Computing second order partial derivatives

Question

I am tasked with the following question:

Consider the wave equation for linearly $x$ polarised waves travelling in the $\pm z$ directions: $$\frac{\partial^2 E_x}{\partial t^2}=c^2\frac{\partial^2 E_x}{\partial z^2}\tag{1}$$ Transform Eq. $(1)$ to the independent variables $q = z − ct$ and $s = z + ct$ and show that $$\frac{\partial^2 E_x}{\partial s \partial q}=0$$

When I am confronted with questions of this type I always use the 'tree diagram for partial derivatives' method which is outlined in this video - Chain rule and tree diagrams of multivariable functions (KristaKingMath) by Krista King.

So writing $$E_{x}=E_{x}(q,s)$$ $$q=q(z,t) \qquad\text{and}\qquad s=s(z,t)$$ then $$\frac{\partial E_x}{\partial z}=\frac{\partial E_x}{\partial s}\cdot\frac{\partial s}{\partial z}+\frac{\partial E_x}{\partial q}\cdot\frac{\partial q}{\partial z}$$

Now since $q = z − ct$ and $s = z + ct$

$$\frac{\partial s}{\partial z}=\frac{\partial q}{\partial z}=1$$ then

$$\frac{\partial E_x}{\partial z}=\frac{\partial E_x}{\partial s}+\frac{\partial E_x}{\partial q}\tag{a}$$

Now with respect to time:

$$\frac{\partial E_x}{\partial t}=\frac{\partial E_x}{\partial q}\cdot\frac{\partial q}{\partial t}+\frac{\partial E_x}{\partial s}\cdot\frac{\partial s}{\partial t}$$

Since, $$-\frac{\partial q}{\partial z}=\frac{\partial s}{\partial z}=c$$ then $$\frac{\partial E_x}{\partial t}=c\biggl(\frac{\partial E_x}{\partial s}-\frac{\partial E_x}{\partial q}\biggr)\tag{b}$$

From $(\mathrm{a})$ (since it holds for all functions)

$$\frac{\partial }{\partial z}=\frac{\partial }{\partial s}+\frac{\partial}{\partial q}\tag{2}$$

So differentiating $(\mathrm{b})$ wrt $z$ using $(2)$

$$\frac{\partial^2 E_x}{\partial z \partial t}=c\left(\frac{\partial }{\partial s}+\frac{\partial}{\partial q}\right)\left(\frac{\partial E_x}{\partial s}-\frac{\partial E_x}{\partial q}\right)=c\left(\frac{\partial^2 E_x}{\partial s^2}-\frac{\partial^2 E_x}{\partial s \partial q}+\frac{\partial^2 E_x}{\partial q \partial s}-\frac{\partial^2 E_x}{\partial q^2}\right)$$ $$=c\left(\frac{\partial^2 E_x}{\partial s^2}-\frac{\partial^2 E_x}{\partial q^2}\right)$$

So the expression that I wanted to show is zero, $$\frac{\partial^2 E_x}{\partial s \partial q}$$ has cancelled out of the expression. Here I have assumed that $E_x$ is sufficiently 'well-behaved' such that $$\frac{\partial^2 E_x}{\partial s \partial q}=\frac{\partial^2 E_x}{\partial q \partial s}$$

---

This is the answer given by the author:

---

Since the solution given by the author doesn't explicitly show that $$\frac{\partial^2 E_x}{\partial s \partial q}=0,$$

could someone please explain how I can show that it is zero?

Answer 1

To show that $\dfrac{\partial^2 E_x}{\partial s \partial q}=0$ compute the second partial derivatives in the original PDE, $\dfrac{\partial^2 E_x}{\partial t^2}-c^2\dfrac{\partial^2 E_x}{\partial z^2}=0$ in terms of the partial derivatives wrt to the new variables. In fact, you did it half the way as you did it for the first derivatives,

$$\frac{\partial E_x}{\partial z}=\frac{\partial E_x}{\partial s}+\frac{\partial E_x}{\partial q}$$

and

$$\frac{\partial E_x}{\partial t}=c\biggl(\frac{\partial E_x}{\partial s}-\frac{\partial E_x}{\partial q}\biggr)\tag{b}$$

So,

$$\frac{\partial^2 E_x}{\partial z^2}=\left(\frac{\partial }{\partial s}+\frac{\partial}{\partial q}\right)\left(\frac{\partial E_x}{\partial s}+\frac{\partial E_x}{\partial q}\right)=\left(\frac{\partial^2 E_x}{\partial s^2}+2\frac{\partial^2 E_x}{\partial q \partial s}+\frac{\partial^2 E_x}{\partial q^2}\right)$$

and

$$\frac{\partial^2 E_x}{\partial t^2}=c\left(\frac{\partial }{\partial s}-\frac{\partial}{\partial q}\right)c\left(\frac{\partial E_x}{\partial s}-\frac{\partial E_x}{\partial q}\right)=c^2\left(\frac{\partial^2 E_x}{\partial s^2}-2\frac{\partial^2 E_x}{\partial q \partial s}+\frac{\partial^2 E_x}{\partial q^2}\right)$$

Substituting:

$$0=\dfrac{\partial^2 E_x}{\partial t^2}-c^2\dfrac{\partial^2 E_x}{\partial z^2}=c^2\left(\frac{\partial^2 E_x}{\partial s^2}-2\frac{\partial^2 E_x}{\partial q \partial s}+\frac{\partial^2 E_x}{\partial q^2}\right)-c^2\left(\frac{\partial^2 E_x}{\partial s^2}+2\frac{\partial^2 E_x}{\partial q \partial s}+\frac{\partial^2 E_x}{\partial q^2}\right)=$$

$$=-4c^2\frac{\partial^2 E_x}{\partial q \partial s}$$

which means,

$$\frac{\partial^2 E_x}{\partial q \partial s}=0$$

Answer 2

Notation: let $E_x:=E$, $\partial_s:=\partial/\partial s$, $\partial_s^2:= \partial^2/\partial s^2$, etc.

You already found the first derivatives in the new variables. Let's write this in a suggestive way of the new conjugate (up to $c$) differential operators. $$\begin{align*} \partial_z E = (\partial_s + \partial_q) E, \\ \partial_t E = c(\partial_s - \partial_q)E\end{align*} $$

Then, just like the familiar ring of polynomials, we have a (non-commutative) ring of differential operators with$$\begin{align*}\partial_z = \partial_s + \partial_q \Rightarrow \partial_z^2 &= \partial_s^2 + 2\partial_{qs}+\partial_q^2, \\ \partial_t = c(\partial_s - \partial_q)\Rightarrow\partial_t^2 &= c^2(\partial_s^2-2\partial_{qs}+\partial_q^2)\end{align*} $$

As $E(t,z)$ satisfies the wave equation, $\partial_{qs}E = 0.$