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I need to solve the following ODE

$$ f''(x) - \zeta f(x) + \zeta\delta(x-b) = 0,$$

where $x\in(-\infty,\infty)$ and where $f(x)\rightarrow 0$ as $x \rightarrow \mp \infty$. The solution ignoring the Dirac impulse is given by

$$f(x) = c_1 e^{\sqrt{\zeta}x} + c_2 e^{-\sqrt{\zeta}x}.$$

Since I have a Dirac impulse at $x=b$, I should be solving for two ODEs, one below $x=b$ and another above $x=b$. Then I have to put together both solutions at $x=b$. This is where I am confused, how can I do this part?

A bit more on the intuition behind the problem. The ODE in question is a steady state Fokker-Planck (or Kolmogorov Forward) Equation. Mass is injected at $x=b$ and dissipates both to the left and right of $x=b$. Then, mass anywhere in $x\in(-\infty,\infty)$ is taken out at a rate $\zeta$ and reinjected back to $x=b$.

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You have two solutions as you noted for: $$ f''(x) - \zeta f(x) + \zeta\delta(x-b) = 0,$$ given by (denoting $\phi^2 = \zeta$). For simplicity, I will let $b = 0$ (you can also do this via a shift). $$ f(x) = \begin{cases} A e^{\phi x} & \text{if } x \le 0 \\ B e^{-\phi x} & \text{if } x > 0. \end{cases} $$ Continuity is required and so we must have $A = B$. Finally, we will see what is required in the neighborhood of $x = 0$ by looking at a small integral containing $0$ of the ODE and take the limit $$ 0 = \lim_{\epsilon \to 0} \int_{-\epsilon}^\epsilon f''(x) - \phi^2 f(x) + \phi^2\delta(x) \ dx = f'(0^+) - f'(0^-) + \phi^2 $$ Hence, $0 = -\phi B - \phi A + \phi^2$. The two conditions imply that $A = B = \phi/2$. Hence the solution: $$ f(x) = \frac{\phi}{2} e^{-\phi |x|}. $$

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  • $\begingroup$ More details are found here: Green's function $\endgroup$
    – Gregory
    Commented Sep 20, 2022 at 21:17
  • $\begingroup$ Thank you! Before asking the question I actually tried integrating the ODE around $x=b=0$, but since I forgot that $f(x)$ is continuous at $b$ I got confused with the integral of $f(x)$ (I didnt recognize that this term would vanish). $\endgroup$
    – Mattewlab
    Commented Sep 20, 2022 at 21:39
  • $\begingroup$ @Snoop The function does integrate to 1. Wouldnt this be a steady state Fokker-Planck equation? $\endgroup$
    – Mattewlab
    Commented Sep 20, 2022 at 21:43
  • $\begingroup$ @Mattewlab I rewrote the code and get a probability density. However the ODE is not a Fokker-Planck equation as per definition. $\endgroup$
    – Snoop
    Commented Sep 20, 2022 at 21:54

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