How to solve a double complex ode using ode45 in MATLAB
조회 수: 5(최근 30일)
Walter Roberson 2021년 10월 16일
The infinite boundary condition is a problem for symbolic solution: dsolve() rejects it.
The 0 initial time is a problem: you divide by time, so you generate a numeric infinity at the initial time, which numeric solvers cannot recover from.
If you set initial time above 0, and set the boundary condition to be finite, then you get a singularity. The time of singularity depends upon the boundary condition you set -- with the failure time being approximately 1 over the boundary condition.
syms x(t) y(t)
dx = diff(x)
dy = diff(y)
odes = [dx == (sqrt(1+x^2)*y)/(x*(t^2));
dy == (x^3)*(t^2)]
ic = [x(0) == 1e8, y(0) == 0]
sol = dsolve(odes, ic)
[eqs,vars] = reduceDifferentialOrder(odes,[x(t), y(t)])
[M,F] = massMatrixForm(eqs,vars)
f = M\F
odefun = odeFunction(f,vars)
xy0 = double(rhs(ic))
tspan = [1e-50 100]
[t, xy] = ode45(odefun, tspan, xy0);
semilogy(t, xy(:,1), 'k+-')
plot(t, xy(:,2), 'k*-')