# Degenerate kernel method to solve Fredholm integral equation of the second kind

$$f(x) + \int_0^1 (xy+x^2y^2) f(y) dy = x^3 +\frac16x^2+ \frac15x$$

I have this fredholm integral equation of the second kind and am not sure how to answer this equation. I know that is has to be written in the form $$\sum a_jx^{j-1}$$

I do not know how to integrate with the function $$f(y)$$ in the integral.

Any help would be appreciated.

Thanks

can you start with: $$f(x)+\int_0^1(xy+x^2y^2)f(y)dy=x^3+\frac16x^2+\frac15x$$ $$f'(x)+(x+x^2)f(1)=3x^2+\frac13x+\frac15$$ $$f'(x)=(3-f(1))x^2+\left(\frac13-f(1)\right)+\frac15$$ so: $$f(x)=\int\left[(3-f(1))x^2+\left(\frac13-f(1)\right)+\frac15\right]dx+C$$
Note that we can write the equation as $$f(x) = x^3 + \left[\frac{1}{6} - \int \limits_0^1 y^2 f(y) \, \mathrm{d} y\right] x^2 + \left[\frac{1}{5} - \int \limits_0^1 y f(y) \, \mathrm{d} y\right] x \, , \, x \in [0,1] \, .$$ Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $$a, b \in \mathbb{R}$$ such that $$f(x) = x^3 + a x^2 + b x \, , \, x \in [0,1] \, .$$ We can then plug this form of $$f$$ into the previous equation and compute the integrals to obtain $$x^3 + a x^2 + b x = x^3 - \left(\frac{a}{5}+\frac{b}{4}\right) x^2 - \left(\frac{a}{4}+\frac{b}{3}\right)x \, , \, x \in [0,1] \, .$$ This yields the linear system $$\begin{pmatrix} \frac{6}{5} && \frac{1}{4}\\ \frac{1}{4} && \frac{4}{3} \end{pmatrix} \begin{pmatrix} a \\b \end{pmatrix} = \begin{pmatrix} 0 \\0 \end{pmatrix} \, ,$$ which only has the trivial solution $$a=b=0$$ . Therefore $$f(x) = x^3 \, , \, x \in [0,1] \, ,$$ is the unique solution to the integral equation.
• Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$\sum a_jx^{j-1}$$ how would I change what you have solved above into this format? – p s Dec 2 '18 at 23:22
• @ps You can write $f(x) = x^3 = \sum_{j=2}^4 a_j x^{j-1}$ with $a_4 = 1$ and $a_3 = a_2 = 0$ . – ComplexYetTrivial Dec 2 '18 at 23:32
• how did you get to $$x^3 - \left(\frac{a}{5}+\frac{b}{4}\right) x^2 - \left(\frac{a}{4}+\frac{b}{3}\right)x \, , \, x \in [0,1] \, .$$ – p s Dec 9 '18 at 19:00