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25

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 ...


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 $$ ...


14

What follows is certainly not a comprehensive answer addressing all of your concerns. It is an answer to the question is there a way to see something clearly pathological like superluminal signals in the heat equation? I would argue that yes, there is. The general solution to the initial value problem $T(x,0) = T_0(x)$ for the heat equation on the ...


12

Your equation is a special case of Riccati equation: $$y'=q_0(x) + q_1(x)y + q_2(x)y^2\!$$ with $q_0(x)=q_1(x)=0$ and a constant $q_2(x)$: $$y'= ky^2\!$$ There are lots of applications for the main Riccati equation in physics, and some of them can be reduced to the special case of $y'= ky^2\!$. (Although, explosive behavior is usually avoided and means ...


12

If I have $n$ objects (say, reactants) colliding, the rate of collisions will be roughly proportional to $n^2$. If the population grows by a fixed amount with each collision, we would find this law. See the rate equation. I think that growth due to sexual reproduction might fit here as well, but I'm not familiar enough with population biology to say.


9

Well, simpler than recognising a differential equation will be to recognise the quantity it describes. Quickly plugging in Solve[y'[t]==k y[t]^2,y[t],t] into wolfram alpha gives $$y(t)=\frac{1}{c-k\ t},$$ which explodes at $t=c/k$. Now let's think of physical quantities which diverge as $\frac{1}{t}$. The Coulomb potential $$\phi(r) = \frac{Q}{4 \pi ...


9

There are a few reasons I can think of: (1) The second order system is that it is time-reversible. If you let $t\to-t$, you get $$ \frac{\partial^2f}{\partial(-t)^2}=\frac{\partial^2f}{\partial t^2}=v^2\frac{\partial^2f}{\partial x^2} $$ whereas the first order system has $$ \frac{\partial f}{\partial(-t)}=-\frac{\partial f}{\partial t}=\pm v\frac{\partial ...


9

There's nothing wrong with the first order wave equation mathematically, but it's just a little boring. If you want to use this equation to describe waves, it basically amounts to having a 1d solid with speed of sound $v$ for left moving waves (say) and speed of sound $0$ for right moving waves. It wouldn't surprise me if such a thing could be constructed ...


9

Firstly, there are a few issues with a time-dependent potential, $V(x,t)$. Namely, if we apply Noether's theorem, the conservation of energy may not apply. Specifically, if under a translation, $$t\to t +t'$$ the Lagrangian $\mathcal{L}=T-V(x,t)$ changes by no more than a total derivative, then conservation of energy will apply, but this resricts the ...


8

Dimensionless equations have the advantage that they work for any value of the parameters. They are scale invariant. So the solution in terms of a single dimensionless variable applies to all values of $D$ and $t$. It also allows the definition of characteristic values for the dynamic variables. In your example, one could say $u_0$ = ...


8

I'm going to give a Material Modelling Example. Rubber has the property that it takes time to adjust to the conditions it is applied to (Visco-Elasticity). The behaviour of these effects for static loads are known (Relaxation). However with the high use of Rubbers for dynamic engineering applications (Example: Car tyres) it has been the new focus to ...


8

I've seen delayed differential equations used in modeling lasers, particularly quantum dot lasers. Here is a nice comparative view of the use of delayed differential equations verses a finite difference model in quantum dot lasers.


8

In fluid dynamics it is often possible for a given geometry to isolate different degrees of freedom and model long range effects by a delayed influence. An example of this is the "delayed action oscillator" for the El Niño/Southern Oscillation phenomenon in oceanography. For more details, see the page on the Azimuth wiki here. Often fast degress of ...


7

Let's say your goal is to describe the shape of some object, such as a box. You could create a completely arbitrary ruler and measure the three axes of the box, coming out for example with lengths of 11.72, 23.44, and 35.16 of your arbitrary ruler units. Or you might look at your results more closely and think hmm, something is going on here, since the ...


6

I don't think "the general solution to this equation is a Gaussian beam" - this is a linear equation, so a superposition of solutions is also a solution, and an arbitrary superposition of Gaussian beams is not always a Gaussian beam.


6

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 ...


6

You can do the following. From $$\frac{1}{\Psi}\frac{\partial \Psi}{\partial x} = Cx,$$ We can write the following interal equation $$\int \frac{1}{\Psi} \mathrm{d}\Psi = C \int x \mathrm{d}x,$$ $$\ln \Psi + \kappa = \frac{1}{2} C x^{2}.$$ where $\kappa$ is our constant of integration. The above can then be simplified to get your Gaussian form ...


6

The flaw in your reasoning seems to be that $C$ is not in fact heat capacity. In Newton's Law of Cooling, the proportionality constant would be related inversely to the heat capacity of the two heated liquids/gasses/materials, and directly to the heat conductance of the object separating the two materials. A material with a higher heat capacity would have a ...


6

The Legendre polynomials occur whenever you solve a differential equation containing the Laplace operator in spherical coordinates with a separation ansatz (there is extensive literature on all of those keywords on the internet). Since the Laplace operator appears in many important equations (wave equation, Schrödinger equation, electrostatics, heat ...


5

Here's my 30 seconds hand waving argument for "Why is it that we always encounter new special functions $f_n$ with orthogonality relations??" $$\int f^*_n\cdot f_m=\delta_{mn}$$ Super broadly speaking, in physics we dealing with the dynamics of certain degrees of freedom. These often employ smooth symmetries, that is we're dealing with Lie groups, which ...


5

The mathematician John Baez recently wrote a long series of blog posts about using quantum techniques for non-quantum stochastic systems, in which chemical reaction networks played a central role as an important special case. This culminated in a paper entitled Quantum Techniques for Reaction Networks, which might be something close to what you're looking ...


5

The ordinary Bessel functions are perfectly well defined for complex arguments. For example, here is a plot of $\Re[J_2(x + i y)]$: The difference between the ordinary and modified Bessel functions is that they satisfy different equations: $$ z^2 y'' + z y' + (z^2 - n^2) y = 0, $$ for the ordinary Bessel functions and $$ z^2 y'' + z y' - (z^2 + n^2) y ...


5

This is an answer by an experimentalist who had been fitting data with mathematical models since 1968. When fitting data one goes to the simplest mathematical models. When the data display variations in time and space the Fourier expansion is extremely useful because it gives the frequencies and amplitudes that will fit a periodic data set. One gets as ...


5

For example, the heat equation is ∂u/∂t−α∇2u=0, but what would a solution to this be? A solution to this equation is a function $u(\vec{x},t)$ that describes the temperature of the material in question at different locations, and how it evolves over time. We say this is a solution to the equation, because out of all possible functions $u(\vec{x},t)$, ...


4

Since the rate of change of $ x $ is the same as the rate of change of $y $ you really only a single equation of with one variable. We write, \begin{equation} x = y + c \end{equation} where the constant $ c $ is determined by your initial conditions, \begin{equation} c = x (0) - y (0) \end{equation} (in your case it is the difference between the ...


4

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 ...


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 ...


4

In practice, the simple answer is that it works when it does. You need to test an unkown situation at hand, by making the assumption of a separated solution $f(x,\,y,\,\cdots)=X(x)\,Y(y)\,\cdots$ and seeing what comes from that. Pretty much all the equations you will need for electromagnetism have been tested for separation in co-ordinates that reflect ...



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