How to Draw This Spiral Solution Curve?

[Freya_the_Goddess]

Hi all,

I read this topic and it is still confusing since the differential equation is a bit different:

Plotting a phase portrait of a differential equation

Hey Folks. I was just wondering if there was an existing package or tool for plotting the phase portrait of a system of differential equations, or a phase portrait plus direction field? I am thinking of a plot like this for the system of equations. \frac{dx}{dt} = -x \\ \frac{dy}{dt} = 2y The corresponding plot would be (taken from Haberman’s ODE textbook): [Selection_002] I did not see any mention of this in the plots documentation for the Julia DifferentialEquations package, but perhaps …

I use only variable of t. That topic use dx/dt and dy/dt, thus I can’t really use it to my problem at hand.

This is the problem, and the solution. What I want to achieve is how to plot the spiral curve along with the arrow? Given the input of matrix A and the initial value x(0).

[dlakelan]

Something is weird with this example. At x0 the derivative is [-2,1] but the spiral curve clearly shows something like [0,1]. I haven’t looked carefully but perhaps this is just a poor drawing?

[Freya_the_Goddess]

The solution curve is for the solution:

\boldsymbol{x} (t) = e^{-2t}[\cos t \ \sin \ t]

with x_{1} (t) = e^{-2t} \cos \ t and x_{2} (t) = e^{-2t} \sin \ t

now if we put the t=0 (at the beginning) we will obtain:
x_{1} (0) = e^{-0} \cos \ 0 and x_{2} (0) = e^{-0} \sin \ 0

which will become the initial coordinate for the solution curve of

\boldsymbol{x} (t) = e^{-0}[\cos 0 \ \sin \ 0] = 1[1 \ 0]

thus the coordinate is (1,0) at the beginning and x-axis is the x_{1} and y-axis is the x_{2} that is why the arrow is pointing to that way. the term e^{-2t} wil goes to 0 as t goes to infinity… that is why it is spiraling to (0,0).

That is from my understanding. Correct me if I am wrong… I know there are lots of silent reader who is an expert / more than PhDs here…

The problem is I want to draw the solution curve…

Can anyone help? Or just pray I can make it alone…
either way thanks folks! or aliens or vampires here…

[dlakelan]

Plotting the solution given the formula is pretty straightforward.

ts=0.0:.01:5.0

x=[exp(-2*t) cos(t) for t in ts]
y=... Etc

plot(x,y)

[Freya_the_Goddess]

@dlakelan

for t in ts

After see this: parametric function plot 3d | [“Julia Plots Gallery”]
I try this code:

using MTH229, ForwardDiff, Plots, LaTeXStrings, SymPy
gr()

t = range(0, stop=10, length=1000)
x = e^(-2t)*cos.(t)
y = e^(-2t)*sin.(t)

plot(x, y)

but it won’t work:

ERROR: LoadError: MethodError: no method matching ^(::Float64, ::StepRangeLen{Float64, Base.TwicePrecision{Float64}, Base.TwicePrecision{Float64}, Int64})
Closest candidates are:
^(::Number, ::Missing) at ~/julia-1.7.3/share/julia/base/missing.jl:124
^(::T, ::Rational) where T<:AbstractFloat at ~/julia-1.7.3/share/julia/base/rational.jl:480
^(::AbstractFloat, ::ForwardDiff.Dual{Ty}) where Ty at ~/.julia/packages/ForwardDiff/eqMFf/src/dual.jl:145
…
Stacktrace:
[1] top-level scope
@ ~/LasthrimProjection/plot.jl:6
[2] include(fname::String)
@ Base.MainInclude ./client.jl:451
[3] top-level scope
@ REPL[1]:1
in expression starting at /home/browni/LasthrimProjection/plot.jl:6

[Ininterrompue]

You’re basically trying to take the exponential of a vector, which is undefined. Try

x = exp.(-2t)*cos.(t) 

or, which is equivalent,

x = @. exp(-2t)*cos(t)

so that the exponential function acts element-wise.

[Freya_the_Goddess]

Wait I fot it works by using GoroPikari example:

using MTH229, ForwardDiff, Plots, LaTeXStrings, SymPy
gr()

xₜ(t) = exp(-2t)*cos(t)
yₜ(t) = exp(-2t)*sin(t)

plot(xₜ, yₜ, 0, 2π, leg=false)

but the question is how to make the small circle goes to (0,0) ? and the quiver or arrow…

[Freya_the_Goddess]

@Ininterrompue

@. exp(-2t)*cos(t)

Alright this code works:

t = range(0, stop=10, length=1000)
x = @. exp(-2t)*cos(t)
y = @. exp(-2t)*sin(t)

plot(x, y)

but how to create the small circle rotating to (0,0) again plus the arrow?

[Freya_the_Goddess]

This looks better:

t = range(0, stop=21, length=10000)
x = @. exp(-2t)*cos(t)
y = @. exp(-2t)*sin(t)

plot(x, y,xlims=(-0.1,1), ylims=(-0.1,0.21))

I see this CalculusWithJulia’ arrow feature, but it is straight line, how to bend it?

Vectors Addition only work with initial (0,0)?

Hi all, I have been learning from this site: https://docs.juliahub.com/CalculusWithJulia/AZHbv/0.0.5/differentiable_vector_calculus/vectors.html# then I change the initial point and see the second vector is not in the place it should be, needs to modify the code… using CalculusWithJulia using Plots gr() # better arrows than plotly() quiver([0],[0], quiver=([1],[2])) unzip(vs) = Tuple(eltype(first(vs))[xyz[j] for xyz in vs] for j in eachindex(first(vs))) u = [2, 1] v = [-2, 1] w = u…

[Ininterrompue]

The solution looks different from the cartoon…

Anyway, you can plot the solution as two parts, one of which has an arrow:

x1(t) = exp(-2t)*cos(t)
x2(t) = exp(-2t)*sin(t)

plot(x1, x2, 0, π/6, legend=false, color=:black, arrow=true)
plot!(x1, x2, π/6, 2π, legend=false, color=:black)

[Freya_the_Goddess]

Yes it works, anyway thanks a lot, now just thinking how to make the small spiraling to (0,0)… or zoom it in

[dlakelan]

@Freya_the_Goddess

spiraling to (0,0)

It reaches 0,0 when t goes to infinity…

The real solution is different from the cartoon. If you want something more like the cartoon try changing the exponential to be a smaller power… exp(-t/2) for example

[Freya_the_Goddess]

Alright then I will try it, so the textbook gives me bias then.