This is the total differential $$df=dx\frac {\partial f}{\partial x}+dy\frac {\partial f}{\partial y}.$$

How do we take higher orders of total differential, $d^2 f=$?

Suppose I have $f(x,y)$ and I want the second order total differential $d^2f$?

  • $\begingroup$ Product and chain rule just how you did to get the first one. Now imagine that $df$ is a function $g(x,y)$, which is formed by the sum of two products, and get the differential of $g$. $\endgroup$ – MyUserIsThis Jun 30 '13 at 14:03
  • $\begingroup$ Like $d^2f=dx\wedge(\frac{\partial^2f}{\partial x\partial x}dx+\frac{\partial^2f}{\partial x\partial y}dy)+dy\wedge(\frac{\partial^2f}{\partial y\partial x}dx+\frac{\partial^2f}{\partial y\partial y}dy)=dx\wedge dy(\frac{\partial^2f}{\partial x\partial y}-\frac{\partial^2f}{\partial y\partial x})$? $\endgroup$ – awllower Jun 30 '13 at 14:05

$d=dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}$ is a differential operator that when applied to $f$ gives $df=dx\frac{\partial f}{\partial x}+dy\frac{\partial f}{\partial y}$

$d^2=\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)$

$d^2f=\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)f=\left(dx^2\frac{\partial^2}{\partial x^2}+dy^2\frac{\partial^2}{\partial y^2}+2dx dy\frac{\partial^2}{\partial x\partial y}\right)f=dx^2\frac{\partial^2 f}{\partial x^2}+dy^2\frac{\partial^2 f}{\partial y^2}+2dx dy\frac{\partial^2 f}{\partial x\partial y}$

  • $\begingroup$ Do you mean, when $f$ is $C^2$, $d^2f=dx^2\frac{\partial^2 f}{\partial x^2}+dy^2\frac{\partial^2 f}{\partial y^2}+2dx dy\frac{\partial^2 f}{\partial x\partial y}$? Even if so, you must have forgotten that $dxdy+dydx=0$. Regards. $\endgroup$ – awllower Jul 1 '13 at 2:41
  • $\begingroup$ @awllower You mean $dx dy\wedge dy dx=0$ not $dx dy+dy dx=0$ $\endgroup$ – llllllllllllllllllllllllllllll Jul 1 '13 at 7:28
  • $\begingroup$ No, I mean $dx\wedge dy+dy\wedge dx=0$, as the wedge product is anti-symmetric. $\endgroup$ – awllower Jul 1 '13 at 12:28
  • $\begingroup$ Yes, but as you saw, I avoided the use of the wedge product because I think @dan only want to use elementary calculus. $\endgroup$ – llllllllllllllllllllllllllllll Jul 1 '13 at 16:34
  • 1
    $\begingroup$ Probably a basic question, but why is $d^2 f$ (ie $d(df)$) the same as multiplying $d$ with $d$ and then applying that result to $f$? Ie how does the composition of differential operators get reduced to a multiplication of those operators? $\endgroup$ – user434180 Nov 3 '16 at 0:55

I will assume that you are referring to the Fréchet derivative. If $U\subseteq\mathbb{R}^n$ is an open set and we have functions $\omega_{j_1,\dots,j_p}:U\to\mathbb{R}$, then $$ D\left(\sum_{j_1,\dots,j_p} \omega_{j_1,\dots,j_p} dx_{j_1}\otimes\cdots\otimes dx_{j_p}\right) = \sum_{j_1,\dots,j_p}\sum_{j=1}^n \frac{\partial\omega_{j_1,\dots,j_p}}{\partial x_{j}} dx_j\otimes dx_{j_1}\otimes\cdots\otimes dx_{j_p}. $$

Here $dx_{i}$ is the projection onto the $i$th coordinate, and if $\alpha,\beta$ are multilinear forms then $\alpha\otimes\beta$ is the multilinear form defined by $(\alpha\otimes\beta)(x,y)=\alpha(x)\beta(y)$.

For example, let $f(x,y)=x^3+x^2 y^2+y^3$. Then \begin{align} Df&=(3x^2+2xy^2)dx+(2x^2y+3y^2)dy; \\ D^2f&=(6x+2y^2)dx\otimes dx+4xy(dx\otimes dy+dy\otimes dx)+(2x^2+6y)dy\otimes dy; \\ D^3f&=6dx\otimes dx\otimes dx+4y(dx\otimes dx\otimes dy+dx\otimes dy\otimes dx+dy\otimes dx\otimes dx)\\ &\qquad+4x(dx\otimes dy\otimes dy+dy\otimes dx\otimes dy+dy\otimes dy\otimes dx) \\ &\qquad+6dy\otimes dy\otimes dy. \end{align}

Since $D^p f(x)$ is always a symmetric multilinear map if $f$ is of class $C^p$, you might want to simplify the above by using the symmetric product (of tensors).

  • $\begingroup$ This answer is good because you're doing the total derivative of the total derivative, rather than just the exterior derivative of the total derivative. $\endgroup$ – Toby Bartels yesterday

Edit: I interpret your $d^2 f$ as $d^2f=d(df)$.

If $f$ is $C^2(\Omega)$, with $(x,y)\in\Omega$, then

$$d(df)=dy\wedge dx \frac{\partial^2 f}{\partial y\partial x}+dx\wedge dy \frac{\partial^2 f}{\partial x\partial y}=\left(\frac{\partial^2 f}{\partial x\partial y}- \frac{\partial^2 f}{\partial y\partial x} \right)dx\wedge dy=0,$$

as $dx \wedge dy=-dy\wedge dx$ and $\frac{\partial^2 f}{\partial x\partial y}= \frac{\partial^2 f}{\partial y\partial x}$ on $\Omega$.

If $f$ is not $C^2$ at $(x,y)$, then we arrive at

$$d(df)=\left(\frac{\partial^2 f}{\partial x\partial y}- \frac{\partial^2 f}{\partial y\partial x} \right)dx\wedge dy. $$

  • $\begingroup$ But OP did not assume that $f\in C^2$, so that the result is not necessarily $0$, right? $\endgroup$ – awllower Jun 30 '13 at 14:22
  • $\begingroup$ @awlower, right. I used the Schwarz lemma. If $f$ is not $C^2$ then we have just the first equality. $\endgroup$ – Avitus Jun 30 '13 at 16:19
  • $\begingroup$ @awlower I add this comment in my answer. $\endgroup$ – Avitus Jun 30 '13 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.