44 lines
1.6 KiB
Matlab
44 lines
1.6 KiB
Matlab
function [y,t]=pepc_nlfode(f,alpha,y0,h,tn,err,key)
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% pepc_nlfode - numerical solution of nonlinear single-term FODEs
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%
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% [y,t]=pepc_nlfode(f,alpha,y0,h,tn,err,key)
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%
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% f - function handles for the nonlinear explicit FODE
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% alpha - the maximum order
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% y0 - initial vector of the output and its integer-order derivatives
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% h - fixed step-size
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% tn - terminate time in the solution
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% err - error tolerance
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% key - identifier, 0 for with predictor, 1 for bypass predictor,
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% 2 for predictor alone
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% y, t - the solution vector and time vector
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% Copyright (c) Dingyu Xue, Northeastern University, China
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% Last modified 28 March, 2017
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% Last modified 18 May, 2022
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arguments
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f, alpha(1,1), y0, h(1,1), tn(1,1){mustBePositive},
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err(1,1)=1e-6; key{mustBeMember(key,1:3)}=0
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end
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t=0; y1=y0; q=ceil(alpha); a1=alpha+1; m=round(tn/h);
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K=h^alpha/alpha/a1; Ga=1/gamma(alpha);
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y=[y1 zeros(size(y0,1),m)]; n=length(y1);
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for k=0:m-1, tk1=t(end)+h; t=[t,tk1]; ii=0:k; y01=0;
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for i=1:q,y01=y01+tk1^(i-1)/factorial(i-1)*y0(:,i); end
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if key==1, yp=y(:,k+1);
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else, yp=y01;
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b=h^alpha/alpha*((k+1-ii).^alpha-(k-ii).^alpha);
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for i=0:k, yp=yp+b(i+1)*f(t(i+1),y(:,i+1))*Ga; end
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end
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if key~=2
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ii=1:k; a2=(k-ii+2).^a1+(k-ii).^a1-2*(k-ii+1).^a1;
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a0=k^a1-(k-alpha)*(k+1)^alpha; a=[a0 a2 1]*K;
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while (1), y1=y01+a(k+2)*f(tk1,yp)*Ga;
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for i=0:k, y1=y1+a(i+1)*f(t(i+1),y(:,i+1))*Ga; end
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if (norm(y1-yp)<err||(key==0&&n==1)||key==3), break;
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else, yp=y1; end
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end
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else, y1=yp; end, y(:,k+2)=y1;
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end, y=y'; t=t(:);
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end
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