Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have an expression such as $\int_0^{x+l}y(z)g(x-z) dz$ and I want to evaluate its Laplace transform w.r.t $x$ in terms of the Laplace transform of $y(x)$. I know that I can substitute $t=x+l$, and coerce it into the standard from to get the Laplace transform w.r.t $t$, but I need its transform w.r.t $x$.

Motivation:

I would like to solve an integral equation: $y(x) = f(x) + \int_0^{x+l} y(z) g(x-z) dz$.

If the integral limit had been to $x$, we would have had $y(x) = f(x) + \int_0^{x} y(z) g(x-z) dz$. This leads to $Y(s) = \frac{F(s)}{1+K(s)}$.

share|improve this question
    
May I ask what made this equation to you? –  B. S. Jun 28 '12 at 7:48
    
@BabakSorouh, as you may guess, I have a certain process where I am trying to calculate some quantity along an axis, the $f(x)$ is an external component and $y(z)g(x-z)$ are local interactions. The applications are related to my research, but it could occur anywhere from EM fields, heat conduction, or chemical and biological processes. –  highBandWidth Jun 28 '12 at 17:00
    
eqworld.ipmnet.ru/en/solutions/ie/ie0217.pdf might be helpful. –  doraemonpaul Jul 11 '12 at 1:15
add comment

1 Answer 1

Hint:

$y(x)=f(x)+\int_0^{x+l}y(z)g(x-z)~dz$

$y(x)=f(x)+\int_x^{-l}y(x-t)g(t)~d(x-t)$

$y(x)=f(x)-\int_x^{-l}y(x-t)g(t)~dt$

$y(x)=f(x)+\int_{-l}^xy(x-t)g(t)~dt$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.