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How to index a value of an anonymous function

조회 수: 7 (최근 30일)
Robert Basler
Robert Basler 2020년 10월 29일
댓글: Ameer Hamza 2020년 10월 30일
I am trying to create a function that uses Euler Method to find the height of water in a tank as a function of time given the equation for dy/dt. The equation dy/dt is to be passed into my function as an anonymous function. The equation for dy/dt depends on both y and t as well as some constants.
My program initilizes a vector called change that I want to store the values of dy/dt as the loop runs. I cannot run the program as it is given because I am trying to store each value of dy/dt (which should be calculated on each iteration based on the changing value of y as the loop runs) to be multiplied by the timestep, but the value of the vector called 'change' is remaining constant. What I essentially want to do is have 'change(k)=dydt(k)' but because I am passing dydt as an anonymous function using 'y' as an input I cannot simply index using '(k)' at the end of the vector name the way I usually would. I'm not sure if there is an easy fix or if I need to re-program the way the way I am using the anonymous function. Any tips and help are appriciated, thank you!!
function [depth_vector] = Depth_Calc(dydt,i,f,n,varargin)
%Calculates the Depth vs Time by evaluating the differential
%equation y with parameters A, Q, alpha from time i to time f
%using timestep n and generates table printout of results
%input:
%dydt= function (given as anonymous function)
%i=initial time
%f=finl time
%n=timestep
t=[i:n:f];
y=ones(1,((f-i)/n));
change=.32635*ones(1,((f-i)/n));
for k=2:((f-i)/n)
change=dydt(varargin{:})
y(k)=y(k-1)+n*change(k-1);
end
depth_vector=y;
output=depth_vector;
  댓글 수: 2
KSSV
KSSV 2020년 10월 29일
You want change to be a matrix? According to your code y also will be a matrix.
Robert Basler
Robert Basler 2020년 10월 29일
y is a function of t
dydt is a function of both t and y
I really only need the output of y as a matrix or vector, as the end result I am looking for are the values of y vs time. The values for dydt are only temporarily needed to calculate the next value of y inside the loop. When I tried to use the function dydt directly in the Euler method (instead of first assigning the value it to variable 'change') I got the error that the number of array elements between y and dydt did not match.

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Ameer Hamza
Ameer Hamza 2020년 10월 29일
The exact solution depends on what are the input arguments of dydt, but if it is defined as an anonymous function, then you should be able to do something like this inside the for-loop
change(k) = dydt(varargin{:})
  댓글 수: 6
Robert Basler
Robert Basler 2020년 10월 29일
This works perfectly thank you so much! I do have one question though, just for my understanding:
If I don't employ 'f=' before calling the function I get an error message of 'too many input arguments'
Why do I have to assign the function to a variable rather than just calling it?
Ameer Hamza
Ameer Hamza 2020년 10월 30일
I don't think that is the issue. Following code works just fine for me
dydt=@(t,y,A,Q,a) 3*(Q/A).*((sin(t)).^2)-(a.*((1+y).^1.5))./A;
A=1300 ; Q=450 ; a=150;
Depth_Calc(dydt, 1, 10, 1, A, Q, a)
Can you show the lines of code that causes this error?

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