I have been studying the Backward Euler Method, and I am have having some of problems representing the stability of the method of some particular ODEs.
If I am thinking correctly this method will be stable if the solutions of the ODE are stable, which means that $Re(λ) < 0$ and $h > 0$, (h= step size).
One of my problems is that I don't know how to calculate the sign of the λ of this non-linear equations.
a)$\ y'=1/(t+y), \quad y(0) = 1$
b)$\ y'=-(1+t)exp(y),\quad y(0) = 1$
If there is any positive λ, then a step size big enough will be stable even though it might not be a good representation of the solution?
Or will the solution be automatically unstable if the $Re(λ) > 0$?
Thank you very much, any help is very appreciated.
You need to consider the $y$-derivative, as the $λ$ are the eigenvalues of the Jacobian. The idea is that the linearization of the equation has solutions that are locally equivalent to the solutions of the non-linear equation.
In the first example the derivative is $-1/(t+y)^2$, always negative on the domain of the ODE. In the second example the derivative is $-(1+t)\exp(y)$, which is also negative in some neighborhood of the initial point.
In the first example there is an asymptote $y=-1-t$ that the numerical solution should also converge to under a stable method. In the second example there is no asymptote, the solution is falling below all boundaries, the stability of the method has no greater influence on the numerical solution.