# Tag Info

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.

10

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 them can be reduced to the special case of $y'= ky^2\!$. (Although, explosive behavior is usually avoided and means failure ...

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 $$\phi(r) = \frac{Q}{4 \pi ... 8 In fluid dynamics it is often possible for a given geometry to isolate different degrees of freedom and model long range effects by a delayed influence. An example of this is the "delayed action oscillator" for the El Niño/Southern Oscillation phenomenon in oceanography. For more details, see the page on the Azimuth wiki here. Often fast degress of ... 8 I've seen delayed differential equations used in modeling lasers, particularly quantum dot lasers. Here is a nice comparative view of the use of delayed differential equations verses a finite difference model in quantum dot lasers. 8 I'm going to give a Material Modelling Example. Rubber has the property that it takes time to adjust to the conditions it is applied to (Visco-Elasticity). The behaviour of these effects for static loads are known (Relaxation). However with the high use of Rubbers for dynamic engineering applications (Example: Car tyres) it has been the new focus to ... 6 The flaw in your reasoning seems to be that C is not in fact heat capacity. In Newton's Law of Cooling, the proportionality constant would be related inversely to the heat capacity of the two heated liquids/gasses/materials, and directly to the heat conductance of the object separating the two materials. A material with a higher heat capacity would have a ... 5 The Legendre polynomials occur whenever you solve a differential equation containing the Laplace operator in spherical coordinates with a separation ansatz (there is extensive literature on all of those keywords on the internet). Since the Laplace operator appears in many important equations (wave equation, Schrödinger equation, electrostatics, heat ... 5 You can do the following. From$$\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}.$$where \kappa is our constant of integration. The above can then be simplified to get your Gaussian form ... 5 The mathematician John Baez recently wrote a long series of blog posts about using quantum techniques for non-quantum stochastic systems, in which chemical reaction networks played a central role as an important special case. This culminated in a paper entitled Quantum Techniques for Reaction Networks, which might be something close to what you're looking ... 5 The ordinary Bessel functions are perfectly well defined for complex arguments. For example, here is a plot of \Re[J_2(x + i y)]: The difference between the ordinary and modified Bessel functions is that they satisfy different equations:$$ 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

I think I understand what you mean when you say that you're not satisfied with the “nontrivial bulk topology argument” when it comes to thinking about edge states. The Chern number (for time-reversal breaking) and $\mathbb{Z}_{2}$ invariant (for time-reversal symmetric) systems, as DaniH suggested, does indeed give you information about the edge states; the ...

4

In principle, yes, you can compute all chemical reactions by Feynman diagrams, since the underlying theory is QED. In practice, to a reasonable accuracy, the mechanism of a chemical reaction can be describe by the transition state theory, ref. Atkins, "Physical Chemistry". The theory uses many assumptions of the transition state related to the equilibrium ...

4

If you take the reaction $$\ |\text{CH}_4,2 \text O_2 \rangle\ \ \overset{\text {burn}}\longrightarrow \ \ |\text{CO}_2, 2 \text H_2\text O\rangle\ "$$ There are 7 nuclei and 42 electrons. In nonrelativistic quantum mechanics, a state of this system is a function on a ~150-dimensional space. It's essentially impossible to do any calculations on such a ...

4

The most difficult part is to actually get a set of consistent boundary conditions in the first place - this requires a combination of educated guessing, physical insights, prior experience with related problems, detailed calculations and trial-and-error. In short, it is a bit of an art. However, once you have a set of boundary conditions (as in your case ...

4

Why do you want to have an understanding of the gapless edge states without using bulk topology? If you allow me to use the bulk topology, an argument is that you can continuously move the edge and consider that as an adiabatic parameter which interpolate two systems. To be more precise, you can consider a sphere with part of it in one topological state A ...

4

The Fourier transform of $y\left(x,t\right)$ from the time to the frequency domain is given by $Y\left(x,\omega\right)=\int_{-\infty}^{\infty}y\left(x,t\right)e^{i\omega t}dt$ and satisfies the differential equation: $$EI\frac{\partial^{4}Y\left(x,\omega\right)}{\partial ... 3 If you want to know why computational physicians like Legendre Polynomials, the answer is rather simple. As the other people has already pointed out, the Legendre Polynomials are orthogonal, they can be a very good basis for many applications. For example, if one tries to construct a function which fits the experiment or simulation data within the estimate ... 3 Here's my 30 seconds hand waving argument for "Why is it that we always encounter new special functions f_n with orthogonality relations??"$$\int f^*_n\cdot f_m=\delta_{mn}$$Super broadly speaking, in physics we dealing with the dynamics of certain degrees of freedom. These often employ smooth symmetries, that is we're dealing with Lie groups, which ... 3 The q equation is a separable ODE that can be directly integrated. To do this, note that it can be written as$$ \frac{dq}{dt} = \frac{1}{2}\alpha q^2 $$so that multiplying both sides by dt and integrating from t_0 to t gives$$ \frac{2}{\alpha}\int_{q_0}^{q(t)}\frac{1}{q^2}dq = \int_{t_0}^t dt' $$which after integration implies$$ ...

3

Causality in the Fourier domain is manifest in the behavior of the transfer function as a complex function of frequency, i.e. the location of poles etc. Your example has a pole at +i/RC, so one may deform the integration countour into the lower complex plane for positive tau and show that it vanishes.

3

I) It is not surprising that a solution that uses time-frequency Fourier transformation can superficially look acausal (without actually being acausal), because the Fourier transform $$V_{\omega}~ :=~ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\! dt~V(t)e^{-i\omega t}$$ depends by definition on all times $t$ from the far past to the far future. II) ...

3

Moshe, do you have access to coursework at the University of Idaho? They have a course listing that has your question in the title of the course. Math WS547 Numerical Analysis of Elliptic PDE's (3 cr) WSU Math 546 However, they don't seem to describe the course itself beyond that. I did find this paper entitled: LECTURES on COMPUTATIONAL NUMERICAL ...

3

If the drop is very much static (in still water) and of similar fluid properties to the water around it (so that the ink just labels some initial region), then this is the correct equation to use. If, however, you want to treat the ink as having distinct properties from the water, then you want the Navier-Stokes equations. Since you are interested in ...

3

Good questions; I'm sure a lot of people are confused on this stuff (as I was the first time I used Jackson). Essentially your confusion boils down to being careful to consider the following fact: The Green's function for a particular boundary value problem depends on the boundary conditions. In particular, let's say you have a Dirichlet boundary value ...

3

The diffusion equation is a partial differential equation. The unknown quantity is a function $C(x,t)$. To complete the problem statement you need to specify an initial condition (at $t=0$) and boundary conditions. I'm guessing that your boundary conditions are at infinity, so we take $$C(x,t) \rightarrow 0,\ x\rightarrow \pm \infty.$$ We take a delta ...

3

With the heat equation : $$\frac{\partial T}{\partial t} = D ~\frac{\partial^2 T}{\partial z^2}$$ You got a kernel : $$K(z,t) = \frac{1}{\sqrt{4\pi Dt}}~ e^{- \large \frac{z^2}{4Dt}}$$ such as : $$T(z',t') = \int dz K(z'-z,t'-t) T(z,t)~~~~~~~~(0)$$ Noting $T(0,t') = T_0(t')$, we have : $$T_0(t') = \int dz K(-z,t'-t) ... 3 The second equation is just a solution of the first one, which is a second order differential equation There are various methods of solving them, one is by simply 'guessing' the solution. You could use the second equation$$F(x) = A \cos(mx) + B \sin(mx)$$and differentiate it twice by x$$\frac{d^2 F(x)}{dx^2}=-Am^2\cos(mx)-Bm^2\sin(mx)=$$... 2 I don't know much about the subject, except that it produces two interesting features that are not possible with local differential equations, mainly that a simple equation like this one$$ \frac{dx}{dt} = b(t) x(t-1)  for suitable choices of the function $b$, this equation might have many initial states that produce the same final state, so they produce ...

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