I know that classical wave equation holds for spherical wave fronts. In addition, Huygens' principle states that any wave front is a superposition of many spherical wavelets, so why does the equation NOT hold for any wave front such as the cylindrical?

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    $\begingroup$ Substitute "holds" for "carries" @AccidentalFourierTransform $\endgroup$ – Danu Mar 11 '16 at 20:21
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    $\begingroup$ The wave equation is true for any shape wave front. Spherical waves and plane waves are the most commonly discussed, but by no means the only solutions. $\endgroup$ – Paul T. Mar 11 '16 at 20:40
  • $\begingroup$ @ Paul T. The wave equation may allow any shape wave front, but Huygens principle does not hold for any shape wave front. For example cylindrical waves do not propagate 'cleanly' without a wake whereas spherical waves and plane waves do. $\endgroup$ – user45664 Oct 11 '16 at 22:42
  • $\begingroup$ See researchgate.net/publication/316994209 $\endgroup$ – user45664 May 19 '17 at 22:37

See https://www.researchgate.net/publication/316994209 This paper shows geometrically why plane and spherical waves propagate without a wake. The same approach could be used to show that some other wave shape would not propagate without a wake.

Basically an impulsive sphere and an infinite impulsive plane are examined. The wave field as a function of time at an external point in space is computed, taking into account 1/r type spherical spreading. Then the time derivative of the wave field is taken to get the wave fronts. The derived wave fronts consist of Dirac Deltas corresponding to the start (and end for the sphere) of the wave fields. (The wave field in between is constant so the derivative there is zero.) So for those sources there are no wakes and the waves propagate cleanly. If the process was repeated for a cylindrical wave the results would probably (or certainly) be different.


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