Solutions for small perturbations I have the following exercise:
Let $f,g\colon \mathbb{R}\rightarrow \mathbb{R}$ be $C^1$ bounded functions and $f(0)=0$. Show that for suitably small $\varepsilon>0$ the following problem has a solution:
$$
\begin{cases}
u''(t)=f(\varepsilon u(t))+\varepsilon g(t) \\
u(0)=0=u(1)
\end{cases}
$$
[Edit: the exercise hint is to use the implicit function theorem]
My attempted solution:
Let's consider the solution map
$$ 
T\colon (\varepsilon,u)\in B_r(0)\times C_0^2([0,1]) \mapsto u''(t)-f(\varepsilon u(t))-\varepsilon g(t)\in C([0,1])
$$
By hypotesis $T(0,0)=0$. We can use the implicit function theorem to prove the statement.
Let's consider its differential in the second variable:
$$D_u T(0,0)[h]=h''$$
which is continuous, injective (since all functions with null second derivative and null in $0$ and $1$ must be zero) and is surjective since for each $k(t)\in C([0,1])$, setting
$$ K(t)=\int_0^t\int_0^\tau k(s)\ ds \ d\tau$$
is such that $K''(t)=k(t)$. Therefore it's an isomorphism and by the implicit function theorem there exist $\delta$ and $\eta$ and a function $G\colon B_\delta(0)\rightarrow B_\eta(0)$ such that for each $\varepsilon \in B_\delta(0)$ we have $T(\varepsilon,G(\varepsilon))=0$.
Is this proof right? Have I not considered the role of $f$ and $g$ making some big mistake?
 A: You learn from $f(εu)\sim f'(0)εu$ and the ODE that $u=O(ε)$. Insert $u=εv$ so that 
$$
v''=\frac{f(ε^2v)}ε+g(t)=g(t)+O(ε)
$$
The reduced equation $v''=g$ has the solution
\begin{align}
v(t)&=v'(0)t+\int_0^t(t-s)g(s)ds  \\
0=v(1)&=v'(0)+\int_0^1(1-s)g(s)ds  \\
v(t)&=-\int_0^ts(1-t)g(s)ds-\int_t^1t(1-s)g(s)ds  \\
 &=\int_0^1G(t,s)g(s)ds,
\end{align}
that is, using the Green kernel for $L[u]=u''$.
Now use that to formulate a fixed-point operator $C([0,1])\to C([0,1])$
$$
T[u](t)=\int_0^1G(t,s)[f(εu)+εg(s)]ds
$$
which is contractive for $ε$ small enough, certainly for $ε=0$ it directly contracts to one point. If $L$ is a Lipschitz constant for $f$ on a sufficiently large segment around $x=0$, then for two functions $u,v$ one gets
$$
\|T[v]-T[u]\|\le εL\|v-u\|\sup_{t\in[0,1]}\int_0^1|G(t,s)|ds=\frac{εL}8\|v-u\|,
$$
so $εL<8$ is sufficient for contractivity.
Now one needs to ensure that if $L$ is the Lipschitz constant on $[-R,R]$ and $\|u\|<R/ε$ then also $\|T[u]\|<R/ε$. Then the Banach fixed-point theorem applies. Continuity of $T$ in $ε$ then implies continuous change of the solutions when $ε$ changes.
