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I have to determine the Laplace transform of $u$, where $u$ is the solution to this Cauchy's problem $$ u'(t) + \int_0^t e^{t-s} \left( \int_0^s u(r) ~ dr \right) ~ ds = 0, t > 0, u(0) = 1$$

$L[u](z) = ~? $

I know the the Laplace transform is $ \int_0^{+\infty} u(t)e^{-zt} \ dt$ but I can't determine the $u$ function. Can you please help me? :(

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    Take the transform on both sides, like you would in simple ODEs. You'll need formulae for the transform of an integral and shifts by exponentials. – 2017-01-29

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$\ds{\mrm{u}'\pars{t} + \int_{0}^{t}\expo{t - s} \bracks{\int_{0}^{s}\mrm{u}\pars{r}\,\dd r}\dd s = 0\,,\quad t > 0. \qquad\mrm{u}\pars{0} = 1}$.

\begin{align} 0 & = \mrm{u}'\pars{t} + \expo{t}\int_{0}^{\infty}\expo{-s}\bracks{s < t} \int_{0}^{\infty}\mrm{u}\pars{r}\bracks{r < s}\dd r\,\dd s \\[5mm] & = \mrm{u}'\pars{t} + \expo{t}\int_{0}^{\infty}\mrm{u}\pars{r} \int_{0}^{\infty}\expo{-s}\bracks{r < s < t}\dd s\,\dd r \\[5mm] &= \mrm{u}'\pars{t} + \expo{t}\int_{0}^{\infty}\mrm{u}\pars{r}\bracks{r < t} \pars{\expo{-r} - \expo{-t}}\,\dd r \\[5mm] & = \mrm{u}'\pars{t} + \expo{t}\int_{0}^{t}\mrm{u}\pars{r}\expo{-r}\,\dd r - \int_{0}^{t}\mrm{u}\pars{r}\,\dd r \end{align}

Multiply both members by $\ds{\expo{-st}}$ and integrate over $\ds{t > 0}$ $\ds{\pars{~\mbox{note that}\ \hat{\mrm{f}}\pars{s} \equiv \int_{0}^{\infty}\mrm{f}\pars{t}\expo{-st}\,\dd t ~}}$:

\begin{align} 0 & = -\ \overbrace{\mrm{u}\pars{0}}^{\ds{=\ 1}}\ + s\,\hat{\mrm{u}}\pars{s} + \int_{0}^{\infty}\expo{-st}\expo{t}\int_{0}^{t}\mrm{u}\pars{r}\expo{-r} \,\dd r\,\dd t - \int_{0}^{\infty}\expo{-st}\int_{0}^{t}\mrm{u}\pars{r}\,\dd r\,\dd t \\[5mm] & = -1 + s\,\hat{\mrm{u}}\pars{s} + \int_{0}^{\infty}\mrm{u}\pars{r}\expo{-r} \int_{r}^{\infty}\expo{-\pars{s - 1}t}\dd t\,\dd r - \int_{0}^{\infty}\mrm{u}\pars{r}\int_{r}^{\infty}\expo{-st}\dd t\,\dd r \\[5mm] & = -1 + s\,\hat{\mrm{u}}\pars{s} + {1 \over s - 1}\int_{0}^{\infty}\mrm{u}\pars{r}\expo{-sr}\dd r - {1 \over s}\int_{0}^{\infty}\mrm{u}\pars{r}\expo{-sr}\dd r \\[5mm] & = -1 + s\,\hat{\mrm{u}}\pars{s} + \pars{{1 \over s - 1} - {1 \over s}}\hat{\mrm{u}}\pars{s} \implies \bbx{\ds{\,\hat{\mrm{u}}\pars{s} = {s^{2} - s \over s^{3} - s^{2} + 1}}} \end{align}

$\ds{s^{3} - s^{2} + 1 = 0}$ has one negative real root $\ds{s_{1} \approx -0.7549}$ and two complex roots $\ds{a \pm b\ic\approx 0.8774 \pm 0.7449\ic}$.

With $\ds{c > a}$, $\ds{\mrm{u}\pars{t}}$ is given by: \begin{align} \mrm{u}\pars{t} & = \int_{c - \infty\ic}^{c + \infty\ic}{s^{2} - s \over s^{3} - s^{2} + 1}\, \expo{st}\,{\dd s \over 2\pi\ic} = {s_{1} - 1 \over 3s_{1} - 2}\,\expo{-\verts{s_{1}}t} + 2\,\Re\pars{{a + b\ic - 1 \over 3\bracks{a + b\ic} - 2}\, \expo{\bracks{a + b\ic}t}} \end{align}

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    Do you use $[s < t]$ as an Iverson bracket, returning $1$ is the bracketed condition is true? – 2017-01-29
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    In the end I prefered to use the convolution product, but this helped me a lot. – 2017-01-29
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    @SeanRoberson Yes. Iverson Bracket is quite fine. It lets easily keep tracks of several conditions. I guess it avoids to make mistakes. Thanks. – 2017-01-30
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    @PeppeDAlterio Thanks. There are a few ways to get the same result. – 2017-01-30
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    @FelixMarin I just wanted to be sure; I don't see it used much outside the text *Concrete Mathematics.* – 2017-01-30