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 ) &lt; 0 and h &gt; 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 ) = 1b) y ′ = − ( 1 + t ) e x p ( y ) , y ( 0 ) = 1If 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 ) &gt; 0?Thank you very much, any help is very appreciated.

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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    )  &amp;lt;  0 and   h  &amp;gt;  0, (h= step size).One of my problems is that I don&#039;t know how to calculate the sign of the   λ of this non-linear equations.a)          y    ′    =  1      /    (  t  +  y  )  ,    y  (  0  )  =  1b)          y    ′    =  −  (  1  +  t  )  e  x  p  (  y  )  ,    y  (  0  )  =  1If 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    )  &amp;gt;  0?Thank you very much, any help is very appreciated.

Shyla Maldonado · Accepted Answer

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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