Continuity for Functions of Several Variables, Partial Derivatives

1. We considered differentiating and integrating functions of a single-variable. How? We began with the notion of a limit, and then considered the derivative. If we have a, e.g., polynomial
(1)\displaystyle  p(x,y) = x^{3}+x^{2}y+xy^{2}+y^{3}
we see
(2)\displaystyle  p(x+\Delta x,y) = x^{3}+x^{2}y+xy^{2}+y^{3}+\bigl(3x^{2}+2xy+y\bigr)\Delta x+ \mathcal{O}(\Delta x^{2})
Again we stop and reflect: this treats {y} as if it were constant. So the derivative formed by
(3)\displaystyle  \lim_{\Delta x\rightarrow0}\frac{p(x+\Delta x,y)-p(x,y)}{\Delta x}=3x^{2}+2xy+y
are “incomplete” or partial. There is some subtlety here due to using multiple variables, and we have to discuss the problems of limits first.

2. Definition. The function {z=f(x,y)} is “Continuous” at {(x_{0},y_{0})} if

(i) {f(x_{0},y_{0})} is defined and finite;

(ii) {\displaystyle\lim_{(x,y)\rightarrow(x_{0},y_{0})}f(x,y)=f(x_{0},y_{0})} is defined;

(iii) {\displaystyle\lim_{(x,y)\rightarrow(x_{0},y_{0})}f(x,y)} is defined (and finite).

Note: this can be determined by picking any curve {\gamma\colon[0,1]\rightarrow{\mathbb R}^{2}} which satisfies
(4)\displaystyle  \gamma(t_{0}) = (x_{0},y_{0})
for some {0\leq t_{0}\leq 1}, then taking
(5)\displaystyle  \lim_{t\rightarrow t_{0}}f\bigl(\gamma(t)\bigr) = \lim_{(x,y)\rightarrow(x_{0},y_{0})}f(x,y).
The subtletly here lies with {\gamma} being arbitrary. If two different curves produce two different results, the limit does not exist. Lets consider some examples and non-examples.

Example 1 (Limit Exists). Find
(6)\displaystyle  \lim_{(x,y)\rightarrow(2,4)}\frac{y+4}{x^{2}y-xy+4x^{2}-4x}.

Solution: for this, we can simply plug in the values
(7)\displaystyle  \begin{aligned} \lim_{(x,y)\rightarrow(2,4)}\frac{y+4}{x^{2}y-xy+4x^{2}-4x} &=\frac{(4)+4}{(2)^{2}(4)-(2)(4)+4(2^{2})-4(2)}\\ &=\frac{8}{16-8+16-8}=\frac{1}{2}. \end{aligned}
This is because the function is sufficiently nice.

Example 2 (Limit Doesn’t Exist). What is
(8)\displaystyle  \lim_{(x,y)\rightarrow(0,0)}\frac{x^{4}}{x^{4}+y^{2}}=?

Solution: Lets first approach it along the {x}-axis, i.e. first setting {y=0}. We find
(9)\displaystyle  \lim_{(x,y)\rightarrow(0,0)}\frac{x^{4}}{x^{4}+y^{2}}=\lim_{x\rightarrow0}\frac{x^{4}}{x^{4}}=1.
Now lets approach it on the {y}-axis, i.e. first setting {x=0}. We see
(10)\displaystyle  \lim_{(x,y)\rightarrow(0,0)}\frac{x^{4}}{x^{4}+y^{2}}=\lim_{y\rightarrow0}\frac{0}{0+y^{2}}=0.
Still, approaching along the curve {y=x^{2}} we see
(11)\displaystyle  \lim_{(x,y)\rightarrow(0,0)}\frac{x^{4}}{x^{4}+y^{2}}=\lim_{x\rightarrow0}\frac{x^{4}}{x^{4}+x^{4}}=\frac{1}{2}.
But we have a problem: this implies {0=1/2=1}. This cannot be! So the limit cannot exist! Very sad.

3. Definition. Let {z=f(x,y)} be defined on a region {R} in the {xy}-plane, and let {(x_{0},y_{0})} be an inerior point of {R}, we just don’t want a boundary point!

(12)\displaystyle  \lim_{\Delta x\rightarrow0}\frac{f(x_{0}+\Delta x,y_{0})-f(x_{0},y_{0})}{\Delta x}
exists, then it is called the “Partial Derivative” of {z=f(x,y)} at {(x_{0},y_{0})} with respect to {x}. It is denoted
(13)\displaystyle  \frac{\partial}{\partial x}f = \frac{\partial}{\partial x}z =f_{x} = \partial_{x}f = \partial_{x}z
evaluated at {(x_{0},y_{0})}. NB: the subscripts in the {\partial_{x}} indicate what variable we are taking the partial derivative of, i.e., it’s shorthand for {\partial_{x}=\partial/\partial x}.

Under similar conditions,
(14)\displaystyle  \lim_{\Delta y\rightarrow0}\frac{f(x_{0},y_{0}+\Delta y)-f(x_{0},y_{0})}{\Delta y}
is the partial derivative of {z=f(x,y)} with respect to {y} at {(x_{0},y_{0})}. We denote this by
(15)\displaystyle  \frac{\partial f}{\partial y}=\frac{\partial z}{\partial y}=\partial_{y}f = \partial_{y}z
among a myriad of different conventions.

4. Higher order partial derivatives are done by taking it one at a time. So if
(16)\displaystyle  z=\mathrm{e}^{xy}
for example, we have
(17)\displaystyle  \partial_{y}z=x\mathrm{e}^{xy}
and taking its derivative again yields
(18)\displaystyle  \begin{aligned} \partial_{y}^{2}z &= \partial_{y}\left(x\mathrm{e}^{xy}\right)\\ &=x\partial_{y}(\mathrm{e}^{xy}) \end{aligned}
Note we factor {x} out in front of the partial derivative with respect to {y} because {x} is constant with respect to {y}. So we then obtain
(19)\displaystyle  \partial_{y}^{2}z = x^{2}\mathrm{e}^{xy}.
We take partial derivatives one at a time, from right to left:
(20)\displaystyle  \partial_{x}\partial_{y}z = \partial_{x}\bigl(\partial_{y}z\bigr).
Question: do partial derivatives commute? I.e., is {\partial_{x}\partial_{y}=\partial_{y}\partial_{x}} always? Lets first consider an example calculation before considering an answer.

Example 3. Consider the function {u=x^{2}-y^{2}}. Find {\partial_{x}^{2}u+\partial_{y}^{2}u}.

Solution: We find that
(21)\displaystyle  \partial_{x}u = 2x\implies \partial_{x}^{2}u = 2.
Similarly, we find
(22)\displaystyle  \partial_{y}u=-2y\implies \partial_{y}^{2}y=-2.
Thus we conclude
(23)\displaystyle  \partial_{x}^{2}u+\partial_{y}^{2}u=2-2=0.

5. Do Partial Derivatives Commute? Answer: not always. The conditions are fairly weak: if {\partial_{x}z}, {\partial_{y}z}, {\partial_{x}\partial_{y}z} and {\partial_{y}\partial_{x}z} are continuous throughout their respective domains, then
(24)\displaystyle  \partial_{x}\partial_{y}z = \partial_{y}\partial_{x}z.

Exercise 1. Let {u(x,t) = f(x+vt) + g(x-vt)} where {v\not=0} is some constant. Prove
(25)\displaystyle  \partial_{t}^{2}u(x,t)=v^{2}\partial_{x}^{2}u(x,t).

Exercise 2. Let {f(x,y)=\ln(x^{2}+y^{2})}. What is {\partial_{x}^{2}f(x,y)+\partial_{y}^{2}f(x,y)}?

Exercise 3. Consider {g(x,y)=1/\sqrt{x^{2}+y^{2}}}. What is {\partial_{x}^{2}g(x,y)}? What is {\partial_{y}^{2}g(x,y)}? Is {\partial_{x}\partial_{y}g(x,y)=\partial_{y}\partial_{x}g(x,y)}?

Exercise 4. Let {f(x,y)=3x^{2}+4y^{3}++x^{2}y^{3}+\sin(xy)}. What is {\partial_{x}f}? What is {\partial_{y}f}?

Exercise 5. Let {z=\arctan(x^{2}\mathrm{e}^{2y})}. What is {\partial_{x}z}? What is {\partial_{y}z}?


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Calculus, Partial Derivative, Vector Calculus and tagged , . Bookmark the permalink.

One Response to Continuity for Functions of Several Variables, Partial Derivatives

  1. Pingback: Chain Rule for Partial Derivatives | My Math Blog

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s