# Using SciML Symbolics to Solve Perturbation Problems

## Background

**Symbolics.jl** is a fast and modern Computer Algebra System (CAS) written in the Julia Programming Language. It is an integral part of the SciML ecosystem of differential equation solvers and scientific machine learning packages. While **Symbolics.jl** is primarily designed for modern scientific computing (e.g., auto-differentiation, machine learning), it is a powerful CAS and can also be useful for *classic* scientific computing. One such application is using the *perturbation* theory to solve algebraic and differential equations.

Perturbation methods are a collection of techniques to solve problems that generally don't have a closed solution but depend on a tunable parameter and have closed-form or easy solutions for some values of the parameter. The main idea is to assume a solution as a power series in the tunable parameter (say ๐), such that ๐ = 0 corresponds to an easy solution.

We will discuss the general steps of the perturbation methods in four examples below. One hallmark of the perturbation method is the generation of long and convoluted intermediate equations, which are subjected to algorithmic and mechanical manipulations. Therefore, these problems are well suited for CAS. In fact, CAS softwares have been used to help with the perturbation calculations since the 1950s.

In this tutorial our goal is to show how to use Julia and **Symbolics.jl** to solve simple perturbation problems. The code for the for examples (`test_*`

functions) and the helper functions are in `src/perturb.jl`

.

## Solving the Quintic

We start with the "hello world!" analog of the perturbation problems: solving the quintic (fifth-order) equations. We want to find a real valued ๐ฅ such that ๐ฅโต + ๐ฅ = 1. According to the Abel's theorem, a general quintic equation does not have a closed form solution. Of course, we can easily solve this equation numerically; for example, by using the Newton's method. We use the following implementation of the Newton's method:

```
using Symbolics, SymbolicUtils
function solve_newton(f, x, xโ; abstol=1e-8, maxiter=50)
xโ = Float64(xโ)
fโโโ = x - f / Symbolics.derivative(f, x)
for i = 1:maxiter
xโโโ = substitute(fโโโ, Dict(x => xโ))
if abs(xโโโ - xโ) < abstol
return xโโโ
else
xโ = xโโโ
end
end
return xโโโ
end
```

In this code, `Symbolics.derivative(eq, x)`

does exactly what it names implies: it calculates the symbolic derivative of `eq`

(a **Symbolics.jl** expression) with respect to `x`

(a **Symbolics.jl** variable). We use `Symbolics.substitute(eq, D)`

to evaluate the update formula by substituting variables or sub-expressions (defined as a dictionary `D`

) in `eq`

. It should be noted that `substitute`

is the workhorse of our code and will be used multiple times in the rest of this tutorial. `solve_newton`

is written with simplicity and clarity in mind and not performance but suffices for our purpose.

Let's go back to our quintic. We can define a Symbolics variable as `@variables x`

and then solve the equation `solve_newton(x^5 + x - 1, x, 1.0)`

(here, `xโ = 0`

is our first guess). The answer is `x = 0.7549`

. Now, let's see how we can solve this problem using the perturbation method.

We introduce a tuning parameter ๐ into our equation: ๐ฅโต + ๐ฅ = 1. If ๐ = 1, we get our original problem. For ๐ = 0, the problem transforms to an easy one: ๐ฅโต = 1 which has a solution ๐ฅ = 1 (and four complex solutions which we ignore here). We expand ๐ฅ as a power series on ๐:

๐ฅ(๐) = ๐โ + ๐โ๐ + ๐โ๐ยฒ + ๐(๐ยณ)

๐โ is the solution of the easy equation, therefore ๐โ = 1. Substituting into the original problem,

(1 + ๐โ๐ + ๐โ๐ยฒ)โต + ๐ (1 + ๐โ๐ + ๐โ๐ยฒ) - 1 = 0

Expanding the equations, we get

๐ (1 + 5๐โ) + ๐ยฒ (๐โ + 5๐โ + 10๐โยฒ) + ๐(๐ยณ) = 0

This equation should hold for each power of ๐. Therefore,

1 + 5๐โ = 0,

and

๐โ + 5๐โ + 10๐โยฒ = 0.

This system of equations does not initially seem to be linear because of the presence of terms like 10๐โยฒ, but upon closer inspection is found to be in fact linear (this is a feature of the permutation method). In addition, the system is in a triangular form, meaning the first equation depends only on ๐โ, the second one on ๐โ and ๐โ, such that we can replace the result of ๐โ from the first one into the second equation and remove the non-linear term. We solve the first equation to get ๐โ = -1/5. Substituting in the second one and solve for ๐โ:

๐โ = (-1/5 + 10(-(1/5)ยฒ) / 5 = -1/25

Finally,

๐ฅ(๐) = 1 - ๐ / 5 - ๐ยฒ / 25 + ๐(๐ยณ)

Solving the original problem, ๐ฅ(1) = 0.76, compared to 0.75487767 calculated numerically. We can improve the accuracy by including more terms in the expansion of ๐ฅ. However, the calculations, while straightforward, become messy and intractable to do manually very quickly. This is why a CAS is very helpful to solve perturbation problems.

Now, let's see how we can do these calculations in Julia (see `test_quintic`

function). Let `n = 2`

be the order of the expansion. We start by defining the symbolic variables:

` @variables @variables ฯต a[1:n] `

Then, we define `x = 1 + a[1]*ฯต + a[2]*ฯต^2`

. Note that in `test_quintic`

we use the helper function `def_taylor`

to define `x`

by calling it as `x = def_taylor(ฯต, a, 1)`

, where the arguments are the expansion variable, an array of parameters, and the constant term. The next step is to substitute `x`

in the problem equation `eq = x^5 + ฯต*x - 1`

. Now, `eq`

is

` ฯต*(1 + aโ*ฯต + aโ*(ฯต^2)) + (1 + aโ*ฯต + aโ*(ฯต^2))^5 - 1`

Or in the expanded form (calculated as `expand(eq)`

):

```
ฯต + aโ*(ฯต^2) + aโ*(ฯต^3) + (aโ^5)*(ฯต^5) + (aโ^5)*(ฯต^10) + 5aโ*ฯต + 5aโ*(ฯต^2) +
10(aโ^2)*(ฯต^2) + 10(aโ^3)*(ฯต^3) + 5(aโ^4)*(ฯต^4) + 10(aโ^2)*(ฯต^4) +
10(aโ^3)*(ฯต^6) + 5(aโ^4)*(ฯต^8) + 20aโ*aโ*(ฯต^3) + 30aโ*(aโ^2)*(ฯต^5) +
20aโ*(aโ^3)*(ฯต^7) + 5aโ*(aโ^4)*(ฯต^9) + 30aโ*(aโ^2)*(ฯต^4) + 20aโ*(aโ^3)*(ฯต^5) +
5aโ*(aโ^4)*(ฯต^6) + 30(aโ^2)*(aโ^2)*(ฯต^6) + 10(aโ^2)*(aโ^3)*(ฯต^8) +
10(aโ^3)*(aโ^2)*(ฯต^7)
```

We need a way to get the coefficients of different powers of ๐. Function `collect_powers(eq, x, ns)`

returns the powers of variable `x`

in expression `eq`

. Argument `ns`

is the range of the powers.

```
function collect_powers(eq, x, ns; max_power=100)
eq = substitute(expand(eq), Dict(x^j => 0 for j=last(ns)+1:max_power))
eqs = []
for i in ns
powers = Dict(x^j => (i==j ? 1 : 0) for j=1:last(ns))
push!(eqs, substitute(eq, powers))
end
eqs
end
```

For example, `collect_powers(eq, ฯต, 1:2)`

means the coefficients of `ฯต`

and `ฯต^2`

in `eq`

and returns a list of expressions `[1 + 5aโ, aโ + 5aโ + 10(aโ^2)]`

. We assign this list to `eqs`

.

`collect_powers`

uses `substitute`

to find the coefficient of a given power of `x`

by passing a `Dict`

with all powers of `x`

set to 0, except the target power which is set to 1. The following experssion returns the coefficient of `ฯต^2`

in `eq`

,

```
substitute(expand(eq), Dict(
ฯต => 0,
ฯต^2 => 1,
ฯต^3 => 0,
ฯต^4 => 0,
ฯต^5 => 0,
ฯต^6 => 0,
ฯต^7 => 0,
ฯต^8 => 0)
)
```

Back to our problem. Having the coefficients of the powers of `ฯต`

, we can set each equation to 0 and solve the system of linear equations to find the numerical values of the coefficients. **Symbolics.jl** has a function `Symbolics.solve_for`

that can solve systems of linear equations. We can start by solving `eqs[1]`

for `aโ`

and then substitute this in `eqs[2]`

and solve for `aโ`

. This process is done by function `solve_coef(eqs, ps)`

:

```
function solve_coef(eqs, ps)
vals = Dict()
for i = 1:length(ps)
eq = substitute(eqs[i], vals)
vals[ps[i]] = Symbolics.solve_for(eq ~ 0, ps[i])
end
vals
end
```

Here, `eqs`

is an array of expressions (assumed to be equal to 0) and `ps`

is an array of variables. The result is a dictionary of *variable* => *value* pairs. For example, `solve_coef(eqs, a)`

returns `Dict(aโ => -0.2, aโ => -0.04)`

. Substituting back in the definition of `x`

, i.e., `x = 1 + a[1]*ฯต + a[2]*ฯต^2`

, we obtain `X = ฯต -> 1 - 0.2*ฯต - 0.04*ฯต^2`

. Therefore, the solution to our original problem becomes `X(1)`

, which is equal to 0.76.

We can use larger values of `n`

to improve the accuracy of estimations:

n | x |
---|---|

1 | 0.8 |

2 | 0.76 |

3 | 0.752 |

4 | 0.752 |

5 | 0.7533 |

6 | 0.7543 |

7 | 0.7548 |

8 | 0.7550 |

Remember the numerical value is 0.7549.

The two functions `collect_powers`

and `solve_coef(eqs, a)`

are used in all the examples in this tutorial.

## Solving the Kepler's Equation

Historically, the perturbation methods were first invented to solve orbital calculations needed to calculate the orbit of the Moon and the planets. In homage to this history, our second example has a celestial theme. Our goal is solve the Kepler's equation:

๐ธ - ๐ sin(๐ธ) = ๐.

where ๐ is the *eccentricity* of the elliptical orbit, ๐ is the *mean anomaly*, and ๐ธ (unknown) is the *eccentric anomaly* (the angle between the position of a planet in an elliptical orbit and the point of periapsis). This equation is central to solving two-body Keplerian orbits. We want to find a function ๐ธ(๐; ๐).

Similar to the first example, it is easy to solve this problem using the Newton's method. For example, let ๐ = 0.01671 (the eccentricity of the Earth) and ๐ = ฯ/2. We have `solve_newton(x - e*sin(x) - M, x, M)`

equals to 1.5875 (compared to ฯ/2 = 1.5708). Now, we try to solve the same problem using the perturbation techniques (see function `test_kepler`

).

For ๐ = 0, we have ๐ธ = ๐. Therefore, we can use ๐ as our perturbation parameter. For consistency, we rename it to ๐. We start by defining the variables (assuming `n = 3`

):

```
@variables ฯต M a[1:n]
x = def_taylor(ฯต, n, M)
```

The problem equation is `eq = E - ฯต * sin(E) - M`

. We further simplify by substituting sin with its power series (using `expand_sin`

helper function):

sin(๐ธ) = ๐ฅ - ๐ฅยณ / 6 + ๐ฅโต / 120 - ๐ฅโท / 5040 + ๐(๐ฅโน).

We follow the same algorithm as before. We collect the coefficients of the powers of ๐:

```
eqs = collect_powers(eq, ฯต, 1:n)
```

and then solve for `a`

:

```
vals = solve_coef(eqs, a)
```

Finally, we substitute `vals`

back in `x`

to get a formula to calculate `E`

:

```
X = substitute(x, vals)
substitute(X, Dict(ฯต => 0.01671, M => ฯ/2))
```

The result is 1.5876, compared to the numerical value of 1.5875. It is customary to order `X`

based on the powers of `M`

instead of `ฯต`

. We can calculate this series as `collect_powers(sol, M, 0:3) `

. The result (after manual cleanup) is

```
E(M, ฯต) =
(1 + ฯต + ฯต^2 + ฯต^3)*M
- (ฯต + 4*ฯต^2 + 10*ฯต^3)*M^3/6
+ (ฯต + 16*ฯต^2 + 91*ฯต^3)*M^5/120
```

Comparing the formula to the one for ๐ธ in the Wikipedia article on the Kepler's equation:

The first deviation is in the coefficient of ๐ยณ๐โต.

## The Trajectory of a Ball!

In the first two examples, we applied the permutation method to algebraic problems. However, the main power of the permutation method is to solve differential equations (usually ODEs, but also occasionally PDEs). Surprisingly, the main procedure developed to solve algebraic problems works well for differential equations. In fact, we will use the same two helper functions, `collect_powers`

and `solve_coef`

. The main difference is in the way we expand the dependent variables. For algebraic problems, the coefficients of ๐ are constants; whereas, for differential equations, they are functions of the dependent variable (usually time).

For the first example on how to solve an ODE, we have chosen a simple and well-behaved problem. The problem is a variation of a standard first-year physics problem: what is the trajectory of an object (say, a ball or a rocket) thrown vertically at velocity ๐ฃ from the surface of a planet. Assuming a constant acceleration of gravity, ๐, every burgeoning physicist knows the answer: ๐ฅ(๐ก) = ๐ฅ(0) + ๐ฃ๐ก - ๐๐กยฒ/2. However, what happens if ๐ is not constant? Specifically, ๐ is inversely proportional to the distant from the center of the planet. If ๐ฃ is large, the assumption of constant gravity does not hold. However, unless ๐ฃ is large compared to the escape velocity, the correction is usually small. After simplifications and reframing in dimensionless variables, the problem becomes ๐ฅฬ(๐ก) = -(1 + ๐๐ฅ(๐ก))โปยฒ, assuming ๐ฅ(0) = 0, and ๐ฅฬ(0) = 1. Note that for ๐ = 0, this equation transforms back to the standard one.

Let's start with defining the variables

` @variables ฯต t y[1:n](t) โโy[1:n]`

Next, we define ๐ฅ (for `n = 3`

):

` x = y[1] + y[2]*ฯต + y[3]*ฯต^2`

We need the second derivative of `x`

. It may seem that we can do this using `Differential(t)`

; however, this operation needs to wait! Instead, we define dummy variables `โโy`

as the placeholder for the second derivatives and define

` โโx = โโy[1] + โโy[2]*ฯต + โโy[3]*ฯต^2`

as the second derivative of `x`

. After rearrangement, our governing equation is ๐ฅฬ(๐ก)(1 + ๐๐ฅ(๐ก))ยฒ + 1 = 0, or

` eq = โโx * (1 + ฯต*x)^2 + 1`

The next steps are the same as before (however, note that we pass `0:n-1`

to `collect_powers`

because the zeroth order term is needed here)

```
eqs = collect_powers(eq, ฯต, 0:n-1)
vals = solve_coef(eqs, โโy)
```

At this stage,

```
vals = Dict(
โโyโ => -1.0,
โโyโ => 2.0yโ(t),
โโyโ => 2.0yโ(t) - (3.0(yโ(t)^2))
)
```

Our system of ODEs is forming. Note the triangular form of the relationship. This is time to convert `โโ`

s to the correct **Symbolics.jl** form by substitution:

```
D = Differential(t)
subs = Dict(โโy[i] => D(D(y[i])) for i = 1:n)
eqs = [substitute(first(v), subs) ~ substitute(last(v), subs) for v in vals]
```

Now, `eqs`

becomes

```
[Differential(t)(Differential(t)(yโ(t))) ~ -1.0,
Differential(t)(Differential(t)(yโ(t))) ~ 2.0yโ(t),
Differential(t)(Differential(t)(yโ(t))) ~ 2.0yโ(t) - (3.0(yโ(t)^2))]
```

We are nearly there! From this point on, the rest is standard ODE solving procedures. Potentially we can use a symbolic ODE solver to find a closed form solution to this problem. However, **Symbolics.jl** currently does not support this functionality. Instead, we solve the problem numerically. We form an `ODESystem`

, lower the order (convert second derivatives to first), generate an `ODEProblem`

(after passing the correct initial conditions), and, finally, solve it.

```
using ModelingToolkit, DifferentialEquations
sys = ODESystem(eqs, t)
sys = ode_order_lowering(sys)
prob = ODEProblem(sys, [1.0, 0.0, 0.0, 0.0, 0.0, 0.0], (0, 5.0))
sol = solve(prob; dtmax=0.01)
```

The solution to the problem can be written as

` X = ฯต -> sol[y[1]] .+ sol[y[2]] * ฯต .+ sol[y[3]] * ฯต^2`

The following figure is generated by running

```
using Plots
plot(sol.t, hcat([X(ฯต) for ฯต = 0.0:0.1:0.5]...))
```

and shows the trajectories for a range of `ฯต`

:

As expected, the higher `ฯต`

is (meaning the gravity is less with altitude), the object goes higher and stays up for a longer duration. Of course, we could have solved the problem directly using as ODE solver. One of the benefits of the perturbation method is that we need to run the ODE solver only once and then can just calculate the answer for different values of `ฯต`

; whereas, if we were using the direct method, we needed to run the solver once for each value of `ฯต`

.

## A Weakly Nonlinear Oscillator

For our final example, we have chosen a simple example from a very important class of problems, the nonlinear oscillators. As we will see, perturbation theory has difficulty providing a good solution to this problem, but the process is instructive. This example follows closely chapter 7.6 of *Nonlinear Dynamics and Chaos* by Steven Strogatz.

The problem is to solve ๐ฅฬ(๐ก) + 2๐๐ฅฬ + ๐ฅ = 0, assuming ๐ฅ(0) = 0, and ๐ฅฬ(0) = 1. If ๐ = 0, the problem reduces to the simple linear harmonic oscillator with the exact solution ๐ฅ(t) = sin(๐ก). We follow the same steps as the previous example.

```
@variables ฯต t y[1:n](t) โy[1:n] โโy[1:n] # n = 3
x = y[1] + y[2]*ฯต + y[3]*ฯต^2
โx = โy[1] + โy[2]*ฯต + โy[3]*ฯต^2
โโx = โโy[1] + โโy[2]*ฯต + โโy[3]*ฯต^2
```

Note that now we also need the first derivative terms. Continuing,

```
eq = โโx + 2*ฯต*โx + x
eqs = collect_powers(eq, ฯต, 0:n-1)
vals = solve_coef(eqs, โโy)
```

Let's inspect `vals`

:

```
vals = Dict(
โโyโ => -yโ(t),
โโyโ => -2.0โyโ - yโ(t),
โโyโ => -2.0โyโ - yโ(t))
)
```

Next, we need to replace `โ`

s and `โโ`

s with their **Symbolics.jl** counterparts:

```
D = Differential(t)
subs1 = Dict(โy[i] => D(y[i]) for i = 1:n)
subs2 = Dict(โโy[i] => D(D(y[i])) for i = 1:n)
subs = subs1 โช subs2
eqs = [substitute(first(v), subs) ~ substitute(last(v), subs) for v in vals]
```

We continue with converting to an `ODEProblem`

, solving it, and finally plot the results against the exact solution to the original problem.

```
sys = ODESystem(eqs, t)
sys = ode_order_lowering(sys)
prob = ODEProblem(sys, [1.0, 0.0, 0.0, 0.0, 0.0, 0.0], (0, 50.0))
sol = solve(prob; dtmax=0.01)
T = sol.t
X = ฯต -> sol[y[1]] .+ sol[y[2]] * ฯต .+ sol[y[3]] * ฯต^2
Y = ฯต -> exp.(-ฯต*T) .* sin.(sqrt(1 - ฯต^2)*T) / sqrt(1 - ฯต^2) # exact solution
plot(sol.t, [Y(0.1), X(0.1)])
```

The result is (compare to Figure 7.6.2 in *Nonlinear Dynamics and Chaos*)

The two curves fit well for the first couple of cycles, but then the perturbation method curve diverges from the true solution. The main reason is that the problem has two or more time-scales.