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DAE_StructuralSimplify.jl
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DAE_StructuralSimplify.jl
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### A Pluto.jl notebook ###
# v0.19.43
using Markdown
using InteractiveUtils
# ╔═╡ 86f2c937-f0f8-4fad-8607-e7b03fa1b46e
using ModelingToolkit, OrdinaryDiffEq, Plots
# ╔═╡ ae4f5b02-ab7d-46e3-ac8c-8fd1fe63ae51
using ModelingToolkit: t_nounits as t, D_nounits as D
# ╔═╡ 1affc344-b742-4e0d-8a21-9dfd46055f98
using LinearAlgebra
# ╔═╡ a298cb80-38f8-4673-8f85-2c00046027a5
using Symbolics
# ╔═╡ bd8835ff-6fad-427a-8042-47d5b1f08b43
md"""
# Understanding Difficult Differential-Algebraic Equations:
#### DAE Index Reduction and Dummy Derivatives
From the acausal modeling example we can see that the underlying foundation of our symbolic-numeric system is not ODEs but differential-algebraic equations, DAEs, or models which include both differential equations and equality relations. In some cases, these DAEs can be simplified down to just ODEs. However, that's not always the case, in some cases the equality relations need to be kept as part of the system definition. That's when we have a fundamental DAE.
ModelingToolkit and DifferentialEquations.jl are made to be agnostic to the underlying form of the solve. If the system is a DAE, it will automatically generate a mass matrix representation of the DAE system, and then choose and appropriate solver via the automated solver algorithm. However, to better use the system, we need to understand a little bit about what the transformations are, what they are doing, and how to better interact with them.
## Our DAE Test Case: Cartesian Pendulum
For our test case for understanding how the DAE tooling works, we will use the cartesian pendulum problem. This is not middle school pendulum where you assumed sin(θ) is approximately θ when the angle is small, we want to represent the whole thing!
The full set of equations is:
"""
# ╔═╡ 70b2cf2f-4f37-40e5-ac19-65f59598258e
@parameters g=1 # normalize gravity to 1
# ╔═╡ 058e5d29-1444-4e27-8c0e-401172fb3d75
@variables x(t) y(t) [state_priority = 10] λ(t)
# ╔═╡ ac027c64-4f39-46e1-950c-207f87ec72b6
eqs = [D(D(x)) ~ λ * x
D(D(y)) ~ λ * y - g
0 ~ x^2 + y^2 - 1]
# ╔═╡ 0058790e-8ef4-4bf0-ab8e-0a5a8b65914b
md"""
Notice that these are second order expressions of the harmonic oscillator in x and y, where there is an another state variable λ for the tension of the rod. Additionally there is an algebraic expression for the constraint on the rod living on the unit circle.
If we build the system, we will notice something immediately peculiar:
"""
# ╔═╡ e5250569-ad65-45f5-8fd1-a153bbeeea40
@mtkbuild pend = ODESystem(eqs, t)
# ╔═╡ 564d8d51-882e-4319-8a10-691a7660ebb7
md"""
So that's the cartesian pendulum right there, QED. You understand that, right? Okay, there was a big jump there. How did we go from our second order expressions to this set of expressions, and why would we solve this set of equations? Let's look at this step by step.
Most numerical ODE solvers cannot solve higher order ODE systems, so instead they need to be lowered to first order systems. This is trivially done by defining a new variable that is equal to the derivative, i.e. D(x) = v. This means D(v) = D(D(x)), and thus we can solve the system D(x) = v, D(v) = D(D(x)), substituing the second derivative expression, and get a two variable system that is in first order form. This can be automated via the `ode_order_lowering` method:
"""
# ╔═╡ 3a003532-1b86-48a6-b439-cd5210e3d4d7
@named pend_nosimp = ODESystem(eqs, t)
# ╔═╡ 716b1138-af82-4db9-a41b-677650cb462d
lowersys = complete(ode_order_lowering(pend_nosimp))
# ╔═╡ bd745912-d04b-4a56-aa65-77a3ae3151b3
md"""
Now `ode_order_lowering` is generally not used for system simulation. Why? Let's try to solve this system:
"""
# ╔═╡ 31fee3ef-fc98-4c82-bc79-6881ca7156bd
odelowerprob = ODEProblem(lowersys, [x => 1, y => 0, D(x) => 0, D(y) => 0, λ => 0], (0.0, 1.5), [g => 1])
# ╔═╡ 200f6915-3fd3-4352-b658-995c09d296b2
loweredsol = solve(odelowerprob, Rodas5P())
# ╔═╡ eb514615-6a7a-4f39-b176-e993207d0d1d
loweredsol2 = solve(odelowerprob, FBDF())
# ╔═╡ 2a36b693-472b-47f0-9355-d6b0129156a5
loweredsol.retcode
# ╔═╡ daf6d7fb-0dd2-41b6-95f1-2b407ac48458
plot(loweredsol)
# ╔═╡ 985bbf12-ba5f-4d92-8a80-bce300b5c7fc
md"""
All of our solvers seem broken, what is going on? The issue can be found by looking at the Jacobian:
"""
# ╔═╡ c44b41e8-30e2-4b3e-bfab-69bf803af063
@variables γ
# ╔═╡ 098a4838-a4c0-443b-ba92-4f0982f7e3d0
stepjac = [1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 0] - γ*calculate_jacobian(lowersys)
# ╔═╡ cfba8a64-329a-49c6-b44b-1ef0be98655f
det(stepjac)
# ╔═╡ 4cf3c74f-f31c-4147-8998-c22754daf27c
substitute(det(stepjac), [γ=>0])
# ╔═╡ 9d830db3-d4f6-4e8c-86f4-7f6137973d9a
md"""
Okay, that needs some explantion. `stepjac` is the Jacobian of the implicit solver in a general time stepping method. When that method is for example Implicit Euler, then `gamma = dt`, but generally it's some constant times `dt`. The point is, this means that as `dt -> 0`, this has the property that the Jacobain becomes singular, and thus the Newton system is not solvable/convergent as `dt -> 0`. This means that the DAE solver is not generally convergent on these types of equations... oh no!
It turns out that DAEs can come in many flavors, and these flavors are known as the index of the DAE. An ODE is an index 0 DAE. An index 1 DAE has the property that all algebraic equations have some of the algebraic variables, and that there's a matching of each algebraic equation to a unique algebraic variable. The point here is that remember we have 5 equations:
"""
# ╔═╡ 522219fb-9efd-4414-95c5-32f1aac1a225
equations(lowersys)
# ╔═╡ 92883016-b30e-4909-a6ef-3d43325793cb
md"""
and we have 5 state variables. We have equations for how x evolves, D(x) evolves, y evolves, D(y) evolves, and a constraint equation. The constraint equation implicitly tells us what the tension λ should be, but the tension variable does not actually show up in the constraint equation. This is the cause of the singularity, and the lack of this property means that it's higher than index one.
The differential index is the number of times the DAE needs to be differentiated in order to become an ODE. Let's see this in action:
"""
# ╔═╡ 60a30f6a-906e-49d6-8e26-41cca46fce06
Symbolics.derivative(x^2 + y^2 - 1, t)
# ╔═╡ 8aa4b09b-ecb9-4555-8370-046b75d3d69e
md"""
And we substitute:
"""
# ╔═╡ bdc3da66-9339-4fc3-8146-2a97b3e66d5d
substitute(Symbolics.derivative(x^2 + y^2 - 1, t), [D(x) => unknowns(lowersys)[1], D(D(x)) => λ * x, D(y) => unknowns(lowersys)[2], D(D(y)) => λ * y - g])
# ╔═╡ b2e711c4-a717-473c-b6b7-2a17a83db15c
md"""
That's still not index 1 because it doesn't have λ still. So let's differentiate again:
"""
# ╔═╡ 97caddef-020c-4ed6-bf2f-a097db91021c
Symbolics.derivative(Symbolics.derivative(x^2 + y^2 - 1, t), t)
# ╔═╡ f7a6a8d7-919c-4529-a815-8824c6b571a2
doublediff = substitute(Symbolics.derivative(Symbolics.derivative(x^2 + y^2 - 1, t), t), [D(x) => unknowns(lowersys)[1], D(D(x)) => λ * x, D(y) => unknowns(lowersys)[2], D(D(y)) => λ * y - g])
# ╔═╡ 34e3e5dc-fb23-4d0d-bd41-e3d5c3071176
md"""
It took differentiating this expression twice and substituting to get an expression with λ, and thus the cartesian pendulum is an index-3 system! This also shows us the equivalent set of equations we could solve that's the index-1 system!
"""
# ╔═╡ ec65dc23-34b7-402f-8dc5-d23a0a0460e6
neweqs = [D(D(x)) ~ λ * x
D(D(y)) ~ λ * y - g
0 ~ doublediff]
# ╔═╡ 60114fc2-6082-4ac8-969d-adb0f496bb70
@named pend_doublediff = ODESystem(neweqs, t)
# ╔═╡ f039750d-b0ce-423f-b9bc-f2b95209317e
lowered_doublediff = complete(ode_order_lowering(pend_doublediff))
# ╔═╡ a00957dc-7ff9-4bf1-928f-62461f490bef
odelowerprob_doublediff = ODEProblem(lowered_doublediff, [x => 1, y => 0, D(x) => 0, D(y) => 0, λ => 0], (0.0, 5.0), [g => 1])
# ╔═╡ a1224a0f-d762-4c50-8255-5d14e898db2c
loweredsol_doublediff = solve(odelowerprob_doublediff, Rodas5P())
# ╔═╡ 8589626c-11fa-4a10-b958-176b166f6a64
plot(loweredsol_doublediff, idxs = (x,y))
# ╔═╡ 45512fc4-b4ae-48b0-9cdd-478e9caca435
md"""
And there you go, plotting x vs y we see that the solution is clearly a pendulum. That's all you had to do, figure out how to change your system and then it would solve well... and okay that's why you want to use ModelingToolkit to handle DAEs instead of writing them yourself. QED. We're done, let's go home.
"""
# ╔═╡ c626ca75-af76-45e2-b5b0-efb4dc8714c9
md"""
## But wait... structural simplify gave us different equations?
If you look at the equations we got from the standard `@mtkbuild`, you will notice that this is not the system of equations it gives you:
"""
# ╔═╡ 58fc7eea-910d-4f33-912d-826da5b2d217
equations(pend)
# ╔═╡ 3eaba3a3-726d-4ee8-9c86-18b8991653d4
md"""
What is going on? Where did our other equations for D(y) and D(D(y)) go? I thought we understood the ODE order lowering, that was the easy part?
It turns out that the transformations we did here were not always a great idea. In particular, let's see what happens if we solve this version of the pendulum over a long time span:
"""
# ╔═╡ feab8583-7853-4c04-aec1-b5e589dff0dd
odelowerprob_doublediff2 = ODEProblem(lowered_doublediff, [x => 1, y => 0, D(x) => 0, D(y) => 0, λ => 0], (0.0, 100.0), [g => 1])
# ╔═╡ 04a0f3fe-8bc9-42c7-9b59-bb3b80e7bdf5
loweredsol_doublediff2 = solve(odelowerprob_doublediff2, Rodas5P())
# ╔═╡ 4def74e5-abea-4a02-80dd-e63719904e23
plot(loweredsol_doublediff2, idxs = (x,y))
# ╔═╡ 4a2889de-f3fb-4e6e-bc29-eeea247be687
md"""
Notice that over time the solution drifts away from having x^2 + y^2 = 1, since we removed that equation! We only have that the second derivative of our constraint is zero, but not that it or its first derivative are zero. So so in order to enforce that our solution truly lives on the manifold, we need to recover these "implicit constraints". When doing so, we need to delete some redundent equations, because now we have 2 more constraint equations, we don't need two of the differntial equations...
etc. etc. etc. and so we end up deleting two of the equations and get more constraint equations, which is why a DAE of index 3 adds 2 two constraints and deletes two differential equations that it understands are redundant.
"""
# ╔═╡ 217fd362-bc89-4a06-9bf7-385ac16e9975
equations(pend)
# ╔═╡ 64c8904d-1e80-4121-b1cb-859ff0309874
prob_ss = ODEProblem(pend, [x => 1, y => 0], (0.0, 1.5), [g => 1], guesses = [λ => 1])
# ╔═╡ a79b15c6-1da3-4abe-9dac-a7bee2b77372
sol_ss = solve(prob_ss)
# ╔═╡ ed61d13e-21a5-4ae8-8f14-da4f041eced1
plot(sol_ss, idxs = (x,y))
# ╔═╡ 7257f24e-6502-485a-bd81-2e5d8d625686
sol[x^2 + y^2]
# ╔═╡ 7c75b89c-7c74-4b45-9a8a-31781d82def7
md"""
And there you go, a pendulum solution which is satisfying the constraints exactly.
But... there's other issues. Etc.
## Conclusion
Let ModelingToolkit fix your DAE equations for you. This is a very deep topic!
"""
# ╔═╡ 00000000-0000-0000-0000-000000000001
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Symbolics = "0c5d862f-8b57-4792-8d23-62f2024744c7"
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Plots = "~1.40.5"
Symbolics = "~5.33.0"
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