Let $u(x, \epsilon, \theta)$ be the solution of $$u''+(\epsilon \cos(x)+\theta-u)u=0$$ with boundary conditions $u'(0)=0$ and $u'(\pi)=0$. Here $\theta\in [0, 1]$.

I tried to put the solution in this form $$ u(x, \epsilon, \theta)=\theta+\epsilon u_1+\epsilon^2u_2+\epsilon^3 u_3.$$ Now the problem is how to prove: $u_3=O(1)$? Moreover, how can we show $u_3$ continuously differentiable in $\epsilon$ and $\theta$? Can any one give me some hint? I do not see a book with proof of perturbation theory in boundary value problems...

I had this: $u_1$ satisfies $$u_1''-\theta u_1+\theta \cos(x)=0;$$ $u_2$ satisfies $$u_2''-\theta u_2+\cos(x)u_1-u_1^2=0;$$ and $u_3$ satisfies $$u_3''-(\theta+2\epsilon u_1+2\epsilon^2 u_2-\epsilon \cos(x)+\epsilon^3 u_3)u_3=2u_1u_2-\cos(x) u_2.$$

I also tried to put the solution in this form $$ u(x, \epsilon, \theta)=\theta+\epsilon u_1+\epsilon^2u_2+\epsilon^3 u_3+\cdots.$$ Then $u_n$ solves $u_n''-\theta u_n=u_{n-1}u_1+u_{n-2}u_2+\cdots+u_1u_{n-1}-\cos(x) u_{n-1}.$ But this does not help me....