I would like to find a bounded $C^{\infty}$ solution of the following equation: $$- \frac{d^2}{dx^2} u + u -\epsilon u^2 = f,$$ where $f$ is a function in the Schwartz space and $\epsilon >0$ is sufficiently small.
My attempt Applying the Fourier transform on both sides, I get $$\widehat{f}(\xi) = (1+ \xi) \widehat{u}(\xi) - \epsilon \widehat{u^2}(\xi).$$ Now, I also tried to express the last term using the convolution theorem, but then I don’t see how this can be helpful in order to define a contraction on a complete metric space (which in this case I think it should be the space $$ C^1_b(\mathbb{R}) := \{u: \mathbb{R} \rightarrow \mathbb{R} \, | \, u\in C^1 \, \, \text{and} \, \, u \, \text{bounded}\}$$ equipped with the $C^1$-norm, which is Banach.
Any suggestions? Thanks in advance!
I will give a sketch of the argument, and you can fill in the details. We can re-write your second order system as the coupled first order system: \begin{equation} \begin{split} u_x&=v\\ v_x&=u-f-\epsilon u^2. \end{split} \end{equation} We next introduce the integral formulation of the above system: \begin{equation} \begin{split} u(x)-u(0)&=\int_0^xu_xds=\int_0^xvds\\ v(x)-v(0)&=\int_0^xv_xds=\int_0^x(u-f-\epsilon u^2)ds \end{split} \end{equation} Therefore, we define the operator $T:C_b^\infty\times C_b^\infty\rightarrow C_b^\infty\times C_b^\infty$ as \begin{equation} T(u,v)=\big{(}u(0)+\int_0^xvds,v(0)+\int_0^x(u-f-\epsilon u^2)ds \big{)}. \end{equation} We equip the space $C_b^\infty\times C_b^\infty$ with the natural product norm $||(u,v)||=\sqrt{||u||^2_\infty+||v||_\infty^2}$, which makes $C_b^\infty\times C_b^\infty$ a Banach space.
\begin{equation} \begin{split} &||T(u_1,v_1)-T(u_2,v_2)||\\ &= ||\Big{(}u_1(0)-u_2(0)+\int_0^x(v_1-v_2)ds, v_1(0)-v_2(0)+\int_0^x(u_1-f-\epsilon u_1^2)-(u_2-f-\epsilon u_2^2)ds\Big{)}||\\ &=||\Big{(}\int_0^x(v_1-v_2)ds, \int_0^x(u_1-u_2)+\epsilon(u_2^2-u_1^2)ds\Big{)}|| \end{split} \end{equation} After a bit of arithmetic, one finds that: \begin{equation} \begin{split} ||T(u_1,v_1)-T(u_2,v_2)||\leq x\sqrt{1+\epsilon^2M}||(u_1,v_1)-(u_2,v_2)||, \end{split} \end{equation} where $M$ is independent of $u_1,v_1,u_2,v_2$. Therefore, choosing $x$ to be suitably small implies that T is a contraction mapping. Hence, there is at least a local solution.
You should also try to extend my argument to see if T is in fact a global contraction mapping and hence there would be a global solution.