Tag Info

New answers tagged

4

You can simply take the semi-derivative of your equation again, which yields $$\begin{align} m\frac{d^2}{dt^2}\underbrace{\frac{d^{\tfrac12}x}{dt^{\tfrac12}}}_{=-\frac mk\frac{d^2x}{dt^2}} &= -k\frac{dx}{dt} \\\Rightarrow m^2\frac{d^4x}{dt^4} &= k^2\frac{dx}{dt} \tag{*} \end{align}$$ and then solve that ODE. But, similarly to squaring an ...


2

So, what you have here is, as others have mentioned, a fractional order differential equation. Since others have provided graphs, there seems to be little point to adding one. Also, you seem more interested in the qualitative aspect than actual analytical solutions, as you've mentioned. In essence, what you have here is some second order system with some ...


2

An attempt for a more explicit solution. Using the definition of the half-derivative given by JamalS, one can transform the differential equation using the Laplace transform and get $$s^2X(s)-sx(0)-x'(0)=-\gamma^3{\sqrt s}\;X(s)$$ where $\gamma=\sqrt[3]{\frac km}$ is a positive constant. Solving for $X$ gives $$X(s)=\frac{sx(0)+\dot x(0)}{s^2+\gamma^3{\sqrt ...


2

One way to try to solve the equation is transforming it in an ODE. Apply the fractional derivative $D^{1/2}$ again to the equation to find $$D^{1/2}[D^2x(t)]=D^{5/2}x(t)-C_1t^{-3/2}-C_2t^{-5/2}-C_3t^{-7/2},$$ and $$D^{1/2}[D^{1/2}x(t)]=Dx(t)-C_4t^{-3/2}$$ Hence we got $$D^{5/2}x(t)=-\frac{k}{m}Dx(t)+C_1t^{-3/2}+C_2t^{-5/2}+C_3t^{-7/2}$$ But, we also have ...


14

I am no mathematician and am a little afraid that my answer is too simple to be true, but here goes: I use Fourier transforms to define the fractional derivative. $x(\omega)$ is defined such that $$ x(t) = \int_{-\infty}^\infty \, \frac{\text{d}\omega}{2\pi} \text{e}^{i \omega t} \, x(\omega) \, .$$ Then any integer derivatives is $$ ...


20

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 ...


0

Does your conductor have any resistivity? In that case the fluctuation-dissipation theorem applies. In the case that your conductor is a perfect superconductor, it would still couple inductively to the electromagnetic field around it, which, per 3rd law of thermodynamics must have a non-zero temperature. To remove those fluctuations, the total field volume ...


1

In the quantum mechanical description of a conductor all energy levels of the conductor are filled up to some specific energy level, called the Fermi level. This is because of the Pauli exclusion principle, which says that electrons with the same spin cannot occupy the same energy level and thus causes higher energy levels to be populated. Therefore, even at ...



Top 50 recent answers are included