Help with deSolveDiffEq

New to Julia
1

Why with deSolveDiffEq I am not receiving all the solutions of the system, as can be seen in the examples?

I only get these results

19:42:41->>retcode: Default
Interpolation: 1st order linear
t: 2-element view(::Array{Float64,2}, :, 1) with eltype Float64:
  0.0
 50.0
u: 2-element Array{Array{Float64,1},1}:
 [0.0, 0.66]
 [-33.0, 15.510000000000002]
### Examples  
using DifferentialEquations
using Plots; gr()
function integrate_ODE!(du,u,t)
    du[1] = -0.4*u[1] - u[2]
    du[2] = u[1] + 0.45*u[2]   
end
u0 = [0.0,0.66]
tspan = (0.0,50.0)
prob = ODEProblem(integrate_ODE!,u0,tspan)
sol = solve(prob)# Euler ImplicitEuler 
plot(sol,linewidth=2,xaxis="t",layout=(2,1))
plot(sol,vars=(1,2), lw=1)

#
using OrdinaryDiffEq
prob = ODEProblem(integrate_ODE!,u0,tspan)
OrdinaryDiffEq.solve(prob, ImplicitEuler())
plot(sol,linewidth=2,xaxis="t",layout=(2,1))
plot(sol,vars=(1,2), lw=1)

#
A=[-0.4 -1.0
    1.0 0.45]
u0 = [0.0; 0.66]
tspan = (0.0,50.0)
f(u,p,t)=A*u
prob = ODEProblem(f,u0,tspan)
sol=solve(prob)
plot(sol,linewidth=2,xaxis="t",layout=(2,1))
plot(sol,vars=(1,2), lw=1)

#
using deSolveDiffEq
u0 = [0.0;0.66]
tspan = (0.0,50.0)
prob = ODEProblem(integrate_ODE!,u0,tspan)
sol = solve(prob,deSolveDiffEq.euler()) 
#
2

you mean integrate_ODE!(du,u,p,t)?

3

In function integr_ODE! (du, u, t), I do not declare parameters. Should I in the declaration of the function, enter it, even though in the body of the function I don’t use them?

function integrate_ODE!(du,u,t)
    du[1] = -0.4*u[1] - u[2]
    du[2] = u[1] + 0.45*u[2]   
end
4 
function integrate_ODE!(du,u,p,t)
    du[1] = -0.4*u[1] - u[2]
    du[2] = u[1] + 0.45*u[2]   
end

using deSolveDiffEq

u0 = [0.0;0.66]
tspan = (0.0,50.0)
prob = ODEProblem(integrate_ODE!,u0,tspan)
sol = solve(prob,deSolveDiffEq.euler())
08:51:07->>retcode: Default
Interpolation: 1st order linear
t: 2-element view(::Array{Float64,2}, :, 1) with eltype Float64:
  0.0
 50.0
u: 2-element Array{Array{Float64,1},1}:
 [0.0, 0.66]
 [-33.0, 15.510000000000002]

Introducing in solve, saveat = 0.1, I get the results!

sol = solve(prob,deSolveDiffEq.euler(),saveat=0.1) 
>retcode: Default
Interpolation: 1st order linear
t: 501-element view(::Array{Float64,2}, :, 1) with eltype Float64:
  0.0
  0.1
  0.2
  0.3
  0.4
  0.5
  0.6
  0.7
  0.8
  0.9
  1.0
  ⋮
 49.1
 49.2
 49.3
 49.4
 49.5
 49.6
 49.7
 49.8
 49.9
 50.0
u: 501-element Array{Array{Float64,1},1}:
 [0.0, 0.66]
 [-0.066, 0.6897]
 [-0.13233, 0.7141365]
 [-0.19845045, 0.7330396425]
 [-0.26381639625000003, 0.7461813814125]
 [-0.32788187854125, 0.7533779039510625]
 [-0.39010439379470624, 0.7544917217747353]
 [-0.4499493902203915, 0.7494334098751279]
 [-0.5068947555990887, 0.7381629742974695]
 [-0.5604352628048721, 0.7206908325809467]
 [-0.6100869355507719, 0.6970783937666021]
 ⋮
 [-4.192619673242593, 17.848340068051428]
 [-5.809748893118055, 18.23225340378949]
 [-7.400584277772191, 18.471729917648197]
 [-8.951733898426145, 18.56289933616515]
 [-10.449954476105635, 18.503056416449965]
 [-11.882261938706426, 18.29069850757965]
 [-13.236041311916154, 17.925553746550086]
 [-14.499155034094445, 17.408599533953254]
 [-15.660048786126008, 16.742071009571696]
 [-16.707853935638152, 15.929459326389809]

My next question,
How to see the saveat, which was used by default in the previous examples? To be able to replicate it in this case

5

I think when you choose a method with fixed time steps, deSolve chooses to set the time step to the maximal value unless you specify a dt which we do implicitly through saveat because of how it mixes those ideas. You can’t make deSolveDiffEq.euler adaptive because their implementation of it is not adaptive.

6

BTW, I’m curious what the purpose of this exercise is. Using Euler from deSolve is a very curious thing :laughing:

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