# Talk:Wave equation

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## Derivation from Hooke's Law

Needs more explanation when converting to N springs. Good example found at: https://physics.stackexchange.com/questions/250983/derivation-of-the-wave-equation-from-hookes-law-generalization-question Jobonki (talk) 20:57, 9 January 2019 (UTC)

Can anyone explain how the moving wave on the string can possibly make sense in 3 dimension? All these wave equations are based on waves in a plane but what plane is defined on the string? The string cannot have a wave in y and z directions like this. At least I can't wrap my mind around that one! Do the waves wind around the string axis?Terence B Allen (talk) 17:42, 12 September 2019 (UTC)

If the stationary wave on the string has a boundary condition dx/dt=0 then how can an equation with an incremental change in x over time make sense?

## Removed

The basic equation is:

$v=f\lambda$ I'm not sure it should be removed. Although "perfectly correct", it's the same as listing the differential-only forms of Newtonian mechanics -- correct, but not useful.
Perhaps you (or I) should add it back in a section describing specific solutions, such a standing wave patters, or in this case, singletons.

Not sure if this is the right way to suggest this,(bit of a beginner with wikipedia) -- surly the above formula should be in the article as it's the basic formula and very widely used, i was looking for it when i searched "wave equation". Sorry again if this is the wrong way of suggesting.

## scalar or vector?

the function in search, u, is scalar or vector-valued? i.e. when is it what?

## An Error?

In the algebraic approach subsection to General Solutions, we see "Another way to arrive at this result is to note that the wave equation may be "factored":

   [ ∂ ∂ t − c ∂ ∂ x ] [ ∂ ∂ t + c ∂ ∂ x ] u = 0 {\displaystyle \left[{\frac {\partial }{\partial t}}-c{\frac {\partial }{\partial x}}\right]\left[{\frac {\partial }{\partial t}}+c{\frac {\partial }{\partial x}}\right]u=0} \left[\frac{\part}{\part t} - c\frac{\part}{\part x}\right] \left[ \frac{\part}{\part t} + c\frac{\part}{\part x}\right] u = 0


and therefore:

   either ∂ u ∂ t − c ∂ u ∂ x = 0 or ∂ u ∂ t + c ∂ u ∂ x = 0 {\displaystyle {\mbox{either}}\qquad {\frac {\partial u}{\partial t}}-c{\frac {\partial u}{\partial x}}=0\qquad {\mbox{or}}\qquad {\frac {\partial u}{\partial t}}+c{\frac {\partial u}{\partial x}}=0} {\mbox{either}}\qquad {\frac {\partial u}{\partial t}}-c{\frac {\partial u}{\partial x}}=0\qquad {\mbox{or}}\qquad {\frac {\partial u}{\partial t}}+c{\frac {\partial u}{\partial x}}=0"


I think this is wrong. See  The real situation is revealed if you plug the full solution f(x+ct) + g(x-vt) into the "factored" wave equation: neither of those sums turns out zero. What is true is that one of those sums must equal some F(x+ct), and the other one, G(x-ct). You can then use that result to get the full solution. All of this gets discussed in the page I've linked to above. HHHEB3 (talk) 12:08, 11 December 2017 (UTC) I suggest the following as a replacement:

Another way to arrive at this result is to note that the wave equation may be "factored":

$\left[{\frac {\partial }{\partial t}}-c{\frac {\partial }{\partial x}}\right]\left[{\frac {\partial }{\partial t}}+c{\frac {\partial }{\partial x}}\right]u=0.$ As a result, if we define v thus,

$\qquad {\frac {\partial u}{\partial t}}+c{\frac {\partial u}{\partial x}}=v$ ,

then

$\qquad {\frac {\partial v}{\partial t}}-c{\frac {\partial v}{\partial x}}=0.$ From this, v must have the form v(x + ct), and from this the correct form of the full solution u can be deduced.

## If c is a constant ...

If c is a constant in the wave equation, the article should state this explicitly.50.205.142.50 (talk) 12:10, 27 March 2020 (UTC)