I am going over the problems in van Dijk's distribution theory book and trying to find the (convolution) inverse of the following distributions:
$$S=\delta''+H,\quad T=He^x +\delta'$$ using Heaviside calculus but I'm not sure the solutions make sense. First (using Heaviside symbolic calculus)
$$S=H+\delta''=\frac{1}{\delta'}+\delta''=\frac{\delta'\delta''+1}{\delta'}=\frac{\delta'''+1}{\delta'}$$ so the inverse of $S$ is given by $$\delta' (\delta'''+1)^{-1}=\delta' H Z$$ where $Z$ solves $(d^3+1)Z=0$ but this doesn't seem correct. For $T$ I get using $(He^x)^{-1}=\delta'-\delta$ $$T=He^x + \delta' = \frac{1}{\delta'-\delta}+\delta'=\frac{\delta''-\delta'}{\delta'-\delta}$$ and so $T^{-1}=(\delta'-\delta)(\delta''-\delta')^{-1}$ Solving the ODE $(d^2-d)Z=0$ with $Z(0)=0,Z'(0)=1$ I get the solution $$(\delta''-\delta')^{-1} =H(e^x-1)=He^x-H$$ and so $$T^{-1}=(\delta'-\delta)(e^x H-H)=H+e\delta-\delta'$$ which again, I'm not sure is correct! Hints/references/Feedback appreciated.
Not sure what you mean with $Z$ solves $(d^3+1)Z=0$.
The Laplace transform of $\delta'''+\delta$ is $1+s^3$ and $\frac1{1+s^3} = \frac1{3\zeta_6^2}\frac{1}{s-\zeta_6}+\frac1{3\zeta_6^6}\frac{1}{s-\zeta_6^3}+\frac1{3\zeta_6^{10}}\frac{1}{s-\zeta_6^5}$ whose (causal) inverse Laplace transform is $U=\frac1{3\zeta_6^2} e^{\zeta_6 t}+\frac1{3\zeta_6^{6}} e^{\zeta_6^3 t}+\frac1{3\zeta_6^{10}} e^{\zeta_6^5 t})1_{t>0}$ which is the convolution inverse of $\delta'''+\delta= S'$ so the convolution inverse of $S$ is $U'$