I want to use the solution of a differential equation into a piecewise function, but when plotting the result Julia returns the error MethodError: no method matching Float64(::LinearAlgebra.Transpose{Float64, Vector{Float64}}). With the code below, you can see the plot works fine until r=50, and that the error shows up for r>50. How can I use the result of the differential equation in a function?
using Plots, DifferentialEquations
const global Ωₘ = 0.23
const global Ωᵣ = 0.0001
const global Ωl = 0.67
const global H₀ = 70.
function lum(du,u,p,z)
du[1] = -u[1]/(1+z) + ((1/H(z))-1/H₀)/(1+z)
end
da0 = [0.]
zspan = (0.,10.)
lumprob = ODEProblem(lum,da0,zspan)
lumsol = solve(lumprob)
α1(r) = π- asin(3*sqrt(3)*sqrt(1-2/r)/r)
α2(r) = asin(3*sqrt(3)*sqrt(1-2/r)/r)
α3(r) = 3*sqrt(3)/r
α4(r) = 3*sqrt(3)/lumsol(r)
function αtot(r)
if 1.99<r<3.0
α1(r)
elseif 3.0<r<=10.0
α2(r)
elseif 10.0<r<=50.0
α3(r)
elseif 50.0<r
α4(r)
end
end
p2=plot(αtot,xlims=(2.0,51.0),xaxis=:log,xlabel="Rₒ/m",label="α")
In your case lumsol returns a 1-element vector but not a number, thus you need to change line
α4(r) = 3*sqrt(3)/lumsol(r)
to the
α4(r) = 3*sqrt(3)/lumsol(r)[1]
(or refactor the code to take into account that fact)
The function plot should take a function that takes a number and returns a number for the case, your αtot function returns a vector if 50 < r.
Related
I'm trying to build a GUI to output plots for control system design when input parameters of transfer function.I got some problems on passing parameters to the function and changing variable type.
I've got a stucture with examples of simulation parameters:
param = [];
param.parameter = "s";
param.dom = "c"; //domain(c for continuous, d for discrete)
param.num = 1; //numerator of transfer function
param.den = "(s+1)^3"; //denominator
param.fmin = 0.01; //min freq of the plot
param.fmax = 100; //max freq
and a function to plot the graphs:
// display function
function displayProblem(param)
parameter = param.parameter;
dom = param.dom;
num = param.num;
den = param.den;
fmin = param.fmin;
fmax = param.fmax;
s = poly(0,parameter.string);
h = syslin(dom.string,num,den);
// Plotting of model data
delete(gca());
//bode(h,fmin1,fmax1);
gain_axes = newaxes();
gain_axes.title.font_size = 3;
gainplot(h,fmin,fmax);
gain_axes.axes_bounds = [1/3,0,1/3,1/2];
gain_axes.title.text = "Gain Plot";
phase_axes = newaxes();
gain_axes.title.font_size = 3;
phaseplot(h,fmin,fmax);
phase_axes.axes_bounds = [1/3,1/2,1/3,1/2];
phase_axes.title.text = "Phase Plot";
phase_axes = newaxes();
gain_axes.title.font_size = 3;
nyquist(h);
phase_axes.axes_bounds = [2/3,0,1/3,1/2];
phase_axes.title.text = "Nyquist Plot";
endfunction
There's something wrong when passing numerator and denominator to the function. The variable type doesn't match what syslin required. If I replace 'num' and 'den' with '1' and '(s+1)^3', everything worked quite well. Also if I try this line-by-line in control panel, it works, but not in SciNotes. What's the proper way to deal with this? Any suggestion will be greatly appreaciated.
There are 3 errors in your code: Replace
param.den = "(s+1)^3"; //denominator
with
param.den = (%s+1)^3;
Indeed, Scilab has a built-in polynomial type. Polynomials are not defined as a string nor as a vector of coefficients. The predefined constant %s is the monomial s.
In addition,
s = poly(0,parameter.string);
is useless (and incorrect: parameter would work, while parameter.string won't since parameter has no string field: It IS the string). Just remove the line.
As well, replace
h = syslin(dom.string,num,den);
with simply
h = syslin(dom, num,den);
Finally, although the figure's layout is not simple, you can use bode() in the function in the following way:
// delete(gca());
subplot(1,3,2)
bode(h, fmin, fmax)
subplot(2,3,3)
nyquist(h)
gcf().children.title.text=["Nyquist Plot" "Phase Plot" "Gain Plot"]
All in one, the code of your function may resume to
function displayProblem(param)
[dom, num, den] = (param.dom, param.num, param.den);
[fmin, fmax] = (param.fmin, param.fmax);
h = syslin(dom, num,den);
// Plotting of model data
subplot(1,3,2)
bode(h,fmin,fmax);
subplot(2,3,3)
nyquist(h);
gcf().children.title.text=["Nyquist Plot" "Phase Plot" "Gain Plot"];
endfunction
Best regards
I have a function solving two ODEs and joining the solution which gives a really nice plot in linear scales, but which has a high drop in quality in log scale depending on the parameters I use. In the code below, I plot the solution for two sets of parameters, in which you can see the first set is not smooth, while the second one is kind of okay. If I try to obtain a smoother visualisation using the saveat option in the second ODEs, solclass = solve(probclass,Tsit5(),saveat=0.001), I get an error when plotting the second set: ArgumentError: range step cannot be zero. Is there a way to obtain smooth linear-log other than manually changing the saveat option? Also, I have tried using a few other backends, but they gave an error at ploting the solution.
using DifferentialEquations, Plots, RecursiveArrayTools
function alpha_of_phi!(s2,d,a0,ϕ₀)
# α in the quantum phase
function quantum!(dv,v,p,ϕ)
s2,d=p
α = v[1]
dv[1] = (ϕ*s2*sin(2*d*α)+2*d*sinh(s2*α*ϕ))/(-α*s2*sin(2*d*α)+2*d*cos(2*d*α)+2*d*cosh(s2*α*ϕ))
end
# When dα/dϕ = 1, we reach the classical regime, and stop the integration
condition(v,ϕ,integrator) = (ϕ*s2*sin(2*d*v[1])+2*d*sinh(s2*v[1]*ϕ))/(-v[1]*s2*sin(2*d*v[1])+2*d*cos(2*d*v[1])+2*d*cosh(s2*v[1]*ϕ))==1.0
affect!(integrator) = terminate!(integrator)
cb = DiscreteCallback(condition,affect!)
# Initial Condition at the bounce
α₀ = [a0]
classspan = (0,ϕ₀)
probquant = ODEProblem(quantum!,α₀,classspan,(s2,d))
solquant = solve(probquant,Tsit5(),callback=cb)
# α in the classical phase
function classic!(du,u,p,ϕ)
αc = u[1]
dαc = u[2]
du[1] = dαc
du[2] = 3*(-dαc^3/sqrt(2)+dαc^2+dαc/sqrt(2)-1)
end
# IC retrieved from end of quantum phase
init = [last(solquant);1.0]
classspan = (last(solquant.t),ϕ₀)
probclass = ODEProblem(classic!,init,classspan)
solclass = solve(probclass,Tsit5())
# α(ϕ) for ϕ>0
solu = append!(solquant[1,:],solclass[1,:]) # α
solt = append!(solquant.t,solclass.t) # ϕ
# α(ϕ) for ϕ<0
soloppu = reverse(solu)
soloppt = -reverse(solt)
pop!(soloppu)
pop!(soloppt)
# Join the two solutions
soltotu = append!(soloppu,solu)
soltott = append!(soloppt,solt)
soltot = DiffEqArray(soltotu,soltott)
end
plot(alpha_of_phi!(10000.0,-0.0009,0.0074847,2.0),yaxis=:log)
plot!(alpha_of_phi!(16.0,-0.1,0.00001,2.0))
```
If you were plotting a solution returned by solve directly, then the Plots recipes for DifferentialEquations enable an optional keyword argument for plot entitled plotdensity, which would let you choose the number of points plotted, and thus smoothness, as described in the docs, e.g.:
plot(sol,plotdensity=10000)
However, this keyword appears to require a solution object, rather than a DiffEqArray. Consequently, your best bet will indeed be manually setting saveat. For this approach, saveat = 0.01 would seem to be plenty to obtain fully smooth lines. However, this still gives the "range step cannot be zero" error you describe.
While I have no deep understanding of the system you are solving, an inspection of the results revealed duplicate timesteps in the results for alpha_of_phi!(16.0,-0.1,0.00001,2.0) run without saveat, suggesting that the classical simulation was being run with over a range of no time. In other words, this hints that last(solquant.t) may well be equal to or greater than ϕ₀ with these parameters, resulting in a timespan of zero. If so, this will quite understandably fail when you request to saveat some finite time within that timespan (last(solquant.t), ϕ₀).
Consequently, working on this hypothesis, if we just rewrite your function to check for this condition
using DifferentialEquations, Plots
function alpha_of_phi!(s2,d,a0,ϕ₀)
# α in the quantum phase
function quantum!(dv,v,p,ϕ)
s2,d=p
α = v[1]
dv[1] = (ϕ*s2*sin(2*d*α)+2*d*sinh(s2*α*ϕ))/(-α*s2*sin(2*d*α)+2*d*cos(2*d*α)+2*d*cosh(s2*α*ϕ))
end
# When dα/dϕ = 1, we reach the classical regime, and stop the integration
condition(v,ϕ,integrator) = (ϕ*s2*sin(2*d*v[1])+2*d*sinh(s2*v[1]*ϕ))/(-v[1]*s2*sin(2*d*v[1])+2*d*cos(2*d*v[1])+2*d*cosh(s2*v[1]*ϕ))==1.0
affect!(integrator) = terminate!(integrator)
cb = DiscreteCallback(condition,affect!)
# Initial Condition at the bounce
α₀ = [a0]
classspan = (0,ϕ₀)
probquant = ODEProblem(quantum!,α₀,classspan,(s2,d))
solquant = solve(probquant,Tsit5(),callback=cb,saveat=0.01)
# α in the classical phase
function classic!(du,u,p,ϕ)
αc = u[1]
dαc = u[2]
du[1] = dαc
du[2] = 3*(-dαc^3/sqrt(2)+dαc^2+dαc/sqrt(2)-1)
end
if last(solquant.t) < ϕ₀
# IC retrieved from end of quantum phase
init = [last(solquant);1.0]
classspan = (last(solquant.t),ϕ₀)
probclass = ODEProblem(classic!,init,classspan)
solclass = solve(probclass,Tsit5(),saveat=0.01)
# α(ϕ) for ϕ>0
solu = append!(solquant[1,:],solclass[1,:]) # α
solt = append!(solquant.t,solclass.t) # ϕ
else
solu = solquant[1,:] # α
solt = solquant.t # ϕ
end
# α(ϕ) for ϕ<0
soloppu = reverse(solu)
soloppt = -reverse(solt)
pop!(soloppu)
pop!(soloppt)
# Join the two solutions
soltotu = append!(soloppu,solu)
soltott = append!(soloppt,solt)
soltot = DiffEqArray(soltotu,soltott)
end
plot(alpha_of_phi!(10000.0,-0.0009,0.0074847,2.0),yaxis=:log)
plot!(alpha_of_phi!(16.0,-0.1,0.00001,2.0))
then we would seem to be in business!
This question has already been asked on another platform, but I haven't got an answer yet.
https://discourse.julialang.org/t/generalizing-the-inputs-of-the-nlsolve-function-in-julia/
After an extensive process usyng the SymPy in Julia, I generated a system of nonlinear equations. My system is allocated in a matrix NxS. Something like this(NN = 2, S = 2).
I would like to adapt the system to use the NLsolve package. I do some boondoggle for the case NN=1 and S =1. The system_equations2 function give me the nonlinear system, like the figure
using SymPy
using Plots
using NLsolve
res = system_equations2()
In order to simulate the output, I do this:
NN = 1
S = 1
p= [Sym("p$i$j") for i in 1:NN,j in 1:S]
res = [ Eq( -331.330122303069*p[i,j]^(1.0) + p[i,j]^(2.81818181818182) - 1895.10478893046/(p[i,j]^(-1.0))^(2.0),0 ) for i in 1:NN,j in 1:S]
resf = convert( Function, lhs( res[1,1] ) )
plot(resf, 0 ,10731)
Now
resf = convert( Function, lhs( res[1,1] ) )
# This for the argument in the nlsolve function
function resf2(p)
p = Tuple(p)[1]
r = resf(p)
return r
end
Now, I find the zeros
function K(F,p)
F[1] = resf2(p[1])
end
nlsolve(K , [7500.8])
I would like to generalize this price to any NN and any S. I believe there is a simpler way to do this.
function [best1, best2] = selection(population)
fitness = zeros(1,length(population));
for i=1:length(population)
fitness(i) = population(i).fitness;
end
[~,index] = max(fitness);
best1 = population(index);
population(index) = [];
fitness(index) = [];
[~,index] = max(fitness);
best2 = population(index);
endfunction
[~,index] = max(fitness);
^
Error: syntax error, unexpected ","
The syntax using "~" when you don't want to assign an output parameter is not available in Scilab. You have to use a dummy variable, e.g.
[dummy,index] = max(fitness)
I'm trying to run a loop over different functions with different number of arguments. The variables are created at runtime inside the loop, and I want to use eval at each iteration to instantiate a Struct using the variable :symbol. However, I can't do this since eval only works in the global scope. This is the MWE for the case that works:
function f1(x); return x; end
function f2(x1,x2); return x1+x2; end
handles = [f1,f2]
args =[:(x1),:(x1,x2)]
x1 = 1; x2 = 1;
for (i,f) in enumerate(handles)
params = eval(args[i])
#show f(params...)
end
f(params...) = 1
f(params...) = 2
However, if I move the variable definitions inside the loop, which is what I actually want, it doesn't work after restarting Julia to clear the workspace.
function f1(x); return x; end
function f2(x1,x2); return x1+x2; end
handles = [f1,f2]
args =[:(x1),:(x1,x2)]
for (i,f) in enumerate(handles)
x1 = 1; x2 = 1;
params = eval(args[i])
#show f(params...)
end
ERROR: UndefVarError: x1 not defined
I've tried several of the answers, such as this one, but I can't seem to make it work. I could write a custom dispatch function that takes[x1,x2] and calls f1 or f2 with the correct arguments. But still, is there any way to do this with eval or with an alternative elegant solution?
EDIT: here are more details as to what I'm trying to do in my code. I have a config struct for each algorithm, and in this I want to define beforehand the arguments it takes
KMF_config = AlgConfig(
name = "KMF",
constructor = KMC.KMF,
parameters = :(mu,N,L,p),
fit = KMC.fit!)
MF_config = AlgConfig(
name = "MF",
constructor = KMC.MF,
parameters = :(mu,N,L),
fit = KMC.fit!)
alg_config_list = [KMF_config, MF_config]
for (i,alg_config) in enumerate(alg_config_list)
mu,N,L,p,A,B,C,D,data = gen_vars() #this returns a bunch of variables that are used in different algorithms
method = alg_config.constructor(eval(method.parameters)...)
method.fit(data)
end
One possible solution is to have a function take all the variables and method, and return a tuple with a subset of variables according to method.name. But I'm not sure if it's the best way to do it.
Here's an approach using multiple dispatch rather than eval:
run_a(x, y) = x + 10*y
run_b(x, y, z) = x + 10*y + 100*z
extract(p, ::typeof(run_a)) = (p.x, p.y)
extract(p, ::typeof(run_b)) = (p.x, p.y, p.z)
genvars() = (x=1, y=2, z=3)
function doall()
todo = [
run_a,
run_b,
]
for runalg in todo
v = genvars()
p = extract(v, runalg)
#show runalg(p...)
end
end
In your example you would replace run_a and run_b with KMC.KMF and KMC.MF.
Edit: Cleaned up example to avoid structs that don't exist in your example.