Skip to main content

plotting - Exporting the solutions of two ODEs sampled over same set of domain values



I have two functions h[r] and c[r] defined by two differentials equation.


operh[h_, c_] := h D[D[h, r, r] + 1/r D[h, r] - 1/lc^2 h, r] + m D[c, r]
operc[h_, c_] := 
  cmt h + ξ ((ϕ - H) (1 - ϕ) )/Sqrt[R^2 - r^2] - 
    1/r D[r h (1 + e^2/1680  h^4  (D[c, r])^2 ) D[c, r], r]

I numerically solve them with NDSolve (with solution called "s" here under). Then I plot them with no problem:



Plot[h[r] /. s, {r, rleft, rright}, PlotRange -> All, AxesOrigin -> {0, 0}]
Plot[c[r] /. s, {r, rleft, rright}, PlotRange -> All, AxesOrigin -> {0, 0}]

With same rright and rleft.


graph for h graph for c


What I want is a .txt or .csv file with 3 columns, containing r, h[r] and c[r]. From what I have read on this site I tried:


ploth = 
  Plot[h[r] /. s, {r, rleft, rright}, PlotRange -> All, AxesOrigin -> {0, 0}]
pts = Cases[ploth, Line[pts_] :> pts, Infinity][[1]]
Export[SystemDialogInput["FileSave", "filename.csv"], pts]


It gave me a list of 157 pairs {r, h[r]}. Then I did the same for c:


plotc = 
  Plot[c[r] /. s, {r, rleft, rright}, PlotRange -> All, AxesOrigin -> {0, 0}]
pts = Cases[plotc, Line[pts_] :> pts, Infinity][[1]]
Export[SystemDialogInput["FileSave", "filename.csv"], pts]

And it gave me a list of 226 pairs {r, c[r]}. This is not the same number of data than for h(r), so I can't merge them.


I tried by plotting the two curves on a single graph but it gave me only one of the two data sets.


I also tried to use Table but it gave me an empty list:



tableh = Table[{r, ploth}, {r, rleft, rright}]


{}



Do you know how to extract h and c data for common r values?


I'm on Mathematica 7.0



Answer



An approach to this issue I like is to rely on the fact that NDSolve found the solution on the same grid points for both functions in the first place:


a simple example:



 sol = First@NDSolve[ {f''[x] + g'[x] + 1 == 0 , g[x] == x^2, f[0] == 0, 
   f'[0] == 1}, {f, g}, {x, 0, 1}];
 Plot[{f[x] /. sol, (g[x] /. sol)}, {x, 0, 1}]

enter image description here


here we do not actually use the interpolation but extract the grid values:


 gridpoints = Flatten[(g /. sol)["Grid"]];
 computeddata = 
    Transpose[{gridpoints, 
              (f /. sol)["ValuesOnGrid"],

              (g /. sol)["ValuesOnGrid"]}];
 ListPlot[{ computeddata[[All, {1, 2}]], computeddata[[All, {1, 3}]]}, 
  Joined -> True, PlotMarkers -> Automatic]

enter image description here


here computeddata is a nice array you can simply export with each row {x,f[x],g[x]}


An advantage to this is the point spacing is naturally refined as was needed by NDSolve


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

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more