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Effacer les filtres

an ode with arguements

11 vues (au cours des 30 derniers jours)
Ray
Ray le 9 Avr 2024
Modifié(e) : Torsten le 10 Avr 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.

Réponses (3)

James Tursa
James Tursa le 9 Avr 2024
Modifié(e) : James Tursa le 9 Avr 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 commentaire
Ray
Ray le 9 Avr 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 le 9 Avr 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 commentaires
James Tursa
James Tursa le 10 Avr 2024
@Ray Can you post an image of the differential equations you are trying to solve?
Ray
Ray le 10 Avr 2024
It has to be as a pdf, the images came out wrong. Hope this makes sense.

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Torsten
Torsten le 10 Avr 2024
Modifié(e) : Torsten le 10 Avr 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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