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Take a look at the notes on lectures 1 and 2 of Geometric Numerical Integration found here. Quoting from Lecture 2 A numerical one-step method $y_{n+1} = \Phi_h(y_n)$ is called symplectic if, when applied to a Hamiltonian system, the discrete flow $y \mapsto \Phi_h(y)$ is a symplectic transformation for all sufficiently small step sizes. From your ...

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Crank-Nicholson method is effectively the average of forward (explicit) Euler $\psi(x,t+dt)=\psi(x,t) - i*H \psi(x,t)*dt$ and backward (implicit) Euler method $\psi(x,t+dt)=\psi(x,t) - i*H \psi(x,t+dt)*dt$ The backward component makes Crank-Nicholson method stable. The forward component makes it more accurate, but prone to oscillations. If you want to ...

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I will not give you the numerical solution, but I will below explain some analytical simplifications that I believe are required to solve the numerical problem. The strategy is simple: try to express all the parameters of the integral in term of dimensionless variables. To achieve a discussion in term of $\delta = \Delta(T) / \Delta(T=0)$ and $\tau = T / ... 7 There are no exact solutions, only approximations and numerical solutions. Don't forget that orbiting black holes will radiate gravitational waves so any solution would have to include those and the corresponding decay of the orbit until the black holes coalesce. 6 "They" are probably talking about symplectic integrators. Most numerical integrators for (partial) differential equations do not specifically consider the energy of the system; they are generic integrators capable of solving any set of DEs, and not all DE's have a concept like "energy". When these are applied to a classical dynamics problem concerning ... 6 It seems like they were able to rigorously prove the existence of N-body choreographies by using interval Krawczyk method to show that a minimum exist to the variational problem solved in the subspace of the full phase space satisfying some symmetry conditions. Following the links given I found this paper where they explain the method. It's not exactly a ... 6 If you look at the Laplacian: $$\nabla^2=\frac{1}{r}\,\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\right)+\frac1{r^2}\frac{\partial^2}{\partial\phi^2}$$ you can clearly see that this diverges at$r=0$so discretization of this should also diverge. There are three solutions to remedying the divergent feature that I can think of: Choose a ... 6 There exists a variety of options for this task but let me stress first that this is an extremely complicated and difficult issue that is still subject of current research because analytical continuation is an ill posed problem! 1) The 'analytical' analytical continuation can be performed when the function$f(\mathrm i\omega)$under consideration is a ... 5 The SIAM 100-Digit Challenge springs to mind. Problem 10 gives the flavor of the type of problems put forward in this challenge: A particle at the center of a 10 x 1 rectangle diffuses until it hits the boundary. What is the probability that it hits at one of the short ends (rather than at one of the long sides)? The answer needs to be accurate to at least ... 5 For a very recent authoritative review of the numerical approach, see Centrella et. al. http://arxiv.org/abs/1010.5260 For the alternate parameterized post Newtonian approach, see Living Reviews of Relativity http://relativity.livingreviews.org/Articles/subject.html and look for articles number 2007-2, 2006-4 and 2003-6. 5 We have observed that the underlying level of nature is quantum mechanical. Quantum means "a definite quantity" of something so definite quantities can be counted and so integral numbers play a role : a) in the number of particles , in the number of energy levels characterized by quantum numbers ( i.e. integer numbers).b) There are the fields which are ... 4 There is rather nice function in Mathematica 7, which allows one to integrate over an arbitrary complicated region. It is Boole:[True,False]$\to${1,0}. Below is just an example taken from Mathematica Documentation Center. If you have a 2D area defined by the inequality$4 x^4-4 x^2+y^2\leq 0$, you can integrate any function$f(x,y)$over this domain as ... 4 The reason that these sorts of libraries don't exist is because the particular algorithm that you use to do the calculation will depend upon the exact details of the light field and the input and output planes you are trying to compute. For example, let me outline the the simplest case for this sort of calcualtion: The input light field$g_0$has a slowly ... 4 According to general relativity, a pair of massive bodies that orbit each other emits gravitational waves - for analogous reasons to the reasons why accelerating charges in electrodynamics emit electromagnetic waves. So there can't be any static solutions resembling binary stars or binary black holes. The solutions have to be non-static and a complicated ... 3 Franz Pretorius has worked on this and developed animations. http://prl.aps.org/abstract/PRL/v95/i12/e121101 The field is numerical relativity. Matthew Choptuik also, I believe, has done work on this. 3 There is no simple way. The "standard" way is to solve Poisson equation with proper boundary conditions (constant$\varphi$at the surface). Out of potential distribution it is easy to extract charge distribution. For simple shapes (infinite plane, sphere, etc) it is possible to solve the problem analytically. For arbitrary shape there is no simple ... 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 A way of physically thinking about this is that a two body problem in general relativity does not generally have closed orbits. If one of the bodies is very large and the other a small satellite the problem is integrable. The periapsis (perihelion) advance of the small satellite is repeated with each orbit, which makes the problem integrable. If the two ... 2 Maybe they do; but Fourier transforms have an inherent flaw which makes them less than useful for such cases. That flaw is that they are transforms of steady state conditions: The initial and final conditions of the system are assumed to be the same; and transients are not considered. The transform for which you seek is the Laplace transform. Laplace is a ... 2 For the broken ones, weight will work. For the dented ones, you have a much harder problem, particularly if you consider arbitrarily small dents. 2 The mechanism you would use to identify the candies would be one of computer vision (which is off topic for this site try http://stats.stackexchange.com/). However, the exact mathematical method you would/could use to determine the broken candies from the okay ones depends on the shape of these candies. Lets say for arguments sake that they were spherical; ... 2 So what should the boundary conditions be for a segment of an infinite line. Let us explore the options ($u$is displacement,$u'$is slope and$u''$is curvature): Fixed Ends:$u(0)=0$,$u(L)=0$Mirror Ends:$u'=0$,$u'(L)=0$Free Ends:$u''=0$,$u''(L)=0$Periodic Displacements:$u(0)=u(L)$,$u'(0)=u'(L)$Periodic Tensions:$u'(0)=u'(L)$,$u''(0)=u''(L)$... 2 Let me take a slightly different perspective to Ron Maimon and say that the answer depends on whether you're after an exact solution of some mathematical model, or whether you want to calculate the exact physical behaviour. Any method for calculating the physical properties of a system rely upon an approximation. If you choose some model you can certainly ... 2 The main problem in your proposed equation is that the electromagnetic equation with the D'Alambertian over the vector potential is not in Hamiltonian form, this means that the separation of solutions in Sturm–Liouville eigenstates of the energy operator is not manifest in the equation. Without that, you cannot find eigenstates of the coupled system. You ... 2 It looks like you integrate by$d\lambda$, whereas, according to the formula, you should integrate by$d\Omega\$ - this may well give you 13 orders of magnitude.

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I think this is just a question of defining what you mean by "solve". All physical problems, most definitely so in classical mechanics, can be posed as differential equations for which solutions (i.e. trajectories of the dependent variables) can be found at least through numerical integration. In this sense, as far as I know, no one has identified any ...

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The most important thing is conservation of momentum to describe the collisions. This part is actually quite straightforward, but before you get to collisions you should model the motion of single balls. Obviously, you will describe the balls classically and probably not at relativistic speeds (though that would be interesting...) so pretty much all you ...

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