93 lines
2.6 KiB
Mathematica
93 lines
2.6 KiB
Mathematica
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function Y = ode5(odefun,tspan,y0,varargin)
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%ODE5 Solve differential equations with a non-adaptive method of order 5.
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% Y = ODE5(ODEFUN,TSPAN,Y0) with TSPAN = [T1, T2, T3, ... TN] integrates
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% the system of differential equations y' = f(t,y) by stepping from T0 to
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% T1 to TN. Function ODEFUN(T,Y) must return f(t,y) in a column vector.
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% The vector Y0 is the initial conditions at T0. Each row in the solution
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% array Y corresponds to a time specified in TSPAN.
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%
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% Y = ODE5(ODEFUN,TSPAN,Y0,P1,P2...) passes the additional parameters
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% P1,P2... to the derivative function as ODEFUN(T,Y,P1,P2...).
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%
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% This is a non-adaptive solver. The step sequence is determined by TSPAN
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% but the derivative function ODEFUN is evaluated multiple times per step.
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% The solver implements the Dormand-Prince method of order 5 in a general
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% framework of explicit Runge-Kutta methods.
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%
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% Example
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% tspan = 0:0.1:20;
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% y = ode5(@vdp1,tspan,[2 0]);
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% plot(tspan,y(:,1));
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% solves the system y' = vdp1(t,y) with a constant step size of 0.1,
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% and plots the first component of the solution.
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if ~isnumeric(tspan)
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error('TSPAN should be a vector of integration steps.');
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end
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if ~isnumeric(y0)
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error('Y0 should be a vector of initial conditions.');
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end
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h = diff(tspan);
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if any(sign(h(1))*h <= 0)
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error('Entries of TSPAN are not in order.')
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end
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try
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f0 = feval(odefun,tspan(1),y0,varargin{:});
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catch
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msg = ['Unable to evaluate the ODEFUN at t0,y0. ',lasterr];
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error(msg);
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end
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y0 = y0(:); % Make a column vector.
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if ~isequal(size(y0),size(f0))
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error('Inconsistent sizes of Y0 and f(t0,y0).');
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end
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neq = length(y0);
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N = length(tspan);
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Y = zeros(neq,N);
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% Method coefficients -- Butcher's tableau
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%
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% C | A
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% --+---
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% | B
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C = [1/5; 3/10; 4/5; 8/9; 1];
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A = [ 1/5, 0, 0, 0, 0
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3/40, 9/40, 0, 0, 0
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44/45 -56/15, 32/9, 0, 0
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19372/6561, -25360/2187, 64448/6561, -212/729, 0
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9017/3168, -355/33, 46732/5247, 49/176, -5103/18656];
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B = [35/384, 0, 500/1113, 125/192, -2187/6784, 11/84];
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% More convenient storage
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A = A.';
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B = B(:);
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nstages = length(B);
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F = zeros(neq,nstages);
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Y(:,1) = y0;
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for i = 2:N
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ti = tspan(i-1);
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hi = h(i-1);
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yi = Y(:,i-1);
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% General explicit Runge-Kutta framework
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F(:,1) = feval(odefun,ti,yi,varargin{:});
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for stage = 2:nstages
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tstage = ti + C(stage-1)*hi;
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ystage = yi + F(:,1:stage-1)*(hi*A(1:stage-1,stage-1));
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F(:,stage) = feval(odefun,tstage,ystage,varargin{:});
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end
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Y(:,i) = yi + F*(hi*B);
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end
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Y = Y.';
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