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I am trying to solve the following linear ODE for $y(x)$:

$$y^{\prime\prime\prime}+y^{\prime\prime}+\mathcal{H}[y^{\prime\prime}]+y^\prime-cy=0$$ subject to the boundary conditions $y\rightarrow0 $ as $x\rightarrow\pm\infty$, where $c$ is a constant and $$\mathcal{H}[y](x)=\frac{1}{\pi}\mathrm{PV} \int_{-\infty}^\infty\frac{y(s)}{x-s}\:\mathrm{d}s $$ is the Hilbert transform and PV means that the integral is taken as the principal value sense.

My attempt:

Noted that the Hilbert transform is a linear operator, I first wrote down its characteristic equation as follows: $$r^3+r^2+\mathcal{H}[r^2]+r-c=0. \tag{1}$$

Then I noticed the Hilbert transform has the property: $\mathcal{H}[y^{\prime\prime}]=(\mathcal{H}[y])^{\prime\prime}$, the ODE may have another characteristic equation: $$r^3+r^2+(\mathcal{H}[r^0])^{\prime\prime}+r-c=0 \Rightarrow r^3+r^2+r-c=0 \tag{2}$$

where the property of $\mathcal{H}[\rm{constant}]=0$ has been used.

I did not know which one is correct or both are wrong. If Eq.(2) is correct, the ODE could be solved with the roots of the characteristic equation. However, if Eq.(1) is right, I even don't know how to write down its roots due to the presence of the Hilbert transform (if it is possible).

Does anyone have any insight into handling this? Thank you.

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  • $\begingroup$ I would call this an ODE but an integro-differential equation. The characteristic equation arises if you plug in exponentials, so what happens when you do this here? $\endgroup$ – Dirk Jun 10 '19 at 15:26
  • $\begingroup$ Hi @Dirk Thanks for the suggestion. But which kind of exponential function is more proper: $y=e^{\sigma x}$ or $y=e^{i\omega x}$, they will give different chara. eqs. $\endgroup$ – jsxs Jun 11 '19 at 3:40
  • $\begingroup$ Well, both are the same (take complex s), but I don't think that this will work. $\endgroup$ – Dirk Jun 11 '19 at 4:59
  • $\begingroup$ I can't follow, sorry. $\endgroup$ – Dirk Jun 11 '19 at 20:17

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