I'm trying to solve the following Fredholm integral equation of the 2nd kind:

$$u(x)=f(x)+\lambda\int_{0}^{1}x^4y^2u(y)dy, 0<x<1$$ where $f(x)$ is given, $u(x)$ is the unknown function.

This is what I've done so far:


$\Rightarrow a=\lambda\int_{0}^{1}y^2u(y)dy=\lambda\int_{0}^{1}y^2[f(y)+ay^4]dy=\lambda\int_{0}^{1}y^2f(y)dy+a\lambda\int_{0}^{1}y^4dy=\lambda\int_{0}^{1}y^2f(y)dy+\frac{a\lambda}{5}$

$\Rightarrow a=\frac{\lambda}{1-\frac{\lambda}{3}}\int_{0}^{1}y^2f(y)dy$

$\Rightarrow u(x)=f(x)+x^4\frac{\lambda}{1-\frac{\lambda}{3}}\int_{0}^{1}y^2f(y)dy$

I'm not really sure where to go from here or even if what I have done so far is correct. If it's correct, how do proceed from here? If it's wrong, what have I done wrong/what should I be doing instead?


$$u(x)=f(x)+x^4\lambda\int_{0}^{1}y^2u(y)dy\tag 1$$ $$u'(x)=f'(x)+4x^3\lambda\int_{0}^{1}y^2u(y)dy=f'(x)+\frac{4}{x}\big(u(x)-f(x)\big)$$ $$u'(x)-\frac{4}{x}u(x)=f'(x)-\frac{4}{x}f(x) \tag 2$$ $f'(x)-\frac{4}{x}f(x)$ is a known function.

Eq.$(2)$ is a first order linear ODE. I suppose that you can take it from here.

  • $\begingroup$ Hey @JJacquelin thank you for your help! I know how to solve 1st order linear ODE's I think, how does this look? $$u'(x)-\frac{4}{x}u(x)=f'(x)-\frac{4}{x}f(x)$$ $$I(x)=\exp(\int\frac{-4}{x}dx)=\frac{1}{x^4}$$ $$\frac{d}{dx}(\frac{u(x)}{x^4})=f'(x)-\frac{4}{x}f(x)$$ $$\frac{u(x)}{x^4}=f(x)-4\int\frac{f(x)}{x}dx+C$$ $$u(x)=x^4f(x)-4x^4\int\frac{f(x)}{x}dx+Cx^4$$ $\endgroup$ – user475596 Oct 11 '17 at 6:41
  • $\begingroup$ The third equation is not correct. $\endgroup$ – JJacquelin Oct 11 '17 at 8:33
  • $\begingroup$ Oh thanks for spotting that @JJacquelin ! How about this? If it's right, is that it or can I simplify it further? I'm kind of worried that the $\lambda$ has disappeared. Also should my integral there be a definite integral? Sorry for asking so many questions $$u'(x)-\frac{4}{x}u(x)=f'(x)-\frac{4}{x}f(x)$$ $$I(x)=\exp(\int\frac{-4}{x}dx)=\frac{1}{x^4}$$ $$\frac{d}{dx}(\frac{u(x)}{x^4})=\frac{1}{x^4}f'(x)-\frac{4}{x^5}f(x)$$ $$\frac{u(x)}{x^4}=\int\{\frac{1}{x^4}f'(x)-\frac{4}{x^5}f(x)\}dx+C$$ $$u(x)=x^4\int\{\frac{1}{x^4}f'(x)-\frac{4}{x^5}f(x)\}dx+Cx^4$$ $\endgroup$ – user475596 Oct 11 '17 at 8:43
  • $\begingroup$ That is correct. But one can simplify to $$u(x)=f(x)+Cx^4$$ This was visible immediately, just by inspection of the ODE. But it's not the end. The Fredholm equation with this result $u(x)$ is not satisfied for any values od $C$.The parameter $C$ have to be determined. Put $u(x)=f(x)+Cx^4$ into the Fredholm equation. $\endgroup$ – JJacquelin Oct 11 '17 at 9:36
  • $\begingroup$ In fact, this is what you did at first. Your result $$u(x)=f(x)+x^4\frac{\lambda}{1-\frac{\lambda}{3}}\int_{0}^{1}y^2f(y)dy$$ ist correct. But the justification of the assumption $u(x)=f(x)+ax^4$ was missing. $\endgroup$ – JJacquelin Oct 11 '17 at 9:53

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