Mercurial > octave
view scripts/ode/private/integrate_const.m @ 22323:bac0d6f07a3e
maint: Update copyright notices for 2016.
author | John W. Eaton <jwe@octave.org> |
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date | Wed, 17 Aug 2016 01:05:19 -0400 |
parents | 9fc91bb2aec3 |
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## Copyright (C) 2013-2016 Roberto Porcu' <roberto.porcu@polimi.it> ## ## This file is part of Octave. ## ## Octave is free software; you can redistribute it and/or modify it ## under the terms of the GNU General Public License as published by ## the Free Software Foundation; either version 3 of the License, or (at ## your option) any later version. ## ## Octave is distributed in the hope that it will be useful, but ## WITHOUT ANY WARRANTY; without even the implied warranty of ## MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU ## General Public License for more details. ## ## You should have received a copy of the GNU General Public License ## along with Octave; see the file COPYING. If not, see ## <http://www.gnu.org/licenses/>. ## -*- texinfo -*- ## @deftypefn {} {@var{solution} =} integrate_const (@var{@@stepper}, @var{@@func}, @var{tspan}, @var{x0}, @var{dt}, @var{options}) ## ## This function file can be called by an ODE solver function in order to ## integrate the set of ODEs on the interval @var{[t0,t1]} with a constant ## timestep @var{dt}. ## ## The function returns a structure @var{solution} with two fieldss: @var{t} ## and @var{y}. @var{t} is a column vector and contains the time stamps. ## @var{y} is a matrix in which each column refers to a different unknown ## of the problem and the row number is the same as the @var{t} row number. ## Thus, each row of the matrix @var{y} contains the values of all unknowns at ## the time value contained in the corresponding row in @var{t}. ## ## The first input argument must be a function handle or inline function ## representing the stepper, i.e., the function responsible for step-by-step ## integration. This function discriminates one method from the others. ## ## The second input argument is the order of the stepper. It is needed to ## compute the adaptive timesteps. ## ## The third input argument is a function handle or inline function that ## defines the ODE: ## ## @ifhtml ## ## @example ## @math{y' = f(t,y)} ## @end example ## ## @end ifhtml ## @ifnothtml ## @math{y' = f(t,y)}. ## @end ifnothtml ## ## The fourth input argument is the time vector which defines the integration ## interval, i.e., @var{[tspan(1), tspan(end)]} and all intermediate elements ## are taken as times at which the solution is required. ## ## The fourth argument contains the initial conditions for the ODEs. ## ## The fifth input argument represents the fixed timestep and the last input ## argument contains some options that may be needed for the stepper. ## @end deftypefn ## ## @seealso{integrate_adaptive, integrate_n_steps} function solution = integrate_const (stepper, func, tspan, x0, dt, options) solution = struct (); ## first values for time and solution t = tspan(1); x = x0(:); direction = odeget (options, "direction", [], "fast"); if (sign (dt) != direction) error ("Octave:invalid-input-arg", "option 'InitialStep' has a wrong sign"); endif ## setting parameters k = length (tspan); counter = 2; comp = 0.0; tk = tspan(1); options.comp = comp; ## Initialize the OutputFcn if (options.haveoutputfunction) if (options.haveoutputselection) solution.retout = x(options.OutputSel,end); else solution.retout = x; endif feval (options.OutputFcn, tspan, solution.retout, "init", options.funarguments{:}); endif ## Initialize the EventFcn if (options.haveeventfunction) ode_event_handler (options.Events, t(end), x, "init", options.funarguments{:}); endif solution.cntloop = 2; solution.cntcycles = 1; cntiter = 0; solution.unhandledtermination = true; solution.cntsave = 2; z = t; u = x; k_vals = feval (func, t , x, options.funarguments{:}); while (counter <= k) ## computing the integration step from t to t+dt [s, y, ~, k_vals] = stepper (func, z(end), u(:,end), dt, options, k_vals); [tk, comp] = kahan (tk,comp, dt); options.comp = comp; s(end) = tk; if (options.havenonnegative) x(options.NonNegative,end) = abs (x(options.NonNegative,end)); y(options.NonNegative,end) = abs (y(options.NonNegative,end)); y_est(options.NonNegative,end) = abs (y_est(options.NonNegative,end)); endif if (options.haveoutputfunction && options.haverefine) SaveVUForRefine = u(:,end); endif ## values on this interval for time and solution z = [t(end);s]; u = [x(:,end),y]; ## if next tspan value is caught, update counter if ((z(end) == tspan(counter)) || (abs (z(end) - tspan(counter)) / (max (abs (z(end)), abs (tspan(counter)))) < 8*eps) ) counter += 1; ## if there is an element in time vector at which the solution is required ## the program must compute this solution before going on with next steps elseif (direction * z(end) > direction * tspan(counter) ) ## initializing counter for the following cycle i = 2; while (i <= length (z)) ## if next tspan value is caught, update counter if ((counter <= k) && (((z(i) == tspan(counter)) || (abs (z(i) - tspan(counter)) / (max (abs (z(i)), abs (tspan(counter)))) < 8*eps))) ) counter += 1; endif ## else, loop until there are requested values inside this subinterval while ((counter <= k) && direction * z(i) > direction * tspan(counter) ) ## add the interpolated value of the solution u = [u(:,1:i-1),u(:,i-1) + (tspan(counter)-z(i-1))/(z(i)-z(i-1))* ... (u(:,i)-u(:,i-1)),u(:,i:end)]; ## add the time requested z = [z(1:i-1);tspan(counter);z(i:end)]; ## update counters counter += 1; i += 1; endwhile ## if new time requested is not out of this interval if (counter <= k && direction * z(end) > direction * tspan(counter)) ## update the counter i += 1; else ## else, stop the cycle and go on with the next iteration i = length (z)+1; endif endwhile endif x = [x,u(:,2:end)]; t = [t;z(2:end)]; solution.cntsave += 1; solution.cntloop += 1; cntiter = 0; ## Call OutputFcn only if a valid result has been found. ## Stop integration if function returns false. if (options.haveoutputfunction) for cnt = 0:options.Refine # Approximation between told and t if (options.haverefine) # Do interpolation approxtime = (cnt + 1) / (options.Refine + 2); approxvals = (1 - approxtime) * SaveVUForRefine ... + (approxtime) * y(:,end); approxtime = s(end) + approxtime*dt; else approxvals = x(:,end); approxtime = t(end); endif if (options.haveoutputselection) approxvals = approxvals(options.OutputSel); endif pltret = feval (options.OutputFcn, approxtime, approxvals, [], options.funarguments{:}); if (pltret) # Leave refinement loop break; endif endfor if (pltret) # Leave main loop solution.unhandledtermination = false; break; endif endif ## Call Events function only if a valid result has been found. ## Stop integration if eventbreak is true. if (options.haveeventfunction) solution.event = ode_event_handler (options.Events, t(end), x(:,end), [], options.funarguments{:}); if (! isempty (solution.event{1}) && solution.event{1} == 1) t(solution.cntloop-1,:) = solution.event{3}(end,:); x(:,solution.cntloop-1) = solution.event{4}(end,:)'; solution.unhandledtermination = false; break; endif endif ## Update counters that count the number of iteration cycles solution.cntcycles += 1; # Needed for cost statistics cntiter += 1; # Needed to find iteration problems ## Stop solving because, in the last 5,000 steps, no successful valid ## value has been found if (cntiter >= 5_000) error (["integrate_const: Solving was not successful. ", ... " The iterative integration loop exited at time", ... " t = %f before the endpoint at tend = %f was reached. ", ... " This happened because the iterative integration loop", ... " did not find a valid solution at this time stamp. ", ... " Try to reduce the value of 'InitialStep' and/or", ... " 'MaxStep' with the command 'odeset'."], s(end), tspan(end)); endif ## if this is the last iteration, save the length of last interval if (counter > k) j = length (z); endif endwhile ## Check if integration of the ode has been successful if (direction * z(end) < direction * tspan(end)) if (solution.unhandledtermination == true) error ("integrate_const:unexpected_termination", [" Solving was not successful. ", ... " The iterative integration loop exited at time", ... " t = %f before the endpoint at tend = %f was reached. ", ... " This may happen if the stepsize becomes too small. ", ... " Try to reduce the value of 'InitialStep'", ... " and/or 'MaxStep' with the command 'odeset'."], z(end), tspan(end)); else warning ("integrate_const:unexpected_termination", ["Solver was stopped by a call of 'break'", ... " in the main iteration loop at time", ... " t = %f before the endpoint at tend = %f was reached. ", ... " This may happen because the @odeplot function", ... " returned 'true' or the @event function returned 'true'."], z(end), tspan(end)); endif endif ## compute how many values are out of time inerval d = direction * t((end-(j-1)):end) > direction * tspan(end) * ones (j, 1); f = sum (d); ## remove not-requested values of time and solution solution.t = t(1:end-f); solution.x = x(:,1:end-f)'; endfunction