5

I know that the chain rule for a function of one variable $y=f(x)$ is written as $$\frac{{\rm d}}{{\rm d}x}=\frac{{\rm d}}{{\rm d}y}\times \frac{{\rm d}y}{{\rm d}x}\tag{1}$$

I also know that if $u=f(x,y)$ then the total derivative is

$${\rm d}u=\frac{\partial u}{\partial x}\cdot{\rm d} x+\frac{\partial u}{\partial y}\cdot{\rm d}y$$

But from $(1)$ is there any plausibility in changing all the derivatives to partial derivatives such that $$\frac{\partial}{\partial x}=\frac{\partial}{\partial y}\times \frac{\partial y}{\partial x}$$ when $y=f(x)$ or does the above formula only hold iff $y=f(x,y)$?

To illustrate my confusion I will add this question and solution to give some context:

Start of Question:

If $f$ is an arbitrary function, show that $\psi(z,t)=f(z-vt)$ is a solution to the wave equation

$$\frac{\partial^2\psi}{\partial t^2}=v^2\frac{\partial^2\psi}{\partial z^2}$$

by writing $f(z-vt)$ as $f(y)$ and using the chain rule to differentiate with respect to $t$ and $z$.

End of Question.

The following is a word for word copy of the solution. I have marked $\color{red}{\mathrm{red}}$ the parts of the solution for which I do not understand and the parts marked with $\color{#180}{\text{green underbraces}}$ are not part of the solution and represent where I think the author has made mistakes:

Start of Solution:

The wave equation is $$\frac{\partial^2\psi}{\partial t^2}-v^2\frac{\partial^2\psi}{\partial z^2}=0$$ Writing $$f^{\prime}(y)=\frac{{\rm d}f(y)}{{\rm d}y}$$ and $$f^{\prime\prime}(y)=\frac{{\rm d}^2f(y)}{{\rm d}y^2}$$ then writing $y=z-vt$, $$\frac{\partial f(y)}{\partial z}=f^{\prime}(y)\underbrace{\color{red}{\frac{{\rm d}y}{{\rm d}z}}}_{\color{#180}{\Large\frac{\partial y}{\partial z}}}=f^{\prime}(y)\tag{A}$$ and $$\frac{\partial^2 f(y)}{\partial z^2}=\frac{{\rm d}f^{\prime}(y)}{{\rm d}y} \underbrace{\color{red}{\frac{{\rm d}y}{{\rm d}z}}}_{\color{#180}{\Large\frac{\partial y}{\partial z}}}=f^{\prime\prime}(y)$$ Similarly, $$\frac{\partial f}{\partial t}=f^{\prime}(y)\underbrace{\color{red}{\frac{{\rm d}y}{{\rm d}t}}}_{\color{#180}{\Large\frac{\partial y}{\partial t}}}=-vf^{\prime}(y)$$ and $$\frac{\partial^2 f(y)}{\partial z^2}=\frac{{\rm d}\left(-v f^{\prime}(y)\right)}{{\rm d}y}\underbrace{\color{red}{\frac{{\rm d}y}{{\rm d} z}}}_{\color{#180}{\Large\frac{\partial y}{\partial z}}}=v^2f^{\prime\prime}(y)$$ Substituting into the LHS of the wave equation gives $$\frac{\partial^2 f(z-vt)}{\partial z^2}-\frac{1}{v^2}\frac{\partial^2 f(z-vt)}{\partial t^2}=f^{\prime\prime}(z-vt)-\frac{1}{v^2}\left(v^2f^{\prime\prime}(z-vt)\right)=0$$

End of Solution.

I have three questions regarding the solution above:

  1. Are the parts marked in $\color{#180}{\text{green underbraces}}$ correct? I think they should be partial derivatives as $y$ is a function of two variables ($z$ and $t$) since $y=z-vt$.
  2. Iff the $\color{red}{{\rm red}}$ text turns out to be correct then from $({\rm A})$ $$\frac{\partial f(y)}{\partial z}=\frac{{\rm d}f(y)}{{\rm d}y}\times \frac{{\rm d}y}{{\rm d}z}$$ How can this possibly be true? Since we have a partial derivative on the LHS and two ordinary derivatives on the RHS. I didn't know that you could 'mix' up derivatives like that. If this is indeed true could someone please explain it to me?
  3. Iff the $\color{#180}{\text{green underbraces}}$ turn out to be correct then from $({\rm A})$ $$\frac{\partial f(y)}{\partial z}=\frac{{\rm d}f(y)}{{\rm d} y} \times \frac{\partial y}{\partial z}$$ But this means that the LHS is partial while the RHS is a mixture. Either way round, I don't understand why the partial derivatives can be mixed with ordinary derivatives.

Is anyone able to explain why you can mix partial and ordinary derivatives in the chain rule?

Feel free to answer using examples if it's easier to explain that way.

BLAZE
  • 1
  • 1
    Where does the $x$ come from in eqn. (A)? I think this is a mistake, you have also carried this down into parts 2 and 3 of your question. With regards to your first question (just under eqn. 2): no it does not make sense to use the chain rule with partial derivatives rather than total derivatives when you are dealing with a function of a single variable, namely $y = f(x)$. Partial derivatives have no meaning in this context, i.e. $\partial y / \partial x$ means differentiate $y$ w.r.t. $x$ holding ALL other variables constant, but $y$ is only a function of the one variable – okrzysik Dec 30 '15 at 08:21
  • @okrzysik Thanks for pointing out those typo's; I appreciate that. Are you able to answer the questions at the end; now that it's been edited? – BLAZE Dec 30 '15 at 13:58

1 Answers1

1

There is abuse of notation going on here, which is quite common in this realm. The wave equation is defined for a function $\psi(z,t)$, a function of two parameters $z,t$ for which it makes perfect sense to evaluate partial derivatives such as $\partial \psi/\partial z$.

When setting $\psi(z,t)=f(z-vt)$, we are contracting $\psi$ to a function $f(y(z,t))$, with $y=z-vt$. Formally, $\partial f(y)/\partial z=0$ since $f$ by definition $f$ is not explicitly dependent on $z$. So what the book intends is something like this:

$$\frac{\partial \psi}{\partial z}=\frac{df(y)}{dy}\frac{d y}{d z}=f'(y)\frac{\partial y}{\partial z}=f'(y).$$

This is perfectly reasonable and justifiable via the chain rule: $f$ depends solely on $y$, and $y=y(z,t)$ admits partial derivatives. You can derive it using the standard definitions.

Summary: the book is abusing notation, when writing $\partial f(y)/\partial z$ they really mean $\partial \psi/\partial z$.

Alex R.
  • 33,289