Why do I receive Jacobian singular error while solving second order differential equation with boundary conditions?

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Vevina Shreya
Vevina Shreya 2024-1-2
编辑： Torsten 2024-1-2
Ran in:
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);
Error using bvp4c
Unable to solve the collocation equations -- a singular Jacobian encountered.
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!

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采纳的回答

Torsten
Torsten 2024-1-2
编辑：Torsten 2024-1-2
Ran in:
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)
sol = 
conds = [y(0)==0,df(0)==0];
sol = dsolve(ode,conds)
sol = 
0

更多回答（1 个）

Ayush
Ayush 2024-1-2
Ran in:
Hi @Vevina Shreya
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

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