Indexing problems using for loop and if statements

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matlabkid602 2017-11-21

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https://ww2.mathworks.cn/matlabcentral/answers/368377-indexing-problems-using-for-loop-and-if-statements

评论： Jan 2017-11-21

I'm trying to solve a large system of ODE's using ode45. I need to assign y(*negative number) = 0 in the function. This is impossible to avoid because I'm using a loop.

My code is:

 k=5
  for p =1:21
    n=p-11;
 shift = k-n;
 nshift= n-k;
 if shift <1 || shift >length
    y(shift) = 0;
 if nshift <1 || nshift >length
    y(nshift) = 0;
 dydt(p) = term1*y(p)*(n-k)^2; + term2*sin((n-k)^2*theta)*
(y(11)*conj(y(shift)) + conj(y(11))*y(nshift))*y(k+11);
 end
 end
 end

I'm having problems with the y(shift) = 0 when shift is negative. How do I fix this?

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Jan 2017-11-21

Your code is easier to read if you apply the standard code indentation (Ctrl-A Ctrl-I) in the editor. Please remove the typo in the line "dydt(p) = ...": I assume the ";" is not wanted.

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Answer 1

Jan 2017-11-21

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Does this appear in the function to be integrated? Then it is not smooth and in consequence ODE45 cannot integrate it reliably. See http://www.mathworks.com/matlabcentral/answers/59582#answer_72047.

What is "length"?

You forgot to explain, what "y" is. If you are calculating "dydt", it might be the value to be integrated. But then indexing it is meaningless, because it is continuous.

If "y" is something else, what about:

 if shift <1 || shift >length
    y_shift = 0;
 else
    y_shift = y(shift);
 end
 if nshift <1 || nshift >length
    y_nshift = 0;
 else
    y_nshift = y(nshift);
 end

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matlabkid602 2017-11-21

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Thanks for that suggestion. I hope this is clearer?

function dydt = abe2(t,y)

 %Defining Parameters For ODE
 k =5;
 theta = pi/50;
 b0 = 20;
 p0 = 1.9e-9;
 wr = 3.77e7 / 2.53e9;
 gamma = 3.77e7;
 term1 = 1i*wr;
 term2 = (1i*b0*p0*gamma)/4;
 nmax = 10; %nmax is the maximum value for n in the subscript dYn/dt
 %Length is the max number of subscript 'n' on RHS of ODE dCn/dt for example
 length = 2*nmax+1;
 dydt = zeros(length,1); %Define the number of Yn solutions
 for p =1:length %To avoid negative indexes, when n ranges from -10 to 10, p-11 adjusts to give correct 'n' value for RHS
    n=p-11;
    shift = k-n; %Positive shift
    nshift= n-k; %Negative shift
    if shift <1 || shift >length
        y_shift = 0;
    else
        y_shift = y(shift);
        if nshift <1 || nshift >length
            y_nshift = 0;
        else
            y_nshift = y(nshift);
            %RHS of the ODE solving for n=-10:10, term1 and term2 defined above with
            %constants of the ODE
            dydt(p) = term1*y(p)*(n-k)^2; + term2*sin((n-k)^2*theta)*(y(11)*conj(y_shift) + conj(y(11))*y_nshift)*y(k+11)
        end %end for loop
    end %end if statement
end %end if statement

ODE45 only returns my initial starting values when I run it.

Jan 2017-11-21

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@matlabkid602: Yes, this is clearer now. One problem remains:

dydt(p) = term1*y(p)*(n-k)^2; + term2*sin((n-k)^2*theta) ...
%       Orphand semicolon:  ^

And

        end %end for loop
     end %end if statement
  end %end if statement

contains confusing comments: The end close the nearest IF or FOR. Then the FOR loop is closed at the bottom.

Redefining important Matlab functions by local variables causes troubles frequently. Avoid using "length" as a name.

Maybe you want to reorder the IF statements:

for p = 1:length
  n = p-11;
  shift = k-n; %Positive shift
  nshift= n-k; %Negative shift
    if shift <1 || shift >length
        y_shift = 0;
    else
        y_shift = y(shift);
    end
    if nshift <1 || nshift >length
        y_nshift = 0;
    else
        y_nshift = y(nshift);
    end
    dydt(p) = term1*y(p)*(n-k)^2 + term2*sin((n-k)^2*theta)* ...
            (y(11)*conj(y_shift) + conj(y(11))*y_nshift)*y(k+11);
  end %end for loop