I have the Fredholm equation,
$$\phi(x)=\sin x+\lambda\int_0^\pi\cos(x/2-3y)\phi(y)dy$$
and would like to solve it.
First, I found using the precondition for contraction of the Fredholm operator: $$|\lambda|<\frac{1}{M(b-a)}$$
That $|\lambda|<\sqrt{2}$ to give a contraction. So I insert 1 as a value for $\lambda$ and get:
$$\phi(x)=\sin x+\int_0^\pi\cos(x/2-3y)\phi(y)dy$$
I solved this numerically using Mathematica. The solution is $\phi(x)=\sin(x)$. However, how can I solve this "by hand"?
Thanks
Compute the second derivative of $\phi(x)$ $$\begin{align} \phi''(x) = \frac{d^2\phi}{dx^2}(x) &= -\sin(x)+\lambda\int_0^{\pi} \underbrace{\left(\frac{\partial^2}{\partial x^2}\cos(x/2-3y) \right)}_{=-\frac{1}{4}\cos(x/2-3y)} \phi(y)dy \\ &=-\sin(x)-\frac{\lambda}{4}\int_0^{\pi} \cos(x/2-3y) \phi(y)dy \\ &=-\sin(x)-\frac{1}{4} \underbrace{\left(\lambda\int_0^{\pi} \cos(x/2-3y) \phi(y)dy \right)}_{=\phi(x)-\sin(x)} \\ &=-\sin(x)-\frac{1}{4} \left(\phi(x)-\sin(x) \right) \\ \phi''(x)&=-\frac{1}{4} \phi(x) -\frac{3}{4}\sin(x) \tag{1} \end{align}$$ The equation $(1)$ can be solved easily, the general solution is $$\phi(x) = \sin(x) + c_1 \sin\left(\frac{x}{2} \right)+ c_2 \cos\left(\frac{x}{2} \right)$$