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Could anyone please explain me how to solve differential equations using Fourier transform, distributions and Schwarz's spaces? I would be so grateful if you could help with a practical example -where i is the imaginary unit, i.e. $\sqrt{-1}$ (in theory I think I understood how it works but cannot actually solve any) \begin{equation} \frac{d^2u}{dx^2}+2i\frac{du}{dx}+4u=0 \end{equation} I tried to understand the passages but get stuck after applying the Fourier transform and transforming derivatives into multiplication by polynomials; in the example above, I get: \begin{equation} \left(-k^2-2k+4 \right) \hat{u}=0 \end{equation}

At this point, I suppose I should think about a linear combination of $\delta$ and derivatives: $ \sum_{n}c_n\delta^{\left(n\right)}$, since I have a distributional equation supported in zero. But I am not sure how to compute the coefficients and whether the method is right.

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  • $\begingroup$ $../dx^2$ ? and what is $i$? $\endgroup$ – G Cab Jan 2 at 17:22
  • $\begingroup$ Sorry, yes, I modified: i is $\sqrt{-1}$! $\endgroup$ – Anna Stone Jan 2 at 17:24
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    $\begingroup$ I see that you are new here. Welcome! Please show us more of what you have done. It will be easier to give better help if you show your first steps and tells us a bit more in detail why you cannot proceed. Again, welcome! $\endgroup$ – mickep Jan 2 at 17:58
  • $\begingroup$ I did, thank you..!! $\endgroup$ – Anna Stone Jan 2 at 18:49
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Factorize your polynomial $$-k^2-2k+4=-(k-a)(k-b)$$ Take $\phi \in C^\infty_c(-r,r),\phi(0)=1$ where $r<|b-a|/2$. For any $\varphi\in C^\infty_c(\Bbb{R})$ then $$\Psi(k)=\frac{\varphi-\varphi(a)\phi(k-a)-\varphi(a)\phi(k-a)}{-(k-a)(k-b)} \in C^\infty_c(\Bbb{R})$$

Thus $$\langle \hat{u},\varphi-\varphi(a)\phi(k-a)-\varphi(a)\phi(k-a)\rangle=\langle -(k-a)(k-b)\hat{u},\Psi\rangle = 0$$ and hence $$\langle \hat{u},\varphi\rangle = \varphi(a) \langle \hat{u},\phi(k-a)\rangle+\varphi(b) \langle \hat{u},\phi(k-b)\rangle$$ ie. $$\hat{u} = A\delta(k-a)+B\delta(k-b)$$ $$ u =\frac{ A e^{ita}+Be^{itb}}{2\pi}$$

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  • $\begingroup$ Very clever. I had the two different pieces in mind that you have but I didn't think to put them together. $\endgroup$ – Cameron Williams Jan 2 at 19:42
  • $\begingroup$ Thank you! But I am sorry, I cannot actually understand how you defined $\Psi$, why in that way, and how to generalize it to other cases. Moreover, why in the definition of $\Psi$ does not appear $b$... $\endgroup$ – Anna Stone Jan 5 at 17:14
  • $\begingroup$ @reuns and if I had a different grade of derivation, for example a third derivative in the equation? $\endgroup$ – Anna Stone Jan 5 at 17:30

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