Why do I receive Jacobian singular error while solving second order differential equation with boundary conditions?
    5 Ansichten (letzte 30 Tage)
  
       Ältere Kommentare anzeigen
    
    Vevina Shreya
 am 2 Jan. 2024
  
    
    
    
    
    Bearbeitet: Torsten
      
      
 am 2 Jan. 2024
            I am trying to solve the following second order differential equation: ηF''-F'+FF'= 0;  The boundary conditions being F(0)=0 , F'(0)=0 and F'(inf)=0.
This is my code:
solinit = bvpinit(linspace(0,10,100),@guess);
solve = bvp4c(@f,@bc,solinit);
plot(solve.x,solve.y(1,:))
xlabel('eta')
ylabel('F')
function dydx = f(x,y)
dydx = [y(2); (1/(x)).*(y(2)-(y(1).*y(2)))];
end
function output = bc(ya,yb)
output = [ya(2)-1; yb(2)];
end
function y = guess(x)
y = [cos(x) -sin(x)];
end
Any help regarding this would be greatly appreciated! Thank you!
0 Kommentare
Akzeptierte Antwort
  Torsten
      
      
 am 2 Jan. 2024
        
      Bearbeitet: Torsten
      
      
 am 2 Jan. 2024
  
      You need two boundary conditions, not three to fix a solution for a second-order ODE. 
Anyhow: in each of your cases, you will get F identically 0 if you use a numerical method for a solution. 
I don't know whether a non-trivial solution for your equation exists.
According to the symbolic approach below, there is no bounded solution at x=0 apart from the trivial one.
syms x y(x)
df = diff(y,x);
d2f = diff(y,x,2);
ode = d2f == df*(1-y)/x;
sol = dsolve(ode)
conds = [y(0)==0,df(0)==0];
sol = dsolve(ode,conds)
0 Kommentare
Weitere Antworten (1)
  Ayush
      
 am 2 Jan. 2024
        I understand that you are receiving Jacobian singular error while solving second order differential equation with boundary conditions. Here is the modified code you can try:
% Set up the initial mesh and initial guess
solinit = bvpinit(linspace(0,10,100), @guess);
% Solve the boundary value problem
solve = bvp4c(@f, @bc, solinit);
% Plot the solution
plot(solve.x, solve.y(1,:))
xlabel('eta')
ylabel('F')
% Define the differential equation as a system of first-order ODEs
function dydx = f(x,y)
    dydx = [y(2); (1/(x+eps)).*(y(2)-(y(1).*y(2)))]; % Add eps to avoid division by zero
end
% Define the boundary conditions
function res = bc(ya,yb)
    res = [ya(1); yb(2)]; % F(0) = 0, F'(inf) = 0
end
% Initial guess for the solution
function y = guess(x)
    % Use a cubic polynomial that satisfies the boundary conditions
    y = [(x/10).^3; 3*(x/10).^2/10];
end
Thanks,
Ayush
0 Kommentare
Siehe auch
Kategorien
				Mehr zu Equation Solving finden Sie in Help Center und File Exchange
			
	Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!

