Skip to main content

plotting - Solving a differential equation involving the square of the derivative


I want to solve the differential equation (drdλ)2=1−L2r2(1−1r)

where L is some parameter. The behavior I am looking for is when r starts "at infinity" at λ=0, reaches some minimum r, and increases out to infinity again. (I'm trying to describe the path that light takes in the presence of a Schwarzschild black hole.)


The issue I'm running into is when I use ParametricNDSolve, I can't tell Mathematica to smoothly transition from negative dr/dλ to positive dr/dλ after reaching the minimum r.


f[r_]:=1-1/r;
soln = ParametricNDSolve[{r'[\[Lambda]]^2 == 1 - L^2/r[\[Lambda]]^2 *f[r[\[Lambda]]], r[0] == 1000}, r, {\[Lambda], 0, 1000}, {L}]

Plot[r[30][\[Lambda]] /. soln, {\[Lambda], 0, 1000}]

In the above code, it plots the graph fine. However if I try to increase the range of λ (say to 1200), it breaks down; the situation where I solve for r′(λ) first and take the negative square root is identical. I'm not sure how to capture this extra information of transitioning from the negative to positive square root in the differential equation.


Sorry if this is a basic question, I'm rather new to Mathematica (and this site).



Answer



You could differentiate to make a second-order equation without square root problems. Actually, it moves the square trouble to the initial conditions, but that is easier to solve.


foo = D[r'[λ]^2 == 1 - L^2/r[λ]^2*(1 - 1/r[λ]), λ] /. Equal -> Subtract // FactorList
(*  {{-1, 1}, {r[λ], -4}, {r'[λ], 1}, {-3 L^2 + 2 L^2 r[λ] - 2 r[λ]^4 (r^′′)[λ], 1}}  *)

ode = foo[[-1, 1]] == 0     (* pick the right equation by inspection *)

(*  -3 L^2 + 2 L^2 r[λ] - 2 r[λ]^4 (r'')[λ] == 0  *)

icsALL = Solve[{r'[λ]^2 == 1 - L^2/r[λ]^2*(1 - 1/r[λ]), r[0] == 1000} /. λ -> 0,
  {r[0], r'[0]}]
(*
  {{r[0]  -> 1000, 
    r'[0] -> -(Sqrt[1000000000 - 999 L^2]/(10000 Sqrt[10]))}, (* negative radical *)
   {r[0]  -> 1000, 
    r'[0] -> Sqrt[1000000000 - 999 L^2]/(10000 Sqrt[10])}}    (* positive radical *)
*)


ics = {r[0], r'[0]} == ({r[0], r'[0]} /. First@icsALL); (* pick negative solution *)

soln = ParametricNDSolve[
   {ode, ics},
   r, {λ, 0, 2000}, {L}];
Plot[r[30][λ] /. soln, {λ, 0, 2000}]

Mathematica graphics


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