Skip to main content

differential equations - singularity in NDSolve for some values of parameter


my equation is:


k[a_?NumericQ, b_?NumericQ, c_?NumericQ] := NDSolve[{r^3 b x''[r] y[r] - 12 x[r] y[r] - 
 4 r^3 b y''[r] x[r] == 0, 4 (2 x'[r]^2 + 
    2 x[r]x''[r])/Sqrt[c] + (2 y'[r]^2 + 
    2 y[r] y''[r])/Sqrt[c] == -4b/r,x[1] == 0, y[1] == Sqrt[-4 b Log[2/b]/(c^(-1/2))],
y'[1] == c^(1/4) (-4 b - 4 b Log[2 /b])/(2 Sqrt[-4 b Log[2/b] ] ), x'[1] == a},
 {x, y}, {r, 1, b/2}, MaxSteps -> 1000000]

and then I plot it:



Plot[Evaluate[{x[r], y[r]} /. k[1, 50, 10^(-7)]], {r, 1, 25}]

enter image description here


this one is okay, cause x[r] and y[r] both equal to 0 at r=b/2


if I set a=0 and the other parameters stay the same


enter image description here


I do not know how comes this sigularity at r=b/2, basing on the second equation in NDSolve, 4 (2 x'[r]^2 + 2 x[r]x''[r])/Sqrt[c] + (2 y'[r]^2 + 2 y[r] y''[r])/Sqrt[c] == -4b/r comes from the second derivative of r in this equation:


c^(-1/2) 4 x[r]^2 + c^(-1/2) y[r]^2 == -4 r b Log[(2 r)/b]


so it's easy to check, if x[r]=0 all the way, y[r] should be 0 at r=b/2


does anyone know how to resolve the infinity at r=b/2 for any values of the parameter? a>=0, b>2, 10^(-3)>c>10^(-7)



thanks a lot



Answer



I think this is just another story of WorkingPrecision, let's try a higher one, say, 32:


k[a_?NumericQ, b_?NumericQ, c_?NumericQ] := 
 NDSolve[{r^3 b x''[r] y[r] - 12 x[r] y[r] - 4 r^3 b y''[r] x[r] == 0,
    4 (2 x'[r]^2 + 2 x[r] x''[r])/
       Sqrt[c] + (2 y'[r]^2 + 2 y[r] y''[r])/Sqrt[c] == -4 b/r, 
   x[1] == 0, y[1] == Sqrt[-4 b Log[2/b]/(c^(-1/2))], 
   y'[1] == c^(1/4) (-4 b - 4 b Log[2/b])/(2 Sqrt[-4 b Log[2/b]]), 
   x'[1] == a}, {x, y}, {r, 1, b/2}, WorkingPrecision -> 32]


Plot[Evaluate[{x[r], y[r]} /. k[0, 50, 10^(-7)]], {r, 1, 25}]

enter image description here


OK, the singularity disappears.


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

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more