Talk:Wave equation

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Removed[edit]

The basic equation is:

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?[edit]

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

Vector laplacian[edit]

I'm pretty sure the equation uses the vector laplacian and not the normal laplacian (a normal laplacian returns a scalar and the left side is a vector) Sonicrs (talk) 12:12, 30 May 2017 (UTC)

This article mainly discusses the scalar wave equation where u is a scalar and thus the scalar laplacian should be used. The section Wave_equation#Elastic_waves briefly discusses the vector wave equation which uses a vector laplacian (without explicitly stating this). Ulflund (talk) 05:00, 31 May 2017 (UTC)

Some explanation of the step that is made in the derivation of the Wave Equation.[edit]

   {\partial^2 \over \partial t^2} u(x+h,t)={KL^2 \over M}{u(x+2h,t)-2u(x+h,t)+u(x,t) \over h^2}

Taking the limit N → ∞, h → 0 and assuming smoothness one gets:

   ∂ 2 u ( x , t ) ∂ t 2 = K L 2 M ∂ 2 u ( x , t ) ∂ x 2 {\displaystyle {\partial ^{2}u(x,t) \over \partial t^{2}}={KL^{2} \over M}{\partial ^{2}u(x,t) \over \partial x^{2}}} {\partial^2 u(x,t) \over \partial t^2}={KL^2 \over M}{ \partial^2 u(x,t) \over \partial x^2 } 

This derivation step is confusing. Not clear from the text how this step was done. Had to refer to another textbook to reveal that this is the very definition of a second derivative. Some reference should be made that this term is a second derivative and therefore can be replaced. — Preceding unsigned comment added by 162.192.4.101 (talk) 05:16, 9 July 2017 (UTC)

Presumably this is what you were saying in this edit, which I reverted because it didn't seem to make the point and used a nonstandard link to another Wikipedia article. I've made an attempt with this change, but the wording feels a little clumsy. Feel free to tweak. I know this is a little intimidating at first! And, as I said on the talk page for your IP address, you're welcome to create your own user account. David Brooks (talk) 13:01, 9 July 2017 (UTC)

I like it. Much better

I am new to this whole wikipedia commenting/changing thing. I will figure out the signup process. Thanks. — Preceding unsigned comment added by 162.192.4.101 (talk) 16:32, 9 July 2017 (UTC)

An Error?[edit]

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 [1]

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)