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Imagine that I am in a lab measuring a certain force that is time dependent, e.g. there is a spring subjected to changes in temperature, which results in a time-dependent stiffness, $$F(t)=k(t)\delta,$$ where $k(t)$ is the time dependent stiffness. The only measurement that I have access to is $k(t)$. From it, I can derive a certain $F(t)$ and, integrating twice, obtain $x(t)$ for any given initial conditions, $$m\ddot{x}(t)=k(t)x(t). \tag{1}$$

However, since my measurement is noisy (e.g. due to limited precision of the apparatus, experimental errors, etc.), $k(t)$ has a certain noise. In fact, what one has is a set of discrete measurements at different steps of time. If $T=[t_0,\;t_f]$, I take $N$ measurements at $t_n=t_0+n\Delta t$, where $\Delta t=\frac{t_f-t_0}{N}$. Each of this measurements has a certain noise, $k_n\equiv k(t_n)+\eta(t_n)$, where for generality I assume that the noise might depend on time. Then, I could integrate $(1)$ via Euler (for simplicity). Defining $p(t)\equiv\dot{x}(t)$, I have $$x_{n+1}=x_{n}+\Delta t p_n,$$ $$p_{n+1}=p_n+\frac{\Delta t}{m}(k_n+\eta_n)x_n.$$

My question is, when using a noisy measurement as a source for an ODE, how is the noise transferred to the solution of the ODE? Does the noise vanish when integrated (since e.g. in Euler the noise term is multiplied by $\Delta t^2$), and therefore the ODE acts as a filter, or is there any more subtle relation between the noise of $x_n$ and the noise of the measurements $k_n+\eta_n$?

Note: the above scenario is just an example, I have no springs or even lab. I am just interested on how the noise behaves when acting as the source of an ODE.

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  • $\begingroup$ You may need smoothing your signal $f(t)$ by some polynomials before integrating. $\endgroup$ Commented May 26 at 10:55
  • $\begingroup$ If you have a noise term, then you no longer have an ODE but an SDE which is solved using Ito calculus. Smoothing may be of no use, depending on the strength of the noise. $\endgroup$ Commented May 26 at 13:03

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In the general case, the differential equation encompassing a noisy force would take the form of a Langevin equation, $$m\ddot{\mathbf{x}}=-\lambda\dot{\mathbf{x}}-\nabla V+\mathbf{f}(t)\tag{1}$$ where $V$ is the particular potential of interest, $\lambda$ the damping coefficient and $\mathbf{f}(t)$ your time-dependent noise term. And while this looks like an ODE, the noise term changes it to a stochastic differential equation which is solved in an entirely different way than ordinary calculus.

In any solution to Eq 1 here, you are going to end up with a term like, $$\int_0^tf(t')\,\mathrm{d}t'$$ which, unfortunately, does not have a closed form solution due to the random nature of $f(t)$. Under general stochastic calculus rules, we really can only say that its mean is 0 and has a variance proportional to $\mathrm{d}t$ (see this article on the Weiner process for instance), from which we can derive features/limits/solutions to $x(t)$.

You also cannot integrate it under forward Euler rules, these have to be adjusted a little bit, as I discuss in this answer, since you have to draw a random number from a distribution at each time step. And even after accounting for that, you have to integrate the system a multitude of times to generate a series of paths and take an average of the solution.

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using a noisy measurement as a source for an ODE, how is the noise transferred to the solution of the ODE?

Adding noise to the equation of motion means that your independent variable becomes an stochastic process (i.e. the variable evolves erratically in time, and every time you repeat the experiment you will obtain different outcomes). The idea of solving an equation with noise is to understand the statistical properties of the process $x(t)$, which will depend on the statistical properties of the noise $\eta(t)$.

Does the noise vanish when integrated ... ?

In general, no. A way to see this is integrating the equation of motion assuming the prescription for the noise that you provided

$$p_n = p_0 +\frac{\Delta t}{m}\sum_{i=0}^{n-i}k_i\,x_i+\frac{\Delta t}{m} y_n,$$ where $$ y_n =\sum_{i=0}^{n-i} \eta_i \,x_i.$$ As you see, $y_n$ is a sum of random variables, so it is a random variable itself. Nothing tell us that $y_n$ should converge to zero as $n\to \infty$.

Typically, noise is removed performing averages. If you have access to different realizations of the noise (i.e. different repetitions of the same experiment $p_n^{(1)},p_n^{(2)},\dots,p_n^{(M)}$), you could study the evolution for the average $$\bar{p}_n= \frac{1}{M} \sum_{j=1}^{M}p_n^{(j)}.$$ If the first moment of the variables $\eta_n$ is zero, then $\bar{p}_n$ will behave as if there was no noise when $M\to\infty$, because $$ \bar{y}_n =\sum_{i=0}^{n-i} \frac{1}{M}\sum_{j=1}^{M} \eta^{(j)}_i \,x_i\approx \sum_{i=0}^{n-i} \langle \eta_i \rangle x_i =0 .$$

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