Let $\lambda\in (-1,1)$. Show that for every $f\in C[0,1]$ there exists a unique solution $u\in C[0,1]$ to
$$-u''(x)+\lambda \int^1_0 \sin(u(y))dy =f(x)$$
With $u(0)=u'(1)=0$.
My work thus far:
$$\langle Lu,v \rangle = \int^1_0 -u''(x)v(x)dx+ \lambda \int^1_0 \sin(u(y))dy\int^1_0 v(x)dx $$
Using IBP, we find $v'(1)=0$, $v(0)=0$, and
$$\langle Lu,v \rangle = \int^1_0 u(x)(-v''(x))dx+\lambda\int^1_0 \sin(u(y))dy\int^1_0 v(x)dx $$
So what is the adjoint operator $L^*$? Orevious experience tempts me to say $L^*v=-v''$
If this is the case $N(L^*)=\left\{0\right\}$, so that there is always a solution to $Lu=f$.
For uniqueness, Let $\lambda=0$. Then $-u''=0$ implies $u=0$ so the solution will be unique.
Let $\lambda\neq 0$. Then boundary conditions tell us
$$-u''(x)+\lambda c =0 \implies u(x)=(x^2-x)\frac{\lambda c}{2}$$
So the question becomes, is there a nonzero $c$ that satisfies
$$-\lambda c+\lambda \int^1_0 \sin\left((y^2-y)\frac{\lambda c}{2} \right)dy =0 $$
Uhh.....?