# Tag Info

## Hot answers tagged differential-equations

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 ...

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 $$... 12 Your equation is a special case of Riccati equation:$$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

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

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 ... 8 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 ... 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 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}.$$where \kappa is our constant of integration. The above can then be simplified to get your Gaussian form ... 6 The Legendre polynomials occur whenever you solve a differential equation containing the Laplace operator in spherical coordinates with a separation ansatz (there is extensive literature on all of those keywords on the internet). Since the Laplace operator appears in many important equations (wave equation, Schrödinger equation, electrostatics, heat ... 6 The flaw in your reasoning seems to be that C is not in fact heat capacity. In Newton's Law of Cooling, the proportionality constant would be related inversely to the heat capacity of the two heated liquids/gasses/materials, and directly to the heat conductance of the object separating the two materials. A material with a higher heat capacity would have a ... 5 The ordinary Bessel functions are perfectly well defined for complex arguments. For example, here is a plot of \Re[J_2(x + i y)]: The difference between the ordinary and modified Bessel functions is that they satisfy different equations:$$ 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

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

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, $$x = y + c$$ where the constant $c$ is determined by your initial conditions, $$c = x (0) - y (0)$$ (in your case it is the difference between 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 $\omega$. $$\omega = \frac{\pm ... 4 You can simply take the semi-derivative of your equation again, which yields$$\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}$$and then solve that ODE. But, similarly to squaring an ... 4 If you take the reaction$$\ |\text{CH}_4,2 \text O_2 \rangle\ \ \overset{\text {burn}}\longrightarrow \ \ |\text{CO}_2, 2 \text H_2\text O\rangle\ "$$There are 7 nuclei and 42 electrons. In nonrelativistic quantum mechanics, a state of this system is a function on a ~150-dimensional space. It's essentially impossible to do any calculations on such a ... 4 In principle, yes, you can compute all chemical reactions by Feynman diagrams, since the underlying theory is QED. In practice, to a reasonable accuracy, the mechanism of a chemical reaction can be describe by the transition state theory, ref. Atkins, "Physical Chemistry". The theory uses many assumptions of the transition state related to the equilibrium ... 4 Here's my 30 seconds hand waving argument for "Why is it that we always encounter new special functions f_n with orthogonality relations??"$$\int f^*_n\cdot f_m=\delta_{mn}$$Super broadly speaking, in physics we dealing with the dynamics of certain degrees of freedom. These often employ smooth symmetries, that is we're dealing with Lie groups, which ... 4 The Fourier transform of y\left(x,t\right) from the time to the frequency domain is given by Y\left(x,\omega\right)=\int_{-\infty}^{\infty}y\left(x,t\right)e^{i\omega t}dt and satisfies the differential equation:$$ EI\frac{\partial^{4}Y\left(x,\omega\right)}{\partial ...

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 ...

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