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Indexing problems using for loop and if statements

조회 수: 1 (최근 30일)
matlabkid602
matlabkid602 2017년 11월 21일
댓글: Jan 2017년 11월 21일
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?
  댓글 수: 1
Jan
Jan 2017년 11월 21일
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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답변 (1개)

Jan
Jan 2017년 11월 21일
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
  댓글 수: 2
matlabkid602
matlabkid602 2017년 11월 21일
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
Jan 2017년 11월 21일
@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
Do I understand correctly, that "y(shift)" means to access a different component of the current y vector?

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