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


\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


\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).


\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 } .


\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 } .


\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. 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

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