Retarded fundamental solution

A retarded fundamental solution is supported in a chosen forward region, such as $[0,\infty )$ for an ordinary differential operator.

For inverse time Fourier transform $\int \hat{G}(\omega )e^{-i\omega t}d\omega /(2\pi )$, a causal stable oscillator has its poles below the real frequency axis. Undamped real poles are therefore displaced by the prescription $\omega \mapsto \omega +i0$ in its frequency polynomial. For $t>0$, closing the contour below encloses poles with clockwise orientation; for $t<0$, closing above gives zero. Changing the transform's sign reverses the corresponding prescription. For a positive oscillator operator ${\partial }_t^2+{\omega }_0^2$, the inverse is $[{\omega }_0^2-(\omega +i0{)}^2{]}^{-1}$ and its time response is $\Theta (t)\sin ({\omega }_{0}t)/{\omega }_0$.

For $P(z)=a_N{z}^N+\cdots +a_0$ with $N\ge 1$ and $a_N\ne 0$, the unique retarded fundamental solution of $P(D)$ is $E=Hu$, whereThe distributional jump formula for a Heaviside product gives $P(D)E={\delta }_0$. Its explicit expression isThe residue theorem supplies the initial derivatives by the coefficients at infinity. Uniqueness follows because a difference of two retarded solutions is an analytic homogeneous solution vanishing on the negative half-line.

With inverse Fourier transform convention $e^{ix\xi }$, the equivalent contour representation integrates $e^{ix\xi }/P(i\xi )$ along $\mathrm{Im}\xi =-c$, with $c$ larger than the real parts of all roots of $P$. This line lies below all frequency poles. This is an inverse Laplace transform expressed in frequency coordinates: for $P(D)=D-a$, $a>0$, the solution $H(x)e^{ax}$ is a distribution but not a tempered distribution, so it has no ordinary tempered Fourier transform. The contour shift accommodates this exponentially growing case.