solve $-u''(x)+\int_0^{\pi}u(y)dy=\lambda u(x)$,where $u(0)=u(\pi)=0$ Consider the eigenvalue problem
$$-u''(x)+\int_0^{\pi}u(y)dy=\lambda u(x)$$for all $ 0\le x\le\pi$, $\lambda\in\mathbb{R}$ and with $u(0)=u(\pi)=0$. 
How to find all the solutions $u$ (in $H^1(0,\pi)$ I think) of this eigenvalue problem?
My first try was to consider $u(x)=a \sin(kx)+b \cos(kx)$ and find out conditions on $a,b,k\in \mathbb{R}$. But I'm not sure if I receive all solutions with this method. Do you know how to proceed?
 A: Assuming you have a solution $u$, then differentiate:
$$
              -u'''-\lambda u'=0.
$$
Ignoring the case where $\lambda=0$ for the moment,
$$
          u(x)= C+D\cos(\sqrt{\lambda}x)+E\sin(\sqrt{\lambda}x).
$$
Assuming $u(0)=0$, $u'(0)=1$,
$$
            C+D=0,\;\;\; E\sqrt{\lambda}=1.
$$
The trial solution becomes
$$
              u(x)=C(1-\cos(\sqrt{\lambda}x))+\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}.
$$
Then
$$
          \int_{0}^{\pi}u(t)dt = C\left(\pi-\frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}}\right) +\frac{1-\cos(\sqrt{\lambda}\pi)}{\lambda}
$$
Plugging into the original equation
$$
    -C\lambda\cos(\sqrt{\lambda}x)+\sqrt{\lambda}\sin(\sqrt{\lambda}x)
   +C\left(\pi-\frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}}\right) +\frac{1-\cos(\sqrt{\lambda}\pi)}{\lambda} \\
    =C\lambda(1-\cos(\sqrt{\lambda}x))+\sqrt{\lambda}\sin(\sqrt{\lambda}x),
$$
the constant $C$ is determined by
$$
    C\left(\pi-\frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}}-\lambda\right)
    +\frac{1-\cos(\sqrt{\lambda}\pi)}{\lambda}=0 \\
     C = - \frac{1-\cos(\sqrt{\lambda}\pi)}{\pi\lambda-\sqrt{\lambda}\sin(\sqrt{\lambda}\pi)-\lambda^2}
$$
The eigenvalue equation for $\lambda$ is determined by $u(1)=0$, which is an equation in $\lambda$ that can only have real solutions:
$$
       C(1-\cos(\sqrt{\lambda}\pi))+\frac{\sin{\sqrt{\lambda}}\pi}{\sqrt{\lambda}}=0.
$$
Once an eigenvalue $\lambda$ is found, the corresponding solution $u_{\lambda}$ is unique up to a multiplicative constant:
$$
     u_{\lambda}(x) = C(1-\cos(\sqrt{\lambda}x))+\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}.
$$
A: Hint: Try fixing $\int_0^\pi u(y)dy=C$ for a constant $C$, and solving this problem as $C$ varies.
