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I want to solve this equation. It reminds me something about Laplace transform.

I am sure that I must use it order to solve it.

$$t-2f(t) = \int_0^t(e^\tau- e^{-\tau})f(t-\tau)d\tau$$

How to do it?

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    Where did you find this? What do you know about $f$?2011-12-13
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    If you make the substitution $g(t) = t-2f(t)$ then $f(t)=(t-g(t))/2$ so you get $$g(t)=\int_0^t \sinh(\tau)(t-\tau -g(t-\tau))d\tau = S * h (t)$$ where $$S(s)=\sinh(s)\cdot \chi_{(0,t)}(s)$$ and $h(s)= s-g(s)$. Which might be useful if you may apply some Fourier transform. (Here $\chi_E$ is the characteristic function of $E$ - i.e $\chi_E(x)=1$ on $E$ and $0$ otherwise).2011-12-13
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    @AD. Wierd thing is that I found this equation in the cover of a book about telecommunications.2011-12-13
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    Ahh... it might be the book of the grand grand grand grand grand grand son of Fermat!2011-12-13
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    Fermat? What Fermat has to do about this?2011-12-13
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    Probably that same book has this equation inside, too. With more context for it. Maybe even a method of solution...2011-12-13
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    A famous anecdote says that when Fermat wrote in the margin of an algebra book that he had a beautiful proof of $x^n+y^n\ne z^n$ for $x,y,z>0$ and $n>2$ integers. (nothing else)2011-12-13

1 Answers 1

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The right hand side of your equation is a convolution of two functions. The Laplace transform of the convolution, $f_1\star f_2$, of $f_1$ and $f_2$ is: $$\tag{1} {\cal L}\bigl[ (f_1 \star f_2)(t)\bigr]= {\cal L}\Bigl[\,\int_0^t f_1(\tau) f_2(t-\tau) \,d\tau\Bigr]=F_1(s)F_2(s),$$ where $F_1(s)$ is the Laplace transform of $f_1(t)$ and $F_2(s)$ is the Laplace transform of $f_2(t)$.

Taking the Laplace transform of both sides of your equation gives

$$ \cal L \bigl[ t-2f(t)\bigr] = \cal L \bigl[\, (e^t-e^{-t})\star f(t)\,\bigr]. $$ By equation (1): $$ \cal L [ \,t -2 f(t)\,\bigr] = \, \cal L[e^t - e^{-t}]\, \cdot \cal L\bigl[f(t)\bigr]. $$ By linearity of the Laplace transform: $$\cal L [ \,t\,] -2\cal L\bigl[ f(t)\bigr] = \bigl(\, \cal L[e^t]-\cal L[e^{-t}]\,\bigr) \cdot \cal L\bigl[f(t)\bigr].$$ So, denoting the Laplace transform of $f(t)$ by $F(s)$: $$ \tag{2}{1\over s^2}-2 F(s) =\Bigl[ { 1\over s-1}-{1\over s+1}\Bigr]F(s). $$

Solving equation (2) for $F(s)$ gives $$ F(s)={s^2-1\over 2s^4}={1\over 2s^2}-{1\over 2s^4}. $$ Applying the inverse Laplace transform now gives $$ f(t)={1\over2} t-{1\over12}t^3. $$

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    Awesome. I was trying to transform it to something that could do this but I was unwaware of the convolution. Thank you so much.2011-12-13