Discussion
A function f(x) satisfies the differential equation \(\frac{d^{2}f}{dx^{2}} - \omega^{2}f(x) = -\delta(x - a)\), where ω is positive. The Fourier transform \(F(k) = \int_{-\infty}^{\infty} f(x)e^{ikx}dx\) and solution f(x) of the differential equations are respectively :
Answer: D
The transform and solution are \(\frac{e^{ika}}{k^{2} + \omega^{2}}\) and \(\frac{1}{2\omega}e^{-\omega|x-a|}\).
Multiply the equation by \(e^{ikx}\) and integrate. Each derivative brings down a factor \(-ik\), so \(f''\) becomes \(-k^{2}F\), and the delta function picks out \(e^{ika}\):
\[ \begin{aligned} -k^{2}F - \omega^{2}F &= -e^{ika} \\ F(k) &= \frac{e^{ika}}{k^{2} + \omega^{2}} \end{aligned} \]
Away from x = a the equation \(f'' = \omega^{2}f\) has solutions \(e^{\pm\omega x}\); keeping the decaying one on each side gives \(Ce^{-\omega|x-a|}\). The jump in slope at x = a must equal -1, which fixes \(C = 1/2\omega\). The options with \(k^{2} - \omega^{2}\) belong to the opposite sign of the \(\omega^{2}\) term, and the option with \(e^{+\omega|x-a|}\) grows without limit, so it is not an acceptable solution.
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