How can I implement this algorithm in matlab?

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Mark N
Mark N 2018-11-20
评论： Rik 2018-11-21
Hi! Would somebody be kind enough to show me how to implement this algorithm as a function to return an array with values of x^n and v^n
Lets just say k=5 m=0.5 x0=1 and v0=0.1 with total run time of T=10
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Mark N
Mark N 2018-11-21
Lets say delta t = 0.1 and the superscripts arent indices, just to count iterations
Mark N
Mark N 2018-11-21
x(1) = x0;
v(1) = v0;
function [v(:,i),x(:,i)] = semi-implicit(k,m,N,T,x0,v0)
for i= 1:N
v(:,i+1)= (v(:,i) - (dt*(k/m)*x(:,i)));
x(:,i+1)= (x(:,i) + (dt*v(:,i+1)));
end
end
This is what i've tried but seems its completely wrong

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回答（1 个）

Robert U
Robert U 2018-11-21
Hi Mark N
There are some issues with the your matlab syntax.
function [v,x] = semi_implicit(k,m,dt,T,x0,v0)
% check inputs
validateattributes(k, {'numeric'},{'column'});
validateattributes(m, {'numeric'},{'column'});
validateattributes(dt,{'numeric'},{'column'});
validateattributes(T, {'numeric'},{'scalar','nonnegative'});
validateattributes(x0,{'numeric'},{'column'});
validateattributes(v0,{'numeric'},{'column'});
if ~isscalar(k) || ~isscalar(m) || ~isscalar(dt) || ~isscalar(x0) || ~isscalar(v0)
dLgnths = [length(k);length(m);length(dt);length(x0);length(v0)];
if ~all((dLgnths == max(dLgnths)) | dLgnths == 1)
errout = find(~((dLgnths == max(dLgnths)) | dLgnths == 1));
error('Please, check input #%d for being either scalar or vector of length %d (depending on max. length of given input vectors).',errout(1),max(dLgnths))
end
end
% initialize
x = x0;
v = v0;
for ik= 1:T
v(:,end+1)= v(:,end) - (dt .* (k./m) .* x(:,end));
x(:,end+1)= x(:,end) + (dt .* v(:,end));
end
% copy end values to output (here, by overwriting variables v and x)
v = v(:,end);
x = x(:,end);
end
Kind regards,
Robert
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Rik
Rik 2018-11-21
Nice write-up. I do have one remark: if you know how many elements an array will have, you should generally not allow them to grow dynamically (the exceptions are often cases where you know there are only a handful of iterations and the size calculation is expensive).
% initialize
x=zeros(size(x0,1),T);
x(:,1) = x0;
v=zeros(size(v0,1),T);
v(:,1) = v0;
for ik= 1:T
v(:,ik+1)= v(:,ik) - (dt .* (k./m) .* x(:,ik));
x(:,ik+1)= x(:,ik) + (dt .* v(:,ik));
end

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