# Ngô Quốc Anh

## March 7, 2010

### Heat equations: The gradient estimates

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 16:37

Let us consider the following problem

$\displaystyle u_t =u_{xx}, \quad x\in \mathbb R, t>0$

and

$\displaystyle u(x,0)=f(x),\quad x \in \mathbb R$.

We assume further that $f \in L^2(\mathbb R)$. We are interested in the gradient estimate, in fact, we are going to find the bound on

$\displaystyle\int_{ - \infty }^{ + \infty } {|{u_x}(x,t){|^2}dx}$

and

$\displaystyle |{u_x}(x,t)|$

in $t>0$.

So far, once we want to derive any estimates like point-wise, $L^2$, gradient, the Green function is very important. Again, since the Green function for this heat equation is already known, we can write down everything via the Poisson formula

$\displaystyle u(x,t) = \frac{1}{{\sqrt {2\pi t} }}\int_{ - \infty }^{ + \infty } {f(y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy}$.

A simple calculation shows that

$\displaystyle {u_x}(x,t) = \frac{1}{{4\sqrt \pi {t^{\frac{3}{2}}}}}\int_{ - \infty }^{ + \infty } {f(y)(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy}$.

Then by using the Holder inequality, we have

$\displaystyle\begin{gathered} {\left( {\int_{ - \infty }^{ + \infty } {f(y)(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)^2} = {\left( {\int_{ - \infty }^{ + \infty } {f(y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}dy} } \right)^2} \hfill \\ \qquad\qquad\leqslant \left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right) \hfill \\ \end{gathered}$

which implies

$\displaystyle |{u_x}(x,t){|^2} \leqslant \frac{1}{{16\pi {t^3}}}\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)$.

In order to do next, the following two results are noticed

Proposition 1. The following identities hold

$\displaystyle\int_{ - \infty }^{ + \infty } {{z^2}{e^{ - \frac{{{z^2}}}{{4t}}}}dz} = 4{t^{\frac{3}{2}}}\sqrt \pi$

and

$\displaystyle\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{z^2}}}{{4t}}}}dz} = 2{t^{\frac{1}{2}}}\sqrt \pi$.

Therefore, it holds that

$\displaystyle\int_{ - \infty }^{ + \infty } {{{(x - y)}^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} \mathop = \limits^{z = x - y} \int_{ + \infty }^{ - \infty } {{z^2}{e^{ - \frac{{{z^2}}}{{4t}}}}( - dz)} = \int_{ - \infty }^{ + \infty } {{z^2}{e^{ - \frac{{{z^2}}}{{4t}}}}dz} = 4{t^{\frac{3}{2}}}\sqrt \pi$

which is independent of $x$. Consequently,

$\displaystyle\begin{gathered} \int_{ - \infty }^{ + \infty } {|{u_x}(x,t){|^2}dx} \leqslant \frac{1}{{16\pi {t^3}}}4{t^{\frac{3}{2}}}\sqrt \pi \int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)dx} \hfill \\ \qquad\qquad= \frac{1}{{4{{\sqrt {\pi t} }^{\frac{3}{2}}}}}\int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)dx}. \hfill \\ \end{gathered}$

The last double integral can be estimated by the Fubini lemma. In fact,

$\displaystyle\int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)dx} \leqslant 2{t^{\frac{1}{2}}}\sqrt \pi \int_{ - \infty }^{ + \infty } {|f(x){|^2}dx}$

which yields

$\displaystyle\int_{ - \infty }^{ + \infty } {|{u_x}(x,t){|^2}dx} \leqslant \frac{1}{{2t}}\int_{ - \infty }^{ + \infty } {|f(x){|^2}dx}$.

Remark. If we estimate as follows

$\displaystyle\begin{gathered} \left( {\int_{ - \infty }^{ + \infty } {f(y)(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right) = \left( {\int_{ - \infty }^{ + \infty } {f(y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}dy} } \right) \hfill \\ \qquad\qquad\leqslant \left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}dy} } \right)^\frac{1}{2}{\left( {\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}}{\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^4}{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}} \hfill \\ \end{gathered}$

we then have

$\displaystyle |{u_x}(x,t)| \leqslant \frac{1}{{4\sqrt \pi {t^{\frac{3}{2}}}}}\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}dy} } \right)^\frac{1}{2}{\left( {\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}}{\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^4}{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}}$.

Note that

Proposition 2. The following result holds

$\displaystyle\int_{ - \infty }^{ + \infty } {{z^4}{e^{ - \frac{{{z^2}}}{{2t}}}}dz} = 3\sqrt 2 {t^{\frac{3}{2}}}\sqrt \pi$

and

$\displaystyle\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{z^2}}}{{2t}}}}dz} = 2{t^{\frac{1}{2}}}\sqrt \pi$.

Thus

$\displaystyle |{u_x}(x,t)| \leqslant\underbrace {\frac{1}{{4\sqrt \pi {t^{\frac{3}{2}}}}}{{\left( {\sqrt {2\pi t} } \right)}^{\frac{1}{4}}}{{\left( {3\sqrt {2\pi } {t^{\frac{3}{2}}}} \right)}^{\frac{1}{4}}}}_{\frac{{\sqrt[8]{6}}}{{4t}}}\sqrt {\int_{ - \infty }^{ + \infty } {|f(x){|^2}dx} } = \frac{{\sqrt[8]{6}}}{{4t}}{\left\| f \right\|_{{L^2}(\mathbb{R})}}$.

This is a pointwise estimate.