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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

I do not know if it is "plausible" (I do not think so), however a trivial model can be constructed for the one-dimensional case with continuous forces depending on velocities, for $c>0$ constant: $$F_{12}(v_1,v_2) = c\sqrt{|v_1-v_2|} \quad\mbox{and}\quad F_{21}(v_1,v_2) = -c\sqrt{|v_1-v_2|}$$ The system of these two particles does not admit a unique ...


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 ...


1

Whether or not a PDE allows separation of variables depends not only on the equation, but also the boundary conditions. The following conditions must be satisfied for the method of separation of variables to work: The differential operator should be separable. An example of the one which is not separable is \begin{equation} \frac{\partial^2u}{\partial x^2} ...


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 ...


3

To add to CuriousKev's answer, another way of looking a linear differential equation is by looking at their eigenvalues. The order of the differential equation determines hoe many eigenvalues there are. So like you stated this is a second-order differential equation, therefore there will be two eigenvalues, which I will call $\lambda_1$ and $\lambda_2$. The ...


4

What you wrote is a solution, it is just not the only form of the solution. Here are a couple ways to proceed: What happens when $4km < b^2$ ? Let's march ahead done this road. The value in the square root becomes negative, and so there is no real solution for $\omega$. However you can find imaginary solutions for $\omega$. $$\omega = \frac{\pm ...


1

A full 3D model might be too complex for this case, because then you need to know everything about the wall, and the tree, and the interaction between the two in great detail. I think a simplified approach might be more suitable. I'm not sure if this is oversimplified, but let me take a swing at it: If we assume the tree is supported by both the ground and ...


1

I interpret your question as What type of equation is the equation $$\int_{t_0}^{t} \vec{F}(\vec{x}(t')) \cdot d\vec{x}(t') = \frac{1}{2} m ((\dot{\vec x}(t))^2 - (\dot{\vec x}_{0})^2) \; \; \;\; (1)$$ where $t$ is variable and $\dot{\vec x}_{0}$ fixed? The answer is, it is a first order integro-differential equation, not a mere ordinary differential ...


1

Intro I'm the original asker of this question (9 months ago); thanks to the comments and answers I've gotten here, I think I've pieced together an answer that I'm happy with. Short forms used in this answer: CoLM = Conservation of Linear Momentum CoAM = Conservation of Angular Momentum KEB = the Kinetic Energy Balance CoE = Conservation of Energy ...


1

I'd look at it as an energy storage vs loss situation. Take a patch of earth (square slab) and neglect rotation of the earth around its axis (days) so that the patch always faces the sun. At any time it's receiving an incident solar flux (assume constant) and emitting due to its own temperature. The slab also has some thermal mass (capturing the ground, ...



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