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Introduction
I think there are several questions on this site about ODEs of the form
$$(x-a)^2 u''(x) = F(x,u,u')$$
with an initial condition at $x=a$.
There is no general guarantee that solutions exist over an interval $(a,b]$, but sometimes it is possible as in this case.
Outline
We transform the equation $u''(x) = F(x,u,u')$ over the infinite interval ...
24
OK, there are a few things going on here. Let me explain them in turn. First, as the message suggests, this should be written in Inactive form (we'll get to the why later). If you click on the three dots in front of the error message
and follow the link to the reference page you will find some information on this error message.
To write the equation in ...
21
An alternative representation is
G = 3.55; ω = 2*Pi*12*10^6;
s = ParametricNDSolveValue[{v'[t] == 2*G*BesselJ[1, v[t - τ]] Cos[ω*τ] - v[t],
v[t /; t <= 0] == 0.001}, {v, v'}, {t, 0, 120}, {τ}];
Manipulate[ParametricPlot[{s[τ][[1]][t], s[τ][[2]][t]}, {t, 60, 120},
AxesLabel -> {v, v'}, AspectRatio -> 1], {{τ, 2}, 1, 4}]
Note that the ...
17
We can solve (approximately) for the initial conditions of solutions that approach an equilbrium by comparing the displacement vector from the equilibrium with the vector field of the ODE. Such trajectory is characterized by the condition that these two vectors become parallel as the solution nears the equilibrium. I used a similar idea before, which is ...
17
The answers of the original questions by Szabolcs:
What does Rescale do when infinities are present? What's the justification for this behaviour? Where is it documented?
were guessed correctly with the comment:
If I may be allowed to speculate, these were picked because they do the job advertised and are "conveniently" algebraic. They certainly work ...
16
Since @hesam asked about a command, and to get a better understanding of @DanielLichtblau's approach, I tried to generalize it and package it in a function. Feedback would be appreciated!
TrackRoot[eqns_List,unks_List,{par_,parmin_?NumericQ,parmax_?NumericQ},ipar_?NumericQ,
iguess_List,opts___?OptionQ]:=
Module[{findrootopts,ndsolveopts,subrule,isol,ics,...
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Please note that though cosmetically appealing this is not rigorous or the best insight into basin of attraction of system as pointed out by bbgodfrey in comment below. I leave it as, perhaps, a road to refinement after exploring the behaviour of equilibrium points in system and apologise for any misconceptions.
Inspired by bbgodfrey's answer (which I have ...
15
Description of the issues
It is a bit unusual to discuss an ODE system as a function of a parameter ν but with a fixed initial condition x[0] == 0, x'[0] == v0, but as such, ν = 1/Sqrt[6], or approx. 0.408248, is a sort of critical value. The phase portraits of the ODEs for ν > 0 all have the same look and features:
There are two equilibria, a center ...
14
(1) Compactification
(2) Compactification
Update
Ok, my original post was rather compact. Let us start from the begining. You functions $A$ and $F$ map $[0,∞)$ into the reals. But you can think of the preimage as the result of another bijective function that maps say $[0,1)$ into $[0,∞)$. Since the domain of that second function can be easily modified to ...
14
I think if you look at the surface defined by the differential equation in λ, λ', λ'' space, you can see that the solution when yp = 2 goes off to infinity. Somewhere in the interval 1.95 < yp < 2, there is a point where the integral curve transitions from being closed to going off to infinity. Three integral curves are plotted on the surface below ...
14
The error message is misleading. NDSolve fails, because not enough boundary conditions in t have been supplied. If, for instance, (D[f[x, t], t] /. t -> 0) == 0 is added, then
sol = First@NDSolve[{D[f[x, t], x, x] - D[f[x, t], t, t] == f[x, t]^3,
f[x, 0] == Sin[2*Pi*x], (D[f[x, t], t] /. t -> 0) == 0,
f[0, t] == 0, f[1, t] == 0}, f, {x, 0, ...
13
As noted by @ChrisK, this works better starting at the top. Reason being there are no real solutions below the parameter value of 48.
Using FoldList one can readily use the prior result to seed the next, that is, providing a starting point. This is a fairly common homotopy approach.
points =
Rest[FoldList[({x, y} /.
FindRoot[{x^2 + y^2 - #2, x*y - ...
13
EDIT #2
My error was useful. It brought me to the conclusion that the difficulties in solving the PDE of the OP are due to the drift term
$$\frac{\partial (x u(x,t))}{\partial x}$$
If the drift term is included, many boundary problems are ill defined. It turns out that there are cases where mathematically there is only a trivial solution u = 0 but ...
13
Denote the disk by $\varOmega$ and its boundary by $\varGamma = \partial \varOmega$. I'd prefer to denote the function residing on the boundary by $u \colon \varGamma \to \mathbb{R}$; the function on the whole disk is called $v \colon \varOmega \to \mathbb{R}$.
Our aim is to solve the system of parabolic equations
$$
\left\{
\begin{aligned}
\partial_t u - ...
12
The problem is with the default starting initial conditions used by the shooting method in NDSolve. The shooting method is where FindRoot is being used internally, so the OP's error message is a strong hint that this is the problem. Getting convergence in a nonlinear system can depend greatly on the starting conditions.
Having luckily solved the system ...
12
Partly NDSolve-based solution
Use a higher even order spatial grid to discretize the PDE to an ODE set seems to be a good approach. The definition of pdetoode can be found here:
bound = 0.25510204081632654;
upper = 99/100; lower = 1 - upper;
range = {L, R} = {-Pi/2, Pi/2};
endtime = 100;
eqn = With[{h = h[t, θ], ϵ = 5/10},
0 == -D[h, t] + D[h^3 (1 - h)^3 ...
11
This is the most difficult of the nearly two dozen nonlinear ODE separatrix computations that I have encountered on Mathematica.SE. Nonetheless, it can be can be solved by a systematically refined search for initial conditions that maximize the range in r over which the ODE system can be integrated before clearly departing from the separatrix.
Series ...
11
Edit: Derivation of symbolic solution
As commented by Mariusz Iwaniuk), the initial conditions in the question are inconsistent. Setting {x[0] == 1, y[0] == 1, z[0] == 1} eliminates the inconsistency but does not cause DSolve to return a solution. So, to understand the desired solution, first solve the equations numerically.
s = NDSolveValue[
{x''[t] =...
11
a) To find equilibria, use Solve:
eq = Solve[{A - B*x - x*y^2, A*(x*y^2 - y)} == {0, 0}, {x, y}]
b) Linearizing around the point $(\hat x,\hat y)$ means making a new, linear system $$\dot{\vec{z}}=J\vec{z}$$ where $\vec{z}=(x-\hat x,y-\hat y)$ and $J$ is the Jacobian matrix $$J=\begin{bmatrix}d\dot x\over dx & d\dot x\over dy \\ d\dot y\over dx & d\...
11
Since I have the code to solve the original problem described in the article GDI-Mediated Cell Polarization in Yeast Provides Precise Spatial and Temporal Control of Cdc42 Signaling, I will give here a modification of this code for 2D. I did not manage to find the solution described in the article, since the system rather quickly evolves to an equilibrium ...
10
The equation is not stiff, despite the claim by Mathematica. It can be solved by using the "Shooting" Method.
NDSolve[{f''[x] == 2 (f[x]^3 - f[x]), f[-3] == 1, f[3] == -1}, f, {x, -3, 3},
Method -> {"Shooting", "StartingInitialConditions" -> {f[-3] == 1, f'[-3] == 0}}][[1, 1]];
Plot[f[x] /. %, {x, -3, 3}, AxesLabel -> {f, x}]
This looks like a ...
10
This not a complete answer, but you ReactionCoefficients do not have to correct shape I think:
nlPdeCoeff =
InitializePDECoefficients[vd, sd,
"ConvectionCoefficients" -> {{{Vx[x, y], Vy[x, y]}, {0, 0}}, {{0,
0}, {Vx[x, y], Vy[x, y]}}}
, "DiffusionCoefficients" -> {{{{-Ga, 0}, {0, -Ga}}, {{0, 0}, {0,
0}}}, {{{0, 0}, {0, 0}}, {...
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You need to use a constraint for c to avoid complex numbers.
fit = NonlinearModelFit[
datatofit, {intensity[t, a, c, i, j, k],
c^2 < 1}, {{a, 3600}, {c, .45}, {i, -.3}, {j, 1}, {k, .4}}, t]
Show[ListPlot[datatofit], Plot[fit[t], {t, 0, 180}]]
Btw, as a matter of style, I would define all parameters as variables. That way you won't run into ...
10
Direct solution of the last equation in the question also is feasible, because the Fourier series converges very rapidly. as will be seen below. The equation for a three term expansion can be written as
Sum[a[n] (n w)^6 Cos[n w t], {n, 1, 5, 2}]^3] - Sum[a[n] Cos[n w t], {n, 1, 5, 2}] == 0
Reducing the cubic terms in terms of Cos[m w t] shows that the ...
10
Because there is a term x^d, I dropped the point {0, 0} from the data (see here).
Next, the terms a 34^b and e 34^f will be just constants - there is no point in taking them as a multiplication of two numbers - there are infinitely many ways to write them. So I changed to just a and e.
Using Manipulate:
Manipulate[
Show[Plot[a x^d + e x, {x, 0, 700}], ...
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We want to find the value of $\lambda$ for which there exists a solution of the differential equation $u'' = \lambda ( - u + u^2)$, subject to the boundary conditions $u(0) = u(1) = 0$. We can do this by a shooting algorithm, i.e., solving the related initial value problem
$$
u'' = \lambda (- u + u^2), \qquad u(0) = 0, \; u'(0) = v.
$$
The value of $u(1) \...
10
What about:
Noting that equations 1 and 3 form a complete set, and solve them first, treating the remaining equation 2 for m afterwards.
Noting that your imposed initial conditions for v do not satisfy boundary conditions, i.e., they violate eq(3). If you insist to use Gaussian distribution, in this particular example the factor in the exponential can be ...
9
You are trying to exactly solve the Ginzburg-Landau equation with potential 1-\frac{1}{r^2}. No wonder that you fail: it has no analytical solution. At least no solution that I would hear about, and I am in this business already for some time. One reason to see that it has no non-trivial exact solution is that when you fix its boundary conditions in another ...
9
These equations are solvable, but the process is exceptionally slow and the output huge. To help Solve, replace the approximate real numbers 1. and 2. by 1 and 2. Then, solve the nine linear equations in terms of p and the parameters.
nine = Simplify[Solve[{eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq9, eq10}, {Ly, Lq, LA,
gA, gB, r, g1, g2, g3}][[1]]];
...
9
This functionality is undocumented, but evident when Spelunking:
Rescale[x, {DirectedInfinity[-1], DirectedInfinity[1]}]
(* (-2 + x + Sqrt[4 + x^2])/(2 x) *)
Rescale[x, {-Infinity, Infinity}] is equivalent to Rescale[x, {-Infinity, Infinity}, {0, 1}]. The relevant entry in the definition when spelunking:
Needs["Spelunk`"]
Spelunk[Rescale]
This returns a ...
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