an ode with arguements
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Here is my function file:
function dfdeta = mufun(eta,f,T)
pr = 1000;
dfdeta = [f(2); f(3); -f(1) * f(3); T(2); -pr*f(:,1)*T(2)];
end
and here is the code to call my function:
clear;
clc;
close all;
guessf = 0.4696;
guessT = .5;
[eta, f, T] = ode45(@mufun, [linspace(0,6,16)], [0; 0; guessf; 0; guessT]);
plot(eta,f);
blasius = table(eta, f(:,1), f(:,2), f(:,3), 'VariableNames',{'eta','f', 'f prime', 'f double prime'})
I was able to figure out the ode45 for just the eta and f variable, but now I have to have f defined in order to solve for T.
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답변 (3개)
James Tursa
2024년 4월 9일
편집: James Tursa
2024년 4월 9일
Create a new function handle with your extra stuff. E.g.,
mufunT = @(eta,f) mufun(eta,f,guessT)
[eta, f] = ode45(mufunT, [linspace(0,6,16)], [0; 0; guessf]);
But, this assumes you know T in advance. What do you mean by "solve for T"?
Star Strider
2024년 4월 9일
You have five differential equations and three initial conditions.
The initial conditions vector must have the same length as the number of differential equations.
Beyond that, you need to pass ‘T’ as an additional parameter:
[eta, f] = ode45(@(eta,f)mufun(eta,f,guessT), [linspace(0,6,16)], [0; 0; guessf]);
.
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James Tursa
2024년 4월 10일
@Ray Can you post an image of the differential equations you are trying to solve?
Torsten
2024년 4월 10일
편집: Torsten
2024년 4월 10일
You have to define your vector of solution variables as
y(1) = f, y(2) = f', y(3) = f'', y(4) = T, y(5) = T'
and your function as
function dydeta = mufun(eta,y)
pr = 1000;
dydeta = [y(2); y(3); -y(1)*y(3)/2; y(5); -pr/2*y(1)*y(5)];
end
Further, your problem is a boundary value problem, not an initial value problem. Use "bvp4c", not "ode45" to solve.
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