Need help solving an Integral Equation

Question

Need help solving:

$$ f(x) = x + \lambda \int_{0}^{1}y(x+y)f(y)dy $$

keeping terms through $\lambda^{2}$,

(a) by using the Fredholm method

(b) by using the Neumann method

Answer 1

Ok, I guess I'll try to help with the Neumann method. (Hopefully it's not too late for you.) First, note that if $\lambda$ was $0$, then $f(x) = x$. If both $\lambda$ and $x$ were small, then we'd hope that $f(x) \approx x$. So let's plug $f(x)\approx x$ into the left-hand side of your equation, to get a new approximation $f_{1}(x)$ to $f(x)$: \begin{align*} f_{1}(x) &= x+\lambda \int_{0}^{1} y(x+y)y\, \mathrm{d}y\\ &= x+\lambda\left(\frac{1}{4}+\frac{x}{3}\right) \\ &= \left(1+\frac{\lambda}{3}\right)x + \frac{\lambda}{4} \end{align*} Ok, so now let's plug $f_{1}(x)$ into the rhs of your equation to get a better approximation $f_{2}$: \begin{align*} f_{2}(x) &= x+\lambda \int_{0}^{1} y(x+y)f_{1}(y)\, \mathrm{d}y\\ &= x+\lambda \int_{0}^{1} y(x+y)\left[ \left(1+\frac{\lambda}{3}\right)x + \frac{\lambda}{4} \right] \, \mathrm{d}y \\ &=x+\lambda \left[\frac{1}{4}+\frac{x}{3} + \frac{\lambda}{6} + \frac{17 \lambda}{72}x \right] \\ &=\frac{\lambda}{4} + \frac{\lambda^{2}}{6}+ \left(1+\frac{\lambda}{3}+ \frac{17\lambda^{2}}{72} \right)x \end{align*} Ok, so we have terms of order $\lambda^{2}$; let's make sure this doesn't change when we refine the approximation: \begin{align*} f_{3}(x) &= x + \lambda \int_{0}^{1} y(x+y)\left[\frac{\lambda}{4} + \frac{\lambda^{2}}{6}+ \left(1+\frac{\lambda}{3}+ \frac{17\lambda^{2}}{72} \right)y \right] \, \mathrm{d}y\\ &= x+\lambda\left[\frac{11 \lambda ^2}{96}+\frac{\lambda }{6}+\frac{35 \lambda ^2 x}{216}+\frac{17 \lambda x}{72}+\frac{x}{3}+\frac{1}{4} \right] \\ &=\frac{\lambda}{4} +\frac{\lambda^{2}}{6} + \left(1+ \frac{\lambda}{3} + \frac{17\lambda^{2}}{72} \right)x +\mathcal{O}(\lambda^{3}) \quad \text{as } \lambda \to 0. \end{align*} So we're good. I'm going to venture a guess that you can obtain the exact solution by assuming $f(x) = c_{0} + c_{1}x$ and then solving for $c_{0}$ and $c_{1}$ in terms of $\lambda$. Perhaps that will help with the Neumann method. When I carried out this procedure, I obtained \begin{align*} f(x) = -\frac{18\lambda}{-72+48\lambda +\lambda^{2}}+ \frac{24(\lambda-3)}{-72+48\lambda + \lambda^{2}} x \end{align*}