Ngô Quốc Anh

March 13, 2010

Wave equations: The gradient estimates and a conservation law

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 0:04

Let us consider the following problem

\displaystyle u_{tt} = \Delta u, \quad x \in \mathbb R^n, t>0


\displaystyle u(x,0)=f(x), \quad u_t(x,0)=g(x).

Assume f, g \in \mathcal S(\mathbb R^n). Note that the Schwartz space \mathcal S(\mathbb R^n) consists of all indefinitely differentiable functions f : \mathbb R^n \to \mathbb R such that

\displaystyle\mathop {\sup }\limits_{x \in {\mathbb{R}^n}} \left| {{x^\alpha }{D^\beta }f(x)} \right| < \infty

for every multi-index \alpha and \beta. We shall prove that the total energy at time t

\displaystyle E(t) = \frac{1}{2}\int_{{\mathbb{R}^n}} {\left( {u_t^2(x,t) + |{\nabla _x}u(x,t){|^2}} \right)dx}

is constant in time.

Our approach is based on the Fourier transform. Note that the Fourier transform of a Schwartz function \varphi is defined by

\displaystyle\widehat\varphi (\xi ) = \int_{{\mathbb{R}^n}} {\varphi (x){e^{ - 2\pi i\xi \cdot x}}dx} .

It is worth noticing that the Fourier transform maps \mathcal S(\mathbb R^n) to itself. Taking the Fourier transform with respect to space variable x we have

\displaystyle \widehat u_{tt}(\xi,t) = |\xi|^2 \widehat u(\xi,t)


\displaystyle \widehat u(\xi,0)=\widehat f(\xi), \quad \widehat u_t(\xi,0)=\widehat g(\xi).

Solving the above initial problem of the ordinary differential equation, we get

\displaystyle\widehat u(\xi ,t) = \widehat f(\xi )\cos \left( {|\xi |t} \right) + \frac{{\widehat g(\xi )}}{{|\xi |}}\sin (|\xi |t)

and therefore

\displaystyle {\widehat u_t}(\xi ,t) = - \widehat f(\xi )|\xi |\sin \left( {|\xi |t} \right) + \widehat g(\xi )\cos (|\xi |t).

As a consequence,

\displaystyle \int_{{\mathbb{R}^n}} {\left( {|{{\widehat u}_t}(\xi ,t){|^2} + |\xi {|^2}|\widehat u(\xi ,t){|^2}} \right)d\xi } = \int_{{\mathbb{R}^n}} {\left( {|\widehat g(\xi ){|^2} + |\xi {|^2}|\widehat f(\xi ){|^2}} \right)d\xi } .

We are now in a position to apply Plancherel’s Theorem to get the desired result.

Theorem (Plancherel). Suppose f \in \mathcal S(\mathbb R^n). Then

\displaystyle\varphi (x) = \int_{{\mathbb{R}^n}} {\widehat\varphi (\xi ){e^{2\pi i\xi \cdot x}}d\xi }.


\displaystyle\int_{{\mathbb{R}^n}} {{{\left| {\widehat\varphi (\xi )} \right|}^2}d\xi } = \int_{{\mathbb{R}^n}} {{{\left| {\varphi (x)} \right|}^2}dx}.

Let us denote by E_0 this common value of the total energy. One can show that

\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^n}} {u_t^2(x,t)dx} = \mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^n}} {|{\nabla _x}u(x,t){|^2}dx} = {E_0}.

The key point is to use the Riemann-Lebesgue lemma from harmonic analysis. We refer the reader to a book entitled Fourier Analysis due to Stein and Shakarchi.

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


\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}


\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


\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


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


\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.

February 28, 2010

Wave equations: The L^2-norm estimates

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 12:06

Now we talk about L^2-norm of solution to homogeneous wave equations. Let u(x,t) be a solution of the Cauchy problem

\displaystyle u_{tt}- \Delta u =0, \quad x \in \mathbb R^3, t>0

with the following conditions

\displaystyle u(x,0)=f(x), \quad u_t(x,0)=g(x), \quad x\in \mathbb R^3.

Assume that f, g \in C_0^\infty(\mathbb R^3). If

\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb R^3} {u{{(x,t)}^2}dx} = 0


\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb R^3}  {g{{(x)}^2}dx} = 0.

Proof. Let R>0 such that

\displaystyle {\rm supp}(f),{\rm supp}(g) \subset B_R = \{x \in \mathbb R^3, |x|<R\}.

It is not difficult to verify that

\displaystyle {\rm supp}(u(\cdot,t)) \subset \{ x\in \mathbb R^3, t-R<|x|<t+R\}

for any t>R.

By using the Green formula, we get the following equality from the equation

\displaystyle\int_{{\mathbb{R}^3}} {{u_{tt}}(x,t)dx} = \int_{{\mathbb{R}^3}} {\Delta u(x,t)dx} = 0.

Therefore, we have

\displaystyle\frac{d}{{dt}}\int_{{\mathbb{R}^3}} {{u_t}(x,t)dx} = 0,


\displaystyle\int_{{\mathbb{R}^3}} {{u_t}(x,t)dx} = \int_{{\mathbb{R}^3}} {g(x)dx},


\displaystyle\int_{{\mathbb{R}^3}} {u(x,t)dx} = \int_{{\mathbb{R}^3}} {f(x)dx} + t\int_{{\mathbb{R}^3}} {g(x)dx}

by the initial conditions. In the other side, by Cauchy inequality, it follows that

\displaystyle\begin{gathered} \int_{{\mathbb{R}^3}} {u(x,t)dx} \leqslant {\left( {\int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{t - R < |x| < t + R} {dx} } \right)^{\frac{1}{2}}} \hfill \\ \qquad= {\left( {\frac{8}{3}\pi R} \right)^{\frac{1}{2}}}{\left( {3{t^2} + {R^2}} \right)^{\frac{1}{2}}}{\left( {\int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} } \right)^{\frac{1}{2}}} \hfill \\ \end{gathered}

for any t>R. This is a contradiction provided that

\displaystyle\int_{{\mathbb{R}^3}} {g(x)dx} \ne 0.

We now try to find the limit

\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb  R^3} {u{{(x,t)}^2}dx}

in general case. For simplicity, we assume f \equiv 0. We will show that there exists a non-negative constant c such that

\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb   R^3} {u{{(x,t)}^2}dx} = c.

Proof. Let \widehat u denote the Fourier transform of u with respect to the variable x. By taking the Fourier transform of the wave equation and the initial conditions we can get

\displaystyle\widehat u(\xi ) = \frac{{\left| {\widehat g(\xi )} \right|}}{{\left| \xi \right|}}\sin (\left| \xi \right|t).


\displaystyle\begin{gathered} \int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} = \frac{1}{{{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {\widehat u{{(\xi ,t)}^2}d\xi } \hfill \\ \qquad= \frac{1}{{{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {|\widehat g(\xi ){|^2}\frac{{{{\sin }^2}(\left| \xi \right|t)}}{{{{\left| \xi \right|}^2}}}d\xi } \hfill \\ \qquad= \frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g(\xi ){|^2}d\xi } - \frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g(\xi ){|^2}\cos (2\left| \xi \right|t)d\xi } . \hfill \\ \end{gathered}


\displaystyle {\left| \xi \right|^{ - 2}}|\widehat g(\xi ){|^2} \in {L^1}({\mathbb{R}^3})

we have

\displaystyle\mathop {\lim }\limits_{t \to \infty } \frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g (\xi ){|^2}\cos (2\left| \xi \right|t)d\xi } = 0.

We then get

\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} = \underbrace {\frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g (\xi )|^2d\xi } }_c.

Note that this constant c is positive if g \not\equiv 0.

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.

February 26, 2010

Wave equations: The first pointwise estimates

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 19:15

Followed by the topic where we discuss a pointwise estimate of solution to a class of heat equations. We are now interested in the case of wave equations.

Consider the wave equation in \mathbb R^3

\displaystyle u_{tt}-\Delta u =0, \quad x\in \mathbb R^3, t>0

together with the following initial data

\displaystyle u(x,0)=0, \quad u_t(x,0)=g(x).

We assume g \in C_0^\infty(\mathbb R^3). We will prove the following estimate

\displaystyle |u(x,t)| \leq \frac{C}{t}, \quad t>0

for some constant C depending only on the given data.

Proof. The assumption on g implies the existence of two positive constants R and M such that

\displaystyle {\rm supp}(g) \subset B_R = \{ x \in \mathbb R^3:|x|<R\}


\displaystyle |g(x)| \leq M, \quad \forall x \in \mathbb R^3.

From the Poisson formula (but is known as Kirchhoff formula) in 3D, the solution to the problem is given as the following

\displaystyle u(x,t)=\frac{1}{4\pi t} \int_{|y-x|=t} {g(y)dS_y}.

It is easy to verify that: the area of the intersection \{ y \in \mathbb R^3:|y-x|=t\} \cap B_R is less than or equals to the area of \partial B_R. Therefore, we obtain

\displaystyle |u(x,t)| \leq\frac{1}{4\pi t} M 4\pi R^2 = \frac{MR^2}{t}, \quad t>0.

Blog at