If a function $f(\vec{x}, t)$ goes to zero faster than $\frac{1}{|\vec{x}|}$, will $\frac{\partial f}{\partial t}$ always decay faster than $\frac{1}{|\vec{x}|}$ too? Assume that $f$ is differentiable.
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Seems reasonable and desirable but without other constraints, such as satisfying a differential equation, the answer is No; here's a counterexample.
In one spatial dimension take $f(x, t) = \sin(x/t) / x^{3/2}$. For $t > 0$, $\vert f(x,t) \vert \leq 1 / \vert x \vert^{3/2}$ but $f_t(x, t) = -\cos(x/t)/(x^{1/2} t^2)$, which, as $x \rightarrow \infty$, doesn't decay at the rate you'd like it to.
A rural reader
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@Bio: next question is Why? – A rural reader Jan 18 '21 at 05:12
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I'm studying Maxwell's equations for electricity and magnetism (In particular the version which neglects the time derivative of the electric field.) I needed the time derivative of the current density to decay fast enough, so that I can justify the use of Helmholtz Theorem. – Bio Jan 24 '21 at 14:47