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?


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.


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.


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

  • $\begingroup$ Can you write for example the first one? Thank's $\endgroup$ – passenger May 3 '12 at 17:36
  • $\begingroup$ How can we map $[s,t]$ to $[0,1]$ ? $\endgroup$ – passenger May 3 '12 at 17:49
  • $\begingroup$ 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! $\endgroup$ – passenger May 3 '12 at 18:11

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