Solving an Integral Equation

Question

Here is the Question;

Solve the integral equation,

$$\int_0^tY(u)Y(t-u)du = \frac12 (\sin t-t\cos t)$$

Really not sure how to go about this, took the Laplace transform of the right side getting,

$$\frac12\left(\frac{1}{s^2+1}-\frac{s^2-1}{(s^2+1)^2}\right)$$

Any suggestions to go about this would be great, thank you in advance.

score 2 · Answer 1 · answered Dec 11 '13 at 00:05

2

Note that, for the left hand side you need the fact, the Laplace of the convolution equals the product of the Laplace, that is

$$ \mathcal{L} (f*f) = \mathcal{L}(f) \mathcal{L}(f)= F(s)^2,$$

where $F(s)$ is the Laplace transform of $f$ and it is given by

$$ F(s) = \int_{0}^{\infty} f(x) e^{-sx}dx . $$

Note:

$$\int_0^tY(u)Y(t-u)du = (Y*Y)(t). $$

answered Dec 11 '13 at 00:05

Mhenni Benghorbal

Making sure I understand, so I will have the convolution of (Y*Y) equal to my right hand side. So I am thinking I could then solve for Y and take it's inverse Laplace transform? – slow_Math Dec 11 '13 at 00:34
@slow_Math: Yes, you are exactly right. – Mhenni Benghorbal Dec 11 '13 at 01:31

1 Answers1