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.

Solve the integral equation

$$ y(t)= f(t) + \lambda \int_{0}^{t} (t-s) y(s) ds $$

where $f$ is continuous using the method of finding the resolvent kernel and Newmann series.

Here it is what I did:

$ K_1 (t,s) \equiv K(t,s) =t-s$

$ K_2 (t,s) = \int_{s}^{t} K(t, \xi) K_1 (\xi ,s) d \xi= \frac{1}{2} (t+s)^2(t-s)-ts(t-s) +\frac{1}{3} (s^3 -t^3) $

From here and on the calculations are too difficult.

Is there any trick?

Any help?

Thank's in advance!

P.S Is there another way to solve it (without using this method) ?

edit: I didn't made any proccess. Some help?

share|improve this question
add comment

2 Answers

A related problem. I am answering your question about the other way to solve the problem. The other technique is to use the Laplace transform technique. Taking the Laplace transform of both sides gives

$$ Y(s)= F(s)+\lambda L(x*y(x))=F(s)+\lambda L(x)L(y(x))=F(s)+\frac{\Gamma(2)\lambda}{s^2}Y(s). $$

Simplifying the above gives

$$ Y(s)= \frac{s^2F(s)}{s^2-\lambda}. $$

Taking the inverse Laplace transform yields the solution

$$ y(x)=f \left( x \right) +\sqrt {\lambda}\int _{0}^{x}\!f \left( t \right) \sinh \left( \sqrt {\lambda} \left( x-{t} \right) \right) {d{t}} .$$

Notes: We used the facts

i)$$ L(\delta(x)+\sqrt {\lambda}\sinh \left( \sqrt { \lambda}x \right) ) = \frac{s^2}{s^2-\lambda}.$$

ii) The Laplace transform of the convolution equals the product of the Laplace.

share|improve this answer
add comment

No need to expand the integrand. Linear change of variables mapping $[s,t]$ to $[0,1]$ reduces integrals for $K_n$ to beta-function. It will be easy to calculate several first ones, guess the formula for $K_n$ and prove it by induction.

Edit

Making change of variables $\xi=s+y(t-s)$ we have $$ \int_s^t(t-\xi)(\xi-s)\,d\xi= (t-s)^3\int_0^1(1-y)y\,dy= (t-s)^3B(2,2). $$

share|improve this answer
    
Can you write for example the first one? Thank's –  passenger May 3 '12 at 17:36
    
How can we map $[s,t]$ to $[0,1]$ ? –  passenger May 3 '12 at 17:49
    
O.K so if my calculations are correct it is $\displaystyle{ K_n(t,s)=(t-s)^{2n-1} B(2,2) \cdot B(4,2) \cdot \cdots \cdot B(2n-2,2)}$. I have one more question: Then how can I find the resolvent kernel $\displaystyle{ \Gamma (t,s,\lambda) =\sum_{n=1}^{\infty} \lambda ^n (t-s)^{2n-1} B(2,2) \cdots B(2n-2,2)}$ ? Thank's! –  passenger May 3 '12 at 18:11
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.