Why does Laplace operator affect only x-coordinates but not time variable in wave equation. Does t variable in U(X, t) is different from x coordinates? I think they are equal.
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Russiancold
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The gradient measures rates of change with respect to spatial position, where a point actually is in space, and the Laplacian is the divergence of the gradient. Time is not a spatial coordinate. If it were, we wouldn't have to ask where and when when meeting somebody, only where. – mallan Jun 03 '18 at 15:04
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It's kinda common explanation, could you please give something more strict? – Russiancold Jun 03 '18 at 15:17
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Oops my brain decided to talk about Fourier transforms but you want Laplace. The simplest reason is probably that your boundary conditions are on $x$, not $t$ so the boundary terms with the Laplace transform don't work out nicely. – Cameron Williams Jun 03 '18 at 15:32
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I would suggest looking at the derivation of the wave equation (for instance a physical derivation from a physics or engineering textbook). Otherwise, from a purely mathematical perspective, the answer is: Because that is the definition of the wave operator. – Jose27 Jun 06 '18 at 21:37
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Often when solving ODE or PDE with Laplace transform (in undergraduate courses), it is about the one-dimensional Laplace transform. In your case, the $t$ is considered constant (a snapshot in time) when we "squeeze" the space $x$ exponentially.
Indeed, $t$ (time) and $x$ (space) are in equal standing mathematically in terms of Laplace transform. You can actually do the Laplace transform on $t$ with $x$ unaffected.
There's of course multi-dimensional Laplace transform that affects more than one variable when operated.
Lee David Chung Lin
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