an ode with arguements

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.

Answer 1

[James Tursa]

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"?

[Ray]

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.

Answer 2

[Star Strider]

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]);

.

[Ray]

So we do have the initial conditions from the second, so the total initial conditions should be as follows: [0; 0; guessf; 0; guessT]

[Star Strider]

I don’t udnerstand what you are doing.

If you define the initial conditions the way you described in your latest Comment (that should work with respect to your differential equations), you likely do not need to pass ‘T’ as an additional parameter. However, since I do not understand what you want to do, I will defer to you to determine that.

[Ray]

I have 2 second order differential equations, the first one we solve for the 3 initial conditions in terms of f. So now i have another question based on the outputs of the first second order differential equation which are in terms of T: T(2); -pr*f(:,1)*T(2) as you can see here I have f(:,1) multiplied by other variables, this is the where the trouble is. I cannot store f in a function file so therefore it is not defined and I cannot solve it. I also tried calling it as an annonymous function and the vector lengths did not align.

[Star Strider]

I can’t even guess what you want to do from what is currently posted.

See ODE with Time-Dependent Terms for one possible approach.

[James Tursa]

@Ray Can you post an image of the differential equations you are trying to solve?

[Ray]

It has to be as a pdf, the images came out wrong. Hope this makes sense.

Answer 3

[Torsten]

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.