Boundary Value Problem with non boundary values
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For solving a Boundary Value Problem with non boundary values (for instance a second order differential defined at [0,1] with values at x=0.2 and x=1) is there a way only using bvp4c except from solving at [0.2,1] and afterwards using bvpxtend for predicting solution's behaviour ?
General suggestions are also welcomed !
6 commentaires
Torsten
le 28 Mai 2021
Your problem description is unclear.
Charalampos Papargyriou
le 28 Mai 2021
Solve your equation on [0,1] by taking the boundary value at x=1 and assuming a boundary value at x=0. Evaluate this solution at x=0.2. Usually, this value won't be the same as the prescribed condition at x=0.2. Use fzero to adjust the boundary condition at x=0 such that both values become the same.
By the way:
Your problem has an analytical solution
y(x)=(x-1)*Ei(x)+6.44286*x-exp(x)-1.72458
where Ei(x) is the exponential integral.
So this example is much suited to test the method I suggested above.
This is a possible implementation (untested !)
The nonlinear equation solver "fzero" tries to find y'(1) such that the solution of the ODE passes through the point (0.2,-1).
function main
y0dot0 = -1.0;
y0dot = fzero(@fun,y0dot0);
tstart = 1.0;
tend = 1e-8;
tspan = [tstart tend];
y0 = [2 y0dot];
[T,Y] = ode15s(@ode,tspan,y0);
plot(T,Y(:,1))
end
function s = fun(y0dot)
tstart = 1.0;
tend = 0.2;
tspan = [tstart tend];
y0 = [2 y0dot];
[T,Y] = ode15s(@ode,tspan,y0);
s = Y(end,1) + 1;
end
function dy = ode(t,y)
dy = [y(2);exp(x)/x^2];
end
Alex Sha
le 19 Juin 2021
Hi, look at your ODE function: y'' = exp(x)/x^2, if x=0, will cause the error of 0/0 .
Alex Sha, If you can reply on my post ( code )in
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