Hot answers tagged differential-equations
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 ...
5
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
$$ z^2 y'' + z y' - (z^2 + n^2) y ...
5
I think I understand what you mean when you say that you're not satisfied with the “nontrivial bulk topology argument” when it comes to thinking about edge states. The Chern number (for time-reversal breaking) and $\mathbb{Z}_{2}$ invariant (for time-reversal symmetric) systems, as DaniH suggested, does indeed give you information about the edge states; the ...
4
You can do the following. From
$$\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}.$$
where $\kappa$ is our constant of integration. The above can then be simplified to get your Gaussian form
...
4
The most difficult part is to actually get a set of consistent boundary conditions in the first place - this requires a combination of educated guessing, physical insights, prior experience with related problems, detailed calculations and trial-and-error. In short, it is a bit of an art.
However, once you have a set of boundary conditions (as in your case ...
3
The diffusion equation is a partial differential equation. The unknown quantity is a function $C(x,t)$. To complete the problem statement you need to specify an initial condition (at $t=0$) and boundary conditions. I'm guessing that your boundary conditions are at infinity, so we take
$$ C(x,t) \rightarrow 0,\ x\rightarrow \pm \infty. $$
We take a delta ...
3
If the drop is very much static (in still water) and of similar fluid properties to the water around it (so that the ink just labels some initial region), then this is the correct equation to use. If, however, you want to treat the ink as having distinct properties from the water, then you want the Navier-Stokes equations. Since you are interested in ...
3
The Legendre polynomials occur whenever you solve a differential equation containing the Laplace operator in spherical coordinates with a separation ansatz (there is extensive literature on all of those keywords on the internet).
Since the Laplace operator appears in many important equations (wave equation, Schrödinger equation, electrostatics, heat ...
3
The $q$ equation is a separable ODE that can be directly integrated. To do this, note that it can be written as
$$
\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
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
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
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 ...
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 ...
2
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 ...
2
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 ...
2
They would be linearly dependent if and only if there exist complex numbers $\alpha$ and $\beta$ such that
$\alpha x_{1}(t) + \beta x_{2}(t) = 0 \forall t$
Clearly, if $\omega_{0}=0$ then this is the case for $\alpha = 1$ and $\beta = -c_{1}/c_{2}$. So then they are linearly dependent. However, if $\omega_{0}\neq0$, you can't find a combination of $\alpha$ ...
2
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 ...
2
Let us reformulate OP's question as
Does a constant of motion always imply that a system has a Hamiltonian formulation (by possibly introducing additional variables)?
Answer: No. Take a system $M$ that has a constant of motion and another system $N$ that doesn't have a Hamiltonian formulation. Then the combined system $M\times N$ (where the two parts ...
2
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 ...
1
Just wanted to add a final point to clear this up for you. The potential of a point charge and the Green's function for your problem are the same, up to the normalization constant. In a comment, you say this is by coincidence; It is not, it's physical!
Take your equation $\nabla^2 \Phi = -\frac{\rho}{\epsilon}$, and suppose you wanted the charge density in ...
1
It is like $\sin(\omega t)$ cannot be reduced to $\cos(\omega t),$ i.e., obtained by multiplying by a coefficient, because they are different functions - they are not proportional to each other.
Existence of two independent solutions is also like existence of two independent integration constants for a second order differential equation.
Using both of them ...
1
It means, that you can't produce one solution out of a linear combination of the other. This is important because if you could gain a solution produced by superposition of other known solutions, it is actually not a new solution. It's information is already stored in the other known solution and therefor not relevant. So here the author wants to tell you: ...
1
The key point of this distinction is the type of initial conditions you have to give for an equation.
The canonical example of a hyperbolic set of equations is the wave equation, where the characteristic polynomical that you get when you do a Fourier transform in all of the variables gives you a graph of a hyperbola in configuration space. For this type ...
1
Here is a link which discuses this problem in some depth. I do not think that this would be a good project for an ODE class because temperature and density are not differentially related to each other i.e. one is not a differential form of the other.
1
I would say the initial conditions are technically the same for the two PDEs, but the PDEs are indeed inequivalent if $s$ does not vanish identically: while $\psi(\overrightarrow{r},t)\equiv 0$ is a solution of the second PDE, it is not a solution of the first PDE. To get the solution of the first PDE, you may wish to use Fourier transform in space and ...
1
The solution is not as simple as you wrote, it is a sum over discrete $\lambda_n$.
To get the right solution, you first introduce a shifted $u$: $u'=u-T_0$. For $u'$ you will get $B=0$ and an equation for finding the discrete spectrum of $\lambda$. Then you make a superposition with different $A_n$ and make it obey the initial condition. This permits to ...
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