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(lambda)<0 and h>0, (h= step size). One of my problems is that I don't know how to calculate the sign of the lambda of this non-linear equations. a) y′=1/(t+y), y(0)=1 b) y′=−(1+t)exp(y), y(0)=1 If there is any positive lambda, 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(lambda)>0?

Vincent Norman

Vincent Norman

Answered question

2022-10-20

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 R e ( 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 ) , y ( 0 ) = 1
b)   y = ( 1 + t ) e x p ( y ) , 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 R e ( λ ) > 0?
Thank you very much, any help is very appreciated.

Answer & Explanation

Shyla Maldonado

Shyla Maldonado

Beginner2022-10-21Added 15 answers

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.

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