mathematical physics - Motion described by $m frac{mathrm{d}^2 x}{mathrm{d}t^2}=-kfrac{mathrm{d}^{frac12 }x}{mathrm{d}t^{frac12}}$

What kind of motion would a (preferably dimensionless for simplicity) body do if the force acted on it was proportional to the semi-derivative of displacement, i.e.

$$m \frac{\mathrm{d}^2 x}{\mathrm{d}t^2}=-k\frac{\mathrm{d}^{\frac12}x}{\mathrm{d}t^{\frac12}} \, \, ?$$

It would be helpful if someone with a copy of Mathematica plotted this for various values of the constants.

Answer

If $D^n$ denotes the $n$th derivative and $D^{-n}$ the $n$th integral, then we have that,

$$D^n f(t) = D^m[D^{-(m-n)}f(t)]$$

providing $m \geq \lceil{n}\rceil$. For our half derivative, we choose $n=1/2$, and $m=2$, in which case we have,

$$D^{1/2}f(t) = D^2[D^{-(3/2)}f(t)]$$

There is a general formula for the $n$th integral of a function, one of my favorite results of Cauchy:

$$f^{-(n)}(t) = \frac{1}{\Gamma(n)}\int_{0}^t (t-u)^{n-1}f(u) \, du$$

which is essentially a convolution $f(t) \ast t^{n-1}$. Applying it, we find,

$$D^{1/2}f(t) = \frac{d^2}{dt^2} \left[ \frac{2}{\sqrt{\pi}}\int_0^t (t-u)^{1/2}f(u) \, du\right]$$

Given the differential equation,

$$\frac{d^2 x(t)}{dt^2} = -\frac{k}{m} \frac{d^{1/2} x(t)}{dt^{1/2}}$$

we can substitute in our definition of $D^{1/2}x(t)$, and conclude,

$$x(t) = -\frac{2k}{m\sqrt{\pi}}\int_{0}^t (t-u)^{1/2}x(u)\, du + c_1t +c_2$$

for $c_1,c_2 \in \mathbb{R}$ which is an integral equation. If we can assume $x(t)$ is supported on $[0,\infty)$ only, then the integral is a convolution $x(t) \ast \sqrt{t}$ and taking the Laplace transform, we find,

$$X(s) = \left( 1+ \frac{k}{ms^{3/2}}\right)^{-1} \left( \frac{c_1}{s^2} + \frac{c_2}{s} \right) = \frac{m(c_1 + c_2 s)}{k\sqrt{s}+ms^2}$$

The solution $x(t)$ is then the inverse Laplace transform of $X(s)$. Formaly, this is given by,

$$x(t) = \frac{1}{2\pi i} \int_{\Gamma} e^{st} \frac{m(c_1 + c_2 s)}{k\sqrt{s}+ms^2} \, ds$$

where the contour $\Gamma$ is in the complex plane; it is a vertical line of infinite length with all poles of $F(s)$ to its left. In practice, we close the contour with an additional contour, ensure the second integral tends to zero (e.g. by the estimation lemma), and use the residue theorem.

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The integrand, which we denote $F(s)$, has three poles located at $s^3 = k^2/m^2$, or equivalently,

$$s_1 = \omega^{4/3}_0, \quad s_2 = \frac{1}{2}(1+i\sqrt{3})\omega^{4/3}_0, \quad s_3 = \frac{1}{2}(i\sqrt{3}-1)\omega^{4/3}_0$$

as well as at $s_0= 0$, where we define $\omega^2_0 := k/m$. The vertical contour should begin after $s_1$ so all poles are to the left. However, doing so analytically is somewhat tedious. I chose to use a numerical method for the evaluation of inverse Laplace transforms due to H.E. Salzer which uses aquadrature formula. With Mathematica, I managed to reconstruct $x(t)$ partially:

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in the simplified case when $c_1 = c_2 = k/m = 1$. It seems, by visual inspection, the solution resembles that of damped harmonic motion, such as when one introduces a damping term $\gamma \dot{x}$ in the equations of motion of a standard harmonic oscillator.