Solvers
For a beginner-friendly guide focused on solving common types of equations, refer to Solving
julia> @syms x, y, z
(x, y, z)
Expand for Python example
>>> from sympy import *
>>> x, y, z = symbols('x y z')
>>> init_printing(use_unicode=True)
A Note about Equations
Recall from the gotchas section of this tutorial that symbolic equations in SymPy are not represented by =
or ==
, but by Eq
.
The infix operator ~
is an elternative to Eq
.
julia> x ~ y
x = y
Expand for Python example
>>> Eq(x, y)
x = y
However, there is an even easier way. In SymPy, any expression not in an Eq
is automatically assumed to equal 0 by the solving functions. Since a = b
if and only if a - b = 0
, this means that instead of using x == y
, you can just use x - y
. For example
julia> solveset(x^2 ~ 1, x)
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 2 elements: 1 -1
julia> solveset(x^2 - 1 ~ 0, x)
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 2 elements: 1 -1
julia> solveset(x^2 - 1, x)
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 2 elements: 1 -1
Expand for Python example
>>> solveset(Eq(x**2, 1), x)
{-1, 1}
>>> solveset(Eq(x**2 - 1, 0), x)
{-1, 1}
>>> solveset(x**2 - 1, x)
{-1, 1}
This is particularly useful if the equation you wish to solve is already equal to 0. Instead of typing solveset(Eq(expr, 0), x)
, you can just use solveset(expr, x)
.
Solving Equations Algebraically
The main function for solving algebraic equations is solveset
. The syntax for solveset
is solveset(equation, variable=None, domain=S.Complexes)
Where equations
may be in the form of Eq
instances or expressions that are assumed to be equal to zero.
Please note that there is another function called solve
which can also be used to solve equations. The syntax is solve(equations, variables)
However, it is recommended to use solveset
instead.
When solving a single equation, the output of solveset
is a FiniteSet
or an Interval
or ImageSet
of the solutions.
Finite sets are turned into Set
containers in Julia
.
The exported 𝑆
objects mirror S
from Python. Using sympy.S
fails, as the underlying object has a __call__
method, so sympy.S
is a callable function, not an object. The construction ↓(sympy).S
works, but is cumbersome and \itS[tab]
seems easy enough to enter.
julia> solveset(x^2 - x, x)
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 2 elements: 0 1
julia> solveset(x - x, x, domain= 𝑆.Reals)
ℝ
julia> solveset(sin(x) - 1, x, domain= 𝑆.Reals)
⎧ π │ ⎫ ⎨2⋅n⋅π + ─ │ n ∊ ℤ⎬ ⎩ 2 │ ⎭
Expand for Python example
>>> solveset(x**2 - x, x)
{0, 1}
>>> solveset(x - x, x, domain=S.Reals)
ℝ
>>> solveset(sin(x) - 1, x, domain=S.Reals)
⎧ π │ ⎫
⎨2⋅n⋅π + ─ │ n ∊ ℤ⎬
⎩ 2 │ ⎭
If there are no solutions, an EmptySet
is returned and if it is not able to find solutions then a ConditionSet
is returned.
julia> solveset(exp(x), x)
∅
julia> solveset(cos(x) - x, x)
{x │ x ∊ ℂ ∧ (-x + cos(x) = 0)}
Expand for Python example
>>> solveset(exp(x), x) # No solution exists
∅
>>> solveset(cos(x) - x, x) # Not able to find solution
{x │ x ∊ ℂ ∧ (-x + cos(x) = 0)}
In the solveset
module, the linear system of equations is solved using linsolve
. In future we would be able to use linsolve directly from solveset
. Following is an example of the syntax of linsolve
.
- List of Equations Form:
julia> linsolve([x + y + z - 1, x + y + 2*z - 3 ], (x, y, z))
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 1 element: (-y - 1, y, 2)
Expand for Python example
>>> linsolve([x + y + z - 1, x + y + 2*z - 3 ], (x, y, z))
{(-y - 1, y, 2)}
- Augmented Matrix Form:
We paas in a symbolic matrix to linsolve
. The variables are passed as a tuple or as, in this case, three variables, but not as a vector. (The ↓
conversion of a vector creates `n\times 1
matrix, not a list, as is expected by the underlying function.)
julia> linsolve(Sym[ 1 1 1 1; 1 1 2 3], (x, y, z))
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 1 element: (-y - 1, y, 2)
Expand for Python example
>>> linsolve(Matrix(([1, 1, 1, 1], [1, 1, 2, 3])), (x, y, z))
{(-y - 1, y, 2)}
- A*x = b Form
julia> M = Sym[ 1 1 1 1; 1 1 2 3]
2×4 Matrix{SymPyCore.Sym}: 1 1 1 1 1 1 2 3
julia> system = A, b = M[:, 1:end-1], M[:, end]
(SymPyCore.Sym[1 1 1; 1 1 2], SymPyCore.Sym[1, 3])
julia> linsolve(system, x, y, z)
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 1 element: (-y - 1, y, 2)
Expand for Python example
>>> M = Matrix(((1, 1, 1, 1), (1, 1, 2, 3)))
>>> system = A, b = M[:, :-1], M[:, -1]
>>> linsolve(system, x, y, z)
{(-y - 1, y, 2)}
The order of solution corresponds the order of given symbols.
In the solveset
module, the non linear system of equations is solved using nonlinsolve
. Following are examples of nonlinsolve
.
- When only real solution is present:
We can pass in equations using a tuple or a vector, but the variables are passed as a tuple or individually, not within a vector.
julia> @syms a::real, b::real, c::real, d::real
(a, b, c, d)
julia> nonlinsolve( (a^2 + a, a - b), (a, b))
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 2 elements: (-1, -1) (0, 0)
julia> nonlinsolve((x*y - 1, x - 2), x, y)
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 1 element: (2, 1/2)
Expand for Python example
>>> a, b, c, d = symbols('a, b, c, d', real=True)
>>> nonlinsolve([a**2 + a, a - b], [a, b])
{(-1, -1), (0, 0)}
>>> nonlinsolve([x*y - 1, x - 2], x, y)
{(2, 1/2)}
- When only complex solution is present:
Again, we use tuples, not vectors, to pass in the variables
julia> nonlinsolve((x^2 + 1, y^2 + 1), (x, y))
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 4 elements: (I, I) (-I, I) (I, -I) (-I, -I)
Expand for Python example
>>> nonlinsolve([x**2 + 1, y**2 + 1], [x, y])
{(-ⅈ, -ⅈ), (-ⅈ, ⅈ), (ⅈ, -ⅈ), (ⅈ, ⅈ)}
- When both real and complex solution are present:
julia> system = (x^2 - 2*y^2 -2, x*y - 2)
(x^2 - 2*y^2 - 2, x*y - 2)
julia> vars = (x, y)
(x, y)
julia> nonlinsolve(system, vars)
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 4 elements: (-sqrt(2)*I, sqrt(2)*I) (sqrt(2)*I, -sqrt(2)*I) (-2, -1) (2, 1)
julia> system = (exp(x) - sin(y), 1/y - 3)
(exp(x) - sin(y), -3 + 1/y)
julia> nonlinsolve(system, vars)
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 1 element: (ImageSet(Lambda(_n, 2*_n*I*pi + log(sin(1/3))), Integers), 1/3)
Expand for Python example
>>> from sympy import sqrt
>>> system = [x**2 - 2*y**2 -2, x*y - 2]
>>> vars = [x, y]
>>> nonlinsolve(system, vars)
{(-2, -1), (2, 1), (-√2⋅ⅈ, √2⋅ⅈ), (√2⋅ⅈ, -√2⋅ⅈ)}
>>> system = [exp(x) - sin(y), 1/y - 3]
>>> nonlinsolve(system, vars)
{({2⋅n⋅ⅈ⋅π + log(sin(1/3)) │ n ∊ ℤ}, 1/3)}
- When the system is positive-dimensional system (has infinitely many solutions):
julia> nonlinsolve((x*y, x*y - x), (x, y))
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 1 element: (0, y)
julia> system = (a^2 + a*c, a - b)
(a^2 + a*c, a - b)
julia> nonlinsolve(system, (a, b))
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 2 elements: (0, 0) (-c, -c)
Expand for Python example
>>> nonlinsolve([x*y, x*y - x], [x, y])
{(0, y)}
>>> system = [a**2 + a*c, a - b]
>>> nonlinsolve(system, [a, b])
{(0, 0), (-c, -c)}
Notes:
The order of solution corresponds the order of given symbols.
Currently
nonlinsolve
doesn't return solution in form ofLambertW
(if there
is solution present in the form of LambertW
).
solve
can be used for such cases:
>>> solve([x**2 - y**2/exp(x)], [x, y], dict=True)
⎡⎧ ____⎫ ⎧ ____⎫⎤
⎢⎨ ╱ x ⎬ ⎨ ╱ x ⎬⎥
⎣⎩y: -x⋅╲╱ ℯ ⎭, ⎩y: x⋅╲╱ ℯ ⎭⎦
>>> solve(x**2 - y**2/exp(x), x, dict=True)
⎡⎧ ⎛-y ⎞⎫ ⎧ ⎛y⎞⎫⎤
⎢⎨x: 2⋅W⎜───⎟⎬, ⎨x: 2⋅W⎜─⎟⎬⎥
⎣⎩ ⎝ 2 ⎠⎭ ⎩ ⎝2⎠⎭⎦
- Currently
nonlinsolve
is not properly capable of solving the system of equations
having trigonometric functions.
solve
can be used for such cases (but does not give all solution):
>>> solve([sin(x + y), cos(x - y)], [x, y])
⎡⎛-3⋅π 3⋅π⎞ ⎛-π π⎞ ⎛π 3⋅π⎞ ⎛3⋅π π⎞⎤
⎢⎜─────, ───⎟, ⎜───, ─⎟, ⎜─, ───⎟, ⎜───, ─⎟⎥
⎣⎝ 4 4 ⎠ ⎝ 4 4⎠ ⎝4 4 ⎠ ⎝ 4 4⎠⎦
solveset
reports each solution only once. To get the solutions of a polynomial including multiplicity use roots
.
The roots
function needs qualification
julia> solveset(x^3 - 6*x^2 + 9*x, x)
Set{SymPyCore.Sym{PythonCall.Core.Py}} with 2 elements: 0 3
julia> sympy.roots(x^3 - 6*x^2 + 9*x, x)
Dict{SymPyCore.Sym{PythonCall.Core.Py}, SymPyCore.Sym{PythonCall.Core.Py}} with 2 entries: 0 => 1 3 => 2
Expand for Python example
>>> solveset(x**3 - 6*x**2 + 9*x, x)
{0, 3}
>>> roots(x**3 - 6*x**2 + 9*x, x)
{0: 1, 3: 2}
The output {0: 1, 3: 2}
of roots
means that 0
is a root of multiplicity 1 and 3
is a root of multiplicity 2.
Note:
Currently solveset
is not capable of solving the following types of equations:
- Equations solvable by LambertW (Transcendental equation solver).
solve
can be used for such cases:
>>> solve(x*exp(x) - 1, x )
[W(1)]
Solving Differential Equations
To solve differential equations, use dsolve
. First, create an undefined function by passing cls=Function
to the symbols
function.
We can use @syms
to define symbolic functions
julia> @syms f() g()
(f, g)
Expand for Python example
>>> f, g = symbols('f g', cls=Function)
f
and g
are now undefined functions. We can call f(x)
, and it will represent an unknown function.
julia> f(x)
f(x)
Expand for Python example
>>> f(x)
f(x)
Derivatives of f(x)
are unevaluated.
julia> diff(f(x), x)
d ──(f(x)) dx
Expand for Python example
>>> f(x).diff(x)
d
──(f(x))
dx
(see the `Derivatives section for more on derivatives).
To represent the differential equation f''(x) - 2f'(x) + f(x) = \sin(x)
, we would thus use
We could use similar notation, as within Python, e.g., f(x).diff(x)
, but we show the use of Differential
which hides some repetition
julia> ∂ = Differential(x)
SymPyCore.Differential(x, 1)
julia> diffeq = ∂(∂(f(x))) - 2 * ∂(f(x)) + f(x) ~ sin(x)
2 d d f(x) - 2⋅──(f(x)) + ───(f(x)) = sin(x) dx 2 dx
Expand for Python example
>>> diffeq = Eq(f(x).diff(x, x) - 2*f(x).diff(x) + f(x), sin(x))
>>> diffeq
2
d d
f(x) - 2⋅──(f(x)) + ───(f(x)) = sin(x)
dx 2
dx
To solve the ODE, pass it and the function to solve for to dsolve
.
julia> dsolve(diffeq, f(x))
x cos(x) f(x) = (C₁ + C₂⋅x)⋅ℯ + ────── 2
Expand for Python example
>>> dsolve(diffeq, f(x))
x cos(x)
f(x) = (C₁ + C₂⋅x)⋅ℯ + ──────
2
dsolve
returns an instance of Eq
. This is because, in general, solutions to differential equations cannot be solved explicitly for the function.
julia> dsolve(∂(f) * (1 - sin(f(x))), f(x))
f(x) = C₁
Expand for Python example
>>> dsolve(f(x).diff(x)*(1 - sin(f(x))) - 1, f(x))
x - f(x) - cos(f(x)) = C₁
The arbitrary constants in the solutions from dsolve are symbols of the form C1
, C2
, C3
, and so on.
The ics
argument of dsolve
allows a dictionary to be passed to specify initiial conditions.