Below code is used to solve a stochastic equation numerically in Mathematica for one particle. I wonder if there is a way to generalize it to the case of more than one particle and average over them. Is there anyone who knows how to do that?
Clear["Global`*"]
{ a = Pi, , b = 2 Pi, l = 5, k = 1};
ic = x#tbegin == 1;
tbegin = 1;
tend = 400;
interval = {1, 25};
lst := NestWhileList[(# + RandomVariate#TruncatedDistribution[interval,
StableDistribution[1, 0.3, 0, 0, 1]]) &, tbegin, # < tend &];
F[t_] := Piecewise[{{k, Or ## #}}, -k] &[# <= t < #2 & ###
Partition[lst, 2]];
eqn := x'[t] == (F#(t)) ;
sol = NDSolveValue[{eqn, ic}, x, {t, tbegin, tend},
MaxSteps -> Infinity];
Plot[sol#t, {t, tbegin, tend}]
First[First[sol]]
Plot[sol'[t], {t, tbegin, tend}]
Plot[F[t], {t, tbegin, tend}]
Related
Does anyone know what does /; mean for example:V[n_][i_/;i<=imax,0];=V[n][i,0]=0 in Mathematica?
It's part of a program to draw the eletrical potential function of a rectangular metal trough using the finite-difference methods.
Here's the improved one, but still cannot work.
V[0][i_, j_] := 0;
tol = 0.0025;
imax = 20; jmax = 20;
V[n_][i_, 0] := V[n][i, 0] = 0;
V[n_][i_, jmax] := V[n][i, jmax] = 100*Sin[0.05*Pi*i];
V[n_][0, j_] := V[n][0, j] = 0;
V[n_][imax, j_] := V[n][imax, j] = 0;
V[n_][i_, j_] := V[n][i, j] = (1/4)*(V[n - 1][i + 1, j] +
V[n - 1][i - 1, j] + V[n - 1][i, j + 1] + V[n - 1][i, j - 1]);
For[m = 1, Max[Table[Abs[V[m][i, j] - V[m - 1][i, j]], {i, 1, imax - 1}, {j, 1, jmax - 1}]] >= tol, m++;
If[m > 20, Break[]]];
Print[m];
Show[ListPlot3D[Table[V[m][i, j], {j, 0, 20}, {i, 0, 20}]], BoxRatios -> {1, 1, 0.85}, PlotRange -> {0, 1}, Axes -> True, AxesLabel -> {"x", "y", "V"}]
I've already get the point through the reference of Wolfram. And here is the web
https://reference.wolfram.com/language/ref/Condition.html. Thanks for #agentp help
I have the following Mathematica Code, But it does not give me anything as the output. Can someone help me.
\[Alpha] = 3;
F[s_] := Exp[-A*s^(2/\[Alpha])];
integral = Re[Assuming[{A > 0, t > 0, {t, A} \[Element] Reals},
Integrate[F[s]*Exp[s*t] /. s -> I*y, {y, 0, Infinity}]/Pi]]
I also want to run the following code:
\[Alpha] = 4;
f[s_] := Exp[-A*s^(2/\[Alpha])];
integral =Re[Assuming[{A > 0, t > 0, {t, A} \[Element] Reals},
Integrate[f[s]*Exp[s*t] /. s -> I*y, {y, -Infinity, Infinity}]/Pi]]
here A is give by
A = Pi*\[Lambda]*P^(2/\[Alpha])*Gamma[1 + 2/\[Alpha]]*Gamma[1 - 2/\[Alpha]];
Lambda and P are known values.
At least under version 10 this seems to work
α = 3;
A = Pi*λ*P^(2/α)*Gamma[1+2/α]*Gamma[1-2/α];
integral = Re[Assuming[{A > 0, t > 0},
Integrate[Exp[-A*(I*y)^(2/α)]*Exp[I*y*t], {y, 0, Infinity}]]/Pi]
α = 4;
A = Pi*λ*P^(2/α)*Gamma[1+2/α]*Gamma[1-2/α];
integral = Re[Assuming[{A > 0, t > 0},
Integrate[Exp[-A*(I*y)^(2/α)]*Exp[I*y*t], {y, -Infinity, Infinity}]]/Pi]
If you can provide more information about the values of P and λ then perhaps Re can do more.
I am trying to plot using piecewise in one of my problems and I have two variables: x and psi. However, the respective functions are only valid for a defined range of "x" and the psi range is the same. I am trying to make a 3D plot of these -- and I basically just have Plot3D[p,{x,0,1},{psi,0.01,1}] ---> These ranges are for the entire plot range and my x range for the respective functions is already defined in the Piecewise function.
I get the following error: saying that Plot::exclul: ...... must be a list of equalities or \ real-valued functions.
Can anyone please help me with this. I am trying to follow the same procedure as: How can I use Piecewise[] with a variable number of graphs/intervals
But, I don't know what to do about the plotting part.
Thanks.
The following is my code:
j = 10;
s = 0; r = 0;
K[x_, psi_] :=
Sum[Sin[n*Pi*x]*
Sin[n*Pi*
psi]*(2*Exp[-(n*Pi)^2*
Abs[s + r]] - (Exp[-(n*Pi)^2*Abs[s - r]] -
Exp[-(n*Pi)^2*(s + r)])/(n*Pi)^2 ), {n, 1, j}];
TL[x_, psi_] = Integrate[K[x - y, psi]*y, {y, -10, 10}];
TU[x_, psi_] = Integrate[K[x - y, psi]*(1 - y), {y, -10, 10}];
eq = {TL[x, psi], TU[x, psi]};
cond = {{0 <= x <= 0.5, 0.01 <= psi <= 1}, {0.5 < x <= 1,
0.01 <= psi <= 1}};
p = Piecewise[{eq, cond}];
Plot3D[p, {x, 0, 1}, {psi, 0.01, 1}]
Here is a working version:
time = AbsoluteTime[];
j = 10; s = 0; r = 0;
K[x_, psi_] :=
Sum[Sin[n*Pi*x]*Sin[n*Pi*psi]*
(2*Exp[-(n*Pi)^2*Abs[s + r]] -
(Exp[-(n*Pi)^2*Abs[s - r]] -
Exp[-(n*Pi)^2*(s + r)])/(n*Pi)^2), {n, 1, j}];
TL[x_, psi_] := Integrate[K[x - y, psi]*y, {y, -10, 10}];
TU[x_, psi_] := Integrate[K[x - y, psi]*(1 - y), {y, -10, 10}];
Plot3D[Piecewise[
{{TL[x, psi], 0 <= x <= 0.5}, {TU[x, psi], 0.5 < x <= 1}}],
{x, 0, 1}, {psi, 0.01, 1}]
ToString[Round[AbsoluteTime[] - time]] <> " seconds"
I am trying to solve an D-equation and do not know y[0], but I know y[x1]=y1.
I want to solve the DSolve only in the relevant xrange x=[x1, infinitny].
How could it work?
Attached the example that does not work
dsolv2 = DSolve[{y'[x] == c*0.5*t12[x, l2]^2 - alpha*y[x], y[twhenrcomesin] == zwhenrcomesin, x >= twhenrcomesin}, y[x], x]
dsolv2 = Flatten[dsolv2]
zsecondphase[x_] = y[x] /. dsolv2[[1]]
I am aware that DSolve does not allow the inequality condition but I put it in to explain you what I am looking for (t12[x,l2] will give me a value only depending on x since l2 is known).
EDIT
t12[j24_, lambda242_] := (cinv1 - cinv2)/(cop2 - cop1 + (h2*lambda242)*E^(p*j24));
cinv1 = 30; cinv2 = 4; cinv3 = 3; h2 = 1.4; h3 = 1.2; alpha = 0.04; z = 50; p = 0.06; cop1 = 0; cop2 = 1; cop3 = 1.3; teta2 = 0.19; teta3 =0.1; co2 = -0.6; z0 = 10;l2 = 0.1;
Your equation is first order and linear, so you can get a very general solution :
generic = DSolve[{y'[x] == f[x] - alpha*y[x], y[x0] == y0}, y[x], x]
Then you can substitute your specific term :
c = 1;
x0 = 1;
y0 = 1;
solution[x_] = generic[[1, 1, 2]] /. {f[x_] -> c*0.5*t12[x, l2]^2}
Plot[solution[x], {x, x0, 100}]
What is wrong with this example?
t12[x_] := Exp[-x .01] Sin[x];
dsolv2 = Chop#DSolve[{y'[x] == c*0.5*t12[x]^2 - alpha*y[x], y[1] == 1}, y[x], x];
Plot[y[x] /. dsolv2[[1]] /. {alpha -> 1, c -> 1}, {x, 1, 100}, PlotRange -> Full]
Edit
Regarding your commentary:
Try using a piecewise function to restrict the domain:
t12[x_] := Piecewise[{{ Exp[-x .01] Sin[x], x >= 1}, {Indeterminate, True}}] ;
dsolv2 = Chop#DSolve[{y'[x] == c*0.5*t12[x]^2 - alpha*y[x], y[1] == 1}, y[x], x];
Plot[y[x] /. dsolv2[[1]] /. {alpha -> 1, c -> 1}, {x, 1, 100}, PlotRange -> Full]
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.