Can one use the IFT on Banach spaces and the simple harmonic oscillator to say that there is a solution for the motion of a pendulum? Let $W^{k,p} (S^1)$ be the Sobolev space of $k$ times 
weakly differentiable periodic functions, all whose weak derivatives 
upto $k$ are in $L^p(S^1)$. Consider the non linear operator $A_t : W^{k,p} \rightarrow W^{k-2,p}$ 
given by 
$$ A_t(u) := \frac{d^2 u}{d x^2} + t^2 \sin (u). $$
Also, consider the linear operator $L_t: W^{k,p} \rightarrow W^{k-2,p}$ 
given by 
$$ L_t(u):= \frac{d^2 u}{d x^2} + t^2 u.$$
My question is the following: given the fact that the ode 
$L_t(u) =0$ does have a solution, can I use that fact to 
conclude that $A_t(u) =0$ has a solution for small $t$, essentially by using the 
Implicit Function Theorem on Banach spaces? 
I basically want to view the 
second equation as a "Linearization" of the first equation. 
My basic question therefore is; what is the "linearization" of 
$A_t$? 
I must emphasize that I am asking if I can conclude that the equation 
$A_t(u) =0$ has a solution from a very specific line of reasoning; there may be  more direct ways to see that this non linear equation has a solution. That is not what I am asking, however.   
 A: This is a pretty standard technique in PDE. Consider the $t$ parameter as the "horizontal axis" for the implicit function theorem.
The linearisation about zero (a known solution) is just the derivative of $A:\mathbb{R} \times W^{k,p} \to W^{k-2,p}$ in the $W^{k,p}$ ("vertical") direction:
$$ dA_t(0)(v) = \frac{d}{ds}\bigg|_{s=0} A_t(s v) = \frac{ds}{ds}\frac{d^2 v}{dx^2}+t^2 \frac{d \sin(s v)}{d(s v)}\bigg|_{s=0}\frac{d(s v)}{ds} = L_t(v),$$
so $L_t$ is indeed the correct linear operator to study in this case. However, having solutions to $L_tv=0$ is not enough for what you want.
The conditions for the IFT to apply are a given solution (in our case $A_{t_0}(0) = 0$) and  that the vertical derivative ($L_{t_0}$) is an isomorphism; this comes down to having unique solutions to $L_{t_0} v = f$ for all source terms $f \in W^{k-2,p}$. If this condition holds then we get a neighbourhood $U$ of $t_0$ and functions $\{u_t\}_{t \in U} \subset W^{k,p}$ such that $L_t u_t = 0$.
As to whether or not the condition on the linearization holds, it certainly doesn't as is - for example at $t=0$ you can add a constant to any solution. I believe you can probably get around this by restricting your function space with some condition like $\int u=0$; give it a shot!
