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What do we actually mean when we say that a certain problem has or does not have an analytical solution? I ask this because some systems that are said to have an analytical solution actually are no better than some systems that are said not to have it. I will give two examples. The standard harmonic oscillator, whose equation of motion is $$m\ddot{x}=-kx$$, has the general solution $$x(t)=A\cos(\omega t+\phi)=A\cos\delta(t),$$ where $$\omega=\sqrt{\frac{k}{m}}$$ and $\delta(t)=\omega t + \phi$. However, that cosine function is defined as $$\cos\delta=\sum_{n=0}^\infty \; (-1)^n \; \frac{\delta^{2n}}{(2n)!}.$$ Which can only be calculated approximately. How is this different with the simple pendulum, where the solution is defined in terms of Jacobi elliptic functions, and can also be calculated only approximately?

Another example, the free particle with linear drag. The equation of motion is $m\dot{v}=-bv$. The solution with initial velocity $u$ is $v(t)=ue^{-bt/m}$. Here, $e$ is an irrational number and thus the velocity cannot be calculated exactly.

Moreover, for any complicated problem, we may define a function to be the solution of the problem, and may be tabulated.

So, why do we say the harmonic oscillator has an analytic solution but the pendulum not? What makes a solution analytical? Also, what means physically that a system has no analytical solution?

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Wikipedia defines an analytic solution as

a mathematical expression constructed using well-known operations that lend themselves readily to calculation

As you say, this simply begs the question of what counts as a “well-known operation”. By convention, the exponential, logarithmic, trigonometric and hyperbolic trigonometric functions and their inverses would certainly qualify. Hypergeometric functions, elliptic integrals, Bessel functions etc. are more of a grey area. Broadly speaking, I would say that if a function has been widely studied enough to give it a name, a notation, and tabulate its values in DLMF or similar references then you can count it as a “well-known operation”.

The distinction between systems that have or do not have an analytic solution is entirely a matter of convention, and has no physical significance whatsoever.

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However, that cosine function is defined as $\cos (\delta)=\sum_{n=0}^{\infty}(-1)^{n} \frac{\delta^{2 n}}{(2 n) !}$. Which can only be calculated approximately.

That's not correct. $\cos\delta$ is an analytical function and, like all analytical functions, is infinitely differentiable, therefore you can rewrite it as an infinite series expansion.

A strict answer could be that an analytical solution can be written in "closed form", that is, using a finite number of operations.

From my experience (which is not much) the election of naming a solution as analytical or not could be in some cases arbitrary. It could depend, in some cases, if you can do some analytical stuff of your interest in the solutions. In that case, you can call it analytical (although they aren't). For example: solving the Schrödinger equation from a free particle in spherical coordinates is the same as solving the Bessel equation which solutions are strictly not analytical, but could be said that are analytical.

Checking my answer on Wikipedia I discovered the term "analytic expression", which can be the origin of the confusion in this terminology topic.

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  • $\begingroup$ Your logic is slightly off here: not all functions which are infinitely differentiable admit a series expansion. This is the distinction between analyticity and $C^\infty$. $\endgroup$ – Richard Myers May 2 at 21:09
  • $\begingroup$ The reason I say it is that often the outputs of the cosine function are irrational numbers. $\endgroup$ – Don Al May 2 at 23:36
  • $\begingroup$ @RichardMyers Thank you for your clarification. I thought all $C^{\infty}$ functions admit a series expansion! $\endgroup$ – LongJohn May 3 at 8:13
  • $\begingroup$ "the election of naming a solution as analytical or not could be in some cases arbitrary" -- that can be the case, but the election of calling a function analytical or not is not arbitrary; it is either complex-differentiable (and thus equal to its Taylor series) or not. $\endgroup$ – Emilio Pisanty May 3 at 10:18
  • $\begingroup$ Regarding Richard's comments, the easiest example is $f(x)=e^{-1/x}$ for $x>0$ (and set to zero for $x\leq 0$). The function is smooth (all derivatives exist, and are zero) at zero, but the Taylor series fails on the right-hand side of that. But, on the other hand, if you impose the stronger condition of being analytical (i.e. having a single complex derivative in a neighbourhood of the point in question) then it does follow that the function's Taylor series converges to the function. $\endgroup$ – Emilio Pisanty May 3 at 10:22

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