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?

  • $\begingroup$ My answer here does not rely on physics. Only the geometric observation of a wave of the form f(x+vt) or f(x-vt) is needed. $\endgroup$
    – user45664
    Commented Apr 25, 2018 at 17:52

2 Answers 2


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 $$u(x,t)=f(x-ct) \tag a$$


$$c\frac{\partial u}{\partial x}=-\frac{\partial u}{\partial t}\tag b$$

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.

  • $\begingroup$ Ok, so now I am thinking of a different question: Suppose you ask me 'what is a wave?' I can either answer 'a function that has the form $ f(x \pm vt) $ or I can say 'a function that is a solution to this differential equation'. Why do we even bother to write the wave equation if the first is a more simple explanation? $\endgroup$
    – ngc1300
    Commented May 10, 2018 at 12:52
  • 1
    $\begingroup$ It looks like they contain the same information. But I think the wave equation is needed to deal with initial conditions and boundary conditions. $\endgroup$
    – user45664
    Commented May 10, 2018 at 15:10

This is a modified version of my answer:


To answer your question: All that is required is that the wave can be expressed in the form $f(x \mp vt)$, using your notation, to obtain the 1D wave equation:

The 1D wave equation can by derived without using physics by examining the geometry of the propagating wave: When the wave propagates as a result of time $t$ increasing, it maintains its shape. The propagated wave is a copy of the original wave shifted to the right, or left, by the distance $ct$. $c$ is the waves propagation speed--the same as your $v$. The wave equation is the mathematical expression of that.

Consider $f(x-ct)$ and consider small changes in $x$ and $t$, ie. $\Delta x$, $\Delta t$. They each cause a small shift or translation of $f(x-ct)$.

Note that $\Delta x$ = $c\Delta t$.


$$\frac{\Delta f}{\Delta x} = \frac{\Delta f}{c\Delta t} = \frac{1}{c}\frac{\Delta f}{\Delta t}.\tag 1$$

Then repeating that we get $$\frac{\Delta^2 f}{\Delta^2 x} = \left(\frac{1}{c}\right)^2\frac{\Delta^2 f}{\Delta^2 t}.\tag 2$$

Letting $\Delta$ become very small we get

$$ \frac{\partial^2f}{\partial x^2}-\frac{1}{c^2}\frac{\partial^2f}{\partial t^2}=0.\tag 3$$

From the geometry alone, it was only needed to note that a change in $t$ multiplied by the velocity yields the same results (as measured by the second derivative) as a change in $x$ --that is, a translation of $f(x-ct)$.

See for a good discussion using a different approach:



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