Talk:Wave equation

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.

Fair enough. as it stood it was confusing and seemed unrealed to the differential equation this article is about -- Tarquin 13:11 Jan 6, 2003 (UTC)

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)