# Fourier and contraction map

First, we prove a lemma for contraction map $T$:

$\displaystyle T(f)(x)=\phi(x)+\int K(x,y)f(y)\mathrm{d}y$

It is easy to prove that when $K(x,y)$ is small or has small $L^2$ norm, then this map is a contraction map. However, if we know that $K(x,y)=K(x-y)$ is a kernel such that

$\displaystyle\int K(x-y)\mathrm{d}x=1$,

then our conclusion still holds.

Consider Fourier transformation $\mathcal{F}$ and non-linear ODE.

For example:

$-u''+u-\epsilon u^2=f(x)$

with sufficient good properties for $f(x)$.

Then use $\mathcal{F}$:

$|\xi|^2\hat{u}+\hat{u}-\epsilon (2\pi)^{-d/2}\hat{u}*\hat{u}=\hat{f}$.

The correspondent functional is:

$T(\hat{u})=\displaystyle\frac{\epsilon}{|\xi|^2+1}(2\pi)^{-d/2}\hat{u}*\hat{u}-\frac{1}{|\xi|^2+1}\hat{f}$,

It seems we have done most work, but still the convolution product is annoying. We should eliminate it in a technical way.

Restrict out solution $\hat{u}$ in some ball $B(R)$ of $L^1$ space.

Then

$|T(\hat{u})-T(\hat{v})|_{L^1}=\displaystyle|\frac{\epsilon}{1+|\xi|^2}\int \hat{u}(\xi-y)\hat{u}(y)-\hat{v}(\xi-y)\hat{v}(y)|_{L^1}$

$\le \displaystyle|\epsilon\int (\hat{u}(\xi-y)-\hat{v}(\xi-y))\hat{u}(y)+\hat{v}(\xi-y)(\hat{u}(y)-\hat{v}(y))|_{L^1}$

$\le 2\epsilon |\hat{u}|_{L^1}|\hat{u}-\hat{v}|_{L^1}$,

when $\epsilon$ is sufficient small, $T$ is a contraction map.

The rest work is to verify that $T:B(R)\rightarrow B(R)$ well-defined.

However, if we encounter a similar problem:

$-u''+u-\epsilon u^k=f(x)$, where $k\ge 3$, and $f(x)$ satisfies some good properties on the boundary.

We can still use Fourier Transformation to solve this. 🙂

$\Box$