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Why are waves expressed as 2nd differentials?


iScience92

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May be a chicken-egg thing. In my experience, it is not that you start out with a wave-like solution and then want to express it as a 2nd order differential equation. But that you arrange conditions on your system, come out with a 2nd order differential equation of a particular kind, and know that the solutions to this are wave-like.

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Because there are two independent variables (time and space) and you have one equation for each variable or alternatively a single second order equation connecting them both.

I don't know what that is supposed to mean, and attempts of making sense of it failed at the stage where I suspect you put a lot of implicit assumptions into your reply. My best interpretation of "two independent variables and one equation for each variable" so far was assuming two first-order differential equations dX/dx = f(x,t) and dX/dt = g(x,t) (with X being the property in question). But my gut feeling is that already this starting point is not what you meant.

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Timo was pretty well there.

 

Whilst we normally resolve vector into x, y and z (or whatever) spatial coordinates, the fact reamins that it can also be (or is) a single entity, which is why I just said 'space'.

 

If you have 2D or 3D space then you need to use a suitable differential operator (nabla or del), but the principle is the same.

 

You have system in space and another in time that cannot solubly combine at the level of the first differentiation.

 

The wave arises as a solution when you equate second differential coeficients.

 

@iScience92, Why not keep it simple to a single space dimension, adding more does not add anything to the discussion.

 

And please answer my question so we have a concrete example to work with.

 

What is your equation for the wave?

 

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