BVP4C for solving an 3rd order ODE

I am stuck in boundary condition which can't be slove with ode45.

to solve this problem can you please help me with boundary condition and convert it BVP4C ?

my progress is -

tRange = [0 10];


function dYdt = odefun(t,Y)

% Extract Y1,Y2,Y3 the element of Y
Y1 = Y(1);
Y2 = Y(2);
Y3 = Y(3);

% Expression for dY1dt, dY2dt,dY3dt.
dY1dt = Y2;
dY2dt = Y3;
dY3dt = -(1/2)*Y1*Y3;

% Create dYdt, column vector containing dY1dt, dY2dt, and dY3dt
dYdt = [dY1dt; dY2dt; dY3dt];

end

Answer 1

[Torsten]

tmesh = linspace(0,10,100);
solinit = bvpinit(tmesh, [0 1 0]);
sol = bvp4c(@odefun, @bcfcn, solinit);
plot(sol.x, sol.y)

function dYdt = odefun(t,Y)
  % Extract Y1,Y2,Y3 the element of Y
  Y1 = Y(1);
  Y2 = Y(2);
  Y3 = Y(3);
  % Expression for dY1dt, dY2dt,dY3dt.
  dY1dt = Y2;
  dY2dt = Y3;
  dY3dt = -(1/2)*Y1*Y3;
  % Create dYdt, column vector containing dY1dt, dY2dt, and dY3dt
  dYdt = [dY1dt; dY2dt; dY3dt];
end

function res = bcfcn(ya,yb)
  res = [ya(1);ya(2);yb(2)-1];
end

[Syed Arafun Nabi]

Thanks a lot. Here doesnot the 1st one will plot y vs t & 2nd one dy/dt vs t?

subplot(2,1,1), plot(sol.x, sol.y)
subplot(2,1,2), plot(sol.x, sol.y(2,:))

[Torsten]

Yes, and sol.y(3,:) is d^2y/dt^2.

[Shriram]

Thanks for your answer!

I was curious about how to implemenet this method for a problem where the boundary conditions are as follows:

y(0) = 0, y'(2) = 0 and y'(10) = 1?

How would we formulate this in MATLAB?

[Torsten]

You need two conditions at t = 2, not only one. Then you can solve the problem first from t=0 to t=2 with a total of 3 boundary conditions (one at t=0, two at t=2) and continue solving from t=2 to t=10, again with a total of three conditions given (two at t=2, one at t=10).

Note that the solution might have some discontinuous derivatives in t=2.

If you only prescribe a total of three conditions, there is no unique solution to the problem. BVP4C cannot be used in this case. Sometimes it's possible to get a parameter-dependent solution with "dsolve".