Partial Differentiability of Real Scalar Fields

Definition: (Continuous) Partial Differentiability of Real Scalar Fields

A real scalar field $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ and let $\varphi$ be a coordinate system for ${\mathbb{R}}^n$.

We say that $f$ is $k$-times (continuously) partially differentiable at some $\mathbf{a}\in \mathcal{D}$ iff all of its $k$-th order partial derivatives with respect to all coordinates of $\varphi$ exist at $\mathbf{a}$ (and are continuous there).

If the above is true for every $\mathbf{x}\in \mathcal{D}$, then we say that $f$ is $k$-times (continuously) partially differentiable on $\mathcal{D}$. We can also omit “on $\mathcal{D}$“.

NOTE

If there is no specific coordinate system mentioned, then we mean that $f$ is $k$-times (continuously) partially differentiable with respect to Cartesian coordinates.

Warning: Partial Differentiability's Dependence on Coordinate Systems

If $f$ is partially differentiable in one coordinate system, then that does not necessarily mean that it is partially differentiable in every coordinate system.

Consider the function $f:{\mathbb{R}}^2\to \mathbb{R}$ defined in Cartesian coordinates as

$$f(x,y)\stackrel{\text{def}}{=}\{\begin{array}{l}{x}^2+y^2\text{ if }y\ne 0\\ 0\text{ if }y=0\end{array}$$

Let’s see what $f$‘s partial derivatives with respect to $x$ and $y$ are.

Partial derivative with respect to $x$:

For $y\ne 0$,

$$\frac{\partial }{\partial x}f(x,y)=\frac{\partial f}{\partial x}{x}^2+y^2=2x$$

For $y=0$, we have $f(x,0)=0$ and thus

$$\frac{\partial }{\partial x}f(x,y)=\underset{h\to 0}{\lim }\frac{f(x+h,0)-f(x,0)}{h}=\underset{h\to 0}{\lim }\frac{0-0}{h}=0$$

So, the partial derivative of $f$ with respect to $x$ exists at every point.

Partial derivative with respect to $y$:

For $y\ne 0$,

$$\frac{\partial }{\partial y}f(x,y)=\frac{\partial }{\partial y}(x^2+y^2)=2y.$$

For $y=0$, again we have $f(x,0)=0$ and so

$$\frac{\partial }{\partial y}f(x,y)=\underset{h\to 0}{\lim }\frac{f(x,0+h)-f(x,0)}{h}=\underset{h\to 0}{\lim }\frac{(x^2+h^2)-0}{h}=\underset{h\to 0}{\lim }\frac{x^2+h^2}{h}$$

However, this limit does not exists, since it depends on whether $h$ approaches $0$ from the left or the right.

Therefore, $f$ is not partially differentiable with respect to $y$ on the points $(x,0)$.

Now, let’s take a look at another coordinate system $u,v$. The transformations between $(u,v)$ and $(x,y)$ are given below.

$$\begin{array}{rl}& u=x+yv=x-y\\ & x=\frac{u+v}{2}y=\frac{u-v}{2}\end{array}$$

When $u=v$, i.e. when $y=0$, the function $f$ is also $0$.

When $u\ne v$, the function $f$ expressed in the coordinate system $(u,v)$ is

$$\begin{array}{rl}f(u,v)& ={\left(\frac{u+v}{2}\right)}^2+{\left(\frac{u-v}{2}\right)}^2\\ & =\frac{u^2+2uv+v^2}{4}+\frac{u^2-2uv+v^2}{4}\\ & =\frac{2u^2+2v^2}{4}\\ & =\frac{u^2+v^2}{2}\end{array}$$

Therefore,

$$f(u,v)=\{\begin{array}{l}\frac{u^2+v^2}{2}u\ne v\\ 0u=v\end{array}$$

Let’s examine the partial derivatives of $f$ with respect to $u$ and $v$.

Partial derivative with respect to $u$:

For $u\ne v$,

$$\frac{\partial }{\partial u}f(u,v)=\frac{\partial }{\partial u}\frac{u^2+v^2}{2}=\frac{1}{2}\times 2u=u$$

For $u=v$, we have $f(u,v)=0$

$$\frac{\partial }{\partial u}f(u,v)=\underset{h\to 0}{\lim }\frac{f(u+h,v)-f(u,v)}{h}=\underset{h\to 0}{\lim }\frac{\frac{(u+h{)}^2+v^2}{2}-0}{h}=\underset{h\to 0}{\lim }\frac{(u+h{)}^2+v^2}{2h}=0$$

Therefore, the partial derivative of $f$ with respect to $u$ exists at every point.

Partial derivative with respect to $v$:

For $u\ne v$,

$$\frac{\partial }{\partial v}f(u,v)=\frac{\partial }{\partial v}\frac{u^2+v^2}{2}=\frac{1}{2}\times 2v=v$$

For $u=v$, we have $f(u,v)=0$

$$\frac{\partial }{\partial v}f(u,v)=\underset{h\to 0}{\lim }\frac{f(u,v+h)-f(u,v)}{h}=\underset{h\to 0}{\lim }\frac{\frac{u^2+(v+h{)}^2}{2}-0}{h}=\underset{h\to 0}{\lim }\frac{u^2+(v+h{)}^2}{2h}=0$$

Therefore, the partial derivative of $f$ with respect to $v$ exists at every point.

This means that $f$ is partially differentiable at every point with respect to the coordinate system $(u,v)$, but it is not partially differentiable with respect to $(x,y)$.