# Tag Info

4

If I understand correctly, you have a point mass on a surface $$z=x^2-y^2$$ with gravitation. But then your equations are wrong. For simplicity we choose $m=1$, then $$L= 1/2 (x'^2+y'^2)-g (x^2-y^2)$$ And this gives the ODE: $$x''+2g x=0$$ and $$y'' - 2 g y =0$$ In MMA: g = 9.81; sol = NDSolve[{x''[t] + 2 g x[t] == 0, y''[t] - 2 g y[t] == 0, x[0] == 1, ...

3

Something like this ? eqn = {(1 + 4 x[t]^2) D[x[t], t, t] - 4 y[t] D[y[t], t, t] x[t] + 4 x[t] D[x[t], t] - 4 x[t] D[y[t], t]^2 - 2 g x[t] == 0, (1 + 4 y[t]^2) D[y[t], t, t] - 4 x[t] D[x[t], t, t] y[t] + 4 y[t] D[y[t], t] - 4 y[t] D[x[t], t]^2 + 2 g x[t] == 0 } /. g -> 10 eqn2 = Solve[eqn, {x''[t], y''[t]}] // FullSimplify // ...

3

Change the MeshFunctions and Mesh and PlotPoints ex1 = ParametricPlot3D[{(3 + Cos[v]) Cos[u], (3 + Cos[v]) Sin[u], Sin[v]}, {u, 0, 2 Pi}, {v, 0, 2 Pi}, Boxed -> False, Axes -> False, MeshFunctions -> Automatic, Mesh -> {{0}}, PlotPoints -> {12, 8}]; reg = DiscretizeGraphics[ex1]; newedges = MeshPrimitives[reg, 1]; Graphics3D[Map[...

5

The FEM package will tend to give you more isotropic triangles as shown in your spherical mesh than other discretization functions in Mathematica. Also, for a torus, an implicit region seems to give a cleaner mesh than a parametric region as can be seen by the FindMeshDefects function. Below, you can see a comparison between ParametricRegionand ...

3

Due to the precedence of nested anonymous functions, the innermost expression: Join[#, Reverse[#, {2}]] & Is essentially identical to: Function[{x}, Join[x, Reverse[x, {2}]] Let's explicitly label this as function f using (yet another mostly equivalent notation): f[x_] := Join[x, Reverse[x, {2}]] So we can rewrite the original expression as: (f@...

0

The definition of the controlpoints is wrong. {s,-Pi,Pi} inside Table creates an s-grid {_pi,-Pi+1,-Pi+2,...} which doesn't contain the last point Pi. Redefine the controlpoints to pts6 = Table[{(2 + 1 Cos[s])*Cos[t], (2 + 1 Cos[s])*Sin[t],1 Sin[s]}, {s, Subdivide[-Pi, Pi, 10]}, {t,Subdivide[-Pi, Pi, 10]}]

1

The equations depend on the variables Hb , Db , kb , Ks , Gb and two parameters K,Ho For given parameters (examplary {K -> 1, Ho -> 1}) NMinimize might solve your problem: eqn = {Hb/Db == K,Ks == Sqrt[(Coth[kb Db])/(1 + Gb)],Gb == (2 kb Db)/(Sinh[2 kb Db]) Hb/Ho == Ks (Ho)^2 == Hb^2 (1 + Gb) Tanh[kb Db]} NMinimize[{1, eqn} /. {K -> 1, Ho -> 1}...

0

I recommend solving for the square of x instead of x itself: L = 31.5*10^(-6); m = 2.6969*10^(-13); f1 = 551715; f2 = 3.90463*10^6; emr1 = 0.46399; emr2 = 0.44793; meff1 = emr1*m; meff2 = emr2*m; α1 = 0.044; α2 = -18.6; G = 2*10^(-3); keff1 = meff1*(2*π*f1)^2; keff2 = meff2*(2*π*f2)^2; Q = 1500; eq = (((ω - (2*π*f1))/(2*π* f1)) - (3*α1*(xsquare)/(8*L^2))^2 + ...

1

First try to solve analytically sol= Solve[x2 == a /(((\[Omega] - \[Omega]R)/\[Omega]R - b x2)^2 + c), x2] and substitute the parameters.

4

The MeshFunctions is a powerful method which I always use. Here we view the Mesh from 3D. So we draw the ParametricPlot3D $$\begin{cases}x=\sqrt{r^2+0.9801} \sin (\theta ),\\ y=r \cos (\theta ),\\ z=f(r,\theta)\end{cases}$$ We set the ViewPoint={0,0,Infinity} and use ViewProjection -> "Orthographic" Clear["`*"]; Σ[r_, a_, θ_] = r^2 + ...

1

@Natas, the thing is with this question that equipotentials are not plotted in polar coordinates, they are plotted in some special coordinates similar to polar coordinates (take a look at the first argument in ParametricPlot, "ParametricPlot[{Sqrt[r^2 + 0.99^2]*Sin[[Theta]], r Cos[[Theta]]},..."). That is why ParametricPlot way was used instead of ...

4

Regarding your question as to why trajectories has four arguments: MeshFunctions for ParametricPlot need three to four arguments. The first two correspond to the coordinates of the plot, the last two (or one) to the parameters of the ParametricPlot. To appreciate this better, consider the following example: funs = { Function[{x, y, u, v}, x], Function[{...

7

Try the following Series[f[InverseFunction[g][y]],{y,0,10}]

2

Integrate[ E^(-\[Alpha] \[Sqrt](x^2 + y^2)), {x, -\[Infinity], \[Infinity]}, {y, -\[Infinity], \ \[Infinity]}, Assumptions -> a > 0] (ConditionalExpression[(2 \[Pi])/\[Alpha]^2, Re[\[Alpha]] > 0]) There is an internal parameter to the notebooks called \$Assumption. There is the documentation page for Assumptions for the built-ins Simplify, ...

2

Here's a numerical complement to my other answer. Just integrate orthogonally to the gradient. The code below produces a parameterization by arc length. It should fail (stop integration) if it reaches a singular point. NDSolve does not allow the domain of integration to be infinite on both sides, that is, {t, -Infinity, Infinity}. If you want to start ...

3

This is usually a tough problem to solve symbolically. The following function doesn't check whether system was solve, nor does it remove singular solutions. ClearAll[parametrize]; (* Polar parametrization centered at basepoint *) parametrize[eqn_, v : {x_, y_}, t_, basepoint : {_, _}] := parametrize[eqn, v, t, Function[{param, elim}, basepoint + ...

2

Integrate[(λ0/Sqrt[2])*Sech[((t - τ) - 1/2*T)/σ] (1 + Tanh[((t - τ) - 1/2*T)/σ]), τ] gives (*1/Sqrt[2]λ0 σ (ArcTan[Sinh[(-2 t+T+2 τ)/(2 σ)]]+Sech[(-2 t+T+2 τ)/(2 σ)]) *) then enter the limits. Numerical evaluation may be difficult for your parameter values

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