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My procedure for solving coupled 1 + 1 (spatial + temporal) PDE system:

(Note: I have graphs of the correct solution with which I compare my result. See figure below text.)

1) I determine the initial conditions (initial functions $u_{ini}$-more functions), numerically (NDSolve) I find the necessary functions. In this solution using NDSolve, I enter the initial condition as $u_{ini} [dx] == du$, where dx is a small value (although in theory there should be $u_{ini} [0] == 0$).

2) I will use the resulting functions as initial conditions for solving the PDE using the "MethodOfLines" method. However, there is uncertainty about how to practically set the boundary conditions at the point $x = 0$, when at this point I do not have a solution of the initial functions $u_{ini}$.

Attempts: a) First I set the initial condition as $u[x, 0] == u_{ini} [x]$ and the boundary condition as $u [dx, t] == ​​u_{ini} [dx]$ (I identified dx as zero, however $dx$ and $u_{ini} [dx]$ were not zero). This led to the right solution, which, however, diverged over time with the right solution and subsequently destabilized (began to oscillate). At first glance, I noticed that the main difference between my solution and the correct solution is only around $x = 0$ when the functions of the correct solution go to zero faster, while my solution always went to the value $u_{ini} [dx]$. I GUESS, that this is the problem with my solution. Do you agree?

b) As a second attempt, I added a point to the initial functions $u_{ini}$ at the beginning, ie point $u_{ini} [0] = 0$, and then when solving, I entered the initial conditions $u [x, 0] == u_{ini} [x]$ and the boundary conditions as $u [0, t ] == u_{ini} [0] (= 0)$. This attempt by NDSolve and "MethodOfLines" did not solve with issue warning that 1/0 infinity appears there.

Does anyone have any tricks on how to set it all up so that I get the right non-oscillating solution?

enter image description here Blue, orange go in time, solid lines are correct solutions, dashed is my solution coming from the procedure a). You can see the beginning of oscillation in orange "time".

CODE:

ClearAll["Global`*"]
c = 2.99792*10^10; gr = 6.67323*10^-8; m0 = 
 1.672621*10^-24*gr/c^2; grc = gr/c^2; Ms0 = 1.98855*10^33; Ms = 
 Ms0*gr/c^2; R0 = 10^8; dr = (1/(4 \[Pi]))^(1/3);
rhs1[r_, h_] = -(gr/
   r^2) (h + p[r]/c^2) (m[r] + 4 \[Pi]*r^3 p[r]/c^2) (1 - (
    2 m[r]*gr)/(r*c^2))^-1;(*TOV*)
rhs2[r_, en_] = 4 \[Pi]*en*r^2; 
rhs3[r_, ro_] = 4 \[Pi]*ro*r^2 (1 - (2*m[r]*gr)/(r*c^2))^(-1/2);
(*____________________________________________________________________\
__*)
(*Guess on kp - destability from equilibrium factor decreasing P due \
to k constant*)
(*----------------------------------------------------------------------*)
\

kp = 36; dkp = kp/100; MP = 10*10^2;
(*g0=5/3;rho0=0.56*10^13;ep0=3.64*10^18;e0=rho0(1+ep0/c^2);pc=kp*(g0-\
1)*rho0*ep0;*)
(*g0=5/3;rho0=8*10^14;ep0=3.64*10^18;e0=rho0(1+ep0/c^2);pc=kp*(g0-1)*\
rho0*ep0;*)
g0 = 5/3; rho0 = 0.557*10^13; ep0 = 3.64*10^18; e0 = 
 rho0 (1 + ep0/c^2); pc = kp*(g0 - 1)*rho0*ep0;
k = pc/rho0^g0; fro1[r_] = (p[r]/k)^(1/g0); 
e[r_] = fro1[r] + (p[r]/c^2)/(
  g0 - 1); rhs10 = -(gr/
   dr^2) (e0 + pc/c^2) (e0 + 4 \[Pi]*dr^3 pc/c^2) (1 - (2 e0*gr)/(
    dr*c^2))^-1;
{fp0, fm0, fmu0} = 
  NDSolveValue[{p'[r] == rhs1[r, e[r]], m'[r] == rhs2[r, e[r]], 
    mu'[r] == rhs3[r, fro1[r]], p[dr] == pc - rhs10, 
    m[dr] == rho0 + (pc/c^2)/(g0 - 1), mu[dr] == rho0, 
    WhenEvent[p[r]/pc < 10^-6, "StopIntegration"]}, {p, m, mu}, {r, 
    dr, R0}];


(*____________________________________________________________________\
______*)
(*Loop for calculation of kp (pc) (careful on initial values-has to \
be same as above!)*)
(*--------------------------------------------------------------------------*)
\

M0 = 21; kp = 6; dkp = kp/10; pocet = 0; M1 = 0; Abs[M0 - M1]/M0;
Do[ep0 = kp*3.64*10^18; pc = (g0 - 1)*rho0*ep0; k = pc/rho0^g0;
  fro1[r_] = (p[r]/k)^(1/g0); e[r_] = fro1[r] + (p[r]/c^2)/(g0 - 1); 
  rhs10 = -(gr/
     dr^2) (e0 + pc/c^2) (e0 + 4 \[Pi]*dr^3 pc/c^2) (1 - (2 e0*gr)/(
      dr*c^2))^-1;
  {fpr0, fmr0, fmur0} = 
   NDSolveValue[{p'[r] == rhs1[r, e[r]], m'[r] == rhs2[r, e[r]], 
     mu'[r] == rhs3[r, fro1[r]], p[dr] == pc - rhs10, 
     m[dr] == rho0 + (pc/c^2)/(g0 - 1), mu[dr] == rho0, 
     WhenEvent[p[r]/pc < 10^-20, "StopIntegration"]}, {p, m, mu}, {r, 
     dr, R0}(*,AccuracyGoal\[Rule]25*)];
  R = fpr0[[1, 1, 2]]; M1 = fmr0[R]/Ms0; MU1 = fmur0[R]/Ms0;
  (*Print[{M1,R,fpr0[R],Abs[M0-M1]/M0,pocet,kp}];*)
  If[Abs[M0 - MU1]/M0 < 
    10^-5, {Print[{"kone", M1, Abs[M0 - M1]/M0, pocet, kp}], Break[]}];
  If[M0 > MU1, kp = kp + dkp, dkp = 0.6 dkp; kp = kp - dkp];, {pocet, 
   1, 200}];
frhor0[r_] = (fpr0[r]/k)^(1/g0); 
e[r_] = frhor0[r] + (fpr0[r]/c^2)/(g0 - 1);
R00 = fpr0[[1, 1, 2]]/100;

(*________________________________*)
(*Initial conditions from M&W with u=0*)
(*--------------------------------*)
mumax = fmur0[R]; dmu = mumax/MP; k = pc/rho0^g0;
(*Parameters scales to normalise parameters*)
{rhoN, rN, mN, eN, uN} = {rho0 // N, 10^5, mumax, 10^-4 c^2, c};

G[mu_] := 4 \[Pi]*(rhoN rN^3)*rho[mu]*r[mu]^2*D[r[mu], mu]/mN(*MW39*);
p[mu_] := k*(rho[mu] rhoN )^g0(*MW40 *);
ep[mu_] := k*(rhoN*rho[mu])^(g0 - 1)/(g0 - 1);
w[mu_] := 1 + ep[mu]/c^2 + p[mu]/(rho[mu] rhoN c^2)(*MW41*);

(*introducing of equation*)
eq = {(4 \[Pi] rN^2*r[mu]^2*G[mu]/w[mu]*
       D[p[mu], mu]/mN + (m[mu]*gr mN/rN^2)/
       r[mu]^2 + (4 \[Pi]*gr rN)/c^2 p[mu]*r[mu]) == 0(*MW33*), 
   D[m[mu], mu]*mN == 
    4 \[Pi]*rhoN*rho[mu] (1 + ep[mu]/c^2) (rN*r[mu])^2*rN*
     D[r[mu], mu], G[mu] == Sqrt[1 - 2 mN*m[mu] gr/(rN*r[mu] c^2)]};
(*Variables,initial and boundary conditions*)
var = {rho, r, m(*,a,ep*)};
{dmu1, dmu3, mumax1} = {dmu, dmu2 = 0, mumax}/mN;
dr1 = FindRoot[fmur0[r] == dmu, {r, 10^6}][[1, 2]];

ic = {r[dmu1] == dr1/rN, m[dmu1] == fmr0[dr1]/mN, 
  rho[dmu1] == frhor0[dr1]/rhoN}; prec = 30;
{frho0, r0, fm0} = NDSolveValue[{eq, ic}, var, {mu, dmu1, mumax1}];
k = k/33;(*for particular calculation of initial data coldata_tov.dat*)
\

frho0 = Interpolation[
  Join[{{fmur0[dr]/mN, frhor0[dr]/rhoN}}, 
   Table[{mu, frho0[mu]}, {mu, dmu1, mumax1, (mumax1 - dmu1)/(MP)}]], 
  InterpolationOrder -> prec];
fmr0 = NDSolveValue[{4 \[Pi]*
      frhor0[r] (1 + k*(frhor0[r]^(g0 - 1)/(g0 - 1))/c^2) r^2 == 
     m'[r], m[dr] == rho0 + (k*rho0^g0/c^2)/(g0 - 1)}, m, {r, dr, R}];
r0 = Interpolation[
   Join[{{fmur0[dr]/mN, dr}}, 
    Table[{mu, r0[mu]}, {mu, dmu1, mumax1, (mumax1 - dmu1)/MP}]]];
fm0 = Interpolation[
   Join[{{fmur0[dr]/mN, fmr0[dr]/mN}}, 
    Table[{fmur0[r]/mN, fmr0[r]/mN}, {r, dr1, R, (R - dr1)/MP}]]];

(*Relativistic hydrodynamical equations-collapse of star*)
G[mu_, t_] := 
  4 \[Pi]*(rhoN rN^3)*rho[mu, t]*r[mu, t]^2*
   D[r[mu, t], mu]/mumax(*MW39*);
p[mu_, t_] := k*(rho[mu, t] rhoN )^g0(*MW40 *);
ep[mu_, t_] := k*(rhoN*rho[mu, t])^(g0 - 1)/(g0 - 1);
w[mu_, t_] := 
  1 + ep[mu, t]/c^2 + p[mu, t]/(rho[mu, t] rhoN c^2)(*MW41*);

a[mu_, t_] := 1/w[mu, t];

(*introducing of equation*)
eq = {D[u[mu, t], 
     t] == (-a[mu, 
         t] (4 \[Pi] rN^2*r[mu, t]^2*G[mu, t]/w[mu, t]*
          D[p[mu, t], mu]/mN + (gr mN/rN^2)*
          m[mu, t]/r[mu, t]^2 + (4 \[Pi]*gr rN)/c^2 p[mu, t]*
          r[mu, t]))/c^2(*MW33*), 
   D[r[mu, t], t] == a[mu, t]*u[mu, t]/rN(*MW34*), 
   D[rho[mu, t] r[mu, t]^2, t] == -a[mu, t]*rho[mu, t]*
     r[mu, t]^2 D[u[mu, t], mu]/D[r[mu, t], mu]/rN, 
   D[m[mu, t], t] == -4 \[Pi]/c^2*rN^3/mN*p[mu, t]*
     r[mu, t]^2 D[r[mu, t], t](*MW12*)};

(*Variables, initial and boundary conditions*)
var = {u, rho, r, m(*,a,ep*)};
ic = {u[mu, 0] == 0., r[mu, 0] == r0[mu ], m[mu, 0] == fm0[mu ], 
   rho[mu, 0] == frho0[mu]};
bc = {u[dmu2, t] == 0, m[dmu2, t] == fm0[dmu2], 
   rho[mumax1, t] == frho0[mumax1]};

tm = 6*10^-1; tm = 3.07*10^7; Dynamic[
 "time: " <> 
  ToString[CForm[{currentTime, Round[currentTime/c, 10^-7]}]]]
(*good results with DiffOr\[Rule]8 for MP=2000, lower than interpol \
initial funcitons*)
AbsoluteTiming[
  sol = Reap[
     NDSolveValue[{eq, ic, bc, 
       WhenEvent[Abs[Max@Im[rho[mu, t]]/Max@Re[rho[mu, t]]] > 1.1, 
        "StopIntegration"]}, var, {mu, dmu1, mumax1}, {t, 0., tm}, 
      InterpolationOrder -> 1, 
      Method -> {"MethodOfLines", 
        "DiscretizedMonitorVariables" -> True, 
        Method -> "StiffnessSwitching", 
        "SpatialDiscretization" -> {"TensorProductGrid", 
          "MinPoints" -> MP, "MaxPoints" -> MP, 
          "DifferenceOrder" -> 10}}, 
      EvaluationMonitor :> (currentTime = t;), 
      StepMonitor :> Sow[t/c]]];];
{fu, frho, fr, fm(*,fa,fm*)} = sol[[1]]; tm = currentTime;
{my1 = Plot[
   Evaluate[Table[-Log10[-Re[fu[mu/21, t]]], {t, {3*10^7, tm}}]], {mu,
     2 dmu1*21, mumax1*21}, Frame -> True, PlotRange -> {All, {0, 1}},
    FrameLabel -> {"fu", currentTime/c}, ImageSize -> Medium, 
   PlotStyle -> Dashed],
 my2 = Plot[
   Evaluate[
    Table[Log10[rhoN Re[frho[mu/21, t]]], {t, {3*10^7, tm}}]], {mu, 
    dmu1*21, mumax1*21}, Frame -> True, PlotRange -> All, 
   FrameLabel -> {"frho"}, ImageSize -> Medium, PlotStyle -> Dashed],
 my3 = Plot[
   Evaluate[Table[Log10[Re[fr[mu/21, t]]], {t, {3*10^7, tm}}]], {mu, 
    dmu1*21, mumax1*21}, Frame -> True, PlotRange -> All, 
   FrameLabel -> {"fr"}, ImageSize -> Medium, PlotStyle -> Dashed],
 my4 = Plot[
   Evaluate[Table[Re[fm[mu, t]], {t, {3*10^7, tm}}]], {mu, dmu1, 
    mumax1}, Frame -> True, PlotRange -> All, FrameLabel -> {"fm"}, 
   ImageSize -> Medium]}

enter image description here Correct solution with times in seconds, time scale in my calculations is t = time*c (c-speed of light)

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  • 2
    $\begingroup$ Without the specific code sample it's hard to give advice. Please show us a minimal working example. $\endgroup$
    – xzczd
    May 4 '20 at 11:50
  • $\begingroup$ @xzczd Thank you for your comment. I added a link to all my code. It seemed like a general problem of solving the initial conditions, so I didn't share the code. $\endgroup$
    – Vrbic
    May 4 '20 at 12:24
  • 2
    $\begingroup$ Please don't provide the code via external link, because the link can easily be broken. If the code is too long to post here, it should be properly simplified. $\endgroup$
    – xzczd
    May 4 '20 at 12:57
  • $\begingroup$ No problem, now there is also code. $\endgroup$
    – Vrbic
    May 4 '20 at 14:00
  • $\begingroup$ 1. Once again, you should simplify your sample as much as possible, for example, why do you include tsteps in your sample? 2. How is the correct solution obtained? 3. Can you add more background information? $\endgroup$
    – xzczd
    May 5 '20 at 2:32

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