894 lines
30 KiB
Python
894 lines
30 KiB
Python
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r"""
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This module contains :py:meth:`~sympy.solvers.ode.riccati.solve_riccati`,
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a function which gives all rational particular solutions to first order
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Riccati ODEs. A general first order Riccati ODE is given by -
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.. math:: y' = b_0(x) + b_1(x)w + b_2(x)w^2
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where `b_0, b_1` and `b_2` can be arbitrary rational functions of `x`
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with `b_2 \ne 0`. When `b_2 = 0`, the equation is not a Riccati ODE
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anymore and becomes a Linear ODE. Similarly, when `b_0 = 0`, the equation
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is a Bernoulli ODE. The algorithm presented below can find rational
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solution(s) to all ODEs with `b_2 \ne 0` that have a rational solution,
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or prove that no rational solution exists for the equation.
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Background
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==========
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A Riccati equation can be transformed to its normal form
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.. math:: y' + y^2 = a(x)
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using the transformation
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.. math:: y = -b_2(x) - \frac{b'_2(x)}{2 b_2(x)} - \frac{b_1(x)}{2}
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where `a(x)` is given by
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.. math:: a(x) = \frac{1}{4}\left(\frac{b_2'}{b_2} + b_1\right)^2 - \frac{1}{2}\left(\frac{b_2'}{b_2} + b_1\right)' - b_0 b_2
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Thus, we can develop an algorithm to solve for the Riccati equation
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in its normal form, which would in turn give us the solution for
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the original Riccati equation.
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Algorithm
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=========
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The algorithm implemented here is presented in the Ph.D thesis
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"Rational and Algebraic Solutions of First-Order Algebraic ODEs"
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by N. Thieu Vo. The entire thesis can be found here -
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https://www3.risc.jku.at/publications/download/risc_5387/PhDThesisThieu.pdf
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We have only implemented the Rational Riccati solver (Algorithm 11,
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Pg 78-82 in Thesis). Before we proceed towards the implementation
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of the algorithm, a few definitions to understand are -
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1. Valuation of a Rational Function at `\infty`:
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The valuation of a rational function `p(x)` at `\infty` is equal
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to the difference between the degree of the denominator and the
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numerator of `p(x)`.
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NOTE: A general definition of valuation of a rational function
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at any value of `x` can be found in Pg 63 of the thesis, but
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is not of any interest for this algorithm.
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2. Zeros and Poles of a Rational Function:
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Let `a(x) = \frac{S(x)}{T(x)}, T \ne 0` be a rational function
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of `x`. Then -
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a. The Zeros of `a(x)` are the roots of `S(x)`.
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b. The Poles of `a(x)` are the roots of `T(x)`. However, `\infty`
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can also be a pole of a(x). We say that `a(x)` has a pole at
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`\infty` if `a(\frac{1}{x})` has a pole at 0.
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Every pole is associated with an order that is equal to the multiplicity
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of its appearance as a root of `T(x)`. A pole is called a simple pole if
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it has an order 1. Similarly, a pole is called a multiple pole if it has
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an order `\ge` 2.
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Necessary Conditions
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====================
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For a Riccati equation in its normal form,
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.. math:: y' + y^2 = a(x)
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we can define
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a. A pole is called a movable pole if it is a pole of `y(x)` and is not
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a pole of `a(x)`.
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b. Similarly, a pole is called a non-movable pole if it is a pole of both
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`y(x)` and `a(x)`.
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Then, the algorithm states that a rational solution exists only if -
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a. Every pole of `a(x)` must be either a simple pole or a multiple pole
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of even order.
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b. The valuation of `a(x)` at `\infty` must be even or be `\ge` 2.
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This algorithm finds all possible rational solutions for the Riccati ODE.
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If no rational solutions are found, it means that no rational solutions
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exist.
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The algorithm works for Riccati ODEs where the coefficients are rational
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functions in the independent variable `x` with rational number coefficients
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i.e. in `Q(x)`. The coefficients in the rational function cannot be floats,
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irrational numbers, symbols or any other kind of expression. The reasons
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for this are -
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1. When using symbols, different symbols could take the same value and this
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would affect the multiplicity of poles if symbols are present here.
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2. An integer degree bound is required to calculate a polynomial solution
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to an auxiliary differential equation, which in turn gives the particular
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solution for the original ODE. If symbols/floats/irrational numbers are
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present, we cannot determine if the expression for the degree bound is an
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integer or not.
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Solution
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========
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With these definitions, we can state a general form for the solution of
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the equation. `y(x)` must have the form -
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.. math:: y(x) = \sum_{i=1}^{n} \sum_{j=1}^{r_i} \frac{c_{ij}}{(x - x_i)^j} + \sum_{i=1}^{m} \frac{1}{x - \chi_i} + \sum_{i=0}^{N} d_i x^i
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where `x_1, x_2, \dots, x_n` are non-movable poles of `a(x)`,
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`\chi_1, \chi_2, \dots, \chi_m` are movable poles of `a(x)`, and the values
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of `N, n, r_1, r_2, \dots, r_n` can be determined from `a(x)`. The
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coefficient vectors `(d_0, d_1, \dots, d_N)` and `(c_{i1}, c_{i2}, \dots, c_{i r_i})`
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can be determined from `a(x)`. We will have 2 choices each of these vectors
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and part of the procedure is figuring out which of the 2 should be used
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to get the solution correctly.
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Implementation
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==============
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In this implementation, we use ``Poly`` to represent a rational function
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rather than using ``Expr`` since ``Poly`` is much faster. Since we cannot
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represent rational functions directly using ``Poly``, we instead represent
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a rational function with 2 ``Poly`` objects - one for its numerator and
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the other for its denominator.
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The code is written to match the steps given in the thesis (Pg 82)
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Step 0 : Match the equation -
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Find `b_0, b_1` and `b_2`. If `b_2 = 0` or no such functions exist, raise
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an error
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Step 1 : Transform the equation to its normal form as explained in the
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theory section.
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Step 2 : Initialize an empty set of solutions, ``sol``.
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Step 3 : If `a(x) = 0`, append `\frac{1}/{(x - C1)}` to ``sol``.
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Step 4 : If `a(x)` is a rational non-zero number, append `\pm \sqrt{a}`
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to ``sol``.
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Step 5 : Find the poles and their multiplicities of `a(x)`. Let
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the number of poles be `n`. Also find the valuation of `a(x)` at
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`\infty` using ``val_at_inf``.
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NOTE: Although the algorithm considers `\infty` as a pole, it is
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not mentioned if it a part of the set of finite poles. `\infty`
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is NOT a part of the set of finite poles. If a pole exists at
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`\infty`, we use its multiplicity to find the laurent series of
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`a(x)` about `\infty`.
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Step 6 : Find `n` c-vectors (one for each pole) and 1 d-vector using
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``construct_c`` and ``construct_d``. Now, determine all the ``2**(n + 1)``
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combinations of choosing between 2 choices for each of the `n` c-vectors
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and 1 d-vector.
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NOTE: The equation for `d_{-1}` in Case 4 (Pg 80) has a printinig
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mistake. The term `- d_N` must be replaced with `-N d_N`. The same
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has been explained in the code as well.
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For each of these above combinations, do
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Step 8 : Compute `m` in ``compute_m_ybar``. `m` is the degree bound of
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the polynomial solution we must find for the auxiliary equation.
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Step 9 : In ``compute_m_ybar``, compute ybar as well where ``ybar`` is
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one part of y(x) -
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.. math:: \overline{y}(x) = \sum_{i=1}^{n} \sum_{j=1}^{r_i} \frac{c_{ij}}{(x - x_i)^j} + \sum_{i=0}^{N} d_i x^i
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Step 10 : If `m` is a non-negative integer -
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Step 11: Find a polynomial solution of degree `m` for the auxiliary equation.
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There are 2 cases possible -
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a. `m` is a non-negative integer: We can solve for the coefficients
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in `p(x)` using Undetermined Coefficients.
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b. `m` is not a non-negative integer: In this case, we cannot find
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a polynomial solution to the auxiliary equation, and hence, we ignore
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this value of `m`.
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Step 12 : For each `p(x)` that exists, append `ybar + \frac{p'(x)}{p(x)}`
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to ``sol``.
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Step 13 : For each solution in ``sol``, apply an inverse transformation,
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so that the solutions of the original equation are found using the
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solutions of the equation in its normal form.
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"""
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from itertools import product
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from sympy.core import S
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from sympy.core.add import Add
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from sympy.core.numbers import oo, Float
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from sympy.core.function import count_ops
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from sympy.core.relational import Eq
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from sympy.core.symbol import symbols, Symbol, Dummy
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from sympy.functions import sqrt, exp
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from sympy.functions.elementary.complexes import sign
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from sympy.integrals.integrals import Integral
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from sympy.polys.domains import ZZ
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from sympy.polys.polytools import Poly
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from sympy.polys.polyroots import roots
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from sympy.solvers.solveset import linsolve
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def riccati_normal(w, x, b1, b2):
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"""
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Given a solution `w(x)` to the equation
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.. math:: w'(x) = b_0(x) + b_1(x)*w(x) + b_2(x)*w(x)^2
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and rational function coefficients `b_1(x)` and
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`b_2(x)`, this function transforms the solution to
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give a solution `y(x)` for its corresponding normal
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Riccati ODE
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.. math:: y'(x) + y(x)^2 = a(x)
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using the transformation
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.. math:: y(x) = -b_2(x)*w(x) - b'_2(x)/(2*b_2(x)) - b_1(x)/2
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"""
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return -b2*w - b2.diff(x)/(2*b2) - b1/2
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def riccati_inverse_normal(y, x, b1, b2, bp=None):
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"""
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Inverse transforming the solution to the normal
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Riccati ODE to get the solution to the Riccati ODE.
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"""
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# bp is the expression which is independent of the solution
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# and hence, it need not be computed again
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if bp is None:
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bp = -b2.diff(x)/(2*b2**2) - b1/(2*b2)
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# w(x) = -y(x)/b2(x) - b2'(x)/(2*b2(x)^2) - b1(x)/(2*b2(x))
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return -y/b2 + bp
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def riccati_reduced(eq, f, x):
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"""
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Convert a Riccati ODE into its corresponding
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normal Riccati ODE.
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"""
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match, funcs = match_riccati(eq, f, x)
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# If equation is not a Riccati ODE, exit
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if not match:
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return False
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# Using the rational functions, find the expression for a(x)
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b0, b1, b2 = funcs
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a = -b0*b2 + b1**2/4 - b1.diff(x)/2 + 3*b2.diff(x)**2/(4*b2**2) + b1*b2.diff(x)/(2*b2) - \
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b2.diff(x, 2)/(2*b2)
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# Normal form of Riccati ODE is f'(x) + f(x)^2 = a(x)
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return f(x).diff(x) + f(x)**2 - a
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def linsolve_dict(eq, syms):
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"""
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Get the output of linsolve as a dict
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"""
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# Convert tuple type return value of linsolve
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# to a dictionary for ease of use
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sol = linsolve(eq, syms)
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if not sol:
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return {}
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return {k:v for k, v in zip(syms, list(sol)[0])}
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def match_riccati(eq, f, x):
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"""
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A function that matches and returns the coefficients
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if an equation is a Riccati ODE
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Parameters
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==========
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eq: Equation to be matched
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f: Dependent variable
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x: Independent variable
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Returns
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=======
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match: True if equation is a Riccati ODE, False otherwise
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funcs: [b0, b1, b2] if match is True, [] otherwise. Here,
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b0, b1 and b2 are rational functions which match the equation.
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"""
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# Group terms based on f(x)
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if isinstance(eq, Eq):
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eq = eq.lhs - eq.rhs
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eq = eq.expand().collect(f(x))
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cf = eq.coeff(f(x).diff(x))
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# There must be an f(x).diff(x) term.
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# eq must be an Add object since we are using the expanded
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# equation and it must have atleast 2 terms (b2 != 0)
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if cf != 0 and isinstance(eq, Add):
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# Divide all coefficients by the coefficient of f(x).diff(x)
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# and add the terms again to get the same equation
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eq = Add(*((x/cf).cancel() for x in eq.args)).collect(f(x))
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# Match the equation with the pattern
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b1 = -eq.coeff(f(x))
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b2 = -eq.coeff(f(x)**2)
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b0 = (f(x).diff(x) - b1*f(x) - b2*f(x)**2 - eq).expand()
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funcs = [b0, b1, b2]
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# Check if coefficients are not symbols and floats
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if any(len(x.atoms(Symbol)) > 1 or len(x.atoms(Float)) for x in funcs):
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return False, []
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# If b_0(x) contains f(x), it is not a Riccati ODE
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if len(b0.atoms(f)) or not all((b2 != 0, b0.is_rational_function(x),
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b1.is_rational_function(x), b2.is_rational_function(x))):
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return False, []
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return True, funcs
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return False, []
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def val_at_inf(num, den, x):
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# Valuation of a rational function at oo = deg(denom) - deg(numer)
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return den.degree(x) - num.degree(x)
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def check_necessary_conds(val_inf, muls):
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"""
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The necessary conditions for a rational solution
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to exist are as follows -
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i) Every pole of a(x) must be either a simple pole
|
||
|
|
or a multiple pole of even order.
|
||
|
|
|
||
|
|
ii) The valuation of a(x) at infinity must be even
|
||
|
|
or be greater than or equal to 2.
|
||
|
|
|
||
|
|
Here, a simple pole is a pole with multiplicity 1
|
||
|
|
and a multiple pole is a pole with multiplicity
|
||
|
|
greater than 1.
|
||
|
|
"""
|
||
|
|
return (val_inf >= 2 or (val_inf <= 0 and val_inf%2 == 0)) and \
|
||
|
|
all(mul == 1 or (mul%2 == 0 and mul >= 2) for mul in muls)
|
||
|
|
|
||
|
|
|
||
|
|
def inverse_transform_poly(num, den, x):
|
||
|
|
"""
|
||
|
|
A function to make the substitution
|
||
|
|
x -> 1/x in a rational function that
|
||
|
|
is represented using Poly objects for
|
||
|
|
numerator and denominator.
|
||
|
|
"""
|
||
|
|
# Declare for reuse
|
||
|
|
one = Poly(1, x)
|
||
|
|
xpoly = Poly(x, x)
|
||
|
|
|
||
|
|
# Check if degree of numerator is same as denominator
|
||
|
|
pwr = val_at_inf(num, den, x)
|
||
|
|
if pwr >= 0:
|
||
|
|
# Denominator has greater degree. Substituting x with
|
||
|
|
# 1/x would make the extra power go to the numerator
|
||
|
|
if num.expr != 0:
|
||
|
|
num = num.transform(one, xpoly) * x**pwr
|
||
|
|
den = den.transform(one, xpoly)
|
||
|
|
else:
|
||
|
|
# Numerator has greater degree. Substituting x with
|
||
|
|
# 1/x would make the extra power go to the denominator
|
||
|
|
num = num.transform(one, xpoly)
|
||
|
|
den = den.transform(one, xpoly) * x**(-pwr)
|
||
|
|
return num.cancel(den, include=True)
|
||
|
|
|
||
|
|
|
||
|
|
def limit_at_inf(num, den, x):
|
||
|
|
"""
|
||
|
|
Find the limit of a rational function
|
||
|
|
at oo
|
||
|
|
"""
|
||
|
|
# pwr = degree(num) - degree(den)
|
||
|
|
pwr = -val_at_inf(num, den, x)
|
||
|
|
# Numerator has a greater degree than denominator
|
||
|
|
# Limit at infinity would depend on the sign of the
|
||
|
|
# leading coefficients of numerator and denominator
|
||
|
|
if pwr > 0:
|
||
|
|
return oo*sign(num.LC()/den.LC())
|
||
|
|
# Degree of numerator is equal to that of denominator
|
||
|
|
# Limit at infinity is just the ratio of leading coeffs
|
||
|
|
elif pwr == 0:
|
||
|
|
return num.LC()/den.LC()
|
||
|
|
# Degree of numerator is less than that of denominator
|
||
|
|
# Limit at infinity is just 0
|
||
|
|
else:
|
||
|
|
return 0
|
||
|
|
|
||
|
|
|
||
|
|
def construct_c_case_1(num, den, x, pole):
|
||
|
|
# Find the coefficient of 1/(x - pole)**2 in the
|
||
|
|
# Laurent series expansion of a(x) about pole.
|
||
|
|
num1, den1 = (num*Poly((x - pole)**2, x, extension=True)).cancel(den, include=True)
|
||
|
|
r = (num1.subs(x, pole))/(den1.subs(x, pole))
|
||
|
|
|
||
|
|
# If multiplicity is 2, the coefficient to be added
|
||
|
|
# in the c-vector is c = (1 +- sqrt(1 + 4*r))/2
|
||
|
|
if r != -S(1)/4:
|
||
|
|
return [[(1 + sqrt(1 + 4*r))/2], [(1 - sqrt(1 + 4*r))/2]]
|
||
|
|
return [[S.Half]]
|
||
|
|
|
||
|
|
|
||
|
|
def construct_c_case_2(num, den, x, pole, mul):
|
||
|
|
# Generate the coefficients using the recurrence
|
||
|
|
# relation mentioned in (5.14) in the thesis (Pg 80)
|
||
|
|
|
||
|
|
# r_i = mul/2
|
||
|
|
ri = mul//2
|
||
|
|
|
||
|
|
# Find the Laurent series coefficients about the pole
|
||
|
|
ser = rational_laurent_series(num, den, x, pole, mul, 6)
|
||
|
|
|
||
|
|
# Start with an empty memo to store the coefficients
|
||
|
|
# This is for the plus case
|
||
|
|
cplus = [0 for i in range(ri)]
|
||
|
|
|
||
|
|
# Base Case
|
||
|
|
cplus[ri-1] = sqrt(ser[2*ri])
|
||
|
|
|
||
|
|
# Iterate backwards to find all coefficients
|
||
|
|
s = ri - 1
|
||
|
|
sm = 0
|
||
|
|
for s in range(ri-1, 0, -1):
|
||
|
|
sm = 0
|
||
|
|
for j in range(s+1, ri):
|
||
|
|
sm += cplus[j-1]*cplus[ri+s-j-1]
|
||
|
|
if s!= 1:
|
||
|
|
cplus[s-1] = (ser[ri+s] - sm)/(2*cplus[ri-1])
|
||
|
|
|
||
|
|
# Memo for the minus case
|
||
|
|
cminus = [-x for x in cplus]
|
||
|
|
|
||
|
|
# Find the 0th coefficient in the recurrence
|
||
|
|
cplus[0] = (ser[ri+s] - sm - ri*cplus[ri-1])/(2*cplus[ri-1])
|
||
|
|
cminus[0] = (ser[ri+s] - sm - ri*cminus[ri-1])/(2*cminus[ri-1])
|
||
|
|
|
||
|
|
# Add both the plus and minus cases' coefficients
|
||
|
|
if cplus != cminus:
|
||
|
|
return [cplus, cminus]
|
||
|
|
return cplus
|
||
|
|
|
||
|
|
|
||
|
|
def construct_c_case_3():
|
||
|
|
# If multiplicity is 1, the coefficient to be added
|
||
|
|
# in the c-vector is 1 (no choice)
|
||
|
|
return [[1]]
|
||
|
|
|
||
|
|
|
||
|
|
def construct_c(num, den, x, poles, muls):
|
||
|
|
"""
|
||
|
|
Helper function to calculate the coefficients
|
||
|
|
in the c-vector for each pole.
|
||
|
|
"""
|
||
|
|
c = []
|
||
|
|
for pole, mul in zip(poles, muls):
|
||
|
|
c.append([])
|
||
|
|
|
||
|
|
# Case 3
|
||
|
|
if mul == 1:
|
||
|
|
# Add the coefficients from Case 3
|
||
|
|
c[-1].extend(construct_c_case_3())
|
||
|
|
|
||
|
|
# Case 1
|
||
|
|
elif mul == 2:
|
||
|
|
# Add the coefficients from Case 1
|
||
|
|
c[-1].extend(construct_c_case_1(num, den, x, pole))
|
||
|
|
|
||
|
|
# Case 2
|
||
|
|
else:
|
||
|
|
# Add the coefficients from Case 2
|
||
|
|
c[-1].extend(construct_c_case_2(num, den, x, pole, mul))
|
||
|
|
|
||
|
|
return c
|
||
|
|
|
||
|
|
|
||
|
|
def construct_d_case_4(ser, N):
|
||
|
|
# Initialize an empty vector
|
||
|
|
dplus = [0 for i in range(N+2)]
|
||
|
|
# d_N = sqrt(a_{2*N})
|
||
|
|
dplus[N] = sqrt(ser[2*N])
|
||
|
|
|
||
|
|
# Use the recurrence relations to find
|
||
|
|
# the value of d_s
|
||
|
|
for s in range(N-1, -2, -1):
|
||
|
|
sm = 0
|
||
|
|
for j in range(s+1, N):
|
||
|
|
sm += dplus[j]*dplus[N+s-j]
|
||
|
|
if s != -1:
|
||
|
|
dplus[s] = (ser[N+s] - sm)/(2*dplus[N])
|
||
|
|
|
||
|
|
# Coefficients for the case of d_N = -sqrt(a_{2*N})
|
||
|
|
dminus = [-x for x in dplus]
|
||
|
|
|
||
|
|
# The third equation in Eq 5.15 of the thesis is WRONG!
|
||
|
|
# d_N must be replaced with N*d_N in that equation.
|
||
|
|
dplus[-1] = (ser[N+s] - N*dplus[N] - sm)/(2*dplus[N])
|
||
|
|
dminus[-1] = (ser[N+s] - N*dminus[N] - sm)/(2*dminus[N])
|
||
|
|
|
||
|
|
if dplus != dminus:
|
||
|
|
return [dplus, dminus]
|
||
|
|
return dplus
|
||
|
|
|
||
|
|
|
||
|
|
def construct_d_case_5(ser):
|
||
|
|
# List to store coefficients for plus case
|
||
|
|
dplus = [0, 0]
|
||
|
|
|
||
|
|
# d_0 = sqrt(a_0)
|
||
|
|
dplus[0] = sqrt(ser[0])
|
||
|
|
|
||
|
|
# d_(-1) = a_(-1)/(2*d_0)
|
||
|
|
dplus[-1] = ser[-1]/(2*dplus[0])
|
||
|
|
|
||
|
|
# Coefficients for the minus case are just the negative
|
||
|
|
# of the coefficients for the positive case.
|
||
|
|
dminus = [-x for x in dplus]
|
||
|
|
|
||
|
|
if dplus != dminus:
|
||
|
|
return [dplus, dminus]
|
||
|
|
return dplus
|
||
|
|
|
||
|
|
|
||
|
|
def construct_d_case_6(num, den, x):
|
||
|
|
# s_oo = lim x->0 1/x**2 * a(1/x) which is equivalent to
|
||
|
|
# s_oo = lim x->oo x**2 * a(x)
|
||
|
|
s_inf = limit_at_inf(Poly(x**2, x)*num, den, x)
|
||
|
|
|
||
|
|
# d_(-1) = (1 +- sqrt(1 + 4*s_oo))/2
|
||
|
|
if s_inf != -S(1)/4:
|
||
|
|
return [[(1 + sqrt(1 + 4*s_inf))/2], [(1 - sqrt(1 + 4*s_inf))/2]]
|
||
|
|
return [[S.Half]]
|
||
|
|
|
||
|
|
|
||
|
|
def construct_d(num, den, x, val_inf):
|
||
|
|
"""
|
||
|
|
Helper function to calculate the coefficients
|
||
|
|
in the d-vector based on the valuation of the
|
||
|
|
function at oo.
|
||
|
|
"""
|
||
|
|
N = -val_inf//2
|
||
|
|
# Multiplicity of oo as a pole
|
||
|
|
mul = -val_inf if val_inf < 0 else 0
|
||
|
|
ser = rational_laurent_series(num, den, x, oo, mul, 1)
|
||
|
|
|
||
|
|
# Case 4
|
||
|
|
if val_inf < 0:
|
||
|
|
d = construct_d_case_4(ser, N)
|
||
|
|
|
||
|
|
# Case 5
|
||
|
|
elif val_inf == 0:
|
||
|
|
d = construct_d_case_5(ser)
|
||
|
|
|
||
|
|
# Case 6
|
||
|
|
else:
|
||
|
|
d = construct_d_case_6(num, den, x)
|
||
|
|
|
||
|
|
return d
|
||
|
|
|
||
|
|
|
||
|
|
def rational_laurent_series(num, den, x, r, m, n):
|
||
|
|
r"""
|
||
|
|
The function computes the Laurent series coefficients
|
||
|
|
of a rational function.
|
||
|
|
|
||
|
|
Parameters
|
||
|
|
==========
|
||
|
|
|
||
|
|
num: A Poly object that is the numerator of `f(x)`.
|
||
|
|
den: A Poly object that is the denominator of `f(x)`.
|
||
|
|
x: The variable of expansion of the series.
|
||
|
|
r: The point of expansion of the series.
|
||
|
|
m: Multiplicity of r if r is a pole of `f(x)`. Should
|
||
|
|
be zero otherwise.
|
||
|
|
n: Order of the term upto which the series is expanded.
|
||
|
|
|
||
|
|
Returns
|
||
|
|
=======
|
||
|
|
|
||
|
|
series: A dictionary that has power of the term as key
|
||
|
|
and coefficient of that term as value.
|
||
|
|
|
||
|
|
Below is a basic outline of how the Laurent series of a
|
||
|
|
rational function `f(x)` about `x_0` is being calculated -
|
||
|
|
|
||
|
|
1. Substitute `x + x_0` in place of `x`. If `x_0`
|
||
|
|
is a pole of `f(x)`, multiply the expression by `x^m`
|
||
|
|
where `m` is the multiplicity of `x_0`. Denote the
|
||
|
|
the resulting expression as g(x). We do this substitution
|
||
|
|
so that we can now find the Laurent series of g(x) about
|
||
|
|
`x = 0`.
|
||
|
|
|
||
|
|
2. We can then assume that the Laurent series of `g(x)`
|
||
|
|
takes the following form -
|
||
|
|
|
||
|
|
.. math:: g(x) = \frac{num(x)}{den(x)} = \sum_{m = 0}^{\infty} a_m x^m
|
||
|
|
|
||
|
|
where `a_m` denotes the Laurent series coefficients.
|
||
|
|
|
||
|
|
3. Multiply the denominator to the RHS of the equation
|
||
|
|
and form a recurrence relation for the coefficients `a_m`.
|
||
|
|
"""
|
||
|
|
one = Poly(1, x, extension=True)
|
||
|
|
|
||
|
|
if r == oo:
|
||
|
|
# Series at x = oo is equal to first transforming
|
||
|
|
# the function from x -> 1/x and finding the
|
||
|
|
# series at x = 0
|
||
|
|
num, den = inverse_transform_poly(num, den, x)
|
||
|
|
r = S(0)
|
||
|
|
|
||
|
|
if r:
|
||
|
|
# For an expansion about a non-zero point, a
|
||
|
|
# transformation from x -> x + r must be made
|
||
|
|
num = num.transform(Poly(x + r, x, extension=True), one)
|
||
|
|
den = den.transform(Poly(x + r, x, extension=True), one)
|
||
|
|
|
||
|
|
# Remove the pole from the denominator if the series
|
||
|
|
# expansion is about one of the poles
|
||
|
|
num, den = (num*x**m).cancel(den, include=True)
|
||
|
|
|
||
|
|
# Equate coefficients for the first terms (base case)
|
||
|
|
maxdegree = 1 + max(num.degree(), den.degree())
|
||
|
|
syms = symbols(f'a:{maxdegree}', cls=Dummy)
|
||
|
|
diff = num - den * Poly(syms[::-1], x)
|
||
|
|
coeff_diffs = diff.all_coeffs()[::-1][:maxdegree]
|
||
|
|
(coeffs, ) = linsolve(coeff_diffs, syms)
|
||
|
|
|
||
|
|
# Use the recursion relation for the rest
|
||
|
|
recursion = den.all_coeffs()[::-1]
|
||
|
|
div, rec_rhs = recursion[0], recursion[1:]
|
||
|
|
series = list(coeffs)
|
||
|
|
while len(series) < n:
|
||
|
|
next_coeff = Add(*(c*series[-1-n] for n, c in enumerate(rec_rhs))) / div
|
||
|
|
series.append(-next_coeff)
|
||
|
|
series = {m - i: val for i, val in enumerate(series)}
|
||
|
|
return series
|
||
|
|
|
||
|
|
def compute_m_ybar(x, poles, choice, N):
|
||
|
|
"""
|
||
|
|
Helper function to calculate -
|
||
|
|
|
||
|
|
1. m - The degree bound for the polynomial
|
||
|
|
solution that must be found for the auxiliary
|
||
|
|
differential equation.
|
||
|
|
|
||
|
|
2. ybar - Part of the solution which can be
|
||
|
|
computed using the poles, c and d vectors.
|
||
|
|
"""
|
||
|
|
ybar = 0
|
||
|
|
m = Poly(choice[-1][-1], x, extension=True)
|
||
|
|
|
||
|
|
# Calculate the first (nested) summation for ybar
|
||
|
|
# as given in Step 9 of the Thesis (Pg 82)
|
||
|
|
dybar = []
|
||
|
|
for i, polei in enumerate(poles):
|
||
|
|
for j, cij in enumerate(choice[i]):
|
||
|
|
dybar.append(cij/(x - polei)**(j + 1))
|
||
|
|
m -=Poly(choice[i][0], x, extension=True) # can't accumulate Poly and use with Add
|
||
|
|
ybar += Add(*dybar)
|
||
|
|
|
||
|
|
# Calculate the second summation for ybar
|
||
|
|
for i in range(N+1):
|
||
|
|
ybar += choice[-1][i]*x**i
|
||
|
|
return (m.expr, ybar)
|
||
|
|
|
||
|
|
|
||
|
|
def solve_aux_eq(numa, dena, numy, deny, x, m):
|
||
|
|
"""
|
||
|
|
Helper function to find a polynomial solution
|
||
|
|
of degree m for the auxiliary differential
|
||
|
|
equation.
|
||
|
|
"""
|
||
|
|
# Assume that the solution is of the type
|
||
|
|
# p(x) = C_0 + C_1*x + ... + C_{m-1}*x**(m-1) + x**m
|
||
|
|
psyms = symbols(f'C0:{m}', cls=Dummy)
|
||
|
|
K = ZZ[psyms]
|
||
|
|
psol = Poly(K.gens, x, domain=K) + Poly(x**m, x, domain=K)
|
||
|
|
|
||
|
|
# Eq (5.16) in Thesis - Pg 81
|
||
|
|
auxeq = (dena*(numy.diff(x)*deny - numy*deny.diff(x) + numy**2) - numa*deny**2)*psol
|
||
|
|
if m >= 1:
|
||
|
|
px = psol.diff(x)
|
||
|
|
auxeq += px*(2*numy*deny*dena)
|
||
|
|
if m >= 2:
|
||
|
|
auxeq += px.diff(x)*(deny**2*dena)
|
||
|
|
if m != 0:
|
||
|
|
# m is a non-zero integer. Find the constant terms using undetermined coefficients
|
||
|
|
return psol, linsolve_dict(auxeq.all_coeffs(), psyms), True
|
||
|
|
else:
|
||
|
|
# m == 0 . Check if 1 (x**0) is a solution to the auxiliary equation
|
||
|
|
return S.One, auxeq, auxeq == 0
|
||
|
|
|
||
|
|
|
||
|
|
def remove_redundant_sols(sol1, sol2, x):
|
||
|
|
"""
|
||
|
|
Helper function to remove redundant
|
||
|
|
solutions to the differential equation.
|
||
|
|
"""
|
||
|
|
# If y1 and y2 are redundant solutions, there is
|
||
|
|
# some value of the arbitrary constant for which
|
||
|
|
# they will be equal
|
||
|
|
|
||
|
|
syms1 = sol1.atoms(Symbol, Dummy)
|
||
|
|
syms2 = sol2.atoms(Symbol, Dummy)
|
||
|
|
num1, den1 = [Poly(e, x, extension=True) for e in sol1.together().as_numer_denom()]
|
||
|
|
num2, den2 = [Poly(e, x, extension=True) for e in sol2.together().as_numer_denom()]
|
||
|
|
# Cross multiply
|
||
|
|
e = num1*den2 - den1*num2
|
||
|
|
# Check if there are any constants
|
||
|
|
syms = list(e.atoms(Symbol, Dummy))
|
||
|
|
if len(syms):
|
||
|
|
# Find values of constants for which solutions are equal
|
||
|
|
redn = linsolve(e.all_coeffs(), syms)
|
||
|
|
if len(redn):
|
||
|
|
# Return the general solution over a particular solution
|
||
|
|
if len(syms1) > len(syms2):
|
||
|
|
return sol2
|
||
|
|
# If both have constants, return the lesser complex solution
|
||
|
|
elif len(syms1) == len(syms2):
|
||
|
|
return sol1 if count_ops(syms1) >= count_ops(syms2) else sol2
|
||
|
|
else:
|
||
|
|
return sol1
|
||
|
|
|
||
|
|
|
||
|
|
def get_gen_sol_from_part_sol(part_sols, a, x):
|
||
|
|
""""
|
||
|
|
Helper function which computes the general
|
||
|
|
solution for a Riccati ODE from its particular
|
||
|
|
solutions.
|
||
|
|
|
||
|
|
There are 3 cases to find the general solution
|
||
|
|
from the particular solutions for a Riccati ODE
|
||
|
|
depending on the number of particular solution(s)
|
||
|
|
we have - 1, 2 or 3.
|
||
|
|
|
||
|
|
For more information, see Section 6 of
|
||
|
|
"Methods of Solution of the Riccati Differential Equation"
|
||
|
|
by D. R. Haaheim and F. M. Stein
|
||
|
|
"""
|
||
|
|
|
||
|
|
# If no particular solutions are found, a general
|
||
|
|
# solution cannot be found
|
||
|
|
if len(part_sols) == 0:
|
||
|
|
return []
|
||
|
|
|
||
|
|
# In case of a single particular solution, the general
|
||
|
|
# solution can be found by using the substitution
|
||
|
|
# y = y1 + 1/z and solving a Bernoulli ODE to find z.
|
||
|
|
elif len(part_sols) == 1:
|
||
|
|
y1 = part_sols[0]
|
||
|
|
i = exp(Integral(2*y1, x))
|
||
|
|
z = i * Integral(a/i, x)
|
||
|
|
z = z.doit()
|
||
|
|
if a == 0 or z == 0:
|
||
|
|
return y1
|
||
|
|
return y1 + 1/z
|
||
|
|
|
||
|
|
# In case of 2 particular solutions, the general solution
|
||
|
|
# can be found by solving a separable equation. This is
|
||
|
|
# the most common case, i.e. most Riccati ODEs have 2
|
||
|
|
# rational particular solutions.
|
||
|
|
elif len(part_sols) == 2:
|
||
|
|
y1, y2 = part_sols
|
||
|
|
# One of them already has a constant
|
||
|
|
if len(y1.atoms(Dummy)) + len(y2.atoms(Dummy)) > 0:
|
||
|
|
u = exp(Integral(y2 - y1, x)).doit()
|
||
|
|
# Introduce a constant
|
||
|
|
else:
|
||
|
|
C1 = Dummy('C1')
|
||
|
|
u = C1*exp(Integral(y2 - y1, x)).doit()
|
||
|
|
if u == 1:
|
||
|
|
return y2
|
||
|
|
return (y2*u - y1)/(u - 1)
|
||
|
|
|
||
|
|
# In case of 3 particular solutions, a closed form
|
||
|
|
# of the general solution can be obtained directly
|
||
|
|
else:
|
||
|
|
y1, y2, y3 = part_sols[:3]
|
||
|
|
C1 = Dummy('C1')
|
||
|
|
return (C1 + 1)*y2*(y1 - y3)/(C1*y1 + y2 - (C1 + 1)*y3)
|
||
|
|
|
||
|
|
|
||
|
|
def solve_riccati(fx, x, b0, b1, b2, gensol=False):
|
||
|
|
"""
|
||
|
|
The main function that gives particular/general
|
||
|
|
solutions to Riccati ODEs that have atleast 1
|
||
|
|
rational particular solution.
|
||
|
|
"""
|
||
|
|
# Step 1 : Convert to Normal Form
|
||
|
|
a = -b0*b2 + b1**2/4 - b1.diff(x)/2 + 3*b2.diff(x)**2/(4*b2**2) + b1*b2.diff(x)/(2*b2) - \
|
||
|
|
b2.diff(x, 2)/(2*b2)
|
||
|
|
a_t = a.together()
|
||
|
|
num, den = [Poly(e, x, extension=True) for e in a_t.as_numer_denom()]
|
||
|
|
num, den = num.cancel(den, include=True)
|
||
|
|
|
||
|
|
# Step 2
|
||
|
|
presol = []
|
||
|
|
|
||
|
|
# Step 3 : a(x) is 0
|
||
|
|
if num == 0:
|
||
|
|
presol.append(1/(x + Dummy('C1')))
|
||
|
|
|
||
|
|
# Step 4 : a(x) is a non-zero constant
|
||
|
|
elif x not in num.free_symbols.union(den.free_symbols):
|
||
|
|
presol.extend([sqrt(a), -sqrt(a)])
|
||
|
|
|
||
|
|
# Step 5 : Find poles and valuation at infinity
|
||
|
|
poles = roots(den, x)
|
||
|
|
poles, muls = list(poles.keys()), list(poles.values())
|
||
|
|
val_inf = val_at_inf(num, den, x)
|
||
|
|
|
||
|
|
if len(poles):
|
||
|
|
# Check necessary conditions (outlined in the module docstring)
|
||
|
|
if not check_necessary_conds(val_inf, muls):
|
||
|
|
raise ValueError("Rational Solution doesn't exist")
|
||
|
|
|
||
|
|
# Step 6
|
||
|
|
# Construct c-vectors for each singular point
|
||
|
|
c = construct_c(num, den, x, poles, muls)
|
||
|
|
|
||
|
|
# Construct d vectors for each singular point
|
||
|
|
d = construct_d(num, den, x, val_inf)
|
||
|
|
|
||
|
|
# Step 7 : Iterate over all possible combinations and return solutions
|
||
|
|
# For each possible combination, generate an array of 0's and 1's
|
||
|
|
# where 0 means pick 1st choice and 1 means pick the second choice.
|
||
|
|
|
||
|
|
# NOTE: We could exit from the loop if we find 3 particular solutions,
|
||
|
|
# but it is not implemented here as -
|
||
|
|
# a. Finding 3 particular solutions is very rare. Most of the time,
|
||
|
|
# only 2 particular solutions are found.
|
||
|
|
# b. In case we exit after finding 3 particular solutions, it might
|
||
|
|
# happen that 1 or 2 of them are redundant solutions. So, instead of
|
||
|
|
# spending some more time in computing the particular solutions,
|
||
|
|
# we will end up computing the general solution from a single
|
||
|
|
# particular solution which is usually slower than computing the
|
||
|
|
# general solution from 2 or 3 particular solutions.
|
||
|
|
c.append(d)
|
||
|
|
choices = product(*c)
|
||
|
|
for choice in choices:
|
||
|
|
m, ybar = compute_m_ybar(x, poles, choice, -val_inf//2)
|
||
|
|
numy, deny = [Poly(e, x, extension=True) for e in ybar.together().as_numer_denom()]
|
||
|
|
# Step 10 : Check if a valid solution exists. If yes, also check
|
||
|
|
# if m is a non-negative integer
|
||
|
|
if m.is_nonnegative == True and m.is_integer == True:
|
||
|
|
|
||
|
|
# Step 11 : Find polynomial solutions of degree m for the auxiliary equation
|
||
|
|
psol, coeffs, exists = solve_aux_eq(num, den, numy, deny, x, m)
|
||
|
|
|
||
|
|
# Step 12 : If valid polynomial solution exists, append solution.
|
||
|
|
if exists:
|
||
|
|
# m == 0 case
|
||
|
|
if psol == 1 and coeffs == 0:
|
||
|
|
# p(x) = 1, so p'(x)/p(x) term need not be added
|
||
|
|
presol.append(ybar)
|
||
|
|
# m is a positive integer and there are valid coefficients
|
||
|
|
elif len(coeffs):
|
||
|
|
# Substitute the valid coefficients to get p(x)
|
||
|
|
psol = psol.xreplace(coeffs)
|
||
|
|
# y(x) = ybar(x) + p'(x)/p(x)
|
||
|
|
presol.append(ybar + psol.diff(x)/psol)
|
||
|
|
|
||
|
|
# Remove redundant solutions from the list of existing solutions
|
||
|
|
remove = set()
|
||
|
|
for i in range(len(presol)):
|
||
|
|
for j in range(i+1, len(presol)):
|
||
|
|
rem = remove_redundant_sols(presol[i], presol[j], x)
|
||
|
|
if rem is not None:
|
||
|
|
remove.add(rem)
|
||
|
|
sols = [x for x in presol if x not in remove]
|
||
|
|
|
||
|
|
# Step 15 : Inverse transform the solutions of the equation in normal form
|
||
|
|
bp = -b2.diff(x)/(2*b2**2) - b1/(2*b2)
|
||
|
|
|
||
|
|
# If general solution is required, compute it from the particular solutions
|
||
|
|
if gensol:
|
||
|
|
sols = [get_gen_sol_from_part_sol(sols, a, x)]
|
||
|
|
|
||
|
|
# Inverse transform the particular solutions
|
||
|
|
presol = [Eq(fx, riccati_inverse_normal(y, x, b1, b2, bp).cancel(extension=True)) for y in sols]
|
||
|
|
return presol
|