Ordinary differential equations: transformation of a non-linear system into a new system that has a point of equilibrium at the origin.

Take a look at the following system

$$
begin {align}
dot {x} _1 & = x_2 \
dot {x} _2 & = -x_1 + x ^ 3_1 – x_2
end {align}
$$

which has three points of equilibrium (0,0), (1,0) and (-1,0). In the book I am reading, the author asks to define a new system that has the equilibrium point as the origin for (1,0) and (-1,0). The procedure in the book is as follows:
leave $ y = x-x_e $ where $ x_e $ is a balance point and we are going to make the transformation for $ x_e = (1,0) $Thus:
$$
begin {align}
y_1 & = x_1 – (1) = x_1 implies dot {y} _1 = dot {x} _1 \
y_2 & = x_2 – (0) = x_2 implies dot {y} _2 = dot {x} _2
end {align}
$$

The new system is now
$$
begin {align}
dot {y} _1 & = y_2 \
dot {y} _2 & = – (y_1 + 1) + (y_1 + 1) ^ 3 – y_2
end {align}
$$

Now we have to show that the new system $ dot {y} = g (y) $ has a point of equilibrium at the origin (ie, $ dot {y} = g (y) implies 0 = g (y_e) $), we obtain

$$
begin {align}
0 & = y_2 \
0 & = – (y_1 + 1) + (y_1 + 1) ^ 3 + y_2
end {align}
$$

$ y_2 = 0, – (y_1 + 1) + (y_1 + 1) ^ 3 = 0 implies y_1 = 0, -2 $. The new system has two equilibrium points and one of them is not at the origin. How to justify this problem. The real question in the book says:

enter the description of the image here

Differential equations – Warnings using NDSolve in the PDE wave. "Using the maximum number of grid points", "Warning: local spatial error estimate scaled"

Version 12 in windows 10.

I can not understand what should be changed in this call to NDSolve to make him happy.

This PDE is solved by DSolve, but NDSolve It gives many warnings. and when trying to trace the solution it gives after a long time, Handle it simply aborts, since each step takes a long time. So there is something wrong with the solution due to these warnings.

This wave PDE is standard, in rectangle, the 4 edges are fixed, with initial position and zero initial velocity.

Clear all[t, U, x, y];
L = 2; (* x dimension *)
H = 3; (* and dimension *)
c = 0.3; (* wave speed *)
f1[x_?NumericQ] : = By parts[{{x, 0 <= x <= L/2}, {L - x, L/2 < x <= L}}];
f2[y_?NumericQ] : = By parts[{{y, 0 <= y <= H/2}, {H - y, H/2 < y <= H}}];
pde = D[U[x, y, t], {t, 2}]== c ^ 2 * Laplaciano[U[x, y, t], {x, y}];
ic = {U[x, y, 0] == f1[x]* f2[y], Derivative[0, 0, 1][U][x, y, 0]    == 0};
bc = {U[x, 0, t] == 0, U[0, y, t] == 0, U[L, y, t] == 0, U[x, H, t] == 0};
numericalSol = First @ NDSolve[{pde, ic, bc}, U, {x, 0, L}, {y, 0, H}, {t, 0, 20}]

Mathematical graphics

enter the description of the image here

Although Manipular shows the starting position correctly, it is very slow to play. Every step takes forever to move.

Mathematical graphics

I tried to use these options as suggested in the comment here

Method -> {"MethodOfLines",
"Spatial discretization" -> {"TensorProductGrid", "MaxPoints" -> 101}}

And they tried to increase the "MaxPoints", but they had no effect. I think NDSolve He does not like something about the previous starting position, given the use of A pieces but I do not see anything wrong with it:

    Plot3D[f1[f1[f1[f1[x]* f2[y], {x, 0, L}, {y, 0, H}]

Mathematical graphics

Here is the manipulation code that is meant to reproduce the solution over time if necessary

Handle

The
Plot3D[{Evaluate[{Evaluate[{Evaluar[{Evaluate[U[x, y, t] /. numericalSol]}, {x, 0, L}, {y, 0, H},
BaseStyle -> 15,
ImageMargins -> 5,
Mesh -> 25,
PerformanceGoal -> "Speed",
BoxRatios -> {1, 1, 0.4},
PlotRange -> {Automatic, Automatic, {-1, 1.4}},
Image size -> 500,
ColorFunctionScaling -> False,
ColorFunction -> ColorData[{"TemperatureMap", {0, 1}}],
AxesLabel -> {"x", "y", "U (r, 0)"},
Spherical Region -> True,
ViewPoint -> {0.796, -2.725, 0.5471}
],
{{t, 0, "time"}, 0, 20, .1, appearance -> "Tagged"}
]

Any suggestions to change in the call to NDSolve up to eliminate these warnings and make the manipulation work better?

differential equations – NBodySimulation – Mathematica Stack Exchange

My question is about the new NBodySimulation package in version 12.
I discovered how to customize the basic functions "PairwisePotential", "PairwiseForce", "ExternalPotetial" …
The next unsolved problem for me is if any of those functions could depend on time (?)
A particular application that I have in mind is the N-body system in, for example, Heat Bath, represented as potential or external force.
Any help is appreciated

Differential equations – You can not get encouraged[] work consitentemente

Wave in the chain [0,L=10]. Fixed endpoints and initial conditions a double sync: Ti1 (x) = Sinc[3*(x – 7)] + Sync[3*(x – 3)]. Initial speed Ti2 (x) = 0. Evaluate as expansion and try to animate fails without an ugly code.

c = 1;
L = 10;
B1 = Sync[3*(0 - 7)] + Sync[3*(0 - 3)];
B2 = Sync[3*(10 - 7)] + Sync[3*(10 - 3)];
r = B1 + (x / L) * (B2 - B1);
lambdan = (n Pi / L) ^ 2;
Ti1 = Sync[3*(x - 7)] + Sync[3*(x - 3)] ;
Ti2 = 0;
Q = (x / L - 1) D[B1, {t, 2}] - (x / L) D[B2, {t, 2}] // FullSimplify;
rhside = Integrate[Q*Sin[n Pi x/L], {x, 0, L}]/ To integrate[Sin[n Pi x/L]^ 2, {x, 0, L}]// FullSimplify;
a1i = Integrate[(Ti1 + (x/L - 1)*(B1 /. t -> 0)  - (x/L)*(B2 /.  t -> 0) )*Sin[n Pi x/L], {x, 0, L}]/ To integrate[Sin[n Pi x/L]^ 2, {x, 0, L}];
a2i = 0;
an = DSolve[{D[an
u = Sum[an*Sin[n Pi x / N[L]], {n, 1, 50}];
u = Re[u]// ComplexExpand;
Encourage[Plot[u,{x,0,10}], {t, 0,100}];

If I do not suppress the output of u, then copy and paste the expansion (very long) explicitly in Animate[] then the animation will work.

As an additional question, I do not like having to add the second line for you, wrapping it in the actual function and applying ComplexExpand. Should not the series evaluate to a real value? Any way to force mathematics to immediately use only real values?

Encourage[Plot[u,{x,0,10}], {t, 0,100}];

differential equations – NDSolve Limit condition Discretization failed

I am trying to solve a PDE system with periodic boundary conditions using NDSolve. This works if I do not specify an initial condition (but it is not interesting, giving the trivial solution u (x, y, t) = 0). I tried to specify random initial conditions, but I get theseFailure discretization limit condition mistakes

Here is my code

eps = 0.1
g1 = 1
random = table[{x, y, RandomReal[]}, {x, 0, 2 [Pi], two [Pi]/ 10}, {y,
0, 2 [Pi], two [Pi]/ 10}];
random[[All, -1, -1]]= random[[All, 1, -1]];
random[[-1, All, -1]]= random[[1, All, -1]];
iniIF = Interpolation[Flatten[random, 1]];
ini[x_, y_] : = 1 + iniIF[x, y]

sols = NDSolve[{D[u[x, y, t], t]== (1 + eps) * u[x, y, t] +
two ([Psi][x, y, t]    + [Phi][x, y, t]) +
Laplacian[[[[[Psi][x, y, t], {x, y}]+
Laplacian[[[[[Phi][x, y, t], {x, y}]+ g1 * u[x, y, t]^ 2 -
you[x, y, t]^ 3,
[Phi][x, y, t]    == D[u, {y, 2}],
[Psi][x, y, t]    == D[u, {x, 2}],
you[x, y, 0] == ini[x, y],
PeriodicBoundaryCondition[or[or[u[u[x, y, t], x == 2 [Pi],
Function[xx-2[xx-2[xx-2[xx-2[Pi]]],
PeriodicBoundaryCondition[or[or[u[u[x, y, t], and == 2 [Pi],
Function[yy-2[yy-2[yy-2[yy-2[Pi]]],
{or, [Psi], [Phi]}
{t, 0, 100},
{x, 0, 2 [Pi]}
{y, 0, 2 [Pi]}
]

Why would I get "Failed to discretize the boundary condition"? How I can avoid this? Thanks in advance!

Differential equations: resolution of a DE dependent on the time of matrix 4 by 4 in terms of given parameters

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Differential equations – Activating WhenEvent as a result of WhenEvent

I'm trying to solve a simple-dead differential equation with events:

Y
   {y[0] == 0, and & # 39;[0] == 0, and & # 39; & # 39;
When the event[t==1y&#39;[t==1y'[t==1y'[t==1y'
When the event[t==2y&#39;[t==2y'[t==2y'[t==2y'
When the event[Y[Y[y[y
When the event[and&#39;[Y'[y'[y'
Y

The result is not what I expected, the Y

$ begin {cases}
0 & 0 leq t leq 1 \
t-1 & 1 <t leq 2 \
1 & 2 <t leq 4 \
text {Undetermined} & text {True}
end {cases} $

I can solve this problem with a trick that includes the change in Y

Y
   {y[0] == 0, and & # 39;[0] == 0, and & # 39; & # 39;
When the event[t==1y&#39;[t==1y'[t==1y'[t==1y'
    WhenEvent[t == 2, y'
When the event[Y[Y[y[y
Y

$ begin {cases}
0 & 0 leq t leq 1 \
frac {t-1} {2} & 1 <t leq 2 \
frac {3-t} {2} and 2 <t leq 3 \
0 and 3 <t leq 4 \
text {Undetermined} & text {True}
end {cases} $

This is the answer I expected.

Is there any way to make the first specification of the differential equation work like the second, and if not, should it be considered an error?

Differential geometry – Geodetic in the helicoid

The problem is the following:

Suppose a geodesic $ alpha $ is intersecting a rulling of the helicoid S given by
$$ x (u, v) = (ucos (v), usin (v), v) $$
at a point p. The angle in p between the sentence and $ alpha $ is $ theta en (0, pi / 2) $. Also, p is not on the z-axis and we call the relative distance $ delta> 0 $.

(a) Show that $ delta> cradle ( theta) $

(b) Show that if $ delta <cot ( theta) $, so $ delta $ does not intersect the z-axis

(c) Find the geodesics for $ delta = cradle ( theta) $

I'm struggling to figure out how to start with this. I do not understand how I can express the angle. $ theta $ and the distance $ delta $. If you could help me explain it, I think I could continue on my own.

Any help appreciated!

Thank you!

P.S.: This question is slightly related to Geodesics of Helicoid, the results obtained could also be useful in this problem.

Differential equations – Different solutions with Mathematica and Wolfram Alpha.

I tried to solve this in a Mathematica notebook:

    DSolve[x

And the results are:

                {{X
Sqrt[-1+Tanh[E^C[-1+Tanh[E^C[-1+Tanh[E^C[-1+Tanh[E^C[1] (t + C[2])]^ 2])},
{X
Sqrt[-1+Tanh[E^C[-1+Tanh[E^C[-1+Tanh[E^C[-1+Tanh[E^C[1] (t + C[2])]^ 2]}}

But if I execute it in Wolfram Alpha the result is:

{{X
{X

And the seconds are the correct results, I needed a real function, while the first ones are complex functions. Why does this happen? What assumptions do Wolfram Alpha make in its entry?

Thanks in advance

Problem in solving the differential equation.

Can someone tell me why I have errors when I try the following code?

`cd
s0 = 10; u = 0.4; w = 0.9; v = 0.01; k = 0.1; p = 5; L = 4; (* I choose them randomly *)
pde1 = w * D[c[x, t], t]+ u * D[c[x, t], X]- v * D[c[x, t], {x, 2}]== -k * c[x, t]* s[x, t];
pde2 = D[s[s[s[s[x, t], t]== -p * k * c[x, t]* s[x, t];
sol = NDSolve[{pde1, pde2, c[0, t] == cd