Where did this term come from? (Fourier)

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SUMMARY

The discussion centers on calculating the Fourier transform of the equation \(\epsilon^2 y''' - y' = \frac{1}{1+x^2}\). The user derives the Fourier transform \(\hat{y}(k,t)\) and identifies a discrepancy in the expected result, which includes an additive constant. The correct form of the transform is \(\hat{y} = -\frac{\pi e^{-|k|}}{ik(1-\epsilon^2 k)} + 2\pi a \frac{k\delta(k)}{k(1-\epsilon^2 k^2)}\). The user realizes that incorporating an arbitrary constant is necessary when transforming higher-order differential equations but seeks clarification on why this principle is not commonly addressed in Fourier transform literature.

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rsq_a
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A simple question.

Suppose I have [tex]\epsilon^2 y''' - y' = \frac{1}{1+x^2}[/tex].

The goal is to calculate the Fourier transform of [tex]y(x,t)[/tex] where we define,

[tex]\hat{y}(k,t) = F[\phi] = \int_{-\infty}^{\infty} y e^{ikx} ds[/tex]

We're also given that,

[tex]F\left[ \frac{1}{1+x^2} \right] = \pi e^{-|k|}[/tex]

Now we take transforms of both sides:

[tex]\rightarrow F[\epsilon^2 y''' - y] = F[\frac{1}{1+x^2}][/tex]

[tex]\rightarrow -i k^3 \epsilon^2 \hat{y} - ik \hat{y} = \pi e^{|k|}[/tex]

[tex]\rightarrow \hat{y} = -\frac{\pi e^{-|k|}}{ik(1-\epsilon^2 k^2)}[/tex]

The answer, however, is supposed to be:

[tex]\hat{y} = -\frac{\pi e^{-|k|}}{ik(1-\epsilon^2 k)} + 2\pi a \frac{k\delta(k)}{k(1-\epsilon^2 k^2)}[/tex]

where 'a' is some constant.

My question is why? I know it has something to do with an additive constant, but I need someone to be explicit with the mistake.
 
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I've figured out a way to 'see' the result:

Instead write the equation like this:

[tex]\epsilon^2 y''' - y' = \frac{1}{1+x^2} + \frac{d}{dx} 2\pi{a}[/tex]

and the result follows automatically.

But what I don't understand is why I 'need' to do this. Why haven't I seen this in any texts on Fourier transforms (that if one is applying the transform to a (for example), third order equation reducible to second order, one needs to add arbitrary constants.
 
Last edited:

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