julia vector differential equation: plot only the first n components of the vector - plot

Question

I am integrating a vector differential equation in Julia (the number of components is of the order of magnitude of 50). When I plot this solution, obviously this plot will be very crowded if I plot all the components, so I want to plot only let's say the first 10 components. I can't find how to do this. Also, instead of plotting the output for each component, I would like to plot not the output of the integration at each time step, but the hyperbolic tangent of this. I also don't find how to do this manipulation. I think the key problem is that I don't understand/find what kind of object the solution of a differential equation is in julia.
Here is a minimal working example that integrates the differential equation and plots the whole solution.
using DifferentialEquations
using Plots
using LinearAlgebra
N=50
J=0.18*randn(Float64,N,N)
g=1
function hDerivative(timederiv,h,p,t)
for i=1:length(h)
timederiv[i] = -h[i]
for j=1:length(h)
timederiv[i]=+timederiv[i]+J[i,j]*tanh(g*h[j])
end
end
end
function pltTimeVolution()
hinit=rand(Float64,N)
tspan=(0.0,50)
prob = ODEProblem(hDerivative,hinit,tspan)
sol=solve(prob)
plot(sol)
print(sol)
end
pltTimeVolution()

Answer 1

For help, take a look at the solution handling page: http://diffeq.sciml.ai/latest/basics/solution.html . The solution is both an abstract array and a continuous function. So you can for example get a continuous solution at 10000 evenly spaced time points like t=range(tspan[1],stop=tspan[2],length=10000), do A = sol(t) and that will give a 50x10000 timeseries of each solution at evenly spaced time points, and then sol[i,:] would be the timeseries of the ith variable at the time points t. You can use that to build the arrays you want an plot them. Full code like this:
t=range(tspan[1],stop=tspan[2],length=10000)
A = sol(t)
ts10 = sol[10,:]
Another way to do this is to solve with something like saveat=t, in which case the solution is an array which saves at those time points. This looks like:
t=range(tspan[1],stop=tspan[2],length=10000)
sol=solve(prob,saveat=t)
ts10 = sol[10,:]
Or you can make use of the plot recipe which does this kind of handling internally. Take a look at http://diffeq.sciml.ai/latest/basics/plot.html#Choosing-Variables-1 . You can do things like plot the first variable and then lay over the plot of the 10th variable like:
plot(sol,vars=(0,1))
plot!(sol,vars=(0,10))
The ! means its the mutating plot function, i.e. modify the previous plot by adding a new series. Then as noted in the documentation there are shortcuts, like
plot(sol,vars = [1, 3, 4])
will plot the 1st, 3rd, and 4th variables as functions of time. In this form, you can also specify what kind of transformation you want as a function. That's mentioned in the same place in the documentation. For example, the two-dimensional plot of time vs tanh would be the transformation function f=(t,y)->(t,tanh(y)), i.e. how you take in the 2D variable and spit out the 2D variable to plot is just pass through the first and tanh the second. Then using this, you'd use the plot command:
f=(t,y)->(t,tanh(y))
plot(sol,vars = (f,0,5))
to plot the tanh of the 5th variable against time.
These are a few different ways to manipulate the solution argument and generate the plots you need. Pick the one that works best for you.