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.πŸ™‚

\Box

About YiMin

This is just a nerd PhD student of Math@UT Austin.