Talk:Wave equation/Archive 1

Culture
I've heard that Laplace was riding his horse next to a river when something big fell in the water. The waves propagated to the riverbanks and down the river. Laplace noticed that there was a wave that persisted, never dying out as long as he followed it. It is said that this inspired him to study waves mathematically. Does anyone else know about this? --Orthografer 16:30, 15 December 2005 (UTC)
 * What he saw is I think a soliton. I heard something similar a while ago. I guess it belongs to the same category of stories as Newton's apple. Might be true. Oleg Alexandrov (talk) 19:52, 15 December 2005 (UTC)


 * It wasn't Laplace, it was a fellow called Russell; it wasn't an object, it was a barge being towed by mules which suddenly stopped. He saw the water gather up at the prow of the boat, and detach itself into a single solitary wave, which then proceeded quite repidly down the canal. He chased the soliton on horseback for 17 2 miles (!). Its all written up in an account, presented to some or another scientific society; its mentioned in many books. The math is given by the KdV equation. linas 18:58, 1 June 2006 (UTC)
 * Why, WP has an article on it: John Scott Russell; see also the external link therein. linas 19:00, 1 June 2006 (UTC)
 * Thank you! Orthografer 20:38, 2 June 2006 (UTC)

I intend to add material on the 3-D case, with emphasis on domain of dependence, Huygens' principle, etc. The 2-D case can then be teated by method of descent. Donludwig 18:03, 31 March 2006 (UTC)

Added Section on In-homogenous W.E. in 1D
I added a section on how to get a solution to an in-homogenous wave equation initial value problem. In one dimension. It's better than none I suppose... :) DAG 10:30, 17 June 2006 (UTC)


 * I did a small copyedit and noticed you called the initial condition f the source function. I believe the correct interpretation is that the inhomogeneous term s is the source function. Someone should correct me if I'm wrong and list a source within this section, perhaps Boyce and DiPrima's book if it has the relevant material. Orthografer 20:22, 17 June 2006 (UTC)


 * No, no, you're completely right, that was my mistake. I initially started by using f as the source function, but to remain consistent with the rest of the article, I wanted to use the same symbols for the initial conditions, so then I had to change f.  Apparently I wasn't completely thorough...  :/  Oops.  DAG 21:26, 17 June 2006 (UTC)

Would movie be useful?
I made a couple of movies where I numerically computed the solution of the wave equation on a circular membrane clamped at the boundary with initial speed set to zero. Would it be worthwhile to add these to the article? Swap 18:04, 5 July 2007 (UTC)

Link explaining 2nd order linear
I added an external link to the term: second order-linear. I couldn't find a wiki article that explained it to my satisfaction. The external article isn't great, but it is simpler than other alternatives. User:Berland reverted the edit and makes the point that external links are discouraged in the body of the text. Yes this is true, but good links that add to the article are encouraged. User:Berland also makes the point that 2nd order is talked about at the bottom of partial dif. People not understanding the terminology of these two articles are going to give up long before they get to the bottom of partial dif article, so that isn't very helpful.

I added a wiki like to the partial dif sub section. I think the article was easier to understand the way I had it before, with the external link. I encourage you to revert to that version. Daniel.Cardenas (talk) 21:29, 25 May 2008 (UTC)
 * From WP:External links I find it quite clear that my revert was in order. From what I read in your comment here, it seems like your point is that the page on partial differential equations should be improved so that the explanation of second order linear becomes easier accessible. That is a better solution than having an in-text external link here. --Berland (talk) 08:04, 26 May 2008 (UTC)

Perhaps a link to "Dirac's coup" should also be considered?
 * $$\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} = (A \partial_x + B \partial_y + C \partial_z + \frac{i}{c}D \partial_t)(A \partial_x + B \partial_y + C \partial_z + \frac{i}{c}D \partial_t)$$195.96.229.83 (talk) 09:24, 19 June 2008 (UTC)

Misleading first sentence
The first sentence (and also other parts of the article) implies that the wave equation exactly describes sound waves and water waves. It doesn't; it's only approximate. It's exact (as far as we know) for light in vacuum, but not for light in matter. There's a lot that can be said about this topic...piling higher-order corrections onto the wave equation is the basis for a large proportion of modern quantum physics. But at the very least, the article could merely mention the fact that the wave equation is approximate in most physical situations. Could anyone make that correction? Thanks!! --Steve (talk) 02:28, 30 June 2008 (UTC)


 * I believe the wave equation is pretty accurate for light in matter, say in uniform medium assuming no absorption or nonlinear effects or anisotropy (which are quite reasonable assumptions in most circumstances). I believe the wave equation is reasonably accurate for sound propagation in air, at least over short distances and with low speed.


 * I like the introduction the way it is. Introductions should be informal, and while it is true that ultimately the wave equation is just a model, and can't be always accurate, I see nothing misleading in the current intro. Oleg Alexandrov (talk) 05:27, 30 June 2008 (UTC)


 * Most objects are not transparent, I think. On the other hand, I think I misunderstood...the article isn't saying that the simplest wave equation in the first equation is THE wave equation, it's saying it's A wave equation, which can be made more accurate. So it's not so bad after all. :-) --Steve (talk) 16:57, 30 June 2008 (UTC)

Is this correct ?
which leads to the general solution

u(\xi, \eta) = F(\xi) + G(\eta) \quad \Rightarrow \quad u(x, t) = F(x - c t) + G(x + c t)

In other words, solutions of the 1D wave equation are sums of a left traveling function F and a right traveling function G. "Traveling" means that the shape of these individual arbitrary functions with respect to x stays constant, however the functions are translated left and right with time at the speed c

Every time I look at this,  the F wave appears to be moving to the right  and the G wave appears to be moving to the left. Whats the catch ? Eregli bob (talk) 07:20, 25 September 2009 (UTC)

Delta, Nabla and Notation
Can someone modify this article to explain what the heck $$\Delta$$ stands for, in that first equation? linas 18:58, 1 June 2006 (UTC)


 * Added identification of $$\Delta$$ in the equation, along with the alternative $$\nabla^2$$ form, as the Laplacian and an added appropriate link. --Nkrupans 16:10, 7 June 2006 (UTC)

A broader question of notation, does anyone else find the $$\nabla^2$$ more used/familiar than the $$\Delta$$ for the Laplacian operator? Looking through the physics articles on Wikipedia, the $$\nabla^2$$ seems dominant. I mention this in the discussion page for the Laplacian and may open up a new topic there about the notation. Let me know what you think. --Nkrupans 16:41, 7 June 2006 (UTC)


 * My two cents: my personal preference is the $$\nabla^2$$ form, though I speak only for myself.  For me, it's simply easier to remember exactly what it is that way ( that is $$\nabla^2$$ suggests to some extent $$\vec \nabla \cdot \vec \nabla$$ ).  Anyway, that's my opinion.  DAG 08:31, 17 June 2006 (UTC)


 * When I first saw that, I thought "What the...?" Particularly in physics, $$ \Delta $$ has its own meaning, and I've never in my life seen it used in that context before.  I've standardised the article to use $$ \nabla $$. Ckerr 03:45, 24 January 2007 (UTC)


 * Old thread I know. From what I understand, some mathematicians use $$\Delta$$, and practically no physicists do. $$\nabla^2$$ is an abuse of notation because operators are not vectors even if they look like them. If most of the people editing these articles are physicists, then that would explain the confusion. —Preceding unsigned comment added by 79.235.185.28 (talk) 17:20, 30 June 2010 (UTC)

Derivation of the wave equation
Is there any "more generic" way to derive the wave equation? I mean, without involving masses, forces etc., but only the common wave properties like propagation velocity and a function u(x,t) [a function of deviation from the equilibrium point for given position and time]? I think the derivation might be simpler then from the account of lesser number of factors. And I would like to know also how to pass from factorized form (d/dt - c*d/dx)*(d/dt + c*d/dx)*u = 0 to those two functions F(x-ct) and G(x+ct). Ah, one more thing: it would be helpful to write somewhere about the d'Alembert operator [], maybe at the beginning. -- SasQ

Yes. You can require the function u{x,t} to maintain its shape as it propagates in either the positive or negative x direction. This will lead to two equations: &part;u/&part;t + c(&part;u/&part;x) = 0 for propagation in the positive direction and &part;u/&part;t - c(&part;u/&part;x) = 0 for negative propagation. The derivation is easy and can be found, for example, on my website (http://prism.texarkanacollege.edu/physicsjournal). Any differentiable function of the form u{x - ct} is a solution of the former and u{x + ct} of the latter. You can get two versions of the wave equation from these two equations. The simpler method is to just multiply the equations together and get (&part;u/&part;t)2 - c2(&part;u/&part;x)2 = 0. This equation is a statement of mechanical energy (potential energy plus kinetic energy) conservation. The first term on the left is related to the kinetic energy of the wave. The second is related to the potential energy where the zero level of potential energy is such to make the total energy zero. This term must be a function of the displacement, u. [For example, for a harmonic wave propagating along a string with a mass per unit length of &mu;, the potential energy per unit length, corrected to make the total energy zero, is (1/2)&mu;k2(u2 - A2), where k is the wave number and A is the wave amplitude.] Any differentiable function of the form u{x &plusmn; ct} is a solution of this equation, just as it is of the usual wave equation. Another method to combine the two equations is to use the equations to define the two operators, &part;/&part;t &plusmn; c(&part;/&part;x). If you operate on the function u{x &plusmn; ct} by first one and then the other operator (in either order) you get the usual wave equation, because whether the function is u{x + ct} or u{x - ct}, one of the operators will give zero. The usual wave equation is, of course, a statement of Newton's second law of motion (F = ma, or, to correspond better to the way the wave equation is usually written, a - F/m = 0.) You can reference the link to my website, mentioned above, to get more details on what I've covered here. P.S. I'm sure this is common knowledge, but I've never seen the statement that a soliton solution to any wave equation has to also be a solution of the linear wave equation. This is obvious, since a soliton solution has to be a function of x &plusmn; ct. - Francis Redfern, Texarkana College Fredfern (talk) 20:53, 27 June 2011 (UTC)

The derivation of the wave equation from Hooke's law needs work. A little more text at the begining of the equation explaining how your equation for Hooke's force is derived would go a long way. From the diagram, I would think the equation should read.. F(hook) =  F(x) + F(x+h) + F(x+2h). --Douglas

Are 1 and 3 spatial dimensions special?
I recall something about the wave equation having special properties in 1 and 3 spatial dimensions that it does not have in other numbers of dimensions; there's something about (hyper)spherical delta function waves existing only for 1 and 3 dimensions. If I am recalling correctly, could someone put together a coherent sentence or two on the topic for somewhere in this article? Thanks — Q uantling (talk &#124; contribs) 23:18, 28 July 2011 (UTC)

Needs defining
Under General solution, just after "For an initial value problem", there are some symbols whose interpretation is not clear:


 * $$u_t(x,0)$$
 * $$f(x)$$
 * $$g(x)$$

If f(x) and g(x) represent the initial conditions mentioned in the text, that we subject the system to, shouldn't that be specified?

Also, if by u_t is meant the once time-derivative of the function u, that could also be somewhat clearer.

80.232.11.13 (talk) 11:31, 11 October 2011 (UTC)

Derivation from the "generic scalar transport equation"
This was put in by "User:Jonsarelli" in August 2008. I do not think it should be here! Either it should be completely removed or at least moved to the very end of the article as a very exotic consideration! It is just confusing making a simple matter rather mysterious! The derivation "from Hook's law" is basically OK although transverse oscillations (in principle exactly the same) should be put more in the focus of the derivation. See "Investigation by numerical methods" that actually contains this derivation.

Stamcose (talk) 08:45, 12 January 2011 (UTC)


 * I agree 100%. I can't see any way that it makes sense. I deleted it. --Steve (talk) 19:25, 20 April 2012 (UTC)

Nonlinear waves and Mach factor
I removed:

"Phenomena in which the speed depends on the amplitude of the wave are modelled by nonlinear wave equations:
 * ${ \partial^2 u \over \partial t^2 } = c(u)^2 \nabla^2 u $

A wave may be superimposed onto another movement (for instance sound propagation in a moving medium like a gas flow). In that case the scalar u will contain a Mach factor (which is positive for the wave moving along the flow and negative for the reflected wave)."

which is unsourced and seems dubious. Non-linear wave equations are unlikely to be of this form, see for example:. And the term "Mach factor" is quite uncommon with respect to the subject of this article. -- Crowsnest (talk) 20:55, 15 September 2012 (UTC)

Elastic eqn
I added the elastic eqn, but network difficulties meant I wasn't logged in.Gwimpey 05:26, Mar 5, 2005 (UTC)

I think there is an error in the last few lines of the Hooke equations. Thee L^2 terms should disappear into the approximation of the second differenttial of the wave function with space so that c^ can equal K/M (propogation speed should not be a function of length) — Preceding unsigned comment added by 109.144.236.215 (talk) 07:47, 17 June 2013 (UTC)

Animation very distracting
The animation is very distracting, even annoying, when reading and trying to understand the not so trivial math. It must be possible to stop it. — Preceding unsigned comment added by 91.53.49.104 (talk) 21:38, 31 August 2013 (UTC)

Generalizations
Some of the generalizations that were in the introduction seemed like quite distant tangents. I've moved them to later in the article, but I'd suggest they simply be deleted. 71.202.141.131 (talk) 04:57, 24 December 2013 (UTC)

Clarification needed re. "one space dimension"
The wave equation in "one space dimension" refers to the independent variables only. The dependent variable can be a second space dimension. For example, a string can exist in two space dimensions, and yet be governed by the wave equation in one space dimension. This really needs to be clarified in the Wikipedia article, to prevent confusion.Anythingyouwant (talk) 18:56, 17 March 2014 (UTC)
 * Done.Anythingyouwant (talk) 02:55, 19 March 2014 (UTC)

Derivation from Hooke's law
In the derivation from Hooke's law, the phrase "The forces exerted on the mass m at the location x+h are:" and equation right after it seem an unusual interpretation of Newton's second law. When one writes F=ma, I wouldn't say "ma" is the "Newton force". Would it make sense to rephrase this part in terms of Newton's second law, rather than equating two forces (where one is actually mass times acceleration)? Or have I misunderstood something? — Preceding unsigned comment added by RobertLipshitz (talk • contribs) 19:44, 8 March 2015 (UTC)

Possible error in the section of spherical waves
The section on spherical waves starts by saying that the wave equation is not invariant under rotations because Laplacian is not invariant. This seems incorrect as the Laplacian is a rotationally invariant operator and also the wave operator (same as the D'Alembertian) is a rotationally invariant object as well. — Preceding unsigned comment added by Jnn 1256 (talk • contribs) 02:57, 14 March 2015 (UTC)

Vector laplacian
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 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.
{\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?
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.

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?

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)