29

This follows from the uniqueness theorem for solutions of ordinary differential equations, which states that for a homogeneous linear ordinary differential equation of order $n$, there are at most $n$ linearly independent solutions. The upshot of that is that if you have a second-order ODE (like, say, the one for the harmonic oscillator) and you can ...


24

You need to be more precise about exactly what problem you're solving and what the inputs are. But if you're considering the general problem of what electromagnetic fields are produced by a given configuration of electric charge and current over spacetime, then the general solution is given by Jefimenko's equations.


9

There are two logical options when you vary $t$: either the value of the left-hand side changes, or it doesn't. If it changes, then the right side must change as well, since they are equal. But the right-hand side can't change when you vary $t$, since it is not a function of $t$! Therefore, since varying $t$ produces no change in the left-hand-side, then ...


9

Look at it as an initial-value problem. If you know the electric and magnetic field throughout space at one instant, and the positions and velocities of all charged particles at that instant, then you can numerically evolve the system forward in time. Two of Maxwell’s equations tell you how fast the fields are changing at each point (and thus their new ...


7

These are all good and correct answers, but I will answer from a different perspective. Any linear-differential equation of degree $n$ has $n$ linearly independent solutions, ie. these $n$ solutions span a vector space, with sets of solutions forming a basis. For simple harmonic motion, the differential equation is: $$m(\dfrac{d^2x}{dt^2})+kx = 0$$ As ...


6

What they have done is focus exclusively on the long-time solution when the system has reached "oscillatory steady state." This solution does not feature any exponentially decaying terms in time. So their solution for the temperature is taken to be of the form: $$T(x,t)-T_0=\alpha(x)\cos(\omega t-\phi)+\beta(x)\sin(\omega t-\phi)+\frac{A}{k}(L-x)$$where ...


5

The equation can be integrated once, then we have second order equation $$x''+x'+\frac {x^2}{2}=C_1$$ Make a substitution $x'=y(x)$, then $x''=y\frac {dy}{dx}$ as a result, we obtain the Abel equation , for which an analytical solution is known, see https://www.hindawi.com/journals/ijmms/2011/387429/#sec2 $$y\frac {dy}{dx}+y=C_1-x^2/2$$ The numerical ...


4

The derivation based on Hooke's law given on Wikipedia is for the 1D case, so it's assumed that the particles can only move horizontally - hence the absence of the square root. In that derivation, $u(x+2h)$ simply means the horizontal displacement of the mass that was initially at $x+2h$. The wave equation then comes from the "usual" trick of considering ...


4

One way of deriving it is using Taylor series (although to be fully rigorous, this requires further justification for restricting to analytic functions). We have that $f(x) = \sum a_n x^t$, so $f''(t)=\sum (n+1)(n+2)a_{n+2}t^n$. If $f''(t)=-\frac k m f(t)$, then $(n+1)(n+2)a_{n+2} = -\frac k m a_n$, so $a_{n+2} = \frac {-k a_n}{(n+1)(n+2)m}$. So $a_0$ ...


4

The linear combination will only solve Laplace's equation. So no. If each $\phi_i$ solves Laplace's equation, then $\Delta\phi_i=0$, and $$\Delta\left(\sum_ia_i\phi_i\right)=\sum_ia_i\Delta\phi_i=0$$


4

Unlike attenuation, scattering is difficult because energy which is scattered out of the beam can be scattered back in. If scattering coefficient is very weak over the path length of interest, or the scattering strongly favours small angular deviations, (green path in diagram) you can add the scattering and attenuation coefficients together to estimate the ...


3

This quasi-linear 1st-order PDE $$\left(\frac{\partial u}{\partial t}\right)_{\!x} + u\left(\frac{\partial u}{\partial x}\right)_{\!t}~=~0\tag{1}$$ is the inviscid Burgers' equation. It can be solved via the method of characteristics. The ODE IVP $$ \frac{dx}{dt}~=~u, \qquad x(t\!=\!0)~=~\xi, \tag{2}$$ (where $u$ is treated as an external parameter and $\...


3

Consider a boundary condition of the form $$ \alpha \psi + \beta \frac{\partial \psi}{\partial x} + \gamma \frac{\partial \psi}{\partial t} =0 $$ on the boundary, where $\alpha$, $\beta$, and $\gamma$ are real coefficients. Standard Dirichlet boundary conditions correspond to $\beta = \gamma = 0, \alpha \neq 0$, while Neumann boundary conditions correspond ...


3

The integral forms of Maxwell's equations are fairly useless unless you have situations with very high degrees of symmetry and/or fields aligned along co-ordinate axes. e.g. The beloved examples of undergraduate physics everywhere of spherical and cylindrical charge and current distributions. Once you move away from these situations then the integral forms ...


3

The simplest model that fits is potential flow around a cylinder (or a circle in 2D). This assumes an an inviscid, incompressible fluid with no vorticity, which is too simple to model the backflow. The backflow occurs because of viscosity produces boundary layer separation. I think the second simplest model possible would be to solve the steady-state ...


3

The equations that describe the flow of fluids like water are the Navier-Stokes equations. These are notoriously intractable. So much so that they are currently the subject of one of the Millenium prizes in mathematics. If you're interested in finding out more about this I recommend Terence Tao's article on the subject.


3

Define $\langle y | E\rangle = \psi_(y)$ and study the asymptotics of your ODE, $$ (\partial_y^2-y^2)~\psi (y)=0. $$ That means the leading behavior of ψ for large y; so , for example, if you had a polynomial in y, you just keep the highest order thereof: If you have an order-m polynomial, you'd just keep the $y^m$ monomial, since it dominates all lower ...


3

This is easiest to see in 1D (where the 'volume' becomes an interval $[a,b]$ and the boundary consists of 2 points $a$ and $b$). Poisson's equation becomes a 2nd-order ODE, which means that the full solution has 2 integration constants. Imposing both Dirichlet and Neumann boundary conditions (BCs) would lead to 4 conditions (2 at $a$ and 2 at $b$), which ...


3

This is why cosmic censorship is considered to be so important -- you are saved from this conclusion if all of the infinite curvature points are hidden behind horizons, and therefore, the exterior of the black holes can still be globally hyperbolic.


3

Assuming from the notation $$ \dot{x}^i~=~f^i(x,p,t), \qquad \dot{p}_i~=~g_j(x,p,t), \tag{1}$$ that the symplectic structure is the standard canonical symplectic structure $$\omega = \sum_{i=1}^n\mathrm{d}p_i\wedge \mathrm{d}x^i,\tag{2}$$ we get that $$\begin{align}\mathrm{d}H(x,p,t)- \frac{\partial H(x,p,t)}{\partial t}\mathrm{d}t ~=~&\sum_{i=1}^n\...


3

The equation comes from adding a source term to the diffusion equation: $$k \frac{\partial^2 T}{\partial x^2} + r (T_a -T(x)) = \frac{\partial T}{\partial t} $$ and then assumes steady state: $$\frac{\partial^2 T}{\partial x^2} + h^\prime (T_a - T(x)) = 0$$ where $h^\prime = r/k$. It is basically the addition of two models (diffusion, cooling) and ...


3

Re-write for legibility: $$u''(\theta)+\alpha u(\theta)-\beta=0$$ Make a substitution: $$y=\alpha u(\theta)-\beta$$ So: $$y'=\alpha u'$$ And: $$y''=\alpha u''$$ $$\Rightarrow u''=\frac{y''}{\alpha}$$ $$\Rightarrow \frac{y''}{\alpha}+y=0$$ $$y''(\theta)+\alpha y(\theta)=0$$ Which is the classic ODE of the SHM. Solve and back-substitute. Don't neglect the BCs!...


2

The hydrogen hamiltonian can be written as $$ H = \frac{p_r^2}{2m} + \frac{L^2}{2mr^2} - \frac{e^2}{r} $$ where $p_r$ is the radial component of the momentum. Since $H$ depends on $L^2$, $p_r$ and $r$, the hamiltonian commutes with every component of $\mathbf L$: $$ [H, \mathbf L] = 0, $$ which in turn means we can simulteneously diagonalize $H$, $L^2$ and ...


2

One cannot transform a $p$ orbital into a $d$ orbital by a rotation. This is because the generators of rotations are the angular momentum operators, and their action can only connect states with the same value $\ell$ of the angular momentum quantum number. Thus, rotations can only connect states with the same $\ell$. In the case of the parity example that ...


2

If you think about the initial conditions in terms of physics, and connect that to your specific DE, you will see that when the velocity is zero ($x'(0)=0$) then your acceleration is zero: $$x''=\frac{\beta}{m}x'.$$ If the acceleration and the velocity are zero, the system won't change position, based on your DE. You need a new DE, or a new initial ...


2

I think that the answer and comments (including references to previous related questions), although correct and also interesting (I have discovered a kind of predictor-corrector variant of the velocity Verlet I was not aware of), are missing a point which could be relevant for your problem. If the force depends linearly on velocity, the implicit step, ...


2

If you make a change of variables, lets say $\omega t'=\omega t - \delta$ you get the same equation without the phase, and you don't lose generality.


2

The equation $\frac{dN}{dt} = -\lambda N$ holds if the rate of the decay of any atom at any point of the time is constant and independent of any external factors, like the number of atoms left there are no other effects influencing the number of atoms other than the radioactive decay The second one can be easily broken, for example by having the atoms to ...


2

Assuming that the model is indeed correct, there is obviously a problem with using $v(t=0)=0$ due to the division by the velocity. There are a few ways I can think of to fix this: Use a non-zero (but small) starting velocity (e.g., $v(t=0)=10^{-10}$). Ignore the $1/v$ term for the first step Add a very small value (e.g., machine-precision) to the ...


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