# Ngô Quốc Anh

## February 28, 2010

### Heat equations: The L^2-norm estimates

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 11:00

In the last topic we consider a pointwise estimate for homogeneous heat equation. The conclusion is the following: if the initial data is $L^2$-bounded then the solution $u$ decays as $t^\frac{1}{4}$. Today, we consider another phenomena. Let assume $\Omega \subset \mathbb R^n$ be an open set with smooth boundary and suppose $u \in C^\infty(\Omega \times [0, \infty))$ is a solution of

$\displaystyle u_t - \Delta u = f$

with

$\displaystyle u=0$

on the boundary $\partial \Omega \times [0, \infty)$. Assume that

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_\Omega {f{{(x,t)}^2}dx} = 0$

then

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_\Omega {u{{(x,t)}^2}dx} = 0$.

Proof. The method used here is very standard when we deal with energy estimate or if we want to estimate $L^2$-norm of the solution.

Multiplying the equation by $u$ and then integrating the resulting equation on $\Omega$, we can get

$\displaystyle \frac{d}{{dt}}\left( {\frac{1}{2}\int_\Omega {u{{(x,t)}^2}dx} } \right) + \int_\Omega {{{\left| {\nabla u(x,t)} \right|}^2}dx} = \int_\Omega {f(x,t)u(x,t)dx}$.

The Poincare inequality gives

$\displaystyle\int_\Omega {u{{(x,t)}^2}dx} \leqslant C\int_\Omega {{{\left| {\nabla u(x,t)} \right|}^2}dx}$

where $C>0$ is constant. For any $\varepsilon>0$, by the Young inequality, it holds that

$\displaystyle\left| {\int_\Omega {f(x,t)u(x,t)dx} } \right| \leqslant \frac{1}{2}\left( {\varepsilon \int_\Omega {u{{(x,t)}^2}dx} + \frac{1}{\varepsilon }\int_\Omega {f{{(x,t)}^2}dx} } \right)$.

Taking $\varepsilon=\frac{1}{C}$ we obtain

$\displaystyle\begin{gathered} \frac{d}{{dt}}\left( {\frac{1}{2}\int_\Omega {u{{(x,t)}^2}dx} } \right) + \frac{1}{C}\int_\Omega {u{{(x,t)}^2}dx} \hfill \\ \qquad\leqslant \frac{d}{{dt}}\left( {\frac{1}{2}\int_\Omega {u{{(x,t)}^2}dx} } \right) + \int_\Omega {{{\left| {\nabla u(x,t)} \right|}^2}dx} \hfill \\ \qquad\leqslant \left| {\int_\Omega {f(x,t)u(x,t)dx} } \right| \hfill \\ \qquad\leqslant \frac{1}{2}\left( {\frac{1}{C}\int_\Omega {u{{(x,t)}^2}dx} + C\int_\Omega {f{{(x,t)}^2}dx} } \right) \hfill \\ \end{gathered}$

which implies

$\displaystyle\frac{d}{{dt}}\left( {\int_\Omega {u{{(x,t)}^2}dx} } \right) + \int_\Omega {u{{(x,t)}^2}dx} \leqslant C\int_\Omega {f{{(x,t)}^2}dx}$.

Solving this differential inequality, we have

Theorem ($L^2$ estimates).

$\displaystyle\int_\Omega {u{{(x,t)}^2}dx} \leqslant {e^{ - \frac{t}{C}}}\int_\Omega {u{{(x,0)}^2}dx} + C\int_0^t {{e^{ - \frac{{t - \tau }}{C}}}\left( {\int_\Omega {f{{(x,\tau )}^2}dx} } \right)d\tau }$.

It is not difficult to verify that

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_0^t {{e^{ - \frac{{t - \tau }}{C}}}\left( {\int_\Omega {f{{(x,\tau )}^2}dx} } \right)d\tau } = 0$.

This completes the proof.

Note that in the above proof, we use a result which is similar to the Gronwall inequality. Clearly, the statement is as follows.

Lemma. If the function $\psi(x,t)$ satisfies

$\displaystyle\frac{d}{{dt}}\psi (x,t) + \alpha (t)\psi (x,t) \leqslant \beta (t)$

we then have

$\displaystyle\psi (x,t) \leqslant {e^{ - \int_0^t {\alpha (\tau )} d\tau }}\psi (x,0) + \int_0^t {{e^{ - \int_\tau ^t {\alpha (s)} ds}}\beta (\tau )d\tau }$.

The proof of this statement is quite simple. We first try to write

$\displaystyle\varphi (x,t) = {e^{\int_0^t {\alpha (\tau )} d\tau }}\psi (x,t)$.

Clearly

$\displaystyle \frac{d}{{dt}}\varphi (x,t) = {e^{\int_0^t {\alpha (\tau )} d\tau }}\left( {\frac{d}{{dt}}\psi (x,t) + \alpha (t)\psi (x,t)} \right) \leqslant \beta (t){e^{\int_0^t {\alpha (\tau )} d\tau }}$

which gives, after integrating with respect to $t$,

$\displaystyle \varphi (x,t) \leqslant \varphi (x,0) + \int_0^t {\beta (\tau ){e^{\int_0^\tau {\alpha (s)} ds}}d\tau }$.

Thus

$\displaystyle {e^{\int_0^t {\alpha (\tau )} d\tau }}\psi (x,t) \leqslant \psi (x,0) + \int_0^t {\beta (\tau ){e^{\int_0^\tau {\alpha (s)} ds}}d\tau }$.

Hence

$\displaystyle \psi (x,t) \leqslant {e^{ - \int_0^t {\alpha (\tau )} d\tau }}\psi (x,0) + \int_0^t {\beta (\tau ){e^{\int_\tau ^t {\alpha (s)} ds}}d\tau }$.

The proof follows.

1. Rất hay. Cảm ơn bạn nhiều.
Bạn có thể đưa thêm bài ứng dụng của bất đẳng thức Gronwall vào cũng rất hay.

Comment by kakro — June 10, 2010 @ 10:08

2. Comment by toan — September 20, 2012 @ 17:05

3. Xin bạn làm rõ thêm được ko ? mình thấy khó hiểu chỗ suy ra kqua dpcm, kqua của ta là tích phân từ 0 -t của e^{ – tich phân từ \tau đến t ..} mà ??? xin giải thích rõ làm sao để có dấu trừ của e^ , vì đang là e^ tích phân từ 0 đến s mà ….

Comment by toan — September 20, 2012 @ 17:30

• Ở dòng nào thế bạn?

Comment by Ngô Quốc Anh — September 20, 2012 @ 17:33