How can I implement this algorithm in matlab?

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

[madhan ravi]

what's delta t value?

[Rik]

And are those superscripts meant as exponents, or as indices?

What have you tried so far on your own?

[Mark N]

Lets say delta t = 0.1 and the superscripts arent indices, just to count iterations

[Mark N]

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

Answer 1

[Robert U]

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

[Rik]

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