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an ode with arguements

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Ray
Ray 2024 年 4 月 9 日
編集済み: Torsten 2024 年 4 月 10 日
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.

回答 (3 件)

James Tursa
James Tursa 2024 年 4 月 9 日
編集済み: James Tursa 2024 年 4 月 9 日
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 件のコメント
Ray
Ray 2024 年 4 月 9 日
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 2024 年 4 月 9 日
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 件のコメント
James Tursa
James Tursa 2024 年 4 月 10 日
@Ray Can you post an image of the differential equations you are trying to solve?
Ray
Ray 2024 年 4 月 10 日
It has to be as a pdf, the images came out wrong. Hope this makes sense.

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Torsten
Torsten 2024 年 4 月 10 日
編集済み: Torsten 2024 年 4 月 10 日
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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