# Textbook Query - Wave Mechanics

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

• 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 $h'(x)$. – aRockStr Oct 28 at 3:13
• Which textbook? – Qmechanic Oct 28 at 4:12
• Rae Quantum Mechanics – PolynomialC Oct 28 at 9:39
• I agree with @aRockStr – user45664 Oct 28 at 17:18