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up Ascher docs, added presets
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‎docs/references.bib‎

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@@ -120,4 +120,15 @@ @article{Asc89
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doi = {10.1137/0910054}
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}
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@article{Pet86,
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author = {L. R. Petzold},
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journal = {SIAM Journal on Numerical Analysis},
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number = {4},
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pages = {837--852},
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publisher = {Society for Industrial and Applied Mathematics},
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title = {Order Results for Implicit Runge-Kutta Methods Applied to Differential/ Algebraic Systems},
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volume = {23},
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year = {1986},
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url = {https://www.jstor.org/stable/2157625}
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}
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‎toolbox/+otp/+ascherlineardae/+presets/Canonical.m‎

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classdef Canonical < otp.ascherlineardae.AscherLinearDAEProblem
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% The original problem defined by Uri Ascher in :cite:p:`Asc89` with
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% $\beta = 0.5$
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% The problem defined by Uri Ascher in :cite:p:`Asc89` (sec. 2)
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% which uses timespan $t \in [0, 1]$ and $\beta = 1 $.
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%
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methods
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function obj = Canonical(beta)
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tspan = [0.0; 1];
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function obj = Canonical(varargin)
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% Create the Canonical CUSP problem object.
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%
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% Parameters
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% ----------
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% varargin
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% A variable number of name-value pairs. The accepted names are
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%
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% - ``Beta`` – The parameter of the Ascher linear DAE problem.
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if nargin < 1
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beta = 0.5;
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end
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p = inputParser;
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p.addParameter('beta', 1);
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p.parse(varargin{:});
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opts = p.Results;
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params = otp.ascherlineardae.AscherLinearDAEParameters;
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params.Beta = beta;
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params = otp.ascherlineardae.AscherLinearDAEParameters;
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params.Beta = opts.beta;
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y0 = [1; beta];
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y0 = [1; params.Beta];
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tspan = [0.0; 1.0];
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obj = obj@otp.ascherlineardae.AscherLinearDAEProblem(tspan, y0, params);
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end

‎toolbox/+otp/+ascherlineardae/+presets/Petzold.m‎

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classdef Petzold < otp.ascherlineardae.AscherLinearDAEProblem
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% The Petzold DAE example :cite:p:`Pet86` as a variant of the
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% Ascher linear DAE problem
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% which uses timespan $t \in [0, 1]$ and $\beta = 0 $.
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%
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methods
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function obj = Petzold
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% Create the Petzold example of the Ascher linear DAE problem object.
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params = otp.ascherlineardae.AscherLinearDAEParameters;
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params.Beta = 0;
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y0 = [1; params.Beta];
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tspan = [0.0; 1.0];
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obj = obj@otp.ascherlineardae.AscherLinearDAEProblem(tspan, y0, params);
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end
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end
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end

‎toolbox/+otp/+ascherlineardae/+presets/Stiff.m‎

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classdef Stiff < otp.ascherlineardae.AscherLinearDAEProblem
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% The Stiff example from :cite:p:`Asc89` (sec. 2. A variant of the
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% Ascher linear DAE problem
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% which uses timespan $t \in [0, 1]$ and $\beta = 100 $.
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%
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methods
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function obj = Stiff
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% Create the stiff example of the Ascher linear DAE problem object.
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params = otp.ascherlineardae.AscherLinearDAEParameters;
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params.Beta = 100;
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y0 = [1; params.Beta];
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tspan = [0.0; 1.0];
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obj = obj@otp.ascherlineardae.AscherLinearDAEProblem(tspan, y0, params);
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end
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end
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end

‎toolbox/+otp/+ascherlineardae/AscherLinearDAEParameters.m‎

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classdef AscherLinearDAEParameters
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% Parameters for Ascher Linear DAE problem
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properties
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% Beta is a scalar parameter in the linear model. It also affects the initial condition
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% of the algebraic variable.
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Beta %MATLAB ONLY: (1,1) {mustBeNumeric} = 0.5
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% Beta is a scalar parameter in the linear model. It affects the stifness
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% of the problem.
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Beta %MATLAB ONLY: (1,1) {mustBeNumeric} = 1
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end
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end

‎toolbox/+otp/+ascherlineardae/AscherLinearDAEProblem.m‎

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% A simple linear differential algebraic problem.
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%
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% The Ascher linear DAE Problem :cite:p:`Asc89` is an index-1 differential agebraic
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% equation given by $M(t) y' = A(t) y + q(t) $ where
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% equation given by
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%
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% $$
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% M=\left(\begin{array}{cc}
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% \begin{bmatrix}
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% 1 & -t \\
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% 0 & 0
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% \end{array}\right), \quad A=\left(\begin{array}{cc}
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% \end{bmatrix} \begin{bmatrix} y'(t) \\ z'(t) \end{bmatrix} = \left[ \begin{array}{cc}
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% -1 & 1+t \\
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% \beta & -1-\beta t
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% \end{array}\right), \quad {q}=\left(\begin{array}{c}
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% \end{array}\right] \begin{bmatrix} y(t) \\ z(t) \end{bmatrix} + \begin{bmatrix}
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% 0 \\
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% \sin t
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% \end{array}\right),
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% \end{bmatrix}.
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% $$
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%
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% defined on timespan $t \in [0,1]$, and initial condition $y_0 = [1, \beta]^T$. The exact solution
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% is given by
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% The problem has the following closed-form solution when the initial condition $y(0) = 1 , z(0) = \beta$ is used:
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%
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% $$
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% y = \begin{pmatrix}
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% \begin{bmatrix} y(t)\\ z(t) \end{bmatrix} = \begin{bmatrix}
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% t \sin(t) + (1 + \beta t) e^{-t}\\
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% \beta e^{-t} + \sin(t)
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% \end{pmatrix}.
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% \end{bmatrix}.
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% $$
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%
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% Due to its stiffness and time-dependant mass
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% matrix, this simple DAE problem can
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% become challenging to solve. This problem is introduced in :cite:p:`Asc89`
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% to study the convergence of implcit solvers applied to DAEs.
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% This DAE problem
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% can be used to investigate
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% the convergence of implcit time-stepping methods due to its stiffness and time-dependant mass
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% matrix.
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%
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% Notes
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% -----
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% +---------------------+-----------------------------------------+
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% | Number of Variables | 2 |
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% +---------------------+-----------------------------------------+
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% | Stiff | typically, depending on $\beta$ |
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% | Stiff | possibly, depending on $\beta$ |
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% +---------------------+-----------------------------------------+
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%
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% Example
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% -------
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% >>> problem = otp.ascherlineardae.presets.Canonical(0.1);
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% >>> sol = problem.solve('MaxStep',1e-5);
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% >>> problem.plotPhaseSpace(sol);
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% >>> problem = otp.ascherlineardae.presets.Canonical('Beta', 50);
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% >>> t = linspace(0,1,100);
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% >>> sol = problem.solveExcatly(t);
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% >>> problem.plot(sol);
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properties (Access = private, Constant)
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NumComps = 2
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% The initial conditions.
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% parameters : AscherLinearDAEParameters
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% The parameters.
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%
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% Returns
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% -------
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% obj : AscherLinearDAEProblem
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% The constructed problem.
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obj@otp.Problem('Ascher Linear DAE', 2, timeSpan, y0, parameters);
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end
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end
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'MassSingular', 'yes');
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end
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function y = internalSolveExactly(obj, t)
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function sol = internalSolveExactly(obj, t)
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sol = [];
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beta = obj.Parameters.Beta;
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if ~isequal(obj.Y0, [1; beta])
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error('OTP:noExactSolution', ...
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y = [t .* sin(t) + (1 + beta * t) .* exp(-t); ...
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beta * exp(-t) + sin(t)];
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sol.x = t;
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sol.y = y;
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end
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function sol = internalSolve(obj, varargin)

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