The following exerpt of a wave mechanics text has me confused:

Worked Example 1.2: Suppose that at time $t=0$, the string is stationary and has shape $y(x,0)=h(x)$, where $h(x)$ is some localized 'bump'. Find a solution to the wave equation that satisfies the initial conditions, and describe the subsequent motion of the string.

Solution: There are properties of differential equations that are important, and provable, but which we will not discuss in great detail in this book. One of these properties is 'uniqueness', namely that if a solution is found which satisfies the initial conditions, then it is the only solution. Therefor, if we can guess this right solution, then we are done. Consider the expression $$y(x,t)=\frac{1}{2}h(x-vt)+\frac{1}{2}h(x+vt).$$ Obviously, $y(x,0)=h(x)$. Also, the inital (vertical) velocity of the string at any point $x$ is $$\frac{\partial y}{\partial t}\bigg|_{t=0}=-v\frac{1}{2}h(x-vt)\bigg|_{t=0}+v\frac{1}{2}h(x+vt)\bigg|_{t=0}=-v\frac{1}{2}h(x)+v\frac{1}{2}h(x)=0.$$

(From Quantum Mechanics by Alastair I. M. Rae, Jim Napolitano)

Surely the correct method would be to use the chain rule here? The author seems to have forgotten to differentiate the function $h$, is this a mistake or a misunderstanding of mine?

Excerpt from Quantum Mechanics by Alastair I. M. Rae, Jim Napolitano

  • 3
    $\begingroup$ I think you're right... unless we could somehow show $h(x) = h'(x).$ Fortunately, I think the expression is still zero even if we make the replacement $h(x)$ with $h'(x)$. $\endgroup$ – aRockStr Oct 28 at 3:13
  • $\begingroup$ Which textbook? $\endgroup$ – Qmechanic Oct 28 at 4:12
  • $\begingroup$ Rae Quantum Mechanics $\endgroup$ – PolynomialC Oct 28 at 9:39
  • $\begingroup$ I agree with @aRockStr $\endgroup$ – user45664 Oct 28 at 17:18

Partial differentiation is used here, with respect to time. Thus, the expression (vt + x) is same as saying (kt + c) where k and c are constants. Thus the expression is absolutely correct.

This is the same as saying that the differential is reduced to zero but that is so with respect to time and we can't say the same for h'(x).

  • $\begingroup$ but h is a function of time? surely you have to use chain rule in this instance? $\endgroup$ – PolynomialC Oct 28 at 3:51
  • $\begingroup$ h is a function which takes both inputs, position (x) and time (t). However, while doing partial differentiation for multi input or multi variable functions, for the variable w.r.t which you're doing it, you treat it as if its single variable function depending on that input only. $\endgroup$ – Mike Karter Oct 28 at 3:58
  • $\begingroup$ then surely you use chain rule with the variable u= x-vt and v' = x+vt, sorry if this seems circular this is just going against everything ive been learning. $\endgroup$ – PolynomialC Oct 28 at 4:00
  • $\begingroup$ Adding, as it's written, when you're looking to find "velocity of the string at any point x" you need to generalise the expression for all x, meaning all x are same, meaning it does not depend on x, if you can see it this way. $\endgroup$ – Mike Karter Oct 28 at 4:02
  • $\begingroup$ Yes, differentiation should have been done, and sorry for the round about. $\endgroup$ – Mike Karter Oct 28 at 4:10

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