - Electrostatics)

   • The gradient of a function $f$, namely $\vec \nabla f = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)$, abuses the notion of scalar multiplication to produce an operator on a function $f$.
   • The divergence of a vector field $\vec F$ is $\vec \nabla \cdot \vec F$, read "del dot F." This operator helps understand whether a vector field $\vec F$ is expanding or contracting. This helps understand gas diffusion. Things that are incompressible (a common assumption about liquids) have zero divergence which means $\vec \nabla \cdot \vec F=0$.
   • The curl of a vector field $\vec F$ is $\vec \nabla\times \vec F$, read "del cross F.". This quantity tracks the rate and direction of rotation that a vector field $\vec F$ causes. Irrotational vector fields have zero curl. We'll see that every gradient field has zero curl, which is precisely related to why we never saw imaginary eigenvalues in the optimization unit.

We've used all the symbols above before, but now we'll use them in a new way. This will be a common theme in the 6th unit, namely we'll review things we've seen all semester, and use them in a slightly new way.

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Brain Gains (Recall/Generation)

Let $$\vec \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right).$$

• For the vector field $\vec F(x,y,z) = (3xy,2x+4y+5z,y+xz^2)$, compute the derivative $D \vec F$ (you'll get a 3 by 3 matrix).

• For the vector field $\vec F(x,y,z) = (3xy,2x+4y+5z,y+xz^2)$, compute $\vec \nabla \cdot \vec F$.

• For the vector field $\vec F(x,y,z) = (3xy,2x+4y+5z,y+xz^2)$, compute $\vec \nabla \times \vec F$.

• For the vector field $\vec F = (4x+2xy,x^2+2y)$, find a function $f(x,y)$ so that $\vec \nabla f = \vec F$. We call $f$ a potential for $\vec F$. In other words, if we know $f_x = 4x+2xy$ and $f_y = x^2+2y$, then what is $f$?

• Does every vector field have a potential?

Group problems

1. Let $f = xy^2+3x$.
   1. Compute $\vec F = \vec \nabla f$. (We'll use capital $F$ for vector fields, and lower case $f$ for real-valued functions, i.e. the output is a single real number, rather than a vector.)
   2. Compute $D^2f$ (so $D\vec F$). It's a square matrix.
   3. Compute $\int_{ (2,1) }^{ (1,3) } df$ (total change in $f$ from $(2,1)$ to $(1,3)$). [Check: Plugging the points into $f$ yields $12-8 = 4$.]
2. Let $\vec F = (2xy+4, x^2+2y)$.
   1. Compute $D\vec F$. It's a square matrix.
   2. Find a real-valued function $f$ so that $\vec \nabla f = \vec F $. In particular, this means $df = \vec F \cdot d\vec r$. [Check: $f = x^2y+4x+y^2$.]
   3. Find the work done by $\vec F$ (so $\int_C \vec F\cdot d\vec r$) to get from $(2,0)$ to $(0,3)$. (Or simpler, just compute $\int_C df$, the total change in $f$ from $(2,0)$ to $(0,3)$, which you can do because $df = \vec F \cdot d\vec r$. ) [Check: $9-8=1$.]
3. Let $\vec F = (2x+3y, 4x+5y)$.
   1. Compute $D\vec F$.
   2. Why it is impossible to find a function $f$ so that $\vec F = \vec \nabla f$.
4. Given a vector field $\vec F$, what condition must be true about $D\vec F$ for there to be a function $f$ such that $\vec\nabla f = \vec F$? We call such function $f$ a potential for $\vec F$. When $\vec F$ has a potential, we say that $\vec F$ is a gradient field.
5. Compute the derivative of each vector field $\vec F$ below (obtaining a square matrix). Then find a potential for $\vec F$ or explain why the vector field has no potential.
   1. $\vec F = (2x,3y)$ [Check: $D\vec F = \begin{bmatrix}2&0\\0&3\end{bmatrix}$ and $f = x^2+\frac{3}{2}y^2$ yields $\vec \nabla f = (2x,3y)$. We can quickly verify that $\vec\nabla f = \vec F$ by a direct computation. ]
   2. $\vec F = (2y,3x)$
   3. $\vec F = (3y,3x)$
   4. $\vec F = (4x,5y,6z)$
   5. $\vec F = (4x,5z,6y)$
   6. $\vec F = (4x,5z,5y)$
   7. $\vec F = (2x-y,-x+4y)$
   8. $\vec F = (y^2+2x,2xy)$
   9. $\vec F = (x+yz,xz+4yz,xy+2y^2)$
   10. $\vec F = (x+yz,4yz,xy+2y^2)$
   11. $\vec F = (x+yz,xz+4yz,xy)$
   12. $\vec F = (yz,xz+4yz,xy+2y^2)$

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Problem Set
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