Filter löschen
Filter löschen

ode45 with boundary conditions

85 Ansichten (letzte 30 Tage)
Rick
Rick am 11 Okt. 2014
Kommentiert: Rick am 11 Okt. 2014
Hello,
If I have boundary conditions such as at x = 0, the temperature is T1, and at x = L, temperature is T2, how do I incorporate that into ode45? can you span two different parameters? I am trying to solve fourier's law of condution
Qdot = -kA dT/dx
Here is my code, but when I run it the temperature at the end is not 300 like I want it to be
k0 = 5;
beta = 10;
k = @(T) k0*(1+beta*T);
xspan = [0 10]; % meters
L = 10; % meters
T1 = 300;
Tf = 100;
A = 10;
Qdot = ((-k0*A)/L)*(Tf + (beta/2)*Tf^2 - T1 + (beta/2)*T1^2);
dTdx = @(x,T) Qdot/(-k(T)*A);
[X, T] = ode45(dTdx,xspan,T1);
plot(X,T)
xlabel('distance (meters)')
ylabel('Temperature (C)')
title('Temperature vs. distance')

Antworten (2)

Marc
Marc am 11 Okt. 2014
Try using bvp4c or bvp5c for boundary value problems.
ode45 is a initial value problem.
  1 Kommentar
Rick
Rick am 11 Okt. 2014
I have never seen this function. How can I incorporate it into this problem? It seems like the second argument for the boundary conditions is a function handle, but I just want the boundary conditions to be at x=0, T=T1, and at x=L, T=T2

Melden Sie sich an, um zu kommentieren.


haMed
haMed am 11 Okt. 2014
Your Problem is discretized on x and you are integrating the resulting ode problem by ode-solver.
In this case I try to include the boundary conditions in the discretized equations. Or to represent the boundary conditions as states.
  1 Kommentar
Rick
Rick am 11 Okt. 2014
I don't understand your response. How can I do whatever you are trying to say?

Melden Sie sich an, um zu kommentieren.

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by