Consider a differential equation \(\left(\frac{d^{4}}{dx^{4}}+2\frac{d^{2}}{dx^{2}}\right)f(x)=g(x)\) Let F(\(k\)) and G(\(k\)) be the Fourier transforms of \(f(x)\) and \(g(x)\) respectively. Then F(\(k\)) is :
\(F(k)=\frac{G(k)}{(k^{4}-2k^{2})}\)
Take \(f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F(k)e^{ikx}dk\). Each derivative with respect to \(x\) brings down a factor \(ik\), so \(\frac{d^{n}}{dx^{n}}\to(ik)^{n}\).
Therefore \(\frac{d^{4}}{dx^{4}}\to(ik)^{4}=k^{4}\) and \(\frac{d^{2}}{dx^{2}}\to(ik)^{2}=-k^{2}\).
Transforming both sides of the equation gives \(\left(k^{4}+2(-k^{2})\right)F(k)=G(k)\), i.e. \((k^{4}-2k^{2})F(k)=G(k)\).
Solving for the unknown transform, \(F(k)=\frac{G(k)}{k^{4}-2k^{2}}\).
Check: with the opposite sign convention the factor is \((-ik)^{n}\), and for the even powers 4 and 2 this gives the same \(k^{4}\) and \(-k^{2}\), so the result does not depend on the convention.
Hence, \(F(k)=\frac{G(k)}{(k^{4}-2k^{2})}\).
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