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So in general for ODE we come up with a numerical method for

$y'=f(t,y)$

Then apply the numerical method to $y'=\lambda y$, where $\lambda$ is some scalar

My question is how can we claim the stability criteria we get for a numerical method from solving $y'=\lambda y$ have any influence in the general case $y'=f(t,y)$

nmasanta
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Vogtster
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1 Answers1

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I wouldn't say that it always does. But, if a numerical method can't even handle a linear problem, then it's probably going to have trouble on a more significant problem. Generally, the different types of stability (zero stability, absolute stability, etc.) are more so a means of crossing methods off the list, so to speak, or to show what context the method is viable. Being absolutely stable, for a given range of $\lambda$, doesn't necessarily say whether a method is good, but more so if it's not (or, to be less cavalier, how restrictive it is).

Take backward Euler. Its stability region is pretty large (it contains the left half-plane; such a method is called $A$-stable), so it's a solid option for the linear problem. Take the explicit midpoint/leapfrog rule. Its region is extremely restrictive (open interval from $-i$ to $i$), so we often won't consider it for even the linear problem, which does not inspire much hope for a more complicated $f$ (it does have applications when $\lambda$ is purely imaginary, which happens for hyperbolic PDEs, actually, but the general point still stands).

If the solution of a non-linear problem $y'=f(y)$ is not changing too much relative to the time step (such as when we have a fixed point), then we can do a linear approximation over a small time interval. In this case, our linear stability results can prove useful (note that it depends on how our method deals with the linear system given by evaluating the Jacobian at the fixed point).

cmk
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  • Hey cmk, thank you for the response. Do you have an example of lets say a nonlinear equation where the stability condition does not work when you try to simulate with it? (even for euler for example) – Vogtster Jun 19 '19 at 22:19
  • Sorry, could you rephrase that? The absolute stability only directly tells me about how my method works in the linear case. Also, I've made an edit to the post. I forgot to mention that they can tell you about autonomous nonlinear systems near equilibrium. – cmk Jun 19 '19 at 22:25
  • Sorry for being not clear. Do you know a simple example of non-linear ode that euler method fails at solving - even though there is a unique solution to the problem? – Vogtster Jun 20 '19 at 04:21
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    @Vogtster : Such an example does not exist, as any consistent method (which all the named methods are) converges towards the exact solution when the step size falls to zero. However, in practical applications there is a lower bound for the step size, as floating point errors accumulate proportional to the number of steps, that is, inverse proportional to the step size. Combinations of ODE and methods where there is (nearly) no room between this minimal step size and the maximal step size required for stability are called "stiff". – Lutz Lehmann Jun 20 '19 at 08:07
  • How is the stability region of the (implicit) midpoint method restricting? It is the full negative half-plane, exactly what you want for A-stability. In more detailed stability terms, the behavior on the imaginary axis is unfavorable for larger step sizes, but then one might question the use of an order 2 method. Or did you mean the explicit method? All explicit methods have a bounded stability region in the negative half-plane. – Lutz Lehmann Jun 20 '19 at 08:14
  • @LutzL I meant explicit (leapfrog), whose stability region is just the open interval from $-i$ to $i$. I should’ve made that more clear, I’ll edit it; thanks for pointing that out. – cmk Jun 20 '19 at 10:31
  • Ah yes, you mean the two step Nyström method, as discussed with the obligatory Hairer et al citation here: https://math.stackexchange.com/q/3074952/115115 – Lutz Lehmann Jun 20 '19 at 11:02
  • @LutzL That's it! Leveque calls it the midpoint rule, I usually call it leapfrog, and as you say, it is a 2-step explicit Nystrom method. – cmk Jun 20 '19 at 12:31
  • Invoking convergence is only valid for linear type problems. If you have a consistent and stable method for finite difference then it is convergence. We have established all methods work for for the problem is it is tested for $y'=\lambda y$, but for nonlinear problems how can you guarantee the convergence? – Vogtster Jun 20 '19 at 20:40
  • @Vogtster Convergence is a general notion, and all common methods (i.e. not pathological methods) you'll see are convergent. For example, linear multistep methods applied to an IVP $y'(t)=f(y(t),t)$ are convergent if and only if they are consistent and zero stable (for $f$ Lipschitz in $y$). – cmk Jun 20 '19 at 21:16
  • @cmk I guess what I was interested in is a nonlinear problem which lets say euler fails at solving, even though we have a unique solution to the problem. Other responder says its not possible - thoughts? – Vogtster Jun 20 '19 at 22:25
  • Yes, I agree with them. We might need a really small time step (which makes it not as viable for stiff equations, just like LutzL said). Note that forward Euler is a linear multistep method, so the theorem that I referenced applies. – cmk Jun 20 '19 at 23:00