Skip to main content

differential equations - Solving nonlinear ODE with boundary conditions


I am trying to solve the below fucntions using a NDsolve, but I can't get the the converges of them at the certain initial variables.I'm using Mathematica 11.3. I appreciate any help.


Z=800;
g= 0.023800000000000000000;
k2= 0.000194519;
R= 1.5472;

ytest0= -13.911917694213733`;
ϵ = $MachineEpsilon ;
y1[ytest_?NumericQ] := 
    NDSolve[
      {y''[r] + 2 y'[r]/r == k2 Sinh[y[r]] , y[1] == ytest, 
       y'[ϵ] == 0}, y, {r, ϵ, 1}, 
       Method -> {"StiffnessSwitching", "NonstiffTest" -> False}];
 y2[ytest_?NumericQ] := 
   NDSolve[
     {y''[r] + 2 y'[r]/r == κ2 Sinh[y[r]], 

      y[1] == ytest, y'[R] == 0}, y, {r, 1, R}, 
      Method -> {"StiffnessSwitching", "NonstiffTest" -> False}];
y1Try[ytest_?NumericQ] := y'[1] /. y1[ytest];
y2Try[ytest_?NumericQ] := y'[1] /. y2[ytest];
f = ytest /. FindRoot[y1Try[ytest] - y2Try[ytest]==-Zg, {ytest, ytest0}]

Answer



The quantity κ2 in the question is undefined. However, the cross-posting identified by Mariusz Iwaniuk in a comment above suggests that k2 actually is meant. With this change, the computation is equivalent to solving the ODE, {y''[r] + 2 y'[r]/r == k2 Sinh[y[r]], y'[R] == 0, y'[ϵ] == 0} with a discontinuity in y'[r] equal to Z g at r == 1. With this modification to the ODE, it can be recast as


Z = 800; g = Rationalize[0.0238, 0];
k2 = Rationalize[ 0.000194519, 0];
ϵ = 10^-4; R = Rationalize[1.5472, 0];

ps = ParametricNDSolveValue[{y''[r] + 2 y'[r]/r == k2 Sinh[y[r]], 
    y[ϵ] == y0, y'[ϵ] == 0, WhenEvent[r == 1, y'[r] -> y'[r] + Z g]}, {y, y'}, 
    {r, ϵ, R}, {y0}, Method -> "StiffnessSwitching", WorkingPrecision -> 30];
sol = FindRoot[Last[ps[y0]][R], {y0, -11.25, -11.}, Evaluated -> False][[1, 2]]
(* -11.1979 *)

Plot[First[ps[sol]][r], {r, ϵ, R}, AxesLabel -> {r, y}, ImageSize -> Large, 
    LabelStyle -> {Black, Bold, Medium}]

enter image description here



Plot[Last[ps[sol]][r], {r, ϵ, R}, PlotRange -> All, AxesLabel -> {r, y'}, 
    ImageSize -> Large, LabelStyle -> {Black, Bold, Medium}]

enter image description here


The computation takes only seconds.


Addendum


In response to a comment below, the solution for k2 = Rationalize[0.05716690559951900467, 0]; is


sol = Quiet@FindRoot[Last[ps[y0]][R], {y0, -12, 0}, Evaluated -> False][[1, 2]]
(* -5.51467 *)


The corresponding plot of y looks very much like that above but shifted approximately by the shift in y0.


enter image description here


The corresponding plot of y' also looks very much like that above. Note that the FindRoot search range has been expanded in this computation, and Quiet has been prepended to avoid printing the numerous errors that FindRoot encounters before finding the correct answer.


Second Addendum


Varying k2, as requested by the OP in a comment below, is accomplished as follows.


psk = ParametricNDSolveValue[{y''[r] + 2 y'[r]/r == k Sinh[y[r]], 
     y[ϵ] == y0, y'[ϵ] == 0, WhenEvent[r == 1, y'[r] -> y'[r] + Z g]}, {y, y'}, 
     {r, ϵ, R}, {y0, k}, Method -> "StiffnessSwitching", WorkingPrecision -> 30];
ytablog = Table[{Exp[k], Quiet@FindRoot[Last[psk[y0, Exp[k]]][R], {y0, -12, 0}, 
    Evaluated -> False][[1, 2]]}, {k, -9, 0, 1/10}];

ListLogLinearPlot[ytablog, PlotRange -> All, Joined -> True, AxesLabel -> {k2, y[0]}, 
    ImageSize -> Large, LabelStyle -> {Black, Bold, Medium}]

enter image description here


A rough fit to this curve is Log[k2] - 2.655. Varying R can be accomplished in a similar manner.


Interestingly, varying R between 5/4 and 2 has minimal effect on the plot just above. Below are plots of y[r] for R = Range[5/4,2,1/4] - {"Blue", "Orange", "Green", "Red"}.


enter image description here


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more