$f(t)=1+t-\dfrac{8}{3}\displaystyle\int_{0}^{t}(\tau-t)^3f(\tau) \ \mathrm d\tau \quad , f(t)=?$

Question

$$ f(t)=1+t-\dfrac{8}{3}\displaystyle\int_{0}^{t}(\tau- t)^3f(\tau) \ \mathrm d\tau $$

According to the convolution theorem, $\displaystyle\int_{0}^{t}(\tau- t)^3f(\tau)d\tau$ = $f(t) * t^3$ (I think)and its transform is equal to $F(s)=\frac{3!}{s^4}$

Finding the transform of the entire equation gives me $$ F(s)=\frac{1}{s}+\frac{1}{s^2}-F(s)\frac{16}{s^4} $$ I got $F(s) = \dfrac{s^5+s^4}{s^2(s^4+16)}$, but I'm not sure if it is right. And even if it is, I don't know how to find the inverse transform.

I was wrong $g(t-\tau) = g(t)$ In the problem $(\tau-t)^3 = -(t-\tau)^3 = -t^3$

In the end the denominator is $s^4 - 16$ which is much easier to deal with. Sorry for my oversight.

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{\fermi\pars{t} = 1 + t - {8 \over 3}\int_{0}^{t}\pars{\tau-t}^{3}\fermi\pars{\tau}\,\dd\tau\,, \qquad\tilde{\rm f}\pars{s} = \int_{0}^{\infty}\expo{-st}\fermi\pars{t}\,\dd t}$

\begin{align} \tilde{\rm f}\pars{s}&= {1 \over s} + {1 \over s^{2}} - {8 \over 3}\int_{0}^{\infty}\dd t\,\expo{-st} \int_{0}^{t}\pars{\tau-t}^{3}\fermi\pars{\tau}\,\dd\tau \\[3mm]&= {1 \over s} + {1 \over s^{2}} + {8 \over 3}\int_{0}^{\infty}\dd\tau\,\fermi\pars{\tau} \int_{\tau}^{\infty}\pars{t - \tau}^{3}\expo{-st}\,\dd t\tag{1} \\[3mm]&= {1 \over s} + {1 \over s^{2}} + {8 \over 3}\int_{0}^{\infty}\dd\tau\,\fermi\pars{\tau} \int_{0}^{\infty}t^{3}\expo{-s\pars{t + \tau}}\,\dd t \\[3mm]&= {1 \over s} + {1 \over s^{2}} + {8 \over 3}\, \overbrace{\int_{0}^{\infty}\expo{-s\tau}\fermi\pars{\tau}\dd\tau}^{\ds{=\ \tilde{\rm f}\pars{s}}}\ \overbrace{\int_{0}^{\infty}t^{3}\expo{-st}\,\dd t}^{\ds{=\ 3!/s^{4}}} \end{align}

$$ \tilde{\rm f}\pars{s} = {1 \over s} + {1 \over s^{2}} + {16 \over s^{4}}\,\tilde{\rm f}\pars{s} \qquad\imp\qquad \tilde{\rm f}\pars{s} = {s^{3} + s^{2} \over s^{4} - 16} $$ The roots of $s^{4} - 16 = 0$ are given by $s_{n} = 2\expo{\ic n\pi/2}$ with $n = 0, 1, 2, 3$: $$ s_{0} = 2\,,\quad s_{1} = 2\ic\,,\quad s_{2} = -2\,,\quad s_{3} = -2\ic \qquad\mbox{where}\quad \pars{s_{n}^{3} + s_{n}^{2}} \not= 0\,,\quad n = 0, 1, 2, 3 $$

With $\gamma > 2$: \begin{align} \fermi\pars{t}&= \int_{\gamma - \ic\infty}^{\gamma + \ic\infty}\tilde{\rm f}\pars{s}\expo{st}\, {\dd s \over 2\pi\ic} = \int_{\gamma - \ic\infty}^{\gamma + \ic\infty}{\dd s \over 2\pi\ic}\, {\pars{s^{3} + s^{2}}\expo{st} \over s^{4} - 16} = \sum_{n = 0}^{3}\lim_{s \to s_{n}}{\pars{s - s_{n}}\pars{s^{3} + s^{2}}\expo{st} \over s^{4} - 16} \\[3mm]&= \sum_{n = 0}^{3}{\pars{s_{n}^{3} + s_{n}^{2}}\expo{s_{n}t} \over 4s_{n}^{3}} = {1 \over 4}\sum_{n = 0}^{3}\pars{1 + {1 \over s_{n}}}\expo{s_{n}t} = {1 \over 4}\sum_{n = 0}^{3}\expo{s_{n}t} + {1 \over 4}\sum_{n = 0}^{3}{\expo{s_{n}t} \over s_{n}} \\[3mm]&= {1 \over 4}\pars{\expo{2t} + \expo{2\ic t} + \expo{-2t} + \expo{-2\ic t}} + {1 \over 4}\pars{{\expo{2t} \over 2} + {\expo{2\ic t} \over 2\ic} + {\expo{-2t} \over -2} + {\expo{-2\ic t} \over -2\ic}} \\[3mm]&= {1 \over 2}\bracks{\cos\pars{2t} + \cosh\pars{2t}} + {1 \over 4}\bracks{\sin\pars{2t} + \sinh\pars{2t}} \end{align}

$$\color{#0000ff}{\large% \fermi\pars{t} \color{#000000}{\ =\ } {1 \over 2}\bracks{\cos\pars{2t} + \cosh\pars{2t}} + {1 \over 4}\bracks{\sin\pars{2t} + \sinh\pars{2t}}} $$