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Let $k_1 \neq k_2$ be positive reals, $t_0 > 0$ and consider the following Cauchy problem in $[0,+\infty)$: \begin{cases} y(t) + k(t)y''(t) = 0 \newline y(0) = 1/\sqrt{k_1} \newline y'(0) = 0, \end{cases} where \begin{equation} k(t) = \begin{cases} k_1 \hspace{1 cm} 0 \leq t \leq t_0 \newline k_2 \hspace{1 cm} t > t_0 \end{cases} \end{equation}

It is clear that there exists a unique $C^1$ solution $y$ in $[0,+\infty)$, obtained by gluing two smooth solutions of the equation in $[0,t_0]$ and $[t_0,+\infty)$. An explicit computation gives: \begin{equation} y(t) = \begin{cases} \dfrac{1}{\sqrt{k_1}} \mathrm{cos} \bigg(\dfrac{t}{\sqrt{k_1}} \bigg) \hspace{1 cm} 0 \leq t \leq t_0; \newline A \ \mathrm{cos}\bigg( \dfrac{t}{\sqrt{k_2}} + \phi \bigg) \hspace{1 cm} t > t_0, \end{cases} \end{equation} where $A$ and $\phi$ are obtained imposing the $C^1$ condition. With further computations one discovers that the amplitude is constant, i.e. $A = 1/\sqrt{k_1}$, if and only if $t_0 = n \pi \sqrt{k_1}$ for some $n \in \mathbb N$.

I am searching for a good numerical method for the problem in the constant-amplitude case. I have tried several classical schemes, but I discovered that when $k_2 >> k_1$ the second amplitude also grows bigger. Any suggestion?

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  • $\begingroup$ Try a symplectic integrator from DifferentialEquations.jl in Julia: docs.sciml.ai/dev then use events to handle the jump. You can also specify that you want to integration steps to include $t_0$ so you hit the jump dead-on $\endgroup$ – whpowell96 Apr 24 at 3:29
  • $\begingroup$ @whpowell96 thank you, I'll have a look. $\endgroup$ – avril_14th Apr 24 at 6:22
  • $\begingroup$ @ChrisRackauckas $\endgroup$ – Alone Programmer Apr 24 at 18:06

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