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Suppose $\mathbf{u}^1\in\mathbb{R}^n$ is an unknown and $\mu^1\in\mathbb{R}$ is a known parameter. Suppose we have solved the non-linear system of equation for $\mathbf{u}^1$ using Newton's iterations.

$$ \mathcal{N}(\mathbf{u}^1;\mu^1)=0 $$

Using $\mathbf{u}^1$ as an initial guess for the Newton iterations, it is possible to accelerate the solution of the system

$$ \mathcal{N}(\mathbf{u}^2;\mu^2) = 0 $$

where $\mu^1$ and $\mu^2$ are sufficiently close. Now, is it possible to extend this to ODEs? For instance, if I have solved

$$ \frac{d\mathbf{u}^1}{dt} = \mathcal{M}(\mathbf{u}^1;\mu^1) $$

Is it possible to accelerate the solution of

$$ \frac{d\mathbf{u}^2}{dt} = \mathcal{M}(\mathbf{u}^2;\mu^2) $$

where $\mu^1$ and $\mu^2$ again are sufficiently close.

The only thing I can think of is to integrate both ODEs implicitly with the same time-step. Solve the first ODE, and use the solution at each time-step to accelerate Newton's iterations for the second. Can anything else be done?

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    $\begingroup$ The start point for the Newton iteration is usually derived from the previous time step. I don't expect a major boost from using the start points from the "neighbouring" solution. I would see a chance to use a close by solution in iterative methods (e.g., multigrid in time, ParaReal, multiple shooting). $\endgroup$
    – cos_theta
    Commented Dec 15, 2022 at 14:00
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    $\begingroup$ If the parameters really only change a small amount, then the solution is a linear function of the parameters and you can solve it for a small additional cost by computing the sensitivity of $u^1$ with regards to the parameters, and then computing $u^2$ as $u^1 + s(\mu^2-\mu_1)$ where $s$ is the computed sensitivity. $\endgroup$ Commented Dec 15, 2022 at 15:51
  • $\begingroup$ @WolfgangBangerth In general the sensitivity will depend on $t$. Consider for example $d_t u = \mu u$. $\endgroup$ Commented Dec 16, 2022 at 7:43
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    $\begingroup$ @MaximUmansky No, you misunderstand: $s(t)$ satisfies an ODE that can be solved alongside the original ODE. ODE solver packages often offer this functionality -- for example, the SUNDIALS package has IDA to solve ODEs and IDAS to solve the sensitivity equations alongside. $\endgroup$ Commented Dec 17, 2022 at 0:05
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    $\begingroup$ @WolfgangBangerth Yes, solving the sensitivity equation alongside the original ODE is a great idea, this seems to be the right way to solve this problem, at least in the limit of a small increment of the parameter $\mu$. You should write this up as an answer. $\endgroup$ Commented Dec 17, 2022 at 4:05

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