I know a similar post has been made about this, but it has been quite some time since then, and none of the answers satisfy me.
I want to derive the 1D-wave equation from the knowledge that what we call a wave takes the form $ \psi = f(x \mp vt)$. Most physics textbooks will derive it from the tension in a string, etc., but I want to be more general than that.
The closest general derivation I have found is in the book Optics by Eugene Hecht. However, he states , "We now derive the one-dimensional form of the wave equation guided by the foreknowledge that the most basic of waves traveling at a fixed speed requires two constants to specify it, and this suggests second derivatives''. It would satisfy me if we did not use this foreknowledge, and instead have everything flow from previous steps. He makes use of a reference frame that moves with the wave form $ \psi = f(x')$, where $ x' = x \mp vt $. His derivation goes:
$$ \frac{\partial \psi }{\partial x} = \frac{\partial f }{\partial x}$$
$$ \frac{\partial \psi }{\partial x} = \frac{\partial f }{\partial x'} \frac{\partial x'}{\partial x} = \frac{\partial f}{\partial x'}$$
$$ \frac{\partial \psi }{\partial t} = \frac{\partial f }{\partial x'} \frac{\partial x' }{\partial t} = \mp v \frac{\partial f }{\partial x'} $$
$$ \frac{\partial^2 \psi }{\partial x^2} = \frac{\partial^2 f }{\partial x'^2} $$
$$ \frac{\partial^2 \psi }{\partial t^2} = \frac{\partial}{\partial t} \left(\mp v \frac{\partial f}{\partial x'}\right) = \mp v \frac{\partial }{\partial x'} \left(\frac{\partial f}{\partial t}\right) = \mp v \frac{\partial }{\partial x'} \left(\frac{\partial \psi}{\partial t}\right) $$
$$ \frac{\partial^2 \psi}{\partial t^2} = v^2 \frac{\partial^2 \psi}{\partial x^2} $$
Also, I am not clear why the $x'$ frame is necessary. $ \psi = f(x \mp vt)$ tells us that it is a wave in the unprimed frame, isn't that all we need?
Best Answer
Here is another way to derive the wave equation using only f(x-ct) and f(x+ct).
See "The Mathematical Theory of Wave Motion", G. R. Baldock 1981, pages 27,28.
Paraphrasing from that source:
Given $$u(x,t)=f(x-ct) \tag a$$
then
$$c\frac{\partial u}{\partial x}=-\frac{\partial u}{\partial t}\tag b$$
or for $$u(x,t)=f(x+ct)\tag c$$ we get $$c\frac{\partial u}{\partial x}=\frac{\partial u}{\partial t}\tag d$$
Therefore $$\frac{\partial^2u}{\partial x^2}-\frac{1}{c^2}\frac{\partial^2u}{\partial t^2}=0 \tag e$$
which is the wave equation.
A lot of steps were left out so I redo the derivation in more detail in the following:
Given $$u(x,t)=f(x-ct) \tag 1$$
then $$\frac{\partial u}{\partial t}=\frac{\partial f(x-ct)}{\partial t} =\frac{\partial f(x-ct)}{\partial (x-ct)} \frac{\partial (x-ct)}{\partial (t)} =-c \frac{\partial f(x-ct)}{\partial (x-ct)} \tag 2$$
and $$\frac{\partial u}{\partial x}=\frac{\partial f(x-ct)}{\partial x} =\frac{\partial f(x-ct)}{\partial (x-ct)} \frac{\partial (x-ct)}{\partial (x)} =\frac{\partial f(x-ct)}{\partial (x-ct)} \tag 3$$
so $$\frac {\partial u}{\partial t}= -c\frac {\partial u}{\partial x} \tag 4$$
and $$u_x+ \frac{1}{c}u_t = 0 \tag 5$$
Next given
$$u(x,t)=f(x+ct) \tag 6$$
then $$\frac{\partial u}{\partial t}=\frac{\partial f(x+ct)}{\partial t} =\frac{\partial f(x+ct)}{\partial (x+ct)} \frac{\partial (x+ct)}{\partial (t)} =+c \frac{\partial f(x+ct)}{\partial (x+ct)} \tag 7$$
and $$\frac{\partial u}{\partial x}=\frac{\partial f(x+ct)}{\partial x} =\frac{\partial f(x+ct)}{\partial (x+ct)} \frac{\partial (x+ct)}{\partial (x)} =\frac{\partial f(x+ct)}{\partial (x+ct)} \tag 8$$
so $$\frac {\partial u}{\partial t}= +c\frac {\partial u}{\partial x} \tag 9$$
and $$u_x- \frac{1}{c}u_t = 0 \tag {10}$$
Finally, multipying the 'factors' gives:
$$(u_x- \frac{1}{c}u_t)(u_x+ \frac{1}{c}u_t) = 0\tag {11}$$ so we get $$u_{xx} = \frac{-1}{c^2}u_{tt} \tag {12} $$ which is the wave equation.