Fourier transform (The 1-D neutron diffusion equation) The $1-D$ neutron diffusion equation with a (plane) source is $-D\frac{\mathrm{d^{2}}\varphi (x) }{\mathrm{d} x^{2}}+K^{2}D\varphi (x)=Q\delta (x)$ where $\varphi (x)$ is the neutron flux, $Q\delta (x)$ is the (plane) source at $x = 0$ and $D$ and $K^2$ are constants. Apply a Fourier transform. Solve the equation in transform space. Transform your solution back into x-space. _ANS_ :$\varphi (x)=\frac{Q}{2KD}e^{-|Kx|}$

$$-D\frac{\mathrm{d^{2}}\varphi (x) }{\mathrm{d} x^{2}}+K^{2}D\varphi (x)=Q\delta (x)$$ Apply Fourier Transform: $$\mathcal {F}\\{ f'(x)\\}=iw \mathcal{F}\\{f\\}(w) \text { , and } \mathcal {F}\\{\delta(x)\\}=1$$ $$-(iw)^2\varphi (w) +K^{2}\hat {\varphi } (w)=\frac Q D$$ $$(w^2 +K^{2})\hat {\varphi }(w)=\frac Q D$$ $$\hat {\varphi } (w)=\frac Q {D(w^2 +K^{2})}$$ $$\hat {\varphi } (w)=\frac Q {2KD} \frac {2K} {(w^2 +K^{2})}$$ Now apply Inverse Fourier Transform: $$\mathcal{F^{-1}} \left \\{\frac {2K} {(w^2 +K^{2})}\right \\}=e^{-K|x|} \text { , with } K>0$$ $$\varphi (x)=\frac Q {2KD} e^{-K |x|} $$