I have intialconditions:
sf = 200;
sm = 100;
p = 40;
betaf = 0.15;
betam = 0.15;
mums = 0.02;
mufs = 0.02;
sigma = 0.20;
mum = 0.02;
muf = 0.02;
and the ODE:
sf' := -muf*sf + (betaf + mums + sigma)*p - HarmonicMean[sf, sm];
sm' := -mum*sm + (betam + mufs + sigma)*p - HarmonicMean[sf, sm}];
p' := p - (mufs + mums + sigma)*p + HarmonicMean[{sf, sm}];
That i want is an abstract solution (sf(t),sm(t),p(t)) with NDSolve to plot it later.
My problem is that all variables are dependet in all 3 equations, so i don't know how to write the NDSolve call.
I could not manage to get an analytic solution, but the numerical one goes like this. Note that not all symbols you listed are variables of the system: those not being dependent of the independent variable t are parameters. (Also note that there are some typos in the OP's code).
variables = {sf[t], sm[t], p[t]};
parameters = {betaf -> 0.15, betam -> 0.15, mums -> 0.02,
mufs -> 0.02, sigma -> 0.20, mum -> 0.02, muf -> 0.02};
equations = {
sf'[t] == -muf*sf[t] + (betaf + mums + sigma)*p[t] -
HarmonicMean[{sf[t], sm[t]}],
sm'[t] == -mum*sm[t] + (betam + mufs + sigma)*p[t] -
HarmonicMean[{sf[t], sm[t]}],
p'[t] ==
p[t] - (mufs + mums + sigma)*p[t] + HarmonicMean[{sf[t], sm[t]}],
sf[0] == 200,
sm[0] == 100,
p[0] == 40
};
sol = NDSolve[equations /. parameters, variables, {t, 0, 100}];
Plot[Evaluate[variables /. sol], {t, 0, 100}]
Related
I am trying to develop a code which calculates Local Density of states of electrons in a material. For which I am using a multiple for loops and multiple tables. it takes 45sec to complete, I need less time for that. any suggestions how to optimize this code.
AbsoluteTiming[Ns=2; \[Eta] = 0.001;
Nx=15;
Ny=15;
NN=Nx*Ny;
Nband=8;
kkmx = Ns*Nx;
kkmy = Ns*Ny;
wmax = 0.2; nw = 800; p = 0;
Print["starting ldos calc"];
nsite = 2;
ldos = 0;
For[kx = 0, kx <= (Ns - 1.)*2*(Pi/kkmx), kx += 2*(Pi/kkmx),
For[ky = 0, ky <= (Ns - 1.)*2*(Pi/kkmy), ky += 2*(Pi/kkmy),
ES = Eigensystem[H];
elist = Table[ES[[1,l]], {l, 1, Nband/2*4*NN}];
ulist = Table[Abs[ES[[2,l,i]]]^2, {l, 1, Nband/2*4*NN}, {i, 388+1, 388+(nsite - 1)*Nband/2 + Nband/2}];
vlist = Table[Abs[ES[[2,l,i + Nband/2*NN*2]]]^2, {l, 1, Nband/2*4*NN}, {i, 388+1, 388+(nsite - 1)*Nband/2 + Nband/2}];
ldossc = Table[Im[Total[Table[ulist[[l,1 ;; All]]*(1/(-wmax + wmax*2*(w/nw) - elist[[l]] + I*\[Eta])) +
vlist[[l,1 ;; All]]*(1/(-wmax + 2*wmax*(w/nw) + elist[[l]] + I*\[Eta])), {l, 1, Nband/2*4*NN}]]], {w, 0, nw}]; ldos = ldos + ldossc;
Export["ldosorb_up_P.dat", Table[{-wmax + wmax*2*(\[Omega]/nw), (-Pi^(-1))*(ldos[[\[Omega] + 1,i]]/Ns^2)}, {\[Omega], 0, nw}, {i, 1,8}]];
(* Export["ldostot.dat", Table[{-wmax + wmax*2*(\[Omega]/nw), (-Pi^(-1))*((ldos[[\[Omega] + 1,i]] + ldos[[\[Omega] + 1,i + 1]] + ldos[[\[Omega] + 1,i + 2]] + ldos[[\[Omega] + 1,i + 3]] + ldos[[\[Omega] + 1,i + 4]])/Ns^2)}, {\[Omega], 0, nw}, {i, 1, (nsite - 1)*Nband/2 + Nband/2 - 4}]]; *)
Print["kx=", kx, " ky=", ky, " nsx=", (kx/(2*Pi))*kkmx + 1.]; ]; ]; ]```
i want to translate my C++ code to wolfram, to improve my calcs.
C++ code
for(int i = 0; i < N - 1; ++i){
matrix[i][i] += L / 3 * uCoef - duCoef / 2 - (double)du2Coef/L;
matrix[i][i+1] += L / 6 * uCoef + duCoef / 2 + (double)du2Coef/L;
matrix[i+1][i] += L / 6 * uCoef - duCoef / 2 + (double)du2Coef/L;
matrix[i+1][i+1] += L / 3 * uCoef + duCoef / 2- (double)du2Coef/L;
}
all this coef are const, N - size of my matrix.
In[1]:= n = 4; uCoef = 2; duCoef = 3; du2Coef = 7; L = 11.;
matrix = Table[0, {n}, {n}];
For[i = 1, i < n, ++i,
matrix[[i, i]] += L/3*uCoef - duCoef/2 - du2Coef/L;
matrix[[i, i+1]] += L/6*uCoef - duCoef/2 - du2Coef/L;
matrix[[i+1, i]] += L/6*uCoef + duCoef/2 + du2Coef/L;
matrix[[i+1, i+1]] += L/3*uCoef - duCoef/2 + du2Coef/L];
matrix
Out[4]= {
{5.19697, 1.5303, 0, 0},
{5.80303, 11.6667, 1.5303, 0},
{0, 5.80303, 11.6667, 1.5303},
{0, 0, 5.80303, 6.4697}}
Each character that has been changed from your original is hinting there is a fundamental difference between C++ and Mathematica
You should use SparseArray for such banded arrays in mathematica:
n = 5; uCoef = 2; duCoef = 3; du2Coef = 7; L = 11.;
matrix = SparseArray[
{{1, 1} -> L/3*uCoef - duCoef/2 - du2Coef/L,
{i_ /; 1 < i < n, i_} -> -duCoef + 2 L uCoef/3 ,
{n, n} -> ( L/3 uCoef - duCoef/2 + du2Coef/L ),
Band[{1, 2}] -> L/6 uCoef - duCoef/2 - du2Coef/L,
Band[{2, 1}] -> L/6 uCoef + duCoef/2 + du2Coef/L}, {n, n}];
MatrixForm#matrix
Even if you insist on the For loop, initialize the matrix as :
matrix = SparseArray[{{_, _} -> 0}, {n, n}];
Clear[x, y, h, k, FirstSlope, SecondSlope];
h = [Pi]; y[[Pi]] = 0;
dy[x_, y_] = (Cos[x] - 3 x^2 y)/x^3;
Do[{x[k] = [Pi] + h*(k - [Pi]),
FirstSlope = dy[x[k], y[k]],
SecondSlope = dy[x[k] + h, y[k] + h*FirstSlope],
y[k + [Pi]] = y[k] + (h*(FirstSlope + SecondSlope))/2}, {k, [Pi],
5[Pi]}] Table[{x[k], y[k]}, {k, [Pi], 5[Pi]}];
MatrixForm[%]
Above image is my error. I'm trying to use Heun's method and my problem is:
1) I want it to stop at y[5 Pi] but it keeps going. I can manipulate it so that it goes to y[5 Pi], but I want to know why exactly it's doing this.
2) y[k] is not evaluating at k=pi,2pi,3pi, etc.
When I was trying to find the maximum value of f using NMaximize, mathematica gave me a error saying
NMaximize::cvdiv: Failed to converge to a solution. The function may be unbounded.
However, if I scale f with a large number, say, 10^5, 10^10, even 10^100, NMaximize works well.
In the two images below, the blue one is f, and the red one is f/10^10.
Here come my questions:
Is scaling a general optimization trick?
Any other robust, general workarounds for the optimizations such
needle-shape functions?
Because the scaling barely changed the shape of the needle-shape of
f, as shown in the two images, how can scaling work here?
thanks :)
Update1: with f included
Clear["Global`*"]
d = 1/100;
mu0 = 4 Pi 10^-7;
kN = 97/100;
r = 0.0005;
Rr = 0.02;
eta = 1.3;
e = 3*10^8;
s0 = 3/100;
smax = 1/100; ks = smax/s0;
fre = 1; tend = 1; T = 1;
s = s0*ks*Sin[2*Pi*fre*t];
u = D[s, t];
umax = N#First[Maximize[u, t]];
(*i=1;xh=0.1;xRp=4.5`;xLc=8.071428571428573`;
i=1;xh=0.1;xRp=4.5;xLc=8.714285714285715;*)
i = 1; xh = 0.1; xRp = 5.5; xLc = 3.571428571428571`;
(*i=1;xh=0.1`;xRp=5.`;xLc=6.785714285714287`;*)
h = xh/100; Rp = xRp/100; Lc = xLc/100;
Afai = Pi ((Rp + h + d)^2 - (Rp + h)^2);
(*Pi (Rp-Hc)^2== Afai*)
Hc = Rp - Sqrt[Afai/Pi];
(*2Pi(Rp+h/2) L/2==Afai*)
L = (2 Afai)/(\[Pi] (h + 2 Rp));
B = (n mu0 i)/(2 h);
(*tx = -3632B+2065934/10 B^2-1784442/10 B^3+50233/10 B^4+230234/10 \
B^5;*)
tx = 54830.3266978739 (1 - E^(-3.14250266080741 B^2.03187556833859));
n = Floor[(kN Lc Hc)/(Pi r^2)] ;
A = Pi*(Rp^2 - Rr^2);
b = 2*Pi*(Rp + h/2);
(* -------------------------------------------------------- *)
Dp0 = 2*tx/h*L;
Q0 = 0;
Q1 = ((1 - 3 (L tx)/(Dp h) + 4 (L^3 tx^3)/(Dp^3 h^3)) Dp h^3)/(
12 eta L) b;
Q = Piecewise[{{Q1, Dp > Dp0}, {Q0, True}}];
Dp = Abs[dp[t]];
ode = u A - A/e ((s0^2 - s^2)/(2 s0 )) dp'[t] == Q*Sign[dp[t]];
sol = First[
NDSolve[{ode, dp[0] == 0}, dp, {t, 0, tend} ,
MaxSteps -> 10^4(*Infinity*), MaxStepFraction -> 1/30]];
Plot[dp''[t] A /. sol, {t, T/4, 3 T/4}, AspectRatio -> 1,
PlotRange -> All]
Plot[dp''[t] A /10^10 /. sol, {t, T/4, 3 T/4}, AspectRatio -> 1,
PlotRange -> All, PlotStyle -> Red]
f = dp''[t] A /. sol;
NMaximize[{f, T/4 <= t <= 3 T/4}, t]
NMaximize[{f/10^5, T/4 <= t <= 3 T/4}, t]
NMaximize[{f/10^5, T/4 <= t <= 3 T/4}, t]
NMaximize[{f/10^10, T/4 <= t <= 3 T/4}, t]
update2: Here comes my real purpose. Actually, I am trying to make the following 3D region plot. But I found it is very time consuming (more than 3 hours), any ideas to speed up this region plot?
Clear["Global`*"]
d = 1/100;
mu0 = 4 Pi 10^-7;
kN = 97/100;
r = 0.0005;
Rr = 0.02;
eta = 1.3;
e = 3*10^8;
s0 = 3/100;
smax = 1/100; ks = smax/s0;
f = 1; tend = 1/f; T = 1/f;
s = s0*ks*Sin[2*Pi*f*t];
u = D[s, t];
umax = N#First[Maximize[u, t]];
du[i_?NumericQ, xh_?NumericQ, xRp_?NumericQ, xLc_?NumericQ] :=
Module[{Afai, Hc, L, B, tx, n, A, b, Dp0, Q0, Q1, Q, Dp, ode, sol,
sF, uF, width, h, Rp, Lc},
h = xh/100; Rp = xRp/100; Lc = xLc/100;
Afai = Pi ((Rp + h + d)^2 - (Rp + h)^2);
Hc = Rp - Sqrt[Afai/Pi];
L = (2 Afai)/(\[Pi] (h + 2 Rp));
B = (n mu0 i)/(2 h);
tx = 54830.3266978739 (1 - E^(-3.14250266080741 B^2.03187556833859));
n = Floor[(kN Lc Hc)/(Pi r^2)] ;
A = Pi*(Rp^2 - Rr^2);
b = 2*Pi*(Rp + h/2);
Dp0 = 2*tx/h*L;
Q0 = 0;
Q1 = ((1 - 3 (L tx)/(Dp h) + 4 (L^3 tx^3)/(Dp^3 h^3)) Dp h^3)/(
12 eta L) b;
Q = Piecewise[{{Q1, Dp > Dp0}, {Q0, True}}];
Dp = Abs[dp[t]];
ode = u A - A/e ((s0^2 - s^2)/(2 s0 )) dp'[t] == Q*Sign[dp[t]];
sol = First[
NDSolve[{ode, dp[0] == 0}, dp, {t, 0, tend} , MaxSteps -> 10^4,
MaxStepFraction -> 1/30]];
sF = ParametricPlot[{s, dp[t] A /. sol}, {t, 0, tend},
AspectRatio -> 1];
uF = ParametricPlot[{u, dp[t] A /. sol}, {t, 0, tend},
AspectRatio -> 1];
tdu = NMaximize[{dp''[t] A /10^8 /. sol, T/4 <= t <= 3 T/4}, {t,
T/4, 3 T/4}, AccuracyGoal -> 6, PrecisionGoal -> 6];
width = Abs[u /. tdu[[2]]];
{uF, width, B}]
RegionPlot3D[
du[1, h, Rp, Lc][[2]] <= umax/6, {h, 0.1, 0.2}, {Rp, 3, 10}, {Lc, 1,
10}, LabelStyle -> Directive[18]]
NMaximize::cvdiv is issued if the optimum improved a couple of orders of magnitude during the optimization process, and the final result is "large" in an absolute sense. (To prevent the message in a case where we go from 10^-6 to 1, for example.)
So yes, scaling the objective function can have an effect on this.
Strictly speaking this message is a warning, and not an error. My experience is that if you see it, there's a good chance that your problem is unbounded for some reason. In any case, this warning is a hint that you might want to double check your system to see if that might be the case.
I'm working with chaotic attractors, and testing some continuous-> discrete equivalences. I've made a continuous simulation of the Rossler system this way
a = 0.432; b = 2; c = 4;
Rossler = {
x'[t] == -y[t] - z[t],
y'[t] == x[t] + a*y[t],
z'[t] == b + x[t]*z[t]-c*z[t]};
sol = NDSolve[
{Rossler, x[0] == y[0] == z[0] == 0.5},
{x, y, z}, {t,500}, MaxStepSize -> 0.001, MaxSteps -> Infinity]
Now, when trying to evaluate a discrete equivalent system with RSolve, Mma doesn't do anything, not even an error, it just can't solve it.
RosslerDiscreto = {
x[n + 1] == x[n] - const1*(y[n] + z[n]),
y[n + 1] == 1 - a*const2)*y[n] + const2*x[n],
z[n + 1] == (z[n]*(1 - const3) + b*const3)/(1 - const3*x[n])}
I want to know if there is a numerical function for RSolve, analogous as the NDSolve is for DSolve.
I know i can make the computation with some For[] cycles, just want to know if it exists such function.
RecurrenceTable is the numeric analogue to RSolve:
rosslerDiscreto = {
x[n+1] == x[n] - C[1]*(y[n] + z[n]),
y[n+1] == (1 - a*C[2])*y[n] + C[2]*x[n],
z[n+1] == (z[n]*(1 - C[3]) + b*C[3]) / (1 - C[3]*x[n]),
x[0] == y[0] == z[0] == 0.5
} /. {a->0.432, b->2, c->4, C[1]->0.1, C[2]->0.1, C[3]->0.1};
coords = RecurrenceTable[rosslerDiscreto, {x,y,z}, {n,0,1000}];
Graphics3D#Point[coords]