Let $f \in L^1(\textbf{R})$ be such that $f'$ is continuous and $f' \in L^1(\textbf{R})$ . Find a function $g \in L^1(\textbf{R})$ such that

$$ g(t) = \int_{-\infty}^{t}e^{u-t}g(u)\,du + f'(t) $$

Im pretty sure I should manipulate this with the fourier transform and that there is a convolution going on but I'm not really sure what to do now when the integral limit goes to $t$ instead of $\infty$.

  • 1
    $\begingroup$ I'm not sure if that helps you, but $\int_{-\infty}^t e^{u-t} g(u)du = \left( e^{-t} \Bbb 1_{t\geq 0} \right) \ast g$ $\endgroup$ – Tryss Dec 11 '15 at 10:52

Using @Tryss hint, let $h(t)=\mathrm e^{-t}u(t)$, where $u(t)$ is the Heaviside step function. With this, $$ (g*h)(t)\equiv\int_{-\infty}^{+\infty}g(u)\,\mathrm e^{-(t-u)}u(t-u)\,\mathrm du=\int_{-\infty}^t g(u)\,\mathrm e^{u-t}\,\mathrm du $$ where $*$ means convolution.

This means that your equation is $$ g(t)=(g*h)(t)+f'(t) $$

Taking the Fourier transform, and using the Convolution Theorem, we find $$ \hat g(\xi)=\hat h(\xi) \hat g(\xi)+2\pi i\xi \hat f(\xi) $$

Next, the Fourier Transform of $h(t)$ is easy: $$ \hat h(\xi)=\frac{1}{1+2\pi i\xi} $$ which means that $$ \hat g(\xi)\left[1-\frac{1}{1+2\pi i\xi}\right]=2\pi i\xi \hat f(\xi) $$

You can solve for $\hat g$, and take the Inverse Fourier Transform to find $g(t)$: $$ \hat g(\xi)=(1+2\pi i\xi)\hat f(\xi) $$ and $$ g(t)=f(t)+f'(t) $$


Explanation of $$ (g*h)(t)\equiv\int_{-\infty}^{+\infty}g(u)\,\mathrm e^{-(t-u)}u(t-u)\,\mathrm du=\int_{-\infty}^t g(u)\,\mathrm e^{u-t}\,\mathrm du $$

My bad, I used the same letter twice: $u$ is both an integration variable and the Heaviside Step function. Im sorry for that. Lets rename the integration variable $w$ instead of $u$: $$ (g*h)(t)\equiv\int_{-\infty}^{+\infty}g(w)\,\mathrm e^{-(t-w)}u(t-w)\,\mathrm dw $$ (this is the definition of the convolution)

Next, $u(x)$ is $1$ for positive $x$ and zero otherwise. This means that $u(t-w)$ is $1$ for $w<t$ and zero otherwise. In the integral above, where $w$ ranges from $-\infty$ to $+\infty$, the function $u(t-w)$ is zero for $w>t$ and $1$ for $w<t$, which means that the interval $w\in(t,\infty)$ doesnt contribute. This implies that we can change the integration interval from $(-\infty,+\infty)$ into $(-\infty,t)$, as the integrand is null from $(t+,\infty)$.

  • $\begingroup$ $$ (g*h)(t)\equiv\int_{\infty}^{+\infty}g(u)\,\mathrm e^{-(t-u)}u(t-u)\,\mathrm du=\int_{-\infty}^t g(u)\,\mathrm e^{u-t}\,\mathrm du $$ Can you explain this step? I know what the heaviside function is but I still can't manage to do this myself. The rest of the solution I understand :) $\endgroup$ – user269620 Dec 11 '15 at 13:21
  • $\begingroup$ @Danny added an explanation. i hope its clear enough now. $\endgroup$ – AccidentalFourierTransform Dec 11 '15 at 14:14

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