an ode with arguements

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Ray
Ray el 9 de Abr. de 2024
Editada: Torsten el 10 de Abr. de 2024
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.

Respuestas (3)

James Tursa
James Tursa el 9 de Abr. de 2024
Editada: James Tursa el 9 de Abr. de 2024
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"?
  1 comentario
Ray
Ray el 9 de Abr. de 2024
We are given a differential equation where these terms: T(2); -pr*f(:,1)*T(2) are needed. We found f previously when we did ode45 without those new terms. But in the differential equation we are given we have to have f(:,1) in order to solve.

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Star Strider
Star Strider el 9 de Abr. de 2024
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]);
.
  6 comentarios
James Tursa
James Tursa el 10 de Abr. de 2024
@Ray Can you post an image of the differential equations you are trying to solve?
Ray
Ray el 10 de Abr. de 2024
It has to be as a pdf, the images came out wrong. Hope this makes sense.

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Torsten
Torsten el 10 de Abr. de 2024
Editada: Torsten el 10 de Abr. de 2024
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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