I'm having trouble making sense on how one arrives to the one-dimensional scalar wave equation.

I'll articulate which steps of the derivation I have issues with, then general questions pertaining to it.

So, the way I was taught to derive it, was to first start with the traveling wave equation:

$$u(x,t) = x - vt$$

Then, define a new variable, $y$, which is a function of $u$ which is a function of $x$ and $t$. Hopefully I'm correct in assuming this variable $y$ is the $y$ displacement of a particle being pushed up and down by a traveling wave.

First, one seeks to arrive to (in order to equate the final result):

$$\frac{\partial ^2 y}{\partial x^2} = \frac{\partial ^2 y}{\partial u ^2}$$

However, there's an intermediate step my lecturer did that confuses me. I figured out what he did, but it requires accepting this:

$$\frac{\partial}{\partial u} * \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \implies \frac{\partial}{\partial x}\left(\frac{\partial y}{\partial u}\right) = \frac{\partial}{\partial u}\left(\frac{\partial y}{\partial u}\right)\frac{\partial u}{\partial x}$$

Hopefully that mess can be made sense of enough. Basically, I have trouble accepting the first part that implies the rest. I'm equating something to an operator? What? This seems illogical. It's as if I treated $\partial u$ as a real number.

In case that's not what he is implying, here is the work itself:

enter image description here

With this, I have two other quick questions:

  • If I'm correct in assuming $y$ is defined the way I think it is, as the $y$ displacement the particle will experience encountering a traveling wave, why can't any variable that is a function of $u$ arrive at this relation? This is just assuming $y$ is a function of $u$, and noting the implications.

  • The final equation: $$\frac{\partial ^2 y}{\partial x^2} = 1/v^2 \frac{\partial ^2 y}{\partial t^2}$$ is confusing to me. It is supposed to relate the transverse displacement of the particle as a function of $x$ and $t$, but even after working out the underlying maths, I am clueless looking for any intuition behind this relationship. Is there anywhere you recommend I can read up on to help understand this equation better?

  • 1
    $\begingroup$ The implication $$\frac{\partial}{\partial u} * \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \implies \frac{\partial}{\partial x}\left(\frac{\partial y}{\partial u}\right) = \frac{\partial}{\partial u}\left(\frac{\partial y}{\partial u}\right)\frac{\partial u}{\partial x}$$ is meaningless. Neither side of $\implies$ is well-defined, so it is impossible for the l.h.s. to imply the r.h.s. $\endgroup$ – AccidentalFourierTransform Jan 30 '18 at 17:27
  • 1
    $\begingroup$ Regarding the intuition behind the wave equation, it basically says that for any point on the wave, the point's transverse acceleration is proportional to the wave's curvature at that point. The proportionality constant (i.e. how quickly a point on a wave accelerates for a given curvature) determines the traveling speed of the wave. For the intuition on that, note that the faster a point on the wave can recoil for a given disturbance, the faster it causes other points in front of it to accelerate; hence, the transverse acceleration is linked to the propagation speed. $\endgroup$ – probably_someone Jan 30 '18 at 17:37
  • $\begingroup$ @AccidentalFourierTransform Odd.. That's what I made sense as his way of deriving it. I've edited to my answer what he actually did. The mess you're referring to is what I theorized he did. $\endgroup$ – sangstar Jan 30 '18 at 17:50
  • $\begingroup$ @probably_someone If you're saying a faster traveling wave will whip particles up and down transversely faster, I'm totally with you there. By curvature, are you saying, for instance, at $\pi$ radians, the transverse velocity will be zero? Just testing to see if my interpretation of what you're saying is fine. $\endgroup$ – sangstar Jan 30 '18 at 17:52
  • 1
    $\begingroup$ @sangstar What I'm referring to here is an intuitive interpretation of the second derivative. Just like the first derivative of a function is related to its slope at that point, the second derivative of a function is related in some way to its curvature. The higher the second derivative, the more sharply curved the function is. $\endgroup$ – probably_someone Jan 30 '18 at 17:56

Your Answer

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

Browse other questions tagged or ask your own question.